I am struggling to come up with a coherent/complete solution to an inhomogeneous wave equation PDE and was hoping someone could shed some light on a few things below in my partial solution. My actual questions are bolded.
- The "original" PDE I am starting with is:
$\begin{cases} \frac{\partial^2}{\partial x^2} u(x,t) - 4\omega^2\frac{\partial^2}{\partial t^2} u(x,t) = 0 \\ \frac{\partial}{\partial x} u(1,t) = 2\pi\omega\sin(\pi t) \\ \frac{\partial}{\partial x} u(0,t) = 0 \end{cases} $
I realize there aren't any initial conditions, only boundary conditions, which is causing me some confusion later on.
- The above PDE is readily transformed into an inhomogeneous PDE with homogeneous boundary conditions by assuming $u(x,t)$ takes the form $u(x,t) = \psi(x,t) + g(x,t)$, where $\frac{\partial}{\partial x}g(1,t)=2\pi\omega\sin(\pi t)$ and $\frac{\partial}{\partial x}g(0,t)=0$. One such $g$ is $g(x,t) = \pi\omega x^2 \sin(\pi t)$.
Substituting this into our original PDE yields a new PDE for $\psi(x,t)$:
$\begin{cases} \frac{\partial^2}{\partial x^2} \psi(x,t) - 4\omega^2\frac{\partial^2}{\partial t^2} \psi(x,t) = (-2\pi\omega - 4\pi^3\omega^3 x^2)\sin(\pi t) \\ \frac{\partial}{\partial x} \psi(1,t) = 0 \\ \frac{\partial}{\partial x} \psi(0,t) = 0 \end{cases} $
From here on I'll let $\alpha = -2\pi\omega$ and $\beta = -4\pi^3\omega^3$ to keep things clean. So the above PDE is $\frac{\partial^2}{\partial x^2} \psi(x,t) - 4\omega^2\frac{\partial^2}{\partial t^2} \psi(x,t) = (\alpha + \beta x^2)\sin(\pi t)$.
- Now, assuming $\psi(x,t) = X(x)T(t)$ (separation of variables), we get the following ODEs:
$\begin{cases} X'' + \lambda X, X'(0) = X'(1) = 0 \\ 4\omega^2 T'' + \lambda T = 0 \end{cases} $
Focusing on the ODE in $X(x)$, we have a family of solutions $X_n(x) = a_n \cos(n\pi x)$ (eigenfunctions) for $n = 0, 1, ...$ with associated eigenvalues $\lambda_n = n^2 \pi^2$.
- I think here we want to have the inner product $<X_n, X_n> = \int_0^1 a_n^2 \cos^2(n\pi x) dx = 1$, so we choose $a_n = \sqrt2$. Thus we have an orthogonal basis $\phi_n(x) = \sqrt2 \cos(n\pi x)$ for functions defined on $[0, 1]$ and we can write:
- $\psi(x,t) = \sum_{n=0}^\infty \psi_n(t) \phi_n(x)$ and
- $(\alpha + \beta x^2)\sin(\pi t) = \sum_{n=0}^\infty f_n(t) \phi_n(x)$, where $f_n(t) = \int_0^1 (\alpha + \beta x^2)\sin(\pi t) \phi_n(x) dx$
I found that:
- $f_0(t) = \sqrt2(\alpha + \frac{\beta}{3})\sin(\pi t)$
- $f_n(t) = \frac{4\sqrt2\beta}{n^2\pi^2}(-1)^n\sin(\pi t)$
- Substituting the above into our PDE for $\psi(x,t)$ yields the following ODEs (by equating the $\cos(n\pi x)$ terms for each n):
- $n = 0$: $-4\omega^2 \psi_0''(t) = \sqrt2(\alpha + \frac{\beta}{3})\sin(\pi t)$
- $n \ne 0: -\lambda_n \psi_n(t) - 4\omega^2 \psi_n''(t) = \frac{4\sqrt2\beta}{n^2\pi^2}(-1)^n \sin(\pi t)$
- Using maple to solve the above ODEs, we get:
- $\psi_0(t) = \frac{\sqrt2(3\alpha - \beta)}{12\omega^2 \pi^2} \sin(\pi t) + c_1 t + c_2$
- $\psi_n(t) = c_3 \sin(\frac{\pi t}{2\omega}) + c_4 \cos(\frac{\pi t}{2\omega}) + \frac{4\sqrt2\beta}{\pi^4}$
So... In theory now we have $\psi(x,t)$, and hence we have $u(x,t)$? I am not so sure. Firstly, because there are no initial conditions, there are arbitrary constants floating around in the solution for $\psi(x,t)$. Is that acceptable? It feels wrong to me when comparing to the below solution from Maple.
As well, asking Maple to solve the PDE for $\psi(x,t)$ yields
$\psi(x,t) = \frac{\sin(\pi t)(\pi\omega \sin(2\pi\omega)x^2 + \cos(2\pi\omega x))}{\sin(2\pi\omega}$,
which looks nothing like my solution and I'm wondering if I went wrong somewhere above, as this solution is so different.
If you read through all of that, thank you.