Mem. Differential Equations Math. Phys. 6(1995), 109–112

I. Kiguradze

ON SOLVABILITY OF FUNCTIONAL EQUATIONS IN THE SPACE OF CONTINUOUS VECTOR FUNCTIONS

(Reported on September 18, 1995)

In the present note, we establish sufficient conditions for solvability of the functional equation

x(t) = p(x)(t) + q(x)(t),  (1)

where p : C([a, b]; R^{n}) → C([a, b]; R^{n}) and q : C([a, b]; R^{n}) → C([x, b]; R^{n}) are, respectively, linear and nonlinear operators.

Before passing to the statement of the basic result, we will give some notation and definitions necessary in the sequel.

R is the set of real numbers, R_{+} = [0, +∞[; R^{n} is the space of n-dimensional column vectors x = (x_{i})_{i=1}^{n} with elements x_{i} ∈ R

(i = 1, …, n) and the norm ∥x∥ = ∑_{i=1}^{n} |x_{i}|;

R^{n×n} is the space of n × n-matrices X = (x_{ik})_{i,k=1}^{n} with elements x_{ik} ∈ R (i, k = 1, …, n);

if x = (x_{i})_{i=1}^{n} ∈ R^{n} and X = (x_{ik})_{i,k=1}^{n} ∈ R^{n×n}, then |x| = (|x_{i}|)_{i=1}^{n} and |X| = (|x_{ik}|)_{i,k=1}^{n};

R_{+}^{n} = {(x_{i})_{i=1}^{n} : x_{i} ≥ 0 (i = 1, …, n)}, R_{+}^{n×n} = {(x_{ik})_{i,k=1}^{n} : x_{ik} ≥ 0 (i, k = 1, …, n)};

if x = (x_{i})_{i=1}^{n} and y = (y_{i})_{i=1}^{n} ∈ R^{n}, then x ≤ y ⇔ x_{i} ≤ y_{i} (i = 1, …, n); r(X) is spectral radius of the matrix X ∈ R^{n×n}; C([a, b]; R^{n}) is the space of continuous vector functions x : I → R^{n} with the norm

∥x∥_{C} = max {∥x(t)∥ : t ∈ [a, b]};
C([a, b]; R_{+}^{n}) = {x ∈ C([a, b]; R^{n}) : x(t) ∈ R_{+}^{n} for t ∈ [a, b]};

An operator g : C([a, b]; R^{n}) → C([a, b]; R^{n}) is said to be uniformly compact if it is continuous and

\{\frac{1}{1 + ∥x∥_{C}} g(x) : x ∈ C([a, b]; R^{n})\}

is a relatively compact subset of C([a, b]; R^{n}).

An operator g : C([a, b]; R^{n}) → C([a, b]; R^{n}) is said to be positively homogeneous if for every x ∈ C([a, b]; R^{n}) and λ ∈ R_{+} we have g(λx)(t) = λg(x)(t) for a ≤ t ≤ b.

Along with (1), we have to consider the functional inequality

|x(t) − p(x)(t)| ≤ g(x)(t).  (2)

Under solution of the functional equation (1) (functional inequality (2)) is meant a vector function x ∈ C([a, b]; R^{n}) which for every t ∈ [a, b] satisfies (1) (satisfies (2)).

1991 Mathematics Subject Classification. 47H10, 34K10. Key words and phrases. Functional equation, functional differential equation, boun-

dary value problem.

Theorem. Let p and q : C([a, b]; R^{n}) → C([a, b]; R^{n}) be, respectively, a linear compact and a uniformly compact operators. Moreover, let there exist a positively homogeneous, continuous operator g : C([a, b]; R^{n}) → C([a, b]; R_{+}^{n}) and a vector h ∈ R_{+}^{n} such that the functional inequality (2) has only trivial solution, and for every x ∈ C([a, b]; R^{n}) the inequality

|q(x)(t)| ≤ g(x)(t) + h.  (3)

is fulfilled on [a, b]. Then the functional equation (1) has at least one solution.

To prove this theorem, we will need the following

Lemma. Let a linear continuous operator p : C([a, b]; R^{n}) → C([a, b]; R^{n}) and a positively homogeneous continuous operator g : C([a, b]; R^{n}) → C([a, b]; R_{+}^{n}) be such that the functional inequality (2) has only trivial solution. Let, moreover, h ∈ R_{+}^{n}, C_{0} be a non-empty subset of the space C([a, b]; R^{n}) such that the set

\{\frac{1}{1 + ∥x∥_{C}} x : x ∈ C_{0}\}  (4)

is relatively compact. Then there exists a positive number ρ such that every vector function x ∈ C_{0} satisfying on [a, b] the functional inequality

|x(t) − p(x)(t)| ≤ g(|x|)(t) + h  (5)

admits the estimate

∥x∥_{C} ≤ ρ.  (6)

Proof. Suppose that the lemma is not true. Then for every natural k there exists x_{k} ∈ C_{0} such that ∥x_{k}∥_{C} ≥ k, and the inequality

|x_{k}(t) − p(x_{k})(t)| ≤ g(x_{k})(t) + h

is fulfilled on [a, b]. Assume \overline{x}_{k}(t) = (1 + ∥x_{k}∥_{C})^{−1}x_{k}(t). Then

lim_{k→∞} ∥\overline{x}_{k}∥_{C} = 1  (7)

and

|\overline{x}_{k}(t) − p(\overline{x}_{k}(t))| ≤ g(\overline{x}_{k})(t) + \frac{1}{k + 1} h.  (8)

Because (4) is relatively compact, without loss of generality we may regard the sequence (\overline{x}_{k})_{k=1}^{∞} to be uniformly convergent on [a, b]. Suppose x(t) = lim_{k→∞} x_{k}(t). By (7) and (8), the vector function x is a solution of the functional inequality (2) satisfying ∥x∥_{C} = 1. But this is impossible for (2) has only trivial solution. The obtained contradiction proves the lemma.

Proof of Theorem. First it should be noted that the linear homogeneous equation (I − p)(x)(t) = 0, where I : C([a, b]; R^{n}) → C([a, b]; R^{n}) is an identical operator, has only trivial solution. From this, by virtue of the Fredholm theorem ([1], Theorem 7.3.7) and the compactness of the operator p it follows that the operator I − p has the bounded inverse (I − p)^{−1}.

Denote by C_{0} the set of those x ∈ C([a, b]; R^{n}) for which there exists α(x) ∈ [0, 1] such that

x(t) = p(x)(t) + α(x)q(x)(t).

C_{0} is non-empty, since 0 ∈ C_{0}. On the other hand, because p is compact and q is uniformly compact, the set (4) is relatively compact.

Let ρ be the number appearing in the above proven lemma,

σ(s) = \begin{cases}1 & for 0 ≤ s ≤ ρ \\ 2 − \frac{s}{ρ} & for ρ < s < 2ρ \\ 0 & for s ≥ 2ρ \end{cases} ,  (9)

_{q}(x) = (∥x∥_{C} )(I − p)^{−1}(q(x)),  (10)
\begin{matrix}ρ_{0} = sup \{∥_{q}(x)∥_{C} : x ∈ C([a, b]; R^{n})\}, \\ K = \{x ∈ C([a, b]; R^{n}) : ∥x∥_{C} ≤ ρ_{0}\}. \end{matrix}  (11)

Because the operator q is uniformly compact, it follows from the equalities (9)–(11) that _{q}is a continuous compact operator transforming the ball K into itself. By Schauder’s theorem, there exists a vector function x ∈ K such that x(t) = _{q}(x)(t) for a ≤ t ≤ b. By the definition of the set C_{0} and owing to the equalities (9)–(11), it is clear that x ∈ C_{0} and

x(t) = p(x)(t) + σ(∥x∥_{C} )q(x)(t).  (12)

From (3) and (12) we obtain the inequality (5). Therefore because of our choice of ρ, the vector function x admits the estimate (6). However, (6), (9) and (12) imply that x is the solution of the functional equation (1).

As an application, let us consider the functional differential equation

\frac{dx(t)}{dt} = f(t, x(τ(t)),  (13)

with the boundary conditions

x(t) = 0 for t ∉ [a, b] and x(a) = ∑_{k=1}^{m} A_{k}(x(b_{k}) − x(a)) + c,  (14)

where f : [a, b] × R^{n} → R^{n} is a vector function satisfying the local Caratheodory conditions, τ : [a, b] → R is a measurable function, b_{k} ∈ [a, b], A_{k} ∈ R^{n×n} (k = 1, …, m), c ∈ R^{n}.

By χ we denote a characteristic function of the interval [a, b].

Corollary. Let the inequality

|f(t, χ(τ(t))y)| ≤ G_{0}(t)|y| + h_{0}(t),  (15)

be fulfilled on [a, b]×R^{n}, where G_{0} : [a, b] → R_{+}^{n×n} and h_{0} : [a, b] → R_{+}^{n} are, respectively, a matrix and a vector functions with summable components, and

r( ∑_{k=1}^{m} |A_{k}| ∫_{a}^{b_{k}}G_{0}(s)ds + ∫_{a}^{b} G_{0}(s)ds) < 1.  (16)

Then the boundary value problem (13), (14) has at least one solution.

Proof. The problem (13), (14) is equivalent to the functional equation (1), where p(x)(t) = 0,

q(x)(t) = c + ∑_{k=1}^{m} A_{k} ∫_{a}^{b_{k}}f(s, χ(τ(s))x(τ_{0}(s))ds + ∫_{a}^{t} f(s, χ(τ(s))x(τ_{0}(s))ds, (17)

τ_{0}(t) = τ(a) for τ(t) ∉ [a, b], τ_{0}(t) = τ(t) for τ(t) ∈ [a, b].

For every x = (x_{i})_{i=1}^{n} ∈ C([a, b]; R^{n}), suppose |x|_{C} = (∥x_{i}∥_{C})_{i=1}^{n}. It is obvious from (16) and (17) that q : C([a, b]; R^{n}) → C([a, b]; R^{n}) is a uniformly compact operator satisfying the inequality (3), where g(x)(t) ≡ G|x|_{C},

G = ∑_{k=1}^{n} |A_{k}| ∫_{a}^{b_{k}}G_{0}(s)ds + ∫_{a}^{b} G_{0}(s)ds, h = |c| + ∑_{k=1}^{m} |A_{k}| ∫_{a}^{b_{k}}h_{0}(s)ds + ∫_{a}^{b} h(s)ds.

To prove the above Corollary, it suffices to determine by using the above proven theorem that the functional inequality

|x(t)| ≤ G|x|_{C}  (18)

has only trivial solution. Indeed, from (18) we have

(E − G)|x|_{C} ≤ 0,  (19)

where E is the identity n × n matrix. However, owing to (16), there exists (E − G)^{−1} ∈ R_{+}^{n×n}. Multiplying both parts of (19) by (E−G)^{−1}, we obtain |x|_{C} ≤ 0, i.e., x(t) ≡ 0.

Example. Consider the problem

\frac{dx(t)}{dt} = G_{0}(t)|x(b)|,  (20)
x(a) = ∑_{k=1}^{m} A_{k}(x(b_{k}) − x(a)) + c,  (21)

where G_{0} : [a, b] → R_{+}^{n×n} is a matrix function with summable components, A_{k} ∈ R_{+}^{n×n}, b_{k} ∈ [a, b] (k = 1, …, m), c = (c_{i})_{i=1}^{n}, c_{i} = 1 (i = 1, …, m). After direct checking we can easily see that the problem (20), (21) is solvable if and only if

r( ∑_{k=1}^{m} A_{k} ∫_{a}^{b_{k}}G_{0}(s)ds + ∫_{a}^{b} G_{0}(s)ds) < 1.

Consequently, the condition (16) in the above proven corollary is optimal and it cannot be weakened.

References

1. V. C. L. Hutson and J. S. Pym, Functional analysis and operator theory. Academic Press, London, New York, Toronto, Sydney, San Francisco, 1980.

Author’s address:

A. Razmadze Mathematical Institute

Georgian Academy of Sciences

1, M. Aleksidze St., Tbilisi 380093

Georgia