Existence of many positive nonradial solutions for a superlinear Dirichlet problem on thin annuli

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1 Nonlinear Differential Equations, Electron. J. Diff. Eqns., Conf. 5,, pp or ftp ejde.math.swt.edu or ejde.math.unt.edu (login: ftp) Existence of many positive nonradial solutions for a superlinear Dirichlet problem on thin annuli Alfonso Castro & Marcel B. Finan Abstract We study the existence of many positive nonradial solutions of a superlinear Dirichlet problem in an annulus in R N. Our strategy consists of finding the minimizer of the energy functional restricted to the Nehrai manifold of a subspace of functions with symmetries. The minimizer is a global critical point and therefore is a desired solution. Then we show that the minimal energy solutions in different symmetric classes have mutually different energies. The same approach has been used to prove the existence of many sign-changing nonradial solutions (see [5]). 1 Introduction In this article we study the existence of many positive nonradial solutions of the equation u + f(u) = in u> in (1) u = on, where = {x R 3 :1 < x <1}. The non-linearity f is of class C 1 (R) and satisfies the following conditions: (C1) f() = and f () <λ 1,whereλ 1 is the smallest eigenvalue of with zero Dirichlet boundary condition in. (C) f (u) > f(u) u for all u. (C3) (Superlinearity) lim u f(u) u =. Mathematics Subject Classifications: 35J, 35J5, 35J6 Key words: Superlinear Dirichlet problem, positive nonradial solutions, variational methods c Southwest Texas State University. Published October 4,. 1

2 Existence of many positive nonradial solutions (C4) (Subcritical growth) There exist constants C > and p (1, 5) such that f (u) C( u p 1 +1), u R. (C5) There exist constants m (, 1) and ρ such that for u >ρ, uf(u) F (u) >, m where F (u) = u f(s)ds. This paper is motivated by the work of Coffman [6], Li [7] and Lin [8]. In 1984, Coffman showed that for f(t) = t+t p,wherep=n+1,n =,theabove problem has many positive nonradial solutions. His result was extended by Li [7] to the N-dimensional case with N 4orN= and for p (1, N+ N ), when the nonlinear term is λt + t p for λ. Our main result (see Theorem 1.1 below) concerns the case N = 3. We will show that for N = 3 our problem has many distinct nonradial solutions. We follow the strategy used in [6, 7] and [8]. That is, we look for the minimizer of the energy functional restricted to the Nehari manifold of a subspace of functions with symmetries. The minimizer is a global critical point and therefore is a solution to (1). Then we show that the minimal energy solutions in different symmetric classes has mutually different energies. We would like to point out here that during the preparation of this article we were unaware of the papers by Byeon [1] and Catrina and Wang [], where the above problem has been solved. However, our approach is different from theirs. Our main result is the following Theorem 1.1 Let conditions (C1) - (C5) be satisfied. Then, for each positive integer k there exists 1 (k) > such that if << 1 (k)then (1) has k distinct positive nonradial solutions. We note that our argument works for N and not only for the three dimensional case. The approach used in proving Theorem 1.1 is similar to the one used in [7], i.e., we consider the functionals J(u) = {1 u F (u) } dx and ( γ(u) = u uf(u) ) dx on H 1(), and for k 1, we consider the space of invariant functions H(, k) := Fix(G k {id R, id R }) = { v H 1 ():v(g(x 1,x ),Tx 3 )=v(x 1,x,x 3 ),

3 Alfonso Castro & Marcel B. Finan 3 (g, T) G k {id R, id R } } = { v H 1 ( ):v(x 1,x,x 3 )=u(x 1,x, x 3 )forsomeu } satisfying u(g(x 1,x ), x 3 )=u(x 1,x, x 3 ) g G k where G k = {g i : i k 1}and gz = e πi/k z, z C R. Also, we consider the Nehari manifold S(, k) ={v H(, k)\{} : γ(v) =}. Similarly, we define the space of functions H(, ) := O() {id R, id R } = { v H 1 ( ):v(g(x 1,x ),Tx 3 )=v(x 1,x,x 3 ), (g, T) O() {id R, id R } } = { v H 1 ():v(x 1,x,x 3 )=u( x 1 +x, x 3 ), for some u }, and the manifold S(, ) ={v H(, )\{} : γ(v) =}, where O() denotes the space of orthogonal matrices. Note that if u H(, ) then uis θ independent. Associated with the above sets, we consider the numbers j k = inf J(v) and v S(,k) j = inf J(v). v S(, ) We prove Theorem 1.1 by establishing the following lemmas: Lemma 1. For k =1,,...,, j k is achieved by some u,k S(, k) and u,k is a critical point of J on H(, k). Lemma 1.3 Let u,k be as in Lemma 1.. Then u,k is a critical point of J on H 1 (). Lemma 1.4 Given a positive integer k, thereexists 1 (k)such that for << 1 (k)we have jk <j. Lemma 1.5 For n =,3,4,..., k =1,,3,..., j kn <j implies j k <j kn. This article is organized as follows: In Section, we prove Lemmas In Section 3, we prove Theorem 1.1.

4 4 Existence of many positive nonradial solutions Proof of Lemmas ProofofLemma1.: The proof follows from [3] combined with the facts that the embeddings H(, k) L p ( )andh(, ) L ( ) are compact (See [9]) Lemma 1.3 is a result of the symmetric criticality principle: if u,k is a critical point of J on H(, k), then u,k is a critical point of J on H 1 ( )(See[9]). To prove Lemma 1.4 we need the following results. Lemma.1 (Poincaré s inequality) Let Ω R N be a smooth bounded domain with diameter d. Then for u H 1 (Ω) we have u dx d u dx. Ω N Ω Lemma. ([3]) is a local minimum of J. Ifu H(, k) {}, then there exists a unique α = α(u) (, ) such that αu S(, k). Moreover, J(αu) = max λ> J(λu) >. Lemma.3 ([4]) For v >ρand s>we have for some constant C>. svf(sv) Cs /m vf(v) Proof of Lemma 1.4: Let k 1 be an integer and >tobechosenbelow. According to Lemma 1., there exist u,k S(, k)andu, S(, ) such that jk = J(u,k) andj =J(u, ). By Lemma 1.3, u,k and u, are solutions to Problem 1. Let Ω k be the set of points x =(x 1,x,x 3 ) (r, θ, x 3 ) such that θ [, π/k]. Let j be a positive integer to be chosen independent of k and. Define ω(r, θ, x 3 )= { u, (r, x 3 )sin(jkθ) for θ [,π/(jk)] for θ [π/(jk),π/k], where r =(x 1+x ) 1/. Extend ω periodically to all of in the θ direction. Let z H(, k) be the resulting extension. Since u, > thenz>. By the chain rule we have ω(r, θ, x 3 ) = (u, ) r sin (jkθ)+ 1 r (u, ) (jk) cos (jkθ) + x3 u, sin (jkθ) if θ [,π/(jk)], otherwise ω =.Thus, ω(r, θ, x 3 ) (u, ) r + 1 r (jk) (u, ) + x3 u,.

5 Alfonso Castro & Marcel B. Finan 5 By Lemma.1 we have Thus, Ω jk u, dx 1 x dx 3 (diam(ωjk )) 3 z dx 1 x dx 3 = k ω dx 1 x dx 3 k k = k Ω jk Ω jk Ω jk Ω jk Ω jk u, dx 1 x dx 3. [(1 + 16(jk) )(u, ) r 7 + x 3 u, ] dx 1 x dx 3 ((u, ) r + x 3 u, ) dx 1 x dx 3 () u, dx 1 x dx 3 provided that <3 3/(4jk). γ(αz) =. Let D = {(r, θ, x 3 ):u, (r, x 3 ) >ρ,θ [ π 4jk, π jk By Lemma. we can find α>such that ]}. Suppose that α>. Then for (r, θ, x 3 ) D we have α sin (jkθ) >. This, the fact that tf(t) is bounded from below, say by E, and Lemma.3 imply αzf(αz) dx 1 x dx 3 = k α sin (jkθ)u, f(α sin (jkθ)u, ) dx 1 x dx 3 Ω jk ( k E Ω jk + Cα /m ) u, f(u, ) dx 1 x dx 3 D ( = k E Ω jk + Cα /m ( u, f(u, )rdrdx 3 )( = ke Ω jk + k C 4 α/m ( u, >ρ u, >ρ u, f(u, )rdrdx 3 )( π/(jk) π/(4jk) π/(jk) dθ) dθ), (3) ) where denotes the volume of. Now, let <1/4. Then K = inf u S(1/4,k) J(u) inf u S(,k) J(u). Let u, also denote the zero extension of u, to all of Ω 1/4. Then u, S(1/4,k). Thus, J(u, ) K and consequently u, dx 1 x dx 3 K +M (4) where M = inf {F (t) :t R}. If we choose so that <K/( M) then (4) implies u, dx 1 dx dx 3 K (5)

6 6 Existence of many positive nonradial solutions and this leads to u, f(u, )rdrdx 3 u, >ρ = u, f(u, )rdrdx 3 u, f(u, )rdrdx 3 u, u, ρ = [ u 1 u, ρ, f(u, )rdrdx 3 ] u u u, f(u, )rdrdx 3,, f(u, )rdrdx 3 u, = [ 1 π u u, ρ, f(u, )rdrdx 3 ] u, dx 1 x dx 3 (1 C) u, f(u, )rdrdx 3 u, u, u, f(u, )rdrdx 3 where the constant C is independent of (, j, k). By choosing in such a way that 1 C > 1/ we conclude u, f(u, )rdrdx 3 > 1 u, f(u, )rdrdx 3. u, >ρ This reduces (3) to αzf(αz) dx 1 dx dx 3 ke Ω jk +kcα /m u, Ω dx 1 dx dx 3. (6) jk On the other hand, using () we have αzf(αz) dx 1 x dx 3 = α z dx 1 x dx 3 Ω kα u, dx 1 x dx 3. (7) Combining (6) and (7) to obtain (Cα /m α ) Hence, Cα /m α Ω jk Ω jk Ω jk u, dx 1 x dx 3 C jk C/(jk) u, dx 1 x dx 3 C u, dx 1 x dx 3 M 1, wherewehaveused(5). Butthefunctiong(α)=Cα /m α satisfies g() = and lim α g(α) =+.Thus,ifg(α) M 1 then there is a constant K such that α K. By letting α max {,K } K (8)

7 Alfonso Castro & Marcel B. Finan 7 we conclude that α is bounded. Let w = αz S(, k). Since F (t) is bounded from below, say by M, then (8) and () imply ( ) w J(w) = F (w) dx 1 x dx 3 α = Ω z dx 1 x dx 3 F (w) dx 1 x dx 3 (K ) u, k dx 1 x dx 3 km Ω jk (K ) j (K ) j Ω jk Ω u, dx 1 x dx 3 + kc jk u, Ω dx 1 x dx u, j Ω dx 1 x dx 3 C u, j Ω dx 1 x dx 3 (9) wherewehaveused(4)withchosen in such a way that C < K and < K/( M). We claim that there exists a constant C such that u, Ω dx 1 x dx 3 CJ(u, ). Indeed, since γ(u, ) = and by (C5) we have u, f(u, ) dx 1 x dx 3 = u, dx 1 x dx 3 Ω = ( u, F (u, )) dx 1 x dx 3 + F(u, ) dx 1 x dx 3 (J(u, )+ m u, f(u, ) dx 1 x dx 3 + C ). It follows that (1 m) u, f(u, ) dx 1 x dx 3 (J(u, )+C ). (1) On the other hand, using (4) and (C5) we have J(u, ) = ( 1 u, F (u, )) dx 1 x dx 3 ( 1 m ) u, f(u, ) dx 1 x dx 3 C ( 1 m )K C.

8 8 Existence of many positive nonradial solutions By choosing in such a way that C < 1 4 (1 m)k, weseethatj(u, ) > ( 1 m 4 )K. Now, we choose such that C < ( 1 m 4 )K. Using this in (1) we obtain u, f(u, ) dx 1 x dx 3 CJ(u, ). Also, using this in (9) we obtain J(w) C j J(u, ). Choosing j such that j>cwe obtain J(w) <J(u, ) and this concludes the present proof. To prove Lemma 1.5 we need Lemma.4 Let v H 1(). Then the function P v (λ) = λvf(λv) F (λv) is increasing on (, ). Proof. Differentiating P v with respect to λ we find ( P v(λ) = λv f (λv) f(λv) ) >, λv where we have used (C). This completes the proof ProofofLemma1.5:Fix k and n. Let << 1 (kn). Lemma 1. guarantees the existence of a minimizer u,kn of J on S(, kn). From Lemma 1.3 we see that u,kn is a solution to (1). Furthermore, from Lemma 1.4 we know that u,kn is nonradial. Now, by the regularity theory of elliptic equations we know that u,kn is a C function. Let x =(r, θ) be the polar coordinates of x R and write u = u,kn (r, θ, x 3 ). Then π u dx 1 x dx 3 = (u r + 1 (r, x 3 ) r u θ + x3 u )rdrdθdx 3 and F (u) dx 1 x dx 3 = Define the function (r, x 3 ) π F (u)rdrdθdx 3. v(r, θ, x 3 )=u(r, θ n, x 3 ), θ π. Then v(r, θ + π k, x 3 ) = u(r, θ n + π kn, x 3 ) = u(r, θ n, x 3 ) = v(r, θ, x 3 ).

9 Alfonso Castro & Marcel B. Finan 9 It follows that v H(, k). On the other hand, it is easy to check that the following equalities hold Therefore, v dx 1 x dx 3 = k v r (r, θ, x 3 )=u r (r, θ n, x 3 ), v θ (r, θ, x 3 )= 1 n u θ(r, θ n, x 3 ), x3 v(r, θ, x 3 )= x3 u(r, θ n, x 3 ). = kn = (r, x 3 ) π/k + x3 u(r, θ n, x 3 ) )rdrdθdx 3 π nk Also F (v) dx 1 x dx 3 = k (u r(r, θ n, x 3 )+ 1 r n u θ(r, θ n, x 3 ) (u r(r, θ, x 3 )+ 1 r n 4u θ(r, θ, x 3 ) + x3 u(r, θ, x 3 ) )rdrdθdx 3 π (u r (r, θ, x 3 )+ 1 (r, x 3 ) r n 4u θ (r, θ, x 3 ) + x3 u(r, θ, x 3 ) )rdrdθdx 3. = (r, x 3 ) π (r, x 3 ) π/k F(u(r, θ n, x 3 ))rdrdθdx 3 F (u(r, θ, x 3 ))rdrdθdx 3. Since u does not belong to S(, ) wehave π u θ (r, θ, x 3 )rdrdθdx 3 >. It follows that γ(v) = (r, x 3 ) = ( v vf(v)) dx 1 x dx 3 π (u r(r, θ, x 3 )+ 1 r n 4u θ(r, θ, x 3 ) (r, x 3 ) < = + x3 u(r, θ, x 3 ) uf(u))rdrdθdx 3 π (u r(r, θ, x 3 )+ 1 r u θ(r, θ, x 3 ) (r, x 3 ) + x3 u(r, θ, x 3 ) uf(u))rdrdθdx 3 ( u uf(u)) dx 1 x dx 3 =.

10 3 Existence of many positive nonradial solutions This yields v dx 1 x dx 3 < vf(v) dx 1 x dx 3. (11) Now, by Lemma. we can find < α < 1 such that αv S(, k). Let w = αv S(, k). Using Lemma.4 and the definition of jk we have jk J(w) =J(αv) =P v (α) < P v (1) = ( 1 vf(v) F(v)) dx 1 x dx 3 = ( 1 uf(u) F (u)) dx 1 x dx 3 = J(u) =jkn. Putting together all the arguments above we conclude a proof of the lemma 3 Proof of Main Theorem For any integer k 1, according to Lemma 1.4 there exists 1 ( k ) such that if << 1 ( k )then j k <j. It follows from Lemma 1.5 that j <j <... < j k <j. (1) According to Lemma 1., there exists u,i S(, i),i=1,..., k, such that j i = J(u,i). Moreover, by Lemma 1.3 u,i is a solution of Problem 1, i = 1,...,k. By (1), {u,i } k i=1, are nonrotationally equivalent and nonradial. This completes the proof of the theorem. References [1] J. Byeon, Existence of many nonequivalent nonradial positive solutions of semilinear elliptic equations on three dimensional annuli, J. Differential Equations, Vol. 136 (1997), [] F. Catrina and Z.Q. Wang, Nonlinear elliptic equations on expanding symmetric domains, to appear in J. Differential Equations. [3] A. Castro, J. Cossio and J.M.Neuberger,A sign changing solution for a superlinear Dirichlet Problem, Rocky Mountain J. Math. 7 (1997),

11 Alfonso Castro & Marcel B. Finan 31 [4] A. Castro, J. Cossio and J.M. Neuberger, A minmax principle, index of the critical point, and existence of sign-changing solutions to elliptic boundary value problems, Electron. J. Differential Equations, Vol (1998), N., [5] A. Castro and M.B. Finan, Existence of many sign-changing nonradial solutions for a superlinear Dirichlet problem on thin annuli, Topological Methods in Nonlinear Analysis Vol. 13, No. (1999), [6] C.V. Coffman, A nonlinear boundary value problem with many positive solutions, J. Differential Equations 54 (1984), [7] Y.Y. Li, Existence of many positive solutions of semilinear elliptic equations on an annulus, J. Differential Equations 83 (199), [8] S.S. Lin, Existence of many positive nonradial solutions for nonlinear elliptic equations on an annulus, J. Differential Equations 13 (1993), [9] M. Willem, Minimax Theorems, Birkhäuser, Alfonso Castro Division of Mathematics and Statistics University of Texas at San Antonio San Antonio, TX 7849 USA. castro@math.utsa.edu Marcel B. Finan Department of Mathematics The University of Texas at Austin Austin, TX 7871 USA. mbfinan@math.utexas.edu