# Wave Equation Solution by Factoring Operators

To solve the Wave Equation

$$u_{tt} - c^2 u_{xx} = 0$$

$$u_{tt} - c^2 u_{xx} = \bigg( \frac{\partial }{\partial t} - c \frac{\partial }{\partial x} \bigg) \bigg( \frac{\partial }{\partial t} + c \frac{\partial }{\partial x} \bigg) u = 0$$

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,

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There is some interesting elementary stuff here www.stanford.edu/class/math220a/handouts/waveequation1.pdf –  BB_ML Nov 21 '11 at 15:41
Properly, this is a wave equation, not heat. –  paul garrett Nov 21 '11 at 21:02

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 $f$.
The condition says that in order for $f$ to satisfy the equation, it must equal $2c$ times an antiderivative of $h$. But the function $h$ was arbitrary, so $f$ is arbitrary too. And adding $g(x-ct)$ corresponds to adding the integration "constant". –  Hans Lundmark Nov 22 '11 at 10:43