RNDr. Petr Tomiczek CSc.

Transcription

1 HABILITAČNÍ PRÁCE RNDr. Petr Tomiczek CSc. Plzeň 26

2

3 Nonlinear differential equation of second order RNDr. Petr Tomiczek CSc. Department of Mathematics, University of West Bohemia

4

5 5 Contents 1 Introduction 9 2 Fredholm alternative Saddle point theorem and the Fredholm alternative Fredholm alternative for a nonlinear equation Fredholm alternative for a linear equation Potential Landesman-Lazer Main result Jumping nonlinearities Main result The solution in regions of type II Main result Duffing equation Main result

6

7 7 Preface This thesis is devoted to the investigation nonlinear differential equations of second order. We use variational methods based on the Saddle point theorem to prove an existence of solution of such equations. The work is divided into six sections. In section 1 we give a motivation problem investigate in this work and we introduce some necessary notions. Each from sections 2-5 being a research paper to be published in some international journals or in proceedings of international conference. In the sixth section we generalize our problem and we investigate so called Duffing equation. I was partially supported by the research plans of the Ministry of education, Youth and Sports of the Czech Republic MSM

8 8 List of notation M N R R n H M N X Y dim X closure of the set M set of all positive integers set of all real numbers real space of the dimension n orthogonal complement of the space H orthogonality of the set M and N algebraic direct sum of the spaces X and Y dimension of the linear space X C(a, b) space of all continuous function on the interval (a, b) L p (M) Lebesgue space on M W k,p Sobolev space on M D(A) R(A) N(A) J u u n u u n u domain of the map A range of the map A null space (kernel) of the map A Frechet derivative of the functional J Laplace operator of the map u strong convergence of the sequence {u n } to the element u weak convergence of the sequence {u n } to the element u

9 9 1 Introduction We will investigate the following type of the nonlinear problem u (x) + m 2 u(x) + g(x, u(x)) = f(x), x (, π), (1.1) u() = u(π) =, where f L 1 (, π), m N and g : (, π) R R is a Caratheodory type function, i.e. g(, s) is measurable for all s R and g(x, ) is continuous for a.e. x (, π). We 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. We set H = W 1,2 (, π) is the Sobolev space of absolutely continuous functions u: (, π) R such that u L 2 (, π), u() = u(π) =. We denote (.,.) a scalar product on the Hilbert space H and for u, v H we define a operator T : H H and an element f H by (u, v) = u v dx, (T u, v) = uv dx, (f, v) = 1 f v dx. m 2 We remark that the operator T is a linear, symmetric, positive and compact operator (with respect to the compact imbedding H into L 2 (, π)). We start with the linear problem for g =. Now we can write a weak formulation of the equation (1.1), i.e. in the form We put λ = 1 m 2 and for f u v dx m 2 uv dx = f v dx 1 (u, v) (T u, v) = (f, v). m2 v H and we denote A = λi T then we have the equation Au = f (1.2) Au =. (1.3) We denote N(A) = {u H u is a solution of (1.3) }.

10 1 We remark that a number λ is called an eigenvalue of the operator T if there exists ϕ λ such that Aϕ λ = λϕ λ T ϕ λ =. Such an element ϕ λ N(A) is called an eigenvector of T. It is well known (see Zeidler [19]) that the operator A satisfies Fredholm alternative: The equation Au = f has a solution for f H if and only if (f, v) = v N(A). In the second section we prove a general saddle Fredholm alternative: Let A : D(A) H, D(A) = H be a linear self adjoint operator with a null space N(A). Let Ĥ, H H be subspaces of H and H = Ĥ N(A) H. If there exist c 3 >, c 4 > such that (Au, u) c 3 u 2 u D(A) Ĥ, (Au, u) c 4 u 2 u D(A) H, then equation Au = f has a solution if and only if (f, v) = v N(A). We show that the classical Fredholm alternative is a consequence of the saddle Fredholm alternative. In the subsection 2.1 we return to the nonlinear problem (1.1) with g. We define a nonlinear operator (S(u), v) = g(x, u)v dx. We write a weak formulation of the equation (1.1) in the form λ(u, v) (T u, v) + (S(u), v) = (f, v) v H. We find the condition on the operator S to prove the existence of a solution of a general problem Au + S(u) = f. We formulate a nonlinear Fredholm alternative (see J.Nečas [1]): Let the equation Au = has only a trivial solution then for each f the equation Au + S(u) = f has at least one solution. We suppose S(u) < min{c 3, c 4 } u γ where γ (, 1 and we prove the nonlinear Fredholm alternative using the Saddle Point Theorem. We also show that the saddle Fredholm alternative is a consequence of the Saddle Point Theorem.

11 11 We note that the operator A generated by problem (1.1) has not a trivial null space. (A(sin mx) = 1 π [(sin m 2 mx) v (sin mx)v] dx =.) It follows from the Fredholm alternative that if g = then boundary value problem (1.1) is solvable for any f H satisfying f(x) sin mx dx =. In the third section we suppose g. We define and G(x, s) = s g(x, t) dt G + (x) = lim inf s + G(x, s) s, G (x) = lim sup s G(x, s) s We prove that (1.1) has at least one solution provided that the following potential Landesman-Lazer type condition holds < [G (x)(sin mx) + G + (x)(sin mx) ] dx (1.4) f(x) sin mx dx < [G + (x)(sin mx) + G (x)(sin mx) ] dx.. In the fourth section 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), (1.5) u() = u(π) =. Here u ± = max{±u, }, λ +, λ R. For g and f the problem (1.5) becomes u (x) + λ + u + (x) λ u (x) =, x (, π), (1.6) u() = u(π) =.

12 12 We define Σ = { (λ +, λ ) R 2 : (1.6) has a nontrivial solution }. This set is called the Fučík spectrum (see [7]) and it can be expressed as Σ = j=1 Σ j where Σ 1 = {(λ +, λ ) R 2 : (λ + 1)(λ 1) = }, ( 1 Σ 2i = {(λ +, λ ) R 2 : i + 1 ) = 1}, λ+ λ Σ 2i+1 = Σ 2i+1,1 Σ 2i+1,2 where ( 1 Σ 2i+1,1 = {(λ +, λ ) R 2 : i + 1 ) λ+ λ + 1 λ+ = 1}, ( 1 Σ 2i+1,2 = {(λ +, λ ) R 2 : i + 1 ) + 1 = 1}. λ+ λ λ We suppose that (λ +, λ ) Σ m, if m N is even or (λ +, λ ) Σ m2, if m N is odd (1.7) and λ < λ + < (m + 1) 2. We denote ϕ m a nontrivial solution of (1.6) corresponding to (λ +, λ ). 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. (1.8) We also suppose that the nonlinearity g is bounded, i.e. p(x) L 1 (, π) : g(x, s) p(x) for a.e. x (, π), s R (1.9) and we prove the solvability of (1.5). The fifth section is devoted the nonlinear boundary-value problem u (x) + αu + (x) βu (x) = f(x), x (, π), (1.1) x() = x(π) =, where α, β R.

13 13 We define two types of regions between the curves of Σ: Type (I) R 1 which consists of regions between the curves Σ 2i and Σ 2i+1, i N. Type (II) R 2 consists of regions between the curves Σ 2i+1,1 and Σ 2i+1,2, i N. If (α, β) R 1 one can solve (1.1) for arbitrary f L 2 (, π) while this is not so for regions R 2 (see [3]). We suppose that the point (α, β) R 2 is between the curves Σ 2i+1,1 and Σ 2i+1,2, α > β, and k > such that λ + = α+k, λ = β+k, (λ +, λ ) Σ 2i+1,2 and λ + < (2i+2) 2. We denote ϕ 2 the solution of (1.6) belonging to (λ +, λ ). In this work we obtain existence results for (1.1) with right hand side f = cϕ 2 + ϕ 2, c > and ( 1 β (2i) + 1 ) (ϕ 2 (2i + 2) 2 2 ) 2 dx< α) where ( ϕ 2 ϕ 2 dx = 1 λ + α 1 ) (cϕ (2i + 2) 2 2 ) 2 dx. α) In the last section we investigate the Duffing equation w (x) + 2cw(x) + g(x, w(x)) = f(x), x (, π), (1.11) w() = w(π) =. We show that we can apply the results of the previous sections on the equation (1.11) and for simplicity we prove the existence theorem under the assumption G(x, s) (c ε) s 2, ε >, s R.

14 14 2 Fredholm alternative We denote H a real separable Hilbert space with a scalar product (.,.) and a norm.. We have a linear operator A: D(A) H with a dense domain D(A) (D(A)=H), a range R(A) and a null space N(A)={v D(A): Av =}. Recall that a linear operator A: D(A) H is said to be bounded if there exists a constant c > such that Au c u u H. An operator A is said to be continuous if u n u in H implies Au n Au in H. For linear operators, these two notions are equivalent. A densely defined linear operator A is said to be closed if {u n } D(A), u n u, Au n z u D(A), Au = z. Note that A continuous implies A closed but not conversely. We define an adjoint operator A : D(A ) H, D(A ) H to an (unbounded) densely defined operator A such that v D(A ) and A v = z, iff (Au, v) = (u, z) u D(A). If A = A, then the operator A is called self adjoint. We denote M = {v H : (u, v) = u M}. If M is a closed subspace of H then the orthogonal decomposition theorem yields H = M M. It is not hard to show that N(A ) is a closed subspace of H and H = N(A ) N(A ). Lemma 2.1 Let A : D(A) H be a densely defined closed linear operator then H = N(A ) R(A). (2.1) Proof : We prove an equivalent assertion R(A) = N(A ). Firstly we suppose that f R(A). Then =(Au, f)=(u, A f) u D(A). Hence it follows f N(A ) and R(A) N(A ). Secondly we suppose f N(A ), i.e. (u, A f) = u H. In particular (u, A f) = (Au, f) = u D(A) f R(A). We have N(A ) R(A). The two inclusions together imply R(A) = N(A ) and the continuity of scalar product yields (R(A) ) = R(A) = N(A ).

15 15 Lemma 2.2 (closed range) Let A : D(A) H be a densely defined closed linear operator and c 1 > such that Au c 1 u u D(A) N(A). (2.2) Then A has a closed range R(A). Proof : Let {f n } R(A) and f n f H, then there exists {v n } D(A) such that Av n = f n. Because the null space of any closed linear operator is a closed subspace of H, we have H = N(A) N(A). Now v n can be written uniquely in the form v n = w n + u n, w n N(A), u n N(A). Then f n f m = Av n Av m = Au n Au m c 1 u n u m. But then {u n } is a Cauchy sequence with a limit point u N(A). And since A is closed we have u D(A) Au = f. Then f R(A) and it follows that R(A) is closed. Lemma 2.3 (Fredholm alternative for a positive definite operator) Let A : D(A) H be a linear self adjoint operator and c 2 > such that then equation (Au, u) c 2 u 2 u D(A) N(A), (2.3) has a solution for every f H if and only if Au = f (2.4) (f, v) = v N(A). (2.5) Proof : We remark that any linear self adjoint operator A = A is closed. An assumption (2.3) and the Schwarz s inequality yield Au u (Au, u) c 2 u 2, hence and from the previous lemma 2.2 it follows that range R(A) is closed. Using lemma 2.1 we get R(A) = N(A ). Since A is the self adjoint operator, so R(A) = N(A ) = N(A) and f R(A) if and only if (f, v) = v N(A).

16 16 Theorem 2.4 (saddle Fredholm alternative) Let A : D(A) H be a linear self adjoint operator, Ĥ, H H are subspaces of H and H =Ĥ N(A) H. If there exist c 3 >, c 4 > such that then equation (Au, u) c 3 u 2 u D(A) Ĥ, (2.6) (Au, u) c 4 u 2 u D(A) H, (2.7) Au = f has a solution for f H if and only if (f, v) = v N(A). Proof : We set u = û + u + ũ, for û Ĥ, u N(A), ũ H. Then (Au, ũ û)=(aû + Aũ, ũ û)=(aũ, ũ) + (Aû, ũ) (Aũ, û) (Aû, û). Since A is self adjoint we get (Aû, ũ) (Aũ, û)=(aû, ũ) (ũ, A û) = (Aû, ũ) (Aû, ũ) =. Hence from the Schwarz inequality and the inequalities (2.6), (2.7) it follows Au ũ û (Au, ũ û )=(Aũ, ũ ) (Aû, û ) c 4 ũ 2 + c 3 û 2. We remark that ũ û 2 = ũ 2 + û 2 for ũ Ĥ, û H and similarly u 2 = ũ + û 2 = ũ 2 + û 2 for u N(A). We set c 2 = min{c 3, c 4 } then Au u = Au ũ û c 2 u 2 u D(A) N(A). To finish proof we use the previous lemma 2.3. Remark It follows from the previous proof that we can suppose that the closed operator A satisfies equality (Aû, ũ) (Aũ, û) = instead the assumption A is self adjoint operator. To prove that the classical Fredholm alternative is a consequence of the theorem 2.4 we need the following theorem (see Dunford-Schwartz [5]).

17 17 Theorem 2.5 (Courant-Fisher Principle) Let the operator T : H H be a linear, compact, self-adjoint and positive operator on a real separable Hilbert space H. Then all eigenvalues of T are positive reals and there exists an orthonormal basis of H which consists of eigenvectors of T. Moreover, if λ 1 λ 2 λ 3 λ n >, denote the eigenvalues of T repeated according to (finite) multiplicity and ϕ 1, ϕ 2,..., ϕ n,... corresponding eigenvectors, then λ 1 = max u H (T u, u) u 2, { (T u, u) } λ n+1 = max (u, ϕ u H u 2 1 ) = = (u, ϕ n ) =, where ϕ 1, ϕ 2,..., ϕ n are the previously obtained eigenvectors and lim n λ n =. Now we formulate the Fredholm alternative for a compact operator (see Zeidler [19]) and we give its new proof. Lemma 2.6 (Fredholm alternative with a compact operator) Let λ R, λ 1 > λ > and T : H H be a linear compact self adjoint positive operator, then equation has a solution if and only if (λi T )u=f (2.8) (f, v) = v N(λI T ). (2.9) Proof : We remark that the linear compact operator T is also continuous and closed. Let ϕ n is an eigenvector corresponding to the eigenvalue λ n of T. We know from the theorem (2.5) that a sequence {ϕ n } create an orthonormal basis of H. We denote Ĥ a subspace of H spanned by ϕ n corresponding to λ n > λ and H a subspace of H spanned by ϕ n corresponding to λ n < λ. Then H = Ĥ N(A) H and dim Ĥ <, dim H =. We define

18 18 λ m = min{λ n λ n > λ} and λ M = max{λ n λ n < λ} (λ m >, λ M > ). We denote A = λi T then from the Courant-Fischer s theorem 2.5 it follows (Au, u) (λ λ m ) u 2 for u Ĥ, (2.1) (Au, u) (λ λ M ) u 2 for u H. (2.11) We set c 3 =λ λ m < and c 4 =λ λ M > and we see that the operator A satisfies the assumptions (2.6), (2.7) of the saddle Fredholm alternative 2.4. Hence it follows that the equation Au = (λi T )u =f has a solution if and only if (f, v) = v N(A) = N(λI T ). 2.1 Saddle point theorem and the Fredholm alternative Let J : H R be a functional such that J C 1 (H, R) (continuously differentiable). We say that u is a critical point of J, if J (u ) v = for all v H. J is a Frechet derivative of J. We say that J satisfies Palais-Smale condition (PS) if every sequence {u n } for which J(u n ) is bounded in H and J (u n ) (as n ) possesses a convergent subsequence. We use the Saddle Point Theorem which is proved in Rabinowitz [11]. Theorem 2.7 (Saddle Point Theorem) Let H = Ĥ H, dim Ĥ < and dim H =. Let J : H R be a functional such that J C 1 (H, R) and (a) there exists a bounded neighborhood D of in Ĥ and a constant α such that J/ D α, (b) there is a constant β > α such that J/ H β. (c) J satisfies Palais-Smale Condition (PS). Then functional J has a critical point in H.

19 Fredholm alternative for a nonlinear equation We again suppose that A : D(A) H be a densely defined linear self adjoint operator, Ĥ, H H are subspaces of H, dim Ĥ <, H = Ĥ N(A) H and A satisfies the inequalities (2.6), (2.7). We take < c 5 < min{c 3, c 4 } and we suppose that there is F s : H R such that F s = S, S : H H be a compact continuous operator and We define a functional J : H H S(u) < c 5 u γ where γ (, 1. (2.12) J(u) = 1 2 (Au, u) + F s(u) (f, u) and we find a critical point u of the functional J, i.e. J (u ) =. We have for all v H (J (u ), v) = (Au, v) + (S(u ), v) (f, v) = Au + S(u ) = f. Now we can formulate the nonlinear Fredholm alternative (see J.Nečas [1]) Lemma 2.8 Let the equation Au = has only a trivial solution. Then for each f the equation Au + S(u) = f has at least one solution. Proof : We verify that the functional J satisfies assumptions of the Saddle Point Theorem 2.7. It is ease to see that J C 1 (H, R) and the inequality (2.12) yields F s (u) F s () = 1 (S(tu), u) dt 1 c 5 tu γ u dt = c 5 γ+1 u γ+1. (2.13) (a) We take an arbitrary α R. We have for u Ĥ the inequality (2.6) and using (2.13) we get J(u)= 1 2 (Au, u)+f s(u) (f, u) c 3 2 u 2 + c 5 γ+1 u γ+1 + f u +F s (). Hence lim J(u) = and there exists a bounded neighborhood D u of in Ĥ such that J/ D α.

20 2 (b) For u H we use the inequalities (2.7) and we have J(u) c 4 2 u 2 c 5 γ+1 u γ+1 f u + F s (). Hence lim J(u) = and the functional J is bounded from below u on H. Then there are constants α, β, β >α and a bounded neighborhood D of such that J/ H β and J/ D α. (c) We prove that J satisfies Palais-Smale Condition. We take a sequence {u n } H and we suppose that there exists a constant c 6 such that and We use (2.6), (2.7), (2.13) and we have J(u n ) c 6 (2.14) lim J (u n ) =. (2.15) n J (u n ) = Au n + S(u n ) f Au n S(u n ) f min{c 3, c 4 } u n c 5 u n γ f. Since γ (, 1, <c 5 <min{c 3, c 4 } this implies that the sequence {u n } is bounded. Then there exists u H such that u n u, S(u n ) S(u ) in H (taking a subsequence if it is necessary) and It yields Au n + S(u n ) f Au + S(u ) f. A(u n u ) (2.3) u n u. This shows that J satisfies Palais-Smale condition and satisfies the assumptions of Saddle Point Theorem (2.7). We have been proved that J satisfies the assumptions of the Saddle Point Theorem 2.7 and there is u a critical point of the functional J.

21 Fredholm alternative for a linear equation We return to the linear problem (2.4) with the nontrivial null space, that means S =, N(A) {}. Then it follows from the proof of the previous lemma 2.8 that the functional J(u) = 1 (Au, u) (f, u) satisfies assumptions 2 of the Saddle Point Theorem on the space N(A) = Ĥ H and there is u N(A) such that (Au f, z)= z N(A). We set w = v + z for an arbitrary w H, where v N(A), z N(A), then (Au f, w) = (Au f, v + z) = (Au, v) (f, v) + (Au f, z) = (u, Av) (f, v) = (f, v). Hence it follows that the equation Au = f has a solution if and only if (f, v) = v N(A). We have been proved that the saddle Fredholm alternative (Theorem 2.4) is the consequence of the Saddle Point Theorem 2.7.

22 3 Potential Landesman-Lazer type condition We again investigate the nonlinear boundary-value problem u (x) + m 2 u(x) + g(x, u(x)) = f(x), x (, π), (3.1) u() = u(π) =, where f L 1 (, π), m N and g: (, π) R R is Caratheodory type function. If g(x, s) = g(s) is a continuous function such that g + = lim s g(s) and g = lim s g(s) exist and are finite numbers then Fučík [7, Th.6.4] proved that (3.1) has at least one solution provided that < [g (sin mx) + g + (sin mx) ] dx (3.2) f(x) sin mx dx < [g + (sin mx) + g (sin mx) ] dx. Intensive study of problem (3.1) started with the paper [8] by Landesman and Lazer in 197. Their result [8] has been generalized in various directions. There is defined in Drábek [4] g + (x) = lim inf g(x, s) and g (x) = lim sup g(x, s) s and the condition (3.2) is generalized as follow < s [g (x)(sin mx) + g + (x)(sin mx) ] dx (3.3) f(x) sin mx dx < [g + (x)(sin mx) + g (x)(sin mx) ] dx. Chun-Lei Tang [15] defined the function F (s) = 2G(s) g(s) and the constants s 22 F + = lim inf s + F (s), F = lim sup F (s) s

23 23 to prove that for m = 1, the problem (3.1) is solvable under the condition [F (sin x) + F + (sin x) ] dx < < f(x) sin x dx (3.4) [F + (sin x) + F (sin x) ] dx. In this section we generalize the Landesman-Lazer type conditions (3.3) and (3.4). By a solution of (3.1) we mean a function u C 1 (, π) such that u is absolutely continuous, u satisfies the boundary conditions and the equations (3.1) holds a.e. in (, π). Let H is the Sobolev space of absolutely continuous functions u : (, π) R such that u L 2 (, π) and u() = u(π) =. We study (3.1) by using of variational methods. More precisely, we look for critical points of the functional J : H R, which is defined by where J(u) = 1 2 [(u ) 2 m 2 u 2 ] dx G(x, s) = s g(x, t) dt. Every critical point u H of the functional J satisfies [u v m 2 uv] dx [G(x, u) fu] dx, (3.5) [g(x, u)v fv] dx = for all v H. (3.6) Then u is also a weak solution of (3.1) and vice versa. The usual regularity argument for ODE yields immediately (see Fučík [7]) that any weak solution of (3.1) is also the solution in the sense mentioned above. We will suppose that g satisfies the growth restriction g(x, s) c s + p(x) for a.e. x (, π) and for all s R, (3.7) with p L 1 (, π) and c >. Moreover, We define g(x, s) lim s s G + (x) = lim inf s + = uniformly for a.e. x (, π). (3.8) G(x, s) s, G (x) = lim sup s G(x, s) s. (3.9)

24 24 Assume that the following potential Landesman-Lazer type condition holds: < Set [G (x)(sin mx) + G + (x)(sin mx) ] dx (3.1) f(x) sin mx dx < [G + (x)(sin mx) + G (x)(sin mx) ] dx. { 2G(x, s) F (x, s) = g(x, s) s, s g(x, ) s =, F + (x) = lim inf F (x, s), F s + (x) = lim sup F (x, s). s We generalize conditions (3.3) and (3.4) by assuming that f and g satisfy at least one of the following set of inequalities: or < < [g (x)(sin mx) + g + (x)(sin mx) ] dx (3.11) f(x) sin mx dx < [g + (x)(sin mx) + g (x)(sin mx) ] dx, [F (x)(sin mx) + F + (x)(sin mx) ] dx (3.12) f(x) sin mx dx < [F + (x)(sin mx) + F (x)(sin mx) ] dx. We shall prove that for any g, g + (x) G + (x), F + (x) G + (x), G (x) g (x), G (x) F (x). (3.13) Therefore, the potential Landesman-Lazer type condition (3.1) is more general than the conditions (3.11) and (3.12).

25 25 Let us prove the first two inequalities in (3.13). The proof of the other two is similar. It follows from the definition of the function g + (x) that ε >, R > such that s > R, g(x, s) g + (x) ε for x (, π). Then for s > R and x (, π) we have G(x, s) s 1 s R g(x, t) dt + (g + (x) ε) s R. s Hence, the inequality g + (x) G + (x) follows. We use the argument from C.-L. Tang [15] to prove that F + (x) G + (x). It follows from (3.12) that F + (x) > for a.e. x (, π). For these x and arbitrary ε >, we set { F+ (x) ε if F + (x) R, F ε (x) = 1/ε if F + (x) =. Then for any ε > there exists K(x) > such that Since F (x, s) F ε (x) for all s K(x). ( G(x, τ) ) = g(x, τ) τ 2 2G(x, τ) τ = τ τ 2 τ 4 for any s > t > K(x), we have s t ( G(x, τ) ) dτ τ τ 2 s t F ε (x) τ 2 dτ ; F (x, τ) τ 2 F ε(x) τ 2 i.e. G(x, t) G(x, s) F t 2 s 2 ε (x)(t 1 s 1 ). Assumption (3.8) implies that G(x, s)/s 2. Since ε > was arbitrary, passing to the limit as s + in the last inequality, we obtain G(x, t)/t 2 F ε (x)/t and so G + (x) F + (x).

26 26 Example We suppose that the nonlinearities g i (x, s) = g i (s) (i = 1, 2, 3, 4) are continuous, odd and for all s ɛ > : g 1 (s) = 1, g 2 (s) = sin s, g 3 (s) = 4 π sin s, g 4(s) = 1 + sin s. We set M 6 = {f L 1 (, π) such that f satisfies (3.6)}, M 7 = {f L 1 (, π) such that f satisfies (3.7)}, M 8 = {f L 1 (, π) such that f satisfies (3.8)}. Then for g 1, one has M 6 = M 7 = M 8, g + = F + = G + = 1; for g 2, one has M 7 =, M 8 M 6, g + = < F + = 4 π 1 < G + = 2 π ; for g 3, one has M 8 =, M 7 M 6, F + = < g + = 4 π 1 < G + = 2 π ; for g 4, one has M 7 = M 8 =, M 6, g + = F + = < G + = Main result Theorem 3.1 Under the assumptions (3.7), (3.8), and (3.1), Problem (3.1) has at least one solution in H. Proof It is known that the spectrum of the linear problem u (x) + λu(x) =, x (, π), x() = x(π) = is the set C = {λ : λ = m 2, m N}. Let Ĥ be the subspace of H spanned by all eigenfunctions corresponding to the eigenvalues 1, 4,..., m 2 and let H be the subspace of H spanned by all eigenfunctions corresponding to the eigenvalues greater or equal to (m + 1) 2. Then H = Ĥ H, dim(ĥ) < and dim( H) =. We denote by the symbols, and 2 the norm in H, and in L 2 (, π), respectively and we note that by (3.8), lim u G(x, u) fu u 2 dx =. (3.14) We shall prove that the functional J defined by (3.7) satisfies the assumptions in Theorem 2.7 (Saddle Point Theorem).

27 27 For Assumption (a), we take an arbitrary α and we prove that there exists a bounded neighborhood D of in Ĥ such that J/ D α. We argue by contradiction, then there is a sequence {u n } Ĥ such that u n and a constant c satisfying lim inf J(u n) c. (3.15) n From the definition of J and from (3.15) it follows that [ 1 π (u lim inf n) 2 m 2 u 2 n n 2 u n 2 For u Ĥ we have dx G(x, u n ) fu n u n 2 ] dx. (3.16) [(u ) 2 m 2 u 2 ] dx = u 2 m 2 u 2 2 (3.17) and the equality in (3.17) holds only for u = k sin mx, k R. Set v n = u n / u n. Since dim Ĥ < there is v Ĥ such that v n v strongly in Ĥ (also strongly in L2 (, π)). Then (3.14), (3.16), and (3.17) yield v = k sin mx, 2 π where k = 1 or k = 1 m m (3.15) by u n to obtain [ 1 π (u lim inf n) 2 m 2 u 2 n n 2 u n 2 π, ( v = 1). Let k = 1 m dx G(x, u n ) fu n u n 2 π. We divide ] dx. (3.18) Because u n Ĥ the first integral in (3.18) is less than or equal to zero and we have [ lim inf G(x, u ] n) v n dx + k f sin mx dx. (3.19) n u n We know that v n k sin mx, k > in Ĥ. Because of the compact imbedding Ĥ C(, π), we have v n k sin mx in C(, π) and we get lim u + for x (, π) such that sin mx >, n(x) = n for x (, π) such that sin mx <.

28 28 We note that from (3.1) it follows that there exist constants R, r and functions A + (x), A (x) L 1 (, π) such that A + (x) G + (x), G (x) A (x) for a.e. x (, π) and for all s R, s r, respectively. We obtain from Fatou s lemma and (3.19) f(x) sin mx dx [G + (x)(sin mx) + G (x)(sin mx) ] dx, (3.2) a contradiction to (3.1). We proceed for the case k = 1 2. m π Assumption a) of Theorem 2.7 is verified. For Assumption (b), we prove that Then lim J(u) = u for all u H and that J 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 (3.7) one has [ ] G(x, u(x)) f(x)u(x) dx Hence J is bounded on bounded subsets of H. Since u H, we have The definition of J, (3.14), and (3.22) yield [ c u2 (x) ] + p(x) u(x) f(x)u(x) dx 2 c c 2 2 u 2 + ( p 1 + f 1 )c 1 u. (3.21) u 2 (m + 1) 2 u 2 2. (3.22) J(u) lim u u lim 1 2 u 2 (2m + 1) u 2 2 u. (3.23) 2 If u 2 2/ u 2 then it follows from the definition of J and (3.14) that J(u) lim u u = (3.24) Then (3.23) and (3.24) imply lim u J(u) = ; therefore, Assumption b) of Theorem 2.7 is satisfied.

29 29 For Assumption (c), we show that J satisfies the Palais-Smale condition. First, we suppose that the sequence {u n } is unbounded and there exists a constant c 3 such that and 1 2 [(u n) 2 m 2 u 2 n] dx [G(x, u n ) fu n ] dx c 3 (3.25) lim J (u n ) =. (3.26) n Let {w k } be an arbitrary sequence bounded in H. It follows from (3.26) and the Schwarz inequality that n lim [u nw k m 2 u n w k ] dx [g(x, u n )w k fw k ] dx k = n lim J (u n )w k (3.27) k lim n k By (3.7) and (3.8) we obtain J (u n ) w k =. lim n k [ g(x, un ) u n w k f ] u n w k dx =. (3.28) 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). One has from (3.27), (3.28) and also lim n m k lim n k [ u n u n w k m 2 u ] n u n w k dx = (3.29) [(v n v m ) w k m 2 (v n v m )w k ] dx =. (3.3) We set k = n and w k = v n, k = m and w k = v m in (3.3) and subtract these equalities we get [ ] lim n m v n v m 2 m 2 v n v m 2 2 =. (3.31)

30 3 Since v n v strongly in L 2 (, π) then v n v m 2. Since (3.31) holds then v n is a Cauchy sequence in H and v n v strongly in H. Hence it follows from (3.29) and from the usual regularity argument for ordinary differential equations (see Fučík [7]) that either v = 1 m 2 sin mx ( v π = 1). Suppose that v = 1 m or v = 1 m w k = sin mx in (3.29), we get lim n 2 sin mx π 2 sin mx. Setting π [ g(x, u n (x)) sin mx + f(x) sin mx] dx =. (3.32) Let H be the subspace of H spanned by the eigenfunctions sin x, sin 2x,..., sin(m 1)x. Then we write v n = v n + a n sin mx + ṽ n, where v n H, ṽ n H and a n R (likewise u n = u n + u n a n sin mx + ũ n ). If we set k = n and w k = v n + a n sin mx + ṽ n in (3.27), we get { lim [u n( v n + a n sin mx + ṽ n ) m 2 u n ( v n + a n sin mx + ṽ n )] dx n =. } [g(x, u n )( v n + a n sin mx + ṽ n ) f( v n + a n sin mx + ṽ n )] dx (3.33) It follows from (3.32) and (3.33) that 1 { lim [ u n u n nu n + m 2 u n u n + u nũ n m 2 u n ũ n ] dx (3.34) [ g(x, un ) ] } u n ( u n + ũ n ) f( u n + ũ n ) dx =. u n For u n H we have u n 2 (m 1) 2 u n 2 2, and for ũ n H we have ũ n 2 (m + 1) 2 ũ n 2 2. It follows from the orthogonality u nũ n dx = that the first integral in (3.34) satisfies [ u nu n + m 2 u n u n + u nũ n m 2 u n ũ n ] dx = u n 2 + m 2 u n ũ n 2 m 2 ũ n 2 2 (3.35) u n 2 m 2 + (m 1) u n 2 + ũ 2 n 2 m2 (m + 1) ũ n 2 2 = 2m 1 (m 1) 2 u n 2 + 2m + 1 (m + 1) 2 ũ n 2.

31 31 It follows from (3.7) and (3.8) that ε >, R > such that for a.e. x (, π) and all s > R, g(x, s) s < ε. Also for a.e. x (, π), and all s R, g(x, s) cr + p(x). It follows from the imbedding H L 2 (, π) ( u 2 u ) that the second integral in the equations (3.34) satisfies [ g(x, un ) ] u n ( u n + ũ n ) f( u n + ũ n ) dx (3.36) u n ε( u n 2 + ũ n 2 ) + (cr + p 1 + f 1 )( u C(,π) + ũ C(,π) ). It follows from the imbedding H C(, π) ( u C(,π) c 1 u ) and (3.34), (3.35), (3.36) that there are constants ϱ 1 >, ϱ 2 > such that 1 lim n u n [ϱ 1 u n 2 + ϱ 2 ũ n 2 (cr + p 1 + f 1 )c 1 ( u n + ũ n )]. Let u n = u n + ũ n. Then it holds u n 2 = u n 2 + ũ n 2 and u n + ũ n 2 u n. Since there are constants ϱ > and β > such that Therefore, 1 lim n u n (ϱ u n 2 β u n ). Now we divide (3.25) by u n. We get lim n { 1 2 [ u n u n u n m 2 u ] n u n u n dx u n 2 = lim n u n. (3.37) We obtain from (3.37) the following equality 1 lim n 2 1 = lim n 2 G(u n ) fu n u n } dx =. (3.38) [ u n u n u n m 2 u ] n u n u n dx (3.39) [ ((u n ) ) 2 u n (u m2 n ) 2 ] dx =. u n

32 32 We know that v n k sin mx, k > in H. Due to compact imbedding H C(, π) we have v n k sin mx in C(, π) and we get lim u + for x (, π) such that sin mx >, n(x) = n for x (, π) such that sin mx <. Using Fatou s lemma, (3.38), and (3.39) we conclude f(x) sin mx dx [G + (x)(sin mx) + G (x)(sin mx) ] dx. (3.4) This is a contradiction to the assumption (3.1). 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), C(, π) (taking a subsequence if it is necessary). It follows from the equality (3.27) that lim n m k { [(u n u m ) w k m 2 (u n u m )w k ] dx (3.41) } [g(x, u n ) g(x, u m )]w k dx =. The strong convergence u n u in C(, π) and the assumption (3.8) imply lim n m [g(x, u n ) g(x, u m )](u n u m ) dx =. (3.42) If we set w k = u n, w k = u m in (3.42) and subtract these equalities, then lim n m [(u n u m) 2 m 2 (u n u m ) 2 ] dx =. (3.43) Hence the strong convergence u n u in L 2 (, π) and (3.43) imply the strong convergence u n u in H. This shows that J satisfies Palais-Smale condition and the proof of Theorem 3.1 is complete.

33 4 Potential Landesman-Lazer type condition and jumping nonlinearities Let us consider the nonlinear boundary-value problem u (x) + λ + u + (x) λ u (x) + g(x, u(x)) = f(x), x (, π), (4.1) u() = u(π) =. Here u ± = max{±u, }, λ +, λ R, nonlinearity g :, π R R is a Caratheodory s function and f L 1 (, π). For g and f problem (4.1) becomes u (x) + λ + u + (x) λ u (x) =, x (, π), (4.2) u() = u(π) =. We define Σ = { (λ +, λ ) R 2 : (4.2) has a nontrivial solution }. This set is called the Fučík spectrum (see [7]), and can be expressed as Σ = j=1 Σ j where Σ 1 = { (λ +, λ ) R 2 : (λ + 1)(λ 1) = }, ( 1 Σ 2i = { (λ +, λ ) R 2 : i + 1 ) = 1 }, λ+ λ Σ 2i+1 = Σ 2i+1,1 Σ 2i+1,2 where ( 1 Σ 2i+1,1 = { (λ +, λ ) R 2 : i + 1 ) + 1 = 1 }, λ+ λ λ+ ( 1 Σ 2i+1,2 = { (λ +, λ ) R 2 : i + 1 ) + 1 = 1 }. λ+ λ λ 33 We suppose that the point (λ +, λ ) Σ m, if m N is even or (λ +, λ ) Σ m2, if m N is odd and λ < λ + < (m + 1) 2. (4.3) Remark 4.1 Assuming that (m + 1) 2 > λ + > λ, if (λ +, λ ) Σ m, m N, then λ > (m 1) 2.

34 34 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 We define the potential of the nonlinearity g and G + (x) = lim inf s + G(x, s) = G(x, s) s s g(x, t) dt, G (x) = lim sup s G(x, s) s We denote ϕ m a nontrivial solution of (4.2) corresponding to (λ +, λ ) (see Remark 4.2). We assume that for any ϕ m the following potential Landesman-.

35 35 Lazer type condition holds: f(x)ϕ m (x) dx < We suppose that the nonlinearity g is bounded, i.e. [G + (x)(ϕ m (x)) + G (x)(ϕ m (x)) ] dx. (4.4) p(x) L 1 (, π) : g(x, s) p(x) for a.e. x (, π), s R (4.5) and we prove the solvability of (4.1) (Theorem (4.7)). This section is inspired by a result in [2] where the authors deal with the case where g(x,s) lies (in some sense) between Σ s 1 and Σ 2 and by a result [1] with the classical Landesman-Lazer type condition (see Corollary 2 in [1]). Remark 4.2 First we note that if m is even then two different functions ϕ m1, ϕ m2 correspond to the point (λ +, λ ) Σ m. For example for m = 2, λ + > λ we have { k1 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 { k2 λ+ sin( λ x), x, π/ λ, ϕ 22 (x) = 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, ϕ 22 (x) = k 2 sin 2x, where k 1, k 2 >. ϕ 21 ϕ 22 Figure 2: Solutions corresponding to Σ 2

36 36 If m is odd then Σ m = Σ m1 Σ m2 and it corresponds only one function ϕ m1 to the point [λ +, λ ] Σ m1, one function ϕ m2 to the point (λ +, λ ) Σ m2, respectively. For m = 3, λ + > λ, λ + > λ we have { k 1 λ sin( λ +x), x, π/ λ +, ϕ 31 (x)= 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 >. { k 2 λ+ sin( λ x), x, π/ λ, ϕ 32 (x)= 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 >. ϕ 31 ϕ 32 Figure 3: Solutions corresponding to Σ 3 For λ + = λ = m 2 we set ϕ m1 (x) = k 1 sin mx, ϕ m2 (x) = k 2 sin mx, where k 1, k 2 > and from the condition (4.4) we obtain and f(x) sin mx dx < f(x)( sin mx) dx < [G + (x)(sin mx) + G (x)(sin mx) ] dx [G + (x)( sin mx) + G (x)( sin mx) ] dx.

37 37 Hence it follows < [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 (3.1)). Remark 4.3 We have v, sin mx = v (x)(sin mx) dx = m 2 v(x) sin mx dx v H. Since and from the definition of the functions ϕ m1, ϕ m2 (see remark 4.2) it follows ϕ m1, sin mx > and ϕ m2, sin mx <. (4.6) We again define 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 (4.1) we mean a function u C 1 (, π) such that u is absolutely continuous, u satisfies the boundary conditions and the equations (4.1) holds a.e. in (, π). We study (4.1) by using of variational method. More precisely, we look for critical points of the functional J : H R, which is defined by J(u) = 1 2 [(u ) 2 λ + (u + ) 2 λ (u ) 2 ] dx Every critical point u H of the functional J satisfies [u v (λ + u + λ u )v] dx [G(x, u) fu] dx. (4.7) [g(x, u)v fv] dx = for all v H. Then u is also a weak solution of (4.1) and vice versa. The usual regularity argument for ODE yields immediately (see Fučík [2]) that any weak solution of (4.1) is also the solution in the sense mentioned above.

38 38 We will use the following variant of Saddle Point Theorem (see [11]) which is proved in Struwe [14, Theorem 8.4]. Theorem 4.4 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 J 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 J(u) > sup u Q J(u) = ν, (iv) J satisfies Palais-Smale condition. Then the number γ = inf defines a critical value γ > ν of J. sup h Γ u Q J(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 J by J λ (u) = [(u ) 2 λ + (u + ) 2 λ (u ) 2 ] dx. Now we present a few results needed later. Lemma 4.5 Let ϕ be a solution of (4.2) with (λ +, λ ) Σ, λ + λ. We put u = aϕ + w, a, w H. Then it holds [(w ) 2 λ + w 2 ] dx J λ (u) [(w ) 2 λ w 2 ] dx. (4.8)

39 39 Proof : We prove the right inequality in (4.8), the proof of the left inequality is similar. Since ϕ is a solution of (4.2) one has and ϕ w dx = (ϕ ) 2 dx = [λ + ϕ + w λ ϕ w] dx for w H (4.9) [λ + (ϕ + ) 2 + λ (ϕ ) 2 ] dx. (4.1) By (4.9) and (4.1) we obtain ] J λ (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. (4.11) For the function (aϕ + ) 2 + 2aϕ + w ((aϕ + w) + ) 2 in the last integral in (4.11) 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 <. Hence and by assumption λ + λ we obtain the assertion of the Lemma 4.5.

40 4 Remark 4.6 It follows from the previous proof that we obtain the equality J λ (u) = [(w ) 2 λ w 2 ] dx in (4.8) if aϕ + w ϕ < w = ϕ then w =. and if w span{sin x,..., sin kx}, k N, 4.1 Main result Theorem 4.7 Under the assumptions (4.3), (4.4), and (4.5), Problem (4.1) has at least one solution in H. Proof : First we suppose that m is even. We shall prove that the functional J defined by (4.7) satisfies the assumptions of Theorem 4.4. Let ϕ m1, ϕ m2 are the solutions of (4.2) corresponding to (λ +, λ ) Σ m (see Remark 4.2). Let Ĥ 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 Ĥ }, V 2 = { u H : u = a 2 ϕ m2 + w, a 2, w Ĥ }. 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,.... Next, we verify the assumptions of Theorem 4.4. We see that S is the closed subset in H and Q is the bounded subset in H. i) Firstly we note that for z Ĥ S it holds 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 Ĥ = a i ϕ mi, sin mx, i = 1, 2. Since and from inequalities (4.6) it follows that a i =, i = 1, 2 and u = w.

41 41 For u = w Q we have u = L > and we obtain a contradiction with u Ĥ S = {o}. ii) We prove that H = V S. We can write a function h H in the form h = m 1 b i sin ix + b m sin mx + b i sin ix = h + b m sin mx + h, b i R, i N. i=1 i=m+1 The inequalities (4.6) 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 = Ĥ {}}{{}}{ ( h b m b m1 ϕ m1 + b m b m1 ϕ m1 ) + ( h b }{{} m b m1 ϕ m1 ). Similarly for b }{{} m : V S h = h + b m b m2 (ϕ m2 ϕ m2 ϕ m2 ) + h {}}{{}}{ = ( h b m b m2 ϕ m2 + b m b m2 ϕ m2 ) + }{{} V ( h b m b m2 ϕ m2 ). We proved that H is spanned by V and S. }{{} S The proof of the assumption S h(q) h Γ is similar to the proof in [14, 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 Ĥ h t (u) = tπ(h(u)) + (1 t)u. 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 was to be shown. iii) Firstly, we note that by assumption (4.5), one has lim u G(x, u) fu u 2 dx =. (4.12) First we show that the infimum of functional J on the set S is a real number. We prove for this that lim J(u) = for all u S (4.13) u

42 42 and J 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 (4.5) one has J(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 2 ( u 2 + λ + c 2 u λ c 2 u 2) + ( p 1 + f 1 ) c 1 u. Hence J is bounded on bounded subsets of S. To prove (4.13), 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 it holds u 2 = lim inf n J(u n) c 3. (4.14) (u ) 2 dx (m + 1) 2 u 2 dx = (m + 1) 2 u 2 2. (4.15) The definition of J, (4.12), (4.14) and (4.15) yield J(u n ) ((m + 1) 2 λ + ) u + n ((m + 1) 2 λ ) u n 2 2 lim inf lim inf. n u n 2 n 2 u n 2 (4.16) It follows from (4.16) and (4.3) that u n 2 2/ u n 2 and from the definition of J and (4.12) we have J(u n ) lim inf u n u n = a contradiction to (4.14). We proved that there is µ R such that inf u S J(u) = µ. Second we estimate the value J(u) for u Q. We remark that u = Kϕ m +w where ϕ m = ϕ m1 or ϕ m = ϕ m2, K >, w Ĥ, w = L. We prove that sup J(Kϕ m + w) = sup J(u) =. (4.17) (K+L) u

43 43 For (4.17), we argue by contradiction. Suppose that (4.17) is not true then there are a sequence (u n ) Q such that u n and a constant c 4 satisfying lim sup J(u n ) c 4. (4.18) n Hence it follows [ 1 π (u lim sup n) 2 λ + (u + n ) 2 λ (u n ) 2 n 2 u n 2 G(x, u n ) fu n dx u n 2 ] dx. (4.19) 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 (4.19) and (4.12) yield 1 2 [(v ) 2 λ + (v + ) 2 λ (v ) 2 ] dx. (4.2) Let v = a ϕ m + w, a R +, w Ĥ. It follows from Lemma 4.5 that [(v ) 2 λ + (v + ) 2 λ (v ) 2 ] dx For w Ĥ it holds [(w ) 2 λ w 2 ] dx [(w ) 2 λ (w ) 2 ] dx. (4.21) [((m 1) 2 λ ) w 2 ] dx. (4.22) Since (m 1) 2 < λ (see Remark 4.1) then (4.2), (4.21) and (4.22) 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 (4.7) by u n then [ 1 π (u lim 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 (4.23) By Lemma 4.5 the first integral in (4.23) is less then or equal to. Hence it follows G(x, u n ) + fu π [ n G(x, un ) ] lim sup dx = lim sup v n + fv n dx. n u n n u n (4.24)

44 44 Because of the compact imbedding Ĥ C(, π), we have v n C(, π) and we get a ϕ m in { + for x (, π) such that lim u ϕm (x) >, n(x) = n for x (, π) such that ϕ m (x) <. We note that from (4.5) it follows that p(x) G + (x), G (x) p(x) for a.e. x (, π). We obtain from Fatou s lemma and (4.24) f(x)ϕ m (x) dx [G + (x)(ϕ m (x)) + G (x)(ϕ m (x)) ] dx, a contradiction to (4.4). We proved that by choosing K, L sufficiently large there is ν R such that sup J(u) = ν < µ. Then Assumption iii) of Theorem 4.4 is verified. u Q iv) Now we show that J satisfies the Palais-Smale condition (see page 18). First, we suppose that the sequence {u n } is unbounded and there exists a constant c 5 such that 1 [(u 2 n) 2 λ + (u + n ) 2 λ (u n ) 2 ] dx [G(x, u n ) fu n ] dx c 5 (4.25) and lim J (u n ) =. (4.26) n Let {w k } be an arbitrary sequence bounded in H. It follows from (4.15) and the Schwarz inequality that n lim [u nw k (λ + u + n λ u n ) w k ] dx [g(x, u n )w k fw k ] dx k = n lim J (u n ), w k n lim J (u n ) w k =. (4.27) 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 (4.27) by u n and using (4.5) we obtain lim n k [v nw k (λ + v + n λ v n ) w k ] dx = (4.28)

45 45 and also lim n i k [(v n v i)w k (λ + (v + n v + i ) λ (v n v i ) ) w k] dx =. (4.29) We set w k = v n v i in (4.29) and we get [ [ lim v n n v i 2 [λ+ (v n + v + i ) λ (vn v i )](v n v i ) ] dx] =. i (4.3) Since v n v strongly in L 2 (, π) the integral in (4.3) 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 (4.28) and the usual regularity argument for ordinary differential equations (see Fučík [7]) 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 Ĥ S. We remark that u = u + u and using (4.9) in the first integral in (4.27) we obtain J 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. (4.31)

46 46 Similarly we obtain J 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. (4.32) We add (4.31) and (4.32) and we have 2J 1 = [(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. (4.33) We set û n = u n + ũ n where u n Ĥ, ũ n S and we put in (4.33) w k = (u n ũ n )/ û n then we have 2J 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. (4.34) Hence 2J 1 1 û n = 1 û n [ (λ + λ ) u n + ũ n u n ũ n ] dx + 2 u n 2 2 ũ n 2 (λ + + λ )( u n 2 2 ũ n 2 2) [ (λ+ λ ) u 2 n ũ 2 n ] dx + 2 u n 2 (λ + + λ ) u n ũ n 2 + (λ + + λ ) ũ n 2 2. Inequality a 2 b 2 max{a 2, b 2 } and (4.35) yield J 1 max{ u n 2 λ u n 2 2, ũ n 2 + λ + ũ n 2 2 } (4.35) 1 û n. (4.36) We note that it holds u n 2 (m 1) 2 u n 2 2, ũ n 2 (m+1) 2 ũ n 2 2. Hence from assumption (4.3) and (4.36) it follows that there is ε > such that J 1 ε max{ u n 2, ũ n 2 } 1 û n. (4.37)

47 47 From (4.27), (4.37) it follows lim εmax{ u n 2, ũ n 2 } n û n [ (g(x, u n ) f) u ] n ũ n û n dx. (4.38) Now we suppose that û n. We note that û n 2 = u n 2 + ũ n 2, we divide (4.38) by û n and using (4.5) we have ε 2 lim εmax{ u n 2, ũ n 2 } π g(x, u n ) f n û n 2 û n u n ũ n û n dx (4.39) a contradiction to ε >. This implies that the sequence {û n } is bounded. Lemma 4.5 with w = û n and we obtain We use (4.8) from Hence [(û n) 2 λ + û 2 n] dx J λ (u n ) J λ (u n ) lim n u n = lim n [(û n) 2 λ û 2 n] dx. [(u n) 2 λ + u 2 n λ u 2 n] dx u n =. (4.4) We divide (4.25) by u n and (4.4) yield lim n [ G(x, un ) + fu ] n dx = (4.41) u n and using Fatou s lemma in (4.41) we obtain a contradiction to (4.4). 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 (4.27) that lim n i k [ (un u i ) w k [λ + (u + n u + i ) λ (u n u i )]w ] k dx =. (4.42) We put w k = u n u i in (4.42) and the strong convergence u n u in L 2 (, π) and (4.42) imply the strong convergence u n u in H. This shows that the functional J satisfies Palais-Smale condition and the proof of Theorem 4.7 for m even is complete.

48 48 Now we suppose that m is odd. We have (λ +, λ ) Σ m2 and the nontrivial solution ϕ m2 of (4.2) 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 4.2). We define the sets Q and S like for m even and the proof of the steps i), ii) of theorem 4.7 is the same. In the step iii) we change inequality (4.21) if v = a ϕ m1 as it follows (4.2) = k (4.22) [ (v ) 2 λ + (v + ) 2 λ (v ) 2] dx [ (v ) 2 (λ + k)(v + ) 2 (λ k)(v ) 2] dx k v 2 dx k v 2 dx + v 2 dx + [ (w ) 2 λ (w ) 2] dx [ ] ((m 1) 2 λ ) w 2 dx. (4.43) Then by (4.43) and assumption (m 1) 2 < λ we obtain k v 2 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 4.7 is complete.

49 49 5 Solution to a semilinear problem on type II regions determined by the Fucik spectrum In this section we investigate the nonlinear problem u (x) + αu + (x) βu (x) = f(x), x (, π), (5.1) x() = x(π) =. Here u ± = max{±u, }, α, β R and f L 2 (, π). For f, α = λ +, and β = λ Problem (5.1) becomes u (x) + λ + u + (x) λ u (x) =, x (, π), (5.2) x() = x(π) =. We define the Fučík spectrum (see page 33) Σ like in the section 4. Note that there are two types of regions between the curves of Σ: Type (I) R 1 which consists of regions between the curves Σ 2i and Σ 2i+1, i N. Type (II) R 2 consists of regions between the curves Σ 2i+1,1 and Σ 2i+1,2, i N. Figure 4: Regions determined by the Fučík spectrum

50 5 If (α, β) R 1 one can solve (5.1) for an arbitrary f L 2 (, π) while this is not so for regions R 2 (see [7]). We suppose that the point (α, β) R 2 is between the curves Σ 2i+1,1 and Σ 2i+1,2, α > β, and k > such that λ + = α+k, λ = β+k, (λ +, λ ) Σ 2i+1,2 and λ + < (2i + 2) 2. We denote ϕ 2 the solution of (5.2) belonging to (λ +, λ ) (see remark 4.2). In this section we obtain existence results for (5.1) with the π right hand side f = cϕ 2 +ϕ 2, c > where ϕ 2 ϕ 1 2 dx = and ( β (2i) π (2i + 2) 2 α) ) (ϕ 2 ) 2 1 dx < ( λ + α 1 π (2i + 2) 2 α) ) (cϕ 2 ) 2 dx. This article is inspired by a result in [13] where the author solves the problem u(x) + αu + (x) + βu (x) + p(x, u(x)) = f(x) with a nontrivial nonlinearity satisfying p(x, u(x)). Remark 5.1 Assuming that (2i+2) 2 > λ + > λ, if (λ +, λ ) Σ 2i+1,1, then λ > (2i) 2. Assuming that (2i + 2) 2 > α, if the point (α, β) is in R 2 between the curves Σ 2i+1,1 and Σ 2i+1,2, then β > (2i) 2. See the illustration in figure 4. We use the same notation like in the previous section 4. We again study (5.1) by using of variational methods. We look for critical points of the functional J : H R, which is defined by J(u) = 1 2 [(u ) 2 α(u + ) 2 β(u ) 2 ] dx + Every critical point u H of the functional J satisfies J (u), v = fu dx. (5.3) [u v (αu + βu )v + fv] dx = for all v H. Then u is a weak solution of (5.1) and vice versa. The usual regularity argument for ODE yields immediately (see Fučík [3]) that any weak solution of (5.1) is also the solution in the sense mentioned above. To find a critical point of the functional J we again use theorem 4.4. Now we present a few results needed later.

51 51 Lemma 5.2 ϕ H such that Let ϕ be a solution of (5.2) with (λ +, λ ) Σ and a function [(w ) 2 λ w 2 ] dx d J(u) = 1 2 ϕϕ dx =. Let d > and w H satisfying w 2 dx. We put u = aϕ + w, a and [(u ) 2 (λ + k)(u + ) 2 (λ k)(u ) 2 2(cϕ + ϕ )u] dx where < k d and c >. Then there is constant ã > such that for u = ãϕ + w it holds J(u) 1 [ 1 2 k (cϕ)2 1 ] d k (ϕ ) 2 dx (5.4) Proof: By (4.8) from Lemma 4.5 and the assumptions on w, we obtain J(u) = 1 ] [ k [(u ) 2 λ + (u + ) 2 λ (u ) 2 dx u2 (cϕ + ϕ )u] dx 1 ] [ k [(w ) 2 λ w 2 dx u2 (cϕ + ϕ )u] dx [ d 2 w2 + k ] 2 u2 (cϕ + ϕ )u dx. (5.5) We substitute u = aϕ + w and we have for the last integral in (5.5) [ d 2 w2 + k ] 2 u2 (cϕ + ϕ )u dx = = [ d 2 w2 + k ] 2 (aϕ + w)2 (cϕ + ϕ )(aϕ + w) dx [ 1 ( k ) ] 2 (d k)w2 + a 2 a c ϕ 2 + (ak c)ϕw ϕ w dx. We put a = c (= ã) and we obtain k [ 1 2 (d k)w2 + c ( k c ) k 2 k c = ] ϕ 2 ϕ w [( 1 ) 2+ (d k)w + ϕ 1 (d k) k (cϕ)2 1 ] d k (ϕ ) 2 dx [ 1 k (cϕ)2 1 ] d k (ϕ ) 2 dx. (5.6) dx