Interval Analytic Method in Existence Result for Hyperbolic Partial Differential Equation ()
1. Introduction
In this paper, we utilize interval analytic methods in the investigation of the existence of solution of the hyperbolic partial differential equation
(1.1)
with characteristic initial values
(1.2)
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.
2. Differential Inequalities and Majorisation of Solution
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
(2.1)
(2.2)
(2.3)
Then we have
(2.4)
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
(2.5)
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
(2.6)
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,
(2.7)
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
(2.8)
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
(2.9)
Combining inequalities (2.8) and (2.9) we have the desired result.
3. 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 
be such that 
on 
and a function 
, defined by
(3.1)
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
(3.2)
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
(3.3)
(3.4)
(3.5)
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:
(3.6)
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
(3.7)
where 
is the constant appearing in Equation (3.1). Then the following hold true.
(3.8)
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:
(3.9)
whenever the functions 
and 
are such that
for constant 
suitably chosen. Then we have
(3.10)
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.