Zeitschrift für Analysis und ihre Anwendungen

Journal for Analysis and its Applications Volume 18 (1999), No. 4, 1101–1106

A Kneser-Type Theorem for the Equation x^{(m)} = f(t, x) in Locally Convex Spaces

A. Szukała

Abstract. We shall give sufficient conditions for the existence of solutions of the Cauchy problem for the equation x^{(m)} = f(t, x). We also prove that the set of these solutions is a continuum.

Keywords: Differential equations, set of solutions, measures of non-compactness AMS subject classification: 34 G 20

Let E be a quasicomplete locally convex topological vector space, and let P be a family of continuous seminorms generating the topology of E. Assume that I = [0, a] and B = {x ∈ E : p_{i}(x) ≤ b (i = 1, …, k)}, where p_{1}, …, p_{k} ∈ P.

In this paper we investigate the existence of solutions and the structure of the set of solutions of the Cauchy problem

x^{(m)} = f(t, x) ; x(0) = 0 ; x′(0) = η_{1} ; ⋮ ; x^{(m−1)}(0) = η_{m−1} }  (1)

where m is a positive integer, η_{1}, η_{2}, …, η_{m−1} ∈ E and f is a bounded continuous function from I × B into E. Our considerations are a continuation of Szufla’s paper [8]. For other results concerning differential equations in locally convex spaces see [4].

Put

M = sup {p_{i}(f(t, x)) : t ∈ I, x ∈ B, i = 1, …, k}.

Choose a positive number d such that d ≤ a and

∑_{j=1}^{m−1} p_{i}(η_{j})\frac{d^{j}}{j!} + M \frac{d^{m}}{m!} ≤ b  (i = 1, …, k).  (2)

A. Szukała: A. Mickiewicz Univ., Fac. Math. & Comp. Sci., J. Matejki 48/49, 60 - 769 Poznań, Poland ISSN 0232-2064 / $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag Berlin

Let J = [0, d]. Denote by C = C(J, E) the space of all continuous functions from J into E endowed with the topology of uniform convergence.

For any bounded subset A of E and p ∈ P we denote by β_{p}(A) the infimum of all ε > 0 for which there exists a finite subset {x_{1}, x_{2}, …, x_{n}} of E such that A ⊂ {x_{1}, x_{2}, …, x_{n}} + B_{p}(ε), where B_{p}(ε) = {x ∈ E : p(x) ≤ ε}. The family (β_{p}(A))_{p∈P} is called the measure of non-compactness of A. It is known [6] that:

1^{∘} X is relatively compact in E ⟺ β_{p}(X) = 0 for every p ∈ P. 2^{∘} X ⊂ Y ⟹ β_{p}(X) ≤ β_{p}(Y). 3^{∘} β_{p}(X ∪ Y) = max{β_{p}(X), β_{p}(Y)}. 4^{∘} β_{p}(X + Y) ≤ β_{p}(X) + β_{p}(Y). 5^{∘} β_{p}(λX) = |λ|β_{p}(X) (λ ∈ ℝ). 6^{∘} β_{p}(X̄) = β_{p}(X). 7^{∘} β_{p}(conv X) = β_{p}(X). 8^{∘} β_{p}(∪_{0≤λ≤h}λX) = hβ_{p}(X).

The following lemma is given in [8].

Lemma 1. Let H be a bounded countable subset of C. For each t ∈ J put H(t) = {u(t) : u ∈ H}. If the space E is separable, then for each p ∈ P the function t ↦ β_{p}(H(t)) is integrable and

β_{p} ({∫_{J} u(s) ds : u ∈ H}) ≤ ∫_{J} β_{p}(H(s))ds.

Moreover, let us recall the following lemma from [9].

Lemma 2. Let w : [0, 2b] ↦ ℝ_{+} be a continuous non-decreasing function and let g : [0, c) ↦ [0, 2b] be a C^{m}-function satisfying the inequalities

\begin{aligned}g^{(j)}(t) & ≥ 0  (j = 0, 1, …, m) \\ g^{(j)}(0) & = 0  (j = 0, 1, …, m − 1) \\ g^{(m)}(t) & ≤ w(g(t))  (t ∈ [0, c)). \end{aligned}

If w(0) = 0, w(r) > 0 for r > 0 and ∫_{0+} (r^{m−1}w(r))^{− \frac{1}{m}} dr = ∞, then g = 0.

We can now formulate our main result.

Theorem. Suppose that for each p ∈ P there exists a continuous non-decreasing function w_{p} : ℝ_{+} ↦ ℝ_{+} such that w_{p}(0) = 0, w_{p}(r) > 0 for r > 0 and

∫_{0+} \frac{dr}{\sqrt[m]{r^{m−1}w_{p}(r)}} = ∞.  (3)

If

β_{p}(f(t, X)) ≤ w_{p}(β_{p}(X))  (4)

A Kneser-Type Theorem for x^{(m)} = f(t, x) 1103

for p ∈ P, t ∈ I and bounded subsets X of E, then the set S of all solutions of problem (1) defined on J is non-empty, compact and connected in C(J, E).

Proof. 1^{∘} Put

r(x) = \{ \begin{matrix}x & for x ∈ B \\ \frac{x}{K(x)} & for x ∈ E \ B \end{matrix}

and g(t, x) = f(t, r(x)) for (t, x) ∈ J × E, where K is the Minkowski functional of B. As B is a closed, balanced and convex neighbourhood of 0, from known properties of the Minkowski functional it follows that r is a continuous function from E into B and

r(X) ⊂ ⋃_{0≤λ≤1} λX for any subset X of E.

Thus β_{p}(r(X)) ≤ β_{p}(X) for any p ∈ P and any bounded subset X of E. Consequently, g is a bounded continuous function from J × E into E such that

β_{p}(g(t, X)) ≤ w_{p}(β_{p}(X))  (4)′

for p ∈ P, t ∈ J and bounded subsets X of E and

p_{i}(g(t, x)) ≤ M  (i = 1, …, k; t ∈ J, x ∈ E).  (5)

We introduce a mapping F defined by

F (x)(t) = q(t) + \frac{1}{(m − 1)!} ∫_{0}^{t}(t − s)^{m−1}g(s, x(s)) ds (t ∈ J, x ∈ C)

where q(t) = ∑_{j=1}^{m−1} η_{j} \frac{t^{j}}{j!} . It is known (cf. [2]) that F is a continuous mapping C ↦ C and the set F (C) is bounded and equicontinuous. It is clear from (1) and (5) that if x = F (x), then

\begin{matrix}p_{i}(x(t)) & ≤ ∑_{j=1}^{m−1} p_{i}(η_{j})\frac{d^{j}}{j!} + \frac{1}{(m − 1)!} ∫_{0}^{t}(t − s)^{m−1}M ds \\ & ≤ ∑_{j=1}^{m−1} p_{i}(η_{j})\frac{d^{j}}{j!} + M \frac{d^{m}}{m!} \\ & ≤ b \end{matrix} (i = 1, …, k)

so x(t) ∈ B for t ∈ J. Therefore, a function x ∈ C is a solution of problem (1) if and only if x = F (x).

2^{∘} For any n ∈ ℕ put

u_{n}(t) = \{ \begin{matrix}0 & if 0 ≤ t ≤ \frac{d}{n} \\ q(t − \frac{d}{n}) + \frac{1}{(m−1)!} ∫_{0}^{t− \frac{d}{n}} (t − s)^{m−1}g(s, u_{n}(s)) ds & if \frac{d}{n} ≤ t ≤ d. \end{matrix}

Then u_{n} is a continuous function J ↦ B and

lim_{n→∞} (u_{n}(t) − F (u_{n})(t)) = 0  (6)

uniformly for t ∈ J. Let V = {u_{n} : n ∈ ℕ}. From (6) it follows that the set {u_{n}−F (u_{n}) : n ∈ ℕ} is relatively compact in C. Since

V ⊂ {u_{n} − F (u_{n}) : n ∈ ℕ} + F (V)  (7)

and the set F (V) is bounded and equicontinuous, we conclude that the set V is also bounded and equicontinuous. Hence for each p ∈ P the function t ↦ β_{p}(V (t)) is continuous on J. Denote by H a closed separable subspace of E such that

g(s, u_{n}(s)) ∈ H (s ∈ J, n ∈ ℕ).

Let (β_{p}^{H})_{p∈P} be the measure of non-compactness in H. Fix t ∈ J and p ∈ P. From (4)′ we have

β_{p}^{H}(g(s, V (s))) ≤ 2β_{p}(g(s, V (s))) ≤ 2w_{p}(β_{p}(V (s))) (s ∈ [0, t]).

By Lemma 1, we get

β_{p}(F (V)(t)) = β_{p} (\{\frac{1}{(m − 1)!} ∫_{0}^{t}(t − s)^{m−1}g(s, u_{n}(s)) ds : n ∈ ℕ\})

≤ β_{p}^{H} (\{\frac{1}{(m − 1)!} ∫_{0}^{t}(t − s)^{m−1}g(s, u_{n}(s)) ds : n ∈ ℕ\})

≤ \frac{1}{(m − 1)!} ∫_{0}^{t} β_{p}^{H} (\{(t − s)^{m−1}g(s, u_{n}(s)) : n ∈ ℕ\}) ds

= \frac{1}{(m − 1)!} ∫_{0}^{t}(t − s)^{m−1}β_{p}^{H}(g(s, V (s)) ds

≤ \frac{2}{(m − 1)!} ∫_{0}^{t}(t − s)^{m−1}w_{p}(β_{p}(V (s)) ds.

On the other hand, from (6) and (7) we obtain

β_{p}(V (t)) ≤ β_{p}(F (V)(t)).

Hence

β_{p}(V (t)) ≤ \frac{2}{(m − 1)!} ∫_{0}^{t}(t − s)^{m−1}w_{p}(β_{p}(V (s)) ds (t ∈ J, p ∈ P).

Putting

g(t) = \frac{2}{(m − 1)!} ∫_{0}^{t}(t − s)^{m−1}w_{p}(β_{p}(V (s)) ds

we see that

g ∈ C^{m}
β_{p}(V (t)) ≤ g(t)
g^{(j)}(t) ≥ 0 for j = 0, 1, …, m
g^{(j)}(0) = 0 for j = 0, 1, …, m − 1
g^{(m)}(t) = 2w_{p}(β_{p}(V (t))) ≤ 2w_{p}(g(t)) for t ∈ J.

Moreover, by (3),

∫_{0+} \frac{dr}{\sqrt[m]{r^{m−1}2w_{p}(r)}} = ∞.

By Lemma 2 from this we deduce that g(t) = 0 for t ∈ J. Thus β_{p}(V (t)) = 0 for t ∈ J and p ∈ P. Therefore for each t ∈ J the set V (t) is relatively compact in E. As the set V is equicontinuous, Ascoli’s theorem proves that V is relatively compact in C. Hence the sequence (u_{n}) has a limit point u. As F is continuous from (6) we conclude that u = F (u), i.e. u is a solution of problem (1). This proves that the set S is non-empty.

3^{∘} Let us first remark that the set S is compact in C. Indeed, as (I − F)(S) = {0}, in the same way as in Step 2^{∘}, we can prove that S is relatively compact in C. Moreover, from the continuity of F it follows that S is closed in C. Suppose that S is not connected. Thus there exist non-empty closed sets S_{0} and S_{1} such that S = S_{0} ∪S_{1} and S_{0} ∩S_{1} = ∅. As S_{0} and S_{1} are compact subsets of C and C is a Tichonov space, this implies (see [3: §41, II, Remark 3]) the existence of a continuous function v : C ↦ [0, 1] such that v(x) = 0 for x ∈ S_{0} and v(x) = 1 for x ∈ S_{1}. Further, for any n ∈ ℕ we define a mapping F_{n} by

F_{n}(x)(t) = F (x)(r_{n}(t)) (x ∈ C, t ∈ J)

where

r_{n}(t) = \{ \begin{matrix}0 & for 0 ≤ t ≤ \frac{d}{n} \\ t − \frac{d}{n} & for \frac{d}{n} ≤ t ≤ d. \end{matrix}

It can be easily verified (cf. [10]) that:

(i) F_{n} is a continuous mapping C ↦ C.

(ii) lim_{n→∞} F_{n}(x) = F (x) uniformly for x ∈ C.

(iii) I − F_{n} is a homeomorphism C ↦ C (I - identity mapping).

Fix u_{0} ∈ S_{0}, u_{1} ∈ S_{1} and n ∈ ℕ. Put

e_{n}(λ) = λ(u_{1} − F_{n}(u_{1})) + (1 − λ)(u_{0} − F_{n}(u_{0})) (0 ≤ λ ≤ 1).

Let u_{nλ} = (I − F_{n})^{−1}(e_{n}(λ)). As e_{n}(λ) depends continuosly on λ and I − F_{n} is a homeomorphism, we see that the mapping λ ↦ v(u_{nλ}) is continuous on [0, 1]. Moreover, u_{n0} = u_{0} and u_{n1} = u_{1}, so that v(u_{n0}) = 0 and v(u_{n1}) = 1. Thus there exists λ_{n} ∈ [0, 1] such that

v(u_{nλ_{n}}) = \frac{1}{2}.  (8)

For simplicity put v_{n} = u_{nλ_{n}} and V = {v_{n} : n ≥ 1}. Since lim_{n→∞} e_{n}(λ) = 0 uniformly for λ ∈ [0, 1], we get

lim_{n→∞}(v_{n} − F (v_{n})) = lim_{n→∞}(e_{n}(λ) + F_{n}(v_{n}) − F (v_{n})) = 0  (9)

and therefore the set (I−F)(V) is relatively compact in C. Using now a similar argument as in Step 2^{∘}, we can prove that the set V is relatively compact in C. Consequently, the sequence (v_{n}) has a limit point z. In view of (9) and the continuity of F, we infer that z ∈ S, so v(z) = 0 or v(z) = 1. On the other hand, from (8) it is clear that v(z) = \frac{1}{2}, which yields a contradiction. Thus S is connected

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Received 28.10.1998