n ' * Supported by the Netherlands Organization for Scientific Research N.W.O. G SWEERS* A sign-changing global minimizer on a convex domain

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1 G SWEERS* A sign-changing global minimizer on a convex domain Introduction: Recently one has established the existence of stable sign-changing solutions for the elliptic problem (1) -.!lu = f(u) in [ u = 0 n ' on an. In [5) there is an example of a sign changing stable solution on a convex domain with f(o) f. 0. Matano [2) shows the existence of a sign-changing stable solution even with f(o) = 0. A next question will be: does a global minimizer have a fixed sign? It has been guessed that the answer is positive if the domain is convex. In this note we will recall a proof for the ball and give a counterexample for a triangle. We will assume that f E CO,l, 0 is bounded with an E CO,l and that (1) has a solution u that minimizes the energy functional J. This functional J : H~(n) -+ IR is defined by (2) J(v) = ~J jvvj 2 dx-f F(v)dx, n n s where F(s) = f f(t) dt. It is classical that if J(u) = min { J(v); v E H~(O) }, then u 0 is a C 2 (n) n C(rl) - solution, (3) J'(u)v := J vu.vv dx - J f(u) v dx = 0 for all v E H~(n) and n n (4) J (u)(v).- J lvvj 2 dx-j f'(u)v 2 dx > o for all v E H~(n). n n * Supported by the Netherlands Organization for Scientific Research N.W.O. 251

2 Proposition 1: If f is antisymmetric or 0 is a ball in IRn, then the global minimizer u has a fixed sign. Remark 1: The result that a (local) minimizer cannot change sign on a ball is due to Lin and Ni, [1]. In their unpublished preprint they also prove the result for 0 being the difference of two balls with the same center. We will sketch their proof. Proof: i) Iffisantisymmetric,f(s)=-f(-s),then J(lul)=J(u).Hence lul isa minimizing solution and I u I E c 2 (n). It follows from x E n and u(x) = 0, that vu(x) = 0. Then the strong maximum principle shows that u = 0. Hence u has a fixed sign. ii) Suppose n is a ball with center 0. Then differentiate the solution u in a tangential direction, that is, apply A = Xi akj - Xj aki Since A and b. commute, the function rp =Au satisfies -b.rp = f'(u)rp in n. Moreover rp = 0 either rp = 0 or rp is an eigenfunction (with eigenvalue 0) of (5) -b.v - f'(u)v =.Xv [ v=o in n I on an. on an. Then From ( 4) one finds that all eigenvalues, except maybe the first, are strictly positive. Hence rp is a multiple of the first eigenfunction. If rp is nonzero this shows rp has a 27!" fixed sign, which contradicts f rp do = 0. 0=0 Since this holds for all i and j, u is radially symmetric. Now suppose changes sign; then there is a positive number r 0 ;uch that u (ro) = 0. Set r Then v E H~(O) and for r < ro, for r ~ ro. 0 < J (u)(v) = J ( lvur1 2 -f'(u) u~) dx = lxl<ro J u/-t..ur - f'(u)ur) dx = lxl <ro which gives a contradiction for nonconstant u. J (n-1) r urdx, lxl <ro 252

3 Proposition~: There is f E c 0 1 (1R), with f(o) = 0, and n c IR 2, bounded and convex, such that the global minimizer changes sign. In this note we will construct just one example. A forthcoming paper of Matano will certainly have a more rigorous approach to sign-changing stable solutions. However, it is not clear if this considers global minimizers. 1 Remark 9: Without the condition f(o) = 0 one can modify the example in [5] to obtain the result of proposition 2. Proof: Set n = {(x1,x2) E IR 2 ; 21 x2 I < X1 < 1} and define the Lipschitz-continuous functions f c: for c: > 0 by fc:(s) = O fc:(s) = -c:- 2 (s+2c:) fc:(s) = c:- 2 s on (-m,-2c:], on (-2c:,-c:], on (-c:,o], fc:(s) = s on (0,2], {/s)=4-s on (2,4], fc:(s) = 0 fl: f: c: on (4,m] Note that f( s) = -2c: f(-tc:s) for s >

4 Let Ao denote the first eigenvalue of (6) [-ti.rp = A'{J in n, 'P = 0 on an, then the bifurcation picture for solutions with fixed sign of (7) [-ti.u = Afiu) in f2, u = 0 on an, looks as follows. extr u I :.:::-;:.:--::.=-~--=-=-:.: =-=--=-=--=-~ E2A..o lo - -& A. ~& Since s- 1 fe(s) is decreasing on [0,4] (and strictly on [2,4]) it is known that there is a unique positive solution for every A > Ao. See [4]. There is no positive solution for A < Ao. Similar arguments hold for negative solutions. Let U A and V1 denote the positive, respectively the negative solution of (7) for A> Ao. Let JE(A,u) denote the energy functional for (7), that is J.. (A,u) = J ( vul 2 -A ff (s) ds) dx. ~ 0 0 Lemma 3: lim A-l JE(A,UA) = and lim A-l J/\,V~) =-IOI, ~m ~m uniformly fore E [0,1], where In I is the Lebesgue measure of n. Proof: We will show the second statement. Since V1 is the only stable solution of the function minimizes -ti.u =A min(fc(u),o) in n, [ u = 0 on an, - f (l 2 u ) J..(A,u)= - 2 1vul -Afmin(f(s),O)ds dx ~ n o E for A > >.o. 254

5 Since we can estimate J~(>.,u) from below by ->.In I : r(>.,u) ~ ->.JI min(f (s),o) ds dx ~ ->. J 1 dx; E; 00 E; 0 it is sufficient to show that for all a> 0 there is cp E 6 H~(O) for c E [0,1] such that uniformly Take cp E C~(O) with cp = -2 in a closed subset of n with measure larger than In I -!a. The result follows for>. large since D Because of Lemma 3 there is >. 1 > >. 0 such that Proof: Fort large enough (8) u >.1(xi,x2) < 3 >.1((x1+t) - 4x2) inn Since -l1>. 1((x1+t) - 4x2) = 6>. 1 > >.1 max f 6 and smce >.1((x1+t) - 4x2) > 0 in ri for t > O, this function is a supersolution for t ~ 0. By the Sweeping Principle [3, Theorem 9] one finds (8) for all t ~ 0. D Finally we will show, for > 0 but small enough, that U >. does not minimize 1 J (>.i, ). We will modify U>. near (0,0) to obtain a 6 H~(O)-function with lower 1 255

6 energy. Hence the solution of (7) for >. = >. 1 that minimizes J {>.i, ) is not U, or c Al vt, which are the only stable solutions with fixed sign. Set Then IO~I = n~ = {{xi,x2) E n ; X1 < 8} ' 08 = {(xi,x2) E 0 ; 8 < X1 < 28} and 0~ = {(xi,x2) E 0 ; X1 < 28 }. Moreover define z E CO,l(IR) by z(s) = 0 z(s) = s-1 [ z(s) = 1 and set for s ~ 1, for 1 < s ~ 2, for 2 < s, Then u 8 E H~(n) and vu 8 (xi,x2) = 8-1 u >. 1 (xi,x2)(l,o) + z(8-1 x1) vu >. 1 (xi,x2) in 08. By using lemma 4 we can estimate the difference in energy as follows: (9) Jc{>.i,u 8 )-Jc{>.i,U>.) ~ ~J (lvu lvu> )dx+>.1j ~utdx ~ n~ n~ The function v 8 defined by 256

7 ' satisfies: -t.v 6 (x) = ~ 6-l (t.u >}(6-1 x) = = -~6-l )qf 6 (u>. 1 (6-1 x)) = 1-1 ( 1 ( -1 )) = 2 6 A 1 26 f U). I 6 X = = >.1 f 6 (v 6 (x)). Hence v 6 is a solution of (7) with e = 6 and n replaced by ng. After extending v 6 by 0 outside of ng we obtain: (10) = i 52 J5(>.i,U >.1). Finally, we set w 6 = u + v and we find, since supp u E 11\ng and supp v6 en~ that by (9) and (10) for 6 sufficiently small: J5(>.i,w5) = J5(>.i,u5) + J5(>.1iv5) < < (1+~2) J5(>.i,U >.1) + C(>.1) 53 < J5(>.i,U >.1). 0 The example uses a triangle for a domain and a piecewise linear right hand side. One can modify both n and f to have the same result on a smooth, strictly convex domain with a Cro -function f. Acknowledgement: I thank Wei Ming Ni for showing me the main part of Proposition

8 References: [1] Chang-Shou Lin and Wei-Ming Ni, On stable steady states of semilinear diffusion equations, preprint [2] H. Matano, in preparation. [3] A. McNabb, Strong comparison theorems for elliptic equations of second order, Journal of Math. Mech. 10, (1961), [4] D.H. Sattinger, Topics in stability and bifurcation theory. Lecture Notes in Math. 309, Springer Verlag, Berlin, [5] G. Sweers, Semilinear elliptic problems on domains with corners, Commun. in Partial Differential Equations 14 (1989), Department of Pure Mathematics, Delft University of Technology, POBox 5031, 2600 GA Delft, The Netherlands 258