__Example 1. .10__

The characteristic equations are

and

and rearranging we obtain

Thus,

On using , (1.15) is now written as

where the constant is actually constant along the characteristic defined by the value of . Thus, is really a function of and we write

since and the function is determined by the initial conditions. At , we have and so that . Replacing by the above expression involving and , we obtain the final solution

__Example 1. .11__Consider the initial-boundary-value problem

Characteristic equations are

and

and (1.16) gives

Hence,

since is constant along the characteristic and dx/dr is the derivative of along the characteristic curve. Therefore, we have

Thus, substituting for into we obtain the

where is determined by the initial condition, namely, , and . Thus, and so

Squaring both sides, we can manipulate this solution into an explicit solution for in terms of and .

and so the final solution is

We can easily check that this is the solution to (1.16) by calculating the partial derivatives,

Therefore,