Basically I got a simple wave equation with an extra twist. The PDE is $$\frac {\partial^2 y}{\partial t^2} = c^2\frac {\partial^2 y}{\partial x^2} + L$$ with homogeneous boundary condition. As usual, I use the ansatz $ Y(x,t) = F(x)G(t) $ and I have $\frac {\partial^2 y}{\partial t^2} =F''G $ and $\frac {\partial^2 y}{\partial x^2} =FG''$. I substitute this in, and now I have $$FG''= c^2F''G +L$$ I don't know how to make this separable--obviously if I divide both sides with $FG$ as usual, then I will have that annoying constant $\frac {L}{FG} $ and I don't know how to make it separable then.
Can anybody help me? I know I have to separate it somehow, but I don't know exactly how to.
{The actual problem is an inhomogeneous boundary condition type of problem, but I have found the steady state solution, so I just have to find the transient solution, thus the homogeneous boundary condition}
The actual question:
- PDE: $$\frac {\partial^2 y}{\partial t^2} = c^2\frac {\partial^2 y}{\partial x^2} + L,\quad 0 \leq x \leq J,\quad t>0$$ where $L$ is constant and $c$ is constant wave speed.
- Boundary Condition: $ u(0,t)=0$, $t>0$ and $u(J,t)=h$, $t>0$
- Initial Condition: $u(x,0) = 0$, $0<x<J$ and $\frac {\partial u}{\partial t} (x,0)=0$, $0<x<J$
I did the steady state solution bit, and got it as $w(x)= \frac{-Lx^2}{c} + \frac {Hx}{L} +\frac{GLx}{c} $. So I modified the boundary condition and initial condition to get the transient part, and I got the boundary condition homogeneous. However when I try to solve it, the question above arises.
Let $y(x,t) = w(x,t) + \phi(x)$ where $\phi(x) = a x^2 + b x + c$. From this it is seen that $\phi'(x) = 2 a x + b$, $\phi''(x) = 2a$ and \begin{align} \partial_{t}^{2} w = c^2 \, \partial_{x}^{2} w + 2a c^2 + L. \end{align} In order to cancel the $L$ term let $2 a c^2 + L = 0$ which leads to \begin{align} \phi(x) = - \frac{L \, x^{2}}{2 \, c^{2}} + b \, x + c_{1}. \end{align} Now it is understood that the transformation \begin{align} y(x,t) = w(x,t) - \frac{L \, x^{2}}{2 \, c^{2}} + b \, x + c_{1} \end{align} will make the p.d.e more solvable without "complications".
How this applies to the original problem is the following: \begin{align} u_{tt} = c^{2} u_{xx} + L \hspace{5mm} u(0,t) = 0 , \, u(a,t) = h\\ u(x,0) = 0, \, u_{t}(x,0)= 0 \end{align} The solution form proposed is \begin{align} u(x,t) = w(x,t) - \frac{L \, x^{2}}{2 \, c^{2}} + b \, x + c_{1} \end{align} for which \begin{align} u(0,t) &= 0 = w(0,t) + c_{1} \\ u(a,t) &= h = w(a,t) - \frac{L \, a^{2}}{2 \, c^{2}} + b a + c_{1} \\ u(x,0) &= 0 = w(x,0) - \frac{L \, x(x-a)}{2 \, c^{2}} + \frac{h \, x}{a} = w(x,0) - \phi_{1}(x) \\ u_{t}(x,0) &= 0 = w_{t}(x,0) \end{align} From the first equation it is seen that $w(0,t) = 0$ and $c_{1} = 0$. From the second let $w(a,t) = 0$ to obtain \begin{align} b = \frac{h}{a} + \frac{L \, a}{2 \, c^{2}}. \end{align} From all of this it can now be said that the original differential set is transformed by \begin{align} u(x,t) &= w(x,t) - \frac{L \, x(x-a)}{2 \, c^{2}} + \frac{h \, x}{a} \end{align} into \begin{align} w_{tt} = c^{2} w_{xx} \hspace{5mm} w(0,t) = w(a,t) = 0 , w(x,0) = \phi_{1}(x), w_{t}(x,0) = 0 \end{align} where \begin{align} \phi_{1}(x) = \frac{L \, x(x-a)}{2 \, c^{2}} - \frac{h \, x}{a}. \end{align} The equation for $w(x,t)$ is then an easier equation to solve. Since the boundaries for $x$ are zero at each end it suggests a sine solution and can be stated as \begin{align} w(x,t) = \sum_{n=1}^{\infty} \left( A_{n} \cos\left( \frac{n \pi c t}{a} \right) + B_{n} \sin\left( \frac{n \pi c t}{a} \right) \right) \, \sin\left( \frac{n \pi x}{a} \right) \end{align}
Note: the boundary conditions $w(x,0) = w_{t}(x,0) = 0$ leads to $A_{1}$ and $B_{1}$ being zero. This suggests that $w_{t}(x,0) \neq 0$.
Returning to the original form the solution is of the form \begin{align} u(x,t) &= \frac{h \, x}{a} - \frac{L \, x(x-a)}{2 \, c^{2}} + \sum_{n=1}^{\infty} \left( A_{n} \cos\left( \frac{n \pi c t}{a} \right) + B_{n} \sin\left( \frac{n \pi c t}{a} \right) \right) \, \sin\left( \frac{n \pi x}{a} \right) \end{align}