In this paper, we utilize interval analytic methods in the investigation of the existence of solution of the hyperbolic partial differential equation

with characteristic initial values

prescribed in a two-dimensional rectangle where and

, and where means that z is continuous on and possesses continuous partial derivatives on

Without the assumption of monotonicity on the function 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 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.

Definition 2.1: A function is said to be an upper solution of the hyperbolic initial value problem (1.1) and (1.2) on if

.

Definition 2.2: A function is said to be a lower solution of the hyperbolic initial value problem (1.1) and (1.2) on if the reversed inequalities hold true with in place of in the specified intervals.

Next, we shall consider some results concerning the upper and lower solutions of Equation (1.1) and conditions (1.2).

Theorem 2.1: Suppose that and

Then we have

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

Suppose that inequality (2.4) is not true at a point and assume that

then by assumption (2.3) and cannot both be zero.

Let be such that then and so

Thus, we have, for (or),

and this contradicts assumption (2.5).

If then (or vice-versa) and for we have

If and a similar argument can be advanced to obtain Hence,

and this is still a contradiction to our earlier assumption (2.5).

Suppose instead that

Then otherwise condition (2.3) would immediately give the required contradiction.

Now for let such that, we have so

This contradicts assumptions (2.6).

Similarly, if we assume that, we would also arrive at a contradiction. At and left hand derivatives are used to obtain the result.

Hence, we conclude that, the assertion (2.4) holds true on and this proves the theorem.

Theorem 2.2: Let and be functions defined on 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 of problem (1.1) and (1.2) together with its derivatives satisfy

On the rectangle, , 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 is replaced by. Therefore and satisfy the hypothesis of Theorem 2.1 and hence

Similarly, replacing by 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.

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 be such that on and a function , defined by

where is the function in Equation (1.1) and is a constant suitably chosen such that Clearly it can be seen that is continuous on

With this new function, 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:

Differentiating with respect to, we obtain

and similarly by differentiating with respect to we obtain

Eliminating the derivatives and by introducing the function and into the integro-differential equations we obtain the system of integral equations

which is equivalent to the problem (3.2) and initial values (1.2).

Denoting the right hand side of these integral equations by and respectively, we have the following:

With these we prove the following result.

Lemma 3.1: Let and satisfy conditions (2.7) of Theorem 2.2. Suppose that for functions with on, we have

where is the constant appearing in Equation (3.1). Then the following hold true.

for all

Proof: We first consider the lower endpoints of the inclusions and differentiating we have, from Equation (3.3)

differentiating again with respect to we obtain

This, by Equation (3.1) and assumption (3.7), gives

Similarly by differentiating Equation (3.3) with respect to, we obtain

By conditions (2.1) and (3.7) we have for and for From these we see that satisfies the assumptions of Lemma 4.2 of [1] since

Thus

It could similarly be proved that

and

Hence the lemma is established.

Theorem 3.1: Let the functions satisfy conditions (2.7). Suppose that the function is such that

and

for function satisfying

and, constant, suitably chosen in Equation (3.1).

Then there exists a convergent nested sequence of interval functions such that the unique solution of Equations (1.1) and (1.2) satisfies

with, degenerate where the initial interval is given by

Proof: 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 solves the integral Equation (3.3) we have that

which by Equations (3.1) and (3.6) gives

with

.

and these imply that 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

Let be an interval function defined on such that for and the interval function an interval extension of the function defined in Equation (3.1). Then the interval integral operator defined by

is an interval majorant of.

Then the problem reduces to solving the interval operator equation

However to determine we need to also determine and which are respectively interval extensions to the function and. This is done by solving the interval operator equations

With and defined respectively by

and

which majorise the real operators and respectively.

Define the sequences by

with

with

and

with

We have the sequence as required.

We shall show that convergences to a limit. But this can only be so if the sequence and also converge.

By Theorem 5.7 of [2] , these sequences converge if

and

Now for

by the first inclusion of Equation (3.8). Hence

Similarly we have by the result given in Equation (3.8) of Lemma 3.1

and

Since these initial intervals satisfy the hypothesis of Theorem 5.7 of [2] , the result of the theorem implies that and converge as sequences and are equally nested. Furthermore, the solution of Equation (1.1) satisfying condition (1.2) belongs to the limit function of the sequence that is,

and this proves the theorem.

Lemma 3.2: Assume that the functions satisfy conditions (2.7) and in addition they also satisfy conditions (2.1) and (2.2). Suppose further that the function f appearing on the right hand side of Equation (1.1) satisfies:

whenever the functions and are such that

for constant suitably chosen. Then we have

for any function satisfying

Proof: From inequality (2.1) we have

Since

From inequality (3.9) we have

and so

which is the first inequality in (3.7).

Also from inequality (2.2) we have

and using inequality (3.9) we have

Therefore

which also is the second inequality in (3.7). Since all the other conditions of Lemma 3.1 are also satisfied, the proof of this lemma follows as for Lemma 3.1 to obtain the desired result.

Remark 3.1: If in inequality (3.9) then we have

for and this implies that is monotone increasing in its domain of definition. Therefore the result of lemma 3.2 also holds for a monotone function

Theorem 3.2: Suppose that the function satisfies conditions (2.1), (2.2) and (2.7). If in addition the function appearing in Equation (1.1) satisfies

whenever

for some constant, suitably chosen.

Then there exists a nested sequence of interval function with each term majorising the unique solution of Equation (1.1) satisfying condition (1.2) such that the limit of this sequence also contains, that is,

Proof: As it has been shown in the proof of Lemma 3.2, the conditions prescribed in this theorem can equally be linked with those of Theorem 3.1. Therefore the proof can be established in a manner similar to that of Theorem 3.1.