POTENTIAL LANDESMAN-LAZER TYPE CONDITIONS AND. 1. Introduction We investigate the existence of solutions for the nonlinear boundary-value problem

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1 Electronic Journal of Differential Equations, Vol. 25(25), No. 94, pp ISSN: URL: or ftp ejde.math.txstate.edu (login: ftp) POTENTIAL LANDESMAN-LAZER TYPE CONDITIONS AND THE FUČÍK SPECTRUM PETR TOMICZEK Abstract. We prove the existence of solutions to the nonlinear problem u (x) + λ + u + (x) λ u (x) + g(x, u(x)) = f(x), x (, π), u() = u(π) = where the point λ +, λ is a point of the Fučík spectrum and the nonlinearity g(x, u(x)) satisfies a potential Landesman-Lazer type condition. We use a variational method based on the generalization of the Saddle Point Theorem. 1. Introduction We investigate the existence of solutions for the nonlinear boundary-value problem u (x) + λ + u + (x) λ u (x) + g(x, u(x)) = f(x), x (, π), (1.1) u() = u(π) =. Here u ± = max{±u, }, λ +, λ R, the nonlinearity g : (, π) R R is a Caratheodory function and f L 1 (, π). For g and f problem (1.1) becomes u (x) + λ + u + (x) λ u (x) =, x (, π), (1.2) u() = u(π) =. We define Σ = {λ +, λ R 2 : (1.2) has a nontrivial solution}. This set is called the Fučík spectrum (see 2), and can be expressed as Σ = j=1 Σ j where Σ 1 = { λ +, λ R 2 : (λ + 1)(λ 1) = }, Σ 2i = { ( 1 λ +, λ R 2 : i + 1 ) = 1 }, λ+ λ Σ 2i+1 = Σ 2i+1,1 Σ 2i+1,2 where Σ 2i+1,1 = { ( 1 λ +, λ R 2 : i + 1 ) + 1 = 1 }, λ+ λ λ+ Σ 2i+1,2 = { ( 1 λ +, λ R 2 : i + 1 ) + 1 = 1 }. λ+ λ λ 2 Mathematics Subject Classification. 35J7, 58E5, 49B27. Key words and phrases. Resonance; eigenvalue; jumping nonlinearities; Fucik spectrum. c 25 Texas State University - San Marcos. Submitted November 26, 24. Published August 29, 25. Partially supported by the Grant Agency of Czech Republic, MSM

2 2 P. TOMICZEK EJDE-25/94 We suppose that λ +, λ Σ m, if m N is even λ +, λ Σ m2, if m N is odd and λ < λ + < (m + 1) 2. (1.3) λ Σ 6 Σ 5,1 Σ 5,2 Σ 4 Σ 3,2 Σ 1 Σ 2 Σ 3, λ + Figure 1. Fučík spectrum Remark 1.1. Assuming that (m + 1) 2 > λ + > λ, if λ +, λ Σ m, m N, then λ > (m 1) 2. We define the potential of the nonlinearity g as G(x, s) = s g(x, t) dt and G(x, s) G(x, s) G + (x) = inf, G (x) = sup. s + s s s We denote by ϕ m a nontrivial solution of (1.2) corresponding to λ +, λ (see Remark 1.2). We assume that for any ϕ m the following potential Landesman-Lazer type condition holds: f(x)ϕ m (x) dx < G+ (x)(ϕ m (x)) + G (x)(ϕ m (x)) dx. (1.4) We suppose that the nonlinearity g is bounded, i.e. there exists p(x) L 1 (, π) such that g(x, s) p(x) for a.e. x (, π), s R (1.5)

3 EJDE-25/94 POTENTIAL LANDESMAN-LAZER TYPE CONDITION 3 and we prove the solvability of (1.1) in Theorem (3.1) below. This article is inspired by a result in 3 where the author studies the case when g(x, s)/s lies (in some sense) between Σ 1 and Σ 2 and by a result in 1 with the classical Landesman-Lazer type condition 1, Corollary 2. Remark 1.2. First we note that if m is even then two different functions ϕ m1, ϕ m2 of norm 1 correspond to the point λ +, λ Σ m. For example for m = 2, λ + > λ we have { k 1 λ sin( λ + x), x, π/ λ +, ϕ 21 (x) = k 1 λ+ sin( λ (x π/ λ + )), x π/ λ +, π, where k 1 >, and ϕ 22 (x) = { k 2 λ+ sin( λ x), x, π/ λ, k 2 λ sin( λ + (x π/ λ )), x π/ λ, π, where k 2 >. For λ + = λ = 4 we set ϕ 21 (x) = k 1 sin 2x and ϕ 22 (x) = k 2 sin 2x, where k 1, k 2 >. ϕ 21 ϕ 22 Figure 2. Solutions corresponding to Σ 2 If m is odd, then Σ m = Σ m1 Σ m2 and it corresponds only one function ϕ m1 od norm 1 to the point λ +, λ Σ m1, one function ϕ m2 of norm 1 to the point λ +, λ Σ m2, respectively. For m = 3, λ + > λ, λ + > λ we have ϕ 31 (x) k 1 λ sin( λ +x), x, π/ λ +, = k 1 λ + sin( λ (x π/ λ +)), x π/ λ +, π/ λ + + π/ λ, k 1 λ sin( λ +(x π/ λ + π/ λ )), x π/ λ + + π/ λ, π, where k 1 >. ϕ 32 (x) k 2 λ+ sin( λ x), x, π/ λ, = k 2 λ sin( λ + (x π/ λ )), x π/ λ, π/ λ + π/ λ +, k 2 λ+ sin( λ (x π/ λ π/ λ + )), x π/ λ + π/ λ +, π, where k 2 >. For λ + = λ = m 2 we set ϕ m1 (x) = k 1 sin mx, and ϕ m2 (x) = k 2 sin mx, where k 1, k 2 > and from the condition (1.4) we obtain f(x) sin mx dx < G+ (x)(sin mx) + G (x)(sin mx) dx

4 4 P. TOMICZEK EJDE-25/94 ϕ 31 ϕ 32 Figure 3. Solutions corresponding to Σ 3 and f(x)( sin mx) dx < Hence it follows G (x)(sin mx) + G + (x)(sin mx) dx G+ (x)( sin mx) + G (x)( sin mx) dx. < f(x) sin mx dx < G+ (x)(sin mx) + G (x)(sin mx) dx. We obtained the potential Landesman-Lazer type condition (see 6). (1.6) Remark 1.3. We have v, sin mx = v (x)(sin mx) dx = m 2 v(x) sin mx dx v H (H is a Sobolev space defined below). Since and from the definition of the functions ϕ m1, ϕ m2 (see remark 1.2) it follows ϕ m1, sin mx > and ϕ m2, sin mx <. (1.7) 2. Preinaries Notation. We shall use the classical spaces C(, π), L p (, π) of continuous and measurable real-valued functions whose p-th power of the absolute value is Lebesgue integrable, respectively. H is the Sobolev space of absolutely continuous functions u: (, π) R such that u L 2 (, π) and u()=u(π)=. We denote by the symbols, and 2 the norm in H, and in L 2 (, π), respectively. We denote, the pairing in the space H. By a solution of (1.1) we mean a function u C 1 (, π) such that u is absolutely continuous, u satisfies the boundary conditions and the equations (1.1) holds a.e. in (, π). Let I : H R be a functional such that I C 1 (H, R) (continuously differentiable). We say that u is a critical point of I, if I (u), v = for all v H. We say that γ is a critical value of I, if there is u H such that I(u ) = γ and I (u ) =. We say that I satisfies Palais-Smale condition (PS) if every sequence (u n ) for which I(u n ) is bounded in H and I (u n ) (as n ) possesses a convergent subsequence.

5 EJDE-25/94 POTENTIAL LANDESMAN-LAZER TYPE CONDITION 5 We study (1.1) by the use of a variational method. More precisely, we look for critical points of the functional I : H R, which is defined by I(u) = 1 2 (u ) 2 λ + (u + ) 2 λ (u ) 2 dx G(x, u) fu dx. (2.1) Every critical point u H of the functional I satisfies u v (λ + u + λ u )v dx g(x, u)v fv dx = for all v H. Then u is also a weak solution of (1.1) and vice versa. The usual regularity argument for ODE yields immediately (see Fučík 2) that any weak solution of (1.1) is also a solution in the sense mentioned above. We will use the following variant of the Saddle Point Theorem (see 4) which is proved in Struwe 5, Theorem 8.4. Theorem 2.1. Let S be a closed subset in H and Q a bounded subset in H with boundary Q. Set Γ = {h : h C(H, H), h(u) = u on Q}. Suppose I C 1 (H, R) and (i) S Q =, (ii) S h(q), for every h Γ, (iii) there are constants µ, ν such that µ = inf u S I(u) > sup u Q I(u) = ν, (iv) I satisfies Palais-Smale condition. Then the number defines a critical value γ > ν of I. γ = inf sup h Γ u Q I(h(u)) We say that S and Q link if they satisfy conditions (i), (ii) of the theorem above. We denote the first integral in the functional I by J(u) = Now we present a few results needed later. (u ) 2 λ + (u + ) 2 λ (u ) 2 dx. Lemma 2.2. Let ϕ be a solution of (1.2) with λ +, λ Σ, λ + λ. We put u = aϕ + w, a, w H. Then the following relation holds (w ) 2 λ + w 2 dx J(u) (w ) 2 λ w 2 dx. (2.2) Proof. We prove only the right inequality in (2.2), the proof of the left inequality is similar. Since ϕ is a solution of (1.2) we have and ϕ w dx = (ϕ ) 2 dx = λ+ ϕ + w λ ϕ w dx for w H (2.3) λ+ (ϕ + ) 2 + λ (ϕ ) 2 dx. (2.4)

6 6 P. TOMICZEK EJDE-25/94 By (2.3) and (2.4), we obtain J(u) = ((aϕ + w) ) 2 λ + ((aϕ + w) + ) 2 λ ((aϕ + w) ) 2 dx = (aϕ ) 2 + 2aϕ w + (w ) 2 (λ + λ )((aϕ + w) + ) 2 λ (aϕ + w) 2 dx = (λ + λ )(aϕ + ) 2 + λ (aϕ) 2 + 2a((λ + λ )ϕ + + λ ϕ)w + (w ) 2 (λ + λ )((aϕ + w) + ) 2 λ ((aϕ) 2 + 2aϕw + w 2 ) dx = {(λ + λ ) (aϕ + ) 2 + 2aϕ + w ((aϕ + w) + ) 2 + (w ) 2 λ w 2} dx. (2.5) For the function (aϕ + ) 2 + 2aϕ + w ((aϕ + w) + ) 2 in the last integral in (2.5) we have (aϕ + ) 2 + 2aϕ + w ((aϕ + w) + ) 2 ((aϕ + w) + ) 2 ϕ < = w 2 ϕ, aϕ + w aϕ + (aϕ + + w + w) ϕ, aϕ + w <. By the assumption λ + λ, we obtain the assertion of the Lemma 2.2. Remark 2.3. It follows from the previous proof that we obtain the equality J(u) = (w ) 2 λ w 2 dx in (2.2) if aϕ + w when ϕ <, and w = when ϕ. Consequently, if the equality holds and if w in span{sin x,..., sin kx}, k N, then w =. 3. Main result Theorem 3.1. Under the assumptions (1.3), (1.4), and (1.5), Problem (1.1) has at least one solution in H. Proof. First we suppose that m is even. We shall prove that the functional I defined by (2.1) satisfies the assumptions in Theorem 2.1. Let ϕ m1, ϕ m2 be the normalized solutions of (1.2) described above (see Remark 1.2). Let H be the subspace of H spanned by functions sin x,..., sin(m 1)x. We define V V 1 V 2 where V 1 = {u H : u = a 1 ϕ m1 + w, a 1, w H }, V 2 = {u H : u = a 2 ϕ m2 + w, a 2, w H }. Let K >, L > then we define Q Q 1 Q 2 where Q 1 = {u V 1 : a 1 K, w L}, Q 2 = {u V 2 : a 2 K, w L}. Let S be the subspace of H spanned by functions sin(m + 1)x, sin(m + 2)x,....

7 EJDE-25/94 POTENTIAL LANDESMAN-LAZER TYPE CONDITION 7 Next, we verify the assumptions of Theorem 2.1. We see that S is a closed subset in H and Q is a bounded subset in H. (i) Firstly we note that for z H S we have z, sin mx =. We suppose for contradiction that there is u Q S. Then u S = u, sin mx u Q = a i ϕ mi + w, sin mx w H = a i ϕ mi, sin mx i = 1, 2. From previous equalities and inequalities (1.7) it follows that a i =, i = 1, 2 and u = w. For u = w Q we have u = L > and we obtain a contradiction with u H S = {o}. (ii) We prove that H = V S. We can write a function h H in the form h = m 1 i=1 b i sin ix + b m sin mx + = h + b m sin mx + h, b i R, i=m+1 b i sin ix i N. The inequalities (1.7) yield that there are constants b m1, b m2 > such that sin mx = b m1 (ϕ m1 ϕ m1 ϕ m1 ) and sin mx = b m2 (ϕ m2 ϕ m2 ϕ m2 ). Hence we have for b m, h = h + b m b m1 (ϕ m1 ϕ m1 ϕ m1 ) + h H {}}{{}}{ = ( h b m b m1 ϕ m1 + b m b m1 ϕ m1 ) + ( h b m b m1 ϕ m1 ). }{{}}{{} V S Similarly for b m, h = h + b m b m2 (ϕ m2 ϕ m2 ϕ m2 ) + h H {}}{{}}{ = ( h b m b m2 ϕ m2 + b m b m2 ϕ m2 ) + ( h b m b m2 ϕ m2 ). }{{}}{{} V S We proved that H is spanned by V and S. The proof of the assumption S h(q) h Γ is similar to the proof in 5, example 8.2. Let π : H V be the continuous projection of H onto V. We have to show that π(h(q)). For t, 1, u Q we define h t (u) = tπ(h(u)) + (1 t)u. The function h t defines a homotopy of h = id with h 1 = π h. Moreover, h t Q = id for all t, 1. Hence the topological degree deg(h t, Q, ) is well-defined and by homotopy invariance we have deg(π h, Q, ) = deg(id, Q, ) = 1. Hence π(h(q)), as needed. (iii) Firstly, we note that by assumption (1.5), one has u G(x, u) fu u 2 dx =. (3.1) First we show that the infimum of functional I on the set S is a real number. We prove for this that I(u) = for all u S (3.2) u

8 8 P. TOMICZEK EJDE-25/94 and I is bounded on bounded sets. Because of the compact imbedding of H into C(, π) ( u C(,π) c 1 u ), and of H into L 2 (, π) ( u 2 c 2 u ), and the assumption (1.5) one has I(u) = 1 (u ) 2 λ + (u + ) 2 λ (u ) 2 dx G(x, u) fu dx 2 1 ( π u 2 + λ + u λ u 2 2 2) + ( p + f ) u dx 1 ( u 2 + λ + c 2 u λ c 2 u 2) + ( p 1 + f 1 ) c 1 u. 2 Hence I is bounded on bounded subsets of S. To prove (3.2), we argue by contradiction. We suppose that there is a sequence (u n ) S such that u n and a constant c 3 satisfying For u S the following relation holds u 2 = inf n I(u n) c 3. (3.3) (u ) 2 dx (m + 1) 2 u 2 dx = (m + 1) 2 u 2 2. (3.4) The definition of I, (3.1), (3.3) and (3.4) yield I(u n ) ((m + 1) 2 λ + ) u + n ((m + 1) 2 λ ) u n 2 2 inf inf n u n 2 n 2 u n 2. (3.5) It follows from (3.5) and (1.3) that u n 2 2/ u n 2 and from the definition of I and (3.1) we have I(u n ) inf u n u n 2 = 1 2 a contradiction to (3.5). We proved that there is µ R such that inf u S I(u) = µ. Second we estimate the value I(u) for u Q. We remark that u Q can be either of the form Kϕ m + w, with w L or of the form a i ϕ mi, with a i K, w = L (i = 1, 2). We prove that sup I(Kϕ m + w) = sup I(u) = for u Q. (3.6) (K+L) u For (3.6), we argue by contradiction. Suppose that (3.6) is not true then there are a sequence (u n ) Q such that u n and a constant c 4 satisfying sup I(u n ) c 4. (3.7) n Hence, it follows 1 π (u sup n) 2 λ + (u + n ) 2 λ (u n ) 2 G(x, u n ) fu n n 2 u n 2 dx u n 2 dx. (3.8) Set v n = u n / u n. Since dim Q < there is v Q such that v n v strongly in H (also strongly in L 2 (, π)). Then (3.8) and (3.1) yield 1 2 (v ) 2 λ + (v + )2 λ (v )2 dx. (3.9)

9 EJDE-25/94 POTENTIAL LANDESMAN-LAZER TYPE CONDITION 9 Let v = a ϕ m + w, a R +, w H. It follows from Lemma 2.2 that π (v ) 2 λ + (v + )2 λ (v )2 dx (w ) 2 λ (w ) 2 dx. (3.1) For w H we have (w ) 2 λ w 2 dx ((m 1) 2 λ ) w 2 dx. (3.11) Since (m 1) 2 < λ (see Remark 1.1) then (3.9), (3.1) and (3.11) yield (v ) 2 λ + (v + )2 λ (v )2 dx = ((m 1) 2 λ ) w 2 2 =. Hence we obtain w = and v = a ϕ m, v = 1. Now we divide (3.7) by u n then 1 π (u sup n) 2 λ + (u + n ) 2 λ (u n ) 2 G(x, u n ) fu n dx dx. n 2 u n u n (3.12) By Lemma 2.2 the first integral in (3.12) is less then or equal to. Hence it follows G(x, u n ) + fu π n G(x, un ) sup dx = sup v n + fv n dx. n u n n u n (3.13) Because of the compact imbedding H C(, π), we have v n a ϕ m in C(, π) and we get u n(x) = n { + for x (, π) such that ϕ m (x) >, for x (, π) such that ϕ m (x) <. We note that from (1.5) it follows that p(x) G + (x), G (x) p(x) for a.e. x (, π). We obtain from Fatou s lemma and (3.13) f(x)ϕ m (x) dx G+ (x)(ϕ m (x)) + G (x)(ϕ m (x)) dx, a contradiction to (1.4). We proved that by choosing K, L sufficiently large there is ν R such that sup u Q I(u) = ν < µ. Then Assumption (iii) of Theorem 3.1 is verified. (iv) Now we show that I satisfies the Palais-Smale condition. First, we suppose that the sequence (u n ) is unbounded and there exists a constant c 5 such that and 1 2 (u n ) 2 λ + (u + n ) 2 λ (u n ) 2 dx G(x, un ) fu n dx c5 (3.14) n I (u n ) =. (3.15)

10 1 P. TOMICZEK EJDE-25/94 Let (w k ) be an arbitrary sequence bounded in H. It follows from (3.15) and the Schwarz inequality that n u n w k (λ + u + n λ u π n ) w k dx g(x, un )w k fw k dx k = I (u n ), w k (3.16) n k n I (u n ) w k =. k Put v n = u n / u n. Due to compact imbedding H L 2 (, π) there is v H such that (up to subsequence) v n v weakly in H, v n v strongly in L 2 (, π). We divide (3.16) by u n and we obtain and n,k i,k We subtract equalities (3.17) and (3.18) we have n,i,k v n w k (λ + v + n λ v n ) w k dx = (3.17) v i w k (λ + v + i λ v i ) w k dx =. (3.18) (v n v i)w k (λ + (v n + v + i ) λ (vn v i ) ) w k dx =. (3.19) Because (w k ) is a arbitrary bounded sequence we can set w k = v n v i in (3.19) and we get v n v i 2 λ+ (v n + v + i ) λ (vn v i )(v n v i ) dx =. (3.2) n,i Since v n v strongly in L 2 (, π) the integral in (3.2) converges to and then v n is a Cauchy sequence in H and v n v strongly in H and v = 1. It follows from (3.17) and the usual regularity argument for ordinary differential equations (see Fučík 2) that v is the solution of the equation v + λ + v + λ v =. From the assumption λ +, λ Σ m it follows that v = a ϕ m, a >. We set u n = a n ϕ m +û n, where a n, û n H S. We remark that u = u + u and using (2.3) in the first integral in (3.16) we obtain I 1 = = = = (an ϕ m + û n ) w k (λ + u + n λ u n )w k dx an ϕ mw k + (û n ) w k ((λ + λ )u + n + λ u n ) w k dx an (λ + ϕ + m λ ϕ m)w k + (û n ) w k ((λ + λ )u + n + λ u n ) w k dx { an (λ + λ )ϕ + m + λ ϕ m w k + (û n ) w k (λ + λ )(a n ϕ m + û n ) + } + λ (a n ϕ m + û n ) w k dx = (λ+ λ )(a n ϕ + m (a n ϕ m + û n ) + )w k + (û n ) w k λ û n w k dx. (3.21)

11 EJDE-25/94 POTENTIAL LANDESMAN-LAZER TYPE CONDITION 11 Similarly we obtain I 1 = (λ+ λ )(a n ϕ m (a n ϕ m + û n ) )w k + (û n ) w k λ + û n w k dx. Adding (3.21) and (3.22) and we have 2I 1 = (3.22) (λ + λ )( a n ϕ m a n ϕ m + û n )w k + 2(û n ) w k (λ + + λ )û n w k dx. (3.23) We set û n = u n + ũ n where u n H, ũ n S and we put in (3.23) w k = (u n ũ n )/ û n then we have 2I 1 = 1 û n (λ+ λ )( a n ϕ m a n ϕ m + u n + ũ n )(u n ũ n ) + 2 (u n) 2 2(ũ n) 2 (λ + + λ )(u 2 n ũ 2 n) dx. Hence 2I 1 1 ( (λ + λ ) u n + ũ n u n ũ n dx û n ) + 2 u n 2 2 ũ n 2 (λ + + λ )( u n 2 2 ũ n 2 2) = 1 ( (λ + λ ) u 2 n ũ 2 û n n dx + 2 u n 2 (λ + + λ ) u n ũ n 2 + (λ + + λ ) ũ n 2 2 The inequality a 2 b 2 max{a 2, b 2 }, (3.25) and (1.3) yield I 1 max { u n 2 λ u n 2 2, ũ n 2 + λ + ũ n 2 2 ). (3.24) (3.25) } 1 û n. (3.26) We note that the following relations hold u n 2 (m 1) 2 u n 2 2, ũ n 2 (m + 1) 2 ũ n 2 2. Hence from assumption (1.3) and (3.26) it follows that there is ε > such that I 1 ε max { u n 2, ũ n 2} 1 û n. (3.27) From (3.16), (3.27) it follows εmax n { un 2, ũ n 2} û n (g(x, un ) f) u n ũ n dx. (3.28) û n Now we suppose that û n. We note that û n 2 = u n 2 + ũ n 2, we divide (3.28) by û n and using (1.5) we have ε { 2 εmax un 2, ũ n 2} g(x, u n ) f u n ũ n n û n 2 dx (3.29) û n û n a contradiction to ε >. This implies that the sequence (û n ) is bounded. We use (2.2) from Lemma 2.2 with w = û n and we obtain (û n ) 2 λ + û 2 π n dx J(un ) (û n ) 2 λ û 2 n dx. Hence J(u n ) n u n = n (u n ) 2 λ + u 2 n λ u 2 n dx =. (3.3) u n

12 12 P. TOMICZEK EJDE-25/94 We divide (3.14) by u n and (3.3) yield n G(x, u n ) + fu n u n dx = (3.31) and using Fatou s lemma in (3.31) we obtain a contradiction to (1.4). This implies that the sequence (u n ) is bounded. Then there exists u H such that u n u in H, u n u in L 2 (, π) (up to subsequence). It follows from the equality (3.16) that n,i,k (un u i ) w k λ + (u + n u + i ) λ (u n u i )w k dx =. (3.32) We put w k = u n u i in (3.32) and the strong convergence u n u in L 2 (, π) and (3.32) imply the strong convergence u n u in H. This shows that the functional I satisfies Palais-Smale condition and the proof of Theorem 3.1 for m even is complete. Now we suppose that m is odd. We have λ +, λ Σ m2 and the nontrivial solution ϕ m2 of (1.2) corresponding to λ +, λ. Then there is k > such that λ + k, λ k Σ m1 and solution ϕ m1 corresponding to λ + k, λ k = λ +, λ (see Remark 1.2). We define the sets Q and S like for m even and the proof of the steps (i), (ii) of theorem 3.1 is the same. In the step (iii) we change inequality (3.1) if v = a ϕ m1 as it follows (v ) 2 λ + (v + )2 λ (v )2 dx = k π (v ) 2 (λ + k)(v + )2 (λ k)(v )2 dx k v 2 dx v 2 dx + (w ) 2 λ (w ) 2 dx. (3.33) Then by (3.9), (3.33) and (3.11) we obtain k v2 dx =, a contradiction to v = 1. The proof of the step (iv) is similar to the prove for m even. The proof of the theorem 3.1 is complete. References 1 A. K. Ben-Naoum, C. Fabry, & D. Smets; Resonance with respect to the Fučík spectrum, Electron J. Diff. Eqns., Vol. 2(2), No. 37, pp S. Fučík; Solvability of Nonlinear Equations and Boundary Value problems, D. Reidel Publ. Company, Holland M. Cuesta, J. P. Gossez; A variational approach to nonresonance with respect to the Fučík spectrum, Nonlinear Analysis 5 (1992), P. Rabinowitz; Minmax methods in critical point theory with applications to differential equations, CBMS Reg. Conf. Ser. in Math. no 65, Amer. Math. Soc. Providence, RI., (1986). 5 M. Struwe; Variational Methods, Springer, Berlin, (1996). 6 P. Tomiczek; The generalization of the Landesman-Lazer conditon, Electron. J. Diff. Eeqns., Vol. 21(21), No. 4, pp Petr Tomiczek Department of Mathematics, University of West Bohemia, Univerzitní 22, Plzeň, Czech Republic address: tomiczek@kma.zcu.cz