# Solving the Wave Equation (PDEs)

1. Oct 15, 2013

### Tsunoyukami

Solve $u_{xx} - 3u_{xt} - 4u_{tt} = 0$, $u(x,0) = x^{2}$, $u_{t}(x,0) = e^{x}$. (Hint: Factor the operator as we did for the wave equation.) (From Partial Differential Equations An Introduction, 2nd edition by Walter A. Strauss; pg. 38)

This is the first of a set of three exercises on the wave equation that I'm unable to solve (the other two are similar). Following the hint, I can factor the operator to obtain the following in a manner similar to the manner in which my text derives D'Alembert's Formula:

$$u_{xx} - 3u_{xt} - 4u_{tt} = 0$$
$$\left( \frac{\partial}{\partial x} + \frac{\partial}{\partial t} \right) \left( \frac{\partial}{\partial x} - 4 \frac{ \partial}{\partial t} \right) u = 0$$

Let $v = u_{x} + u_{t}$. Then $v_{x} - 4v_{t} = 0$. We now have two first order PDEs. Solving $v_{x} - 4v_{t} = 0$ yields $v(x,t) = q(4x + t) = f(x + \frac{1}{4} t)$ (this may be more obvious if we write $v_{x} - 4v_{t} = \frac{1}{4}v_{x} - v_{t} = 0$. Then we must solve $v = f(x + \frac{1}{4} t) = u_{x} + u_{t}$ - we can solve this by reducing it to an ODE in x' using the change of variables $x' = ax + bt = x + t$ and $t' = bx - at = x - t$; then (after some algebra) we are left with the following expression:

$$f(\frac{5x' + 3t'}{8}) = u_{x'}$$

$$u(x',t') = \int f(\frac{5x' + 3t'}{8}) dx' + g(y')$$

So this is roughly where I get stuck. First and foremost I'm not sure whether or not what I've done have is correct. (Someone please confirm?) Secondly, I'm not sure how to do the integral in my last expression - how do I determine f? or g? (I would expect f and g to be determined by the initial conditions but I'm not sure how to do so...)

Any help in understanding this problem would be greatly appreciated!

(Furthermore, after solving this problem would anyone be able to shed light on how to solve an inhomogeneous PDE of this type? Thanks again!)