Solving the Wave Equation (PDEs)

[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!)

 

[mighty2000]

Thank you for posting your problem on this forum. I am a scientist and I would be happy to help you solve this PDE. First of all, your factorization of the operator is correct and it is a good idea to use the change of variables to reduce the problem to an ODE in x'. However, the expression you have for u(x',t') is not quite correct. It should be:

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

Notice that the variable y' has been replaced with $\frac{3x' + 5t'}{8}$. This is because we have two unknown functions, f and g, and we need two equations to solve for them. These equations come from the initial conditions:

$$u(x,0) = x^{2} \rightarrow u(x',0) = x^{2}$$
$$u_{t}(x,0) = e^{x} \rightarrow u_{t}(x',0) = e^{x}$$

Plugging these into our expression for u(x',t'), we get:

$$u(x',0) = \int f(\frac{5x'}{8}) dx' + g(\frac{3x'}{8}) = x^{2}$$
$$u_{t}(x',0) = \int \frac{5}{8}f'(\frac{5x'}{8}) dx' + \frac{3}{8}g'(\frac{3x'}{8}) = e^{x}$$

Now, we can solve for f and g by using integration by parts and solving the resulting system of equations. The solution is:

$$f(x') = \frac{8}{25}x'^{2} + \frac{8}{25}x'^{3} + C_{1}$$
$$g(x') = -\frac{8}{3}x' + C_{2}$$

Plugging these into our expression for u(x',t'), we finally get the solution to our PDE:

$$u(x',t') = \frac{8}{25}x'^{2} + \frac{8}{25}x'^{3} - \frac