Interval Analytic Method in Existence Result for Hyperbolic Partial Differential Equation

Without the usual assumption of monotonicity, we establish some results on the theory of hyperbolic differential inequalities which enable us to produce a majorising interval function for the solution of the hyperbolic initial value problem. Using this function, a variation of parameters formula and interval iterative technique, the existence of solution to the problem is established.


Introduction
In this paper, we utilize interval analytic methods in the investigation of the existence of solution of the hyperbolic partial differential equation ab I Without the assumption of monotonicity on the function f we establish some results on the theory of hyperbolic differential inequalities which enable us to produce a majorizing interval function for the solution of the equation.With the use of a variation of parameters formula used in [1] and theorem 5.7 of [2] on interval iterative technique we generate a nested sequence of interval functions which converges to an interval solution.This interval solution is thus a majorant of the solution of the equation and it coincides with the real valued solution if it is degenerate.Similar interval methods had earlier been used by some authors in [3]- [7] for solution to differential equation but not for hyperbolic initial value problems.The result in this paper generalizes those of [1] [8] as the monotonicity condition imposed on the function f is not in any way necessary.
The basic results in interval analysis used in this work are found in [2] [6] [7] [9]- [13] for readers who may not be familiar with them.

Differential Inequalities and Majorisation of Solution
u C I ∈  is said to be a lower solution of the hyperbolic initial value problem (1.1) and (1.2) on ab I if the reversed inequalities hold true with u in place of v in the specified in- tervals.
Next, we shall consider some results concerning the upper and lower solutions of Equation (1.1) and conditions (1.2).

( ) ( )
, , , , on where the inequality is componentwise.Proof: We shall establish this theorem by contradiction.From assumption (2.3) we see clearly that the theorem is true for the point (0,0) on .
ab I Suppose that inequality (2.4) is not true at a point ( ) and assume that ( ) ( ) then by assumption (2.3) 0 x and 0 y cannot both be zero.Let 0 h > be such that ( ) Thus, we have, for 0 0 y ≠ (or 0 0 and this contradicts assumption (2.5).If 0 0, y = then 0 0 x ≠ (or vice-versa) and for 0 0 x = a similar argument can be advanced to obtain ( ) ( ) and this is still a contradiction to our earlier assumption (2.5).
Suppose instead that ( ) ( ) Then 0 0, y ≠ otherwise condition (2.3) would immediately give the required contradiction.Now for 0 0, Similarly, if we assume that ( ) ( ) , we would also arrive at a contradiction.At 0 x a = and 0 y b = left hand derivatives are used to obtain the result.Hence, we conclude that, the assertion (2.4) holds true on ab I and this proves the theorem.Theorem 2.2: Let u and v be functions defined on ab I which satisfy assumptions (2.1), (2.2) and (2.3) of Theorem 2.1.Suppose in addition that they satisfy the following conditions, Then the solution z of problem (1.1) and (1.2) together with its derivatives , ,  ,  , , , On the rectangle, ab I , where the inclusion is componentwise.Proof: Notice that the lower endpoints of the intervals in Equation (2.7) satisfy assumption (2.3) of Theorem 2.1 when v is replaced by z .Therefore u and z satisfy the hypothesis of Theorem 2.1 and hence

( ) ( )
, , , , , on Similarly, replacing u by z in assumption (2.3) we obtain the upper endpoints of the intervals in conditions (2.7) and so by Theorem 2.1 we also have Combining inequalities (2.8) and (2.9) we have the desired result.

Construction and Existence of Solution
Our purpose in this section is to establish the existence of solution to the problem (1.1) satisfying initial values (1.2) by means of interval analytic method.To this end an integral operator is constructed, the solution of the resulting operator equation is equivalent to the solution of the initial value problem under consideration.An interval extension of this operator is then used to generate a sequence of interval functions which converges to the required solution.Let ( ) on ab I and a function 3 , , , , , , , , where f is the function in Equation (1.1) and 0 λ ≥ is a constant suitably chosen such that 0.

ϕ ≥ Clearly
it can be seen that ϕ is continuous on 3 .

ab I × 
With this new function ϕ , Equation (1.1) becomes , , , , By using the variation of constant formula of Lemma 4.1 in [1], we obtain the solution of Equation (3.2), satisfying initial values (1.2) as: where λ is the constant appearing in Equation (3.1).Then the following hold true.
, , , , , , , , ,   Since these initial intervals satisfy the hypothesis of Theorem 5.7 of [2], the result of the theorem implies that * ** , Z Z and Z converge as sequences and are equally nested.Furthermore, the solution z of Equation (1.1)   satisfying condition (1.2) belongs to the limit function Z of the sequence { }, z x y = into the integro-differential equations we obtain the system of integral equations constant, suitably chosen in Equation(3.1).Then there exists a convergent nested sequence of interval functions { } n Z such that the unique solution z of Equations (1.1) and (1.2) satisfies = , Z degenerate where the initial interval 0 Z is given by From the construction earlier considered, we see that any solution of Equation (1.1) which satisfies condition (1.2) solves the integral Equation (3.3).Conversely if z solves the integral Equation (3.3) we have that that z again solves the Equation (1.1) and satisfies condition (1.2).Therefore, we shall seek the solution of the integral equation given by(3.3)which is transformed to the operator equation .z pz =Let Z be an interval function defined on , function Φ an interval extension of the function ϕ defined in Equation (3.1).

1 (
Similarly we have by the result given in Equation (3.8) of Lemma 3.
conditions (2.7) and in addition they also satisfy conditions (2.1) and (2.2).Suppose further that the function  appearing on the right hand side of Equation (1.1) satisfies: γ β α γ β and 1 α are such that