Page 236 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 15: Cauchy’s Problem and Characteristics for First Order Equations

So that F(x, y, z, p, q) = k, is a constant along the strip. Notes
Theorem 3: If a characteristic strip contains at least one integral element of F(x, y, z, p, q) = 0, it is
an integral strip of the equation F(x, y, z, p, q) = 0.

We are now in a position to solve Cauchy’s problem. Suppose we want to find the solution of the
partial differential equation (1) which passes through a curve whose freedom equations are
x = ( ), y = ( ), z = ( ) ...(19)

then in the solution
x = x(p , q , x , y , z , t , t) etc., ...(20)
0 0 0 0 0 0
and in the characteristic equations (18) we may take
x = ( ), y = ( ), z = ( )
0 0 0
as the initial values of x, y, z. The corresponding initial values of , , are determined by the
relations
= p ( ) + q ( )
0 0
F( ( ), ( ), ( ), p , q ) = 0
0 0
We substitute these values of x , y , z , p , q and the appropriate value of t in equation (20), and
0 0 0 0 0 o
find that x, y, z can be expressed in terms of two parameters t, to give
x = X( , t), y = Y( , t), z = Z( , t) ...(21)
Eliminating , t from these equations, we get a relation

y
z
x
( , , )= 0
which is the equation of the integral surface of equation (1) through the curve . We shall
illustrate this procedure by an example.
Example: Find the solution of the equation

1
2
2
F = (p q ) + (p x) (q y) z ...(1)
2
that passes through the x-axis.
It is readily shown that the initial values are
x = , y = 0, z = 0, p = 0, q = 2 , t = 0, ...(2)
0 0 0 0 0 0
The characteristic equations of this partial differential equations are
x (t) = F , y (t) F , z (t) = p F + q F
p q p q
p (t) = F p F , q (t) = F q F ...(3)
x z y z
F F
F = = p + q y, F = = q + p x
p p q q

F F
F = = q + y, F = = p + x, F = 1 ...(4)
x x y y z
Substituting these values of partial derivatives of F in equations (3) we have

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