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## Method of Characteristics : Other Examples

Example 1. .10

 (1.15)

with the initial condition

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. .11Consider the initial-boundary-value problem

 (1.16)

with

Characteristic equations are
 (1.17)

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 implicit solution

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,

Next: Partial Differential Equations of Up: Method of Characteristics Previous: Initial-boundary-value Problems
Prof. Alan Hood
2000-10-30