Georgian Mathematical Journal Volume 9 (2002), Number 2, 287–294

ORDINARY DIFFERENTIAL EQUATIONS WITH NONLINEAR BOUNDARY CONDITIONS

TADEUSZ JANKOWSKI

Abstract. The method of lower and upper solutions combined with the monotone iterative technique is used for ordinary differential equations with nonlinear boundary conditions. Some existence results are formulated for such problems.

2000 Mathematics Subject Classification: 34A45, 34B15, 34A40. Key words and phrases: Lower and upper solutions, monotone iterations, convergence, existence of solutions.

1. Introduction

In this paper, we shall consider the following differential problem

\begin{cases}x′(t) = f(t, x(t)), t ∈ J = [0, T], T > 0, \\ x(0) = g(x(T)), \end{cases}  (1)

where f ∈ C(J × ℝ, ℝ), g ∈ C(ℝ, ℝ).

It is well known that the monotone iterative technique is a powerful method used to approximate solutions of several problems (see, for example [5]). The purpose of this paper is to show that it can be applied successfully to problems of type (1). Assuming one-sided Lipschitz condition on f (with respect to the second variable) combined with the corresponding monotonicity conditions on g, it is shown that linear iterations converge to a solution of problem (1). Some comparison results are also formulated.

In many papers, the monotone iterative technique was applied to problem (1) when g(u) = λu + k. If λ = 0, then (1) reduces to the initial value problem for differential equations and this case is considered, for example, in [6], [7], [9], [10]. If λ = 1 and k = 0, then we have a periodic boundary problem considered, for example, in [8], [10], while if λ = −1 and k = 0, then we have an antiperiodic boundary problem, see [10], [11], [12]. A general case where λ, k ∈ ℝ is discussed in [1]. Nonlinear problems, more general than (1), are studied, for example, in [2], [3], [4].

2. Some General Facts

From Theorem 1.1 proven in [2] follows

ISSN 1072-947X / $8.00 / \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag www.heldermann.de

Theorem 1. Let f ∈ C(J × ℝ, ℝ), g ∈ C(ℝ, ℝ). Moreover, we assume that there exist functions y_{0}, z_{0} ∈ C^{1}(J, ℝ) such that

y_{0}(t) ≤ z_{0}(t), y′_{0}(t) ≤ f(t, y_{0}(t)), z′_{0}(t) ≥ f(t, z_{0}(t)), t ∈ J,
y_{0}(0) ≤ g(s) ≤ z_{0}(0) for y_{0}(T) ≤ s ≤ z_{0}(T).

Then problem (1) has at least one solution in Δ = {w ∈ C^{1}(J, ℝ) : y_{0}(t) ≤ w(t) ≤ z_{0}(t), t ∈ J}.

On the basis of Theorem 1 one can easily prove

Theorem 2. Let the conditions of Theorem 1 be satisfied and, moreover, let the function g be nondecreasing. Then problem (1) has, in the set Δ, the minimal and the maximal solution.

As for the uniqueness of the solution of problem (1), the following result holds.

Theorem 3 (Kiguradze). Let the conditions of Theorem 1 be fulfilled and, moreover,

f(t, v) − f(t, u) ≤ h(t)(v − u) for t ∈ J, y_{0}(t) ≤ u ≤ v ≤ z_{0}(t), (2)
g(v) − g(u) ≤ L(v − u) for y_{0}(T) ≤ u ≤ v ≤ z_{0}(T),  (3)

where h : J → ℝ is an integrable function and L is a nonnegative constant such that

L exp ( ∫_{0}^{T} h(s) ds) < 1.  (4)

Then problem (1) has, in the set Δ, a unique solution.

Proof. The existence of a solution of (1) follows from Theorem 1. Thus it remains to prove the uniqueness. Let x, \overline{x} ∈ Δ be arbitrary two solutions of (1). We distinguish two cases.

Case 1. x(t) ≠ \overline{x}(t) for all t ∈ J. Indeed, without the loss of generality, we can assume that p(t) = x(t) − \overline{x}(t) > 0 for t ∈ J. Hence

p′(t) = f(t, x(t)) − f(t, \overline{x}(t)) ≤ h(t)p(t), t ∈ J,
p(0) = g(x(T)) − g(\overline{x}(T)) ≤ Lp(T),

by assumptions (2) and (3). Therefore,

0 < p(0) ≤ Lp(T) ≤ L exp ( ∫_{0}^{T} h(s) ds)p(0).

By condition (4), this inequality yields p(0) = 0 and thus p(t) = 0 on J, which is a contradiction.

Case 2. There exists t_{0} ∈ J such that x(t_{0}) = \overline{x}(t_{0}). If t_{0} = T or t_{0} = 0, then

x(0) = g(x(T)) = g(\overline{x}(T)) = \overline{x}(0).

This and condition (2) prove that x(t) = \overline{x}(t) on J, which is a contradiction. If t_{0} ∈ (0, T), then x(t) = \overline{x}(t) on [t_{0}, T]. Hence x(T) = \overline{x}(T), so x(0) = \overline{x}(0) showing that x(t) = \overline{x}(t) on J. It is a contradiction. This proves that problem (1) has, in Δ, a unique solution. It ends the proof.

In the next two sections we are going to construct the solution of problem (1).

3. Case where g is Nondecreasing

A function u ∈ C^{1}(J, ℝ) is said to be a lower solution of problem (1) if

\begin{cases}u′(t) ≤ f(t, u(t)), t ∈ J, \\ u(0) ≤ g(u(T)), \end{cases}

and an upper solution of (1) if the inequalities are reversed.

Let Ω = {u : y_{0}(t) ≤ u ≤ z_{0}(t), t ∈ J} be a nonempty set.

We introduce the following assumptions for later use. (H_{1}) f ∈ C(J × ℝ, ℝ), g ∈ C(ℝ, ℝ); (H_{2}) y_{0}, z_{0} ∈ C^{1}(J, ℝ) are a lower and an upper solution of (1), respectively,

such that y_{0}(t) ≤ z_{0}(t), t ∈ J; (H_{3}) there exists M ≥ 0 such that f(t, u)−f(t, v) ≤ M[v−u] for t ∈ J, y_{0}(t) ≤

u ≤ v ≤ z_{0}(t); (H_{4}) g is nondecreasing on the interval [y_{0}(T), z_{0}(T)].

Lemma 1. Let Assumption H_{1} hold. Assume that u, v ∈ Δ are a lower and an upper solution of problem (1), respectively, and u(t) ≤ v(t) on J. Let Assumptions H_{3}, H_{4} hold. Let y, z ∈ C^{1}(J, ℝ) and

y′(t) = f̃(t, y(t)) ≡ f(t, u(t)) − M[y(t) − u(t)], t ∈ J, y(0) = g(u(T)), (5)
z′(t) = \overline{f}(t, z(t)) ≡ f(t, v(t)) − M[z(t) − v(t)], t ∈ J, z(0) = g(v(T)). (6)

Then

u(t) ≤ y(t) ≤ z(t) ≤ v(t), t ∈ J,  (7)

and y, z are a lower and an upper solution of problem (1), respectively.

Proof. It is easy to see that problems (5) and (6) have their unique solutions y, z ∈ C^{1}(J, ℝ). Put p = u − y, so p(0) ≤ g(u(T)) − g(u(T)) = 0, and p′(t) ≤ f(t, u(t)) − f̃(t, y(t)) = −Mp(t). It gives p(t) ≤ 0 on J so u(t) ≤ y(t), t ∈ J. Similarly, we get z(t) ≤ v(t), t ∈ J. Now let q = y − z. Then q(0) = g(u(T)) − g(v(T)) ≤ 0, by Assumption H_{4}. Moreover, Assumption H_{3} yields

q′(t) = f(t, u(t)) − f(t, v(t)) − M[q(t) − u(t) + v(t)] ≤ −Mq(t).

This and the condition for q(0) prove that q(t) ≤ 0 on J, so (7) holds.

Now we need to show that y, z are a lower and an upper solution of (1), respectively. Indeed, in view of Assumptions H_{3} and H_{4}, we have

y′(t) = f̃(t, y(t)) − f(t, y(t)) + f(t, y(t))
≤ f(t, y(t)) + M[y(t) − u(t)] − M[y(t) − u(t)] = f(t, y(t)),
z′(t) = \overline{f}(t, z(t)) − f(t, z(t)) + f(t, z(t))
≥ f(t, z(t)) − M[v(t) − z(t)] − M[z(t) − v(t)] = f(t, z(t))

and

y(0) = g(u(T)) ≤ g(y(T)), z(0) = g(v(t)) ≥ g(z(T)).

This ends the proof.

Note that if Assumptions H_{1} to H_{4} are satisfied, then by Lemma 1 the sequences y_{n}, z_{n} ∈ C^{1}(J, ℝ) (n = 0, 1, ⋯) are defined uniquely so that for every natural n we have

\begin{cases}y′_{n+1}(t) = f(t, y_{n}(t)) − M[y_{n+1}(t) − y_{n}(t)], & t ∈ J, y_{n+1}(0) = g(y_{n}(T)), \\ z′_{n+1}(t) = f(t, z_{n}(t)) − M[z_{n+1}(t) − z_{n}(t)], & t ∈ J, z_{n+1}(0) = g(z_{n}(T)). \end{cases}

Theorem 4. Assume that Assumptions H_{1}, H_{2}, H_{3} and H_{4} hold. Then

y_{0}(t) ≤ y_{1}(t) ≤ ⋯ ≤ y_{n}(t) ≤ z_{n}(t) ≤ ⋯ ≤ z_{1}(t) ≤ z_{0}(t),  (8)
t ∈ J, n = 0, 1, …,

and uniformly on J we have

lim_{n→∞} y_{n}(t) = y(t), lim_{n→∞} z_{n}(t) = z(t),

where y and z are the minimal and the maximal solution of problem (1) in Δ.

Proof. Using Lemma 1, by mathematical induction we can get (8). Indeed, the limits of sequences {y_{n}}, {z_{n}} are solutions of problem (1). By Theorem 2, problem (1) has, in Δ, extremal solutions. To finish the proof, it is enough to show that y and z are the minimal and the maximal solution of (1). To do it, we need to show that if w is any solution of (1) such that y_{0}(t) ≤ w(t) ≤ z_{0}(t), t ∈ J, then y_{0}(t) ≤ y(t) ≤ w(t) ≤ z(t) ≤ z_{0}(t), t ∈ J. Assume that for some k, y_{k}(t) ≤ w(t) ≤ z_{k}(t), t ∈ J. Put p = y_{k+1} − w, q = w − z_{k+1}, so p(0) = g(y_{k}(T)) − g(w(T)) ≤ 0, q(0) = g(w(T)) − g(z_{k}(T)) ≤ 0. Moreover,

p′(t) = f(t, y_{k}(t)) − M[y_{k+1}(t) − y_{k}(t)] − f(t, w(t))
≤ M[w(t) − y_{k}(t)] − M[y_{k+1}(t) − y_{k}(t)] = −Mp(t),
q′(t) = f(t, w(t)) − f(t, z_{k}(t)) + M[z_{k+1}(t) − z_{k}(t)] ≤ −Mq(t),

by Assumptions H_{3}. It shows p(t) ≤ 0, q(t) ≤ 0 on J, so y_{k+1}(t) ≤ w(t) ≤ z_{k+1}(t) on J. Hence, by the method of mathematical induction, we have that y_{n}(t) ≤ w(t) ≤ z_{n}(t) for t ∈ J and for all natural n. Taking the limit n → ∞, we get the assertion of Theorem 4.

4. Case where g is Nonincreasing

Functions u, v ∈ C^{1}(J, ℝ) are called weakly coupled (w.c.) lower and upper solutions of problem (1) if

\begin{cases}u′(t) ≤ f(t, u(t)), & t ∈ J, u(0) ≤ g(v(T)), \\ v′(t) ≥ f(t, v(t)), & t ∈ J, v(0) ≥ g(u(T)). \end{cases}

Let us introduce the following assumptions. (H_{5}) y_{0}, z_{0} ∈ C^{1}(J, ℝ) are w.c. lower and upper solutions of problem (1) and

y_{0}(t) ≤ z_{0}(t), t ∈ J; (H_{6}) g is nonincreasing on the interval [y_{0}(T), z_{0}(T)]; (H_{7}) there exist nonnegative constants A, M and an integrable function K :

J → ℝ such that

−M[v − u] ≤ f(t, v) − f(t, u) ≤ K(t)[v − u] for t ∈ J,  (9)
y_{0}(t) ≤ u ≤ v ≤ z_{0}(t),
−A[v − u] ≤ g(v) − g(u) ≤ 0 for y_{0}(T) ≤ u ≤ v ≤ z_{0}(T), (10)

and

A exp ( ∫_{0}^{T} K(s) ds) < 1.  (11)

Lemma 2. Assume that Assumption H_{1} is satisfied. Let u, v ∈ Δ be w.c. lower and upper solutions of (1) and u(t) ≤ v(t), t ∈ J. Let Assumptions H_{3}, H_{6} hold. Let y, z ∈ C^{1}(J, ℝ) and

\begin{cases}y′(t) = f(t, u(t)) − M[y(t) − u(t)], & t ∈ J, y(0) = g(v(T)), \\ z′(t) = f(t, v(t)) − M[z(t) − v(t)], & t ∈ J, z(0) = g(u(T)). \end{cases}

Then

u(t) ≤ y(t) ≤ z(t) ≤ v(t), t ∈ J,  (12)

and y, z are w.c. lower and upper solutions of problem (1).

Proof. Note that y and z are well defined. Put p = u − y, q = z − v. Then p(0) ≤ g(v(T)) − g(v(T)) = 0, q(0) ≤ g(u(T)) − g(u(T)) = 0, and

p′(t) ≤ f(t, u(t)) − f(t, u(t)) + M[y(t) − u(t)] = −Mp(t),
q′(t) ≤ f(t, v(t)) − M[z(t) − v(t)] − f(t, v(t)) = −Mq(t), t ∈ J.

It yields p(t) ≤ 0, q(t) ≤ 0, t ∈ J, so u(t) ≤ y(t), z(t) ≤ v(t), t ∈ J. Now let p = y − z, so p(0) = g(v(T)) − g(u(T)) ≤ 0, by Assumption H_{6}. In view of Assumption H_{3}, we see that

p′(t) = f(t, u(t)) − f(t, v(t)) − M[y(t) − u(t) − z(t) + v(t)] ≤ −Mp(t).

Hence p(t) ≤ 0 on J, so (12) holds.

By Assumptions H_{3} and H_{6}, we have

y′(t) = f(t, u(t)) − M[y(t) − u(t)] − f(t, y(t)) + f(t, y(t))
≤ f(t, y(t)) + M[y(t) − u(t)] − M[y(t) − u(t)] = f(t, y(t)),
z′(t) = f(t, v(t)) − M[z(t) − v(t)] − f(t, z(t)) + f(t, z(t)) ≥ f(t, z(t)),

and

y(0) = g(v(T)) ≤ g(z(T)), z(0) = g(u(T)) ≥ g(y(T)).

It proves that y, z are w.c. lower and upper solutions of problem (1). The proof is complete.

We see that if the assumptions of Lemma 2 are satisfied then the sequences y_{n}, z_{n} ∈ C^{1}(J, ℝ) (n = 0, 1, ⋯) are defined uniquely so that for every natural n we have

\begin{cases}y′_{n+1}(t) = f(t, y_{n}(t)) − M[y_{n+1}(t) − y_{n}(t)], & t ∈ J, y_{n+1}(0) = g(z_{n}(T)), \\ z′_{n+1}(t) = f(t, z_{n}(t)) − M[z_{n+1}(t) − z_{n}(t)], & t ∈ J, z_{n+1}(0) = g(y_{n}(T)). \end{cases}

Theorem 5. Assume that Assumptions H_{1}, H_{5}, H_{7} hold. Then problem (1) has a unique solution x ∈ Δ,

y_{0}(t) ≤ y_{1}(t) ≤ ⋯ ≤ y_{n}(t) ≤ z_{n}(t) ≤ ⋯ ≤ z_{1}(t) ≤ z_{0}(t),  (13)
t ∈ J, n = 0, 1, …,

and uniformly on J we have

lim_{n→∞} y_{n}(t) = lim_{n→∞} z_{n}(t) = x(t).

Proof. It is easy to verify that all assumptions of Theorem 3 are satisfied, so problem (1) has a unique solution x ∈ Δ. To finish the proof, it is enough to show that the sequences y_{n}, z_{n} converge to x. Using Lemma 2, we can prove (13), by mathematical induction. It follows from the standard argument that {y_{n}}, {z_{n}} converge uniformly to their respective limit functions. Let y(t) = lim_{n→∞} y_{n}(t), z(t) = lim_{n→∞} z_{n}(t). Then

\begin{cases}y′(t) = f(t, y(t)), & t ∈ J, y(0) = g(z(T)), \\ z′(t) = f(t, z(t)), & t ∈ J, z(0) = g(y(T)) \end{cases}  (14)

and y_{0}(t) ≤ y(t) ≤ z(t) ≤ z_{0}(t), t ∈ J. Put p = z − y, so p(t) ≥ 0 on J. Hence p(0) = g(y(T)) − g(z(T)) ≤ Ap(T), by Assumption H_{6}. Moreover,

p′(t) = f(t, z(t)) − f(t, y(t)) ≤ K(t)p(t), t ∈ J.

This inequality yields

0 ≤ p(0) ≤ Ap(T) ≤ Ap(0) exp ( ∫_{0}^{T} K(s) ds)

since

0 ≤ p(t) ≤ p(0) exp ( ∫_{0}^{t} K(s) ds), t ∈ J.

By condition (11), p(0) = 0, and thus p(t) = 0, t ∈ J. This proves that y(t) = z(t) on J, so y and z are solutions of problem (1). Since problem (1) has a unique solution x, we have y = z = x.

It ends the proof.

Remark. The results of this paper remain true if the constant −M is replaced by an integrable on J function M.

Acknowledgment

The author expresses thanks to Prof. I. Kiguradze for useful comments and suggestions.

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(Received 10.04.2001; revised 11.02.2002)

Author’s address: Technical University of Gdańsk Department of Differential Equations 11/12 G. Narutowicz Str., 80–952 Gdańsk Poland E-mail: tjank@mifgate.mif.pg.gda.pl