Elliptic equations with one-sided critical growth

Transcription

1 Electronic Journal of Differential Equations, Vol. 00(00), No. 89, pp ISSN: URL: or ftp ejde.math.swt.edu (login: ftp) Elliptic equations with one-sided critical growth Marta Calanchi & Bernhard Ruf Abstract We consider elliptic equations in bounded domains R N with nonlinearities which have critical growth at + and linear growth λ at, with λ > λ 1, the first eigenvalue of the Laplacian. We prove that such equations have at least two solutions for certain forcing terms provided N 6. In dimensions N = 3, 4, 5 an additional lower order growth term has to be added to the nonlinearity, similarly as in the famous result of Brezis-Nirenberg for equations with critical growth. 1 Introduction We consider the superlinear problem u = λu + u g(x, u + ) + f(x) in u = 0 on (1.1) where R N (N 3) is a bounded domain with smooth boundary, = N/(N ) is the critical Sobolev exponent, g(, s + ) C( R, R + ) has a subcritical growth at infinity, and s + = max{s, 0}. Furthermore, we assume that λ > λ 1, the first eigenvalue of the Laplacian. This means that the function k(s) = λs + s g(x, s + ) interferes with all but a finite number of eigenvalues of the Laplacian, in the sense that k(s) k(s) λ 1 < λ = lim < lim = + s s s + s For subcritical nonlinearities, such problems have been treated by Ruf-Srikanth [11] and de Figueiredo [4], proving the existence of at least two solutions provided that the forcing term f(x) has the form f(x) = h(x) + te 1 (x), (1.) Mathematics Subject Classifications: 35J0. Key words: Nonlinear elliptic equation, critical growth, linking structure. c 00 Southwest Texas State University. Submitted March 01, 00. Published October 18, 00. 1

2 Elliptic equations with one-sided critical growth EJDE 00/89 where h L r (), for some r > N, is given, e 1 is the (positive and normalized) first eigenfunction of the Laplacian, and t > T, for some sufficiently large number T = T (h). We remark that the search of solutions for forcing terms of the form (1.) is natural if the nonlinearity crosses all eigenvalues, i.e. λ < λ 1 (then the problem is of so-called Ambrosetti-Prodi type); indeed, in this case there exists an obvious necessary condition of the form fe 1dx < c. In the present case there does not seem to exist a necessary condition for solvability, and therefore one might expect solutions for any forcing term in L (). This is an open problem; the only positive result known is for the corresponding ODE with Neumann boundary conditions, cf. [7]. Equation (1.1) with nonlinearities with critical growth have recently been considered by de Figueiredo - Jianfu [6], proving a similar existence result as the one stated for the subcritical case, however with the restriction that the space dimension N satisfies N 7. While it is known that in problems with critical growth the low dimensions may show different behavior (see Brezis - Nirenberg [3], where different behavior occurs in dimension N = 3), it is somewhat surprising to encounter difficulties already in the dimensions 4, 5 and 6. The present work is motivated by the mentioned work of de Figueiredo and Jianfu [6], who treated equation u = λu + u f(x) in u = 0 on. (1.3) They established the existence of at least two solutions under suitable conditions on f = h + te 1, more precisely they proved: Theorem 1.1 (de Figueiredo - Jianfu [6]) Suppose that i) λ > λ 1 ii) h L r (), r > N, is given, with h ker( λ) if λ is an eigenvalue. Then there exists T 0 = T 0 (h) > 0 such that if t > T 0 then problem (1.3) has a negative solution φ t W,r W 1,r 0 C 1,1 N/r. Suppose in addition that iii) λ is not an eigenvalue of (, H0 1 ()) iv) N 7 Then problem (1.3) has a second solution for t > T 0. We remark that in [6] only h L () is assumed, however this seems not sufficient to get the stated results. In this paper we improve and extend the result of de Figueiredo and Jianfu in the following ways: Main Results: 1) λ > λ 1 and λ is not an eigenvalue: N 6: if h L r () (r > N), then equation (1.1) has a second solution for f = h + te 1, with t > T 0 (h).

3 EJDE 00/89 Marta Calanchi & Bernhard Ruf 3 N = 3, 4, 5: if h L r () (r > N) and g(x, s + ) cs q for some q = q(n) > 1 and c > 0, then equation (1.1) has a second solution for f = h + te 1, with t > T 0. ) λ is an eigenvalue, i.e. λ = λ k, for some k : If h L r () (r > N) satisfies h ker( λ k, H 1 0 ()), then equation (1.1) has a second solution for f = h + te 1 with t > T 0 (in dimensions N = 3, 4, 5 the assumptions made in 1) have to be added). 3) λ in a left neighborhood of λ k, k : There exists a δ > 0 such that if λ (λ k δ, λ k ) and h L r () (r > N) satisfies h ker(, H 1 0 (), then equation (1.1) has at least three solutions for f = h+te 1, with t > T 0 (in dimensions N = 3, 4, 5 the assumptions made in 1) have to be added). For proving these statements, one proceeds as follows: the first (negative) solution is easily obtained (see [11], [4], [6]). To obtain a second solution, one uses the saddle point structure around the first solution and applies the generalized mountain pass theorem of Rabinowitz [10]. To prove the Palais-Smale condition, one proceeds as in [3], using a sequence of concentrating functions (obtained from the so-called Talenti function). However due to the presence of the first solution, lower order terms appear in the estimates. To handle these estimates, an orthogonalization procedure based on separating the supports of the Talenti sequence and the (approximate) first solution is used (this approach was introduced in [8]). With this method we obtain the results 1) and ). To prove 3), one shows that the branch of solutions with λ (λ k, λ k+1 ) can be extended to λ = λ k if h ker( ). Actually, this branch can be extended slightly beyond λ k, i.e. to λ (λ k δ, λ k ], and the corresponding solutions are clearly bounded away from the negative solution. Since in λ k starts a bifurcation branch (λ, u) emanating from the negative solution and bending to the left (as shown by de Figueiredo - Jianfu [6]), we conclude that for λ to the left of and close to λ k there exist at least three solutions. Statement of theorems In this section we give the precise statements of the theorems. Furthermore, the notation and basic properties are introduced. We consider problem (1.1) under the following conditions on the nonlinearity g: (g 1 ) g : R R + is continuous; (g ) g(x, s) 0 for s 0, i.e. g(x, s) = g(x, s + ) with g(x, 0) = 0; (g 3 ) There exist constants c 1 > 0 and 1 < p < (N + )/(N ) such that g(x, s) c 1 s p, for all s R; For N = 3, 4, 5 we assume in addition:

4 4 Elliptic equations with one-sided critical growth EJDE 00/89 { (g 4 ) There exist c > 0 and q with max g(x, s + ) c(s + ) q, for all s R. We prove the following results: N (N 1), 3 } < q+1 < 1, such that Theorem.1 N 6: Let h L r (), r > N, be given, and let T 0 = T 0 (h) the number given by Theorem 0. Assume that λ > λ 1, and that g satisfies (g 1 ), (g ), (g 3 ). If λ is an eigenvalue, say λ = λ k, assume in addition that h ker( λ k ). Then problem (1.1) has a second solution for f = h + te 1, with t > T 0. Theorem. 3 N 5: Let h L r, r > N, be given, and suppose that all other assumptions of Theorem.1 are satisfied. If g satisfies also (g 4 ), then problem (1.1) has a second solution for f = h + te 1, with t > T 0. Theorem.3 Assume the hypotheses of Theorems.1 and., and assume in addition that h ker( λ k ), for some k. Then there exists δ > 0 such that for λ (λ k δ, λ k ) problem (1.1) has at least three solutions for f = h+te 1, with t > T 0. The first solution of equation (1.1) is a negative solution, and its existence, for t sufficiently large, is not difficult to prove (see [4], [6]): first note that a negative solution satisfies the linear equation y λy = h + te 1 y = 0 on in (.1) The solution of this equation is unique, and we denote it, in dependence of te 1, by φ t (we remark that if λ = λ k then h ker( λ k ) is required, and the solution is unique in (ker( λ k )) ). Note that we may assume that he 1dx = 0, and then the solution φ t can be written as φ t = w + s t e 1, with we 1dx = 0 and s t = t/(λ 1 λ) (and with w ker( λ k ) if λ = λ k ). Since h L r, r > N, we have w C 1,1 N/r (), and it is known (see [9]) that on the (interior) normal derivative n e 1(x) is positive; hence φ t < 0 for t sufficiently large. To find a second solution of equation (1.1), we set u = v + φ t ; then v solves v = λv + (v + φ t ) g(x, (v + φ t ) + ) in v = 0 on (.) Clearly v = 0 is a solution of this equation, corresponding to the negative solution φ t for equation (1.1). To find a second solution of equation (.) one can look for non trivial critical points of the functional J(v) = 1 ( v λv )dx 1 (v + φ t ) + dx G(x, (v + φ t ) + )dx, where G(x, s) := s g(x, ξ) dξ. 0

5 EJDE 00/89 Marta Calanchi & Bernhard Ruf 5 This was the approach of de Figueiredo-Jianfu in [6]. For applying the Generalized Mountain Pass theorem of Rabinowitz [10] one needs to prove some geometric estimates. Furthermore, since the nonlinearities have critical growth, one needs to show that the minimax level avoids the non-compactness levels given by the concentrating sequences u ɛ (see [3] and below). In these estimates, the terms of the form v + φ t + u ɛ are not easy to handle. In this paper we apply a method introduced in [8] to make such estimates easier. The idea consists in separating the supports of v + φ t and u ɛ by concentrating the support of the functions u ɛ in small balls, and cutting small holes into the functions v + φ t such that the respective supports are disjoint. These manipulations create some errors, but these are easier to handle than to estimate the mixed terms arising in expressions like (v + φ t + u ɛ ) p +. Moving these small holes near where the first solution φ t is small allows to further improve the estimates. 3 Variational setting and preliminary properties We begin by replacing equation (.) by an approximate equation. We denote by B r (x 0 ) a ball of radius r and center x 0. Choose m N so large that B /m (x 0 ), and let η m C0 () such that 0 η m (x) 1, η m (x) m and { 0 in B 1/m (x 0 ) η m (x) = 1 in \ B /m (x 0 ) Define the functions φ m t = η m φ t and set f m = φ m t λφ m t. Setting as before u = v + φ m t solves the equation in equation (1.1), we see that then v = φ t φ m t v = λv + (v + φ m t ) g(x, (v + φ m t ) + ) + (f f m ) in v = 0 on (3.1) Clearly v = φ t φ m t corresponds to the trivial solution of this equation; for finding other solutions of (3.1) we look for critical points of the functional, J : H R, J(v) = 1 ( v λv ) 1 (v + φ m t ) + G(x, (v + φ m t ) + ) (f f m )v, where H denotes the Sobolev space H = H 1 0 (), equipped with the Dirichlet norm u = ( u dx) 1/. We begin by estimating the error given by the term f f m.

6 6 Elliptic equations with one-sided critical growth EJDE 00/89 Lemma 3.1 For N 3, as m + we have: φ t φ m t cm N/ ; (3.) (f f m )ψ dx c ψ m N/, for all ψ H. (3.3) Proof. Note first that by the regularity assumption h L r, r > N, it follows that φ t C 1,1 N/r (), and hence in particular that there exists c > 0 such that for any point x φ t (x) c x x. Furthermore we may choose for every large m N a point x 0 at distance 4/m from the boundary point x, such that φ t (x) c 1 m, x B 4/m(x 0 ). (3.4) We may assume that x 0 = 0 for every choice of m; from now on we write B r = B r (0). Thus, we can estimate (φ t φ m t ) = φ t (1 η m ) φ t η m = φ t 1 η m φ t (1 η m ) φ t η m B m B m \B 1m + φ t η m B m \B 1m c 1 m N + c m N + c 3 m N = cm N, hence (3.). The estimate (3.3) is now obtained as follows: (f f m )ψ dx = (φ t φ m t ) ψ λ(φ t φ m t )ψdx c φ t φ m t ψ c ψ m N/, for all ψ H. Let λ 1 < λ... the eigenvalues of and e 1, e..., the corresponding eigenfunctions. Take m as before and let ζ m : R be smooth functions such that 0 ζ m 1, ζ m 4m and { 0 if x B /m ζ m (x) = 1 if x \ B 3/m We define approximate eigenfunctions by e m i = ζ m e i. Then the following estimates hold.

7 EJDE 00/89 Marta Calanchi & Bernhard Ruf 7 Lemma 3. As m, we have e m i e i in H. Moreover, in the space H j,m = span{em 1,..., e m j }, we have } max { u : u H j,m, u = 1 λ j + c j m N and e m i e m j dx = δ ij + O(m N ). Proof. See [8] and observe that, since is of class C 1, also for the eigenfunctions e i an estimate as (3.4) holds. Consider the family of functions [ u N(N )ε ε(x) = ε + x which are solutions to the equation u = u u u(x) 0 ] (N )/ in R N as x and which realize the best Sobolev embedding constant H 1 (R N ) L (R N ), i.e. the value u H S = S N = inf. u 0 u L Let ξ C 1 0(B 1/m ) be a cut off function such that ξ(x) = 1 on B 1/m, 0 ξ(x) 1 in B 1/m and ξ 4m. Let u ε (x) := ξ(x)u ε(x) H. For ε 0 we have the following estimates due to Brezis and Nirenberg Lemma 3.3 (Brezis-Nirenberg, [3]) For fixed m we have (a) u ε = S N/ + O(ε N ) (b) u ε = SN/ + O(ε N ) (c) u ε K 1 ε + O(ε N ) (d) u ε s N N s K ε s. Moreover, for m and ε = o(1/m), we have (see [8]) (e) u ε = S N/ + O((εm) N ) (f) u ε = SN/ + O((εm) N ) while (c) and (d) hold independently of m.

8 8 Elliptic equations with one-sided critical growth EJDE 00/89 Proof. We only prove (d); for the other estimates, see [3, 8, 1]. u s ( ε ) N s ε dx c dx ε + x c which completes the proof. B 1 m ε 4 The linking structure 0 ( ε ε + ρ ) N s ρ N 1 dρ cε N s ε N. In this section we prove that the functional J has a linking structure as required by the Generalized Mountain Pass Theorem by P. Rabinowitz [10]. For the rest of this article, we assume λ [λ k, λ k+1 ). Let H + = [span{e 1,..., e k }], S r = B r H +, Hm = span{e m 1,..., e m k } and Qε m = (B R Hm) [0, R]{u ε }, where m N is fixed. Define the family of maps H = {h : Q ε m H continuous : h Q ε = id}, and set m c = inf sup J(u) (4.1) h H u h(q ε m ) Then the Generalized Mountain Pass theorem of P. Rabinowitz states that if 1) J : H R satisfies the Palais-Smale condition (PS) ) there exist numbers 0 < r < R and α 1 > α 0 such that J(v) α 1, for all v S r (4.) J(v) α 0, for all v Q ε m, (4.3) then the value c defined by (4.1) satisfies c α 1, and it is a critical value for J. First note that for v H m R{u ε }, v = w + su ε, we have by definition supp(u ε ) supp(w) =. It is easy to prove that this implies that J(v) J(w + su ε ) = J(w) + J(su ε ). We begin by showing that the functional J satisfies condition (4.). Lemma 4.1 There exist numbers r > 0 and α 1 > 0 such that J(v) α 1 for all v S r = B r H + Proof. Let v H +. From the variational characterization of λ k+1 and the Sobolev embedding theorem, we have, using (g 3 ) and Lemma 3.1, J(v) 1 ( λ ) 1 v 1 λ k+1 v+ c v p+1 + (f f m )v c 1 v c v c 3 v p+1 c 4 m N/ v

9 EJDE 00/89 Marta Calanchi & Bernhard Ruf 9 Let k m (s) = c 1 s c s c 3 s p+1 c 4 m N/ s. Clearly, there exists m 0 such that max R k m (s) = M m M m0 > 0 for all m m 0. Thus there exist α 1 > 0 and r > 0 such that which completes the proof. Next we prove condition (4.3). J(v) α 1 > 0 for v = r. Lemma 4. There exist R > r and α 0 < α 1 such that for ε sufficiently small J Q ε m < α 0. Proof. Let v = w +su ε (Hm B R ) [0, R]{u ε ]. Since J(v) = J(w)+J(su ε ) we can estimate J(w) and J(su ε ) separately. J(w) 1 w dx λ w dx 1 (w + φ m t ) + dx, since (f f m)w dx = (φ t φ m t ) w dx λ (φ t φ m t )w dx = 0. J(su ε ) s u ε B dx λs u ε dx 1m B 1m s u ε dx f su ε dx. B 1m Let Q ε m = Γ 1 Γ Γ 3, where Γ 1 = {v H : v = w + su ε, w Hm, w = R, 0 s R}, Γ = {v H : v = w + Ru ε, w Hm B R }, Γ 3 = Hm B R. Note that it follows by Lemma 3. that w dx (λ k + c k m N ) B 1m w dx, for all w H m 1. Suppose v Γ 1 ; then v = w + su ε with w = R and 0 s R. (i) if λ (λ k, λ k+1 ), we choose m 0 such that c k m n < λ λ k, for m m 0. Then, using Lemma 3.3 J(v) 1 ( λ ) 1 λ k + c k m N w dx 1 (w + φ m t ) + dx + s u ε B dx λs u 1m B ε dx s 1m ε dx + s f u ε B 1m u cr + S N/[ s s ] + R O(ε N ) + R O(ε N ) + cs cr + c 1 + c R + c 3 R ε N.

10 10 Elliptic equations with one-sided critical growth EJDE 00/89 Thus J(v) 0 for R R 0 and ε > 0 sufficiently small. (ii) if λ = λ k, for w = R w Hm with w = 1 we write w = αy + βe m k, with y span{e m 1,..., e m k 1 } and y = 1. Then J(w) = R ( α y + β e m k λ k αy + βe m k ( φ ) dx R t ) w + R + dx. Using Lemma 3., we can estimate the first integral as follows ( α y + β e m k λ k αy + βe m k ) dx α ( λ k ) 1 + β ( λ k ) λ k 1 + cm N 1 λ k + cm N +αβ ( y e m k λ k ye m k ) dx cα + c 1 (β + αβ)m N. Note now that if α δ > 0, for some δ > 0, then J(w) cδ R + c 1 m N and hence J(v) 0 for R R 1 (δ). We show now that there exists δ > 0 such that if α δ, then there exist constants c > 0 and R > 0 such that ( φ t ) w + R c + > 0 for all R R and for all w H m B R. To this aim we prove that there exist δ > 0 and η > 0 such that max(αy + βe m k ) η > 0 for all y H k 1,m, y = 1, α δ. By contradiction assume that there exist sequences α n 1/n, y n H k 1,m with y n = 1 such that max{α n y n + β n e m k } 0 as n +. Then α n y n 0, β n = 1 α n + O(m N ) β = 1 + O(m N ) 1/, for m m 0. Therefore, we conclude that max(βe m k ) = 0, with β 1 for m m 0 i.e. (e m k ) + = 0. This is a contradiction since e m k e k in H implies that e m k for m large. Therefore, there exist δ > 0, η > 0 such that must change sign, max { w, α δ} η > 0, w H k,m, w = 1, m m 0.

11 EJDE 00/89 Marta Calanchi & Bernhard Ruf 11 Denoting w = {x : w(x) η/}, then w ν > 0, w H k,m, with w = 1 and α δ, m m 0, since the functions w H k,m are equicontinuous. Moreover φ t R > η for R sufficiently large. 4 Then ( w + φ t + ( η 4 ) w. R ) Thus, we can conclude that there exists R > 0 such that J(w) cr R ( w + φ t + cr R ( η ν 0, 4 ) R ) for all R R. In particular J(w) as R +.. Let v Γ, i.e. v = w + Ru ε with w R. Then J(v) =J(w) + J(Ru ε ) cm N w + R u ε B dx R 1m u ε dx + R f u ε < 0, for R sufficiently large. Now fix R > 0 such that the previous estimates hold. 3. Let v Γ 3, i.e. v = w H m B R. Hence J(v) c 1 m N w 1 if m is sufficiently large. (w + φ t ) + α 0 5 Existence of a second solution In this section we prove Theorems.1 and.. By the Linking Theorem we construct a Palais Smale sequence {v n } H at the minimax level c; the sequence {v n } satisfies J(v n ) = 1 ( v n λv n) dx 1 (v n + φ m t ) + dx (5.1) G(x, (v n + φ m t ) + ) dx + (f f m )v n dx = c + o(1) and J (v n ), z = for all z H. ( v n z λv n z) dx (v n + φ m t ) 1 + z dx g(x, (v n + φ m t ) + )z dx + (f f m )z dx = o(1) z (5.)

12 1 Elliptic equations with one-sided critical growth EJDE 00/89 Lemma 5.1 Under the hypotheses of Theorem.1 or Theorem., the sequence {v n } is bounded in H. Proof. From (5.1) and (5.), it follows that J(v n ) 1 J (v n ), v n = 1 (v n + φ m t ) + 1 (v n + φ m t ) 1 + φ m t N G(x, (v n + φ m t ) + ) + 1 g(x, (v n + φ m t ) + )v n 1 (f f m )v n = c + o(1) + o(1) v n Therefore, using that φ m t 0 and Lemma 4, 1 (v n + φ m t ) + dx G(x, (v n + φ m t ) + ) dx N 1 g(x, (v n + φ m t ) + )(v n + φ m t ) dx +c + (o(1) + dm N ) v n Then by (g 3 ), we get (v n + φ m t ) + dx c 1 (v n + φ m t ) p+1 + dx + c + (o(1) + dm N ) v n ( c 1 (v n + φ m t ) + dx ) p+1 + c + (o(1) + dm N ) v n Since p + 1 <, we obtain (v n + φ t ) + dx c + (o(1) + dm N ) v n c 1 + c v n (5.3) (i) First we consider the case λ (λ k, λ k+1 ). Let v n = v n + + vn (as in [6]), with vn H k = span{e 1... e k } and v n + (H k ). We obtain J (v n ), v n + = ( v n + λ(v n + ) ) dx (v n + φ m t ) 1 + v n + dx g(x, (v n + φ m t ) + )v n + dx (f f m )v mdx + = o(1) v n +

13 EJDE 00/89 Marta Calanchi & Bernhard Ruf 13 From the variational characterization of λ k+1 Young inequalities and (5.3), we get, using the Hölder and ( λ ) 1 v + λ n k+1 (v n + φ m t ) 1 + v n + + c (v n + φ m t ) p + v n + + o(1) v n + + dm N v n + ( ε v n + ) / + c ε ( (v n + φ m t ) ( +c (v n + φ m t ) + ε v + n + c ε ( +c ε ( ) p/ ( (v n + φ m t ) + v + n + ) ( 1) p ) ( 1) + ε ) p/ (v n + φ m t ) + + c v n + ) p + c v + n ( v + n By (5.3) and by the Sobolev embedding theorems, we obtain v n + c + o(1) ( v n N+ N ) p p + vn p ) + c v + n (5.4) For vn H, we have ( λ 1) vn λ k (v n + φ m t ) 1 + vn + In the same way we obtain g(x, (v n + φ m t ) + ) vn + f f m vn vn H c + o(1)( v 1 n N+ N Joining (5.4) and (5.5), we find so v n is bounded in H. v n c + c( v n N+ N + vn p ) + c v n (5.5) + vn p ) + c v n, (ii) If λ = λ k we write v n = vn + v n + + β n e k = w n + β n e k, where we denote with vn and v n + the projections of v n onto the subspace H k 1 = span{e 1,..., e k 1 } and H + k = (H k ) respectively. With a similar argument as above we obtain the following estimate w n c + c ( v n N+ N We can assume v n 1. Then, from (5.6), we have + vn p ) + c wn (5.6) w n c + c v n p c + c( w n + β n ) p (5.7)

14 14 Elliptic equations with one-sided critical growth EJDE 00/89 If β n is bounded we conclude as above. If not, we may assume β n + and w n + (we neglect the case w n c which is much easier). Therefore, from (5.7), w n [ 1 ( w n + βn) ] p/. Then w n cβn p/ and wn β n 0 since p < 1. Therefore, possibly up to a subsequence, w n /β n 0 a.e. and strongly in L q, q <. Therefore, for all q (, ) Moreover, since ( w n + φ m t β n + e k ) q +e k dx o(1) = J (v n ), e k = (v n + φ m t ) 1 + e k (e k ) q+1 + dx. (5.8) g(x, (v n + φ m t ) + )e k + (f f m )e k we get, using (g 3 ) (w n + φ m ) t + e 1 c (w n + φ m ) t p k e β + k o(1) + n β 1 p + e k n β e + k. n Finally, by (5.8) we get (e k ) + 0 which is a contradiction. Thus (v n ) is bounded. Returning to relation (5.), we may therefore assume, as n + : v n v weakly in H0 1, v n v in L q q < and v n v a. e. in. In particular, it follows that v is a weak solution of v = λv + (v + φ m t ) g(x, (v + φ m t ) + ) + f f m in v = 0 on (5.9) To conclude the proof, It remains to show that v φ t φ m t, the trivial solution of (5.9). First, we estimate J(φ t φ m t ) = 1 (φ t φ m t ) λ φ t φ m t (f f m )(φ t φ m t ) = 1 [ (φt φ m t ) λ φ t φ m t ]. From Lemma 3.1 we get, taking ε β = 1 m 0 < β < 1, J(φ t φ m t ) cm N := cε βn. (5.10)

15 EJDE 00/89 Marta Calanchi & Bernhard Ruf 15 Since v is a weak solution of (5.9), we have ( v λv ) dx (v + φ m t ) + dx + (v + φ m t ) 1 + φ m t dx g(x, (v + φ m t ) + )v dx (f f m )v dx = 0 By the Brezis Lieb Lemma ([]) (v n + φ m t ) + dx = (v n v) + dx + (v + φ m t ) + dx + o(1) (5.11) Since v n v in L q q <, we have G(x, (v n + φ m t ) + ) dx G(x, (v + φ m t ) + ) dx = o(1). (5.1) Since v n v in H, v n = v + (v n v) + o(1). (5.13) Then, by (5.1), (5.11) and (5.13), we find c + o(1) = J(v n ) (5.14) = J(v) + 1 (v n v) dx 1 (v n v) + dx + o(1) Similarly, since J (v) = 0, we obtain J (v n ), v n = (v n v) (v n v) + (v n v) 1 + φ m t g(x, (v n + φ m t ) + )v n + g(x, (v + φ m t ) + )v + o(1). Since and we get Now, let (v n v) 1 + φ m t dx = o(1) g(x, (v n + φ m t ) + )v n dx g(x, (v + φ m t ) + )v dx = o(1), (v n v) dx = K = (v n v) + dx + o(1). (5.15) lim (v n v) dx. n +

16 16 Elliptic equations with one-sided critical growth EJDE 00/89 If K = 0, then v n v strongly in H and in L ; then by (5.14) and (5.10) J(v) = c α 1 > cε βn J(φ t φ m t ) so that v φ t φ m t. If K > 0, using the Sobolev inequality and (5.15), we have (as in [6]) ( ) / ( ) / v n v S v n v dx S (v n v) + dx ( ) / S (v n v) dx + o(1) This implies that K SK N N, that is K S N/. To complete the proof we use the following Lemmas which will be proved below. Lemma 5. Under the hypotheses of Theorem.1 one has, for ε β = 1/m, with α/n < β < (N 4)/(N ), c < 1 N SN/ cε. (5.16) Lemma 5.3 Suppose that the hypotheses of Theorem. are satisfied. Then for ε β = 1/m, with we have max { 1 N (N 1) (q + 1), N N (q + 1)} < β < 1 (q + 1) N, c < 1 N N SN/ N cε (q+1). By (5.14) and (5.15) we get { J(v) + K N = c 1 cε (Theorem.1) N SN/ (q+1) (Theorem.) N N cε Assume now by contradiction that v φ t φ m t. Then we get by (5.10) K N + J(v) 1 N SN/ cε βn which is impossible, due to the choice of β. Proof of Lemma 5.. Let ε > 0 and m N be fixed such that the hypotheses of the Linking Theorem are satisfied. For v = w + su ε, we have J(v) = J(w) + J(su ε ) J(w) + s u ε B s 1m +s f f m u ε B 1m u ε B 1m λs B 1m u ε

17 EJDE 00/89 Marta Calanchi & Bernhard Ruf 17 As above, we have J(w) cm N. To estimate the last inequality we use the argument developed in [8] (Lemma 3.1). We have from (e) and (f) in Lemma 3.3 s u ε B1/m s u B 1/m ε ( s s )[ S N/ + O((εm) N ) ] ( 1 1 )[ S N/ + O((εm) N ) ] SN/ N + c(εm)n. (since s s attains its maximum at s = 1) We make use of the Hölder inequality to estimate the last term. Let α be such that 1 α + 1 r + 1 = 1, i.e. 1 α = 1 1 r > 1 1 N = 1, in particular > α. Since supp f m \B 1/m we have B 1/m f f m u ε ( ) 1/r ( ) 1/(µ(B1/m f r u ε )) 1/α B 1/m B 1/m cεm N/α. If ε β = 1/m with 0 < β < 1 we have, by (c) in Lemma 3.3, J(v) 1 N SN/ + c 1 m N + c (εm) N c 3 ε + c 4 εm N/α We choose β such that = 1 N SN/ + c 1 ε Nβ + c ε (1 β)(n ) c 3 ε + ε 1+βN/α (5.17) α N < β < 1 N. Such a choice is possible only for N 6. This then implies that for ε > 0 sufficiently small in particular c < 1 N SN/ cε. J(v) < 1 N SN/ cε, Proof of Lemma 5.3. With a similar argument as above and using (g 4 ) and Lemma 3.3(d), we obtain J(v) = J(w) + J(su ε ) J(w) + s u ε dx λs u ε dx s G(x, (su ε + φ m t ) + ) dx s (f f m )u ε dx u ε dx

18 18 Elliptic equations with one-sided critical growth EJDE 00/89 Reasoning as in Lemma 5. from (d) in Lemma 3.3 we obtain as in (5.17) J(v) SN/ N + c 1ε Nβ + c ε (1 β)(n ) c 3 s q+1 SN/ N u q+1 ε dx + c 4 ε 1+βN/α + c 1ε Nβ + c ε (1 β)(n ) c 3 ε N (q+1)+n + c 4 ε 1+βN/α Using that α <, we get the following condition on β (0, 1) N N (q + 1) < which is equivalent to the system 1 N N (N 1) N βn 1 + βn (1 β)(n ) (q + 1) < β < N + 1 (q + 1) (q + 1) < β < N + 1 (q + 1) This choice is possible if q + 1 > { N (N 1)(N ) 3 N N Therefore the result follows. 6 Existence of a third solution We consider only the case N 6 and g 0; the other cases follow with small changes. Let φ t (λ) denote the negative solution of (1) for λ (λ k δ, λ k ). Since h ker( λ k ), φ t (λ) is uniformly bounded when λ λ k. Consider the functional J λ (v) = 1 ( v λv ) dx 1 (v + φ t (λ)) + dx (f f m )v dx. We prove the main geometrical properties of the functional J λ. Lemma 6.1 Let H + = span{e 1,..., e k }. There exist ᾱ > 0, r > 0 such that J λ (v) ᾱ > 0 for all v S r = B r H +, λ (λ k 1, λ k ).

19 EJDE 00/89 Marta Calanchi & Bernhard Ruf 19 Proof. With a similar argument as in Lemma 4.1, since λ < λ k, we have J λ (v) 1 ( λ ) 1 v c v c 3 m N/ v λ k+1 1 ( λ k ) 1 v c v c 3 m N/ v λ k+1 Then there exist m 0, ᾱ and r such that for m m 0, J λ (v) ᾱ > 0 for v = r and all λ (λ k 1, λ k ). Now let Hm = span{e m 1,..., e m k } as in the proof of Theorems.1 and.. Lemma 6. Let Q m = (Hm B R (0)) [0, R]u ε, where ε β = 1/m. Then, if R is large enough, there exists m m 0 such that: (i) J Q m < ᾱ (ii) sup Q m J < SN/ N c ε. Proof. (i) Let Q m = Γ 1 Γ Γ 3 as in Lemma 4.. (1) For v = w + su ε Γ 1 we have J(v) = J(w) + J(su ε ). Arguing as in Lemma 4. (ii) we have J(su ε ) c 3 R ε N + c 1 + cr (6.1) Writing w = R w H m with w = 1, w = αy + βe m k, y span{em 1,..., e m k 1 }, y = 1, it was shown in Lemma 4. (ii) that there exists a number σ > 0 such that for α σ, J(w) c 1 R c R. (6.) On the other hand, if α > σ, one deduces as in Lemma 4. (ii) (using that δ < λ k λ k 1 ) that J(w) R (α y + β e m k ) λ αy + βe m k [ R α λ (1 λ k 1 + cm N ) + β (1 ] +αβ ( y e m k λye m k ) [σ R λ k 1 (λ k δ) + cm N λ k 1 + cm N R [ cσ + c 1 δ + c m N ]. λ λ k + cm N ) + β δ + cm N + αβcm N] λ k + cm N Thus, we get in this case, for δ sufficiently small and m sufficiently large J(w) c σr. (6.3)

20 0 Elliptic equations with one-sided critical growth EJDE 00/89 Joining (6.1), (6.) and (6.3) we have J(v) = J(w) + J(su ε ) 0, for δ small, R sufficiently large and ε sufficiently small. () Let v Γ, i.e. v = w + Ru ε with w R; then J(v) = J(w) + J(Ru ε ) c(m N + δ)r + R R ε dx + R f u ε < 0 B 1m u for R sufficiently large. (3) Let v Γ 3 = B R H m. Then B 1m u ε dx J(v) c 1 (m N + δ) w 1 (w + φ t (λ)) + dx B 1m c 1 (m N + δ) w < α 0 if m m 1 and δ > 0 sufficiently small. ii) For v Q m we have as in Lemma 5. inequality (5.17), J(v) 1 N SN/ + c 1 (m N + δ)r + c (εm) N c 3 ε c 4 εm N/ If ε β = 1/m, with β as in Lemma 5., then J(v) SN/ N cε + c 4 δr Then there exists δ 1 such that if 0 < δ < δ 1, then J(v) SN/ N c ε. Arguing as in Theorem.1, we find a critical point v λ of the functional J λ at a level c λ ᾱ > 0 with ᾱ independent of λ (λ k δ, λ k ). It was shown in [6, Theorem 1.3] that in every λ k, k 1, starts a bifurcation branch (λ, k) emanating from the negative solution and bending to the left (Prop. 4.) (and thus corresponding to our second solution). We conclude that for λ to the left and close to λ k there exist at least three solutions for equation (1.1), for t > T 0 with T 0 = T 0 (h) sufficiently large. References [1] A. Ambrosetti, G. Prodi, On the inversion of some differentiable mappings with singularities between Banach spaces, Ann. Mat. Pura Appl., 93, 197,

21 EJDE 00/89 Marta Calanchi & Bernhard Ruf 1 [] H. Brezis, E. Lieb, A relation between pointwise convergence of functions and convergence of integrals, Proc. Amer. Math. Soc., 88, 1983, [3] H. Brezis, L. Nirenberg, Positive solutions of nonlinear elliptic equations involving critical Sobolev exponents, Comm. Pure Appl. Math., 4, 1983, [4] D.G. De Figueiredo, On superlinear elliptic problems with nonlinearities interacting only with higher eigenvalues, Rocky Mountain J. Math., 18, 1988, [5] D.G. De Figueiredo, Lectures on boundary values problems of Ambrosetti Prodi type, Atas do 1 Seminario Brasileiro de Análise, São Paulo, [6] D.G. De Figueiredo, Y. Jianfu, Critical superlinear Ambrosetti Prodi problems, Topological Methods in Nonlinear Analysis, 14, 1999, [7] D.G. De Figueiredo, B. Ruf, On a superlinear Sturm Liouville equation and a related bouncing problem, J. reine angew. Math. 41 (1991), 1. [8] F. Gazzola, B. Ruf, Lower order perturbations of critical growth nonlinearities in semilinear elliptic equations, Advances in Diff. Eqs.,, n. 4, 1997, [9] D. Gilbarg, N.S. Trudinger, Elliptic partial differential equations of second order, Springer Verlag, New York, [10] P. Rabinowitz, Minimax methods in critical point theory with applications to differential equations, 65 AMS Conf. Sec. Math, [11] B. Ruf, B., P.N. Srikanth, Multiplicity results for superlinear elliptic problems with partial interference with spectrum, J. Math. Anal. App., 118, 1986, [1] M. Struwe, Variational methods, Springer Verlag, Berlin, Marta Calanchi ( calanchi@mat.unimi.it, ph ) Bernhard Ruf ( ruf@mat.unimi.it, ph ) Dip. di Matematica, Università degli Studi, Via Saldini 50, 0133 Milano, Italy