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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