To solve the Wave Equation
$$ u_{tt} – c^2 u_{xx} = 0$$
One method is to start with operator factorization
$$ u_{tt} – c^2 u_{xx} =
\bigg( \frac{\partial }{\partial t} – c \frac{\partial }{\partial x} \bigg)
\bigg( \frac{\partial }{\partial t} + c \frac{\partial }{\partial x} \bigg)
u = 0
$$
Then it is claimed the solution is
$$u(x,t) = f(x+ct) + g(x-ct)$$
where $f$ ve $g$ are arbitrary functions of single variable.
The proof goes by letting $v = u_t + cu_x$
then
$$ v_t – cv_x = 0 $$
has to be true. Then simultaneously the two equations are solved
$$ v_t – cv_x = 0 $$
$$ u_t + cu_x = v $$
We know the solution for the top equation above is
$$ v(x,t) = h(x+ ct) $$
where $h$ is an arbitrary function. Now wave equation is
$$ u_t + cu_x = h(x + ct) $$
Here is what I do not understand. At this point that a solution is guessed as $u(x,t) = f(x+ct)$ and the book says "it is easy to check by differentiation"
$f'(s) = h(s) / 2c$.
What do I do with $h(s) / 2c$? Is the proof complete with this conclusion? When I plug in $f(x+ct)$ for $u(x,t)$ yes, I do get this equality but how does this help?
Also, after proving $f(x+ct)$ is a solution, then magically a $g(x-ct)$ is added, where does this come from? I do understand linear independence, but why didnt we add $g(x+2ct)$ for example?
Thanks,
Best Answer
It's easier to understand if you change to new variables $\xi=x+ct$ and $\eta=x-ct$. Then the PDE becomes $\partial^2 u/\partial \xi \partial \eta = 0$. Integration with respect to $\eta$ gives $\partial u/\partial \xi = f(\xi)$ with $f$ an arbitrary function (the "constant" of integration, constant here meaning "independent of $\eta$"). Integrating again, now with respect to $\xi$, gives $u=F(\xi)+g(\eta)$ with $g$ an arbitrary function and $F$ the antiderivative of $f$.