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0 on Bp(x) in the sense of //*(Q). The set {x e Q: u(x) > 0} is open. Let u be the solution of Problem 6.1 with "obstacle" ^. We divide Q into the sets {x E Q: u(x) > ^(x)}, which is open, and its complement / = /[«], which is closed in Q. Formally, / is the set of points x where u(x) = i^(x). Definition 6.8. The set / is called the coincidence set of the solution u. Consider a point x0 € Q - /. Then there is a ball Bp(x0)andq>(x)e CoW*o)X 0 on Bpl2(x0\such that Hence for any £ e C0X3(6p/2(x0)), we may find an £ > 0 such that Consequently, /; = w + e£ e IK. Substituting this v in the variational inequality and dividing by e, we find that Since this is particularly true of — C, we obtain In other words, At this point we have obtained the properties and, formally,
44
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VARIATIONAL INEQUALITIES IN HILBERT SPACE
Of course (6.7) does not characterize the solution u. On the other hand, Lu —/is nonnegative, that is, Consequently, by the Riesz-Schwarz theorem, Schwarz[l],«(u, () — is given by a nonnegative Radon measure u on Q, namely,
In view of (6.7) the support of /i is contained in I. To summarize: Theorem 6.9. Let u be the solution to Problem 6.1. Then there exists a nonnegative Radon measure \i such that
with
In particular,
We want to establish a relation between the measure u and the capacity. Given a compact subset E a O, we define the closed convex of Hp(Q)
Definition 6.10. The capacity ofE with respect to £1, capn £, or simply cap E, is defined by
A set F c Q is of capacity zero if cap E = 0 for every compact E c= F. The minimum problem (6.8) admits a solution a(x). Indeed, a(x) is the unique solution of the variational inequality
6
THE OBSTACLE PROBLEM I FIRST PROPERTIES
45
and is called the conductor potential or capacitary potential of E (with respect to Q). If £ e H10(&) with £ > 0, then a + £ e IK£, so
Hence a is a — A supersolution. By Theorem 5.7, a > 0 in Q. On the other hand, min(a, 1) £ IK£ and
(where {x a(x) < 1} is understood in the pointwise sense appropriate to the application of Theorem A.I). Since a is the unique element of smallest Dirichlet integral in IK £ , a = min(a, 1). Thus
Suppose, in fact, that vn e KE n H1' °°(Q) satisfy vn -> a in //'(Q). Define
and set wn = 6(vn) e K£ n // I( °°(Q) with the property that wn = 1 on E. Now
It follows from the parallelogram law that {vvn} also enjoys the property that wn ->• a in /fo(^)- By Definition 5.1, a = 1 on E in the sense of Hl(£i). Finally, if x e Q — £, there is a neighborhood t/ c Q — E of x and a + C e IKE for any £ e C^(t/). Consequently
We have shown that a e Ho(Q) satisfies
46
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VARIAT1ONAL INEQUALITIES IN HILBERT SPACE
Again applying the theorem of Riesz-Schwarz, there is a measure m = m£ such that — Aa = m in Q, namely,
Denote by 0(x, y) = gn(x, y) the Green's function of — A for Q. It is a part of the Riesz representation theorem that
and hence, formally,
This formula is easily justified by a direct computation, with due regard to the singularity of the Green function. Theorem 6.11. Letf0,fi,..., fNe L2(fi)and \je H^Q). Lef u feethe solution of Problem 6.1 m IK^/or/ = /0 + ^ (d/dx^ff e H ~l (Q), anrf let p denote the measure determined by u. Then there exists a constant C > 0 such that In particular, if cap E = 0, it/jen /^(£) = 0. Proof. We assume, of course, that IK^ 7^ 0, that is, that i^ < 0 on dQ. Let 0 < C e C£(Q) satisfy C > 1 on £. Then
7
THE OBSTACLE PROBLEM IN THE ONE DIMENSIONAL CASE
47
where Poincare's inequality was used to estimate jn /0 £ dx. In particular, since Co(Q) n IK£ is dense in IK £ ,
Corollary 6.12. Lef w be the solution to Problem 6.1 and let I denote its set of coincidence. If cap I = 0, then
One might observe that the hypotheses of Theorem 6.11 may be weakened since the best constant C satisfies where C0 depends on the bilinear form a and Q, and therefore C depends on \l/ only through ||wJL2(n). In particular, the conclusion holds for obstacles \l/ e L°°(fi) for which IK, ^ 0. Once again we may raise the question of the smoothness of the solution u. In Chapter IV we provide an extensive discussion of this topic. For the present, let us note that the solution cannot be expected to be in C2(Q). It suffices to consider N = 1,Q = (— 3,3), \{/(x) = 1 — x2, and the bilinear form
The solution to Problem 6.1 for/ = 0 is given by a function which is partially the parabola i^ and partially its tangents, because such a function is the smallest supersolution in IK. We discuss this problem from a different viewpoin in the next section.
7.
The Obstacle Problem in the One Dimensional Case
Before consideration of the general question of the regularity of the solution of the obstacle problem, we shall discuss the one dimensional case. Let us remark, to begin, that when Q = (a, /?) is an open interval of R, the functions of Hl(Q) are the absolutely continuous functions u(x) on (a, ft) whose derivatives u'(x) are in L2(Q). The functions of //o(Q) are those functions of H^Q) which vanish at a and ft. Let
48
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VARIATIONAL INEQUALITIES IN HILBERT SPACE
where
Define
Obviously, a(u, v) is a coercive bilinear form on //o(Q). Given/e H~ *(Q), we know from Theorem 2.1 that there exists a unique solution of the variational inequality From Section 6, there exists a nonnegative measure n whose support is in the coincidence set / = {x e Q: u(x) = i/K*)} and in the sense of distributions. Since ij/(a) < 0 and \l/({$) < 0, / is compact in Q and /i(Q) = /*(/) < oo by Theorem 6.11. Now // is nonnegative so there is a nondecreasing function (p(x) such that q)' = \i
(in the sense of distributions).
Of course, it is clear that Moreover, since/e H~ !(Q), we may write Therefore Eq. (7.1) may be written which implies that We remark that if F(x)eLp(Q), p> 1; i.e., if/e/f 1 ''^), then w e H ' (Q). In fact, (p(;t), being a nondecreasing function, is bounded in Q. It then follows that M(X) e C°-A(Q), A = 1 - (1/p). Henceforth assume that F is continuous. Then it follows that u'(x) is continuous on the set Q — / = {x e Q: u(x) > iK*)}, or in other words, the discontinuities of u'(x) can occur only at points of /. A discontinuity of cp is a jump discontinuity, i.e., 1 P
APPENDIX A
SOBOLEV SPACES
49
inasmuch as
(0 and (G) in the z = xl + ix2 plane given by This maps (— 1, 1) onto the curve We shall find, for e sufficiently small, a function such that 0. Note that %ff(0) = 0, 1 < a < N — 1. The terms of order three in v do not occur in the linearized equations, nor do the terms of order two in v in the last boundary condition occur. For variations v, ~ are analytic near 0 in U u S. This proves the theorem. Q.E.D. oo, hence V admits an expansion 0 and satisfies the boundary conditions of Problem 1.2. In addition, a subsequence of the uf. converges weakly to u in //2>p(0, R), 1 < p < oo, for each fixed t,Q < t < T. Since, however, u&- -» u uniformly in D, it follows that the entire sequence UK- -> w weakly in //2>p(0, #), 0 < r < T. In particular, uxx e L °°(Z>). To complete the proof that u is a solution to Problem 1.2, let u e L°°(£>) satisfy i; > <5 > 0. Multiplying (2.2) by v — ue and noting that (i£(v)= 0 if e < (5, we see that Integrating with respect to x e [0, K\ and observing that we obtain
If x < £, then thus while for x > £, £ e I, so
Theorem 7.1. /« addition to our previous hypotheses, assume now that ij/'(x) has only discontinuities of the form Then u'(x) is continuous. Indeed, Consequently, by (7.2), and this proves that u'(x) is continuous.
Appendix A. Sobolev Spaces A systematic investigation of the real variable properties of elements of Hm's(Q) may be found in Morrey [1, Chapter 3] but for the convenience of the reader we will provide selected proofs. We first discuss real variable properties giving two proofs of Theorem A.I below. The first employs the notion
50
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VARIATIONAL INEQUALITIES IN HILBERT SPACE
of "absolutely continuous representative," Lemma A.2, and the second utilizes an appropriate choice of test function, Lemmas A.3 and A.4. Some generalizations of Theorem A.I are mentioned. The trace operator and a "matching lemma" are considered. We then state and prove some Sobolev and Poincare type lemmas. To conclude this appendix we state the well-known Rellich theorem, whose proof may be easily read in Morrey [1]. Theorem A.I. Let Q c IRN, ueH 1>s (Q), 1 < s < oo. Then max(w, 0)e H (Q) and/or 1 < i
in the sense of distributions. We emphasize the meaning of this theorem: the derivatives of max(«, 0) are determined almost everywhere by the formula (A.I). In particular, for any test function £,
It suffices to prove the theorem in the case «e//o >s (Q), 1 < s < +00. Since £ has compact support, the case s = oo corresponds to Rademacher's theorem about the differentiability of Lipschitz functions, which we assume known. Lemma A.2. Let i;eH£' s (Q), 1 < s < oo. Then for each i, 1 < i < N, there is a representative vt of v which is absolutely continuous on almost every parallel to the x, axis in Q. Moreover, where hj is a representative ofh, € LS(Q), the ith distributional derivative ofv. Proof.
Let vm e Co(Q) be a sequence such that
We extend vm, v to be zero outside Q so that
APPENDIX A
SOBOLEV SPACES
51
that is, the integral is taken over a parallel to the x, axis passing through x. For C6C?(Q),
Now y mx . converges in LS(Q) to function hit the ith distributional derivative of v. Consequently,
This implies that v is equal almost everywhere to a function v(x) which is absolutely continuous on almost every parallel to the x, axis and whose derivative dv/dXj = /j,, a representative of /i, eLS(Q). In this way, we can understand with equality being understood either in LS(Q) or a.e. in Q for a subsequence of the i;™,. Q.E.D. Proof of Theorem A.I. First observe that max(M, 0)e// 1>s (^X 1 < s < oo. For let «„ 6 C1^) satisfy un -> u in Hl >s(^)- Then max(wn, 0) is Lipschitz and satisfies for some constant and, recalling that max(w, 0) e Z/(Q), From (A.2),'it follows that max(wn, 0) converges weakly to an element of H 1>S(Q) and from (A.3) we conclude that this element is equal to max(w, 0) a.e. Now we apply Lemma A.2 to v = max(u, 0). Let vt and u, be the representatives of v and M, respectively, with /j, and 0, the representatives of their distributional derivatives determined by the lemma. First, let F c: Q be any subset such that v = 0 on F. We want to show that fit = 0 a.e. on F, or what is the same thing, that
First, we replace F by a subset of equal measure where t^ = 0. By Fubini's theorem, viz., (A.4), we need only prove /i; = 0 a.e. on F when F c= R, that is,
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VARIATIONAL INEQUALITIES IN HILBERT SPACE
for functions of a single variable x,. Now F c U may be written F = F' u N, where F' is the set of density points of F and meas N = 0. On F' c (R, yf is differentiable a.e. and its derivative on F' may be computed from its value there. Hence Finally, let F <= O be any set where v = u. We want to show that h( — g,, = 0 on F. We argue as before. After replacement of F by a set of equal measure where M{ — vh we need only prove that /i, = & a.e. on F when F <= R by Fubini's theorem. Again, this may be verified by consideration of ht and #, on the set of density points F' of F. The theorem is proved. Q.E.D. We can deduce Theorem A.I from the following lemmas: Lemma A.3. Let 9(t) be a Cl(U) function such that 0' is bounded; \9'(t)\ < M. Ifu e Hl's(Q\ Q a bounded open set ofRN, 1 < s < + oo, then
and
Proof. Since u e Hl >S(O), there exists a sequence {«„} of C J (Q) functions such that un tends to u in H li;s (Q) and also un tends to w a.e. in £1 Obviously and since 9(un) converges to 9(u) in LS(O). On the other hand, In fact,
and /4n converges to zero in Z/OQ) while Bn converges to zero a.e. in Q and Therefore, by the Lebesgue theorem, Bn converges to zero in LS(Q).
APPENDIX A
SOBOLEV SPACES
53
Since the derivatives in the sense of distribution of Q(u) are the limit in LS(O) of (dldx$(un\ (A.5) follows. If u e H * • *(Q) is such that u: Q ^ Q' -> (a, 0) and 0 is C' ((a, 0)), then (A. 5) holds a.e. in Q'. Lemma A.4. Lef w e / / 1 ' S(O). T/ren
Proof.
For each y e [— 1, 1] define
so ay£(0) = y and
Define
Then 0 c y eC ! (R)and |0;e(t)| < 1. Hence 0£y(w) e Hl-S(O) and
. In addition, and
Hence Thus In particular,
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VARIATIONAL INEQUALITIES IN HILBERT SPACE
for every y £ [— 1, 1], so
A slightly simpler proof may be based on Exercise 15. A particular consequence of the last lemma is that for any u e H1>S(Q), and
In other words, (A.I) holds, which proves Theorem A.I. We give two generalizations of Theorem A.I. Corollary A.5. Let 9(t\ — co
A more subtle refinement of Lemma A.4 is Corollary A.6. Let 9(t), —co
APPENDIX A
SOBOLEV SPACES
55
and
The trace operator is a continuous linear mapping with the property Here we denote a point of UN by x = (x', XN). Moreover, denoting by TEv the trace of t; on the subset ZE = G n {x: XN = e}, it is easy to check that in the obvious way. Assume now that £ e H 1>S (G) and T£ = 0. Identify £ with one of its representatives absolutely continuous on segments parallel to the XN axis. Existence of such a representative follows from Lemma A.2. By Lebesgue's theorem, there is a subsequence e, -* 0 such that £(x', EJ) -> 0 for a.e. x' e £ as EJ -> 0. Now extend £(x) to be zero in G~ = {x: |x| < 1, XN < 0). For any x' with lim £(x', e,-) = 0, the function of a single variable £(x', XN), |x w | < 1, is absolutely continuous and
by Fubini's theorem, (,XNE LS(B\ B = { x : \x\ < 1}. It is easy to check that (x. € LS(B) for i < N. Hence a function £ with T£ = 0 on Z may be extended as 0 in G~ to become an element of H 1<S(B). From this it is obvious that £ = 0 on E in the sense that £ = lim £„ in //^(G), where £„ e //'• °°(G) and £„ = 0 for XN < l/n. Finally, observe that the converse is also true: if £ e H 1 ' °°(G) is the limit of smooth functions £„ vanishing near Z, then T£ = lim T£n = 0. We have shown Lemma A.7. Let £ e // 1>5 (G), 1 < s < oo. T/ie« T£ - 0 on E i/am* on/y //"£ is in the closure in Hl>s(G) norm of the smoothfunctions which vanish near E. From this we deduce a "matching" lemma. Lemma A.8. Let B denote the unit ball in UN, G+ = {xeB:xN> 0}, G" = {x e B: XN<0}, and E = B n {XN=0}.Letv+eH I ! S (G + )and tT e //1>i(G")safis/3'
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VARIATIONAL INEQUALITIES IN HILBERT SPACE
Then
isinHl's(B). Proof. Set w(x) = iT(x', -X B ) for x e G + . Then weH 1 > s (G + ) and Tv = Tw on E. Consequently, +
Let £,. e H 1 ' °°(G) be a sequence of Lipschitz functions satisfying
and Also let i;£+ e /f 1 -°°(G + ) converge to t;+ in H 1 ' S (G + ). Define
which is Lipschitz in B. Obviously
Moreover,
soiJe// l i S (B). Q.E.D. We now state and prove some Sobolev lemmas. Lemma A.9. (Sobolev). T/J^/7
w/iereas
where C depends only on
Let dQ. be Lipschitz and u e // liS (n), 1 < s < oo.
APPENDIX A
SOBOLEV SPACES
57
The first assertion of Lemma A.9 is a consequence of the following useful inequality about functions in Ho's(Q). Lemma A.10. for s < N,
There exists a constant C, depending only on N, such that
Poincare's inequality follows from this lemma and Holder's inequality. We state a form of particular use to us. Corollary A.ll.
Let u e Hl0(Br\ B, = {x e RN : \x\ < r}. Then
where C depends only on N. Proof of Lemma A. 10. We begin by proving that the inequality is true for 5 > 1 if we assume it true for s — 1. Let u(x) be a function of Co(ft); the function |w|s*/1* belongs to HQ' H^) and therefore we have
Using Holder's inequality and noting that (1/1*) — (1/s*) = 1/s', we have
and thus
It remains to prove the inequality for s = 1. To this end let x = (jc 1 ? ..., XN) be any point of RN. From the inequalities
where the right-hand side means that the integral is over the straight line
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VARIATIONAL INEQUALITIES IN HILBERT SPACE
we infer
Integrating with respect to xl, then with respect to x 2 , and so on and making use of the Holder inequality, 1 we get
In such a way the inequality is proved also for s = 1 with the constant C = 1/21*. Q.E.D. Proof of Lemma A.9(ii). We first show that
for u€Hl's(£l). By extending u to a function, still called u, of Ho's(Q0)» Q c: Q0, so that
we see that it is sufficient to consider functions u e Ho's(fi). Since u e /fo's(Q) is the limit of functions un e C^(Q), we need only show (A.8) for u 6 C*(Q). 1
We use the inequality
APPENDIX A
SOBOLEV SPACES
59
Given x, we expand u in a Taylor expansion about each y e Q and integrate with respect to y. This gives
We now apply Holder's inequality to each term. To estimate the last term
set z = x + t(y — x) for a fixed t and Q(f) = {z: x + t(y — x)eQ}. Note that meas Q(r) = f N meas Q. Then
From (A.9) we now deduce
which proves (A.8). A particular consequence of (A.8) is that functions of H1>s(^), s > N, are continuous. Lemma A.12.
LetveH1' s(Br(x))
satisfy
r/ie« vv/iere C depends only on N and s. The demonstration of this lemma is left as an exercise. To complete the proof of Lemma A.9(ii), given u e //0>S(Q) and x, y e Q, \x - y\ = d, we shall apply Lemma A.12 to
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VARIATIONAL INEQUALITIES IN HILBERT SPACE
whose average over Bd(x) is zero. Consequently,
employing Lemma A. 12 to estimate the first term of (A.8). Here C = C(s, N). Q.E.D. Another useful inequality concerns functions which vanish on a subset ofB r . Lemma A.13. Let E a Br satisfy and let UE Hl> l(Br) vanish almost everywhere on E. Then for each a > 0
where C depends only on $. We shall use Lemma A.14. Letf(x) E L\UN) and define
the potential of order 1 generated byf. There exists a constant C > 0 such that
Proof. Therefore Hence where
Now
APPENDIX A
SOBOLEV SPACES
61
On the other hand, it is easy to check that KefeLl(RN).Indeed, by Fubini's theorem
Thus
Consequently
Lemma A. 15. Let veH1- °°(J5r) vanish on E c Br, where meas E > & meas Br,
Q < S < 1.
Then there exists a constant ft > 0 depending only on $• and N such that
Proof. Let xe Br. If x e E, then the conclusion is valid for any /?. For x££, consider the function r-> t;(x + rf) with £eSN~l, S"'1 = dBj(O). Let I = {^ e S N ~ 1 : there exists R > 0 such that x + R£ e E}. Then
and if x + K£e£,
Letrfcodenote the measure on SN~l, so that We have that
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VARIATIONAL INEQUALITIES IN HILBERT SPACE
In order to prove the lemma, it is sufficient to show that the measure |Z | of Z is bounded below. Now
Hence and thus
Proof of Lemma A. 13. It suffices to verify the conclusion for functions u £ Hl'co(Br) which vanish on E, where For such u, by the preceding lemma. Therefore, by Lemma A. 14, meas{x: |w(x)| > a} < meas{x: K(\ux\) > a/j8}
We state without proof the well-known Rellich compactness theorem (Morrey [1]). Theorem A. 16. Suppose that Q is a bounded domain with smooth boundary
Appendix B. Solutions to Equations with Bounded Measurable Coefficients We shall now prove the estimate for solutions of the Dirichlet problem,
APPENDIX B
EQUATIONS WITH L°° COEFFICIENTS
63
that was stated in Section 5. This estimate will be useful to us in Chapter IV. We need this lemma: Lemma B.I. Let q>(t\ k0 < t < GO, be nonnegative and nonincreasing such that
where C, a, ft are positive constants with ft > 1. Then
where
Proof.
Consider the sequence
By assumption, we have
We now prove by induction that
For r = 0, (B.5) is trivial. Assuming it holds up to r, we shall show that it holds for r + 1. Indeed, by (B.4) we obtain
Using (B.3),
As r -> +00, the right-hand side tends to zero, and so
Theorem B.2. Let a,/x) e L°°(Q) safis/y w/iere v > 0. Letf0,fi, . . . , f N € L s ( Q ) f o r s
> N and let
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VARIATIONAL INEQUALITIES IN HILBERT SPACE
Then
where K is a constant independent of v. Proof.
For k > 0 define
According to Corollary A.5 and Proposition 5.3, ( e //o(O). Note that £*,. = ux. on the set /l(/c) = {x e Q: |w(x)| > k} whereas C Xi = 0 elsewhere. Thus
where we have used the Poincare inequality to control the first term, allowing the constant b > 1 as a factor in the inequality. Therefore,
By Holder's inequality,
and hence
APPENDIX B
EQUATIONS WITH L°° COEFFICIENTS
65
From Sobolev's inequality, Lemma A.9,
where 1/2* = (1/2) - (1/N). If /i > fe > 0, /l(/j) c /l(fc) and we have that
Consequently
That is, we have obtained that
where Now we apply the Lemma B. 1 to derive that meas A(d) = 0,
This implies that
Definition B.3.
A bounded open set Q <=. UN is of class S if there exist two
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We have the following theorem: Theorem B.4.
Let u(x) e Ho(Q) be a solution to the equation
where a,-/*)^,- > v|£| 2 (v > 0), |a,.,{x)| < M, and j ] e LS(Q) with s > AT; let Q. be oj class S. Then u(x) is Holder continuous in Q; indeed, there exist two constants k and A with 0 < A < 1, depending on v, M, such that
Observe that the right-hand side may be of the form
In fact let F be the solution of the equation then /0 = — ^ (5/5x,-)Fx.. Since from the Calderon-Zygmund inequality (recalled in Section 2 of Chapter IV) Vx.x.e Lr(D), it follows, by the Sobolev inequality, that VXl e Lr*(Q) with r* > N.' The proof of the theorem will be given in Appendix C.
Appendix C. Local Estimates of Solutions
We recall Definition C.I. An open set Q is said of class S if there exist two numbers a (0 < a < 1) and p0 > 0 such that for all XQ e Q. Here Q(x, p) denotes the set Q n Bp(x). Consider the operator
APPENDIX C
LOCAL ESTIMATES OF SOLUTIONS
67
and assume that The aim of this and of the next sections is to prove the following theorem: Theorem C.2. Let u e Ho(Q) be a solution of the equation
where ff e Ls,s > N, and Q is of class S. Then u is Holder continuous in Q, i.e., there exist two constants K > 0 and A (0 < A < 1), depending only on v, M, Q, N, such that
In Appendix B we have already proved that the solution u is bounded. We need to establish the inequality (C.4). This will be done in several steps. First of all we need some local estimates for solutions, supersolutions, and subsolutions relative to the operator L. Definition C.3. A function u E Hioc(Q) is said to be a local L subsolution if
A function u e #{OC(Q) is said to be a local L supersolution if the function — u is a local L subsolution, i.e., if the inequality in (C.5) is reversed. Therefore a local solution is at the same time a local L supersolution and a local L subsolution. The following theorem holds, and it will be proved in the sequel. Theorem C.4. Let u be a local L subsolution; then there exists a constant K = K(v, M, N) such that, for xeQand Q(x, 2p) c Q,
Corollary C.5. Let u(x) be a local L supersolution; then there exists a constant K = K(v, M, N) such that, for x E Q and Q(x, 2p) e= Q,
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VARIATIONAL INEQUALITIES IN HILBERT SPACE
In fact — u is a subsolution and thus (C.7) follows from (C.6). Theorem C.6. Ifu(x) is a solution ofLu = 0 which belongs to Hl(Q(x, 2p)} and vanishes ondfinB2p(x), then (C.6) and (C.7) are valid. If x e Q and 2p < dist(x, <3Q), the statement is a consequence of Theorem C.4 and corollary C.5. If x e 3Q, this will follow from the proof of Theorem C.4. In order to prove Theorem C.4 and Theorem C.6 we need some lemmas. First of all we prove the following one, which is analogous to Lemma B.I. Lemma C.7. If
for h > k > k0 and p < R < R0, where C, a, /?, y are positive constants with ft > 1, then i/O < a < 1, where
Proof.
We consider the sequences
and we prove by induction that (C.I2) is true for r — 0; assume that it holds true for r and we shall prove that it is true for r + 1. Indeed, from (C.8), (C.10), and (C.I2) we get Passing to the limit as r -> + oo, we get (C.9).
Q.E.D.
Lemma C.8. Ifv is a local nonnegative subsolution relative to the operator L and i f a e Co(Q), then there exists a constant K = K(v, M, N) such that
APPENDIX C
Proof.
LOCAL ESTIMATES OF SOLUTIONS
69
By assumption we have
with
Setting
Under the same assumptions of Lemma C.8 we have
It is enough to make use of the Sobolev inequality for functions of HQ(Q). Corollary C.10.
Under the same assumptions of Lemma C.8 we have
Use Corollary C.8 and the Holder inequality. Remark C.ll. If v is a subsolution in Hl(Q,(x, R)\ vanishing on d£l n BR(x\ then we have, for x e dQ,
It is enough in Lemma C.8 and in the Corollaries C.9 and C.10 to choose a e CJ(Q) such that a = 1 in Bp(x\ a = 0 outside of B2p(x) and in such a way that | a, | < 21 p. Proof oj Theorem C.4.
Since w is a local L subsolution the function
is a local L subsolution for any constant k (see Theorem 6.6). Notice that the same function is a subsolution in Hl(Q(x, 2p)) vanishing on dQ nB 2p(x) ifk >0.
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Let a e Co(Q) with a = 1 in Q(x, p) and a = 0 outside BR(x\ where p < R with |ax| < 2/(R — p). Making use of Corollary C.10, we have
where A(k, p) = {y: u(y) > k} n Q(x, p). For h > kwe have
If we set
we have
Let <^, ^ be two positive numbers that will be fixed later. From (C.22) we get
Choose £ and 77 in such a way that then 0 needs to be a positive solution of the equation
So and we can fix rj = 1, £ = JV0/2. Setting
(C.23) may be written
APPENDIX C
LOCAL ESTIMATES OF SOLUTIONS
71
where h > k, p < R and 9 > 1. We are now in condition to use Lemma C.I for a = 2, k0 = 0, R0 = 2p; then where
(C.6) follows from an easy computation. Theorem C.6 follows with the same proof taking into account Remark C. 11. Q.E.D. With the same proof of Theorem C.4 we can establish the following: TheoremC.12. Under the same assumption of Theorem C.4 (or of Theorem C.5) the following inequality is true:
where
and k0 is any real number i f x e Q , (or k0 is any nonnegative number i f x e dQ). The proof proceeds exactly as in the proof in Theorem C.4 in order to obtain (C.24); then we make use of Lemma C.7 with kQ instead of 0 and we obtain (C.25). Consider now the solution of the elliptic equation
with/- e LS(Q), s > 2. Then we have the following: Theorem C.13. // u(x) is a local solution of (C.26), then there exists a constant K = K(v, M, N) such that for BR(x) = Q(x, R) c Q and s > N
Theorem C.14. Let xedQ and let u(x) be a solution of (C.26), vanishing on dQ. n BR(x); then the inequality (C.27) holds.
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Proof of Theorems C.I 3 and C.14. Let v be the solution in //o(Q(x, /?)) ofEqn.(C.26)andset then w is a solution in Hl(Q(x, /?)) of the equation Lu = 0, which vanishes on dQ. r\ BR(x) in the case of Theorem C.14. By Theorem C.4, Corollary C.5, and Theorem C.6 we have
But, from Theorem B.2
Therefore
Theorem C.I3 and Theorem C.14 are proved. Q.E.D.
Appendix D.
Holder Continuity of the Solutions
In this section we assume that the open set Q is "of class S" (see Definition C.I). We prove the following lemmas. Lemma D.I. Let w(x)e Hl(Q,(x, R)) and denote by A(k, R) the set
APPENDIX D
HOLDER CONTINUITY OF THE SOLUTIONS
73
A(k, R) = {ye Q(x, R): u(y) > k}; if there exist two constants k0 and 0, with 0 < 0 < 1, such that meas A(k0, R) < 6 meas Q(x, R) then, for h.>k>k0,we have
where K = K(0, N}. Moreover, making use ofCauchys inequality, we get
Proof. If k > k0, then v = min{u, h} — mm{u, k} eH1(5R(x)) and meas{xe6R(x): t;(x) = 0} = meas{BR(x) — A(k,R)} > (1 — 9) meas BR(x). From the formula of Lemma A. 13 we have
and (D.I) and (D.2) follow. Q.E.D. We need the following: Lemma D.2. Let h -» (p(h) be a nonnnegative and nonincr easing function defined for h e [/c0, M] such that the inequality for k < h < M holds, where a, ft, and C are positive constants. Then
Moreover, if we set kn = M — (M — /c0)/2", we have
Proof. Since kn - kn^ = (M - k0)/2", M - kn., - (M - kQ)l2n~\ we have from (D.3)
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Adding for n — 1 , . . . , v and noting that (p(kn)>(p(kv),we find and thus which implies (D.4).
Q.E.D.
If u is a solution of the equation Lu = 0, we set
Theorem D.3. Let u(x) be a local solution of the equation Lu = 0. Assume that for x 0 e Q
wJim? /c0 = i(M(2K) + m(2R)); then
Theorem D.4. Let u(x) be a solution of the equation Lu = 0, vanishing on dQ n BR(x0), where Q. is of class S and x0 e 5Q. Then the same conclusion as Theorem D.3 holds true. Proof of Theorems D.3 and D.4. Let a e Co(O(x0, 2/?)), where a = 1 in Q(x0, R); then from Lemma C.8 and Remark C.ll with v = max(u, k) — k, we have
here k is any number if x0 e O and k > 0 if x0 e dQ. From Lemma D. 1 we have that when h > k,
and thus Lemma D.2 implies (D.7). Theorem D.5. there exist
Q.E.D.
Under the same assumption of Theorems D.3 and D.4,
w/iere a>(p) denotes the oscillation M(p) — m(p) of the solution u in O(x0, p}.
APPENDIX D
HOLDER CONTINUITY OF THE SOLUTIONS
75
Proof. We can assume that (D.6) is satisfied, indeed, for x0 e d£l if (D.6) does not hold, the solution v = — u satisfies (D.6) since
Otherwise, if x0 e 5Q, we can assume that (D.6) is satisfied since for one of the two solutions u or -u we have /c0 = ^M(2R) + m(2R)) > 0. If k0 > 0 for u, the function v = max(w, k) — k vanishes outside 3Q n Bp(x0) and therefore, extended by zero in BR(XQ) — Q(x0, R\ Remark C.ll holds. Making use of Theorem C.I2 with k0 there given by with v large enough so that /cv > /c0, we have with 6 > 1. Because of Theorems D.3 and D.4, we can fix v in such a way that and thus
Moreover, noticing that m(R/2) > m(2R), we have
and Theorem D.5 is proved. Q.E.D. It is known that from Theorem D.5 it follows that the solution is Holder continuous (Exercise 18). At this point we may conclude in particular that a local solution of the equation Lu = 0 satisfies a Holder condition in each compact C of Q; in other words, there exist two constants K and /I with 0 < A < 1, depending on v, M, N and the compact, such that
Let us now consider the solution of the equation Lu = ]Tf= t (/)x., vanishing
on an.
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Let i;e//o(Q(x, 8p)) be the solution of the equation Lv = £J=1 (fyxi and set u = v + w; w is a solution of the equation Lw = 0 vanishing on 5Q n B8p(x). Since
we deduce from Theorems D.5 and B.2 that for/] e L*(Q) with s > N there exist constants K > 0, p0 > 0, and >/, 0 < q < 1, such that And from this estimate we may deduce analogously that u is Holder continuous in Q. Thus we deduce Theorem C.2. Remark D.6. Under the same assumptions of Theorem C.2 there exist constants K > 0, p0 > 0, and A with 0 < A < 1 such that
Proof. It is enough to remark that u — M(X O ) is still a solution of the same equation in Q(x0, p0) vanishing on dQ n BPO(XQ) and that | u(x) — u(x0) | < KpK for x e Q(x0, 2p) because of Theorem C.2. Making use of an inequality similar to (C.16) for the inhomogeneous equation and the solution w(x) — w(x0), we get (D.I 1). Q.E.D. The inequality (D.I 1) implies, because of a theorem of Morrey, that u is Holder continuous. It is worth noticing that, while in two variables it is possible to prove (D. 11) and from this deduce Holder continuity of u, in more than two variables we had to prove first Holder continuity of u and from this deduce the validity of (D.ll).
COMMENTS AND BIBLIOGRAPHICAL NOTES Theorem 2.1 was first proved in Stampacchia [3]. A different proof was given in Lions and Stampacchia [1]. The special case of a symmetric form had been treated previously. We also mention the papers of Littman et al. [1] and Fichera [1] cited in the bibliography. The first example of a variational inequality given in Section 3 introduces the truncation of an L2 function. This basic procedure connects the variational
EXERCISES
77
approach of boundary value problems to classical potential theory because it leaves the Sobolev spaces //1>S(Q) invariant. In addition, the distributional derivatives of the truncated function may be computed explicitly (cf. Theorem A.I and Corollaries A.5 and A.6). This fact was well known for the functions which are absolutely continuous in the sense of Tonelli and/or of Morrey but requires a careful proof when the spaces //liS(Q) are defined as completion of smooth functions in a suitable topology. Many results of second order elliptic differential equations in divergence form depend on this result, for instance, the weak form of the maximum principle, proved in Section 5, and those connected with the celebrated theorem of E. De Giorgi on the Holder continuity of solutions of a second order elliptic differential equation in divergence form. The results of Section 6 on the obstacle problem make use of the paper of Lions and Stampacchia [1], mentioned above and of that of Lewy and Stampacchia [1]. The theorems of Section 6 depend on this paper. We do not treat the problem of regularity of the solutions, which is postponed to Chapter IV. But in Section 7 we treat the obstacle problem in the one dimensional case. The appendices to this chapter contain some facts relevant to the content of the book. Besides the results about Sobolev spaces and inequalities, the L°° estimates and Holder continuity of solutions of second order equations with bounded measurable coefficients are proved. The presentation of the latter material is based on Stampacchia [2]. The proof of Lemma A.4 is due to M. Crandall and that of Lemma A. 13 to E. Fabes.
EXERCISES 1. Let a(u, v) be a bilinear coercive form on a real Hilbert space H, i.e., \a(u, v)\ < C\\u\\ • \\v\\ and a(u, u) > a||u||2, for all u, v e H (see Definition 1.1). Let p be a real number such that 0 < p < 2a/C2. Show that there exists a 3, 0 < S < l,such that 2. Prove Theorem 2.1 by making use of the result of the previous problem (Lions and Stampacchia [1]). 3. Let F be a convex function from H in R u { + 00} which is proper, i.e., not merely the constant +00 and lower semicontinuous. Then there exists a unique ueH such that
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Deduce from this statement Theorem 2.1 by choosing
where/e //' and IK is a closed convex subset of H. 4. Let E c Br be a subset of the ball Br of UN. The points belong to the unit sphere S: K — x|| = 1. The N — 1 dimensional measure of the set {£ e Z : y e E} is called the angle under which the set E is seen from x in Br and is denoted by ang{x, E}. Assume that there exists a constant m > 0 such that Let u e tf1' l(Br) vanish on E. Show that, for each a > 0,
where C depends only on E. (Hint: Adapt the proof of Lemma A.7 to this situation. Let T be a bi-Lipschitz mapping of O onto Br, i.e., T: Q -» Br and 0 < a < || Tx - TJ/llx - y|| < j? < + oo for all x, y € Q.) l Show that if u e H • '(Q) vanishes on T(E) with £ satisfying the condition above, then the inequality
C constant, holds true. 5. Let Q be an open bounded set of IRN and E a subset of Q. Let u be a function of f/ 1 ' *(Q) vanishing on E. How must the set £ be chosen in order that there exist a constant ft such that
(see Lemma A. 15.) Is the condition of Lemma A. 15 necessary? 6. Use Exercise 4 in order to study in the framework of Section 4 the existence of the solution of the mixed problem
where
EXERCISES
7. 8.
79
Prove Lemma A. 12. Study the variational inequality
where
9. Let Q c K N be a domain, //o(Q) : v > I/A on Q} and consider
Suppose that/e L2(Q) and u e /f 2(Q). Show that w is a solution of (*) if and only if
The form (**) is called the "complementarity form" by analogy with Complementarity Problem 1.5.4. 10. Let Q be a bounded domain with smooth boundary dQ and suppose given/e H~ l(Q) and g e L2(dQ). For A e R, study the existence and uniqueness of solutions of the variational inequality
where IK = { v E H l ( Q ) : v > 0 on dQ}. Write the solution in the "complementarity form." 11. Again let Q be a bounded domain with smooth boundary dQ and suppose that/e H~ l(^). Let
and
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Show there exists a unique solution Write this problem in "complementarity form." Show that there is a constant c > 0 such that where u, is the solution of the variational inequality corresponding to/. 12. Show that a solution to the Neumann problem for/e L2(Q) and g e L2(dfi) exists provided the compatibility condition
is fulfilled. Find a compatibility condition for this problem when/e // '(^) and g e L2(Q). 13. Let fi be a bounded domain of UN with smooth boundary dQ and let Let IK be the closed convex set of pairs Consider the bilinear form
where A, // e IR. Study the variational inequality
imposing restrictions on A, /u if necessary. 14. Let Q be a bounded domain of UN with smooth boundary dQ. and let F, P be open nonempty submanifolds of dQ satisfying (i) F n F = 0, (ii) F u F' = 3Q, (iii) F n F' is a smooth (N — 2) dimensional manifold. Given/,,/ 2 e H~ '(Q), consider the problem: To find u{, u2 e // J (Q) such that
where n is the normal to F. Find a weak formulation for this problem and show that a solution exists.
EXERCISES
[Hint: space
81
Formulate the problem for a bilinear form defined on the Hilbert
15. Let u 6 Hl'"(Br), 1 < p < + oo, where Br is the ball of RN of radius r. Show that there exists a constant j such that
where C is a suitable constant. [Hint:
We can write
for any constant y. Choose y in such a way that the measures, where the first and the second term on the right are zero, are > \ meas Br. Then make use of Lemmas A. 15 and A. 14.] 16. Prove Corollary A.6 [Hint: Assume that \u(x)\ < M and choose /„ e C( - M, M) such that fn(i) -» 0'(0 for | r | < M, f ^ ait ..., aM. Prove first that fn(u(x))ux.(x) -* O'(u(x))ux.(x) a.e. in H. Now write that (assuming that 0(0) = 0)
For/n(?) it is always possible to choose
17. Suppose that «„ e L5(Q) and 0 < wn e LS(Q) satisfy
and
Then un -> u in LS(Q). Use this fact to show that if un e Hl's(£i) and un -> w in/f ^'(Q), then |wj -> |M| in //^'(Q). As a corollary, max(« n , 0) -> max(u, 0) in H1>S(Q). Can you also use this in Exercise 14 or Lemma A.4? 18. Let a)(p), 0 < p < R, be a continuous function which satisfies
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where C > 0, 0 < r\ < 1, and a > 0. Then where K is a universal constant and
[Hint'.
By induction there is a constant K such that
if 4" < R/p < 4" +1 , then »f < r/9 < 77"+ 1 where ^ = (log R/p)/log4 and w(p) < o>(4'"/?).]
CHAPTER III
Variational Inequalities for Monotone Operators
1. An Abstract Existence Theorem
We have already encountered monotone operators in connection with the uniqueness of the solution to a variational inequality in a finite dimensional space. In this chapter, dedicated to the study of variational inequalities in general spaces, the property of monotonicity becomes important for the existence of a solution. In the last chapter we saw that a variational inequality associated to a bilinear form could be solved provided the form was coercive. In fact, the property of coerciveness implies strict monotonicity for a bilinear form, but certainly not vice versa. In general if the convex IK is unbounded it will be necessary to add hypotheses of coerciveness to achieve the existence of a solution; cf., for example, Corollary 1.8. Let X be a reflexive Banach space with dual X'. Let <•, •> denote a pairing between X' and X. Let IK <= X be a closed convex set. Definition 1.1. A mapping A: K -> X' is called monotone if The monotone mapping A is called strictly monotone if
83
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VARIATIONAL INEQUALITIES FOR MONOTONE OPERATORS
Definition 1.2. The mapping A from IK -»X' is continuous on finit dimensional subspaces if for any finite dimensional subspace M c X the restriction of A to IK n M is weakly continuous, namely, if
is weakly continuous. Definition 1.3. q> e IK such that
Finally, A is coercive on IK if there exists an element
We now prove Theorem 1.4. Let IK be a closed bounded convex subset of X(=£0) and let A: IK —> X' be monotone and continuous on finite dimensional subspaces. Then there exists a
Note that if A is strictly monotone the solution u to the variational inequality (1.4) is unique. We first prove a lemma due to Minty [1]. Lemma 1.5. Let IK be a closed convex set of X and let A: IK -* X' be monotone and continuous on finite dimensional subspaces. Then u satisfies if and only if it satisfies
Proof.
First we show that (1.5) implies (1.6). By monotonicity of A,
0 < (Av — Au, v — u) ==
for
u, u e IK.
Thus
Now we show that (1.6) implies (1.5). Let we IK, and set, for 0 < t < 1, v = u + t(w — u) e IK since IK is convex. Hence by (1.6) for t > 0,
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AN ABSTRACT EXISTENCE THEOREM
85
or
Since A is weakly continuous on the intersection of IK with the finite dimensional subspace spanned by u and w, we may allow t -> 0 to obtain
Proof of Theorem 1.4. To begin, let M <= A' be a finite dimensional subspace of X of dimension, say N < oo. We may assume without loss of generality that 0 e IK. Define
to be the injection map and to be its dual. The pairing between M' and M, <•, -> M , is chosen, of course, so that
We set toM = to n M = to r\jM and consider the mapping j'Aj: KM -» M'. Now toM is a compact convex set of M and j'Aj is continuous by hypothesis from to into M'. Hence there exists an element UM e toM such that or what is the same, since juM = UM andjvM=v, By Minty's lemma 1.5,
In view of this we define
Obviously, S(v) is weakly closed for each u e to. Moreover, since to is bounded, to is weakly compact. Consequently n«eK^( u )» a closed subset of to, is weakly compact. To conclude that it is nonempty, we employ the finite intersection property. Let {vi,.. .,vm} <= to. We claim that
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Let M be the finite dimensional subspace of X spanned by {vlt..., vm} and define IKM = IK n M as before. According to the argument given earlier there is an element UM £ KM such that viz., (1.7). In particular, Hence for any finite collection v1? ..., vm, (1.8) holds. Therefore there exists an element
which means that By Lemma 1.5, once again
The solution to (1.4) need not be unique, but it is easy to verify, using Lemma 1.5, that its set of solutions is a closed convex subset of IK. Corollary 1.6. Let H be a Hilbert space and IK c H a closed bounded convex set (^ 0). Suppose that F: IK -» IK is nonexpansive. Then F posseses a nonempty closed convex subset K c IK of fixed points. The definition of nonexpansive mapping is given in Definition 1.1. of Chapter I. Proof. The proof of this Corollary is very simple. It suffices to observe that we may take H' = H and pairing (•, •) with the scalar product of H. Now if F is nonexpansive, / — F is monotone, so we may apply Theorem 1.4. Any solution to the variational inequality for / — F is a fixed point for F. Q.E.D. Theorem 4.2 and Corollary 4.3 of Chapter I have analogs here. Theorem 1.7. Let K be a closed convex subset of X and let A: IK ->• X' be monotone and continuous on finite dimensional subspaces. A necessary and sufficient condition that there exists a solution to the variational inequality
2
NONCOERCIVE OPERATORS
87
is that there exist an R > 0 such that at least one solution of the variational inequality
satisfies the inequality
We also state Corollary 1.8. Let K c X be a closed convex set (^0) and A: K -* X' be monotone, coercive and, continuous on finite dimensional subspaces. Then there exists
The proofs of Theorem 1.7 and Corollary 1.8 are identical to those for the case of finite dimension, so are omitted.
2. Noncoercive Operators In this section we shall consider some cases when the assumption of coerciveness is relaxed. We confine ourselves to the cases of a linear operator in a Hilbert space, for simplicity. The bilinear form associated to it is assumed only to be nonnegative. This leads to a method which is called elliptic regularization. We explain it in the next two theorems. Then we consider the case when the form is semicoercive, for which we prove an existence theorem provided the right-hand side, i.e., the inhomogeneous term, of the variational inequality satisfies certain conditions. Consider the following case: when X = H, a Hilbert space, (•, •) denotes the inner product in H, and < •, • > denotes the pairing between H and H', let /e H'. Let K be a closed cone in H and let/e H'. Let a(u, v) be a bilinear form on H x H such that a(v, v) > 0. Define the operator A: H -» H' by
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From Theorem 1.4 it follows that there is a solution to the variational inequality or equivalently provided IK is bounded or IK in unbounded but A — /satisfies the condition o Theorem 1.7. However, if neither of these conditions is satisfied, such a solution MO e IK may not exist. In order to recognize whether such a solution exists the following theorem is useful. Its proof makes use of a procedure that is called elliptic regularization. Let us define We know that x is a closed, convex set. It should be noted, however, that x may be empty. Now let /?(M, v) be a coercive bilinear form on H x H and let g e H'. For each £ > 0 we consider the form Clearly this form is bilinear and coercive on H, so in view of Theorem 2.1 of Chapter II, there exists a unique solution ue of the variational inequality We have the following theorem. Theorem 2.1. Using the notation given above, % =£ 0 if and only if there exists a constant L, independent ofE, such that
Proof. Assume first that x ^ 0- Since / is a closed convex subset of IK c: H, there exists a unique solution of the variational inequality Moreover, since u0 e #, we have Setting v = M£ in this variational inequality and v = MO in the variational inequality defining ut and adding the two inequalities, we get
2
NONCOERCIVE OPERATORS
89
and by coerciveness of /?(•, •),
where L is independent of £. Assume now that \\us\\ < L, independently of e. Then there is a subsequence un of M£ such that un -> w weakly in /f. Thus w e K and we wish to show that Using Lemma 1.5 we have
Letting >/ -» 0, since un -» w weakly, we have
and using Lemma 1.5 again, we get
and therefore w e %. Q.E.D. We have an additional theorem. Theorem 2.2. With the same assumption of Theorem 2.1, if the sequence M£ is uniformly bounded, then ue -* M O as e -»• 0 strongly in H and
Proof . Since ||M£|| < L (independently of e), then % =f 0 from the previous theorem (and vice versa). Thus there exists a unique solution of the variational inequality Since ||u£|| < L there exists a subsequence un of M£ such that un -*• w weakly inH.
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VARIATIONAL INEQUALITIES FOR MONOTONE OPERATORS
In the first part of the proof of previous theorem we have proved that
Since ($(v, v) is lower semicontinuous we have
and therefore Next we show that ue -» u0 strongly in H, proving in succession that (i) w = M 0 , (ii) UE -> M 0 weakly in H, and (iii) UE -> u0 strongly in H.
Since M O e / and j8(w0, v — M O ) >
and from coerciveness of ft hence w = M O . Since M O is the unique limit of any weakly convergent subsequence of M £ , we have ue -> M O weakly in H. Therefore it remain to be shown that the convergence is strong. But Now as £ -> 0, J3(u0, ME — M0) ~* 0- On the other hand,
Hence M£ -> M O strongly in H.
Q.E.D.
In the remainder of this section, we apply Theorem 1.7 to show the existence of a solution to a variational inequality for a certain type of noncoercive bilinear form on a Hilbert space H. The result may be used to solve an obstacle problem with Neumann boundary conditions (Lions and Stampacchia [1]) (cf. Exercise 12) or the problem in Fichera [1].
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NONCOERCIVE OPERATORS
91
We consider a Hilbert space H with norm ||-|| and pairing <•, •> and assume that || • || is equivalent to p 0 (-) + PI(-)» where p
is a norm under which H is a pre-Hilbert space
and is a seminorm on H. Suppose that
is finite dimensional and that there exists GI > 0 such that
Suppose that a(u, v) is a continuous bilinear form on H x H such that
In particular, note that a(v, v) = 0 implies that v e M but not that v = 0. Let IK c= H be a closed convex set with 0 e IK. Finally, let/e H' admit the representation / = /0 +/i with
and
Theorem 2.3. Assume r/ie hypotheses (2.1). T/ien r/iere exists
Proof. Let IKR = IK n {v: \\u\\ < R}.SinceIKRisbounded,we know from Theorem 1.4 that there exists a UR e KR such that
By Theorem 1.4, there exists a solution to (2.2) if and only if ||WK|| < R for some R. Assume that (2.2) does not admit a solution. We shall show that this leads to a contradiction. In this circumstance the solution UR to (2.3) satisfies \\UR\\ = R. Let WR = (l/R)uR, R > 1. Then \\WR\\ = 1. Moreover, since K is convex, UR e IK
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VARIATIONAL INEQUALITIES FOR MONOTONE OPERATORS
and 0 e K, we see that WR e K. Once again, since 0 e KR <= K, we may take v = Oin(2.3)soby(2.1,v)
Therefore
so Since ||WK|| = 1, R > 1, there is a subsequence, again called W K , and an element w e H such that WR -» w weakly in #. Because IK is closed, w £ IK. I addition, pv is lower semi-continuous, so PI(W) = 0. Therefore w e IK n M. Consider the projection P: H -* M given by Pu = ^/, i.e.,
We know i/ exists because M is finite dimensional. We prove at this point that vtt-+ v weakly in H implies Pvn -*• Pv (strongly) in M. At this point, let (-, -)M and (•, -)M^ denote the inner products on M and its perpendicular complement M"1, respectively. Observe that we may take and the inner product (•, •) on all of H to be In particular, if C e M, then (v, 0 = (v, 0w- Since
and
for some constant C > 0. Now since M is finite dimensional there is a subsequence of vn, still called vn, for which Pi;,, -> QeM for some element 5. But by the weak convergence of vn to v in H, (Pv, 0 = (B, 0 for all CeM. Therefore & = Pv. Note that there is a constant c3 > 0 such that p0(Pwie) > c3 > 0 for R
3
SEMILINEAR EQUATIONS
93
sufficiently large. Otherwise there is a sequence of R for which PQ(P\VR) -» 0 as R -> oo. But since PI(WK) = 0(1A/R). Since the norm || • || on H is equivalent to p 0 (-) + Pi('X \\WR\\ -* 0- This contradicts ||WK|| = 1. Finally, since weM, we have Pw = w and hence If M n K = {0}, then w = 0, which contradicts (2.5), so M n IK ^ {0}. Now < 0 since 0 ^ w € M n IK. Thus Now .PwR -> Pw = w strongly in M by the foregoing and hence there is an jR0 > 0 such that Employing (2.4),
Therefore and, since ap^w^)2 > 0, But - > ft > QforR > R0, which implies that fiR < O(^R).This is impossible. The proof of the theorem is complete. Q.E.D.
3.
Semilinear Equations
As an example of the application of Theorems 1.4 and 1.7 and Corollary 1.8, we solve a problem for a semilinear equation. Let Q c UN be a bounded connected domain with smooth boundary dfl. Set where
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VARIATIONAL INEQUALITIES FOR MONOTONE OPERATORS
Let F(x, u), x eQ, — oo < u < oo, be bounded and measurable on compact subsets of H x R and continuous and nondecreasing in u. Define by
Then
because F is nondecreasing in u. So L is a coercive and strictly monotone operator on IK. Note that it is not defined on all of HQ(^), but that it is defined on the closed convex IK. It is easy to see that L is continuous on finite dimensional subspaces of Ho(H) intersected with IK. Hence there exists a unique
4.
Quasi-Linear Operators
Let a(£) = («i(£)> • • • > £N(£)) be a continuous monotone mapping from R N to (R*)', which we identify with R v , as defined in Chapter I, Section 4. In this section we shall be concerned with some special vector fields satisfying a condition about their growth at infinity and their monotonicity. More precisely, we introduce Definition 4.1. a(£) is a strongly coercive monotone vector field if for |<^| sufficiently large, i.e., |^| > R 0 ,
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95
and
for some c0 > 0. When a(£) is a strongly coercive monotone vector field
is well defined in Ho(Q) and it maps Hl0(ty into //" X^) since a£ux) e ^2(^X 1 < i < N, for w e //o(O) due to (4.1). It is strictly monotone (Definition 1.1), coercive in the sense of Definition 1.3, and continuous on finite dimensional spaces (see Definition 1.2). An example of such an operator is obtained by choosing a(£,) = £ with the associated operator
It satisfies all the assumptions above. Let /=/<> - ZyCd/dXfX/ieH'Hfl) be given and let IK be a closed convex set in HQ(£}). Then by Theorem 1.7 and Corollary 1.8, there exists a unique solution u to the variational inequality
As a second example, we consider the vector field
and the operator
Using again Theorem 1.7 and Corollary 1.8 we may solve a variational inequality relative to A for 1 < a < 2 and the convex set Indeed, since |a(OI < 1 + I^T" 1 , the operator A maps #£'4(Q) into H~ (Q). It is obviously monotone since a(£) is the gradient of the convex 1>a
96
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function (l/a)(l + £ 2 ) a/2 , continuous on finite dimensional subspaces, and coercive in the sense of Definition 13; that is, for
In fact, we first note that
as llwllf/^^jj) -* +00. Second,
where the first term on the right tends to infinity as ||u|| H i,=. (n) tends to infinity while the second and the third are bounded. Now applying Theorem 1.7, given fe H~1>a (Q), there exists a unique
For a = 1, the operator above is the minimal surface operator, or the operator of prescribed mean curvature. It is worthwhile to note that this operator is not included in the abstract theorems of Section 1. This fact will be clarified later. The operator of prescribed mean curvature is related to the vector field
It is strictly monotone because it is the gradient of the strictly convex function
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97
However, it is not a strongly coercive vector field, that is, condition (4.2) does not hold and, in fact, the variational inequality (4.6) need not admit a solutio even when/ = 0 (c.f. Serrin [1]). The case a = 1 suggests the introduction of the notion of a locally coercive vector field. Definition 4.2. A continuous vector field a(t;) = (fli(£), ..., aN(£,)} is locally coercive if for every compact C <= UN there exists a constant v = v(C) such that We remark that a locally coercive vector field is monotone and that the vector field (4.7) is locally coercive. In the sequel this lemma will be useful: Lemma 4.3. Let a(£,) be a locally coercive vector field on UN and let M > 0 be given. Then there exists a strongly coercive vector field a(£) such that More precisely, for |£| > 3M for a suitable constant K > 0. In other words the lemma affirms that we may modify a locally coercive vector field outside a given ball to obtain a strongly coercive field which satisfies (4.1). Proof of the Lemma. Let a(£) be the given vector field which we suppose locally coercive. Thus there exists v > 0 such that Let \l/ be defined on Ul such that
nonnegative and smooth for 0 < t < +00. Let g(i) be a smooth and increasing function such that
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Assume that g(i) > c0 > 0 for t > 2M. Define
where k is a positive constant we will fix later. We wish to show that a is strongly coercive and satisfies the statement of the lemma. Let
Consider first the following cases.
obvious that
We show that
In fact But | ^ | < | Y] I implies ^( | ^ \) < g( \ r\ \) and, as we have seen above, r\ • (£ — t]) < 0. Therefore the left-hand side is nonnegative. (4) 2 M < K | < \rj\ <3M; then and
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99
Hence
since, as in case (3), [0(|f|) - 0(M )]»?(£ - »/) > 0. Also
since «^ is smooth. Hence and therefore
Choosing now k > max(v, (K + v)/c0) also in case (4), we have Now let £ and v\ be any two points in IR v. If | ^ | = | Y\ \, then the previous arguments apply. Assume |£\ < \rj\. There exist at most six points £, on the line joining £ and tj intersecting the spheres of radius M, 2M, and 3M. Denote them by <^, i = 1,2,... ,7 < 6. Then and
Therefore, since two consecutive ^ belong to same ball, we have
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We remark in connection with Lemma 4.3 that if a(£) is C1 and a locally coercive vector field on UN, then the vector field a(£) may be chosen to be in Cl(RN). By the way, we remark that a C1 vector field a(£) in UN is coercive if and only if
In fact, if a(£) is locally coercive, for any compact C there is a constant v = v(C) such that and vice versa. This implies that for each £ e int C and ff e IRN, there is an e > 0 such that
or
Allowing r -> 0, we obtain
The above clearly holds for ^ € C. On the other hand, assuming (4.8),
where v is the constant of the convex hull of C. Moreover, if \da-Jdpf£)\ are bounded, then (4.1) is satisfied but (4.1) does not imply boundedness of (dajdpj)(£).
COMMENTS AND BIBLIOGRAPHICAL NOTES Although it is difficult to ascertain the first person to consider monotone operators, among the many modern initiators of the field are F. Browder
EXERCISES
101
[1,2], Goulomb, Minty [1,2], Vishik [1], and Zarantonello [1]. Certainly it was Browder who systematically employed the monotonicity of operators in the study of nonlinear elliptic problems. Studying variational inequalities in this framework is due to Hartman and Stampacchia and, independently, to Browder. The proofs of Section 1 follow Hartman and Stampacchia [1]. The material of Section 2 is from Lions and Stampacchia [1]. A special case of Theorem 2.3 had been previously proved by Fichera [1]. The presentation in Section 4 employs the results of the paper of Brezis and Stampacchia [4]. The special case when the monotone vector field is the gradient of a convex function was considered in Stampacchia [1]. The theory of monotone operators also has interesting applications to the theory of ordinary differential equations and systems (Schiaffino and Troianello [1]; Vergara-Caffarelli [1]).
EXERCISES 1. Let a(u, v) be a bilinear from on a Hilbert space H and let Au defined by be the associated linear transformation from H to H'. Show that Au is coercive on H in the sense of Definition 1.3 of Chapter III if and only if a(u, v) is coercive in the sense of Definition 1.1 of Chapter II. 2. Let H be a Hilbert space and a(u, v) a nonnegative bilinear form on H. Prove that a(u, v) is lower semicontinuous with respect to weak convergence in H, i.e., weakly in H.
3. Prove that the set of solutions to the variational inequality (1.4) is a closed convex subset of the given convex IK. 4. Let Q be a bounded domain in UN and let Jf be a closed linear subspace of L2(Q). Let y(r) be a continuous increasing function satisfying A,B constants. Finally, let F e L2(Q). Show that there exists a unique
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where Jf1 is the orthogonal complement of Jf. (Hint: Solve the variational inequality
5. Let IK be a closed bounded subset of a Hilbert space H and let F: IK -» IK be a nonexpansive mapping. Let W be a contraction on H such that W(IK) c IK. Show that for 0 < e < 1, is a contraction mapping from IK to IK. Let XE denote the unique fixed point of Fe and show that xe converges to a fixed point of F as £ tends to 0. 6. Let Q be a bounded open domain in R'v and let E and F be closed nonempty subsets of Q. Let \l/ £ Hl(ty and denote by IK the convex set Let
Show that there exists a solution to the variational inequality (cf. lordanov [1]). 7. Let T: L2(Q) ->• L2(Q) be a coercive monotone operator, where Q c= IR* is a bounded open set. Show that the problem: Find u E L2(£l) such that
or equivalently,
can be formulated as the variational inequality: Find
where (cf. H. Brezis and L. Evans [1]). 8.
Show that the form
EXERCISES
103
is coercive on #o(Q) n H2(£l), Q c R N bounded. Generalize this result for the form
where ,4* and A2 are linear elliptic operators of the form (cf. Brezis and Evans [1], Ladyzhenskaya and Ural'tseva [1]). 9. Let Q c R w be a bounded domain and let /I1 and A2 be two elliptic operators given by Show that a solution of the problem of Bellman \\~\\find where fe L2(Q) is given, is a solution of the variational inequality: Find
where Establish conditions which insure the existence of a solution to this variational inequality (Brezis and Evans [1]). 10. Let IK be a closed convex subset of a reflexive Banach space X with dual X' and let A: IK -> X' be monotone and continuous on finite dimensional subspaces. Then a necessary and sufficient condition for the existence of a solution of the variational inequality is that there exists a nonempty bounded subset C of IK such that for each u1 6 IK — C, there exists v e C such that Moreover, assume that IK is strictly convex in the sense that for all u, v e IK, the points
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are in the interior of IK, and also that A'1^) consists of no more than one element. Then the solution of (I) is unique. This condition says that uniqueness for the equation Av — 0 implies uniqueness for the variational inequality (I) provided IK is strictly convex (Coppoletta [1]). 11. Show that the operator of mean curvature is not strongly coercive, but that the vector field p/(\ + p2)1''2islocally coercive. 12 Let Q c: UN be a smooth domain with boundary 3Q. Let and
with
Show that there exists a solution of the variational inequality
Write this problem in its complementarity form and discuss the uniqueness of its solutions.
C H A P T E R IV
Problems of Regularity
1. Penalization We were able to give an abstract existence theorem for the solution of a variational inequality in part because our notion of solution was very weak. In this chapter we study the smoothness of the solution in some detail. The solution of an elliptic boundary value problem enjoys a degree of smoothness depending on that of its data. However, we have already seen, for instance, that the solution of an obstacle problem associated to a second order equation cannot in general be of class C2 regardless of the smoothness of the data. This illustrates that our regularity theory will have to be especially invented for variational inequalities. To initiate our discussion, we shall briefly describe a method called penalization. We want to emphasize that this method, often used to demonstrate the regularity of solutions, may sometimes be used to attain their existence when the assumptions of general theorems do not apply. This we shall see later. The method of penalization consists in substituting the variational inequality by a family of nonlinear boundary value problems and demonstrating that their solutions converge to the solution of the variational inequality. The principal difficulty lies in obtaining suitable a priori estimates. We stress that there are different choices of penalization. 105
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2. Dirichlet Integral Let Q c (RN be a bounded connected domain with smooth boundary
We shall suppose that/and — Ai/> are measures with the properties /£LS(Q)andmax( —A»// — /, 0)€LS(Q)for someN < s < oo.
(2.2) Under these assumptions it will be shown that the solution u of (2.1) is in /f 2 ' a (ft)nC 1>A (fi), A = 1 ~(N/s). A major tool will be the following estimate, which is based on a theorem of Calderon-Zygmund. Suppose that
Then
Formally, it is evident that the solution u of Problem 2.1 satisfies, with / = {xeQ:i/Oc) = ^(x)},
so we would like to obtain u as a solution to the Dirichlet problem
Recalling Section 6 of Chapter II, however, we know that
2
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107
(in the sense of distributions), where n is a certain nonnegative measure whose support is in 7. Combining the two representations leads to the formal description
Denoting by $ the function
the formula above may be rewritten Unfortunately, we have no way to ensure a priori that n is absolutely continuous, which is certainly implied by (2.6), nor can we determine that a solution v to (2.4), where / = I(u\ u the known solution of (2.1), is an element of (K. This notwithstanding, (2.6) leads to consideration of the problems
where #£ is a sequence of Lipschitz functions which tend to 9 defined in (2.5) almost everywhere on K. Note that since u is a supersolution of — A — / we know that so we replace — Ai/^ —/by max( —A«^ — f, 0). The boundary value problem
is called the penalized problem. Lemma 2.2. Let B(t), — oo < £ < oo, fee uniformly Lipschitz, nonincreasing, and satisfy 0 < B(t) < 1. Assume (2.2). Then f/iere exists a unique
Moreover, where C0(s, Q) is tne constant of (2.3).
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Proof. First note that #(w -
is in H~ H^)- We claim that it is strictly monotone and coercive on HQ(&). Since 3(0 is nonincreasing,
We compute
This proves that L is strictly monotone and coercive. Furthermore, wn -*• w in HO(^) implies Lwn -» Lw weakly in H"1^)* which certainly implies that L is continuous on finite dimensional subspaces of Ho(fi). We may apply Corollary 1.8 of Chapter III to obtain the existence of w. The estimate follows from (2.3). Q.E.D. We now specify
and consider the Dirichlet problems, for e > 0,
We turn to the main result of this section. Theorem 2.3. Assume (2.2) and let u denote the solution to Problem 2.1. Then u e H2>S(Q) n Cll>l(D), A = 1 - (N/s). In addition, let ue, e > 0,
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109
Proof. We first claim that uee IK. Set ( = u£ — max(w£, i}/) < 0, an element of H£(Q). We want to show that £ = 0. By (2.9)
and by Green's theorem
Subtracting these two equations,
Now
so
At this point we observe that C(x) < 0, which is the same as ue(x) — \j/(x) < 0, implies that &e(ue(x) —
Hence C = 0. By Lemma 2.2, the family {wj is bounded in H2>S(Q) and hence contains a subsequence ut, which converges weakly in /f 2>s (Q) and uniformly in H to a ti e //2>S(Q) c C1"1^), A = 1 - (AT/s). Since M£ e IK, u e IK. To show that u is a solution to Problem 2.1 we use Minty's Lemma 1.5 of Chapter III, applied to the monotone operator defined by (2.9). Let v e IK and suppose v > \l/ + d for some positive 5. Then
For e < <5, &e(v — \l/) = 0. Now allowing £ -> 0 we obtain, by weak convergence,
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We now let 6 -> 0, whence it follows that
Employing Minty's lemma once again, we conclude that u = u, the unique solution to Problem 2.1. The solution being unique, the entire family UE converges weakly to u in H2'S(Q). Q.E.D. The approximations we have employed possess some interesting properties. Theorem 2.4. Assume (2.2). Let u denote the solution to the variational inequality (2.1) and w £ , e > 0, denote the solution to the penalized problem (2.7) with 9K defined by (2.8). Then {wj is a nondecreasing sequence and
Proof. To prove that (wj is a nonincreasing sequence as e decreases observe, first of all, that the sequence «9£(r) decreases as e decreases. So given E < e' we write, for £ 6 #o(Q),
Since the function
is decreasing,
2
DIRICHLET INTEGRAL
111
and since #£(f) < &e-(t) for e < e' and ( < 0, Therefore
Since u£ -» u, it follows that u£ > u for e > 0. Now observe that u + e is a supersolution to the approximating equation. Namely,
because u is a solution to the variational inequality. Moreover, From (2.7) and (2.10), choosing C = min(w + £ - «£, O)eH^(Q), < 0, we conclude that
Similarly, one may choose a sequence of approximating functions u£ to be solutions to the problem
with
The family M£ has the property that it increases as e decreases. For ease of exposition we chose a convex set associated to homogeneous boundary values to illustrate our method of penalization. A general guideline is that the solution of the variational inequality is of class //2>S(Q) whenever the solution of the associated boundary value problem enjoys this property. It is easy to check that the conclusion of Theorem 2.3 is valid for the convex IK = {v 6 H 1 ^): v > \l/ in Q, v = g on dfi}, where g e //2-s(Q) and \l/ < g on
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d£l. We offer a further example useful in the study of filtration (cf. Chapter VII.) Let Q c UN be a domain with Lipschitz boundary d£l which may be written 5Q = d^l u d 2 Q, where c^Q and d 2 Q are disjoint open Lipschitz hypersurfaces. In particular, the trace from H *(Q) to L 2 (djQ) is well defined. Set and assume that the Ho(Q) norm is also a norm on V, that is, that the form
is coercive on V. Let/e L2(Q), g, t// e H J (^) with \j/ < g on c^Q, and set From Theorem 2.1 in Chapter II we conclude that there exists a unique solution u of the variational inequality
Written in the complementarity form, which we do not justify, w is a solution of
where v is the exterior normal to d 2 Q. Thus (2.12) is related to a mixed boundary value problem (cf. Chapter II, Mixed Problem 4.9). Our object i to study the smoothness of u for which we impose the hypothesis (2.14) below. We stress that (2.14) does not always hold: its validity depends on djQ, d2£l, and also g. Let/e LS(Q) and g e H2 S(Q), some s, N < s < oo, and determine w by
Assume that for a constant C0 > 0 independent of/
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113
In the formulation of (2.13) we intend of course that
Theorem 2.5. LetfeLS(Q) and \l/,geH2-S(Q), \l/ < g on d&for some s,N<s
where Be is defined by (2.8). The estimate (2.14) applies to « £ . Details are left to the reader.
3.
Coercive Vector Fields
Coercive vector fields were discussed in Section 4 of Chapter III. In order to obtain regularity of the solution of a variational inequality, we must restrict our attention to vector fields which are at least C1. In this situation Definition 4.1 of Chapter III becomes Definition 3.1. A vector field a(p) is C1 strongly coercive if a,-(p) e Cl(RN\
1 < i < N,
for some constants 0 < v < M < oo. Let Q be a bounded open subset of UN with smooth boundary <3Q. Using (3.1) it is easy to verify that the quasi-linear operator
is well defined, monotone, and coercive. If vn -> v in Ho(O), then, by the smoothness of a,(p), Avn -»• Av weakly in f/"1^), s° it is clear that A is
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continuous on finite dimensional subspaces of Ho(Q). In addition we pointed out at the end of Section 4 of Chapter HI that
so the operator A is uniformly elliptic in the ordinary sense. Consider a function \l> e C2(Q) which satisfies
and define We consider again Problem 3.2. Given fe U°(&\find
The existence of a solution to (3.2) has already been shown in Section 4 of Chapter III, indeed, under much weaker assumptions about ^ and/ Here we discuss its smoothness. As before, we define
and consider the Dirichlet problem As in the previous section, one checks that the operator is strictly monotone, coercive, and continuous on finite dimensional subspaces of Ho(Q). Therefore by Corollary 1.8 of Chapter III there exists a unique solution to (3.3) for each e > 0. We now state a theorem which wil serve us in our proof of regularity. Its proof is discussed in the appendix. Theorem 3.3.
Let a(p)beaC1vectorfieldandletg6L°°(Q). Suppose tha
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115
and that
for some numbers 0 < v < M < oo. Then for each s there is a c0 = c0(s, v, M, ft) such that
As an immediate corollary we have Corollary 3.4. Ifa(p) is aC1 strongly coercive vector field (see Definition 3.1) and g e L°°(Q), then the solution u of (3.4) satisfies
In particular we state Corollary 3.5. Letfe L°°(Q) and let ue, a > 0, denote the solution to (3.3). Then « £ eH 2 ' s (Q), 1 < s < oo, and
These considerations lead us to Theorem 3.6. Let u denote the solution to Problem 3.2. Then u e //2>S(Q) n C1 "*(£!)/or 1 < s < oo am/ 0 < A < 1. /n addition, w/jere c0 = c0(s, Q, v, M). // w£ denotes the solution to (3.3), ffce« w£ -»• u weakly in #2>S(Q) as e -» 0. Proo/ Of course the spirit of this theorem is the same as that of Theorem 2.3. Because of its utility, we explain again the highlights of the arguments. The first step is to show that ut e IK. This is accomplished by truncation. Th function £ = ue — max(w£, «/^)e//o(Q) and satisfies £ < 0. Our object is to show that C = 0. From (3.3)
and by definition, since C e HQ(&) and ij/ is smooth,
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IV PROBLEMS OF REGULARITY
Now if C(x) < 0, then #£(«£(x) —
On the other hand,
(in the sense of distributions), as we recall from Appendix A of Chapter II, Theorem A.I. Consequently
so C = 0.
From Corollary 3.5, {uf} form a bounded set of H2>S(Q), from which we may extract a weakly convergent subsequence ue. The limit u of this subsequence is shown to be a solution of (3.2) in a manner identical to that of the proof of Theorem 2.3. Namely, one applies Mirny's Lemma 1.5 of Chapter III twice. We leave the details to the reader. Since the solution u of (3.2) is unique, the only limit point of the {ue} is u. Q.E.D.
4.
Locally Coercive Vector Fields
The existence theory of Chapter III cannot be applied directly, for example, to a variational inequality associated with the first variation of the area integrand, because this vector field is not strongly coercive; cf. Chapter III, Section 4. To achieve the existence and the regularity of the solution to such a problem, we shall prove an a priori estimate for the gradient of the solution. We then modify the vector field and avail ourselves of the previously developed theory.
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LOCALLY COERCIVE VECTOR FIELDS
117
This will require hypotheses about the nature of Q. The reader will recall that the mere existence of a solution to the boundary value problem
rests on certain geometrical conditions between / and dQ. Recall that a vector field a(p) = (av(p\ ..., aN(pJ) e Cl(UN) is called locally coercive (see Definition 4.2 of Chapter III) if for each compact C <= UN there exists a v = v(C) > 0 such that Let \l/ e C2(Q) satisfy
and set Note that the convex is now of Lipschitz functions, not a reflexive Banach space. We shall assume that Q is convex. Consider Problem 4.1.
To find
It is easy to see, by (4.1), that a solution of Problem 4.1 is unique provided it exists. We now state our a priori estimate. Lemma 4.2. Let Q be convex and let a(p) be a C1 locally coercive vector field. Suppose that u e C^H) n #2(Q) is a solution to Problem 4.1. Then
Proo/ First we note that the continuously differentiable function u — ^ attains its minimum at any point x e /. Hence Now u > \l/ in Q — /, from which it follows that
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IV PROBLEMS OF REGULARITY
in the sense of distributions. For any direction ^, we choose £ 6 C£"(Q — /) and compute
where Recalling that a(p) is locally coercive and \ux\e L°°(Q), we have and
for some numbers 0 < v < M < GO. Hence we may apply the weak maximum principle to (du/d£)(x\ Theorem 5.5 of Chapter II, which yields
in view of (4.3). Again using (4.3), we infer that
By convexity of Q, given x0 e 5Q there is an affine function TT(X) with the properties
We now show that
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119
To this end, set w = min(M, n) e IK; then
Also, w — u e //o(Q) and
so
Subtracting, we obtain
Since the vector field a(p) is strictly monotone, w = u, i.e., u < n. In a similar way it follows that u > 0 in Q. With n an exterior normal direction to Q at x0 e dQ, so
Consequently, (4.4) assumes the form
Finally, given £ > 0, we may choose x0 e Q and £ e R^ such that
whence it follows that
Our main result is Theorem 4.3. Let a(p) be a C1 locally coercive vector field on IR* and /ef Q fee convex wirli smooth boundary in UN. Let \l/ e C2(H) satisfy maxn ^ > 0 and ij/ < Q ondQ and set
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Then there exists a unique
Moreover, u e H2-S(Q) n C1>A(Q)/or l < s < o o , 0 < A < l . Proof. Let M = supn|^x| and consider a strongly coercive vector field d(p) with the property
Such a vector field was constructed in Lemma 4.3 of Chapter III. According to Theorem 3.6, there exists a unique solution u e #o(Q) to the problem
Since by Theorem 3.6 ueC 1 ^), we may apply the preceding lemma to conclude Hence ^(uj = a,(«x) and M is a solution to (4.2). As we have already remarked, u is unique in IK. Q.E.D.
5.
Another Penalization
In this section we introduce a different form of penalized equation, based on the concept of a monotone graph, and show that its solutions converge to the solution of the variational inequality. This will also provide us with a second proof of regularity for the solution. Regularity is further studied in successive sections. We always assume that Q <= R'v is a bounded connected domain with smooth boundary dQ. Choose a function a(f) e C°°(IR) such that
and define
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121
Note that
Let ij/ e C2(H) be an obstacle, that is, maxn ty > 0 and \l/ < 0 on dQ, and let /e L°°(Q). Let a(p) = (a^p),..., MP)) be a strongly coercive C1vectorfield. We shall approximate the solution u to Problem 3.2 by solutions ut of the equation Note that the operator
is strictly monotone, coercive, and continuous on finite dimensional sub spaces because f$e(i) is increasing. Hence there exists a unique solution to (5.2). It is smooth in Q by a well-known theorem about elliptic equations analogous to Theorem 3.3. Lemma 5.1. Let UE, 0 < £ < 1, denote the solution to (5.2). Then
where c 4 = 2(2||/||L.(n) + M^||L-(n) + II^IL« ( n)) ««^ c0 = c0(v, M, s, Q) is the constant of Theorem 3.3. Proof. Let ae(0 = (l/e)a(f) = 0t(t) - £t. We first show that a£(w£ - i/O is bounded independently of e. Now a£(r) = 0 for t > 0 so a£(«£ — i/O = 0 on dQ. Subtracting A^ from each side of (5.2) and multiplying the resulting equation by [ae(u£ — i/0]p~1, P > 0 even, we obtain
We integrate the first term by parts. There results
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PROBLEMS OF REGULARITY
by monotonicity of a(p), evenness of p, and our assumption that v!E(i) > 0. Consequently
Each term on the left-hand side is nonnegative because p is even and f«(0 > 0, so
Applying Holder's inequality on the right-hand side, we see that or
In the limit as p -> GO, We next show that ue itself is bounded. We write (5.2) in the form where g e Lao(Q). Multiplying both sides of the above by (u e ) p ~ 1 , p even, and integrating, we deduce that
The first term is nonnegative, so
Applying Holder's inequality as before, it follows that
whence it follows that
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123
Returning now to Eq. (5.2) and applying (5.3) and (5.4), we conclude that for 0 < £ < 1
The conclusion now follows from Theorem 3.3. Q.E.D. Theorem 5.2. Let u denote the solution to Problem (3.2) and «£, 0 < e < 1, the solution to (5.2). Then
Proof. Let {ut<} be a subsequence of {u£} which converges weakly to a function u e //2>S(Q). We appeal to a familiar argument to show that u is the solution of the variational inequality. Since is monotone, Minty's lemma implies that
Suppose now that v > \l/ + d for some d > 0. Then so by weak convergence
If u(x0)—
and some constant C > 0. Hence This is a contradiction. Hence w e IK.
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IV PROBLEMS OF REGULARITY
Since any v e K may be approximated by functions v e IK with v > \l/, we see that
Applying Minty's lemma once again, it follows that u = u, the solution to Problem 3.2. Q.E.D.
6. Limitation of Second Derivatives We prove now that second derivatives of the solution to a variational inequality are bounded if some smoothness assumptions are satisfied by a(p),»//, and/ This result serves in the analysis of the coincidence set of the solution u. The theorem is first proved for the approximations defined by (5.2). Although we prove a theorem relative to a strongly coercive vector field, we observe that whenever a Lipschitz function is the solution to a variational inequality of the type (3.2) relative to a locally coercive vector field, the vector field may be altered by Lemma 4.3 of Chapter III so that the function is also the solution to a problem for a strongly coercive field; cf. Section 4. As usual, Q c= (RN is a smooth bounded connected domain with boundary dQ. Suppose that is strongly coercive,
Let a(r) e C°°(R) satisfy
and set pe(t) = et + (l/e)a(r). Note that &(f) > 0 and &'(0 < 0 for t e R. We denote by UE the solution to the problem
6
LIMITATION OF SECOND DERIVATIVES
125
We have observed in Section 5 that the solution UE exists and is unique. Moreover, the functions [UE : 0 < £ < 1} lie in a bounded set of #2>S(Q), for each 5, 1 < s < oo. Theorem 6.1. Assume (6.1) and (6.2). Then for each compact K a Q there is a constant C = C(K) > 0 such that
The proof depends on estimating pure second order partial derivatives from below. To accomplish this we shall consider an inequality obtained by differentiating (6.3) twice. This approach does not appear promising at first glance; however, the higher order derivatives of u£ may be expressed as divergences of functions in LS(Q), 5 > N. To this situation we apply Theorem B.2 of Chapter II. Instrumental in obtaining the differential inequality is the concavity of j8. Lemma 6.2. Let UE, 0 < £ < 1, be the solution of (6.2) and K <= Q be compact. Then there exists a C2 = C2(K) such that
where d/d£ is any directional derivative. Proof. Let ee (0, 1] and set u = ue. In the proof, we shall depress the dependence of the various quantities on e and denote by const any constant independent of e, e < 1. It will be convenient for us to assume that u e C4(Q), for which we temporarily impose the conditions that \l/,f, and a(p) be very smooth. Since the final estimate for C2 will depend only on \\(d/d£)f\\L<*,(n}, \\^\\H^{ci), and sup Uik | [d2a/(dp,-dpk)](Mx)|at w ju be ciear that it is valid assuming only (6.1). We observe that for each 7, 1 < j < N,
Therefore
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Since ft"(t) < 0,
Combining these two relations, we obtain
Let us denote by L = Lt the linear operator
and note that
and
for some 0 < v < M < oo independent of e. This is a consequence of the strong coerciveness of a(p) and the boundedness of ux, of Lemma 5.1. Restricting our attention to a compact K cr Q, let C 6 Cg'(Q) satisfy C = 1 on K, C > 0 in Q. One verifies that for any w e C2(Q),
Multiplying (6.4) by C and setting w = u^ in (6.6) yields
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LIMITATION OF SECOND DERIVATIVES
127
We rewrite this in the form
With exception of the terms involving/, any term on the right-hand side is the product of second derivatives of u multiplied by a bounded function or else the derivative of such a quantity. So we can rewrite (6.7) in this manner:
where
for any s, 1 < s < oo. According to Theorem 5.5 of Chapter II there is a solution w e Ho(Q) to the problem
and by Theorem B.2 of Chapter II
Hence by (6.8),
Finally, let x0 e Q be a point where C"^ — vv achieves its minimum. If C(x0) = 0, then
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PROBLEMS OF REGULARITY
whence it follows that
Now, ((x0) > 0 and /?'(/) > 0 imply that u^(x0) > t/^(x0), so
Consequently, In particular, The lemma is proved.
Q.E.D.
Proof of Theorem 6.1. As before, we fix a K c: Q, compact, and depress the dependence of the various quantities on e, 0 < e < 1. At a given point x0 e K, we assume uXiX.(x0) = 0 for i 7^ 7. Let us observe that this entails no loss in generality. Let C be an orthogonal matrix and set and
where 'C denotes the transpose of C. It follows that u is a solution to the equation
where ij/'(y) = «/^(x), /'(y) =/(x), and Q' = x0 4- C(fi — x0). The vector field a'(p) is given by the formula where the vectors a, a, and p are written as columns. From (6.1) it is clear that
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LIMITATION OF SECOND DERIVATIVES
129
a'(p) is strongly coercive; indeed,
where v is the constant of coerciveness of the original a. Also
Hence the conclusions of Lemmas 5.1 and 6.2 hold for u with constants C,, i = 1,2, increased in modulus at most by a function of the dimension N alone. We choose C so that u'y.y.(xo) = 0 for i ^ j. In other words, having fixed beforehand x0 e X, we may as well assume that ux.x.(xQ) — 0 for i ^ j. At this point with Hence
We may now choose an arbitrary coordinate system. It follows that in any coordinate system (x 1 ? ..., XN), for an appropriate C(K).
Q.E.D.
Theorem 6.3. Assume (6.1) and let u denote the solution to Problem 3.2. Then for each compact K c= Q there is a C(K) such that The proof is evident from the weak convergence of ut to u in //2>S(Q) and Theorem 6.1. Corollary 6.4. Assume (6.1) and let u denote the solution to Problem 3.2. //, in addition, \l/ < 0 on dQ, then
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IV PROBLEMS OF REGULARITY
Proof. Since ij/ < 0 on <3Q there is a neighborhood of 5Q in Q where u > ^ and hence where Au == f. It follows from the standard regularity theory that u e C2 of this neighborhood and from Theorem 6.3 that it is in H 2 - °° of its complement. Q.E.D.
7.
Bounded Variation of Au
In this section we show that Au is a function of locally bounded variation when (6.1) is satisfied and if/ e C3(Q). This will then provide us with a geo metric property of the coincidence set. Precisely, for co C Q open, we say that/eL^co) is of bounded variation in co, or/eBK(co), if there exists a constant C > 0 such that
If/e BF(co), we define its variation Vmftobe
If Ko/ < °° f°r eacn co c= co c Q, we say that/e BKloc(Q). It is easy to see that this coincides with the classical definition of bounded variation when N = 1. Indeed, the functions/of BF(co) are those functions whose distributional derivatives are (signed) measures on co. This follows from the theorem of Schwarz because the distributions 7] e &'(&), 1 < i < N, given by
is nonnegative. Finally, if/n -*/in L1 weakly and/n e BK(co), then
Now suppose that E c fi is a Borel set and that cpE e BK(Q). Then E is said to be of finite perimeter (in Q) and its perimeter P(E) is defined to be If £ cz Q is a compact manifold with smooth boundary dE, then P(E) is the surface area of dE in the classical sense. So the finiteness of P(E) for a general E is an indication of the area of its boundary. In particular, let us suppose that
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BOUNDED VARIATION OF Au
131
£ is a set of finite perimeter. Let us denote by uh i = 1,..., N, the measures which are the distributional partial derivatives of
If supp C n d£ = 0, then, obviously,
so Formulas (7.2) and (7.3) give us a version of Green's theorem for a set of finite perimeter. Finally observe that if £ has finite perimeter, F c: £, and meas(£\F) = 0, then P(£) = P(F). This is evident from (7.1). As in the previous section, we shall prove a local result. The reader who wishes to visualize the analytic ideas of the proof is encouraged to neglect the function £ below together with all boundary integrals. Theorem 7.1. Assume (6.1) and (6.2). In addition suppose that ^ e C3(Q). Let ue, 0 < e < 1, denote the solution to (6.3). Then for each o> <= co c: Q, open, there is a constant C(a>) > 0 independent o/e, 0 < e < 1, such that Before proving this theorem, let us mention its consequences. Theorem 7.2. Assume (6.1) and assume ty 6 C3(fl). Let u denote the solution to Problem (3.2). Then Au e BKioc(Q). Proof. This follows immediately from the lower semicontinuity of the variation. Q.E.D. Corollary 7.3. Assume (6.1) and that ^ e C3(Q). Suppose that Let u denote the solution to Problem (3.2). Then the coincidence set I = {x 6 Q: u(x) = i/f(x)} ofu is of locally finite perimeter in Q. Proof.
Since 4n e BKloc(Q) and Q* A\J/ -fis C1^),
is of locally bounded variation in Q. Q.E.D.
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IV PROBLEMS OF REGULARITY
Proof of Theorem 7.1. Again we suppress the dependence of the various quantities on e, 0 < e < 1, and denote by const any constant independent of e. We select an approximation to sgn t, that is, a sequence of smooth functions y,)(t), 6 > 0, satisfying
Given co c= m c Q, we select ( e Co'(Q) satisfying C = 1 on CD and 0 < C < 1 inQ. Now differentiating (6.3) with respect to the direction
For y = yd(u^ — (f/s), we compute that
The first term on the right-hand side is nonnegative, that is,
7
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BOUNDED VARIATION OF Au
because a(p) is monotone and / > 0. We infer the inequality
Observe that
to conclude that
for any fixed s < oo. This estimate is independent of e, 6 so we use (7.3) in order to derive that
cpmst Recall now that
pointwise in Q, so by the bounded convergence theorem
Returning now to (7.2),
The theorem follows after an integration by parts. Q.E.D.
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8. Lipschitz Obstacles In this section we shall refine our study of obstacle problems to include Lipschitz obstacles and their important subclass of polyhedra. For simplicity we confine ourselves to the conditions of Section 4, where Q <= R* is a convex domain with smooth boundary dQ and a(p) is a locally coercive C1 vector field. Given an obstacle ij/ e H1' °°(Q), namely,
we shall consider Problem 8.1. To find w e IK: J n a,(M,c)(i; — u)x. dx > 0 for y e IK, where IK = IK, = {v e HIQ °°(Q): v > \j/ in Q}. Theorem 8.2.
There exists a unique solution u to Problem (8.1). Moreover,
ll"*llL«°
Proof. This theorem is really just a corollary of Theorem 4.3. Let M = supn | \l/x | and let \l/a e C2(Q) satisfy
and \l/a-+\j/
uniformly in
Q
as
with M^ the solution of (8.1) for ij/a, we know from Theorem 4.3 that C independent of a for a small. Hence we may extract a subsequence, still called ua, such that ug-* u uniformly and weakly in Hl(£i) to some u e C(Q) To show that u is a solution, we avail ourselves once again of Minty's Lemma 1.5 of Chapter III. Let y e IK satisfy y > (// in Q. Then v > \l/a for ff < a', some a', whence it follows that
As a -> 0 we obtain
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LIPSCHITZ OBSTACLES
135
It is easy to see that (8.2) holds for any v e IK, so again by Minty's lemma, we conclude that u is a solution to (8.1). Uniqueness is a consequence of the strict monotonicity of a(p\ as usual. Q.E.D. Theorem 8.3. Let G denote the class of all Lipschitz super solutions of Av = — (ai(vx))x. which exceed i/> in Q and are nonnegative on dQ. Let u denote the solution to Problem 8.1. Then
The proof of this is almost identical to that of Theorem 6.4 of Chapter II and is omitted. As a corollary, however, we state Corollary 8.4. Let \l/ and
Proof.
Let m = ||
implies that Now u + m is a supersolution, so by Theorem 8.3 Similarly, u < u' + m. Hence \u — u'\ < m. Q.E.D. Envisage now this situation. Suppose \l/ e // 1>00 (Q) is an obstacle which satisfies The sequence \l/a -> i/^ of Theorem 8.2 may be chosen so that \l/a -»\li uniformly together with its first and second derivatives on compact subsets of U. For a fixed
136
IV
PROBLEMS OF REGULARITY
field considered here is only locally coercive is of no importance since we may replace a(p) by a strongly coercive a(p) independent of
Given open subsets V" c: V cz U it is possible to localize the estimate of Corollary 3.4 by replacing UE by CW E , C an appropriate cutoff function. There results an estimate Passing to the limit as e -» 0. so
keeping in mind that HwJI/^n) < \\i(/a\\L«,(n). Since ua -> u weakly in H^Q) we conclude that and
We have proven Theorem 8.6. Let \l/ be a Lipschitz obstacle and u the solution to Problem &1. //> e C2(LO for some open U c Q, then w e H 2 ' S ( U " ) for every U" c 17" c L/. As a consequence we state this corollary, referred to as the principle of the face of contact. Corollary 8.7. Let a(p) be a C2 locally coercive vector field and \jj a Lipschitz obstacle satisfying ij/EC2(U)for some U c Q open. Suppose that A\l/ = 0 in U. Then Au = 0 in U and either or
where u denotes the solution to Problem S.I for i/f.
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LIPSCHITZ OBSTACLES
137
Proof. According to the previous theorem, ueH2<s(U") for any U" <= 0" c= U open. Hence we may compute Au by Theorem 8.6. So Au = 0 in U" n (Q - /), obviously. In U" n /, and so
Therefore, (in the sense of distributions). Smoothness of a(p) implies that u, ^ e C 2> A((7") for some A > 0. By the mean value theorem, where
So M — \l> is a solution to the equation where Owing to the smoothness of u, i^, and a^p), a u e C 1>A (L/"). We may apply a classical maximum principle to infer that unless For example, if \j/ is the height function of a polyhedron, the coincidence set consists of a number effaces and parts of edges and may well consist only of portions of edges. We may use the preceding theorem to study the obstacle problem for obstacles defined only on a portion of Q. Suppose that E c= Q is a compact subset and \j/ e C(E) admits an extension to H^Q). Let a(p) be a strictly coercive C2 vector field and, as before, set
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PROBLEMS OF REGULARITY
From Theorem 1.7 and Corollary 1.8 of Chapter III, we infer the existenc of a solution u of the variational inequality
Now let w e //*(Q\£) denote the solution of the problem
and define
We shall impose the hypothesis ij/(x)
is Lipschitz in Q.
Hence there is a Lipschitz solution u of the problem
We shall prove that u = M. In particular, (8.3) admits a Lipschitz solution. Since A^ = 0 in Q\£, by the previous theorem, u — \js or u > \j/ in Q\£. In either case Given v e K^, and e > 0, define and note that v'e = 0 near 5Q u E. Hence i£ e Ho(Q\£). Set Then by (8.5). Now allowing e -> 0, Hence by (8.4)
9
MIXED BOUNDARY CONDITIONS
139
A frequently studied class of obstacles like \l/ above is the class of thin obstacles, or obstacles defined on lower dimensional submanifolds of Q.
9.
A Variational Inequality with Mixed Boundary Conditions
Here we show how the preceding theory leads to the resolution of a problem which bears a formal resemblence to a mixed boundary value problem. Namely, we consider a variational inequality for a linear elliptic second order operator on a bounded domain fi of UN whose solution lies above a given obstacle and assumes prescribed boundary values only on a portion of dQ. Let Q be a bounded open set with smooth boundary 5Q, for sake of simplicity, and let c^Q, d 2 Q be two smooth and disjoint open subsets of d£l such that dQ = d^Q u d^Q. Set Let (^ e //J(Q) satisfy \j/ < 0 on djQ and let Now consider a uniformly elliptic second order operator where the coefficients a f /x) satisfy the usual condition of ellipticity. The operator A defines the bilinear form
Let T be a distribution determined by
where/0,/j, . . . , f N , g belong to suitable Lp spaces. If c^Q is sufficiently large so that the Poincare inequality holds for the functions of V, then the form a(u, v) is coercive on V. Therefore the theorem of existence, Theorem 2.1 of Chapter II, applies and we may conclude that there exists a unique solution of the variational inequality Our discussion centers about the regularity of the solution to the variational inequality (9.1):
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Theorem 9.1. I f u e K is the solution of the variational inequality (9.1), where
and then we have, for a suitable constant C,
almost everywhere in Q. For any real number k > k0 = max(maxn \j/, 0), let v = min(w, k) e IK. Substituting this v in the variational inequality (9.1) and repeating the arguments similar to that used in Theorem B.2 we get for the function an inequality of the type Employing Lemma B.I the theorem follows.
Q.E.D.
Theorem 9.2. If ueh. is the solution of variational inequality (9.1), where the assumptions of the theorem above hold, then where A, 0 < A < 1, depends only on Q, N, p, q. The proof of this theorem is similar to that of Theorem C.2 of Chapter II and we omit it. The aim of the theorem above is to show the Holder continuity of u. When \l/ is absent, that is, in the case of an equation we are led to a regularity theorem for the mixed boundary value problem which can be treated by the same method of truncation as in Theorem C.2 of Chapter II. One could attempt to extend to this problem the regularity theorems proved in Sections 4 and 5. However, with this in mind, it is easy to see that the solution does not belong to H 2>P (Q) for all p < oo. Indeed, this result is known not to be true even if ^ is absent unless the data of the problem satisfy some compatibility conditions. We refer here to the result of Shamir [1] on mixed boundary value problems.
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MIXED BOUNDARY CONDITIONS
141
We consider now the solution of the variational inequality (9.1) with /i = 0 for i = 1, 2,..., N and g = 0. It can be approximated in C 0 > A n V by solutions of certain quasi-linear mixed boundary value problems associated with the elliptic operator A. Assume that A\J/ is a measure on Q and that the outward conormal derivative di/s/dv — dijil/x vit
v = ( v ! , . . . , VN)
the outward normal,
is a measure on c^Q which fulfills the conditions
Let «9(r) be a nonincreasing Lipschitz function on R, 0 < $(f) < 1. Consider the nonlinear mixed boundary value problem:
The variational formulation of the mixed problem (9.2) can be defined by means of the quasi-linear form
Indeed, (9.2) is equivalent to the problem of finding a solution of
Then the general theory of monotone operators, which we have treated in Chapter III, yields the existence of a solution of (9.4). We consider two sequences of functions of the type 9(t) defined as follows:
and
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PROBLEMS OF REGULARITY
Then .9^(0 is a nondecreasing sequence of functions each of which is Lipschitz and not increasing while ^(r) is a nonincreasing sequence of functions with the same properties. Both sequences "converge" to the multivalued function &(t) defined by
Let \l/ be a function such that \l/ < 0 on d^l and let $ be the unique solution of the Dirichlet problem and assume that $ e /f2?p(Q). Consider the closed convex subsets of V: then we have the following: Theorem 9.3.
Ifu is the solution of the variational inequality
then u resolves the variational inequality
Proof. H e IK since IK c: IK. It is enough to show that (9.6) holds true f all u e IK. If t; £ IK, then either v > \j/ in Q or v < $ in some open subset of Q of positive capacity. In the first case (9.5) is nothing but (9.6). So we have only to consider the case in which v > $ in Q does not hold true. In this case, we write v as a sum by defining v = vt + v2, where v± = max(t;, $) and v2 — v — vl. Then Vi e IK and v2 £ V with
Since y e IK by assumption, so that v > $ =• \jt on 5Q, it follows that supp v2 <= Q and thus v2 e f/o(H) c: V. We can now write
APPENDIX A
PROOF OF THEOREM 3.3
143
and we shall show that
In fact, u is the weak limit in Vr\ C°'\^l) of the sequence (a subsequence) u'm of solutions of the mixed boundary value problems
Since A\j/ = fin Q and since supp v2 c Q we find that
which, passing to the limit, proves (9.7). Finally, since u is the solution of (9.5) and since t^ e IK, we find
Since the last inequality holds for all v e IK, Theorem 9.3 is proved. Q.E.D.
Appendix A. Proof of Theorem 3.3
The main purpose of this appendix is to sketch the proof of the following theorem stated in Section 3. Theorem A.I. Let Q be a smooth bounded open set o/lR v . Let a(p) be a C vector field such that l
and Let g e L°°(Q) and suppose that
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PROBLEMS OF REGULARITY
Then for each s > 1, there is a C0 = C0(s, v, M, Q) such that
The proof of this theorem is based on several steps. Lemma A.2. Under the assumption of Theorem A.I, u e H2(O). Recall that by assumption
Now let V be the solution of the Dirichlet problem
It is known by the Calderon-Zygmund theorem (see Section 2) that for #eL s (Q) it follows that KeH 2 ' s (Q) and thus Vx.eHl'*(ty, where 1/s* = (1/s) - (1/N). So (A.3) can be written
(a) First of all we shall prove that u e H2OC(Q). Denote by q>s>h the first difference quotient
of the function (p(x). Let C(x) be any function in H1^) with compact support in Q. Let Q' be an open set such that Q' <= Q; then if x € Q' and h is small enough we can insert the function
instead of cp in Eq. (A.3'). By changing variables and using the integral form of the mean value theorem, we obtain an identity of the form
where the functions a,/x, k) satisfy uniformly with respect to h
APPENDIX A
PROOF OF THEOREM 3.3
145
Let a be a C°° function with compact support in a ball BR(x) <= Q' and equal to 1 in Bp(x\ p < R. Taking ( = a 2 w s>/1 , by a standard computation we get
with p < JR, where the constant in front is independent of h. Passing to the limit and remembering that fi' is compact, we obtain
From this estimate we can infer that u e H2OC(Q) and that u satisfies a.e. in Q the equation
(b) In order to prove that u e //2(Q) we straighten a portion d t Q of dQ and obtain in the new coordinates an identity of the type of the one above. In a neighborhood of ^jQ we examine only these difference quotients u s f l which are taken in directions parallel to ^Q. These difference quotients vanish on d,Q. Using the same device as above we can deduce that the // J (Q) norms of these difference quotients are bounded uniformly with respect to h. The derivative remaining in the normal direction may then be estimated directly from Eq. (A.6). Thus u e f/2(Q). Proof of theorem A.I. Passing to the limit in (A.4) as h -> 0 (through a subsequence) we get
Then from Theorem C.2 of Chapter II we get that H ^') for a suitable X with 0 < A < 1. Hence we have a.e. in Q 1
If we straighten a portion djQ of <5Q we have in the new variables an identity like the one above and we examine only those difference quotients w s ,, which are taken in directions parallel to c^Q; we deduce that these derivatives u Xj are Holder continuous up to the boundary.
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PROBLEMS OF REGULARITY
Moreover, we know from (b) above that From Remark D.6 of Chapter II we may deduce that
where z is any second derivative of u except u v v , i.e., except the second pure normal derivative. But u vv may be expressed by mean of the other derivatives through Eq. (A.6). So we deduce that
for s, i = 1, 2 , . . . , N and for a A such that 0 < A < 1. From a well-known lemma due to Morrey we infer that the derivatives uXs are Holder continuous up to the boundary of Q. Since Eq. (A.6) can considered as a linear equation with continuous coefficients(ddi/dp^Ux),using again the result of Calderon-Zygmund we deduce that (A.2) holds.
COMMENTS AND BIBLIOGRAPHICAL NOTES It is unfortunately beyond the scope of these brief notes to survey the extensive regularity theory associated to variational inequalities. The principal references for the material of this chapter are the papers Lewy and Stampacchia [1] (Section 2), [2] (Sections 3, 4, 8); Brezis and Kinderlehrer [1] (Sections 5,6, and Appendix); Murthy and Stampacchia [1] (Section 9); and Kinderlehrer [1] (Section 9). A detailed presentation of the subject may be found in Brezis [1] (cf. also Brezis and Stampacchia [1]). A concise treatment of penalization and some of its consequences, especially in regard to interpolation theorems, appears in Lions [1]. The notion of finite perimeter and its properties, due to De Giorgi, is developed in Miranda [1]. Existence and smoothness of the solution of Problem 4.1 when a(p) = />/(! + p 2 ) 1 ' 2 was also considered by Giaquinta and Pepe [1] and Miranda [2], who studied the parametric problem. The question of lower dimensional obstacles for the Dirichlet integral was considered in Lewy [6]. Additional results about the smoothness of the solution have been obtained in Lewy [7], Frehse [3, 4], and Caffarelli [3].
EXERCISES
147
General existence theorems about the minimal surface case may be found in Giusti [1], Kinderlehrer [1], and Stampacchia and Vignoli [1]. For problems with mixed boundary data, we note also the work in Beirao da Veiga [1]. The relationship between coercivity of the vector field a(p) and the smoothness of the solution was studied by Mazzone [1]. The special case of assigned mean curvature, namely, the inhomogeneous problem with a(p) = p/(l + p2)1''2 was considered by Mazzone [2] and Gerhardt [1]. Gerhardt [2] also gave a different proof of Theorem 6.3.
EXERCISES 1. Supply the details for Theorem 2.5. 2. Let Q denote the unit cube (x e IR A ': 0 < x; < 1, i = 1, . . . , N} and let /e LS(Q), g e H2>S(Q), for some s, N < s < oo. Suppose that
Show that weH 2 > s (Q). [Hint: After superposition, we may take g = 0. Extend w to Q = {x e UN: -1 < XN < 0, 0_< xa < 1, a = 1, . . . , N - 1} by w(x', X A ) = — w(x', — XN), x = (x', XN) e Q. Show that the extended w is the solution to a weak equation in Q u Q. This may be used to show that w is smooth on the faces XN = xa = 0, 1 < a < N — 1. Now continue this procedure, performing further reflections.] 3. With O,/, g as in Exercise 2, set and let ^jQ denote the union of the remaining faces of d£l. Consider
Assume that g = \gx\ — 0 on the boundary of 52Q. Prove that w e //2>S(Q). [Hint: First extend w to XN < 0 by w(x', X N ) = w(x', — XN), — 1 < X N < 0.] 4. Let Q be convex with smooth boundary dQ and let ^e// l i 0 0 (Q) with \$i — 0 on dQ. Let IX denote the set of Lipschitz functions v in Q which
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PROBLEMS OF REGULARITY
satisfy v > ^ in Q and v = 0 on dd. Given a locally coercive vector field a(p) let u = T(a) denote the solution of
Now define
Show that U exists and that U is the height function of the convex hull of max(i//, 0) in Q. 5. Consider Exercise 13 of Chapter II. Show that under suitable hypotheses, the solution u = (ul,u2) has components w'eH 2 > s (Q) n H?0f(Q), 1 < s < oo. (Hmf: formulate a penalized problem.)
CHAPTER V
Free Boundary Problems and the Coincidence Set of the Solution
1.
Introduction
In the previous chapters our study emphasized the existence and smoothness of the solution of a variational inequality. The smoothness of the solution, we have seen, was limited by the constraints which defined the convex set under consideration and, indeed, this necessitated the development of a special regularity theory. For an obstacle problem, in particular, the effect of these constraints is also manifested by the presence of the coincidence set. Now we shall investigate some of its properties, with our major goal the establishment of criteria assuring the regularity of its boundary. Such problems are of interest for several reasons. From one point of view, they illustrate how a set with certain topological properties is necessarily smooth. For example, consider Problem 4.1 of Chapter IV when Q c= (R2 is strictly convex and the obstacle ^ is strictly concave and smooth. In this case we shall be able to conclude that dl is a smooth Jordan curve. In addition, the formulation of many physical problems requests an analytic free boundary as part of the solution. Examples of this are steady cavitational flow or the flow of water through a homogeneous porous medium. 149
150
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FREE BOUNDARY PROBLEMS: COINCIDENCE SET
Let us begin by suggesting formally what constitutes a free boundary. We consider an obstacle problem. Let O c IRN be a bounded, open, connected set with smooth boundary dQ, and \ft e C°°(Q) an obstacle satisfying
We set IK = {ue HQ(Q) : v > & in Q}. Consider the solution u of
and its coincidence set
As we have observed, u — \j/eCl(fi) point x€ /; hence
and attains its minimum at any
Also, Au = 0 in £1 — /. In this fashion we may regard u as the solution of the Cauchy problem
Since the Cauchy problem for A is not well posed even for a smooth initial surface dl, the mere existence of a solution suggests restrictive conditions on dl. The set dl is what we call a "free boundary." We want to indicate the reasoning which leads to the conclusion that dl is analytic for the solution of (1.1) with an analytic ij/ and also some of the difficulties encountered in adapting it to other problems. First we discuss a two dimensional situation. Our attention is limited to the analytical questions. A discussion of the topological ones is delayed until Section 6. Before we begin let us remark about how we regard functions of two variables g(x) = g(x^, x2). Introducing the complex variable z = xl + ix2 and its conjugate z = xl — ix2 we frequently write G(z, z) = #(xj, x 2 ) = g(%(z + z), (l/2i)(z — z)). We also write G(z) for G(z, z) without necessarily intending that G be holomorphic in z. Suppose, for instance, that g(x) is real analytic near x = c; thus,
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INTRODUCTION
151
where the series is absolutely and uniformly convergent for |x — c\ small. Then, with y = c^ + ic2,
for suitable ajfc where the series is absolutely and uniformly convergent for |z — y\ and |z — y\ sufficiently small. Let ( = £1 + '£2 denote a second complex variable. The series £ ajfc(z — 7)J(C — y)fc is absolutely and uniformly convergent for small \z — y\ and |C — y\ and is holomorphic in both z and (. Thus if g(x) is real analytic near x = c we may regard G(z, z) to be the restriction to C = z of the holomorphic function of two complex variables
Theorem 1.1. Suppose that co is a simply connected bounded domain in open U <= IR2. Consider functions w, i/' satisfying
(I/ real analytic in a neighborhood o/co u F,
T/ie« F admits an analytic parametrization. Proof. We shall show that there is a function z*(z) analytic in o> such that z*(z) = z on F. From the existence of such a function, we then conclude that F is analytic. Suppose that z = 0 6 F. The complex gradient /(z) = uXi — iuX2 is holomorphic in co and has continuous boundary values F(z, z) = if/Xl — iif/X2 for z e F. In a neighborhood of z = 0, consider the equation
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keeping in mind that z*(z) = z is a solution of (1.5) whenever z E F. Recalling that
we infer the existence of a solution z* = C(z,/) of (1.5) holomorphic in (z,/) near (0,/(0)). In particular is holomorphic in z. By the uniqueness statement of the implicit function theorem At this stage, we have represented z on F as the boundary value of a function holomorphic in ct>. This implies that F is analytic by a reflection principle. To verify this conclusion, let
be a conformal mapping of G onto co which transforms the real segment ( — 1, 1) onto F. We know that cp exists and q> e C(G u (— 1, 1)). Define
a holomorphic function in G u {f: \t\ < 1, Im t < Q}. When t is real, (1.6) ensures that O is continuous. By Morera's theorem is holomorphic in a neighborhood of t = 0. Consequently, exhibits an analytic parametrization of a portion of F. Q.E.D. The topological requirement that F be a Jordan curve will be weakened considerably. The proof fails, evidentally, when \j/ is not analytic. Indeed, the definition z* = ((z,/) is suspect in this situation. From a slightly different viewpoint, (1.7) offers an extension of (p to a function O satisfying the homogeneous analytic equation (d/di)® = 0 in | f | < 1. In general, it is not possible to achieve an extension of this nature, so an alternative idea has been devised which might be interpreted as a combination of classical potential theory and some ideas about Sobolev spaces, especially their trace classes.
2
THE HODOGRAPH AND LEGENDRE TRANSFORMATIONS
2.
The Hodograph and Legendre Transformations
153
We introduce here another method for determining the smoothness of dl, the boundary of the coincidence set of (1.1). Although this method will serve us in any dimension, some initial regularity of dl is required for its application. It has the effect of "straightening" the free boundary at the expense of replacing the equation by a highly nonlinear one. Suppose that u(x) is a solution of (1.1) with
In addition, assume that
where uu = ux.x.. We assume that O e T and that the inward normal to Q\7 at 0 is in the direction of the positive Xj axis. Set w = u — ^. Then, analogously to (1.3), suppose that
where a = — A^ is analytic in a neighborhood of x = 0. We introduce the change of variables and the function We refer to (2.5) as a partial hodograph transformation and to v in (2.6 as the Legendre transform of w. Note that (2.5) and (2.6) differ from the customary definitions by a change in sign. It is easy to see that (2.5) is 1:1 near x = 0. In fact, since w, = 0 on F and (1,0,..., 0) is normal to T at 0,
so from (2.3)
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Hence dy/dx(Q) is nonsingular. Under the mapping (2.5), a neighborhood of x = 0 in Q\/ is mapped onto a set and a neighborhood of x = 0 in F is mapped onto The property of the Legendre transform is that
or
Here the subscripts of v refer to differentiation with respect to y while those of w refer to differentiation with respect to x. In particular, a portion P of F admits the parametrization
The smoothness of F' becomes a question of that of v in [/ u Z. We calculate the equation satisfied by v in U. Set / = ( _ v 2 , . . . , yN). First observe that
and from (2.7),
Since
we obtain that
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Hence
Now we see from (2.3) and (2.4) that
Note that -y n (0) = l/w n (0) > 0. The equation of (2.9) is elliptic and analytic near y = 0 in U; hence, by a well-known theorem (cf. Morrey [1], Section 6.7), v(y) is analytic near y = 0 in U u I. Therefore, in view of (2.8), a portion of F is analytic near x = 0. In summary, once (2.2) is assumed, the free boundary is analytic provided the obstacle is. Similarly, if «A e Cm>a(Q), then F is of class Cm~1>a, 0 < a < 1. This idea may be implemented in many other problems. We shall discuss this in Chapter VI. So once some regularity is assumed, we may conclude higher regularity of F. But the hypothesis (2.2) is not always satisfied. For example, dl may exhibit cusps, as we shall illustrate in Section 4. We shall summarize what has been recently developed about the first stages of regularity of F in Chapter VI.
3.
The Free Boundary in Two Dimensions
In two dimensional problems, the smoothness of the free boundary may be determined by a method analogous to our analytic extension of Section 1. Here we shall prove a theorem useful in a variety of situations. It is more general than the discussion of Section 1 in two ways: first, the data and the equation are not required to be analytic; second, the free boundary need not be a Jordan arc. We then apply our theorem to a case of Problem 1.1. The argument entails the extension of a conformal mapping, so let us review this subject first. Throughout we denote by co a bounded, connected, and simply connected domain in the z = Xj + ix2 plane and we set G = {t = ti + it2'.\t\ < l,Imt > 0}. Suppose that
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is a conformal mapping of G onto co. Then q>(t) is an analytic function bounded in G; but moreover we may express the area A of CD by the formula
Since a> is bounded, A < co, so (p e H1(G). By our trace theorem (Chapter II, Appendix A),
and(?(ri)eL 2 (-l, 1). Now set
y — {zEdco:there exists a sequence £„ e G, It is clear that y contains the graph of
Suppose that (p is a conformal mapping ofG onto co and that y c da> admits the representation (3.2). Then To interpret the hypotheses of the theorem, consider the solution u of (1.1) when N = 2. Set
which is Lipschitz on compact subsets of Q by Chapter IV, Theorem 6.3, when \l/ e C2(Q). We compute that
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hence, the hypotheses about \j/ and u are satisfied in Q — / if «^e C 2>A (Q), 0 < A < 1, and has a nonvanishing Laplacian. Conditions which provide us with an appropriate choice of y c dl will be discussed in the sequel. We shall begin by demonstrating that q>' has a suitable integrability property. We then extend (p to a function which is differentiate at a prescribed point of the ti axis. Given z0 e y, we fix a solution 0*(z) = #*(z, z0) to the equation dg*/dz = a for z 6 U such that
Then g*eC1-\U) and
with a constant C > 0 independent of z0 £ y. Such a choice of g* is easy to find. Indeed, assuming without loss in generality that a(z)e C°'\V) for a neighborhood V of £/, one may set
Then yl 6 Cl'*(U) and for any z0 e I/ admits the development where R satisfies (3.7) and we may take A complex valued function/ e Hl(D), D c [R2 is called quasi-conformal in D provided that supD |/f(z)//z(z) j = q < 1. Lemma 3.2. Assume the hypotheses of Theorem 3.1. Then q> admits a quasi-conformal extension into a neighborhood of the real axis and
Proof. To prove the lemma it suffices to verify the conclusion at a given point t0 e (— 1, 1). We may assume that t0 = 0, and that z0 = lim^o (p(i£) exists. Choose r > 0 such that #*(z) defined in (3.5) satisfies
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This is possible by (3.5). Let ( = g*(z) denote the mapping of Br(zQ) into another plane under g*. Observe that the Jacobian of this mapping so there is a neighborhood Br-(zQ) in which g* is 1:1 and has a C 1 > A inverse (g*)~l. We suppose that Br(z0) <= Br>(zQ). It is possible to choose r independently of z0 e y for z0 in compact subsets of j contained in y n
Let us check that O(Oe/f1(J5(J(0)). Since ^p is a conformal mapping onto a domain of finite area, cp e Hl(Bs(Q) n G), as we have remarked at the beginning of this section. Now (g*)~l is smooth and h((p(f)) is the composition of a Lipschitz function with a very smooth function. Hence g*~l(h(
so the traces of O from above and below are equal. By Lemma A.8 of Chapter II, We now compute that ®(0 is a quasi-conformal function. Indeed, and
\QM/®t(t)\ = \g*-l/gr1\ = \gt/gi\<* It follows (cf. Bers er al [1], p. 276) that
for
I m r < 0 , |t| «5.
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with
In particular, cp e H 1>S (G n Bg/2(0)). This proves the lemma.
Q.E.D.
Consider now a function, with B = { t : \t\ < 1}, w e H1>S(B), such that
satisfying
Lemma 3.3. Let w satisfy (3.10). Then there exists a complex number c such that
w/r/i
where CR, C'R depend on R, s, r. Although this lemma is straightforward to prove, condition (3.10) is not an obvious one. Proof. Note that w e C°- \B\ A = 1 - (2/s), by Sobolev's lemma. We recall the formula of Green in complex form. Let g = u + iv and dz = dx{ + i dx2. For any E <= Bl with dE smooth and g € H1<S(E),
In particular, if £(f) is analytic in a neighborhood of E and weH1>s(B)is given as above,
We choose C(0 = l/[f(f - z)] f°r a fixed z e B and the sets £e = {r: \t\ > e, \t — z\> e,\t\ < R}. Performing the integration above and passing to the
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limit as e -> 0, which exists because w is Holder continuous, yields the wellknown formula
Hence, formally,
Observe that, in fact,
Now a > 2/s implies that (a — 2)s' + 1 > — 1 so the right-hand side of the inequality above is finite. We set
and subtract it from (3.11). This gives
Now for z ^ 0,
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The last integral on the right-hand side may be estimated in a familiar way which we indicate here. Clearly,
For \z\ < l,let
and set
Assume now that o < 2. Then \
On the other hand,
which leads to the estimate
Finally, for / 3 , so
Therefore
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since (o — 2}s — s + 2 = s'[o — 3 + (2/s')] < 0 if cr < 2. Consequently,
or
The second integral is easy to control, so the lemma follows in the case cr < 2. But if a > 2, then
so the conclusion holds with exponent 1. Q.E.D. Proof of the Theorem. To prove the theorem we show that (p, known to be in H1>S(G n BR), 0 < R < 1, admits an extension to BR which satisfies (3.10). By change of scale, we may assume cp e Hl's(G). We exhibit this extension at t = 0, the situation being the same at other points of (— 1, 1). Let us recall again the function g*(z) of (3.5) and the difference h(z) — g*(z) — Q(Z\ which is analytic in co and satisfies Write and determine
so
by (3.7).
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163
According to Lemma 3.2, cp' e LS(G) for every s, 1 < s < GO, so given T, 0 < T < A, we may choose s so large that Furthermore,
Since w is continuous in |r| < 1, it follows that we// 1>s (fi) by the Hl, or H l i S , matching lemma, and by (3.12) it satisfies (3.10). Therefore we may apply Lemma 3.3 to infer the existence of a constant c such that
When Im t > 0, w(r) = cp(t\ so we have achieved the estimate
Since 0 was an arbitrary point of ( — 1, 1), what we have really shown is that given R < 1, for each t0 e [ — R, K] there is a complex number c(t0) such that
where C R , C^ depend on R, T, s, and some Hl's norms of (p. It is easy to check by Cauchy's theorem that a function
ami Let
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and set
I ••= |zeQ:«(z) = i//(z)}. Then F = dl is a Jordan curve which admits a C 1 > T parametrization. Proof. In Section 6 we shall discuss the topological nature of /. For the moment, let us assume the conclusions of Theorem 6.2, namely, that / is a connected and simply connected domain whose interior int / is also connected. Moreover, Q\7 is of the topological type of an annulus. This implies the existence of a conformal mapping (p of the upper half t = t { + it2 plane, from which an appropriate disk has been deleted, onto Q\/ that maps [—00, oo] onto dl. This means that dl enjoys the property dl = (ze U2: there exists a sequence tn e { t : Im t > 0} such that (p(tn) -» z and Im tn -» 0}. Given a particular z 0 , we may assume that the points tn with (p(tn) -> z satisfy
Kl
We may apply Theorem 3.1 to a) =
and
In a neighborhood of /, g(z) is Lipschitz because the second derivatives of u are bounded there (cf. Theorem 6.3 in Chapter IV). The strict concavity of i// implies | a(z) | ^ 0 on compact subsets of Q. Hence dl admits the parametrization, which is of class C 1>T ,
and is thereby a curve. Suppose
for a point on the arc C x would not be the limit of interior points of/. Hence dl has no double points, and so it is a Jordan curve. Q.E.D. The last argument, we note, did not employ the fact that the parametrization of dl was smooth, but only that it was continuous.
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We remark that it is also possible to show that dl is a "regular" curve, that is, it is free of cusps, at least if \J/ e C3>A(fi). This requires a deeper study of the extension w(t) defined by (3.13). Indeed, w satisfies the differential inequality
where q E LS(B) for some s > 2. Such functions enjoy properties analogous to those of analytic functions. We state here without proof a description of these properties: Theorem 3.5.
Let w satisfy (3.15). //
for an integer n > 0, then
exists. Moreover, j/lim r _ >0 z~"w(z) = 0 for every n > 0, f/ie« w(z) = 0 in a neighborhood of z = 0. With reference to Theorem 3.4, the restriction of w to Im t > 0 is a nonvanishing conformal mapping; hence there is a smallest interger n for which
It then follows from the 1 : 1 nature of the conformal mapping that n = 1 or w = 2. It is natural to ask about higher differentiability of the curve y. We state here a result about this; its proof is left as an exercise. Theorem 3.6. With the hypotheses of Theorem 3.1, assume that ae C"lA(tf), rn > 0 and 0 < A < 1. T/je« J/u? conformal mapping (peC m + 1 ' T (G n BR)for each r < X and R < 1. This theorem is slightly stronger than the result we obtained with the aid of the Legendre transform because here dl is not required to be a regular curve.
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A Remark about Singularities
Theorems 3.1 and 3.5 show that the only singularities displayed by the free curve y are cusps. Such cusps may arise. We give here a simple example. Consider the mapping
provided that /^ = 4/c + 1, k = 1, 2, 3, to the variational inequality
It follows that u is the solution
where In view of (4.1), the set of coincidence of w, will have free boundary F n Bt in B£. That F has an analytic parametrization as the boundary values of a conformal mapping suggests the existence of a holomorphic/(z), z e co, such that Indeed, it is given explicitly by
To describe its behavior, we find z as a function of t. Clearly
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167
where • • • represents terms of degree higher than jU/2, so F(t) = t2 — it* = — z + 2t2 and Our function u is found by integration:
for e sufficiently small. Since/(z) = z on F, Now u admits the expansion
from which we see that u^) > 0 when x j < 0 provided // = 4k + 1, fc = 1, 2 , . . . . To complete the proof that w satisfies (4.1), we must show that u > 0 in co n Be for some e > 0. Since (///2) + 1 > f,
Also, MX2(z) = -u X2 (z\ zeBE, u(z) = u(z\ z€Be. On each vertical line xl = 6, \6\ < e, u(d + ix2) is a convex C 1 ' 1 function whose minimum is attained at a point where ^((5 + ix2) = 0. If <5 < 0, the unique point which this property is x2 = 0, and we have seen that u(xv) > 0 for *! < 0. Suppose <5 > 0. Then u(<5 + ix2) = uX2(d + ix2) = 0 for |x2| < d^2 whereas uX2X2(6 + ix2) > 0 for |x 2 | > <5M/2. Hence u is strictly increasing for \x2\ > ^"/2, so it is positive there.
5. The Obstacle Problem for a Minimal Surface
To commence, we review the geometry of a two dimensional nonparametric minimal surface in IR3. A function u e C2(U\ U cr R2 open, is a solution to the minimal surface equation in U if Its graph
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is called a minimal surface. More generally, the mean curvature H of the surface defined by v e C2(U) is given by
From this, and the obvious fact that (5.1) is the Euler equation of the nonparametric area integrand
we infer, formally speaking, that the surface which minimizes area among all those having a given boundary is a surface of vanishing mean curvature. The metric of the surface is defined by the symmetric positive definite matrix
To the point x e M we associate n(x) e S2, the upward normal of M at x, This mapping induces another, possibly degenerate, metric on M via the bilinear form
The matrix / is symmetric and nonnegative. A minimal surface S enjoys the property that the matrices g and / are proportional at each point. This is the statement that the normal mapping of S to the sphere S2 is conformal, or anticonformal, as this function of proportionality is positive or negative. We shall verify this in the ensuing discussion. Replacing a point (c1} c 2 , c3)eS2 by its (negative conjugate) stereographic projection onto the equatorial plane from the south pole, and applying this to «(x), we obtain what we shall refer to as the Gauss mapping of the surface S, whether it is minimal or not,
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THE OBSTACLE PROBLEM FOR A MINIMAL SURFACE
169
Since the mapping from S2 -> C given by stereographic projection induces a metric on S2 proportional to its original one, we need only check that | dC, \2 for C = f(z) is proportional to the metric g. Lemma 5.1. Let utC2(U\ U <= U2 open, be a solution to the minimal surface equation. Then where
The proof of the lemma is a direct, but tedious, computation. See also Kinderlehrer [3] for a more geometric proof. The matrix g, expressed in terms of dxl and dx2, namely, written as a tensor, is Rewriting this with dz = dx^ + i dx2 and dz = dx^ — i dx2, we obtain
Now we compute that
Observe that
and
hence
We have demonstrated Proposition 5.2. Let u e C2(U), U c IR2 open,feea solution to the minimal surface equation. Then the Gauss mapping of is conformal.
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These considerations suggest that it is feasible to exploit the Gauss mapping of the surface S in place, say, of the complex gradient of u to ascertain the smoothness of the free boundary. First we recall the theorem of Korn and Lichtenstein (cf. Courant and Hilbert [1].) Theorem 5.3. Let Q c R2 be a simply connected domain with smooth boundary and set B = {C: 1C I < 1 }• Suppose that p, e C°* A(Q) satisfies Then there exists a homeomorphism C = h(z)from Q to B such that h £ C1'A(Q),
and h(zo) = Ofor a preassigned z0 € Q. ///e Cl(U), U c Q open, satisfies
then F(Q =f(z) is a holomorphic function of(, in h(U). The second statement of the theorem is elementary inasmuch as so
Now \u\ < 1 and hz ^ 0, so F?- = 0, which means that F is a holomorphic function of (,. Fix a domain Q <= [R2 strictly convex with smooth boundary <3Q and a strictly concave obstacle \l/ e C 2>A (Q). Consider the problem: To find
for v e IK with
and let
We know w exists and moreover u e C1>a(Q) n H2- °°(Oo) f°r and Q0 c= Q0 c Q.
eacrl a 6
(0> 1)
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THE OBSTACLE PROBLEM FOR A MINIMAL SURFACE
171
Theorem 5.4. Let I = l(u) denote the coincidence set of the solution u to the problem (5.4). Then dl is a Jordan curve which admits a C 1>r parametrizationfor each i < A. Proof. In the course of the proof we shall use Theorem 6.2 as we did in the proof of Theorem 3.4. Define the Gauss mapping/(z), z e Q, by (5.3). Then/(z) is Lipschitz in compact subsets of Q and satisfies the equation
since
according to Lemma 5.1. In addition, with
f*eCl'\fX)&nd since uXj(z) = $Xj(z) in /. The next lemma derives from the concavity of the obstacle. Lemma 5.5. Withf* defined by (5.5), |/*(z)| < |/*(z)| for z e ft. Proof. We may rotate coordinates so that iA*lX2(z0) = 0 for a given z0. This does not change the modulus of/* or/?. One then computes directly that
and where aj > 0 and b is real. Since \l/ is concave,
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and suppose z0 e d/ is given and that h(z0) = 0. Define the functions and set
a function that is Lipschitz in a neighborhood of fi(/). Now F(() is holomorphic in B — h(I\ whence
Evidently, a(() defined by (5.7) for all C e B is in C0t A(B), since h and its inverse h~l are in C 1 > A and/* e C'-^Q). We must check that |oc(C)| ^ 0. From (5.6) one sees immediately that
where \a\ = 1. Keeping in mind that dh~l/d£ ^ 0 and |/i| < 1, we use the preceding lemma to calculate that
At this point we wish to apply Theorem 3.1. Assuming Theorem 6.2, we argue as in the proof of Theorem 3.4. Given £0 e ^/, consider a conformal mapping
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173
Theorem 5.6. Let I = I(u) denote the coincidence set of the solution u to the problem (5.4). Suppose in addition that if/ is real analytic. Then dl is a Jordan curve which admits an analytic parametrization.
6.
The Topology of the Coincidence Set When the Obstacle is Concave
For a general obstacle \l/ and domain Q the topology of the coincidence set for a solution of the problem (1.1) is very difficult to ascertain. In the exercises we show that for any preassigned integer k there is an obstacle whose coincidence set has at least k components. However, if the obstacle is strictly concave and Q c IR2 is strictly convex, it is possible to prove that I is a simply connected domain equal to the closure of its interior. We have already used these facts in the proofs of Theorems 3.4 and 5.4. Throughout this section let us assume Q is a strictly convex domain with smooth boundary in the plane, ty e C2(Q) is a strictly concave obstacle, namely, and a
(p) = (a\(p\ fl2(p)) is a locally coercive analytic vector f
Consider the solution u to the variational inequality
and its set of coincidence
There is a unique solution u to (6.2) and it satisfies
and for all subdomains Q0 c: Q0 o Q. Since \l/ < 0 on 5Q, / is a compact subset of Q (Chapter IV, Theorems 4.3 and 6.3).
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As usual, we define
by
The concavity of \l/ implies that In fact,
= — trace a1?, where
and
Now a is positive definite and symmetric and NP is negative definite symmetric. Hence —trace a1? > 0. Lemma 6.1.
The set Q — I is connected.
Proof. Since u(z) = 0 > \f/(z) for z e dti and 5Q is connected, there is only one component of Q - / whose closure intersects dfl. Suppose w! to be another component. Then dco' c: /. For any ( e C^(co'), £ > 0,
where
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175
is a positive definite form owing to the local coercivity of a(p). Hence \l/ — u is a supersolution to an elliptic equation, which implies
But dco' c /, so \l/ — u > 0 in co', a contradiction to a/ c Q — /.
Q.E.D.
Consider for the moment a solution y(z) to the equation
By a well-known theorem, v(z) is a real analytic function in Br(zQ). Writing v as a series of homogeneous polynomials,
unless v linear, one checks that Pm, the polynomial of lowest degree, is a solution to the elliptic equation with constant coefficients
Consequently, Pm(0 is affinely related to the harmonic Re a(m, for some complex number a, and dPm(()/d( is affinely related to awC m ~ 1 , m > 1. From this we conclude that given a solution v oj (2.2) which is not linear in a neighborhood Uofz0, (dv/dz)(z) is an open mapping of U. We define two mappings of Q onto another plane. The first is the homeomorphism of Q
The second is the continuous mapping of Q,
which by (6.4) is open on Q — /, since otherwise u would be linear.
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Theorem 6.2. Then
FREE BOUNDARY PROBLEMS: COINCIDENCE SET
Let 1 denote the coincidence set of the solution u to (6.2).
(i) / is a connected simply connected domain which is the closure of its interior,
(ii) /(Q) = /*(/),
(iii) (iv) Proof. (*,
Q — / is homeomorphic to an annulus, and int / is connected. Let 1^ be the set of points z e Q for which the tangent plane at
does not meet Q. Since the point z e Q where i/f assumes its maximum has this property, / j is not empty. Since \l/ e C1, /! contains a neighborhood of z and is closed. Also
because \j/ is strictly concave. We show that /! c 7. Indeed, t?(j;) is a solution to Xi> = 0 in Q and v > 0 on dQ. Hence by Theorem 8.3 of Chapter IV, u < v in Q. Therefore
Now z e /! is an interior point unless y(y) = 0 for some y e 5Q, in which case the intersection of FIr with the plane containing Q is tangent to dQ at some point. The boundary of/! consists of all such points. The converse also holds. Let P 6 dQ and construct a plane through the tangent to dQ at P which is tangent to \fy at some z e Q. Then z e /, by the preceding argument, and the correspondence P -»z(P) is a 1:1 continuous mapping in view of the strict concavity of \lt and strict convexity of 3Q. With z = z(P), let w(y) = u(y) — t>(y), which is a solution to the linear elliptic equation
in Q near P. The maximum principle is applicable, so because w(y) < 0 near P, and attains its maximum value zero, at P,
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177
where v is outward directed normal. That u(y) > 0 in Q implies that
where 6 denotes the angle from the positive x^ axis to the inward normal at P. Since v is a linear function and the tangent to (f/ at z = z(P),
The tangential derivative of u vanishes on c*Q so that
and therefore
With (6.5) this shows that maps 5Q onto a star shaped curve Fj with origin as reference point. On the other hand, Oz is tangent to 5Q at P; hence,
Therefore
maps 5Q onto a star shaped curve F 2 . According to (6.6), Fj lies inside F 2 , and F! cannot intersect F 2 . Our interest is in the outer boundary /? of/(Q), which is by definition the boundary of the infinite component of the complement of/(Q). Since /(Q) is connected, P is connected. Now f(d£i) — Fj, which lies inside F2 =f*(dll) =f(dll); hence, p n/(dQ) = 0. Since/is open on Q - 7,/(Q - /) n /? = 0. Hence if z e Q and/(z)€j8, then zel and/(z) =/*(z). So/*' 1 ^) c / and is the outer boundary of/*~ 1 (/(£I)) since/* is a homeomorphism of Q. Consider a component V of/(Q) — /(/). By openness of/on Q and the fact that/(5Q) <= int/(/), if/(z) e dV, then z e 57. Hence any component U off~l(V) satisfies U a Q — / and dU c /. By the previous lemma this is impossible; therefore /(H) -/(/) is empty, so/(n)_=/(/) =/*(/)• We infer that / = /* ~ X/C^)) is connected since/* is a homeomorphism.
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At this point let z0edl and z n e Q — 7 satisfy zn -+ z 0 . Since/is open, /(zn) e int/(Q) = int/(/). Moreover,/* is a homeomorphism, so/* ~ X/foi)) € int /. Now
because/ = /* in /. Therefore / = int /. Similarly, Q — 7 is open and connected; hence/(Q — /)<=: int/(7) = int/*(/) by openness of/ Moreover,/(Q — /) is connected, so/*~ 1 (/(Q — /)) cr / 0 > a component of int /. Let 7j be any other component of int / and zl ed/i, zj $dI0. Again we find a sequence z n e Q — 7 with zn -» Zj and f * ~ l ( f * ( Z n ) ) e 7 0 . But then zl e 7 0 . This contradiction implies that 1^ — 0. That Q — 7 is topologically an annulus follows by a standard argument. Q.E.D.
7.
A Remark about the Coincidence Set in Higher Dimensions
The analysis of the coincidence set has proceeded up to this point by developing an appropriate function theory and thus has been limited to two dimensional problems. A first step toward smoothness of the coincidence set in higher dimensions was taken in Chapter IV when we provided a criterion under which 7 was a set of finite perimeter. Here we continue this development. As in Section 1, let Q c RN be a bounded open set with smooth boundary <5Q and \l/ an obstacle satisfying
Suppose that an inhomogeneous term/is given with
Our efforts are directed toward the analysis of Problem 7.1. Let u e IK satisfy
With minor technical modifications, the proofs given here extend to include Problem 4.1 of Chapter IV.
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Theorem 7.2. Let u be the solution of Problem 7.1 under the hypotheses (7.1) and (7.2) and let I be its set of coincidence. Suppose that for some r\ > 0. T/ien f/jere exist r0 > 0 ant/ A, 0 < /I < 1, such that for any x€dl, there is a yeBr(x)for which Moreover, for some a > 0
This density property of Q — / leads immediately to Corollary 7.3. Let « 6e the solution of Problem 7.1 under the hypotheses (7.1) and (7.2) and let I be its set of coincidence. Suppose that for some n, > 0. Then
Proof of Corollary 7.3. Since / is closed, 57 <= /. Now
with con = meas #i(0). Thus no point of dl can be a density point of dl. Hence meas dl = 0. Q.E.D. Proof of Theorem 1.2. Let Q0 be a neighborhood of/ in fi, which exists because / is compact, and choose r0 > 0 so that for each x e /, Br(x) <= Q0 for r < 2r 0 . Note that w € //2'°°(Q0) by Corollary 6.4 of Chapter IV. For ft > 0 set where x0 e 57 is given. Then for x e Q — 7
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Hence From the maximum principle, the continuous w attains its maximum on d(Br(x0)n (Q — /)) c=dBr(x0)udlSupposeforthemomentthatx 0 satisfies an interior sphere condition with respect to Q — /. By this we mean that there is some ball Bd(£) c Q - / with x0 e dBd(E,). Now so by the Hopf boundary point lemma, x0 is not an extremum of w. Furthermore,
For x e dl, w(x) = [l/(2N)]j8|x - x012 < 0, so there exists y e 3Br(x0), y$I, such that
or To extend this inequality to a ball, we use the fact that uXiXj.EL°°(Q0). Our choice of r ensures that Br(y) c= Q0. Since (u — ^)x is Lipschitz in Q0, we may write
where K(x, y) is a remainder, and
Consequently, recalling that
Therefore
for |x — y\ < rr, i sufficiently small but independent of
COMMENTS AND BIBLIOGRAPHICAL NOTES
181
If x0 e dl is an arbitrary point, we may find a sequence xfc e dl, xk -» x 0 , with xfc possessing the interior sphere property. By continuity, Consequently, BT,(y) <= Q — /. Changing the names of the variables a little, i.e., replacing r by r + n < 2r < 2 r 0 a n d r b y A = t/(l + T), we conclude that A portion of dl may be negligible in the sense that it is not part of the boundary of int /. Indeed, consider the open set in Q
and note that Q — / is also open in Q. Now CD — (Q — /) c: dl, so since - Aw = /in fi — /, we may say that — Aw = /a.e. in CD. However, co is open, so the regularity theory for elliptic equations guarantees that In other words co is, in some sense, the largest open set in which — AM = / In this way we regard dl — d int / inessential or negligible.
COMMENTS AND BIBLIOGRAPHICAL NOTES The discussion of free boundary problems connected with variational inequalities began in Lewy and Stampacchia [1]. For the case of the Dirichlet integral, N = 2, the boundary of the coincidence set was shown to be an analytic Jordan curve under the hypotheses of Theorem 3.4 with if/ analytic. The reader may prove this theorem directly from Theorems 1.1 and 6.2. Section 2 is a preview of Chapter VI, where the argument is developed and references are provided. To discuss more general problems where, for example, the Schwarz reflection principle was not directly applicable, it was useful to obtain some preliminary regularity of the free boundary prior to demonstrating, say, its analyticity. This motivated the development of Section 3 (cf. Kinderlehrer [3]). We have adapted the proof of Kinderlehrer [4]. Lemma 3.3 and Theorem 3.5 are related to a result of Hartman and Wintner [1]. So to analyze the behavior of the free boundary in the minimal surface case, dl was first shown to be continuously differentiable. This we have explained in Sections 3 and 5. The analyticity of dl when ij/ is also analytic was achieved by the resolution of a system of differential equations which
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connects a conformal representation of the minimal surface {(xt, x2, x 3 ): x3 = u(x), (*!, x 2 )eQ — /} with its harmonic, or analytic, extension. This idea is due to Hans Lewy, who used it to study minimal surfaces with prescribed or partly free boundaries (Lewy [2, 4]). Examples of the method are given in the exercises. Theorem 3.6 is due to Caffarelli and Riviere [2]. Schaeffer [1] has studied the existence of singularities of dl. An asymptotic description was devised by Caffarelli and Riviere [3]. The content of Section 4 was adapted from Kinderlehrer and Nirenberg [1]. The results of Section 7 are due to Caffarelli and Riviere [1]. They serve as the starting point of the investigation of the coincidence set in higher dimensions. Perturbation of the free boundary has been considered by Schaeffer [1] and Lewy [9].
EXERCISES 1. Let cp(f) be analytic in G = {t = tl + it2 '• \t\ < 1, Im t > 0} and continuous in G u (— 1,1). Suppose that for each £ 0 e [ — R, R],R < 1, there is a complex number c = c(r0) such that
for some fixed ju, 0 < /j. < 1. Use Cauchy's representation to show that (peCl'*(GnBR). 2. Give a proof of Theorem 3.6. 3. Under the hypotheses of Theorem 3.4 show that dl is a "regular" curve, that is, it has a continuous tangent. [Hint: The only possible singularities of dl are cusps, in view of the remarks in the text. Given z0 e dl, consider the mapping from Bs(z0) n (Q — /) into Bs(z0) r\ int / given by
and define
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183
Let uk be the solution of the problem (1.1) for the obstacle \l/k and let Ik be its coincidence set. Show that Ik has at least k components. Show that \}ik may be modified to a C°° function $k with the same solution uk. 5. Let G = {z = x + iy: \z\ < 1, y > 0} and a = ( — 1, 1). Suppose that u€Cl(G) satisfies
where A(x, w, p) is an analytic function of the variables x, w, p near x = 0, u = w(0), p = wx(0). Show that u may be extended harmonically into a neighborhood of z = 0 and hence u has analytic boundary values near z = 0 on
where A is an analytic function of x l 5 x2 near z = 0, ,4(0, 0) ^ 0. Show that F is analytic near z = 0. 7. (Kellogg's Theorem) Let Q be a simply connected bounded domain in the z = X j + ix2 plane whose boundary contains the arc Assume that/^ C 1 ' A ([ —a, a]), 0 < A < 1, and let (p be a conformal mapping ofG = {|t| < l:Imt > 0} onto Q which maps (-1,1) onto T with 9(0) = 0. Prove that q>ECl'\Brr\G) for each T < A and r < 1. [Him: Consider z*(z) = z — 2 i/((z + z)/2) for |z| small and apply Theorem 3.1.] Extend this conclusion slightly by proving that cp € Cl'\Br n G), r < 1.
C H A P T E R VI
Free Boundary Problems Governed by Elliptic Equations and Systems
1.
Introduction
In this chapter we shall illustrate the use of hodograph methods and their generalizations to ascertain the smoothness of free boundaries. A hint of this technique was described in Section 2 of Chapter V, where new independent and dependent variables were defined in terms of the solution w of the variational inequality. Here we shall extend that idea, proposing, for example, in the case of a single equation, the selection of a combination of derivatives of w or even w itself as a new independent variable. The object, in all situations, is to straighten the free boundary at the expense of introducing a very nonlinear equation. To the nonlinear problem which results, we may apply a known regularity theorem to deduce smoothness. Several complications will arise. At the purely formal level, the free boundary problem may suggest a system instead of a single equation. Or the original dependent variable or variables may be defined on both sides of the free boundary as in the variational inequality of two membranes (Chapter II, Exercise 13). To surmount the first difficulty we shall study nonlinear elliptic systems. A brief review of this topic is included in the text, Section 3, as much to establish our conventions as to inform the reader unfamiliar with the theory of its basic elements. To confront problems where the functions are
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185
defined on both sides of the free boundary, we introduce a reflection mapping defined by means of the hodograph transformation in use. This will always lead to a system of equations. Another, perhaps more delicate, question is to decide whether the hypotheses of the elliptic regularity theory are satisfied once the problem has been rewritten with hodograph transforms, reflections, and the like. This question, at times extremely difficult and technical, with rare exceptions is beyond the scope of this book when N > 2. We discuss this briefly and without proof in Section 2. Our considerations will be local. Again, our theme will be that once some smoothness of the free boundary is assumed, it must be as smooth as the data permit. In this way we wish to explore the conditions at a free boundary which ensure its regularity. This chapter is independent of Chapter V.
2.
Hodograph and Legendre Transforms: The Theory of a Single Equation
Let u e C2(Q), where Q <= (RN is a bounded domain. Throughout this chapter we set u, = ux., utj = ux.x., etc. The transformation of Q defined by
is called a first order (partial) hodograph transformation and the function v(y) defined by is called the first order Legendre transform of u. Let us assume for the moment that (2.1) defines a 1:1 mapping of Q onto a domain U. The property of the Legendre transform is that which means that
Here and in what follows we shall understand subscripts of u to denote differentiation with respect to x and those of v to indicate differentiation with respect to y. The inverse mapping to (2.1) is given by
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Moreover, if u satisfies a second order equation, then v does also. To check this, merely note that
and
imply that
In other words, with dk = d/dyk,
In particular,
so that, for example,
More generally, if u satisfies an elliptic equation, then v does also (cf. Exercise 1). Another change of variables is given by
which we refer to as a zeroth order hodograph transformation. To it we associate the new dependent variable
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Again let us suppose that (2.8) defines a 1:1 mapping ofQ onto a domain U. The property of the zeroth order transformation is that
or
Also, note that the inverse of (2.8) is
From (2.10) it is obvious that
We also note that
so that, in particular,
Again, if u satisfies an elliptic equation, then ij/ does also (Exercise 2). We now apply these transformations to the study of a simple free boundary problem. As before, let Q c R w be a domain. Let OedQ and suppose that near 0 c>Q is a C1 hypersurface F with inner normal at 0 in the direction of the positive XN axis. Let u(x) e C2(Q u F) satisfy
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where a(x) is smooth in a full neighborhood of x = 0 and, for simplicity, c1!, . . . , CN, beU. The nondegeneracy condition CN =£ 0 will always be required. More generally, it is necessary to know that the direction c = (c!,..., CN) is not tangential to F. We shall consider two cases of (2.14). First assume that b = 0, CN ^ 0, and a(0) < 0. Here we seek to apply (2.1). Indeed, since CN ^ 0, uk(x) = 0, 1 < k < N, in a neighborhood of 0 in F, which we may take to be F, so (2.14) may be restated as
To see that (2.1) is 1:1 near 0 in fi observe that since uk = 0 on F and (0,..., 0, 1) is normal to F at x — 0,
so from (2.15)
It follows that (2.1) maps a neighborhood of 0 in Q, say Q, onto a domain
and F onto a portion S of the hyperplane 3;^ = 0. Consequently, v(y) defined by (2.2) satisfies
and F admits the representation
Theorem 2.1. Let u(x)£C 2 (QuF) sar/s/y (2.15) w/r/i a(0) ^ 0. // a e Cm>a(Br(°))> w > 0,0 < a < 1, then there is a neighborhood Bp(0) such that
«w/ J3P(0) n F is o/c/ass C m+ 1>a . If a is analytic in Br(Q\ then F n BP(G) is also analytic.
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Proof. The function v is a solution of the (Dirichlet) boundary value problem (2.16). According to Agmon et al. [1, Theorem 11.1], Hence y, e C m+ 1>a (L7 u S) and, in particular, (2.17) is a C m+ 1 > a parametrization of F. To complete the proof note that since the mapping jc = (/, vN(y)) is of class C m+ 1>a , so is its inverse y = y(x). Thus and
If a(x) is analytic, it follows that (2.17) gives an analytic parametrization of T (Morrey [1] or Friedman [1]). Q.E.D. We encounter a different situation when b ^ 0. So, suppose now that 6 ^ 0 and CN ^ 0 and introduce the new dependent variables (2.8). Again, since u = 0 on F and (0,..., 0, 1) is normal to F at x = 0, whence Assuming that b/cN > 0, (2.8) is a 1:1 mapping of a portion of Q near x = 0, which we take to be all of Q, onto a domain U a {y: yN > Q}. Also F is mapped onto a portion S of {y: yN = 0}. We now have from (2.10), (2.13), and (2-14) that
Also we may represent F as We state this next conclusion only for the analytic case. Theorem 2.2. Let u e C2(Q u F) satisfy (2.14), where a is analytic in a full neighborhood of the origin, and suppose that b ^ 0, and CN ^ 0. Th'en F is analytic near x = 0.
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Proof. Again we appeal to Theorem 11.1 of Agmon et al. [I]. In this case the boundary condition is not a Dirichlet condition, but a Neumann one. The regularity theorem applies nonetheless because the boundary condition is of coercive type. We shall discuss this property in more detail in the next section. Q.E.D. When can we verify the hypothesis of the theorems? With respect to Theorem 2.2, once u is assumed merely in C*(Q u F), F is also C1 and the conclusion applies. For Theorem 2.1 we state this criterion: Theorem 2.3. Let Q c: RN be a bounded open set with OedQ and let UGH2- °°(Q) satisfy (2.15) near x = 0 in Q and 3Q. Suppose that a(x) e C°'a(Q) for some a, 0 < a < 1, and a(0) ^ 0. //
then (i) 5Q is a C1 hyper surf ace F near x = 0, and (ii) w e C 2 ( Q u F ) .
3. Elliptic Systems This section is a brief introduction to the theory of elliptic systems. For the greater part, our treatment recounts well-known definitions and properties, but the reader unfamiliar with "weights" and their use will find here a self-contained explanation. Let Q c: [RN be a domain, which in our applications will usually be bounded or 1R+ = {y e UN:yN > 0} and set
Let Lfe/y, D), 1 < j, k < n, be linear differential operators with continuous complex valued coefficients and consider the system of equations in the dependent variables u1,..., u":
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where the/ k are given, say smooth, functions. To each equation we assign an integer weight sk < 0 and to each dependent variable an integer weight tk > 0 such that
Such an assignment of weights is called consistent. By convention, Lkj(y, D) = 0 if sk + tj < 0. Let Lkj(y, D) denote the part of LkJ{y, D) of precisely order sk + ty, it is either zero or homogeneous in D of degree sk + tj. Definition 3.1. The system (3.1) is elliptic (with weights sk and t) provided that and, in addition, for each pair of independent vectors £, r\ e UN and y e Q the polynomial in z, p(z) = det(LkJ{y, £ + z//)), has exactly fj. = % deg p roots with positive imaginary part and /^ = i deg p roots with negative imaginary part. The condition about the roots is automatic if N > 3, but when N = 2 an equation may satisfy (3.3) but not the root condition. As an example consider the Cauchy-Riemann equation
or any of its powers. The matrix L'kj(y, £) is called the principal symbol matrix. Suppose, for fixed y0 e Q, we search for solutions to the homogeneous equations with constant coefficients
which have the form uj(y) = cV^, cj e C, 0 =^ £ e UN. Then (3.3) holds if and only if c1 = • • • = c" = 0 for every 0 ^ £ e UN, y0 € Q. Such solutions are called exponential. Thus, equivalent to (3.3) is the condition that the principal part of (3.1), which is (3.4), admits no nontrivial exponential solutions. A general system of equations
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where here D stands for all derivatives and not merely first order ones, is called elliptic along the solution u = (u1, ..., u") provided that the variational equations
constitute an elliptic system in the sense of Definition 3.1. We now discuss boundary conditions which we need only consider on portions S of dQ. contained in the hyperplane yN = 0. Let Bhj(y, D\ 1 < h < M» 1 < . / < « , be linear differential operators with continuous coefficients. Definition 3.2. The set of boundary conditions
is coercive for the system (3.1) provided that (i) the system (3.1) is elliptic and 2p = £" (Sj + tj) > 0 is even, (ii) there exist integers rh, 1 < h < n, such that order Bhj(y, D) < rh + tj on S, and (iii) with B'hj the part of Bhj of exactly order rh + tj, for each y0 e S the homogeneous boundary value problem
admits no nontrivial bounded exponential solutions of the form [Here, as usual, / = ( y 1 ? . . . , yN_i) and % = (^,..., ^ N _i).] Similarly, we may extend the notion of coercivity to nonlinear systems. A set of boundary conditions is coercive for the system (3.5) a/om? u = (it1, . . . , u") provided there exist weights r l 5 . . . , r^ such that the set of linearized boundary conditions
is coercive for (3.6) on S.
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In the sections which follow, our concern will be the use of hodograph and Legendre transforms to rephrase free boundary problems as nonlinear elliptic systems with coercive boundary conditions. Although such systems display many interesting properties of existence and uniqueness, our interest is in the regularity of their solutions. We state here the theorem to which we shall appeal. Let U be a neighborhood of 0 in IR+ and S = dU n {yN = 0}. Theorem 3.3. Suppose that u = (u1,..., un) is a solution of
where (3.11) is elliptic and coercive along u with weights sk,tj,rh,i < j, k < n, 1 < h < ft. Suppose also that Fk and 3>h are analytic functions of y and the derivatives of u. Let r0 = maxA(0, 1 + rh). If ujeCti+r°'\V u S) for some a > 0, then the uj are analytic in U u S, 1 < j < n. As a first example of an elliptic system, consider
where (akj) is a u x /z constant matrix and, as usual, U is a neighborhood of 0 in K+ and S = dU n {yN = 0}. Assign to each equation the weight sk = 0, to each w j the weight tj = 2, and to each boundary condition the weight rh = — 2. The system corresponding to (3.8) is
A bounded exponential solution of the differential equations is given by
for each 0 ^ £ e UN~1. Hence
namely, (3.12) is coercive if and only if dei(ahj) ± 0. In particular, the Dirichlet problem for the Poisson equation is coercive and, indeed, the Dirichlet problem for any elliptic equation is coercive (Exercise 3).
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Choosing the weights for a system can be a delicate matter, as we shall soon observe. For the moment let us examine the system
where 0 ^ X e IR. Our question is what may be decided about the Cauchy data which is shared by two different equations on the hyperplane yN = 0? A preliminary assignment of "obvious weights" s{ = s2 = 0, t1 = t2 = 2, TI = — 2, r2 — — 1 leads to the "linearized system"
which is not coercive. Now introduce w = u2 — u1 so that (3.13) becomes
Set sl = 0, s2 = —2,ti = 2,t2 = 4, and rl = — 4, r2 = —3. The principal symbol of (3.14) is
With so C is a solution of
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with the additional stipulation that supKI < °o. Consequently, C(0 = C 0 £?-'<'" + Cite-K']t and, by (3.16) and (3.17),
So C0 = C, = c = 0; hence (3.15) admits no bounded exponential solutions. We may now apply Theorem 3.3 to deduce Theorem 3.4. //A and A + A, A ^ 0, share Cauchy data on yN = 0/or solutions defined in a neighborhood U of y = 0 in yN > 0, then the data is analytic. For a second proof of the theorem, apply A to the second equation of (3.14). Then w is a solution of the fourth order problem
This problem is coercive with the obvious weights s = 0, t = 4, rl = — 4, r 2 = —3. Note that t = t2. Hence w is analytic in U u S. Since A ^ 0,
is also analytic. This technique suggests that altering what appear to be the "obvious weights" of a system-so it becomes coercive is analogous to differentiating the equation. It is interesting to compare a simple transmission problem with the one we have just considered. Set U+ = {yN > 0} n ^(O) and U~ = [yN < 0} n fi^O) and suppose that w 1 e C 1 (L r+ ), v € Cl(U~) satisfy
where A e R. Defining M2(y) 6 C^t/-") by
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we find that
It is an elementary matter to check that this system is coercive with the "obvious weights" s x = s2 = 0, r t = t2 = 2, r± = — 2, r2 = — 1, whether or not A is zero. Hence w and v are analytic in a full neighborhood of 0. The equations and boundary conditions which occur in subsequent sections will be nonlinear. To test the ellipticity and coerciveness of such a system it will be convenient to have a criterion which may be applied at a single point. The remainder of this section is devoted to proving that a system with a consistent choice of weights which is elliptic and coercive "at a single point" is indeed elliptic and coercive in the sense of Definitions 3.1 and 3.2 in neighborhood Q. <= K+ and S = Q n {yN = 0}. Recall the system of equations its associated system of variational equations
the boundary conditions and its associated system of variational equations
Suppose that sjt tk,rh,\ <j,k
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Theorem 3.5. Suppose that u1,..., u" is a solution 0/(3.5), (3.9) in Q u S and that Fk, 1 < k < n, O fc , 1 < h
The coefficients of LkJ(y, D) are continuous in Q u S. Hence if
and _y = 0, it also holds for y e Q u S, \ y \ small. Now suppose that has exactly u = | deg P roots with positive imaginary part and u = ? deg P roots with negative imaginary part for each pair of independent vectors
Consequently the condition about the roots of P(y, £ + zrj) is fulfulled for arbitrary independent vectors £, rj if and only if it is fulfilled for independent unit vectors £, r\. From this point, a simple compactness argument shows that if (3.22) holds for y = 0, it holds for y e Q u S, \ y \ small. Q.E.D. To show that the coerciveness property is also open, we present an argument based on the inverse mapping theorem for Hilbert spaces. This proof is technically simple and has the advantage of exposing some of the underlying ideas in the study of elliptic boundary problems. Lemma 3.7. Suppose that (3.6) is elliptic at y = y0. Then for each k there is at least one j such that the degree in £N ofL'kj(y0, £) equals sk + tj. Proof. This is an elementary lemma. Obviously the degree in £ N of L'kj(y0, £), degjy L'kJ{y0, £), does not exceed sk + tj. Suppose that for some /, degN L',/y0, £) < s, + tj, 1 < j < n. We may calculate P(y0, 0 by expanding
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in cofactors across the /th row. Now the cofactor of Li/y0, £), Llj(y0, £)> is a homogeneous polynomial of degree £k,t, sk + ^h*jth, so since
This is inconsistent with the coerciveness hypothesis as P(yQ, (
The system of ordinary differential equations with con-
Dt = —i d/dt, and weights s k , fj, rh is coercive provided (i) deg pkj(r) < sk + t j t deg qhk(i) < rh + f^-, 1 < k, j < n, 1 < h < n, (it) for each k there is at least one; such that deg pfc/t) = sk + tj, (iii) p(r) = det(pk/t)) is of degree m = £ (s, + fj) = 2^ and has exactly H roots with positive imaginary part and ^ roots with negative imaginary part, and finally, (iv) the only solution (y1, . . . , vn) of the homogeneous system (3.23), (3.24) bounded for t > 0 is vl = • • • = v" = 0. As usual, pfcj = 0 if sk + tj < 0, etc. It is evident that if (3.5), (3.9) is coercive at y 0 , then for each 0 ^ Q e UN~1
is a coercive system of ordinary differential equations.
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Let us consider for a moment a solution of an arbitrary system of equations with constant coefficients
Observe that det(Pk/Dr))u'(f) = 0 for each /, ! < / < « , and hence the dimension of the kernel of (3.25) is finite. To check the assertion, simply multiply the matrix (Pk/D,)) on the left by its classical adjoint. By Ha(U + ) we understand the completion of the space of complex valued C°°((R+) functions in the H" norm. We take the inner product in H"(R+) to be
Given a coercive system (3.23), (3.24), for any a > 0 continuously and because of (ii) we may assert that
is continuous. Recall that the "trace mapping"
is continuous. Hence
is continuous for all h, j provided that a > r0 = maxh(rh + 1, 0). Let us arrange our formalism. Set
and define the continuous linear mapping by TV =f, for v = (0 1 ,..., V")E V,f= ( f l t . . . ,/„,
fll ,...,
aJeW, if
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and
Lemma 3.9. With T: V -> W defined by (3.26), ker T = {0} am* the dimension of the cokernel oj T is finite.
Proof. The first assertion is part (iv) of the definition of a coercive system. Indeed, any solution ( of the homogeneous equation of (3.23) has components £J'(r) which are solutions of where p(t) = det(pfcj(.T)) [cf. (3.25)], and hence (J(r), t > 0, is bounded if and only if it belongs to each H"(U+) inasmuch as i = 0 is not a root of p(-r). Now let / = (/!, ...,/„, a l 5 . . . , dp) e coker T. This means that for all v = (v1,...,v")eV
In particular, choose uj e C^(1R+). Then
Note that it is no loss in generality to assume that/ k e C°°(IR+). Indeed, each fk(t) is the limit in //~ Sk+r °(lR + ) of a sequence cpe *fk(t) as e -* 0, where q>E is a sequence of symmetric mollifiers with compact support. Replacing fk by q>£ *fk and rearranging the integrals in (3.27) gives (3.27) for the original fk and new test functions
hence/!, ...,/„ are solutions of the system
As we have remarked, the dimension of this space of solutions is finite, say M.
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Hence dim coker T < M + dim C" = M + p. Q.E.D. Theorem 3.10. Suppose that (3.23), (3.24) is a coercive system. Then whenever vj E Hro+tj(U + \ I <j
we have the estimate
/or a constant C > 0, independentofvj. Proof. The mapping T" defined by (3.26) is 1:1 and has a finite dimensional cokernel E according to the last lemma. Hence T induces a continuous isomorphism T: V-> W/E. By the inverse mapping theorem, T~i is continuous. Hence if TV =/and [/] denotes the equivalence class of/in W/E,
for some C > 0, since g = 0 e E. This proves the theorem. Q.E.D. Completion of the Proof of Theorem 3.5. Since tj + sk < r0 + tj, 1 < j, k < n, and rh + fj- < r0 + tj, 1 < h < n, \ <j < n, the linearized equations (3.6), (3.10) have continuous coefficients. By Lemma 3.6, the system is elliptic near y = 0. Consider the system of ordinary differential equations, for 0^'eIR"- 1 fixed,
It remains to verify part (iv) of Definition 3.8 for this system when \y.'\ is small. Given e > 0, the continuity of the coefficients of the L'kj, B'hk insures that for some 6 > 0, |/| < d implies the coefficients of il of the L'kj(y, £', T) and pkj(?) differ by less than e > 0 for all / and the same for the coefficients of T' in B'hj{y, £', T) and qhj{i} where pkj(i) = L^/O, £', T), qh]{i) = B'hj{Q, £', T).
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Hence if v e K, the previous theorem implies that the solution of (3.28), (3.29) satisfies
Hence for £ sufficiently small
for functions v j , f ksatisfying(3.28) and (3.29). In particular, any bounded exponential solution of the homogeneous equations of (3.28), (3.29) vanishes. Q.E.D.
4.
A Reflection Problem
Our applications to free boundary problems begin with a system which arises in the study of confined plasmas in physics. It resembles the transmission problem (3.19). Suppose that F is a C1 hypersurface in B^O) c 1RN passing through x = 0 which separates fi^O) into two components, Q + and Q~. We assume that the normal to F pointing into Q + at 0 is (0,..., 0, 1). Suppose now that satisfies
where A(x) is an analytic function of x e Bi(0). Theorem 4.1. Let u be a solution of (4.1). Then F is an analytic hypersurface. Proof. In this situation we apply our zeroth order hodograph transform (2.8). We assume that u > 0 in Q + and u < 0 in Q". Then the mapping
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is a 1:1 mapping of ££(0) onto a neighborhood U of y = 0. We may assume that (4.2) is 1 : 1 in a neighborhood V of B^O). Then (4.2) maps Q + onto U+ = { y ( = U : y N > 0}, Q~ onto U~ = {yeU:yN< 0}, and T onto S = {y£ U: yN = 0}. As before, we define
Consequently, from (2.13)
As usual, the sums extended on 0- are 1 < a < N — 1. The system (4.3) resembles a transmission problem and we treat it in the same fashion. Introduce This leads to the system
with the boundary conditions
To prove the theorem it suffices to check that this system is coercive. In view of our local criterion Theorem 3.5, we need only check this at y = 0. By our choice of coordinates
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hence the linearized equations for (4.4) with the obvious weights s1 = s2 = 0, t! = r 2 = 2, are, for variations \j/, (p~,
where a = uN(0) > 0. The linearized boundary conditions are just
It is easy to see that (4.6), (4.7) admits no bounded exponential solutions; hence, the system (4.3), (4.4) is coercive. Q.E.D.
5.
Elliptic Equations Sharing Cauchy Data
Suppose we are given solutions to two different elliptic equations defined in the same domain which share Cauchy data on a portion of its boundary. What can be said of this boundary portion? This question, a free boundary analog of (3.13) and Theorem 3.4, will be discussed here in its simplest form. The problem certainly has some interest in its own right and it will also prepare us for the study of the free boundaries which occur in the double membrane variational inequality and in variational inequalities of higher order. In order to exhibit the overdetermined nature of our free boundary with a minimum amount of technical detail we shall assume that the various quantities possess a good deal of initial smoothness. Let Q, c UN be a domain whose boundary near 0 e d£l is a smooth hypersurface F passing through x = 0 with inward pointing normal (0,..., 0, 1) at x = 0. Suppose that
where/i,/ 2 are analytic functions in a neighborhood of Q u F and A e R. It is not sufficient to know merely that u1 = u2.
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ELLIPTIC EQUATIONS SHARING CAUCHY DATA
205
We shall consider two cases of this problem, each suggesting a new choice of variables defined in terms of w(x) = ul(x) — u2(x). In the first case, we suppose that/jCO) -/2(0) ^ 0. Here a first order hodograph transform will be appropriate, leading to a coercive elliptic system. In the second case we suppose that/! = /2. This will suggest a fourth order problem for w alone, which, after suitable transformation, also leads to an elliptic system, although one rather different from the first case. Theorem 5.1. Let u 1 , u1 be a solution of (5.1), (5.2) and suppose that /i(0) ^ /2(0). Then Y is analytic near x = 0. Proof.
Setting w = w 1 — u2 permits us to express (5.1), (5.2) in the form
The boundary condition for w means that w,- = 0 on F, 1 < i < N. Let us introduce the hodograph transform and the new dependent variables
Since the normal to F at x = 0 is in the XN direction, so
Assuming that wNN(0) < 0, we see that (5.5) maps a neighborhood of x = 0 in Q, which we may take to be all of Q, onto a domain L/C: {y: yN > 0} and maps F onto a'portion S of yN = 0. According to (2.7), vl satisfies a second order nonlinear equation which we shall write down later ((5.8)). Let us turn our attention to the equation satisfied by v2(y). According to (2.6),
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and
Thus it is apparent that our equation, though of second order in v2, is of third order in v1. Some care in choosing the weights will be necessary! Our new system for v1, v2 is [cf. (2.7)]
with
The summations above are on cr, 1 < cr < N — 1. We choose weights sl = —2, s2 = 0, r, = 4, t2 = 2, r: = —4, and r2 = —2. In this way, r : + s2 = 4, so the principal symbol will not contain any third derivatives of v1. To complete the proof, it remains to show that (5.8), (5.9), (5.10) is elliptic and coercive with respect to our chosen weights. By our localization principle, it suffices to check that at y = 0, where we may assume fjU(O) = — l/wNJV(0) = 1. Also, from (5.7), For variations y1 and v2 we find the system
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207
The principal symbol of (5.11) is
which is of rank 2; hence the system is elliptic. It is easy to check that (5.11), (5.12) has no bounded exponential solutions. Hence Theorem 3.3 applies, so vl and v2 are analytic in U u S. As a consequence F is also analytic inasmuch as it permits the representation [cf. (2.17)]
We turn now to the discussion of (5.1), (5.2) when/^x) =/2(x). Here (5.3) assumes the more concise form
Applying A to the first equation and substituting the second in it gives
To assess the behavior of w on F we state this elementary lemma. Lemma 5.2.
Let w, u2 be a solution o/(5.13) with the boundary conditions
Then w,/x) = 0 on V for 1 < 1,7 < N. Note that the condition on F is simply (5.4). Proof. At a given point x° e F, choose coordinates £j, . . . , £N with the £N axis normal to F at x°. Then since wx. = 0 on F for all i, w^f = 0 on F for all j, and hence
Now from (5.13)
The lemma follows. Q.E.D.
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Our new transform is based on the choice of — WNN as a new independent variable. For this to be valid, we shall assume that A ^ 0 and du2/8v(Q) = dul/dv(G) 7* 0. We shall first state our theorem in terms of the system (5.1), (5.2) and then in terms of the single function w. Theorem 5.3. Let ul,u2 be a solution of the system
where f is analytic in a neighborhood of x = 0 and Ae IR. Suppose that A ^ 0 and du1 /dv(0) =£ 0. Then F is analytic in a neighborhood ofx = 0. Proof. We reduce the proof of this theorem to the one below. With w = u1 — w 2 , as usual, we have that and
for any multi-index a, |aj < 2, in view of Lemma 5.2. Now note that
whenever i + j < 2JV because (d2/dxN dxj)w = 0 on F. Hence from (5.13),
Theorem 5.4.
Let
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209
where A and g are analytic in a neighborhood ofx = 0. Then F is analytic near x = 0. Proof. Recall that the XN axis is normal to F at x = 0. Hence (5.18) insures that the mapping
is 1:1 in a neighborhood of 0 in Q, which as usual we take to be all of Q, and maps it onto a region U which we may assume to be in yN > 0. The hypersurface F is mapped in this way onto a portion S of yN = 0. We set
We determine equations for q>1 and (p2 from (5.17). Now (p1 is merely the Legendre transform of WN with respect to (5.19) so
where, as usual, subscripts of (p1 refer to differentiation with respect to y and those of w to differentiation with respect to x. In addition,
Also
Now we write (5.17) as
so that
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where
in U. The second equation is obtained by differentiating (p2 in two ways:
and
So
We have deduced that *, ^j2 is a solution to (5.21), (5.23) subject to the boundary conditions
Choose weights s x = 0, s2 — — 1, f i = 3, t 2 = 2, rj = — 3, and r 2 = —2. To linearize the equations at j; = 0, notice that for 1 < 0,1: < N — l,j = 1,2. For variations (pl and ^ 2 we find from (5.22) and (5.23) that
in R + , with
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211
The symbol of (5.25) is
where a =
Hence the system is elliptic near y = 0. Finally, we check the coerciveness of this system. Writing we obtain for Ci and (2 tne system of ordinary differential equations
for t > 0. Differentiating the first equation and substituting the second into it gives, after multiplication by a3, Hence, assuming a > 0, From the boundary condition, c2 = 0. Now so Ct =• 0. Hence Ci — C 2 = 0. The theorem is proved. Q.E.D. We offer some comments about the hypotheses of Theorems 5.1 and 5.3. The transformation (5.5) is defined and the resulting equations have continuous coefficients if we only assume that ul,u2 e C 2>a (Q u F), say. To invoke Theorem 3.3, however, we must know that y1 6 C 4>a (C/ u 5). Because of our choice of weights, this may be achieved by first applying Theorem 2.1 to the equation for w alone in (5.3). Since
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it follows that weC 4 > a (QuF), FeC 3 > a , and the transformation y(x', Wjv) and its inverse are C 3>a . Since v^(y) — wff(x), 1 < a < N — 1, and 4 = xjv, all derivatives oft' 1 are in C 3>a , that is, ve C4'"(17 u S). In a similar way we may weaken the hypotheses of Theorem 5.3. Differentiating the equation for w in (5.13) with respect to XN , we obtain the problem
subject to the restriction wNNN(0) ^ 0 provided the hypotheses of Theorem 5.3 are in force. Now WN, u^ e C 1 > a (fi u F), so Theorem 2.1 may be applied. After some manipulations (Exercise 13) we conclude that w e C4>a(Q u F) and (p1 and (p2 defined by (5.20) are in C 3>a (t/ u S) and C2'a(U u 5), respectively. Indeed, in this fashion it is possible to "bootstrap" to a C°° result, in distinct contrast to the problem of section 6, which we are about to consider.
6. A Problem of Two Membranes In the previous section we learned how to assess the behavior of a free boundary when several functions were defined on the same side of it. The question here consists in the determination of the smoothness of a hypersurface when two functions are defined on one side of it and a third defined on the other side. This suggests incorporating into our argument a reflection mapping, as in Section 4. First we state our problem, then we illustrate how it arises from a variational inequality. Let F be a smooth hypersurface in UN passing through x = 0 whose normal at x = 0 is in the x^ direction. Suppose that F separates a neighborhood of the origin into domains Q + and Q~ with, say, (0,..., 0, 1) the inward pointing normal to fi+ at x = 0. Let w 1 , u 2 eC 2 (Q + u F) n C2(Q~ u F) n C1(Q+ u F u Q~) satisfy
and
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A PROBLEM OF TWO MEMBRANES
213
where/j,/ 2 are given smooth functions in Q + u F u Q~ and A e R . The "boundary" conditions satisfied by uj implied by (6.1) and (6.2) are that the uj are C1 across F. Now let Q c RN be a bounded domain with smooth boundary dQ and suppose given h1, h2 e C2(dQ) with hl > h2. Denote by IK the convex set of pairs of functions
For given, say smooth, functions/ l5 / 2 in Q let (w1, u 2 ) be the solution of the variational inequality
The existence and smoothness properties of the uj are the subject of Exercise 13, Chapter II, and Exercise 5, Chapter IV. We suppose that t/ j eH 2>s (Q), 1 < s < oo, and define the coincidence set of u1 and u2. One easily checks that
LetC e C*(Q)andset follows that (u1, y 2 )elK. Substituting this (v1, v2) into (6.3) leads to the equation Hence, since u\ = u2 in /, 1 < i < N, whence u\j = w,2 a.e. in /, we obtain
From these considerations we infer that (6.4), (6.5) may be recast as (6.1) (6.2) near suitably smooth and small portions F of dl. However, it is far from obvious that (6.1), (6.2) is overdetermined. In fact, neglecting the equation in Q~ temporarily gives the problem
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and
where v is the normal direction to F, which is generally coercive, and hence solvable when F has only a limited degree of smoothness. Evidently, we must utilize the information given by (6.2) in Q~. To assist us in the resolution of this difficulty, let us introduce w = ul u2 and w' = w 1 + u2. Then
From the second equation The elliptic regularity theory applied here informs us that Now we impose the nondegeneracy criterion Then w is a solution of
with fl(x, w) analytic in w and C 3>0[ in x, a(0, 0) ^ 0. Thus we are led to Lemma 6.1.
Let
satisfy (6.1), (6.2), where F is of class C1. Assume that (6.7) /jo/ds. T/ien r/iere is a neighborhood B,(0) SMC/I t/iaf w = u1 - u2 e C5'a((Q+ u F) n Bp(0)) a«^ F is of class C 4>a near x = 0. Proof. This is direct application of Theorem 2.1 to w which satisfies (6.8). Q.E.D. Unfortunately this technique cannot be iterated to prove that F e C°°.
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A PROBLEM OF TWO MEMBRANES
215
Let us proceed to introduce the first order hodograph mapping which, we may assume, maps Q + in a 1 : 1 manner onto a region U in yN > 0 and F 1:1 onto a portion S of yN = 0. Set The inverse mapping to (6.9) is given by which carries U into O + , of course. Now let C be any constant larger than l|0Mi.°°(i;)anddefine When y = (/, 0)eS, g~(y) = (/, %(/, 0)) = g+(y)eT. But due to our choice of constant C, g~ maps a neighborhood of 0 in U, say all of U, into Q~. Using this new reflection mapping we define new dependent variables
We shall prove that the three functions v, cp+, and
Suppose that
is a solution of (6.1), (6.2). Suppose thatft andf2 are analytic in a neighborhood ofx = 0 and that T/zen F is analytic in a neighborhood ofx = 0. Proof. We begin by calculating the system of equations satisfied by v, cp +, and (p~. Our point of departure is the system (6.6). With dk = d/dyk,
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and
Reserving for the moment our calculation of the various equations, let us determine the boundary conditions obeyed by v, cp + , (p~ on yN = 0. As usual for a Legendre transform, v = QonyN — 0. Since w' is continuous across T,
Summarizing,
The first equation of (6.6) transforms in a familiar manner. From (2.7) we have that
where is an analytic function of its arguments. We continue in a manner analogous to the proof of Theorem 5.1. Considering the second equation of (6.6) and restricting our attention to x e Q + , we obtain, as in (5.9),
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A PROBLEM OF TWO MEMBRANES
217
where Evidently, F2 is an analytic function of its arguments. For x e Q~ note that w(x) = 0. Consequently,
where We assign the weights Sj = — 2 to (6.15) and s2 = s3 = 0 to (6.16) and (6.17), tv = 4, t+ = 2, r_ = 2 to y,
It is easy to see that this system is elliptic and has no nonzero bounded exponential solutions, keeping in mind that C — a > 0.
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The coerciveness of our problem, (6.15), (6.16), (6.17) with boundary conditions (6.14), is established. By Lemma 6.1, v, q> + , and (p~ are sufficiently differentiable to apply Theorem 3.3. Hence v,
COMMENTS AND BIBLIOGRAPHICAL NOTES The use of hodograph and Legendre transforms is well established in topics as diverse as fluid dynamics and convex bodies. For example, they occur in Friedrichs' study of cavitational flow (Friedrichs [1]) and Lewy's study of the Monge-Ampere equation (Lewy [1]). Our treatment here is an adaption of (Kinderlehrer and Nirenberg [1]). The regularity criterion Theorem 2.3 is due to Caffarelli [2]. Boundary value problems for elliptic equations as well as elliptic systems have been extensively studied. We refer to Agmon, Doughs, and Nirenberg [1,2], Lions and Magenes [l],and Morrey [1] for example. The topics in Sections 4-6 are drawn from Kinderlehrer, Nirenberg, and Spruck [1], For more details about plasma type problems, we refer to Temam [1, 2] and Kinderlehrer and Spruck [1]. Lewy's elegant original proof of Theorem 3.4 consists in extending analytically u^ — iulX2 by employing the Riemann function of the equation A + A, when N = 2, in the spirit of Chapter V (Lewy [5]). This method may also be used to prove Theorem 5.3 when N = 2. Other generalizations and techniques for the study of higher order free boundary problems are given in (Kinderlehrer, Nirenberg, and Spruck [2]). The existence and regularity theory for the problem of two membranes was developed by G. Vergara-Caffarelli [2, 3, 4]. The latter papers deal with nonlinear membranes.
EXERCISES 1. Let v be the Legendre transform of u given by (2.2) with respect to (2.1). Show that if F(D2u, Du, M, x) is elliptic along u and G(D2v, Dv, v, y) = F(D2u, Du, u, x), then G is elliptic along v. 2. Let \l/ be the zeroth order Legendre transform (2.9) of u given with respect
EXERCISES
219
to (2.8). Show that if F is elliptic along u and H(D2\l/> Zty, \j/, y) = F(D2u, Du, u, jc), then H is elliptic along ty. 3. Let L(D) be a linear elliptic operator with smooth coefficients of order 2/i, n > 1, defined in a neighborhood U of x = 0. Show that the problem
is coercive. Study this question for the problem
where 0 < hl < h2 < • • • < h^ < 2m - 1. In Exercises 4-7 determine whether or not or under what conditions the systems below are elliptic.
under various assumptions about m, p, n.
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Check the coercivity of problems 8-10 below.
11.
The boundary value problem
fails to meet the definition of coerciveness because the operator d/dz = ^[d/dxi) + i(d/dx2)~] does not satisfy the condition about roots. Nevertheless, the mapping satisfies ker T = {0}. Show, however, that dim coker T = oo. [Hint: (/, g) e coker T if and only if
12. Suppose that F is a C1 hypersurface which divides a neighborhood of the origin into Q + and Q~ as in Section 6. Assume
EXERCISES
221
where u,+ , u~ are the normal derivatives of u evaluated on approach to Q + and Q~, respectively, and v is taken pointing into Q + . Assume that u* ^ 0 on F. What can be said of the regularity of F? 13. Prove that the hypotheses of Theorem 5.3 imply that 14. Let p(z) be a polynomial of degree 2fj. with roots a 1 ; ..., a^ in the upper half-plane and a^ + 1, . . . , a2t, in the lower half-plane. Show that a general solution of
is given by
where F is any contour enclosing the roots a l 5 . . . , a 2M of p and g(z) is a polynomial with deg q < 2\i — 1. If u(t) -» 0 as f -> +00, then
where p + (z) = (z — a t ) • • • (z — ^), deg q < \i — 1, and F is a contour enclosing a 1 ? . . . , a^. Show also that 15. Consider this variation of the problem of the confined plasma: Given A > A 0 , A0 = smallest eigenvalue of + A for the Dirichlet problem in Q, find w e H1^) such that
Reformulate this question in terms of "Bellman's problem," Chapter III, Exercises 8-10.
C H A P T E R VII
Applications of Variational Inequalities
1.
Introduction
One of the attractions of the theory of variational inequalities is its application to many questions of physical interest. Several of them are described here; specifically, a problem of lubrication (Section 2), the steady filtration of a liquid through a porous membrane in both two and three dimensions (Sections 3-6), the motion of a fluid past a given profile (Sections 7 and 8), and the small deflections of an elastic beam (Section 9). The formulations of these problems are all variational inequalities involving elliptic operators and convex sets of admissible functions of obstacle type. In the next chapter we study the Stefan problem, which is connected to the heat operator. In several cases the problem as first given requires finding the solution of a free boundary problem which itself is not the solution of a variational inequality. A new dependent variable which is the solution of a variational inequality is then introduced by means of an auxiliary transformation. We show that this new unknown may be adopted as the solution of a Cauchy problem involving the original one. 222
2
2.
A PROBLEM IN THE THEORY OF LUBRICATION
223
A Problem in the Theory of Lubrication
In this section we present a simple mechanical problem, the lubrication of a complete journal bearing, whose resolution is facilitated by the use of variational inequalities. One seeks the pressure distribution of a thin film of lubricant, a fluid of given viscosity t], contained in the narrow clearance G between two surfaces, the shaft I.s and the bearing Sb, which are in relative motion. In the region G, p must satisfy Reynolds' equation, which will be written below in a form appropriate for us, together with certain boundary conditions. Let us direct our attention to a case of special interest in engineering (see Fig. 1). Consider a full cylindrical bearing where Zs and Zb are portions of circular cylinders with parallel, though distinct, axes of height 2b. We assume that Zb is inside Z s , Is is fixed, and Zb rotates with constant angular velocity co. Under normal conditions of operation the pressure reaches a certain minimum value pc related to the vapor pressure of the lubricant below which the film separates from Ls or £b forming a region of cavitation. This region of cavitation is occupied by a vapor at constant pressure pc. We take pc = 0, an approximation to atmospheric pressure. Our naive formulation of the problem must be extended to account for this phenomenon. We seek a
Figure 1. The cylindrical journal bearing.
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function p > 0 in G satisfying Reynolds' equation where p > 0. Since no mass is transferred across the boundary of the set / of cavitation, dp/dv — 0 on dl n G, v the normal to dl. Introduce cylindrical coordinates r, 0, z with origin at the center of Zs and 0,0 < 0 < 2n, measured from the line of maximum clearance of the centers (Fig. 1). Since the gap between the bearing and the shaft is assumed very small, we take p to be a a function of 9 and z alone, which permits a formulation in the 6, z plane. To write Reynold's equation, we introduce some parameters. Let rs = 1 be the radius of Zs, rb the radius of Z b , e the distance between the axes of Zs and Z b , and e = e/(l — rb), 0 < e < 1, the eccentricity ratio of the bearing. Set Q = {(0, z): 0 < 0 < 2n, \z\ < b}. The question we are discussing takes the form Problem 2.1. Find a function p(6,z) e Cl(&) and a region I c Q such
and
where
This description, i.e., (2.3), presumes that dl n O is smooth, an issue to which we shall return momentarily. The first boundary condition of (2.4) means that the lubricant can flow freely at both ends from reservoirs at atmospheric pressure. The second is an obvious periodicity condition. Suppose that dl n Q is indeed smooth and (2.3) holds. Then setting p = 0 in / extends it as a Cl(Cl) function in (Q — /) u / c Q. It would seem an intelligent move to replace (2.3) by In this way, Q — / becomes precisely the set where p is positive. The conditions (2.1), (2.2), (2.3) may be summarized as
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A PROBLEM IN THE THEORY OF LUBRICATION
225
A solution to this problem may be found by solving a variational inequality, more specifically, Problem 2.2 below. It will enjoy the additional property that Ap -/> O i n Q . Denote by //^(Q) the subspace of Hl(Q) whose elements v satisfy
and set A solution to Problem 2.1 is given by the solution p of Problem 2.2. where
Find peK:
a(p, v - p) > jn f ( v — p)d9dzfor all ve IK,
and y. and fare defined by (2.5).
It is easy to check that a(u, v) is coercive on //^(Q). Therefore by Theorem 2.1 of Chapter II we deduce the existence of a unique solution p(0, z) to Problem 2.2. Moreover, p is smooth in Q. To verify this in the most elementary manner, transform Q into an annulus O in the x = (xlt x 2 ) plane according to
for a fixed a > 0. Thus and the segments z = — b and z = b are mapped onto the circles | x \ — a and |x| — a + 2b, respectively, which bound O. The periodicity condition (2.6) merely ensures that (2.9) induces a continuous mapping of H^(Q) onto HQ(O). In this way IK is transformed to the closed convex set IK = {ue Hl0(O):v > O i n O } . For a function g(9, z) let us set g(x) = g(9, z) and define
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In lieu of Problem 2.2 we consider the variational inequality
whose solution is u(x) = p(x). Now p e f/ 2>s (0), 1 < s < oo, by Chapter IV, Theorem 2.3; hence p e H2'S(Q), 1 < s < oo. To summarize: Theorem 2.3. There exists a unique solution p(9, z) to Problem 2.2. In addition, p e H 2 - S (Q) n C1-A(Q), 1 < s < oo, 0 < A < 1. We briefly describe some properties of the solution p. One might note in particular that regardless of how slowly the inner cylinder Sb rotates, cavitation is always present, namely, 1*0. Theorem 2.4.
Let p(9, z) denote the solution to Problem 2.2 and define
the set of cavitation. Then (i) p(9,z) = p(9,-z)and 00 1*0. Proof. To prove (i) simply observe that u(9, z) = p(9, — z) is a solution of Problem 2.2. By uniqueness, we conclude that p(9, z) = p(9, — z). To prove (ii) assume 1 = 0. Then p is the unique solution of the Dirichlet problem
An elementary calculation shows that u(9, z) = —p(2n — 9, z) is also a solution of (2.11), whence u = p in Q. In this event, however, p $ K since there are points where it is negative. Hence 1^0. Q.E.D. Other interesting features of p are given in the exercises. Let us note especially the dependence of the solution on the eccentricity parameter e. To this end, we rewrite a and /in (2.5) as and
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THE FILTRATION OF A LIQUID THROUGH A POROUS MEDIUM
227
and let p0 denote the solution of
Theorem 2.5. defined in (2.13),
With pF the solution of Problem 2.2 for a.f and ft and p0
for a constant C > 0. The function p0 may be regarded as the first term in an asymptotic expansion for p£. The proof is left as an exercise.
3. The Filtration of a Liquid through a Porous Medium In this section we shall describe a problem of the filtration of a liquid through a porous dam assuming that the geometry of the dam is very simple. A more general problem is studied in Section 7. Consider two reservoirs of water separated by an earthen dam. This dam is to be infinite in extent and of constant cross section consisting of permeable vertical walls and a horizontal impermeable base. We are thus led to a two dimensional problem defined in the rectangle R which is the cross section of the physical configuration. The unknowns of the problem are (the cross section of) the wet portion Q c= R of the dam and the pressure distribution of the water. We shall offer a precise description in terms of a free boundary problem and then transform it to a variational inequality by the introduction of a new unknown function (see Fig. 2). A mathematical model of this situation leads to Problem 3.1. Let a, h, H e R satisfy a > 0 and 0 < h < H. Find a decreasing function-y = cp(x), 0 < x < a, with and a function u(x, y), (x, y) e Q, where
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VII APPLICATIONS
Figure 2. The porous dam in two dimensions.
satisfying (i)
w(x, y) is harmonic in Q and continuous in Q,
where v is the outward normal to the curve y — (p(x). The relation of M, the piezometric head of the fluid, to the pressure p is given by where the constant of gravitation is suitably normalized and y is the specific weight of the fluid. It must be remarked that both the domain Q and the function «(x, y) are unknown in this problem. In the unknown part of dQ, namely,
the function u(x, y) has to fulfill two requirements, Cauchy conditions, so F is a free boundary. Our aim is to reformulate this free boundary problem as a variational inequality. Condition (iii) imposes a smoothness criterion on F. Let us assume first of all that Problem 3.1 admits a solution pair {
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THE FILTRATION OF A LIQUID THROUGH A POROUS MEDIUM
229
where
The solution u of Problem 3.1 satisfies
for all £e Cl(Q) such that £ = 0 in a neighborhood of the segments 5Q n {x = 0} and <9Q n (x = a}. The proof is a consequence of the Green formula
We now prove Lemma 3.3.
Let u E Hl(Q) be a solution of Problem 3.1. Then
Proof. Note that physically this is nothing more than the statement that the pressure of the fluid is nonnegative. Observe first that
and Choose
which is nonpositive and vanishes for x — 0 and x = a. Moreover,
and
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We apply the preceding lemma, observing that ( = 0 on F, so that
using here that ( < 0. It follows that £ = 0 in Q, so that u > y in fi. Q.E.D. For a fixed x e (0, a), the flux of the dam, or its discharge, is given by
Lemma 3.4.
Let u be a solution of Problem 3.1. Then
Proof. In fact, choosing £ = £(x), a function of x alone which vanishes at x = 0 and x = a, we have by Lemma 3.2
Thus
but
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231
Unlike the solution p(9, z) of the lubrication problem, u(x, y) does not itself resolve a variational inequality. What properties can it enjoy in the rectangle The function v(x, y) = u(x, y) — y is continuous in Q and vanishes on F. Let us define v continuously R — Q by setting it equal to 0 there. The extended v is in Hl(R), so for £ e C£(R)
by Lemma 3.2, where / n denotes the characteristic function of Q. What we have calculated is that
in the sense of distributions. This suggests introducing a new function w(x, y), the solution of the Cauchy problem
The problem may be integrated explicitly, so
From (3.6) and the assumed smoothness of F, wx = 0 on F, so the function
is of class C^/?) and satisfies This property suggests that w might be the solution of a variational inequality. Continuing this formal line of argument, (3.5) leads us to suspect that
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so that whenever v > 0 in R. This is a compelling statement, for we have identified the convex set IK of competing functions for our variational inequality, namely, nonnegative v, modulo the appropriate boundary conditions. Let us identify these boundary values and demonstrate precisely that vv is the solution of a variational inequality. From (3.6) or (3.7),
By (iii) and Lemma 3.4,
so
Let us denote by g(x, y) the function defining the boundary values of w(x, y) in Q and extend it to an H2' °°(R) function, still called g, by
Note that W E Hl(Q) since u e Hl(Q) by assumption. Also wy = y — u < 0 in Q, so
Lemma 3.5. Let w be defined by (3.8), where w denotes the solution of the Cauchy problem (3.6). Then, with /n the characteristic function o/Q,
for all C e CCC(R) which vanish near x = 0, x = a, and y = 0.
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233
Proof. Set C = Xy»where % and /x vanish on the bottom and on the lateral sides of R; for instance, set
Then
after two integrations by parts, using that at least one of the two factors involved vanishes at each point of d£l. Continuing, using Lemma 3.2 and (3.6),
This proves the lemma.
Q.E.D.
Keeping in mind that Q is an unknown of our formulation, the characterization of w just given cannot be used to solve Problem 3.1. We have shown, however, that
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Now set
a nonvoid closed subset oiHl(R) whether or not we assume the existence of u because g e IK. By (3.9)-(3.12), w e IK also. For any v e IK, v — w can be approximated by functions £ satisfying the hypotheses of Lemma 3.5; hence
since v > 0 in R. We have proved Theorem 3.6. Let {cp, u} be a solution of Problem 3.1 with u e Hl(£T) n C°(Q) and
and define
Then w satisfies the variational inequality
where K is defined by (3.13). Moreover,
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RESOLUTION BY VARIATIONAL INEQUALITIES
235
A consequence is Corollary 3.7. // {(p, u} is a solution of Problem 3.1 with ueHl(fy C°(Q) and cp smooth, then the solution of Problem 3.1 is unique.
r\
For otherwise there would be more than one solution of the variational inequality (3.14).
4.
The Resolution of the Filtration Problem by Variational Inequalities
From this point, our object will be the discovery of a solution to our original problem by studying the variational inequality (3.14). The bilinear form associated to the Dirichlet integral
which we term for brevity the Dirichlet form, is coercive on Ho(R). So we deduce from Chapter II, Theorem 2.1 the existence of a unique solution of (3.14), which we call w(x, y). In what follows, we shall prove that u = y — wy and the set F = d£l n jR, Q = {(x, y): w(x, y} > 0}, constitute a solution to Problem 3.1. A study of the free boundary F itself will be a part of this project. The proofs we give here are two dimensional in nature. Their higher dimensional generalizations appear in Section 6. Our initial concern is the smoothness of the solution w. To apply the regularity theory of Chapter IV we must prove that a solution of the Dirichlet problem
satisfies u e H2- P(R) whenever/e L"(R) and h e H2>P(R) for 2 < p < oo. This result is not immediate in a neighborhood of a vertex of R since dR lacks smoothness there. This problem admits a unique solution ueH1(R), for example, by Chapter II, Theorem 2.1. Replacing u by u — h we may take h = Q. Extend u to the region R* = {(x, y ) : 0 < x < a, — H < y < 0} by setting
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With F! = {(x, 0):0 < x < a}, the base of the dam, we have that tie H*(R u r t vj /?*) by the H1 matching lemma (Chapter II, Lemma A.8). Moreover, for any x 0 , 0 < x0 < a, for K small by the elliptic regularity theory. Note that
where
We claim that (4.2) is valid in the sense of distributions. To check this, select Ce C$(BE(xQ, 0)), 0 < x0 < a and e > 0 small, write J5e+ = jR n B£(x, 0), B~ = R* n Be(x0, 0), and calculate that
Here we have used that the traces of uy on rl n Bg(x0, 0) are well defined since u e H\Bt n /?) n //2(BE n /?*). Hence (4.2) is valid in the sense of distributions, so u is a solution of
Now x = 0 is a smooth surface, so the regularity theory now informs us that ueH2>p(Bf(Q, 0) n (R u /?*)). Thus our criterion is established at (x, y) -
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RESOLUTION BY VARIATIONAL INEQUALITIES
237
(0, 0). The treatment at the other vertices of R is identical. This ensures that the solution w to (3.14) satisfies
We define and observe that To facilitate our description of Q, we set
and F4 = {(x, //): 0 < x < a}. The first lemma follows by direct calculation and by noting that w assumes its minimum on F3 u F4. Lemma 4.1.
With the preceding notations
and
wy < 0
on F0,
where w denotes the solution o/(3.14). Next, we prove Lemma 4.2. The solution w o/(3.14) is continuous together with its second derivatives in a neighborhood o/F0 u F t u F2 in R and
Proof. By continuity of w and the positivity of g on F0 u F t u F 2 , there is a neighborhood in Q of F0 u r1 u F 2 . Recalling that Aw = 1 in Q and geC2-\r0 u F! u F2), the regularity theory permits us to conclude that there is a ball BE(x0, y0), F, sufficiently small, for each (x0, y0) e F0 u F, u F 2 ,
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such that w€C2'\Be(x0, y0) n 0), 0 < A < 1. Hence wxx = 1 - wyy = 1 ~ yyy = 0 on F0 u F2 while wyy = 1 — wxx = 1 — gxx = 1 on F^ Q.E.D. The next lemma leads to intriguing conclusions about the free boundary F. Lemma 4.3.
The solution w to (3.14) satisfies
Proo/. We shall rely on the maximum principle. We know from our regularity theorem that w,,, w y eC°' A (JR), 0 < A < 1, and wx = wy = 0 in R - Q. Since Awx = Aw y = 0 in D,
Consider wx. By the preceding lemma, w^ = 0 on F0 u F 2 , so wx does not achieve its maximum there in view of the Hopf maximum principle. Therefore
By continuity, wx = 0 on F whereas wx < 0 on F t u F3 u F4. Hence wx < 0 in Q. Turning to consideration of wy, its maximum cannot lie on r1 since (Wy)y = 1 > 0 on Fj. It is easily seen that wy < 0 at other points of d£l. Q.E.D. For any point P0 = (x0, y0) e R, set and
Lemma 4.4. IfPeR-Q, then Q$ c R - Q, a«J (/ P e R n dQ, f/ien <2p c Q. Proof. Let P0 e R — Q, so that w(x () , y0) = 0. Since wx < 0 and w,, < 0 in R by Lemma 4.3, w(x, y) = 0 for x > x0 or y > y0, namely, (5p c R — ^« Hence Q$ c K - Q. If P e K n dQ and Q^ were not contained in n, there would be a point P0 e Qp where w(x 0 , y0) = 0. From the above Qp0 c R — Q. But (x, y) € Qp0 which is disjoint from R n dQ. This is a contradiction. Q.E.D. Here is our final preparatory lemma.
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RESOLUTION BY VARIATIONAL INEQUALITIES
Lemma 4.5.
239
The solution w of (3.14) satisfies
Moreover, Proof. Let 0 < x0 < a and choose A small enough that 0 < x0 < x0 + A < a. Then
since w^ < 0 in ^. Hence wy(x, H) is a nondecreasing function of x. Since it vanishes at x = 0 and x = a, necessarily, wy(x, H) = 0 for 0 < x < a. A different proof of this statement may be adapted from Lemma 6.3. Now if (x0, H) e dfl, then necessarily the interval
for otherwise (x0, H)eQp for some Pe R n dQ. For (x t , H)ea, Aw,, = 0 near (x t , //) and wx assumes its minimum value 0 at (x t , H). Hence w^x^ H) < 0. But from the first statement of the lemma, w^x^ //) = w^x^ //) = 0. So dQ n T4 = 0, whence 5Q n 7? ^ 0. Q.E.D. At this stage we are in a position to define
The point (x,
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VII APPLICATIONS
the limits above exist. Finally,
where we recall that Q is defined by (4.3), because wy < 0. Lemma 4.6. The set 3Q n R does not contain segments parallel to the x or y axes. Hence (p is continuous and strictly decreasing. Proof. For a proof by contradiction, assume there exists a segment a parallel to the y axis in 5Q n R. Then there is an open neighborhood U c: Q with a c= dU such that w e C°°(l/ u a) and
by Lemma 4.3. Consequently the nonpositive harmonic w v attains its maximum on a so, availing ourselves of the Hopf maximum principle, we deduce
However, (wy)x = wyx = wxy = 0 on a by the conditions above and the smoothness of w. Similarly, there are no segments parallel to the x axis in <3Q n R. Q.E.D. So the set F = <3Q n R = dQ. — (F0 u F t u F2 u F3 u F4) may be described as follows:
with (p defined by (4.7). Theorem 4.7.
Let w denote the solution o/(3.14), and set
Then F = 5Q n R is an analytic curve. This theorem is a direct application of Chapter V, Theorem 1.1, with u = w — i|2| 2 and ^ = — i|z|2, z = x + iy. We conclude our discussion of Problem 3.1 with
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RESOLUTION BY VARIATIONAL INEQUALITIES
241
Theorem 4.8. Problem 3.1 admits a unique solution pair {u,
where K — {v £ Hl(R): v > 0 in R and v — g on dR}, with g defined in (3.11), and set
Define cp(x), 0 < x < a, by (4.7). Then {u,
Proof. It remains to verify (3.1), from which (i), (ii), and (iii) easil follow. Let C be a smooth function in R which vanishes near jc = 0 and x — a. Then
We shall not verify that
242
5.
VII
APPLICATIONS
The Filtration of a Liquid through a Porous Medium with Variable Cross Section
We renew our attention to the filtration problem by considering a question which requires three independent variables to formulate owing to the complex geometry of the dam. Specifically, we shall assume the dam to be fashioned of vertical walls of variable thickness. The major difficulty we shall encounter in posing the variational inequality is determining the boundary values at the base of the dam of the elements of the convex set of competing functions. In the previous case they were linear, but this will no longer be the case. To resolve the problem requires a deeper investigation of the free boundary. To begin we describe the classical problem (see Figs. 3 and 4). Consider a dam M given by where a > 0, H > 0, and gj e C 2> A([0, a]) for some A > 0 satisfy
Let B be the base and T the top of M:
Assume given h, 0 < h < H, and let cp: B -> [0, #] be continuous and satisfy For any such (p we set it will correspond to the wet part of the dam. We adopt these notations for the portions of its boundary:
5
THE FILTRATION PROBLEM IN THREE DIMENSIONS
Figure 3. The three dimensional dam M.
Figure 4. The base of the three dimensional dam, M = B x (0, //).
243
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VII APPLICATIONS
and finally, Now we can state our problem. Problem 5.1. Find a function g>(x'}, x' e J5, satisfying (5.2) and a function u defined in Q such that
and
As is customary, in the statement of the problem we suppose that q> is smooth enough that (iii) makes sense. We shall rephrase this problem as a variational inequality in which (p does not appear explicitly, analogous to our treatment of Problem 3.1. Indeed, just as in (3.6), we assume the existence of a solution u e Hl(Q) of Problem 5.1 and define a new function w(x), the solution of the Cauchy problem
Although w(x) may be expressed as an integral, its explicit form is not required. We extend w to all of M by setting it equal to zero in M — Q. Our assumption that F is smooth implies that the resulting w e C1(M) and
Before proceeding it will be helpful to review a maximum principle for a mixed boundary value problem. Let Q c= |RW be a bounded domain with a Lipschitz boundary dQ and suppose that Fj and F2 are open subsets of dQ satisfying
5
THE FILTRATION PROBLEM IN THREE DIMENSIONS
245
Finally, suppose that F t is "sufficiently large" that the Dirichlet form is coercive on the functions which vanish on F l 5 namely, C € //'(Q), for a constant C > 0 (independent of (). It is easy to verify the existence of a unique solution u eHl(Q) of the problem
given/e L2(Q), ge/f^fi), and /ieL 2 (F 2 ), where v denotes the outward normal of F 2 . Indeed, with Vg and K0 the subspaces of H*(Q) functions u such that i> = 0 on Fj and u = 0 on F1} respectively, u is the solution of the weak problem
Existence is a consequence of Chapter II, Theorem 2.1 in view of (5.4,ii). Proposition 5.2. Assume that Q satisfies (5.4) ami let u be the solution of (5.5), where fe L2(Q), 0 e //!(Q), ami /i 6 L2(F2). ///< 0 m Q and h < 0 on F 2 ,then
The proof of the proposition follows upon taking £ = max(u — k, 0) e VQ in (5.6), where /c = max Fl g. Lemma 5.3. wX3 < 0 in M. Proof. satisfies
Let w be defined by (5.3). Then w e Hl(M) and w > 0 and
It is obvious that w e Hl(M). Now w X3 = x 3 — u is in //'(Q) and
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whereas
and
So wX3 < 0 in Q by the proposition. The lemma follows.
Q.E.D.
To delineate the convex set of admissible functions which will appear in the statement of the variational inequality, we investigate the boundary conditions which pertain to w. It is no difficulty to verify that
In addition we prove Lemma 5.4. Let a0(x'), x' e B, denote the solution oj the mixed problem
T/ien and the solution w of (5.3) satisfies w — a0 in B. Proof. First note that there is a Lipschitz function in B which attains the boundary values asked of «0 on the curves x 2 = g^x^ and x 2 = g 2 (xj), 0 < Xj < a. Hence (5.8) admits a unique solution a0 e H1(B). The inequality (5.9) follows by applying Proposition 5.2 to a0 and to oc0 — %h2. Observe that
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THE FILTRATION PROBLEM IN THREE DIMENSIONS
247
whence
Let i//(x') e C°°(B) vanish near the curves x2 = g\(xi) and x 2 = # 2 ( x iX 0 < X j < a. Then with ((x) = \l/(x'\
by Fubini's theorem and because u is a solution of Problem 5.1. So w(x', 0) is a solution of (5.8). The conclusion now follows since (5.8) admits a unique solution. Q.E.D. By an abuse of notation, let us henceforth adopt the symbols S+ and S~ to mean S~ = {x E U3 : x: = 0, ^(O) < x 2 < 02(0), 0 < x3 < H} c dM
and S+ = {x € (R3 : x t = a, g^a) < x2 < g2(a\ 0 < x 3 < H} c 5M.
The proof of the next statement is similar to that of Lemma 3.6 and is left to the reader. Lemma 5.5.
Let w be defined by (5.3). Then
where /n is the characteristic function o/Q, whenever £ £ C°°(M) and vanishes neardM - (S+ vj S~).
Set
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and define
The set IK = {y e //^M): v > 0 in M and u = a in dM - (S+ u S~)}
(5.11)
is a closed convex nonempty subset of Hl(M). To see that it is not void, merely extend a to a nonnegative Hl(M) function. For example, let 6(x')e C°°(IF82) satisfy 0 < & < 1, 0 = 1 near the curve x2 = ^(xj), 0 < xl < a, and 0 = 0 near the curve x2 = ^a( x iX 0 < x t < a. Let i/f(x3) e C°°(1R) satisfy 0 < i/f < 1, i/<0) = 1, and ij/ = 0 for x 3 > h. Now extend a to M by
Such an extension has the advantage that aeH 2 > s (M) if a 0 eH 2i %B), a property which will assist us in our regularity considerations. Also, IK =£a whether or not u exists. Theorem 5.6. Let (M(X),
anrf extend w by zero in M — Q. Then
where K is defined by (5.11). Moreover,
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THE FILTRATION PROBLEM IN THREE DIMENSIONS, CONTINUED
Proof.
249
By Lemmas 5.3 and 5.4, w 6 IK. Now let u e IK. From Lemma 5.5,
since v — weH^M) and vanishes on dM — (S+ u S~). This proves the theorem. Q.E.D. Corollary 5.7. Problem 5.1 admits at most one solution. This is because, of course, the variational inequality (5.12) has a unique solution.
6. The Resolution of the Filtration Problem in Three Dimensions As in Section 4, we reverse our strategy to seek the solution of Problem 5.1 by examining the variational inequality (5.12). Properties of the free boundary will also be discussed. Once again, our first step is to affirm the smoothness of w(x). Theorem 6.1. There exists a unique solution w(x) to the variational inequality (5.12). Moreover, we// 2 < s (M) n C 1>A (M) for 1 < s < oo and 0 < A < 1. The existence and uniqueness of w is a consequence of Chapter II, Theorem 2.1. To assure the smoothness of w we must verify that the solution of the problem
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APPLICATIONS
satisfies we// 2 > s (M) whenever /e L A (M) and heH2's(M) in addition to verifying that out choice of boundary values <x(x) defined by (5.10) is the restriction of an // 2>S (M) function. The proofs of both of these statements are analogous to our discussion of (4.1): the difficulties occur at the edges of M and the vertices of B, but in both cases the functions may be extended to be solutions of equations in H 2 ' 00 domains. Here (5.1) plays an influential role. This is sufficient to apply the elliptic regularity theory. We state these facts as lemmas, the proofs of which are left as exercises. Lemma 6.2. Let <x0(x'), x' e B, be the solution of(5.8). Then a0 e H2'S(B) and hence a e // 2 ' 4 (M)/ora defined by (5.11')Lemma 6.3. Let u be the solution of(6.1) for /e L\M) and h e H 2>S (M), 1 < s < oo. Thenu€H2's(M). We define, for w the solution of (5.12),
thus u e C°- A(H) and Aw = 0 in Q. Lemma 6.4.
Let w be the solution o/(5.12). Then
and consequently,
Proof. It is convenient to observe that, since «0 > 0 in B and w is continuous, there is a "slab" B x (0, <5) cr Q for some 5 > 0. Thus B c: <9Q. In Q, w^3 is harmonic. For xedCl — M, w(x) = | grad w(x) | = 0, so W x3(x) = 0- Now suppose x e dO. n M. If x E Gj u G 2 u G^ + u T, then w^3(x) < 0 either by direct evaluation or, in the case that x e T, because w > 0 in M and w(x) = 0. If x e B c dQ, then Thus x cannot be a point where wX3 achieves its maximum. Finally, suppose that x e S+ u 5~, say x € S~, and w^x) ^ 0. Then, by continuity of wX3, there is a neighborhood of x in M where wX3 ^ 0 and per forza a neighborhood of x in M where w > 0. Say that e > 0 is small enough that BE(x) n M c O, so B£(x) n S~ c dO. Now Aw = 1 in B£(x) n M and w = 0 on B£(x) n S~, so w is smooth in J5£(x) n M. We may calculate now
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THE FILTRATION PROBLEM IN THREE DIMENSIONS, CONTINUED
251
that wXlX3 = 0 in BK(x) n S~, so, in particular, x cannot be a point where wX3 attains an extremum. Therefore
By the maximum principle, wX3(x) < 0 in il The lemma follows.
Q.E.D.
From the lemma, w is decreasing along each line parallel to the x 3 axis. Introduce
Continuity of w implies that
Define the point set F by
Note that O admits the description
Lemma 6.5.
Let w denote the solution of (5.12). Then
Proof. Our proof must be based on an idea different from that used in Lemma 4.5. Let v(x) = \(H — x 3 ) 2 . We claim that v — w > 0 in Q. Since this difference is harmonic in Q, we need only check its values on 8Q. Thus
and
recalling (5.9). Finally consider (S+ u S~) n dQ. We argue similarly to the last lemma. In view of the description of Q given in (6.5), for each xe(S + u S ~ ) n dQ, we may find a ball B» in Q with x e dB&, unless x e (r)Q n M), where (6.6) holds
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by continuity. From the Hopf maximum principle, at any such x where v — w attains an extremum, its normal derivative does not vanish. But since w is a solution of (5.12),
Hence v — w > 0 in Q. For x E M — Q, v(x) — w(x) = u(x) > 0. So
and
whence
We cannot yet show that {u, cp} is a solution to our problem, but we are able to achieve this intermediate step. Theorem 6.6. The pair {u, q>} defined in (6.2), (6.3) is a weak solution of Problem 5.1. Namely,
for all C e H l ( M ) which vanish near
anJ
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THE FILTRATION PROBLEM IN THREE DIMENSIONS, CONTINUED
253
Proof. In fact, for any £ which vanishes near G ! u G 2 u G 2 + we calculate that
since w| B = <x0 is harmonic and w = 0 on T. Recall that wxi = 0 on S + u S and £ vanishes near G t u G 2 u Gj + . The other properties of u are immediate. Note especially that
To extend the last theorem, we must show that {q>, u} is a classical solution to Problem 5.1, at least insofar as the smoothness of F is concerned. We begin our analysis with Theorem 6.7. Let w be the solution of (5.12) and define the free boundary rby
Then (p is a Lipschitz function in B.
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Knowing only the definition of F, it is not even certain that F n M is closed in M. The theorem tells us that d£l n M c: F. That F = dQ n M, or alternatively that cp(x') < H for x e B', will be a corollary of the smoothness of q>. To prove the theorem we first discuss several lemmas. Lemma 6.8. Extend w to B x (0, 2H) by setting
ThenweH&(B x (0, 2H)). Proo/ We know that w e H,20'C°°(M). Our object is to show that w^ e L°° near T, the top of M. The proof of this fact is quite easy because of Lemma 6.5. Indeed, we assert that w is the solution of the variational inequality
where D = B x (0, 2H) and
where a is defined by (5.10) as usual. For v E IK, the product
This follows from the boundary conditions on dM — T, but wv(v — w) = w X3 (y — w) = 0 on T by Lemma 6.5. Therefore
As we suggested at the beginning of the proof, w e /f ,20'C°°(D) by Chapter IV, Theorem 6.3. Q.E.D.
6
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255
Lemma 6.9. Let x0 e F and Br(x0) c B x (0, 2/f). Then there is a cone A, c IR3 = {x € IR3 : x 3 > 0} SMC/I £/iaf
and
whenever £ £ A, n S2. Proof.
Define £(x) € C1 •l(BJx0y> = H2' °°(Br(^o)) by
and for 0 ^ ^ e IR3 and B > 0 set
Since Ay = Aw — 8 A£ = 1 — e A( > 0 in Q n Br(x0) f°r ficiently small,
8= £ r
( ) suf-
Suppose x e F. Then if x e M, w(x) = | grad w(x) | = 0 whereas if x e T, then w(x) = jgrad w(x)| = 0 by Lemma 6.5. Thus v < 0, xeF. For any other x e d(Q n Br(QJ) we must have x e dBr(0) n Q. Consider two cases. First suppose that w(x) < e/2. Since wXj(x) is Lipschitz in Br(x0) and vanishes at x = x 0 , applying here the preceding lemma,
When c^3 > 0, £3wXi(x) < 0 by Lemma 6.4. Choosing £\ + & sufficiently small, namely,
we obtain
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VII APPLICATIONS
On the other hand, if w(x) > e/2 > 0, then wX3(x) < 0 so that
when £3 is sufficiently large. Therefore In particular, ((x) = 0 for p < r/2, so
Proof of Theorem 6.7. Given x0 £ F, choose r > 0 with Br(x0) c: 5 x (0, 2/7) as in the previous lemma and let x e F n 5r/4(xo). Then w is decreasing on any ray x + t£, Q < t < r/4, £ e Ar n S2. We may take Ar to be open in IR3, so, since w(x) = 0, the set Similarly, suppose that w(y) = 0 and Then which contradicts x e F, Hence This shows that (p is Lipschitz. Q.E.D. Corollary 6.10. Let w denote the solution of (5.12) and define u, (p by (6.2), (6.3) and F by (6.4). T/ien F is an analytic surface and {u, cp} is a classical solution of Problem 5.1. Proof. The criterion Chapter VI, 2.20 applies since F is Lipschitz. Hence F is of class C1 and w^. e C(Q u F). The conclusion follows from Chapter VI, Theorem 2.1. Q.E.EX Corollary 6.11.
With the previous notations,
Proof. Suppose
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257
boundary point lemma may be applied at (x'Q, H). Hence
But each term vanishes. This contradicts that (p(x'o) = H. Q.E.D. 7.
Flow past a Given Profile: The Problem in the Physical Plane
The determination of the flow of a fluid past a given profile consists in ascertaining the fluid velocity at each point of the region subject to various conditions at the profile and at infinity. We illustrate here how an especially simple case may be regarded as a variational inequality. This case is the motion of a steady, irrotational, and incompressible fluid about a symmetric convex profile in IR2. The unknown function in the variational inequality will be a transform of the original stream function, similar to the situation encountered in our study of the filtration problem, but its domain of definition will be the hodograph plane of the flow. The parameters which occur in the variational inequality will depend only on the given profile, the given fluid speed at infinity, and any estimate for the maximum of the fluid speed in the entire region. This last, in turn, may be estimated a priori (cf. Theorem 8.7). To give a mathematical formulation of the physical problem, let there be given a closed convex profile & in IR2 which is symmetric with respect to the x axis and contains the origin in its interior. Let G = U2 — £P, an open set, and let v be the outward normal to £P, which we assume defined everywhere on d& except perhaps at the two points A and B where j; = 0. Problem 7.1. (i)
Find the velocity q = (qlt q2) defined in G and satisfying
q is of class C1 in G and continuous in G,
where qx > 0 is given.
258
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Assume for the present that a solution q of Problem 7.1 exists. Then there exist locally in G two functions
Because of the symmetry condition (vi), so that an e U for n > 1. Choosing a counterclockwise orientation for d£P = dG, we have by Cauchy's theorem
Also,
The boundary condition (iv) tells us that ql dy — q2 dx = 0 on dG, so the integral is real. This reveals that which is the statement that symmetric flows have no circulation about &. Consequently, there is a single valued holomorphic function (z) in G such that and with the expansion
7
FLOW PAST A GIVEN PROFILE
259
The functions q> = Re O and (// = Im O are the velocity potential and the stream function, respectively. They are harmonic in G and in Cl(G) by virtue of the continuity of q. In addition,
Theorem 7.2. Let and assume that d0> + is of class C2'* up to and including its end points. Then there exists a unique solution to Problem 7.1. We are assuming, of course, that ^ is a convex profile. Proof. To check the uniqueness it suffices to check that (f/ is unique. Note that for any two stream functions \l/ and \j/, the difference «/r — \ji is harmonic in G, vanishes on d&, and tends to 0 at oo. By the maximum principle ij/ — # = 0. Existence is a consequence of the Riemann mapping theorem. Let/be a conformal mapping from G onto D = {£ : |£| > 1} such that/(oo) = oo and /'(oo) > 0, so/is a 1:1 holomorphic mapping of G,/'(z) ^ 0, |/(z)| -> oo a |z| -» oo, and/'(z) converges to a positive limit as |z| -> oo. Also/maps G continuously onto D. Since G is symmetric and / thus specified is unique, f ( z ) = f(z). Now define
and set V(z) = <&'(z). Then V is holomorphic in G and V(z) = V(z) so (ii) (iii), and (vi) hold. Since
V(z) -> qx as \z\ -> oo. Moreover hence $ = 0 on 5^.
260
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It remains to show that q = (i^ v , — i/O is continuous in G. Once this is known, (iv) follows from the vanishing of i// on dG. We leave this detail of the behavior of conformal mappings to the reader. Q.E.D. Let A and B denote the end points of d&+ , defined in the statement of the theorem. Note that because g2(x, 0) = 0 and
but the one-sided limits of v as z -> A are different from (0, ± 1) by hypothesis. We establish some properties of \{/ which will be useful later. Proposition 7.3. Let (j/ denote the stream function of Problem 7.1. Then \l/(z) > 0/or z e G+ = {z € G : Im z > 0}. Proof. Inspecting (7.5), we see that i//(z) and Im/(z) have the same sign. Now/maps G n {z: Im z = 0} onto D n {( : Im ( = 0} with/'(oo) > 0. It follows that/(G + ) = { C e D : I m C > 0}, namely, Im/(z) > 0 for z e G + . Q.E.D. Proposition 7.4. Let \l/ denote the stream function of Problem 7.1. Then i/fj, = Re V > 0 in G u dG except at z = A, B (where V = 0). Proof. The harmonic ^ achieves its minimum on dG + , where i/f = 0. Hence by the maximum principle, d\l//dv > 0 on dG+ except at z = A, B where it is not defined. Hence \l/y(x, 0) > 0 for (x, G)EdG + , (x, 0) ^ A, B, and since i// = 0 on d&>+, it follows that \l/y > 0 on d&+, (x, 0) ^ A, B. Now \l/y > 0 on dG+ and harmonic in G+ , so ij/y > 0 in G+ . Q.E.D.
8.
Flow past a Given Profile: Resolution by Variational Inequalities
We wish to reinterpret Problem 7.1 in terms of new variables and new unknown functions. The new independent variables will be obtained via a hodograph transform and, as one would suspect, the new unknown will be
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FLOW PAST A GIVEN PROFILE: RESOLUTION
261
Figure 5. Physical plane.
related to its Legendre transform. Consider the hodograph mapping z -* V(z). We state the theorem below, whose proof is reserved for the exercises. Theorem 8.1. (a) The function q^x, 0) decreases from qx to 0 as x increases from — oo to A and increases from Qtoq^asx increases from B to oo. (b) The set V(d0>+) is a simple closed curve. //£+ denotes the bounded open component ofR2 — V(d^ + ), then (0, g^] <= L + and the hodograph mapping z -> V(z) is 1: I from G+ onto S + — (0, g^]. Refer to Figs. 5 and 6. Slightly different hodograph variables will be more convenient. Let 9A e(0, ^n] (6BE [ — %n, 0)) be the angle determined by the x axis and the tangent to d& + at A (B). Assuming d& + to be strictly convex and smooth, for each Oe(6B, 9A), there is a unique point Ped£P+ where the tangent to d& + at P makes an angle 0 with the positive x axis. We denote the coordinates of P by (X(6), Y(9)\ C1-* functions of 0e[0B,0J under our hypothesis that d&+ is of class C2"1. In terms of 9, the radius of curvature R(ff) of 80>+ at (X(9\ Y(9)) is given by
Figure 6. Hodograph plane.
262
VII APPLICATIONS
For z e G + , z 7^ A, B, we define where the range of the argument is taken so that For a point z = X(0) + iT(0) e d.0» + , in view of (iv); thus
The curve F defined by F = W(d& + ) thereby admits the representation where 1(0) = -log) K(X(0) + iT(0))|. Observe that /(0) e C1-^, 0^) and
It follows from Theorem 8.1 that the mapping z -> W(z) is 1 : 1 from G + onto an open domain & defined by
with cr^ = -log^.
We define the stream function in 3D in terms of our new variables:
Since i// is harmonic and W is antiholomorphic, $ is harmonic in Si. In addition, \j/(0, a) = 0 on F, i^(0,CT)= 0 for a > a^, and $ -> 0 as 0 + /
Although we also eschew this approach, we do wish to emphasize that the solution of the free boundary problem provides the velocity distribution
8
FLOW PAST A GIVEN PROFILE: RESOLUTION
263
of the fluid along the profile d& + by (8.3) and (8.4). Namely, so
Introduce u(9, a), the solution of
and
Method II.
Introduce u(6, a\ a modified Legendre transform of\l/ by
u(9, a) = MX, y) - x\j/x(x, y) - yif/y(x, y) - e~°\X(Q) sin 9 - Y(0) cos 0].
Theorem 8.2. There is a unique solution u(9, a) to the Cauchy problem (8.5) and it is the Legendre transform (8.6). In addition,
where R(9) is the radius of curvature (8.1). Consequently, the two methods are equivalent. Assume the theorem for the moment and choose any m < min f l B < e < ^ 1(9) and let Q = (0B, 9A) x (m, oo). By (8.5), extending u as 0 for 9 + ia e O — 3> gives rise to a function in C J (Q — {ia: a > a^}) which fulfills the relation
for v > 0 in Q, vanishing near {ia : a > a^}, and having compact support.
264
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APPLICATIONS
We conclude that our new u is a likely candidate to be the solution of a variational inequality. We now turn to the Proof of Theorem 8.2. Proof.
First observe that if u(0, a) is any solution of the equation
then (d/da)(e"v) = eaf. So if v = u(0, a) and/ = $ > 0, it follows that e"u(9, a} is a strictly increasing function of a for each 0. Since eau(6, a) = 0 on F, «(0, a) > 0 in Q). On the other hand, if v is the difference of two solutions of (8.5) so/ = 0, then e"v(0, a) — £(0), a function of 6 alone, and may be evaluated on F, where ((0) = 0. Hence (8.5) admits a unique solution. We now show that this solution is obtained from (8.6). From the definition of 0, a given by (8.2), or
For the moment, set duQ = -xd\l/x - yd\l/y or
Now u = u0 - e~a(X(0)sin0 - 7(0) cos 0), so by (8.10), which displays duQ/da, Recalling that x = X(0) and y = 7(0) on F, it is evident that u = 0 on F. Hence ua — 0 on F since $ vanishes there and (8.5') holds. Finally
since tan 0 = Y'(0)/X'(0) was the defining property of 0. The function \j/ — xi]/x — yif/y is harmonic in G+ and z -» 0 + ia is anticonformal, so ^ — x\l/x — y\l/y = n0 is a harmonic function of 0, a in ^. Consequently,
keeping in mind that X'(&) = R(0) cos 0 and 7'(0) = R(0) sin 0. Q.E.D.
8
265
FLOW PAST A GIVEN PROFILE: RESOLUTION
To complete our project we shall provide a legitimate meaning for (8.9) by determining the conditions satisfied by u "at infinity" and on the ray [iff: a > a^} and by showing that u has the requisite integrability properties. Theorem 8.3. The function u(9, a) defined by (8.5) or (8.6) is continuous in 13> provided we define
where H = 7(0) is the height of the profile Proof.
Recalling the function
large.
As 9 + ia - » i f f ^ , the corresponding z e G+ tends to oo; thus
Now consider a point /<TI with a^ > a^. It has two preimages in dG+, (x l5 0) with X j > 0 and (x 2 , 0) with x 2 < 0. In either case, when 0 + i
recalling here that Corollary 8.4. Let u be defined by (8.5) or (8.6) and set H = 7(0), the height of the profile ^. Then and in particular, u e L2(Q). P/-00/ We have to show that (8.11) holds in 2. Set v(9, a) = He~"-cos 0, which is harmonic and positive in Q since — n/2 < 9B < 9A < n/2. Thus
266
VII
APPLICATIONS
hence
according to the maximum principle, provided u is bounded. Reserving this detail for the end of the proof, note that whereas
Consequently (8.11) holds once we verify that u is bounded. By continuity of u, that is, Theorem 8.3,
Now for some M > 0,
since the points 9 + ia with a > 2ax refer to points z with | z | bounded. From (8.5),
so
Lemma 8.5.
Let u be defined by (8.5) or (8.6). Then u E H10(Q.).
Proof. We only need to verify that grad u e L2(fi). To this end, let v e C°°(Q) n HQ(&) be any function satisfying
and
this last condition has been imposed for convenience.
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FLOW PAST A GIVEN PROFILE I RESOLUTION
267
Let f O(
and set £n(a) = £0(a - n). Also let ^0 e C°°([R2) satisfy 0 < r]Q < 1 and
and set ^(0, a) = f/0(«0, n(
Also we calculate that
Combining these two we have that
The right-hand side is finite by (8.11). But \& A(n d9 da remains bounded as n -> oo. Indeed, for n sufficiently large, the supports of grad £„ and grad r\n are disjoint, so
268
VII APPLICATIONS
which leads us to the estimate
We are now at the final stage of our investigation. Set IK = {v e //£(Q): v > 0 in Q and y(0, a) = He'a for a > O, (8.12) which is a closed convex subset of f/o(Q) by virtue of the continuity of the trace (cf. Chapter II, Appendix A). Theorem 8.6. The function u defined by (8.5) or (8.6) satisfies
for all veH. Proof. By Theorem 8.2 and Lemma 8.5 we know that u e IK. It is sufficient to show that (8.13) holds for i?e IK n L°°(Q); the general case follows by a truncation argument. For any v e IK n Lao(Q),
Now
since R(9) < 0, u = 0 on Q - ^, and u > 0 in Q.
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FLOW PAST A GIVEN PROFILE: RESOLUTION
269
Thus to prove (8.13), it suffices to demonstrate that
For n sufficiently large, grad („ = grad £„ + grad v\n. Considering the contribution from <^n first,
as n -> oo since | grad u \ e L2(Q). Now for rjn we have
as n -> oo, again because |grad u\ e L2(Q). This proves (8.14) and concludes the demonstration of the theorem. Q.E.D. We may now consider the variational inequality (8.14), in agreement with our usual policy, which may be formulated once the parameters Q, ax, H, and R(0) are known. Of these, H and R(0) are functions of the profile & and CTOO = — log q oo is given. The set Q = (0B,9A)x (m, oo) depends on ^throug 0A and 9B and on m < min 1(0). Recall that
because the speed of the fluid is bounded in G, and in particular, along the profile d&. A careful estimate for m may be useful in computing the solution.
270
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APPLICATIONS
The variational inequality (8.13) may be solved to yield the desired solution of Problem 7.1. To this end observe that the Dirichlet form
is coercive on //o(Q). In conclusion, we observe that an estimate for max|q| occurring in (8.15) may be found a priori. Theorem 8.7.
Let r = m\neB<0<6A\R(6)\ > 0. Then
where aQ > 0 is the unique solution of
The proof of this is left to the reader.
9. The Deflection of a Simply Supported Beam A variational inequality of higher order concerns the small deflections of a stiff beam constrained to lie above a given obstacle. We shall investigate this problem restricting ourselves to the case of a one dimensional beam. Given a, 0 < a < oo, set Q = ( —a, a) c: R 1 and let \j/ be a smooth function in Q satisfying
Let V = //2(Q) n tf£(Q) and set a closed convex subset of V. Define the bilinear form
It is easy to check that a(v, w) is coercive on V. The norm on V is given by ll y 'IL 2 <$i) + ll y "llL 2 (n)> so checking the coerciveness means showing that
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THE DEFLECTION OF A SIMPLY SUPPORTED BEAM
271
for some C > 0. Indeed, for v e V, v e C!(Q) and y(a) = v( — a) = 0. Suppose y e Q is a point where v'(y) = 0. Then
Integrating over Q and applying Schwarz' Inequality yields
Our project is to study Problem 9.1.
Given /e K', ./im*
Theorem 9.2.
There exists a unique solution to Problem 9.1.
This is a direct application of Chapter II, Theorem 2.1. Note that UE C 1 ' 1/2(Q) since Q c= IR1. We examine some properties of the solution. For C e C^(Q), C > 0, the function u + ( € IK, so the mapping is a nonnegative distribution. By the Schwarz theorem, there is a measure H > 0 such that
and moreover, From this characterization we infer some regularity properties of u. For example: Theorem 9.3. Let u denote the solution of Problem 9.1 withf= 0. Then (d3/dt3)u is a function of bounded variation and (d2/dt2)u is continuous in Q. To prove this observe that the distribution (d4/dt4)u = du by (9.5), where li > 0. Thus we may find an increasing function m(t) such that
272
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APPLICATIONS
The result follows. The astute reader will observe that u" is actually Lipschitz in a. We now direct our attention to a means of deriving smoothness properties of u by penalization. The particular penalization will involve u itself. Define ut(t) to be the solution of the second order problem
where u is the solution of Problem 9.1. Lemma 9.4. Let ue be the solution of (9.6). Then
Proof. Since u e H2(£l\ u£ e //4(O). In particular, ue e C3(O). At a point x E Q where u£(x) = min w£(0, «e(x) > 0, so It follows that u£ e IK. Q.E.D.
Lemma 9.5. Suppose thatfeH~l(&) and ut is defined by (9.6). Then for a constant C > 0
Proo/
Write/ = /0 - (d/A)/i,/o./i eL2(Q). Then
or, invoking Minty's lemma (Chapter III, Lemma 1.5),
Choosing y = u £ , we have v — u = eu", so
ntegrating by parts in the first term gives
COMMENTS AND BIBLIOGRAPHICAL NOTES
273
since ul(a) = ul(— a) = 0. Therefore Inasmuch as u" e //o(Q), we may appeal to Poincare's lemma, whence Therefore The lemma follows observing that UE G V. Q.E.D. The theorem below has essentially the same content as Theorem 9.1 in this one dimensional case. However, the method of proof (Lemmas 9.4 and 9.5) are available for use in higher dimensions too. Theorem 9.6. Let u be the solution of Problem 9.1 withfe H ~ l(Q). Then weH 3 (Q)nC 2 - 1 / 2 (Q). Proof. We need only show that ut -» u in some sense. Indeed, by the previous lemma, \\ul||L2(n) < C < oo for 0 < e < 1, so by (9.6) There exists a subsequence u£k which converges weakly in H3(Q) to some function which by (9.7) must be u. Hence u e H3(Q). Q.E.D.
COMMENTS AND BIBLIOGRAPHICAL NOTES The problem of hydrodynamic lubrication was discussed by O. Reynolds in 1886 and by A. Sommerfeld in 1904. Its formulation as a variationa inequality is due to Cimatti [1]. The asymptotic expansion of Theorem 2.5 also appears there and its extension to an asymptotic series is derived in Capriz and G. Cimatti [1]. The filtration problem has an extensive literature both before and after the advent of variational inequalities. An interesting idea about its solution was proposed by Renato Caccioppoli (see the article by C. Miranda [I]). The introduction of the transformation (3.7) is due to C. Baiocchi [1]. This ha stimulated much research in the subject. We refer to the papers of Baiocchi [2, 4]. For extensions and additional literature, there is the recent book of
274
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APPLICATIONS
Baiocchi and A. Capelo [1]. The existence of the seepage interval (h, (p(a)) with (p(d) > h was shown by A. Friedman and R. Jensen [1]. The higher dimensional problem of Sections 5 and 6 was formulated in Stampacchia [6]. The analysis of the free boundary in this case is due to Caffarelli [1]. The proof of Theorem 6.7 is adapted from Alt [2], There are other methods to approach the filtration problem. We refer to Alt [1] and Brezis, Kinderlehrer, and Stampacchia [1]. A general bibliography about fluid flow may be found in the book of Bers [1]. Here we have followed Brezis and Stampacchia [2]. See also Stampacchia [8]. Although the present method is limited to symmetric flow it applies to compressible fluids (Brezis and Stampacchia [3]) cavitation (Brezis and Duvaut [1]), and flow in a channel (Tomarelli [1]) (cf. the exercises.) We have just barely touched on the theory of higher order variational inequalities. For examples of work in this general field we cite P. Villaggio [1, 2], Frehse [2] Caffarelli and Friedman [1], and Brezis and Stampacchia [4] and Stampacchia [7]. Among the applications we have not had space to consider we note the problem of elastoplastic torsion. Here the convex set of competing functions has the form IK — {veHQ(Q):\vx\ < 1 a.e. in Q}. The reader may wish to consult Brezis [2], Brezis and Stampacchia [1], Caffarelli and Riviere [4], and Ting [1]. We have briefly touched on the problem of the confined plasma and its formulation as a variational problem (see Chapter VI, Section 4 and Exercise 15). In addition to the references mentioned there we cite Berestycki and Brezis [1]. There are questions in steady vortex flow which lead to similar considerations; see Fraenkel and Berger [1].
EXERCISES 1. Let Q c IRN be a domain and if/ e C1^) an obstacle, namely, maxn \l> > 0 and \l/ < 0 on 5Q. With IK = {v E //o(Q) • v > ^ in Q} consider the variational inequality
where /e L°°(Q) is given. Illustrate that this problem is not necessarily equivalent to the free boundary problem:
EXERCISES
275
Find a closed subset / c: Q such that
and
Consider a one dimensional case. When are the two problems the same? What implications does this have for the lubrication of a journal bearing treated in Section 2? 2. Prove Theorem 2.4. {Hint: First take v = 0 in (2.8) to establish that ||grad p£||L2(n) < CE. Then multiply (2.13) by e2 and set v = p £ e IK and add the result to (2.8) with v = ep0.] 3. For the solution of Problem 2.1 show that
(Hint: (d/dz)p obeys a maximum principle in Q — /.) 4. Let M_c UN be a domain and let £eSN~l a UN, and finally, let/be smooth in M. Consider the problem: Find a pair {Q,«}, Q c M and u G C(Q) such that
Introduce a new unknown and write this problem as "nearly a variational inequality," by which we mean that boundary conditions of elements of IK are left unspecified. Under what additional hypotheses about u on dM n dQ. can the problem be rewritten as a variational inequality? To begin an investigation, write the problem in weak form. 5. Let w be the solution of the variational inequality (3.14) and assume that
276
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APPLICATIONS
and the boundary conditions of Problem 3.1, where
and
Introduce the new unknown w(x, y) as the solution of the Cauchy problem
(Benci [1].) 7. Prove Lemma 6.2. 8. Prove Lemma 6.3. 9. (Example of Fluid Flow) Suppose the profile in Problem 7.1 is &> — {z: \z\ < 1} and qx = 1. Then the conformal mapping / of Theorem 7.2 reduces to /(z) = z. Calculate (z), (j/(z), V(z), and the solution u(0, a) of the variational inequality. In particular, study the singularity of \j/(G, a) at 9 + ia = iox — 0. 10. Suppose the profile in Problem 7.1 is the ellipse and q^ = 1. Find the solution to Problem 7.1 explicitly. 11. Study the steady motion of a compressible flow about a symmetric convex profile. In this case there is a given density function p(x, y) such that (ii') div(pq) = 0, and (iii') curl q = 0 in G together with (i), (iv), (v) and (vi) of Problem 7.1. One assumes that P - M l q D - There are a velocity potential (p and a stream function \j/ which satisfy (Brezis and Stampacchia [3] and Stampacchia [8]). 12. Suppose in Problem 7.1 that the fluid is in motion about the same convex profile situated in the middle of a channel. Find a variational inequality for this problem (Tomarelli [1]). 13. Study the possibility of cavitation in Problem 7.1. (Brezis and Duvaut [1]). 14. (Detail of the Proof of Theorem 7.2) Let L: z = x(s) + iy(s),
EXERCISES
277
0 < s < 1, be a closed arc of class C1U, some X > 0, in the second quadrant of the z = x + iy plane with z(0) = 0, and for e > 0 small set Suppose that ^ e C(V) satisfies
Then for some ^ > 0, i^eC 1>M ([/ n {z: |z| < <5}) for every 6 < e. [Hint: Suppose the tangent to L at z = 0 meets the positive x axis at an angle a, Ti/2 < a < TT, and consider the mapping z -» zn/a; (Brezis and Stampacchia [2]). 15. Prove Theorem 8.1. (Hint: Use the principle of the argument, i.e., the index of the mapping z -> F(z); (Brezis and Stampacchia [2]). 16. Prove Theorem 8.7. [Hint: Let
and prove that the solution u(0, a) of the variational inequality satisfies u < h. Hence u = 0 forCT< 0^ + cr0, so 1(6) > ax + <70, 6B < 0 < 0 M ] 17. Solve Problem 9.1 for V = {y 6 f/ 2 (Q): t? = y' = 0 on dQ}. 18. Consider Problem 9.1 for dimension N > 1. Show a solution u exists and that u e H3(Q). 19. Suppose in Problem 9.1, IK = {v e H2(Q) n H^(Q): 11/'(01 < ]9}. Show that a solution exists and, moreover, it is in //4>S(Q). 20. Describe the dual space V for Problem 9.1 and its variants above. 21. Suppose N > I and analyze formally the free boundary of Problem 9.1 in the spirit of Chapter VI. Although the equations are of higher order, they may be written as second order systems when a(u, y) = Jn AM Au dx. 22. Let Y:y — ip(x\ 0 < x < a, be the free boundary of Problem 3.1. Check that q>(x) is an analytic function of x. (Refer to Theorem 6.7).
C H A P T E R VIII
A One Phase Stefan Problem
1.
Introduction
A one phase Stefan problem is the description, typically, of the melting of a body of ice maintained at 0°C in contact with a region of water. In general, the Stefan problem refers to the change of phase in a thermoconductive medium when energy, that is, heat, is contributed to the system. Unknown are (i) the temperature distribution of the water as a function of space and time and (ii) the free boundary consisting of the ice-water interface. The temperature distribution is required to solve the heat equation in the aqueous region and energy is conserved across the interface. By limiting ourselves to the case of one space dimension we shall be able to explore many of the features of this problem while avoiding much of the technical complication. This will be especially true in our investigation of the free boundary by means of the Legendre transform. Let T > 0 and s0 > 0 be given. This is the classical Stefan problem. Problem 1.1. such that
278
To find a function &(x, t) and a curve T:t = s(x\ x > s0,
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INTRODUCTION
279
where h(x) > 0 is the initial temperature, g(t) > 0 is the heat contributed at time t, and k > 0 is the heat of fusion. The function 0(x, f) is interpreted as the temperature of the water at the point x at time t and the curve F represents the interface between the ice and the water. At time t the water occupies the subset defined by {x: s(x) < t}, where we understand that s(x) = 0 when x < s 0 . Note that 0(x, f) = min 0 = 0 for (x, t) e FI so 0X < 0 on F. Hence which means that the curve F is monotone. This is the property of the one phase problem which permits its transformation without difficulty to a variational inequality. At this point we transform Problem 1.1 to a variational inequality. Let R > s0 > 0. Later we shall exhibit this R as a function of the initial data /c, and so on. Set D = (0, R) x (0, T). Our variational inequality will be for u(x, t) defined by
We compute the differential equation satisfied by u assuming that © and s are smooth:
Also
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STEFAN PROBLEM
In the same manner,
So let us define
and
where we recall that k > 0 and g and h are smooth functions positive in [0, T] and [0, s0), respectively. With IK the set of nonnegative functions in L2(D\ the conditions on u may be expressed as the alternatives
This leads to the variational inequality, for T > 0 given: Problem 1.2.
To find u e L2 (0, T; H2(0, R)) n IK such that
Here we have used the notation L2(0, T; H2 (0, K)) to denote the functions M(X, r) satisfying
2
2.
EXISTENCE AND UNIQUENESS OF THE SOLUTION
281
Existence and Uniqueness of the Solution
A solution to Problem 1.2 will be found by approximation with a suitable penalty function. The particular choice of penalty is intended to simplify the existence and regularity proofs. For e > 0 we choose /J£(f) e C°°(IR) with the properties
Next choose a sequence /£(x) of smooth functions in [0, R] which are uniformly bounded and decrease to/(x) defined by (1.3) as e -» 0. Finally choose TJ(X) e q?((R) to satisfy
and 0 < rj(x) < 1, x e R. For T > 0 and £ > 0 given, consider the initial boundary value problem Problem 2.1.
To find u(x, t), (x, t) e D, such that
where k > 0 is the constant of Problem 1.1 and \l/ is defined by (1.4). It is well known that Problem 2.1 admits a classical solution u = ue; cf. Friedman [2]. Lemma 2.2. Let ut, e > 0, denote the solution to Problem 2.1 in D, Then there is an £0 > 0 such that
where K > 0 is independent ofe,Q<e<e0.
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STEFAN PROBLEM
Proof. We differentiate (2.2) with respect to t and apply the maximum principle. For w = dujdt, we obtain
Since /%u£) > 0 in D we infer from the maximum principle that
where d p D denotes the parabolic boundary of D, i.e., First we evaluate the left-hand member of this expression. Recalling that / £ >/and we deduce that
for e < e0, e0 sufficiently small, because h(x) is positive in [0, s0). On the other hand and
Hence On the vertical sides of the boundary, and
Consequently, min(mine n w, 0) = 0.
2
EXISTENCE AND UNIQUENESS OF THE SOLUTION
283
We now inspect the right-hand inequality of (2.5). We need only note that when t = 0, for a Kt independent of £ < £0 show that
The assertion of the lemma now follows from (2.5). Q.E.D. Lemma 2.3. Then
Let u£, 0 < e < eQ, denote the solution to Problem 2.1.
Proof. From the previous lemma, dujdt > 0 in D so ue(x, t) > 0 in D by (2.3). Recalling our choice of /?£ in (2.1),
At this point we address the existence and uniqueness of a solution to the variational inequality. Theorem 2.4. There exists a unique solution u to Problem 1.2. It enjoys the properties that and Let ue denote the solution to Problem 2.1. Then as e -> 0 and
ybr eac/i f, 0 < t < T 0 , and /jence UE-> u uniformly in D and u£X -*• MXuniformly in (0, /?)/or eac/i t e (0, T). Proof. In view of the preceding lemmas, the solution «£, 0 < e < e 0 , of Problem 2.1 satisfies
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STEFAN PROBLEM
Hence UEX(X, t) is a family of bounded functions, because the UE are uniformly bounded by Lemma 2.2. Therefore
independent of e, 0 < e < e 0 . This implies the existence of a sequence e' -» 0 and a function u e Hltp(D),1
Taking e -»• 0, owing to the weak convergence we get
By approximation, (2.7) holds for all v > 0, v e L2(0, R). It follows that u is a solution to Problem 1.2. We can easily check that the solution is unique in L2(0, T; #2(0, R)). This implies, in particular, that the entire sequence UE converges weakly to u in Hl-p(D)and in /f2-"(0, /?) for eacht, 0 < t < T. Thatwf > 0 follows from Lemma 2.2. Q.E.D. Corollary 2.5. Let u denote the solution to Problem (1.2) and define Then Q(r) c Q.(t')for t < t', t, t' e [0, T].
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EXISTENCE AND UNIQUENESS OF THE SOLUTION
285
Proof.This is obvious because u is Lipschitz withu We shall now show that given T > 0, R > 0 may be chosen so that u = 0 in a neighborhood ofx=R.This implies the existence of a free boundary. To begin we prove a comparison theorem analogous to our weak maximum principle of Chapter II. Lemma 2.6.Suppose that/
Proof.Forv e IK*,thesetofnonnegativeH1(D)functions obeying the boundary conditions of Problem 1.2, the integrated form of the problem is valid. Namely,
In this expression we may take v = max(w, M) e IK* since \j/ < $. Hence
Further observe that — uxx+u, — fa.e. inA sinceu>u >0 there. Consequently, for £ = max(w — ii,0), which is zero ond
Writing (2.8) in terms of (, we see that
Adding (2.9) and (2.10) and again using the definition of (, we obtain
Hence C = 0 a.e. in D, i.e., meas A = 0. Q.E.D.
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We may now construct an explicit solution of the Stefan problem which we employ to dominate our solution u. Denote byN where Mis a positive constant to be determined. Set
and
where
with C, C satisfying
It is easy to check that 0(x, f) and the curve f defined by 4>(x, t) — 0 is a solution to the classical Stefan Problem 1.1 with and
Hence 0 determines a solutionft to Problem 1.2 via (1.2). Note that the heat of fusion corresponding to 0 and f is the samek as that pertaining tou. Now
whereas
is bounded as M -» oo. Hence we may find M so large.that
2
EXISTENCE AND UNIQUENESS OF THE SOLUTION
287
Now we apply Lemma 2.6 to conclude Sinceu(x, t) =0 forx>M(t +1) In sum:
Proposition 2.7.Given T >0wemaychooseR>Q so that the solution uoProblem1.2vanishesinaneighborhood ofx= R. Proof.We choose M as above and R > M(T + 1) It will always be assumed that R has been chosen so large that the conclusion of Proposition 2.7 is satisfied. Set
Theorem 2.8.Let ubea solution of Problem 1.2. Then F defined by (2.13) admits the representation where a is a continuous increasingfunctionoftwiths0=o-(O) < ff(f)for t > 0. Proof. We first show that Q(?) is connected, which implies that a(t) is well defined. The interval (0,s 0)c Q(t)forallbecauseut>0. Suppose that 2 )is a component of the open set Q(t) which does not contain (0,s0). Then
so, by the maximum principle, This is a contradiction; hence, Q(f) is connected. We define
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Since Q(0 <= Q(r') for t < t', it is obvious that a is monotone. Suppose now that
for somet, 0
By the regularity theory for parabolic equations,uis smooth inQ so From the equation, u,(x, t) = — /c for X j 4- e < x <x2— £, whence w,(x, T) < 0 in a neighborhood in Q of C^q +x2),t). This contradicts the terms of our existence theorem, Theorem 2.4. In the same way we may show that u is continuous from below. Finally we point out thata(t) > S >0. By monotonicity of a, a(i) =s0for0 0, small, and observe that
xiscontinuousinDanduattainsitsminimumatx=s 0 ,
Introduce the solution v(x, t) to the Dirichlet problem
pQdenotesheparabolicoundaryofQ.Since
h > 0,
3
SMOOTHNESS PROPERTIES OF THE SOLUTION
289
Hence by the Hopf-Friedman maximum principle, On the other hand,u —vis a solution to the heat equation inQ with u — v >0ond pQHence,since u and v are not identical, because u — v attains its minimum there. Combining this with (2.14) and (2.15) we obtain the desired contradiction.
Q.E.D.
It is useful to point out that F is a Lipschitz curve in the coordinates obtained from (x, r) by rotation throughn/4.Hence a function Ce/f^Q) admitsaraceinL2(F).
3. Smoothness Properties of the Solution To discuss the free boundary F of our Stefan problem it is useful to know (x,0andu xx xf (x, r) are continuous in Q near F. One feature of the development is the proof that u,eHl(D n {t > t use of the penalizations. Another is the application of a weak maximum principleandBernstein'smethodtoshowthatutisLipschitzinQnearF. e, 0<e<£ 0 ,denote
the solution to Problem 2.1. Then
and
where C >0is independent0/e, 0 < e < £ Proof.From the initial conditions (2.4),
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STEFAN PROBLEM
so,sinceuet(x, t) > 0, in D, £(w£(x, r)) = 0 in 0 < x <s0/3, 0 < f < T, and u, is a solution to the equation
The conclusion now follows from the standard Schauder theory of the heat equation (Friedman [2]). Q.E.D. £ ,0
< e < £ 0 ,denotethesolutiontoProblem2.1.Then
where C > 0 is independent of&, 0 < £ <e Proof.For the duration of the proof, we set u —u Differentiate Eq. (2.3) with respect to t and multiply by w to obtain which, integrated over (0, R) for fixed t, 0 < t < T, yields
The first term may be integrated by parts. We observe that
because w(JR, r) =ut(R, t) = 0. By the previous lemma Therefore, from (3.1) and recalling that /?'(«) > 0,
This, integrated over (0, T), gives
3
SMOOTHNESS PROPERTIES OF THE SOLUTION
291
According to Lemma 2.2, | w(x, 01 ^ ^ independent of e, whence
Lemma3.3.Let w£ ,0 < e <e 0 denotethesolutiontoProblem2.1.Then for0
where C >0isaconstant independento/£, 0 < e < £ This lemma is somewhat more difficult. Clearly it leads to the integrability criterion for the solutionu we mentioned at the beginning of the paragraph. Incidentally, we use here the assumption thatP" Proof. As before, we set u = u Eq. (2.2) with respect tot and multiplying by w which, integrated over (0, R\ yields
We begin as usual with the first term, so,
The boundary term atx=Rdoes not appear because w = w, = 0 there. Whenx= 0, we may use Lemma 3.1 and our assumptions about\l/to conclude independent of e, 0 < e <e 0.Consequently, by (3.2),
292
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STEFAN PROBLEM
To estimate the last term, we write
At this point we employ Lemma 2.2 in an essential way to show that w > 0 andp"(u) <0 imply
Therefore
Using this estimate in (3.3) we obtain
Fori
Now this estimate holds for each T < t, so we may integrate it with respect to T over the interval (0,
3
SMOOTHNESS PROPERTIES OF THE SOLUTION
293
Here we have used the fact that ft' and w are nonnegative and Lemma 2.3. In view of Lemma 3.2, our lemma is proved. Q.E.D. Theorem 3.4.Let u be the solution to Problem 1.2. Then there is a constant C > 0 such that for each & > 0
Proof.This follows immediately from Lemma 3.3 by the weak convergenceofu atou,inLp.n particular, note that(D n {(x, f ) : t >
Proof.This is elementary. In view of Theorem 2.8, there is a neighborhood U of (x0, f 0 )e F such that U <= {(x, t): x >s0,t > 0}. Hence/(x) = - k inU and Therefore
Lemma 3.6et Fdenotethe free boundary associated to the solution u of Problem 1.2 and let Q = {(x, t): \x — X 1 0 , f 0 ) r and e > 0, (5 > 0. Suppose that w,0eH QnQ)san's/y
w/iere<9 boundary ofQr\Q. Then
294
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STEFAN PROBLEM
The proof of this maximum principle is analogous to our weak maximum principle of Chapter II. We relegate it to an exercise. Lemma3.7.Let u denote the solution to Problem 1.2 and F its free boundary. Then for each (x
wherec l5 c 2 andBarepositiveconstantsindependent0/(x0,£0)eF,t0>d >0, and
Proof.Let us first summarize our knowledge aboutu,. We know there is a neighborhood Q = {(x, f ) : | x — x Hl(Q)by Theorem 3.4. Since F is a Lipschitz curve,u on F n Q. To see that trace is zero simply note that
and thatu,—0 inQ\d.SincethetraceisacontinuousmappingromH1(Q) toL 2 (F nQ),tracer =0. Also note that
Let us set andchoosec 1>0 so large that
This is possible, clearly, because w, = 0 on F.Nowchoosec2>0 so large that
According to Lemma 3.5, w(x, 0 > 0 in Q n Q, for suitable Q, so
By (3.4), (3.5) we may apply our maximum principle Lemma 3.6, whence
The conclusion follows.
Q.E.D.
3
SMOOTHNESS PROPERTIES OF THE SOLUTION
295
Aparticularconsequenceofthelemmaisthatu tiscontinuous. Theorem 3.8.LetudenotethesolutiontoProblem1.2 andFitsfree boundary. Then for each (x 0 , t0)e F, there is a neighborhood U of(x0, t0)such that
Proof. It suffices to prove that u,eH1>00(17 n Q). CAhoose Q as in the previous lemmas so the conclusion of Lemma 3.7 holds, and set 0 =u We shall apply the method of S. Bernstein, but to an approximation of 0 in place of © because its smoothness at F is in question (Bernstein [1]). Let a(C) e Co'(R) be a mollifier and set
and
Observe that
and
We now transform the estimate of Lemma 3.7 to an estimate for& hxonFnQ. First note that when (x, t)e3QhnQ,here is a y with | x — y \ = h and (y, t)F. Now by Lemma 3.7
recallingherethatu x(x, t) is a Lipschitz function of x. So we have established that
296
VIII
STEFAN PROBLEM
We compute that
in view of (3.7). Now choose a function £e C^(0, 0 < ( < 1, with ( = 1 near (x 0 , t0) and set Applying the heat operator to w we compute
for a )U > 0 which depends only on (. Hence by the maximum principle,
keeping in mind (3.8) and (3.6). Restricting ourselves to a neighborhoodU of (x0, t0)where C = 1 and passing to the limit ash -» 0 we conclude that As a particular consequence of this fact,uxsaLipschitzfunctionof,so, employing the estimate of Lemma 3.7, we deduce that when (JCG,? 0 ) eTand(x, 0 e [/ n Q. To complete the proof of the theorem, introduce the mollification*¥hofutinthetimecoordinateonlyndderivea differential inequality analogous to (3.9) for the function The details are left to the reader.
Q.E.D.
We conclude this section with Theorem3.9.Let u denote the solution to Problem 1.2 and F its free
4
THE LEGENDRE TRANSFORM
297
boundary.Thenforeach(x0,t0)e F,there is a neighborhood U of(x0, t0)such that
Proof.
Choose U as in the previous theorem. Since
uxxisnotonlycontinuousbutLipschitzinUnQ.Differentiating with respect tor,
so withC independent oft. Moreover,uxasolutionoftheheatequationin Un Q, is continuous there. Finally, let (x, r) e F nU and(x, t')eUn Q. Given e > 0 we estimate
Hence
which proves the theorem. Q.E.D.
4.
The Legendre Transform
Applying the Legendre transform to the free boundary value problem for ugives rise to a new problem with a smooth "free" boundary but with a highly nonlinear equation, even in the simplest case. Throughout this section, let u denote the solution to Problem 1.2 and F its free boundary. Given (JCQ, t0)€ F, let U be a neighborhood of (x 0 , f 0 ) Since F is a Lipschitz curveitiseasytocheckthatuxx,uxteC(Q nU) imply uxeC J (H nU)andindeedthatwxdmitsaC1extensionintotheneighborhood U.
298
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STEFAN PROBLEM
Now we introduce the transformation
ThismappingisC1with
Sinceux(x0,IQ) = k >0,d(£,t)/d(x, f) is nonsingular at (;c0, f 0 )and maps a neighborhood, say U n Q, 1:1 onto a region
Recallherethatu x<0 in Q near T. Also, U n T is mapped onto a subset
inasmuchasux0onF. The Legendre transform of u is
It enjoys the property that
orv$=xandyt=ut.In particular,
on the basis of (4.2). Our free boundary problem is
In terms of v, this boundary value problem becomes
COMMENTS AND BIBLIOGRAPHICAL NOTES
299
The new problem (4.4) is parabolic because its linearized operator L evaluated at C(£, T) is
a parabolic operator with continuous coefficients. It follows from the theory of nonlinear parabolic equations (Ladyzhenskaya et al [1]) that so F nUx=t^(0, T), 11—TO|small,isaC00parametrizationofaportion of F. We have shown Theorem 4.1.Let u be the solution of Problem 1.2 and F its free boundary. ThenTisaCxurve. Note that in general F cannot be analytic.
COMMENTS AND BIBLIOGRAPHICAL NOTES The classical Stefan problem, for arbitrary dimension, is discussed in Friedman [3]. The treatment of the one phase problem as a variational inequality was proposed in Duvaut [1]. It was further developed and the properties of its solution explored in Friedman and Kinderlehrer [1]. For a general discussion of the free boundary we refer to Kinderlehrer and Nirenberg [2,3]. An interesting proof of the infinite differentiability of F (N = 1) is due to Schaeffer [1]. Also in this case, it was shown that F is analytic provided the supplied heat is analytic (Friedman [4]). Recently, Caffarelli and Friedman [2] have shown that the temperature in theN dimensional one phase problem is continuous. The Stefan problem is a paradigm for many variational and quasivariational inequalities of parabolic type. These concern, for example, stopping times, optimal control, and impulse control. We refer to Bensoussan and Lions [1] and Friedman [5] as examples.
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Index
A
Angle under which set is seen, 78 Asymptotic behavior, lubrication problem, 227, 275 Stefan problem, 286
B Baiocchi transformation, see Cauchy problem Banach-Saks theorem, 35 Beam, deflection of variational inequality, 271 penalization, 272 smoothness of solution, 271, 273 Bellman-Dirichlet problem, 103 Beltrami equation, 169-172 Bernstein's method, 295 Bilinear form coercive, 24 symmetric, 23 Bounded measurable coefficients, see Linear second order equations Bounded variation, 130 lower semicontinuity, 130 Brezis and Evans theorem, 102 Brouwer fixed point theorem, 8, 10, 18
C
Calderon-Zygmund inequality, 106 Capacity, 44 Cauchy data, see Cauchy problem; Linear second order equations, sharing Cauchy data Cauchy problem, 150, 228 of filtration, 231, 244 of flow past profile, 263 Cavitation flow past a profile, 276 journal bearing, 223, 226 Circulation, 258 Coercive bilinear form, 24 Coercive boundary conditions, 192 Coercive monotone operator, 84, 85, 113 Coercive system of ordinary differential equations, 198 Coercive vector field, 15 Coincidence set, 5, see also Free boundary filtration of liquid, constant cross section, 235-241 variable cross section, 251-256 lubrication of journal bearing, 226 obstacle problem, 43 capacity, 46
309
310
INDEX
Conincidence set (cont.) concave, smoothness, Dirichlet integral, 163, 182 minimal surface, 171 topology, 173 density property, 179 finite perimeter, 131 local smoothness criterion, 190 measure of boundary, 179 polyhedral, 137 two membranes problem, 212 Comparison of solutions of variational inequalities elliptic, 135 parabolic, 285 Complementarity form boundary constraint problem, 79 obstacle problem, 79 Complementarity problem, 17 Conductor potential, 45 variational inequality, 44 Conformal mapping, see also Coincidence set; Extension; Free boundary; Reflection boundary behavior, 157, 183, 277 flow past profile, 259 integrability of derivative, 156 Continuous on finite dimensional subspaces, 84 Contraction mapping, 7 Convex function, 16 proper, 77 Coppoletta's theorem, 103 D
Dam problem, see Filtration of liquid Deflection of simply supported beam, see Beam DerivativeofHl'sfunction,50, 53 Dirichlet problem for elliptic operator, coerciveness, 193, 219 linear second order operator classical, 30 for monotone coercive vector field, 114 smoothness of solution, 114, 115, 143 weak form, 31 existence of solution, 31, 38 smoothness of solution, 34 Dynamic programming problem,see Bellman-Dirichlet problem
E
Elastic-plastic torsion, 274 Elliptic equations, see Elliptic system; Linear second order equations Elliptic regularization, 88 Elliptic system, 191 coercive boundary conditions, 192 linear, 191 nonlinear, 192 open condition, 197 principal symbol matrix, 191 smoothness of solution, 193 weights, 191, 192 Exponential solution, see Elliptic system Extension analytic, 152 harmonic, 183 ofH1-'function, 55, 169 of Lipschitz function, 29 quasi-conformal, 158 for smoothness, 169
F
Face of contact principle, 136 Finite perimeter, 130 of coincidence set, 131, 132 Green's theorem, 131 Fixed point, 7 Filtration of liquid constant cross section, 227 classical formulation, 227 coincidence set, 237-241 flux, 230 free boundary, 240, 241 variational inequality, 234, 235 variable cross section, 242 classical formulation, 242 variational inequality, 248 convex set, 246, 248 free boundary, 251 analyticity, 256 Lipschitz continuity, 253 Flow past profile, 257 existence of solution, 259 variational inequality, 260 with cavitation, 276 in channel, 276 compressible, 274, 276 solution
311
INDEX continuity, 265 integrability, 266 Free boundary, 150, see also Coincidence set; Filtration of liquid; Flow past profile; Plasma problem; Stefan problem analytic parametrization(N -2), 151 (N>2),153, 188, 1 cusps, 166, 167 higher order, 277 smooth parametrization (N - 2), 155 (vV>2), 153, 188, 189 Stefan problem, 287 Free boundary problem, 150, 221 parabolic, 299 G
Gauss mapping, 168 conformality for minimal surface, 169
H
Harmonic extension, 183 Heat of fusion, 279 Higher order hodograph transformation, 209 Higher order variational inequality, 270 Hodograph transformation, 153, 185 first order, 153, 185, 205, 215 flow past profile, 262, 263 reflection determined by, 215 second order, 209 Stefan problem, 298 zeroth order, 186, 202 Holder continuous derivative, 28 Holder space, 28 Holder's inequality extended, 58j5^ Hyperplane of support, 12 I
Impulse control, 299 InequalityinHl,35, 41
J
Journal bearing, see Lubrication of journal bearing
K
Kellogg's theorem, 183 Knaster-Kuratowski-Mazurkiewicz theorem, 19 Korn and Lichtenstein theorem, 170 Kuhn-Tucker condition, 21 L
Legendre transformation, 153, 185 first order, 185, 205, 214 flow past profile, 263 Stefan problem, 298 zeroth order, 186, 203 Lewy's theorem on elliptic equations sharing Cauchy data, 194, 218 Linear second order equations with bounded measurable coefficients, 62,seealsoDirichlet problem global estimate of solution, 63, 64, 127 Holder continuity of solution, 66, 72-75 maximum principle, weak, 38 sharing Cauchy data on free surface, 204 on hyperplane, 194 Lipschitz function, 28 composedwithH1''function, 54, 81 obstacle, 134 Lubrication of journal bearing, 223 variational inequality, 224 asymptotic behavior, 227, 275 cavitation, 223, 226 M
Maximum principle, weak, 38 mixed problem, 245 parabolic, 293 Melting of ice,seeStefan problem Minimal surface, 167 Minty's lemma, 84, 109 Mixed problem, 33, 78 filtration of liquid, 246, 249 variational inequality, 111, 112, 139 Monotone graph, 121 Monotone operator, 83 coercive on convex set, 84 defined by coercive vector field or quasi-linear, 113
INDEX
312 Monotone operator(cont.) defined by locally coercive vector field, 117
strictly, 83 variational inequality for, 84 solution, existence, 84-86 necessary and sufficient condition for existence, 86, 87 Morrey's lemma, 76, 146 N
Neumann problem, weak formulation, solution, 33 compatibility condition, 80 Noncoercive operator, 87 Nonexpansive mapping, 7, 102 fixed points, 86 O
Obstacle problem, variational inequality, 40 for coercive vector field, 95, 114 smoothness of solution, 115 coincidence set of solution, see Coincidence set complementary form, 79 Dirichlet integral, 106 existence of solution, 40 for locally coercive vector field, 117 with Neumann boundary conditions, 91, 104 one dimension, 47 penalization, 107, 110, 120 second derivative limitation, 124, 129 smoothness of solution, 108, 141, 113, 115, 119, 129 Obstacle problem, variational inequality (by property of obstacle) Lipschitz obstacle, 134 mixed conditions, 139, 142 obstacle defined on a portion of O, 138 polyhedral obstacle, 134, 137 thin obstacle, 139 Optimal control, 299
P N
Pairing,U ,11 Banach space with dual, 83 Penalized problem, 107, 110, 120 mixed boundary conditions, 141
simply supported beam, 272 Stefan problem, 281 Piezometric head, 228, see also Filtration of liquid Plasma problem, 202, 218, 220, 221 Poincare inequality, 57, 59, 81 weak type, 60 Potential conductor, 45 order one, 60 Pressure liquid, 228 lubricant, 223 Projection onto convex set, 8 characterization, 9 variational inequality, 9
Q Quasi-conformal function, 157, 158 extension, 158 Quasi-variational inequality, 299 R
Radius of curvature, 261 Reflection, analytic (AT = 2), 151, 152, see also Extension by inverse hodograph, 215 Rellich compactness theorem, 62 Reynolds' equation, 224 Rod, see Beam, deflection of S
S, set of class, 65, 66 Singularity, see Free boundary Sobolev lemma, 56, 57 Sobolev, space, 28, 29, 49 matching lemma, 55 trace operator, 55 Spaces, function, xiv Stefan problem, one phase, 278 classical formulation, 278, 279 explicit solution, 286 free boundary, 278, 287 monotone interface, 279, 287 smoothness, 299 temperature, 279 smoothness, 289 variational inequality, 280, 283 penalization, 281 smoothness of solution, 296, 297
313
INDEX Stopping times, 299 Stream function, 258 free boundary problem, 262 Subdifferential, 20 Subsolution of second order operator, 39 aprioriestimate, 67 gradient estimate, 68 local estimate, 67 Summation convention, notation, xiv Supersolution, of second order operator, 39, see also Comparison of solutions aprioriestimate, 67 L-f.4l local, 67 minimum principle, 39 minimum of two, 42 Symmetric flow,seeFlow past profile T
Temperature, 279, see also Stefan problem Transmission problem, 195 Truncation in L\ 27 intf1'17 of solution, 64, 69 U
Unique continuation property, extended, 165 V
Variational inequality, 1 with boundary constraint, 79 for coercive vector field, 95, 113 of conductor potential, 44 of deflection of beam, 270, see also Beam, deflection of for Dirichlet integral, 106
elliptic regularization, 88 example, 1, 5 of filtration of liquid,eealsoFiltration of liquid constant cross section, 227 variable cross section, 242 for locally coercive vector field, 116 of lubrication of journal bearing, 223, see also Lubrication of journal bearing of mixed problem, 113, 139 for monotone operator, 84, see also Monotone operator of obstacle problem, 40, see also Obstacle problem penalization, 107, 120, see also Penalized problem of projection, 9 for quasi-lineaar operator, 95,see also Obstacle problem inRw,13 for semilinear operator, 94 with seminorm, 91 of Stefan problem, 278, see also Stefan problem of two membranes, 3, 80, 148, 212 Vector field, monotone, 15 see also Monotone operator; Obstacle problem coercive, 15 coerciveC\ 100 coercive locally, 97 coercive strongly, 94 mean curvature, 96 modification, 97 Velocity distribution, 257 a priori estimate, 270 along profile, 262 Velocity potential, 258 W
Weights of elliptic system, 191, 192
