On a k-dimensional nonlocal boundary value problem at resonance Katarzyna Szyma´ nska-D¸ebowska Abstract. In this paper we study the existence of at least one solution to the following system of nonlocal resonant boundary value problem Z 1 x00 = f (t, x), x0 (0) = 0, x0 (1) = x0 (s)dg(s), 0 k k k where f : [0, 1] × R → R , g : [0, 1] → R . Mathematics Subject Classification (2010). Primary 34B10; Secondary 34B15. Keywords. nonlocal boundary value problem; boundary value problem at resonance; the perturbation method; Neumann BVP. 1. Introduction In the paper the following system of ordinary differential equations Z 1 x00 = f (t, x), x0 (0) = 0, x0 (1) = x0 (s)dg(s), (1.1) 0 where f = (f1 , . . . , fk ) : [0, 1] × Rk → Rk is continuous, g = (g1 , . . . , gk ) : [0, 1] → Rk has bounded variation, is studied. Speaking precisely, (1.1) is the system of k BVPs 00 xi (t) = fi (t, x(t)), x0 (0) = 0, R1 i0 xi (1) = 0 x0i (s)dgi (s), R1 where t ∈ [0, 1], i = 1, . . . , k and the integrals 0 x0i (s)dgi (s) are meant in the sense of Riemann-Stieltjes. Our main goal is to show that the problem (1.1) has at least one solution. We impose on the function f a sign condition, which we called: the asymptotic 2 Katarzyna Szyma´ nska-D¸ebowska integral sign condition. The idea comes from [21], where the author shows that the following problem x0 = f (t, x), x(0) = 0, x(T ) = 0, has periodic solutions. The method can be successfully applied to other BVPs (not necessarily only for differential equations of the first or second order but, for instance, involving p-Laplacians), for which the function f does not depend on x0 . As far as we are aware, the BVP (1.1) has not been studied in this generality so far. Note that a special case of (1.1) is the Neumann BVP x00 = f (t, x), x0 (0) = 0, x0 (1) = 0. Until now, under suitable monotonicity conditions or nonresonance conditions, some existence or uniqueness theorems or methods for Neumann BVPs have been presented (see, for instance, [1, 4, 18, 23, 22, 24, 26, 27, 25] and the references therein). For example, in [10], the authors study the Neumann boundary value problem x00 + µ(t)x+ − ν(t)x− = p(t, x), x0 (0) = 0 = x0 (π), where µ, ν lie in L1 (0, π), p(t, x) is a Carathodory function, p ≥ 0, x+ (t) = max(x(t), 0), and x− (t) = max(−x(t), 0). They obtain several necessary and sufficient conditions on p so that the Neumann problem has a positive solution or a solution with a simple zero in (0, π). In [11], the author uses phase plane and asymptotic techniques to discuss the number of solutions of the problems −x00 = f (t, x), x0 (0) = σ1 , x0 (π) = σ1 . It is assumed that f : [0, π] × R → R is a continuous jumping nonlinearity with nonnegative asymptotic limits: x−1 f (t, x) → α as x → −∞ and x−1 f (t, x) → β as x → ∞. The limit problem where f (t, x) = αx− +βx+ plays a key role in his methods. The authors describe how the number of solutions of the problem depends on the four parameters: α, β, σ1 , σ2 . The results differ from those obtained by various authors who were mainly concerned with forcing the equation with large positive functions and keeping the boundary conditions homogeneous. BVP x00 = f (t, x, x0 ), x0 (0) = 0, x0 (1) = 0, is considered in [8]. The authors obtain some results of existence of solutions assuming that there is a constant M > 0 such that yf (t, x, y) > 0 for |y| > M and the function f satisfies the Bernstein growth condition (or the BernsteinNagumo growth condition). The generalization of the Neumann problem (1.1) is a nonlocal problem. BVPs with Riemann-Stieltjes integral boundary conditions include as special cases multi-point and integral BVPs. The multi-point and integral BCs are widely studied objects. The study of multi-point BCs was initiated in 1908 by Picone [20]. Reviews on differential equations with BCs involving Stieltjes measures has been written in 1942 by Whyburn [29] and in 1967 by Conti. On a k-dimensional nonlocal BVP at resonance 3 Since then, the existence of solutions for nonlocal nonlinear BVPs has been studied by many authors by using, for instance, the Leray-Schauder degree theory, the coincidence degree theory of Mawhin, the fixed point theorems for cones. For such problems and comments on their importance, we refer the reader to [3, 5, 6, 7, 12, 13, 14, 15, 16, 28, 30, 31] and the references therein. 2. The perturbed problem First, we shall introduce notation and terminology. Throughout the paper | · | will denote the Euclidean norm on Rk , while the scalar product in Rk corresponding to the Euclidean norm will be denoted by ( · | · ). Denote by C 1 ([0, 1], Rk ) the Banach space of all continuous functions x : [0, 1] → Rk which have continuous first derivatives x0 with the norm ( ) kxk = max sup |x(t)| , sup |x0 (t)| . t∈[0,1] (2.1) t∈[0,1] The Lemma below, which is a straightforward consequence of the classical Arzel` a−Ascoli theorem, gives a compactness criterion in C 1 [0, 1], Rk . Lemma 2.1. For a set Z ⊂ C 1 [0, 1], Rk to be relatively compact, it is necessary and sufficient that: (1) there exists M > 0 such that for any x ∈ Z and t ∈ [0, 1] we have |x (t)| ≤ M and |x0 (t)| ≤ M ; (2) for every t0 ∈ [0, 1] the families Z := {x | x ∈ Z} and Z 0 := {x0 | x ∈ Z} are equicontinuous at t0 . Now, let us consider problem (1.1) and observe that the homogeneous linear problem, i.e., Z 1 x00 = 0, x0 (0) = 0, x0 (1) = x0 (s)dg(s), 0 has always nontrivial solutions, hence we deal with a resonant situation. The following assumptions will be needed throughout the paper: (i) f = (f1 , . . . , fk ) : [0, 1] × Rk → Rk is a continuous function. (ii) g = (g1 , . . . , gk ) : [0, 1] → Rk has bounded variation on the interval [0, 1]. (iii) There exists a uniform limit h(t, ξ) := lim f (t, λ ξ) λ→∞ k with respect to ξ ∈ R , |ξ| = 1. 4 Katarzyna Szyma´ nska-D¸ebowska (iv) Set Z 1 1 Z s Z h(u, ξ)du − h0 (ξ) := h(u, ξ)dudg(s). 0 0 0 For every ξ ∈ Rk , |ξ| = 1, we have (ξ | h0 (ξ)) < 0. The problem (1.1) is resonant. Hence, there is no equivalent integral equation. The existence of a solution will be obtained by considering the following perturbed boundary value problem: x00 = f (t, x), t ∈ [0, 1], (2.2) x0 (0) = 0, Z 0 x (1) = (2.3) 1 x0 (s)dg(s) + αn x(0), αn ∈ (0, 1), αn → 0. (2.4) 0 Notice that the problem (2.2), (2.3), (2.4) is always nonresonant. Now, let us consider the equation (2.2) and integrate it from 0 to t. By (2.3), we get Z t x0 (t) = f (u, x(u))du. (2.5) 0 By (2.4) and (2.5), we obtain Z 1 Z f (u, x(u))du = 0 1 0 s Z f (u, x(u))dudg(s) + αn x(0), 0 so x(0) = 1 αn Z 1 Z 1 Z f (u, x(u))du − 0 0 s f (u, x(u))dudg(s) , 0 Moreover, by (2.5), we have Z tZ x(t) = x(0) + s f (u, x(u))duds. 0 0 Now, it is easily seen that the following Lemma is true: Lemma 2.2. A function x ∈ C 1 ([0, 1], Rk ) is a solution of the problem (2.2), (2.3), (2.4) if and only if x satisfies the following integral equation Z tZ s x (t) = f (u, x(u))duds + 0 0 Z 1 Z 1Z s 1 + f (u, x(u))du − f (u, x(u))dudg(s) . αn 0 0 0 To search for solutions to the problem (2.2), (2.3), (2.4), we first reformulate the problem as an operator equation. On a k-dimensional nonlocal BVP at resonance 5 Given x ∈ C 1 ([0, 1], Rk ) and fixed n ∈ N let Z tZ s f (u, x(u))duds + (An x) (t) = 0 0 Z 1 Z 1Z s 1 + f (u, x(u))dudg(s) . f (u, x(u))du − αn 0 0 0 Then t Z 0 (An x) (t) = f (u, x(u))du. (2.6) 0 It is clear that An x, (An x)0 : [0, 1] → Rk are continuous. It follows that the operator An : C 1 [0, 1], Rk → C 1 [0, 1], Rk is well-defined. By assumption (iii), function f is bounded and we put M := sup |f (t, x)|. t∈[0,1],x∈Rk By (2.6), we have sup |(An x)0 (t)| ≤ M. (2.7) t∈[0,1] Moreover, we get sup |(An x)(t)| ≤ M + t∈[0,1] 1 (M + M Var(g)) , αn (2.8) where Var(g) means the variation of g on the interval [0, 1]. From (ii), L := Var(g) < ∞. Put Mn := M + α1n (M + M L), then kAn xk ≤ Mn for every n ∈ N. Moreover, (An x)00 (t) and (An x)0 (t), t ∈ [0, 1], are bounded, hence the families (An x)0 and (An x) are equicontinuous. Now, by Lemma 2.1, the following Lemma holds: Lemma 2.3. The operator An is completely continuous. Let Bn := x ∈ C 1 [0, 1], Rk kxk ≤ Mn . Now, considering operator An : Bn → Bn , by Schauder’s fixed point Theorem, we get that the operator An has a fixed point in Bn for every n. We have proved the following Lemma 2.4. Under assumptions (i) − (iii) problem (2.2), (2.3), (2.4) has at least one solution for every n ∈ N. 6 Katarzyna Szyma´ nska-D¸ebowska 3. The main result Let ϕn be a solution of the problem (2.2), (2.3), (2.4), where n is fixed. Lemma 3.1. The sequence (ϕn ) is bounded in C 1 [0, 1], Rk . Proof. Assume that the sequence (ϕn ) is unbounded. Then, passing to a subsequence if necessary, we have kϕn k → ∞. We can proceed analogously as in (2.7) to show that sup |(ϕn )0 (t)| ≤ M, t∈[0,1] for every n. Hence, supt∈[0,1] |ϕn (t)| → ∞, when n → ∞. ϕn ) ⊂ C 1 [0, 1], Rk and noLet us consider the following sequence ( kϕ nk tice that the norm of the sequence equals 1. Hence, the sequence is bounded. ϕ0 ϕn Moreover, the family ( kϕ ) (and simultaneously ( kϕnn k )) is equicontinuous, nk ϕ0 (t) ϕ00 (t) since kϕnn k (or kϕnn k ) is bounded. By Lemma 2.1, there exists a convergent ϕn subsequence of ( kϕ ). To simplify the notation, let us denote this subsenk ϕn quence as ( kϕn k ). First, observe that ϕ0n (t) kϕn k → 0 ∈ Rk . Now, we shall show that ϕn (t) → ξ, kϕn k (3.1) where ξ = (ξ1 , . . . , ξk ) does not depend on t and |ξ| = 1. Indeed, notice that ϕn (t) kϕn k ϕn (t) kϕn k is given by RtRs = f (u, ϕn (u))duds + kϕn k R1Rs R1 f (u, ϕn (u))du − 0 0 f (u, ϕn (u))dudg(s) 0 + . αn kϕn k 0 0 (3.2) Since f is bounded, we obtain RtRs lim n→∞ 0 0 f (u, ϕn (u))duds = 0 ∈ Rk . kϕn k (3.3) Now, by (3.2) and (3.3), we can easily observe that the limit (3.1) does not ϕn depend on t. The norm of the sequence ( kϕ ) equals 1. Hence ϕkϕnn(t)k → ξ, nk where |ξ| = 1. On a k-dimensional nonlocal BVP at resonance 7 On the other hand, we get RtRs f (u, ϕn (u))duds ϕn (t) ξ = lim = 0 0 + n→∞ kϕn k kϕn k R1 R1Rs f (u, ϕn (u))du − 0 0 f (u, ϕn (u))dudg(s) + 0 = αn kϕn k R 1 n (u) f (u, kϕn k ϕkϕ )du 0 nk − = lim n→∞ αn kϕn k R1Rs n (u) )dudg(s) f (u, kϕn k ϕkϕ k 0 0 n . − αn kϕn k Now, observe, that there exist a uniform limits of Z 1 ϕn (u) )du f (u, kϕn k kϕn k 0 and 1 Z Z s f (u, kϕn k 0 0 ϕn (u) )dudg(s) kϕn k Moreover, by (iv), the sum of the limits is different from zero. Hence, since (3.1) holds, there exists γ ∈ (0, ∞) such that γ := limn→∞ 1/(αn kϕn k). Now, by assumption (iii), we obtain ξ ϕn (t) lim = kϕn k Z 1 Z = γ h(u, ξ)du − = n→∞ 0 0 1 Z s h(u, ξ)dudg(s) . Finally, by (3.4) and (iv), we get Z 1 Z 1 = (ξ | ξ) = γ ξ h(u, ξ)du − 0 = (3.4) 0 0 1 Z s h(u, ξ)dudg(s) 0 γ(ξ | h0 (ξ)) < 0 a contradiction. Hence, the sequence (ϕn ) is bounded. Now, it is easy to see that the following lemma is true: Lemma 3.2. The set Z = {ϕn | n ∈ N} is relatively compact in C 1 [0, 1], Rk . By the above Lemmas, we get the proof of the following Theorem 3.3. Under assumptions (i) − (iv) problem (1.1) has at least one solution. 8 Katarzyna Szyma´ nska-D¸ebowska Proof. Lemma 3.2 implies that (ϕn ) has a convergent subsequence (ϕnl ), ϕnl → ϕ. We know that ϕnl (ϕ0nl ) converges uniformly to ϕ (ϕ0 ) on [0, 1]. Since (ϕnl ) is equibounded and f is uniformly continuous on compact sets, one can see that f (t, ϕnl ) is uniformly convergent to f (t, ϕ). Since ϕ00nl (t) = f (t, ϕnl (t)), the sequence ϕ00nl (t) is also uniformly convergent. Moreover, ϕ00nl (t) converges uniformly to ϕ00 (t). Note that we have actually proved that function ϕ ∈ C 1 [0, 1], Rk is a solution of the equation of problem (1.1). By (2.3) and (2.4), it is easy to see that ϕ satisfies boundary conditions of problem (1.1). This ends the proof. 4. Applications To illustrate our result we shall present some examples. Example 1. Let us consider the Neumann BVP x00 = f (t, x), x0 (0) = 0, x0 (1) = 0. In this case gi (t) = constant, i = 1, . . . , k, t ∈ [0, 1] and condition (ii) always holds. Moreover, we have Z 1 h0 (ξ) = h(s, ξ)ds. 0 Hence for any f which satisfies conditions (i), (iii) and (iv) the Neumann BVP has at least one solution. Example 2. Let k = 1, g(t) = t and f (t, x) = t−|x|x x2 +1 . We have −1, ξ=1 h(t, ξ) = lim f (t, λ ξ) = . 1, ξ = −1 λ→∞ Then h0 (1) = − 12 and h0 (−1) = 12 and we get (ξ|h0 (ξ)) < 0. Hence, problem (1.1) has at least one nontrivial solution. Example 3. Let k = 3, g(t) = (t, t, t) and f1 (t, x1 , x2 , x3 ) = f2 (t, x1 , x2 , x3 ) = f3 (t, x1 , x2 , x3 ) = −x1 p x21 + x22 + x23 + sin2 t + 1 −x2 − t p , 2 x1 + x22 + x23 + 1 −x3 + arctan(x2 − t) p . x21 + x22 + x23 + 1 For every ξ = (ξ1 , ξ2 , ξ3 ), |ξ| = 1, we get ξ1 ξ2 ξ3 h(t, ξ) = lim f (t, λ ξ) = − , − , − λ→∞ |ξ| |ξ| |ξ| , On a k-dimensional nonlocal BVP at resonance and 9 ξ2 ξ3 ξ1 ,− ,− . h0 (ξ) = − 2|ξ| 2|ξ| 2|ξ| Then ξ2 ξ2 ξ12 + 2 + 3 = − 12 |ξ| < 0. |ξ| |ξ| |ξ| Hence, problem (1.1) has at least one nontrivial solution. (ξ|h0 (ξ)) = − 21 References [1] G. Anichini and G. Conti, Existence of solutions of a boundary value problem through the solution map of a linearized type problem, Rend. Sem. Mat. Univ. Politec. Torino (1990) 149–159. [2] R. Conti, Recent trends in the theory of boundary value problems for ordinary differential equations, Boll. Un. Mat. Ital. (3) 22 (1967) 135178. [3] Z. Du, W. Ge and X. 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W´ olcza´ nska 215, Poland e-mail: [email protected] 11

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