Electronic Journal of Differential Equations, Vol. 2020 (2020), No. 107, pp. 1–16. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu DIRICHLET PROBLEM FOR SECOND-ORDER ABSTRACT DIFFERENTIAL EQUATIONS GIOVANNI DORE Abstract. We study the well-posedness in the space of continuous functions of the Dirichlet boundary value problem for a homogeneous linear second- order differential equation u′′ + Au = 0, where A is a linear closed densely defined operator in a Banach space. We give necessary conditions for the well- posedness, in terms of the resolvent operator of A. In particular we obtain an estimate on the norm of the resolvent at the points k2, where k is a positive integer, and we show that this estimate is the best possible one, but it is not sufficient for the well-posedness of the problem. Moreover we characterize the bounded operators for which the problem is well-posed. 1. Introduction We consider the Dirichlet boundary value problem u′′(t) +Au(t) = 0 , t ∈ [0, π] , u(0) = x0 , u(π) = xπ . where A is a linear closed densely defined operator in a (real or complex) Banach space. We are interested in the uniform well-posedness of the problem in the sense of continuous functions, that is we ask that for every x0, xπ in the domain of A there exists a unique solution u such that Au and u′′ are continuous; moreover we require that the solution depends continuously on the boundary values. Unlike most of the articles about abstract Dirichlet problems, we do not suppose that −A is a positive operator. We give some necessary conditions for the uniform well-posedness in terms of the resolvent operator of A, in particular we prove that if the problem is uniformly well-posed then k2 belongs to the resolvent set of A for every positive integer k and (k2I − A)−1 has norm bounded by C/k for a suitable C ∈ R+. We give examples showing that this is the best possible estimate of the resolvent, but it is not sufficient for the well-posedness. Finally we show that if A is bounded then there is uniform well-posedness if and only if k2 belongs to the resolvent set of A for every positive integer k. The Dirichlet problem for an abstract second order equation (homogeneous or non-homogeneous) has been studied in various papers. Usually it is supposed that 2010 Mathematics Subject Classification. 34G10. Key words and phrases. Boundary value problem; differential equations in Banach spaces. c©2020 Texas State University. Submitted August 19, 2018. Published October 29, 2020. 1 2 G. DORE EJDE-2020/107 −A is a positive operator, we refer to Section 4 of the review paper [2], and the articles quoted therein. In [1] and [3] maximal regularity in the Lp sense for the Dirichlet problem for the non-homogeneous equation u′′ +Au = f is characterized. In [3] it is supposed that A = −B2, where B is the generator of an exponentially stable analytic semigroup. They prove (see Corollary 3.4) that, if there is maximal Lp regularity for the Cauchy problem for the first order equation u′ − Bu = f , then there is maximal Lp regularity for the Dirichlet problem. In UMD spaces the converse implication holds. In [1, Theorem 6.3] the authors prove that if A is an arbitrary closed operator in a UMD space, then there is maximal Lp regularity for the Dirichlet problem if and only if k2 belongs to the resolvent set of A for every positive integer k and{ k2(k2I −A)−1 ∣∣ k ∈ N } is R-bounded. We recall also [6] and [5] where the existence of solutions of the Dirichlet problem for particular boundary values is studied under the hypothesis that A is a positive self-adjoint operator in a Hilbert space. This paper is organized as follows. In Section 2 we show that the uniform well-posedness of the Dirichlet problem is equivalent to the existence of a suitable strongly continuous operator valued function. In Section 3 we give necessary condi- tions for the uniform well-posedness. Section 4 contains two examples showing that the estimate on the norm of the resolvent operator of A obtained in Theorem 3.1 is the best possible one, but it is not sufficient to ensure the uniform well-posedness. In Section 5 we characterize uniform well-posedness in case A is a bounded operator. 2. Well-posed problems In what follows X will be a Banach space over the field K (real or complex numbers) and A a closed linear operator from D(A) ⊆ X to X with dense domain; ρ(A) will denote the resolvent set of A. We denote with L(X) the space of linear bounded operators in X. Finally N will denote the set of positive integers and N0 the set of non-negative integers. We study the second-order abstract differential equation u′′(t) +Au(t) = 0 , t ∈ [0, π] , (2.1) and the Dirichlet boundary value problem for this equation u′′(t) +Au(t) = 0 , t ∈ [0, π] , u(0) = x0 , u(π) = xπ . (2.2) We call solution of equation (2.1) a function v : [0, π]→ X such that (1) v ∈ C2 ( [0, π], X ) ∩ C ( [0, π],D(A) ) ; (2) for all t ∈ [0, π], v′′(t) +Av(t) = 0. We call solution of problem (2.2) a solution v of equation (2.1) such that v(0) = x0 and v(π) = xπ. Obviously a solution of this problem can exist only if x0, xπ ∈ D(A). We say that problem (2.2) is uniformly well-posed if (1) for all x0, xπ ∈ D(A), problem (2.2) has solution; EJDE-2020/107 DIRICHLET PROBLEM FOR ABSTRACT DIFFERENTIAL EQUATIONS 3 (2) there exists C ∈ R+ such that for any solution v of equation (2.1) we have sup t∈[0,π] ‖v(t)‖ ≤ C ( ‖v(0)‖+ ‖v(π)‖ ) . Condition (2) implies the uniqueness of the solution of problem (2.2). Theorem 2.1. Let X be a Banach space and A be a linear closed densely defined operator in X. Problem (2.2) is uniformly well-posed if and only if there exists S : [0, π]→ L(X) such that: (1) for all x ∈ X, the function S(·)x is continuous; (2) for all x ∈ X, we have S(0)x = 0, S(π)x = x; (3) for all x ∈ D(A), the function S(·)x is solution of equation (2.1); (4) if v : [0, π]→ X is solution of equation (2.1), then v(t) = S(t)v(π) + S(π − t)v(0) . Proof. Suppose that problem (2.2) is uniformly well-posed. If x ∈ D(A), let v be the unique solution of the problem u′′(t) +Au(t) = 0 , t ∈ [0, π] , u(0) = 0 , u(π) = x ; for t ∈ [0, π] put S̃(t)x = v(t). Then S̃(t) is a linear operator from D(A) to X and, because of the uniform well-posedness, there exists C ∈ R+ such that∥∥S̃(t)x ∥∥ ≤ C‖x‖, for every x ∈ D(A). Since D(A) is dense in X, S̃(t) can be ex- tended to a bounded linear operator S(t) from X to X and ∥∥S(t) ∥∥ ≤ C. Obviously S satisfies conditions (2) and (3). If v is solution of equation (2.1) then it is solution of the Dirichlet problem u′′(t) +Au(t) = 0 , t ∈ [0, π] , u(0) = v(0) , u(π) = v(π) . The function t 7→ S(t)v(π) + S(π − t)v(0) is solution of the same problem, hence, by the uniqueness of the solution, we have v(t) = S(t)v(π) + S(π − t)v(0) . Therefore (4) is satisfied. If x ∈ D(A) the function S(·)x is solution of equation (2.1), hence it is continuous. If x ∈ X there is a sequence (xn)n∈N in D(A) converging to x. The functions S(·)xn are continuous and sup t∈[0,π] ∥∥S(t)xn − S(t)x ∥∥ ≤ C‖xn − x‖ −−−−→ n→∞ 0 . Since it is the uniform limit of continuous functions S(·)x is continuous. Therefore condition (1) is satisfied. Conversely suppose that there exists S satisfying conditions (1)–(4). Since, for all x ∈ X, the function S(·)x is continuous, it is bounded, hence by the uniform boundedness principle { S(t) ∣∣ t ∈ [0, π] } is bounded, therefore there exists C ∈ R+ such that ∥∥S(t) ∥∥ ≤ C, for all t ∈ [0, π]. 4 G. DORE EJDE-2020/107 If x0, xπ ∈ D(A), let v : [0, π]→ X , v(t) = S(t)xπ + S(π − t)x0 . By (3) v ∈ C2 ( [0, π], X ) ∩ C ( [0, π],D(A) ) , and for all t ∈ [0, π] we have v′′(t) +Av(t) = S′′(t)xπ + S′′(π − t)x0 +AS(t)xπ +AS(π − t)x0 = 0 ; moreover, by (2) v(0) = S(0)xπ + S(π)x0 = x0 , v(π) = S(π)xπ + S(0)x0 = xπ . Hence problem (2.2) has solution. From (4) and the estimate ∥∥S(t) ∥∥ ≤ C it follows that if v is solution of equation (2.1) then ‖v(t)‖ = ∥∥S(t)v(0) + S(π − t)v(0) ∥∥ ≤ ∥∥S(t) ∥∥ ‖v(0)‖+ ∥∥S(π − t) ∥∥ ‖v(π)‖ ≤ C ( ‖v(0)‖+ ‖v(π)‖ ) . Hence problem (2.2) is uniformly well-posed. � Theorem 2.2. Let X be a Banach space and A be a linear closed densely defined operator in X. If problem (2.2) is uniformly well-posed, then, for all m ∈ N \ {1}, the same problem is uniformly well-posed for the operator m−2A. Proof. Let S be the strongly continuous operator valued function whose existence is guaranteed by Theorem 2.1. Let T : [0, π] → L(X) defined as follows. If m is even T (t) = m/2∑ j=1 ( S ( (2j − 1)π + t m ) − S ( (2j − 1)π − t m )) , (2.3) if m is odd T (t) = (m−1)/2∑ j=1 ( S (2jπ + t m ) − S (2jπ − t m )) + S ( t m ) . (2.4) It is easy to verify that T satisfies conditions (1)–(3) of Theorem 2.1 with respect to the operator m−2A. To prove that also condition (4) is satisfied it is sufficient to show that the unique solution of the problem u′′(t) + 1 m2 Au(t) = 0 , t ∈ [0, π] , u(0) = 0 , u(π) = 0 , is the identically zero function. Suppose that v ∈ C2 ( [0, π], X ) ∩ C ( [0, π],D(A) ) is solution of this problem. Let w : R → X be the 2π-periodic repetition of the odd extension of v. Since v(0) = v(π) = 0, w ∈ C1(R, X) ∩ C(R,D(A)). Moreover w is twice continuously differentiable in R \ {jπ | j ∈ Z}, with second derivative equal to −m−2Aw. In the points jπ the second derivative from the left and the second derivative from the right are both equal to −m−2Aw(jπ) = 0, hence w is twice differentiable also EJDE-2020/107 DIRICHLET PROBLEM FOR ABSTRACT DIFFERENTIAL EQUATIONS 5 at these points and w ∈ C2(R, X) ∩ C(R,D(A)). Let z : [0, π] → X be defined by z(t) = w(mt). Then z is solution of the problem u′′(t) +Au(t) = 0 , t ∈ [0, π] , u(0) = 0 , u(π) = 0 ; hence it is identically zero. Therefore v = 0. � 3. Necessary conditions In this section we give necessary conditions for the uniform well-posedness of problem (2.2) in terms of the resolvent operator of A. First of all we have a condition on the resolvent set and an estimate of the resolvent operator of A. Theorem 3.1. Let X be a Banach space and A be a linear closed densely defined operator in X such that problem (2.2) is uniformly well-posed. Then for all k ∈ N, k2 ∈ ρ(A) and there exists C ∈ R+ such that, for all k ∈ N,∥∥(k2I −A)−1 ∥∥ ≤ C k . (3.1) The fact that k2 does not belongs to the point spectrum of A is a particular case of [4, Theorem 1]. Proof. Let S be the operator valued function whose existence is guaranteed by Theorem 2.1 and C = supt∈[0,π] ∥∥S(t) ∥∥. For k ∈ N the operator k2I − A is injective. Indeed if x ∈ D(A) is such that k2x = Ax, it is easy to check that the function t 7→ sin(kt)x is solution of the problem u′′(t) +Au(t) = 0 , t ∈ [0, π] , u(0) = 0 , u(π) = 0 . The unique solution of this problem is the identically zero function, hence x = 0. For k ∈ N and x ∈ X let Rkx = (−1)k+1 k ∫ π 0 sin(ks)S(s)x ds . In this way an operator Rk ∈ L(X) is defined; moreover ‖Rkx‖ ≤ 1 k ∫ π 0 | sin(ks)| ∥∥S(s)x ∥∥ ds ≤ 1 k ∫ π 0 | sin(ks)|C‖x‖ ds = 2C k ‖x‖ , hence ‖Rk‖ ≤ 2C/k. For all x ∈ D(A) we have ARkx = (−1)k+1 k ∫ π 0 sin(ks)AS(s)x ds = (−1)k k ∫ π 0 sin(ks)S′′(s)x ds = (−1)k k [ sin(ks)S′(s)x ]π 0 − (−1)k ∫ π 0 cos(ks)S′(s)x ds = −(−1)k [ cos(ks)S(s)x ]π 0 − (−1)kk ∫ π 0 sin(ks)S(s)x ds = −x+ k2Rkx . 6 G. DORE EJDE-2020/107 If x ∈ X let (xn)n∈N be a sequence in D(A) converging to x. Then Rkxn → Rkx and ARkxn → −x + k2Rkx. Since A is closed this proves that Rkx ∈ D(A) and ARkx = −x+ k2Rkx. Hence, for all x ∈ X, we have (k2I −A)Rkx = x. Therefore Rk is a right inverse of k2I −A. To prove that it is also a left inverse, observe that if x ∈ D(A) then (k2I −A)Rk(k2 −A)x = (k2I −A)x , hence Rk(k2 −A)x = x since k2I −A is injective. Therefore k2I −A is invertible and (k2I −A)−1 = Rk. This proves the theorem. � To prove another necessary condition, we need a result about Fourier series. Lemma 3.2. The series of functions ∑∞ k=1 ( (−1)k+1/k ) sin(kt) converges pointwise for t ∈ [0, π], the sequence of partial sums is uniformly bounded and ∞∑ k=1 (−1)k+1 k sin(kt) =  1 2 t , if 0 ≤ t < π , 0 if t = π . Proof. We have ∞∑ k=1 (−1)k+1 k sin(kt) = ∞∑ k=1 1 k sin ( k(π − t) ) and the convergence of this series is proved in [9, Chapter I, (2.8)]; the uniform boundedness of the partial sums of this series is proved in [9, Chapter II, Section 9]. � Theorem 3.3. Let X be a Banach space and A be a linear closed densely defined operator in X such that problem (2.2) is uniformly well-posed. Then for all m ∈ N the series ∑∞ k=1 ( (mk)2I −A )−1 converges in operator norm. Moreover, if S is the operator valued function introduced in Theorem 2.1, then ∞∑ k=1 ( k2I −A )−1 x = 1 2 ∫ π 0 tS(t)x dt and, for m ∈ N \ {1}, ∞∑ k=1 ( (mk)2I−A )−1 x = 1 2 ∫ π 0 tS(t)x dt− (m− 2j + 1)π 2m [m/2]∑ j=1 ∫ (m−2j+2)π/m (m−2j)π/m S(t)x dt Proof. From the proof of Theorem 3.1 we know that for all k ∈ N and for all x ∈ X (k2I −A)−1x = (−1)k+1 k ∫ π 0 sin(kt)S(t)x dt , Then ∥∥∥ k∑ j=1 ( j2I −A )−1 x− 1 2 ∫ π 0 tS(t)x dt ∥∥∥ = ∥∥∥∫ π 0 k∑ j=1 (−1)j+1 j sin(jt)S(t)x dt− 1 2 ∫ π 0 tS(t)x dt ∥∥∥ EJDE-2020/107 DIRICHLET PROBLEM FOR ABSTRACT DIFFERENTIAL EQUATIONS 7 ≤ ∫ π 0 ∥∥∥( k∑ j=1 (−1)j+1 j sin(jt)− 1 2 t ) S(t)x ∥∥∥ dt ≤ ∫ π 0 ∣∣∣ k∑ j=1 (−1)j+1 j sin(jt)− 1 2 t ∣∣∣ dt C‖x‖. By Lemma 3.2 the series ∑∞ j=1 ( (−1)j+1/j ) sin(jt) converges pointwise to t/2 and the partial sums are uniformly bounded. Therefore, by the dominated convergence theorem, the integral tends to 0 as k tends to∞. This proves the theorem if m = 1. By Theorem 2.2 the same is true for the operator m−2A. Since( k2I −m−2A )−1 = m2 ( (mk)2I −A )−1 , the series ∑∞ k=1 ( (mk)2I − A )−1 converges in operator norm. The expression of the sum of the series can easily be obtained from what we have just proved and equalities (2.3) and (2.4). � 4. Examples In this section we see two examples showing that the estimate of the resolvent stated in Theorem 3.1 cannot be improved but it is not sufficient to guarantee the uniform well-posedness. For the first example we need two lemmas. Lemma 4.1. Let K : [0, π]2 → R be such that K(t, s) = { (t−π)s π , if s ≤ t, t(s−π) π , if s > t. (1) If f ∈ C ( [0, π], X ) then the function z : [0, π]→ X such that z(t) = ∫ π 0 K(t, s)f(s) ds belongs to C2 ( [0, π], X ) , with z′′ = f and z(0) = z(π) = 0. (2) If z ∈ C2 ( [0, π], X ) , with z(0) = z(π) = 0 then, for all t ∈ [0, π], we have z(t) = ∫ π 0 K(t, s)z′′(s) ds . The proof of the above lemma is an easy consequence of the fundamental theorem of calculus. Lemma 4.2. Let (vk)k∈N be a sequence in C2 ( [0, π], X) such that (1) (vk)k∈N converges pointwise to v ∈ C ( [0, π], X); (2) (v′′k )k∈N converges uniformly to w ∈ C ( [0, π], X). Then v ∈ C2 ( [0, π], X) and v′′ = w. Proof. Let zk : [0, π]→ X , zk(t) = vk(t)− π − t π vk(0)− t π vk(π) . We have zk(0) = zk(π) = 0 and z′′k = v′′k . Hence, by Lemma 4.1, for all t ∈ [0, π], we have vk(t) = zk(t) + π − t π vk(0) + t π vk(π) 8 G. DORE EJDE-2020/107 = ∫ π 0 K(t, s)v′′k (s) ds+ π − t π vk(0) + t π vk(π) . By (2) we can pass to the limit under the integral sign, hence v(t) = ∫ π 0 K(t, s)w(s) ds+ π − t π v(0) + t π v(π) . Therefore, by Lemma 4.1, v is twice differentiable and v′′ = w. � Example 4.3. Let X = `p(N,K), with 1 ≤ p <∞. Choose α such that 0 < α < 1 and let A be the operator in X defined by D(A) = { x ∈ X ∣∣ (n2xn)n∈N ∈ X } , (Ax)n = (n− α)2xn . The operator A is linear, it is easy to prove that it is closed and has dense domain, since the domain contains the eventually zero sequences. For t ∈ [0, π] let S(t) : X → X , ( S(t)x ) n = sin ( (n− α)t ) sin ( (n− α)π ) xn . We prove that S satisfies conditions (1)–(4) of Theorem 2.1, hence the Dirichlet problem for the operator A is uniformly well-posed. Since ∞∑ n=1 ∣∣∣ sin ( (n− α)t ) sin ( (n− α)π ) xn∣∣∣p ≤ ∞∑ n=1 1( sin(απ) )p |xn|p , S(t) ∈ L(X) with ∥∥S(t) ∥∥ ≤ (sin(απ) )−1 . (4.1) A similar argument shows that if x ∈ D(A) then S(t)x ∈ D(A). Moreover we have AS(t)x = S(t)Ax. The function ( S(·)x ) n is continuous, for every x ∈ X and n ∈ N; therefore if x is an eventually zero sequence then S(·)x is continuous. If x ∈ X then it is the limit of a sequence of eventually zero elements of X, hence, by estimate (4.1), S(·)x is the uniform limit of a sequence of continuous functions, therefore it is continuous. This proves (1). Property (2) is obvious. The function ( S(·)x ) n is of class C2, for every x ∈ X and n ∈ N, and d2 dt2 ( S(t)x ) n = d2 dt2 ( sin ( (n− α)t ) sin ( (n− α)π ) xn) = −(n− α)2 sin ( (n− α)t ) sin ( (n− α)π ) xn . Therefore if x ∈ D(A) then d2 dt2 ( S(t)x ) n = − ( AS(t)x ) n . Hence if x is eventually zero then S(·)x is solution of equation (2.1). Now let x ∈ D(A) and, for m ∈ N, let x(m) ∈ X such that( x(m) ) n = { xn , for n ≤ m, 0 , for n > m . We have sup t∈[0,π] ∥∥S(t)x(m) − S(t)x ∥∥ ≤ sup t∈[0,π] ∥∥S(t) ∥∥∥∥x(m) − x ∥∥ EJDE-2020/107 DIRICHLET PROBLEM FOR ABSTRACT DIFFERENTIAL EQUATIONS 9 ≤ 1 sin(απ) ( ∞∑ n=m+1 |xn|p )1/p −−−−→ m→∞ 0 , and sup t∈[0,π] ∥∥S(t)Ax(m) − S(t)Ax ∥∥ ≤ sup t∈[0,π] ∥∥S(t) ∥∥ ∥∥Ax(m) −Ax ∥∥ ≤ 1 sin(απ) ( ∞∑ n=m+1 (n− α)2p|xn|p )1/p −−−−→ m→∞ 0 . Hence the sequence of functions S(·)x(m) converges uniformly to S(·)x and the se- quence S′′(·)x(m) = −AS(·)x(m) = −S(·)Ax(m) converges uniformly to the function −S(·)Ax = −AS(·)x. By Lemma 4.2 S(·)x ∈ C2 ( [0, π], X ) and S′′(·)x = −AS(·)x. Therefore (3) is satisfied. If v : [0, π]→ X is solution of equation (2.1), then for all n ∈ N we have v′′n(t) + (n− α)2vn(t) = 0 . The functions t 7→ sin ( (n − α)t ) and t 7→ sin ( (n − α)(π − t) ) are two linearly independent solutions of this equation, therefore there exist c1, c2 ∈ K such that vn(t) = c1 sin ( (n− α)t ) + c2 sin ( (n− α)(π − t) ) . By putting t = π or t = 0 in this equality we get c1 = vn(π) sin ( (n− α)π ) , c2 = vn(0) sin((n− α)π ) , hence vn(t) = sin ( (n− α)(π − t) ) sin ( (n− α)π ) vn(0) + sin ( (n− α)t ) sin ( (n− α)π ) vn(π) , that is v(t) = S(π − t)v(0) + S(t)v(π) . Therefore (4) is satisfied. If k ∈ N it is easy to check that k2I − A is invertible and, for all x ∈ X, for all n ∈ N, ( (k2 −A)−1x ) n = 1 k2 − (n− α)2 xn . If we denote with ek the k-th element of the canonical basis of X, we have∥∥(k2 −A)−1 ∥∥ ≥ ∥∥(k2 −A)−1ek ∥∥ = 1 k2 − (k − α)2 = 1 2kα− α2 ≥ 1 2kα . Therefore estimate (3.1) is the best possible one with respect to the dependence on k. Moreover the best constant C can be arbitrarily large. The next example is based on the results of [8]. We recall some facts from that article. Let X be a Banach space and (yn)n∈N be a basis of X. We denote with Pn the n-th projection operator associated with this basis, i.e. Pn ∈ L(X) is such that, for all x ∈ X, we have x = ∑∞ n=1 Pn(x)yn. By the uniform boundedness principle M0 = supn∈N ∥∥∑n k=1 Pn ∥∥ <∞. 10 G. DORE EJDE-2020/107 We associate with every scalar valued sequence a = (an)n∈N a linear operator A in X defined by D(A) = { x ∈ X ∣∣ ∞∑ n=1 anPn(x)yn is convergent } , Ax = ∞∑ n=1 anPn(x)yn . Let a be a scalar valued bounded sequence; we set ‖a‖∞ = supn∈N |an| and V (a) = ∑∞ n=1 |an+1 − an|. We denote with BV the space of the sequences of bounded variation, i.e. such that V (a) <∞. Lemma 4.4 ([8, Lemma 2.4]). Let a be a scalar valued sequence. Then the operator A associated with a is densely defined and closed. Moreover if a ∈ BV , then A ∈ L(X), with ‖A‖ ≤M0 ( ‖a‖∞ + V (a) ) . Lemma 4.5 ([8, Lemma 2.5]). Let a be a scalar valued sequence, A be the oper- ator associated with a and λ ∈ K \ {an |n ∈ N}. Then λI − A is one-to-one and (λI −A)−1 is the operator associated with the sequence ( (λ− an)−1 ) n∈N. In par- ticular λ ∈ ρ(A) if and only if for all x ∈ X the series ∑∞ n=1(λ− an)−1Pn(x)yn is convergent. We say that the basis (yn)n∈N of the Banach space X is unconditional if, for all x ∈ X, the series ∑∞ n=1 Pn(x)yn is unconditionally convergent. Otherwise we say that the basis is conditional (see [7, Chapter II, Definition 14.1]). Every Banach space with a basis has an unconditional basis (see [7, Chapter II, Theorem 23.2]). Lemma 4.6 ([7, Chapter II, Theorem 16.1, 1⇔ 8]). Let (yn)n∈N be a conditional basis of the space X. Then there exists x ∈ X and a sequence (εn)n∈N in {−1, 1} such that sup n∈N ∥∥∥ n∑ j=1 εjPj(x) ∥∥∥ =∞ . Example 4.7. Let X be a Banach space with a conditional basis (yn)n∈N and let Pn be the n-th projection operator associated with this basis. Let (`n)n∈N be a sequence of non-negative integers, unbounded and such that `1 = 0 and `n+1 − `n ∈ {0, 1}. Set an = ( 2`n + 1 2 )2 and let A be the operator associated with the sequence (an)n∈N. By Lemma 4.4, A is closed and densely defined. We prove that A satisfies the conditions of Theorem 3.1. Let k ∈ N. For every n ∈ N we have an 6= k2, hence, by Lemma 4.5, (k2I − A) is one-to-one and, if we set bn = 1/(k2−an), then (k2I−A)−1 is the operator associated with the sequence (bn)n∈N. For every ` ∈ N we have∣∣∣k2 − (2`+ 1 2 )2∣∣∣ ≥ ∣∣∣k2 − (k − 1 2 )2∣∣∣ = k − 1 4 , hence ‖b‖∞ = 1 infn∈N |k2 − an| ≤ 1 k − (1/4) . EJDE-2020/107 DIRICHLET PROBLEM FOR ABSTRACT DIFFERENTIAL EQUATIONS 11 Let n = max{n ∈ N | 2`n < k}; the sequence bn is positive and non-decreasing for 1 ≤ n ≤ n while is negative and non-decreasing for n ≥ n+ 1. Hence V (b) = ∞∑ n=1 |bn+1 − bn| = n−1∑ n=1 (bn+1 − bn) + bn − bn+1 + ∞∑ n=n+1 (bn+1 − bn) = bn − b1 + bn − bn+1 + lim n→∞ bn − bn+1 ≤ 2bn − 2bn+1 ≤ 4 k − (1/4) . Therefore, by Lemmas 4.4 and 4.5, k ∈ ρ(A) and there exists C ∈ R+ such that∥∥(k2I −A)−1 ∥∥ ≤ C/k. Now we prove that A satisfies the condition of Theorem 3.3 with m = 1, that is the series ∑∞ k=1(k2I −A)−1 converges in operator norm. First of all for all k ∈ N and all n ∈ N0 we have ∞∑ j=k 1 j2 − ( n+ (1/2) )2 = lim p→∞ p∑ j=k 1 j2 − ( n+ (1/2) )2 = 1 2n+ 1 lim p→∞ p∑ j=k ( 1 j − n− (1/2) − 1 j + n+ (1/2) ) = 1 2n+ 1 lim p→∞ ( p−n∑ `=k−n 1 `− (1/2) − p+n+1∑ `=k+n+1 1 `− (1/2) ) = 1 2n+ 1 lim p→∞ ( k+n∑ `=k−n 1 `− (1/2) − p+n+1∑ `=p−n+1 1 `− (1/2) ) = 1 2n+ 1 k+n∑ `=k−n 1 `− (1/2) . Hence ∞∑ j=k 1 j2 − ( n+ (1/2) )2 = 1 2n+ 1 k+n∑ `=k−n 1 `− (1/2) ; (4.2) in particular ∞∑ j=1 1 j2 − ( n+ (1/2) )2 = 1 2n+ 1 1+n∑ `=1−n 1 `− (1/2) = 2 (2n+ 1)2 , where the last equality can be easily proved by induction on n. Therefore, for all n ∈ N, we have ∞∑ j=1 1 j2 − ( 2`n + (1/2) )2 = 2 (4`n + 1)2 . Let B be the operator in X associated with the sequence ( 2/ ( 4`n + 1 )2) n∈N. This sequence is non-negative and decreasing, hence it is bounded and has bounded 12 G. DORE EJDE-2020/107 variation, therefore, by Lemma 4.4, B is bounded. We prove that ∞∑ k=1 (k2I −A)−1 = B . The operator B − ∑k−1 j=1 (j2I − A)−1 is associated with the sequence whose n-th term, taking into account (4.2), is 2( 4`n + 1 )2 − k−1∑ j=1 1 j2 − ( 2`n + (1/2) )2 = ∞∑ j=k 1 j2 − ( 2`n + (1/2) )2 = 1 4`n + 1 k+2`n∑ j=k−2`n 1 j − (1/2) , If we put Ck,` = 1 2`+ 1 k+∑̀ j=k−` 1 j − (1/2) , we have to prove that the sequence (Ck,2`n)n∈N has least upper bound and variation converging to 0 as k → 0. Obviously sup n∈N |Ck,2`n | ≤ sup `∈N0 |Ck,`| . Moreover, since `n+1− `n ∈ {0, 1} each difference Ck,2`n+1 −Ck,2`n either is null or is equal to Ck,2`+2 − Ck,2` for a suitable ` ∈ N0. Hence ∞∑ n=1 |Ck,2`n+1 − Ck,2`n | ≤ ∞∑ `=0 |Ck,2`+2 − Ck,2`| ≤ ∞∑ `=0 |Ck,`+1 − Ck,`| . For all k ∈ N and ` ∈ N0 we have Ck,` > 0. Indeed, if ` < k then Ck,` > 0, since each term in the sum is positive. If ` ≥ k then k+∑̀ j=k−` 1 j − (1/2) = `−k+1∑ j=k−` 1 j − (1/2) + k+∑̀ j=`−k+2 1 j − (1/2) = k+∑̀ j=`−k+2 1 j − (1/2) > 0 . If we put ck,j = 1/ ( k + j − (1/2) ) , we have ck,−j + ck,j = 2k − 1( k − (1/2) )2 − j2 Hence it is easy to prove that 2ck,0 < ck,−1 + ck,1 < ck,−2 + ck,2 < · · · < ck,−k+1 + ck,k−1 and, if k ≤ j, ck,−j + ck,j < 0. We observe that Ck,` is the mean of ck,−`, ck,−`+1, . . . , ck,`−−, ck,` and Ck,`+1 is the mean of the same numbers plus ck,−`−1 and ck,`+1. If ` + 1 < k the mean of ck,−`−1 and ck,`+1 is greater then the mean of ck,−`, . . . , ck,`, hence Ck,`+1 > Ck,`. If `+1 ≥ k the mean of ck,−`−1 and ck,`+1 is negative, while the mean of ck,−`, . . . , ck,` is positive, hence Ck,`+1 < Ck,`. Therefore sup{Ck,` | ` ∈ N0} = Ck,k−1. Moreover ∞∑ `=0 |Ck,`+1 − Ck,`| = k−2∑ `=0 (Ck,`+1 − Ck,`) + ∞∑ `=k−1 (Ck,` − Ck,`+1) EJDE-2020/107 DIRICHLET PROBLEM FOR ABSTRACT DIFFERENTIAL EQUATIONS 13 = Ck,k−1 − Ck,0 + Ck,k−1 − lim `→∞ Ck,`+1 < 2Ck,k−1 . Since Ck,k−1 = 1 2k − 1 2k−1∑ j=1 1 j − (1/2) ≤ 1 2k − 1 ( 2 + ∫ 2k−1 1 1 x− (1/2) dx ) = 1 2k − 1 ( 2 + log(4k − 3) ) −−−−→ k→∞ 0 , the statement about the sequence (Ck,`n)n∈N is proved. Let v : [0, π]→ X be a solution of equation (2.1). Then for all n ∈ N we have d2 dt2 Pn ( v(t) ) = Pn ( v′′(t) ) = −Pn ( Av(t) ) = −anPn ( v(t) ) . Hence the function Pn ◦ v is solution of the equation u′′(t) + anu(t) = 0, that is u′′(t) + ( 2`n + (1/2) )2 u(t) = 0. Hence there exist c1, c2 ∈ K such that Pn ( v(t) ) = c1 sin (( 2`n + 1 2 ) (π − t) ) + c2 sin (( 2`n + 1 2 ) t ) . Obviously we have c1 = Pn ( v(0) ) , c2 = Pn ( v(π) ) . Now we choose a particular sequence (`n)n∈N. By Lemma 4.6, there exist x ∈ X and a sequence (εn)n∈N, in {−1, 1} such that sup n∈N ∥∥∥ n∑ k=1 εkPk(x) ∥∥∥ =∞ . The sequence (εn)n∈N can be chosen such that ε1 = 1. We put ε0 = 1 and, for n ∈ N, `n = (1/2) ∑n k=1 |εk−εk−1|. In this way we have defined a non-decreasing sequence of integer numbers, with `1 = 0 and `n+1− `n ∈ {0, 1}. Since the sequence (εn)n∈N must have an infinite number of changes of sign, the sequence `n is unbounded. Moreover it is easy to show, by induction, that, for all n ∈ N, we have (−1)`n = εn. If v is solution of the problem u′′(t) +Au(t) = 0 , u(0) = 0 , u(π) = x , (4.3) then Pn ( v(t) ) = sin (( 2`n + 1 2 ) t ) Pn(x) , hence Pn ( v (π 2 )) = sin ( `nπ + π 4 ) Pn(x) = (−1)`n√ 2 Pn(x) = εn√ 2 Pn(x) . Since the series ∑∞ k=1 εkPk(x) doesn’t converge, problem (4.3) doesn’t have solu- tion. 14 G. DORE EJDE-2020/107 5. Bounded operators If the operator A is bounded there is a simple characterization of uniform well- posedness of the Dirichlet problem. For the proof of the theorem we need a result about Fourier series. Lemma 5.1. The series of functions ∑∞ k=1 ( (−1)k+1/k3 ) sin(kt) converges uni- formly for t ∈ [0, π] and 2 π ∞∑ k=1 (−1)k+1 k3 sin(kt) = π 6 t− 1 6π t3 . Proof. The uniform convergence of the series is obvious, since it is uniformly esti- mated by the harmonic series of exponent 3. Let f : [−π, π]→ R , f(t) = π 6 t− 1 6π t3 . It is easy to show that, for all k ∈ N, we have∫ π 0 f(t) sin(kt) dt = (−1)k+1 k3 . Hence the Fourier coefficients of the odd function f are equal to 2(−1)k+1/(πk3) (see [9] Chapter I, (4.8)). Since f ∈ C1 ( [−π, π],R ) and its 2π-periodic repetition has the same regularity, by [9, Chapter II,Theorem (8.1)] its Fourier series converges to f . � Theorem 5.2. Let X be a Banach space and A ∈ L(X). Problem (2.2) is uniformly well-posed if and only if {k2 | k ∈ N} ⊆ ρ(A). Proof. The necessity of the condition is a consequence of Theorem 3.1. We prove that, if {k2 | k ∈ N} ⊆ ρ(A), there exists S : [0, π]→ X that satisfies conditions (1)–(4) of Theorem 2.1; hence problem (2.2) is uniformity well-posed. It is well known that if λ ∈ K is such that |λ| > ‖A‖ then λ ∈ ρ(A) and (λI −A)−1 = ∑∞ n=0 λ −n−1An, hence∥∥(λI −A)−1 ∥∥ ≤ ∞∑ n=0 |λ|−n−1‖A‖n = 1 |λ| − ‖A‖ . Therefore there exists M ∈ R+ such that, for all k ∈ N, we have∥∥(k2I −A)−1 ∥∥ ≤ M k2 . For t ∈ [0, π] and x ∈ X put S(t)x = 2 π ∞∑ k=1 (−1)k+1 k3 sin(kt) (k2I −A)−1A2x+ (π 6 t− 1 6π t3 ) Ax+ t π x . We have∥∥S(t)x ∥∥ ≤ 2 π ∞∑ k=1 1 k3 | sin(kt)| ∥∥(k2I −A)−1 ∥∥ ‖A‖2‖x‖+ (π 6 t− 1 6π t3 ) ‖A‖ ‖x‖+ t π ‖x‖ ≤ 2 π ∞∑ k=1 M k5 ‖A‖ ‖x‖+ π2 6 ‖A‖ ‖x‖+ ‖x‖ = C‖x‖ . EJDE-2020/107 DIRICHLET PROBLEM FOR ABSTRACT DIFFERENTIAL EQUATIONS 15 Hence S(t) ∈ L(X). This estimate shows also that the series is uniformly conver- gent, therefore the function S(·)x is continuous. Hence (1) is satisfied. Obviously (2) is satisfied. It is easy to check that the series of first and second derivatives are uniformly convergent, hence S(·)x ∈ C2 ( [0, π], X ) and, for all t ∈ [0, π], S′′(t) +AS(t) = 2 π ∞∑ k=1 (−1)k k3 sin(kt)k2(k2I −A)−1A2x− t π Ax + 2 π ∞∑ k=1 (−1)k+1 k3 sin(kt)A(k2I −A)−1A2x+ (π 6 t− 1 6π t3 ) A2x+ t π Ax = 2 π ∞∑ k=1 (−1)k k3 sin(kt)A2x+ (π 6 t− 1 6π t3 ) A2x = 0 , the last equality follows from Lemma 5.1. Therefore condition (3) is satisfied. To prove (4) we first show the uniqueness of the solution of problem (2.2). Let v ∈ C2 ( [0, π], X ) be a solution of u′′(t) +Au(t) = 0 , t ∈ [0, π] , u(0) = 0 , u(π) = 0 . For k ∈ N let ak = ∫ π 0 sin(kt)v(t) dt . Then Aak = ∫ π 0 sin(kt)Av(t) dt = − ∫ π 0 sin(kt)v′′(t) dt = − [ sin(kt)v′(t) ]π 0 + ∫ π 0 k cos(kt)v′(t) dt = [ k cos(kt)v(t) ]π 0 + ∫ π 0 k2 sin(kt)v(t) dt = k2ak . Hence (k2I−A)ak = 0. Since k2I−A is injective, this proves that ak = 0. Therefore all the Fourier coefficients of the 2π-periodic repetition of the odd extension of v are null, hence v(t) = 0 for a.e. t. Since v is continuous v(t) = 0 for all t. If x0, xπ ∈ X, then the function t 7→ S(π − t)x0 + S(t)xπ is solution of problem (2.2), but the solution is unique, hence every solution of problem (2.2) coincides with that function. Therefore (4) is satisfied. � Acknowledgments. The author wants to thank Alberto Venni for the useful dis- cussions about this article. References [1] W. Arendt, S. Bu; The operator-valued Marcinkiewicz multiplier theorem and maximal regu- larity, Math. Z., 240 (2002), 311–343. 16 G. DORE EJDE-2020/107 [2] A. Ashyralyev, J. Pastor, S. Piskarev, H. A. Yurtsever; Second order equations in functional spaces: qualitative and discrete well-posedness, Abstr. Appl. Anal., 2015, Art. ID 948321, 63 pp. [3] Ph. Clément, S. Guerre-Delabrière; On the regularity of abstract Cauchy problems and bound- ary value problems, Atti Accad. Naz. 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Giovanni Dore Department of Mathematics, University of Bologna, Piazza di Porta San Donato 5, I-40126, Bologna, Italy Email address: giovanni.dore@unibo.it 1. Introduction 2. Well-posed problems 3. Necessary conditions 4. Examples 5. Bounded operators Acknowledgments References