ON THE EXPONENTIAL GROWTH OF SOLUTIONS TO NON-LINEAR HYPERBOLIC EQUATIONS

Existence-uniqueness theorems are proved for continuous solutions of some classes of non-linear hyperbolic equations in 5ounded and unbounded regions. In case of unbounded region, certain conditions ensure that the solution cannot grow to infinity faster than exponentially.

In this paper we study the existence of a unique solution to some non-linear partial differential equations of hyperbolic type.These equations appear in a mathematical model for the dynamics of gas absorption [i], and the main interest is to find solutions of exponential growth to a non-linear hyperbolic equation with characteristic data.It is possible to investigate such problems by the method of successive approximations, after reducing the differential equation to a Volterra integral equation in two variables.However, here we use the method of equivalent (weighted) norms, which considerably reduces the volume of computations.It should be noticed that in [i], an asymptotic investigation of corresponding linear equations has been conducted as tm.Periodic and almost-periodic solutions of a similar class of non-linear hyperbolic equations have been studied in [2].The method of successive approximations has been applied in [3] and [4] to find bounded solutions of non-linear hyperbolic equations with time delay, which arise in control theory and in certain biomedical models.
We consider the equation Uxt(X,t F(x,t,u(x,t),Ux(X,t )), (i) and pose for (I) the following initial and boundary conditions: where u0(t and e(x) are given functions in the domain A [0,6] [0,T], and we are interested in existence-uniqueness to problem (1)-(2).
Two norms llxIl, IIxll,on a Banach space are called equivalent if there exist two positive numbers p and q such that For example, if. the function x(t) belongs to the space of continuous functions on [0,T], it is easy to see that the norms (3) 0tKT are equivalent.In order to prove the existence of a unique continuous solution to our problem, we use a norm similar to (3) and choose L 1 so that a certain integral operator becomes a contraction. 2.

MAIN RESULTS
We prove our first result for equation (i) with the initial and boundary conditions (2) as follows.THEOREM i. Assume the hypotheses: (i) The function u0(t is continuously differentiable on [0,T] and e(x) is continuously differentiable on [0,6].
(ii) The function F(x,t,u,v) is continuous in A 2 and satisfies the Lipschitz condition for u, v, u, v e uniformly with respect to x, t.
Then poblem (i)-( 2) has a unique continuous solution in A. Proof.
We change equation and introduce the operator on the space CI(A) of all functions w(x, t) continuously differentiable in A.
We define a weighted norm in C I(A) by the formula: w II.max e w(x,t) + Wx(X,t) (5) where the constant L 1 > 0 will be chosen later.Since u0(t), e(x), are continuously differentiable and F(x,t,u,v) is a continuous function of its variables, operator (4') maps C I(A) into C I(A) C 1 Now, we want to show that A is a contraction on (A).Consider the difference Aw(x,t) Aw(x,t) F(,,w( ),w (,))-F(,,w( ),w ( )) dd 0 0 for w, w E el(A) and apply the Lipschitz condition, then Aw(x,t)-Aw(x,t) < L w(,)-w(,) + w(,)-w(,) Consider the derivative of Aw(x,t) and Aw(x,t) with respect to x, then (Aw(x,t)) (Aw(x,t)) x x < F(x,,w(x,),Wx(X ,))-F(x,,w(x,),wx(x, )) If we pick L 1 > L(t+I) and define q L 1 with 0 < q < i.This shows that the operator A is a contraction and proves the theorem.
The following proposition concerns the solution behaviour of an equation linear with respect to u (x,t) in an unbounded region as x t.Although this result is generalized in Theorem 3, its proof is given for instructive purposes.For equation Uxt(X,t + a(x,t)Ux(X,t f(x,t,u(x,t)), ( and the initial and boundary conditions u(0,t) u0(t); [0,m) and satisfies the condition a(x,t) 2 m, where m is a constant, the function e(x) is continuously differentiable on [0,].
(ii) The function f(x,t,u) is continuous in n and satisfies the Lipschitz condition I(x.t,u) f(x,t,v) < L Isv for u, v 6 , uniformly with respect to x, t; the function f(x,t,0) satisfies the inequality f(x,t,0) < K 1 e Llt where (x,t) e , K is a constant, and From here we see that if w [** is bounded, then As [** is bounded, which proves that A maps the space B() into itself.Now, we evaluate Au Av for u, v 6 B() ] m + L I +mim l-e (m+L I)T Since the above limit is i, one can write As Av ,, L u v ,, m + L 1 which shows that A is a contraction on and proves that problem (6)-(7) has a unique continuous solution in n which is bounded in the sense of no (ii).
THEOREM 3. Assume for problem (1)-(7) the following hypotheses: (i) The function u0(t is continuously differentiable and eLlt for t e [0,) where K is a satisfies [u0(t) (ii) The function F(x,t,u,v) is continuous in n x 2and satisfies Lipschitz condition (L), uniformly in x, t. ddE.
Then problem (1)-( 7) has a unique continuous solution u(x,t) in n and -Llt sup e lu(x,t) < Proof.