After you have found the general solution of the partial differential equation as described in the previous sections, you probably want to find the remaining undetermined function that it involves by applying a given boundary or initial condition. To do so:
Question: (5.30 continued) Solve
with initial condition
The general solution to the partial differential equation for this example is given by
Plug that into the initial condition on the line in order to figure out what function must be:
Don’t try to deduce one-parameter function directly from an expression involving two different parameters. Instead convert to a single parameter by expressing one parameter in terms of the other. In this case, you can use to express in terms of as . That gives
or cleaned up
Now you have an expression for function in terms of a single parameter. To get the function itself, give some name to its parameter that is not already used. Call the argument, say, . So . According to the initial condition above
Solve this for in terms of ,
and plug that into the expression for to get an expression for function in terms of only:
or worked out
Now that function has been identified, plug it into the general solution, valid everywhere,
to get the final solution
Note that this solution is only valid in the grey region of figure 3.3; the characteristics in the white region never intersect the line . To find the solution there, you would need an initial condition on, say, the line .
You see how important it is to graph the characteristics.