On a problem of Bieberbach and Rademacher

Nonlinear Analysis. Theory, Methods &Applications. Printed in Great Britain.

Vol. 21, No. 5, pp. 327-335, 1993. 0

0362-546X/93 $6.00+ .oO 1993 Pergamon Press Ltd

ON A PROBLEM OF BIEBERBACH AND RADEMACHER A. C. LAzEat and P. J. MCKENNAS TDepartment of Mathematics and Computer Science, University of Miami, Coral Gables, FL 33124, U.S.A.; and SDepartment of Mathematics, University of Connecticut, Storrs, CT 06268, U.S.A. (Received 15 December 1991; received for publication 1 March 1993) Key words and phrases: Eigenfunction, condition, uniqueness.

subsolution, supersolution, star-shaped region, exterior sphere

1. INTRODUCTION

bounded domain in IR”(n 2 1) and let f(x, S) be defined for (x, S) E Q x IR and strictly increasing in s. Perhaps the simplest known uniqueness theorem in the theory of semilinear elliptic boundary value problems is the following: if u1 and u2 are in C’(Q) Il C(a), u1 and u2 are solutions of Au(x) = f(x, u(x)) for x E a, and ui(x) = u,(x) for all x E aQ, then u1 = z+. This follows from the observation that u1 - u2 can attain neither a positive maximum nor a negative minimum at a point in a since at such a point A@, - u?) would have the wrong sign. The uniqueness question is a whole new ball game for the following problem. Find u E C2(sZ) such that LET Q BE A

Au(x) = f(x, u(x))

XEQ

= 00 i ujan where the last condition means that u(x) + a0 uniformly as d(x) + 0, where d(x) denotes the distance from a point x E Q to aa. A problem of this type was first considered by Bieberbach [l] in 1916 wheref(x, U) = e” and n = 2. In this case the problem plays an important role in the theory of Riemannian surfaces of constant negative curvatures and in the theory of automorphic functions. In [2] Rademacher, using the ideas of Bieberbach, showed that if n is a bounded domain in IR3such that ?JsZis a C2 submanifold of IR3,then there exists a unique u E C2(n) such that Au = e” in Sz and [U(X)- ln(d(x)-2)1 is bounded on Sz. According to [2], this problem arises in the study of the electric potential in a glowing hollow metal body. In [2] the methods used depend heavily on the assumption that aa is of class C2 since it is required that d(x) be of class C2 near %2. In this note we shall consider a slightly more general problem but with much less smoothness assumed for i3Q. Moreover, in the uniqueness problem considered, it is not assumed beforehand that U(X)becomes infinite like ln(d(x)-2) as d(x) + 0, as in [l, 21. Let Q be an open subset of II?“.Recall that Sz satisfies a uniform external sphere condition (hereafter referred to as condition E) if there exists a number r > 0 such that for any z E an there exists a closed ball B of radius r with B fl fi = (z). We prove that if Q is a bounded domain in lR”which satisfies condition E and p is continuous and strictly positive on a, then there is at most one solution of the problem

Au(x) = p(x)

I

e’(‘),

ulasz= 00 321

XECI

(1.1)

328

A. C. LAZER and P. J. MCKENNA

and for any such solution [U(X)- ln(d(x)-‘)I is bounded on a. We show that if Sz is as above andp is locally Holder continuous with exponent (YE (0, l), strictly positive on Q, and bounded above, then there exists a solution of (1 .l). We also give a different uniqueness proof for bounded domains Q which are star-shaped with no smoothness assumption on XX The existence of solutions of the problem

(1*a with f monotone, but not uniqueness, was studied by Keller in [3]. For the case where f(@ = cU(“+2)/(“-2) (n = dim Q > 2, c > 0) Loewner and Nirenberg [4] proved that if X2 consists of the disjoint union of finitely many compact C” manifolds, each having codimension less than n/2 + 1, then there exists a unique positive solution of (2.2). Pohozaev [5] proved existence but not uniqueness for (2.2) when f(u) = u 2. More recently, uniqueness was established for the casef(u) = uk, with k 1 3, when aSJ is a C2-manifold and A is replaced by a more general second order elliptic operator, by Kondrat’ev and Nikishkin in [6]. We have recently learned that Diaz and Letelier [7] have proved existence and uniqueness when A is replaced by the p-Laplacian, f(u) = uk with k > 1, and aQ is of class C2. The authors have established uniqueness for the casef(u) = &, k > 1 with less regularity on the boundary but, since the method is fundamentally different than those used here, the proof will appear elsewhere. 2. PRELIMINARIES

In order to treat regions of the type considered in the Introduction, we first consider two special types of regions with smooth boundaries: disks and annuli. It takes no extra effort to establish a preliminary result for general domains with smooth boundaries. Existence could be treated by the method of [3], but, since we need to establish the existence of solutions of Au = p e” which not only “blow up” on %2 but tend to infinity like In d(~)-~, we give a different short proof based on the method of upper and lower solutions. LEMMA

2.1. Let Sz be a bounded, connected, open subset of rbv whose boundary aa is a smooth (C”) submanifold of dimension N - 1. Let &(x) be a positive eigenfunction corresponding to the smallest eigenvalue A1 of the problem Au + Au = 0 I u

in &2

lai-2 = 0.

If for x E 0, d(x) denotes the distance of x to aS& then there exists a constant k. such that Iln $i(x) - In d(x)/ I k.

(2.1)

for all x E Sz. Proof. Pucci [8] has shown that if an is of class C2, then there exist positive constants bi and b2 such that &d(x) 5 41(x) 5

~,m~

for all x E a, so (2.1) follows with k,, = max(Iln b,l, Iln b,l).

n

Bieberbach and Rademacher problem

329

LEMMA2.2. Let n be as in the statement of lemma 2.1 and let p E C*(a), for some cx E (0, l), - such that Au = p e” in Sz and lu(x) - ln(d(x)-2)1 withp(x) > 0 on a. There exists u E C2+u(a) is bounded on a. Proof. Let +i be as in lemma 2.1. Let p1 and p2 be positive constants p1 I p(x) I p2 for x E Q. If c is a constant and W(X)= -2 In&(x) + c, then

such that

on Sz. Since by the strong maximum principle [9, pp. 69-701 IV+,(x)( > 0 on %2, if c1 is a large negative constant, c, is a large positive constant and Wj(X)

=

-2ln+,(x)

+

Cj,

j=

1,2,

(2.2)

then Aw,(x) c p(x) e”@),

VXEsz,

Awl(x) > p(x) ewl@),

VXE52,

and wi(x) < w2(x) on Sz. Let (Q,JT be an increasing sequence of open sets with smooth boundaries such that fii, c Q and the union of these sets is Sz. For each nz 2 1 the restrictions of wi and w2 to ai, are a subsolution and a supersolution of the problem Au =peU

in Sz,

ulasz,= w,Ian,,

(2.3)

respectively. By the well-known method of subsolutions and supersolutions [lo], there exists a solution u, of (2.3) with wi(x) I u,(x) I w2(x) on Q2,. Since u,(x) 2 u,+i(x) for x E X2, and both are solutions of Au = p e” in Q,, it follows from the maximum principle that u,(x) I u,+~(x) for all x E a,. It follows that if A is any compact subset of Q, u, is defined on A for all sufficiently large m and wt(x) I u,+~(x) for x E A. By a more or less standard argument (see [l I]), u(x) = lim u,(x) is a smooth solution m-‘co of Au = p e” on 52 and wi(x) I u(x) 5 w,(x). Therefore, from (2.1) and (2.2), the assertion of the lemma follows. n COROLLARY 2.3. If Q is an annulus (x I rl < 1x1 c r2] and p(x) = p, a positive constant, then there exists a radially symmetric u that satisfies the assertion of lemma 2.2. Proof.

Since the eigenfunction +i is radially symmetric, the functions w1 and w2 in the proof are radially symmetric. If 0 < E < (r2 - r,)/2 and we let S&, = {xlr, + c/m < 1x1c r, - e/m) form = 1,2,..., then these domains are radially symmetric. Since the maximal solution of (2.3), which is between w1 and w, , is obtained as the limit of the decreasing sequence (Q): defined by ui = w2 I CJ,,,and -Avk+i + cuk+i = cuk - p evr, for a large constant c (see [IO]), uk+i I an, = w,,and these functions are radially symmetric, we can assume u,,, is radially symmetric. Hence, u is radially symmetric. preceding

A. C. LAZERand P. J. MCKENNA

330

3. UNIQUENESS FOR STAR-SHAPED DOMAINS Before considering the uniqueness problem for domains satisfying the uniform exterior sphere condition, we prove a uniqueness result for star-shaped domains. Although this limits the shape, we do not need any assumptions concerning smoothness of the boundary. Moreover, we only assume that u(x) -+ 00 as x -+ an uniformly, which is much weaker than the assumption that [u(x) - ln(d(x)-2)1 is bounded on Sz, which is made in [2]. THEOREM3.1. Let n be a bounded star-shaped domain in IRN. If p is continuous and strictly positive on fi and u1 and u2 are two solutions of the boundary blow-up problem

uniformly.

Au =pe”

in Sz

u(x) + 00

as d(x) -+ 0

(3.1)

Then ui = u2.

Proof. Without loss of generality we can assume that 0 E Sz and that n is star-shaped with respect to 0. Given r > 1 we let

Sz, = ((l/r)xlx

E M] C a

and P(X) d(r) = In rm2 min X E ii,p(rx) ( >*

If for x E Sz, we define w(x, r) = ul(rx) - d(r) then, for x E Q2,, we have Aw(x, r) = r2 Au,(rx)

= r2p(rx) eU1@) = r’p(rx) e(w(x,r)+d(r))I p(x) ewcxsr).

Moreover, w(x, r) + a0 as x + aM, uniformly. We claim that for r > 1 u2(x) 5 w(x, r)

vx E n,.

(3.2)

Assuming the contrary, there would exist x,, E a, such that u2(x0) > w(x,, r). Since u2(x) is bounded on fii, C Q and w(x, r) becomes infinite on aa,, there would exist an open set W with x,, E W and w C Or such that u2(x) - w(x, r) > 0 in W and u2(x) - w(x, r) = 0 on 8 W. But for x E W c Q, Au,(x) - Aw(x, r) 2 p(x)(e”z(“) - ew(X9r) ) > 0 contradicting the maximum principle. This contradiction establishes the claim (3.2). Let x0 E Sz be fixed. For r > 1 and r sufficiently close to 1 we have x0 E S&, and, according to (3.2), u2(x0) 5 ul(rxO) - d(r). Since d(r) -+ 0 as r -+ 1, letting r + 1, we obtain u2(x0) I ul(xO). Since x0 was arbitrary u2(x) 5 z+(x) for all x E Sz. Interchanging the roles of u1 and u2, we obtain the reverse inequality and the lemma is proved. w

331

Bieberbach and Rademacher problem 4. DOMAINS

SATISFYING

CONDITION

E

LEMMA

4.1. Let ki > 0. There exists a constant c = c(k,) such that if Q is any proper open subset of @, p is continuous on 0, p(x) r k, for all x E Q and Au = p e” on Q, then U(X) 5 c + ln(d(x)-2)

(4.1)

for all x E M. Proof.

Let B1 be the open ball of radius 1 centered at 0 in&? and u be the solution of Av = k, e”

in B,,

v + 00 as x + dB,

uniformly. If x,, is arbitrary and B,(x,) denotes the open ball of radius r > 0 centered at x0, then, as may be quickly verified, w(x) 3 v(r-l(x - x0)) + In rm2

(4.2)

is the solution of the problem Aw = ki e’” in B,(x,), w(x) -+ 00 as x -+ aBAxo) uniformly. Let x0 E Q, 0 < r < d(x,), and let B,(x,) and w be as above. We claim that U(X) 5 w(x)

v x E &(x0).

(4.3)

Clearly U(X) < w(x) near aB,(x,) so, if the claim were false, there would exist an open set W which is nonvacuous such that w C B,(x,), u(x) - w(x) > 0 for x E W and U(X) - w(x) = 0 for x E a W. Therefore, for x E W A(u - w)(x) = p(x) eucX)- k, e”‘@) > 0 which contradicts the maximum principle. This establishes the claim (4.3). In particular, we have that u(xo) 5 w(xo) = v(0) + In r-‘. Letting r + d(x), we obtain the assertion of the lemma.

(4.4)

n

Remark. The above proof shows that if p is continuous on RN and p(x) L kl for all x, then there cannot exist a solution of Au = p e” defined on all of RN. Indeed if u were such a solution and x0 E RN, then since (4.4) would hold for arbitrary large r, u(xo) would be less than any number,

which is absurd. In the rest of this paper we assume that n is a bounded domain in rdv which satisfies condition E. (See Introduction.) In what follows we let rl > 0 be such that for any z E 8.2, there exists a closed ball B of radius rl such that ZEBCP-0. If r, > rl + diam Sz, x1 is the center of B, and A(z) = lx I rl < lx - x11 < rd,

(4.5)

then n c A(z). In what follows if z E X2, then A(z) will denote an annulus with inner radius rl and outer radius r2 of the type just described. Of course, A(z) is not in general unique-for

example, if S2 is a rectangle in R2 and z is a corner of a.

332

A. C. LAZERand P.

J. MCKENNA

Let ri and r, be as above, let A, = (xl rl < 1x1 < rJ, and let k2 > 0. By corollary 2.2, there exists u0 such that AU, = kz eUo on AO, lu,(x) - ln(d(x)-2)1 is bounded on AO, and Mx) = vdxl) where ly:(rl, r2)+ IR is smooth. Since for x E A,, and 1x1 = rl + T, with 0 < r < (r2 - r,)/2, we have d(x) = T, and since ty(r, + 7) - ln(te2) is bounded below for (r2 - r&2 5 7 < r, - rl , we infer the existence of a constant c, such that c, + ln(re2) 5 y(r, + 7) LEMMA

for 0 <

7 <

r, - rl .

(4.6)

4.2. Let 0 be as above, and let p be continuous on Q and satisfy&x) I k2 on !2, where and w is as above, then

k2 > 0. If Au = p e” on Q, U(X) + 00 as d(x) + 0 uniformly,

u(x) 2 v(r, + d(x))

for all x E a.

(4.7)

Proof. Let x0 E Q and let z E Xl be such that d(x,) = Ix, - zl. If A(z) is as above and B = (x I Ix - xi] < rt], then since B c RN - S& z is the point on aB which is closest to x0. Therefore,

Ix, - x1 I = Ix, - ZI + Iz - x11 = d(x,) + rl . If s > 1 and ut(x) = J&SIX- x11) - ln(sm2) = u,(s(x - xi)) - ln(K2), quickly verified, Au, = k, eul

(4.8) then,

as may be

in Sz,

where Q, = (x I smlrl < Ix - x11 < s-‘r,). From the way in which A(z) is defined it is clear that if s is sufficiently close to 1 and greater than 1, then ficcn,. We assume that s is so close to 1 that this is the case. We claim that

w

2

u&4,

VXEQ.

(4.9)

Clearly U(X) > #i(x) for points x in Sz which are sufficiently close to 82, so, if the claim were false, there would exist a nonvacuous open set Vsuch that, v C Q, u,(x) > U(X)for x E Sz, and U,(X) = U(X) for x E aV. But then for x E V, A@, - u)(x) = k2 e’l@) - p(x) eucx) > 0 which contradicts the maximum principle. This proves the claim. In particular, (4.9) holds at x0 so for s > 1 and s - 1 sufficiently small we have that u(xo) 2 y/(slx, - x,1) - In sm2. Letting s tend to 1 and using (4.8) we obtain Mxo) 2 w(lxo - x1 I) = w(d(xo) + ri). Since x0 E Sz was arbitrary,

this proves the lemma.

We can now establish our main result.

n

Bieberbach and Rademacher problem

333

THEOREM4.1. Let Q be a bounded open subset of IRNwhich satisfies condition E. Let p(x) be continuous on Sz and satisfy 0 < k1 I p(x) I k2 for all x E n where ki and kz are constants. If U(X)is a solution of the boundary blow up problem Au =pe’

in Q (4.10)

i U(X) -+ 00

as d(x) + 0 uniformly,

then there exists a constant m such that 1u(x) - ln(d(x)-‘)I

5 m,

VXEQ.

(4.11)

Moreover, if u1 and u2 are two solutions of (4.10), then u1 = u2. Proof. Let u be a solution of (4.10) and let c be as in the statement of lemma 4.1 and v/ and rl be as in the statement and proof of lemma 4.2. From (4.1), (4.6), and (4.7), we have c, + ln(d(x)-2) 5 ly(r, + d(x)) 5 u(x) 5 c + ln(d(x)-2), and from these inequalities we obtain (4.11) with m = max(lcl, 1~~1). To prove the other part of the theorem we use the following calculus result. LEMMA4.3. If t I s < 1 and u L 2 In 2, then se” 2 es”.

Proof. Assuming that s and u have the indicated ranges, the inequality holds if and only if r.4+ Ins 2 su or -In s ull-s=

1 + (1 - s) + (1 - s)2 2 3 + -***

Since the right-hand side is decreasing on the interval * I s < 1, the assertion follows. Returning to the proof of theorem 4.1, suppose that under the hypotheses of the theorem that u1 and u2 are two solutions. Clearly u1 and u2 are bounded below on Q. Let b be a constant so large that Q(X) + b 2 2 In 2,

VXEQ

(4.12)

for k = 1,2 and let Q(X) = Q(X) + b for k = 1,2. If b1 = emb, then Auk = blpeuk

in Q

G(X) + O”

as d(x) + 0

(4.13) uniformly for k = 1,2. Let t I s < 1 and let w(x) = sul(x). We claim that w(x) 5 for all x E M. Since, according to (4.11)

u2w

(4.14)

334

A.C. LAZER and P.J.McKENNA

as d(x) --t 0 uniformly, w(x) < Q(X) for all points x E 51 which are sufficiently close to %2. If the claim were false there would exist an open set V with v c Q such that w(x) > v,(x) in I/ and w(x) = u2(x) on 8 V. For x E I/ it would follow from lemma 4.3 that A(w - Q)(X) = ~(x)(se~l@) - e”‘“) 2 p(x)(esUl(*) - e”@)) = p(x)(e”@) - e”z@))> 0 contradicting the maximum principle, This contradiction proves the claim (4.14). Letting s + 1 we have vi(x) 5 Q(X) for all x E Q and, hence, U,(X) I UJX) on Q. By symmetry uZ(x) I ur(x) on Q so u1 = u2. This proves the theorem. n Finally, with more smoothness assumed for p, we prove the existence of a solution of the blow up problem (4.10). We emphasize that unlike the conditions of theorem 4.1, we do not need p(x) to be bounded below by a positive constant but only sufficiently smooth and positive on a. In this case the question of uniqueness remains open. THEOREM 4.2.Let n satisfy the assumptions of theorem 4.1. Let p E C”(L2) for some CYE (0, 1). If p(x) > 0 for all x E Sz and there exists k, > 0 such that p(x) I k2 for all x E a, then there

exists a solution of the boundary blow up problem (4.10). Proof. Let (Q,JT be a sequence of open subsets of s2 such that a,,, c SJ,,, c i2 and the boundary %2, of Q, is a smooth (N - 1)-dimensional submanifold of IT? for m L 1, and Q is the union of the members of this sequence. To obtain the existence of such a sequence, let (Urn): be a sequence of open sets such that u, C U,,,,,c S2 for all m and Q is the union of the members of this sequence. For each m let em be a C” function such that 0 I 0,(x) I 1 for all x,&(x)=lforx~Si,,andB,(x)=Oforx~lR”-Q,+r. By Sard’s theorem, there exists a number c, E (0, 1) which is a regular value of 0,. If Q, = 8,;l’((c,, l]), then the sequence (QJ will fulfil the conditions given above. (See, for example, [12].) Let m I 1. According to lemma 2.2 and our assumptions on p(x), for each component W of Q, there exists w defined on W such that Aw(x) = p(x) e”‘@) for x E W and w(x) + co as x + a W uniformly. It follows that we can define U, on Q,, such that Au,(x) = p(x) cum@)for x E Q,, and u,(x) + COas x + an,,, uniformly. We claim that u,(x) 2 u,+~(x) for all x E Sz,. Since u,+r is bounded on fii, and u,(x) tends to 00 as x + asz,uniformly, u,(x) > u,+r (x) for all points x E Sz, sufficiently close to asz, . If the claim were false, there would exist a subdomain 8 of a, such that u,+i = u, on a0 and u,+i > U, in 8. But A(u,+r - U,)(X) = p(x)(e”m+lcx) - eumcX))> 0 in 13 contradicting the maximum principle. This proves the claim. To avoid confusion concerning domains, let do(x) denote the distance from a point x E Q to 82. Let m2 1 andx,EQ2,. Let x1 E as2 be such that &(x0) = (x,, - x11. Referring to the proof of lemma 4.2, let I,Yand c0 have the same meanings as there. Replacing u and Sz in the proof of lemma 4.2 by U, and Sz,, respectively, we see that u, (x0) 2 w( Ix0 - x1 I) = W-, + &(x0)). Since x0 E 0, was arbitrary we see that u,(x) 2 v(r, + d,(x)) for all x E 0,. u,(x) 2 c, + ln(&(x)-2),

VXEQ2,.

Hence, by (4.6) (4.15)

Bieberbach and Rademacher problem

335

If A is a compact subset of Sz then there exists m, such that A C QnO and the sequence (%I (x)1;, is decreasing for x E A. By (4.15), the sequence is bounded below on A so U(X) = lim u,(x) exists for all x E 0. By standard arguments u E C2+Ol(n) and Au = p e” on a. (See, m+m for example [l l] .) Since u(x) 2 c0 + In&(x)-‘), we see

VXECJ

that u is a solution of (4.10) and the proof is complete.

Acknowledgement-The DMS-9102632.

authors

gratefully

acknowledge

the support

of NSF by grants

DMS-910223 and

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BIEBERBACHL., Au = e” und die automorphen Funktionen, Math. An&n 77, RADEMACHER H., Einige besondere probleme partieller Differentialgleichungen,

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