Solving non-homogeneous wave equation by separating method I am studying PDEs by myself. I need your help to know that if I do correct or not.
Some days ago I asked a question on a wave equation with non-homogeneous boundary conditions.(Here) After some discussion, we reached to a non-homogeneouss PDE with homogeneous B.Cs.
I tried to solve the new equation in my free-times, but I couldn't. The new equation is:
\begin{cases}
v_{tt}=v_{xx}+(\frac{-2}{\pi}x+2)\sin(t) & 0\leq x\leq\pi\\
v(x,0)=x\\
v_t(x,0)=\frac{2}{\pi}x-2\\
v_x(0,t)=v_x(\pi,t)=0
\end{cases}
I know I should solve the equation for a homogeneous case, and then find the answer... But how to do that?
Let $v(x,t)=F(x,t)+G(x,t)$ where $F(x,t)$ is the solution of homogeneous case. Then we have:
$$
v_{tt}=F_{tt}+G_{tt}
$$ and
$$
v_{xx}=F_{xx}+G_{xx}
$$
So from the PDE:
$$
F_{tt}+G_{tt}=F_{xx}+G_{xx}+(\frac{-2}{\pi}x+2)\sin(t)
$$
Since we look for homogeneous case:
$$
G_{xx}-G_{tt}=-(\frac{-2}{\pi}x+2)\sin(t)
$$
It looks like a loop! Because this new one is as same as our main equation :|
Could you please help me to find out the answer?
 A: HINT:  Go back to your original equation:
$$
v_{tt}=v_{xx}+(\frac{-2}{\pi}x+2)\sin(t)
$$
Define $u = x + t$ and $w = x - t$.  Write the derivatives $v_{xx}$ and $v_{tt}$ in terms of derivatives with respect to $u$ and $w$ instead.  The resulting equation will be much easier to solve.
A: It looks like you are misunderstanding "homogeneous case" (I would say "associated homogeneous equation").  Here the associated homogeneous equation is $G_{xx}- G_{tt}= 0$.  Separating, write G(x, t)= X(x)T(t) so the equation becomes $X''(x)T(t)- X(x)T''(t)= 0$.  Dividing both sides by X(x)T(t) that becomes $X''/X= T''/T$.  Since those are to be equal for all x and t, they must be the same constant. $\frac{X''}{X}= \lambda$ so $X''= \lambda X$ and $\frac{T''}{T}= \lambda$ so $X''= \lambda X$ for constants $\lambda$.  You have probably done that before and recognize that the solutions can be written as a Fourier series in t. Now write the "non-homogeneous" part, $\left(-\frac{2}{\pi}x+ 2\right)sin(t)$ as a Fourier series in t- that is just Bsin(t) where $A= -\frac{2}{\pi}x+ 2$ and all other coefficients are 0.
A: According to the above hint:
\begin{cases}
r=x+t\\
s=x-t
\end{cases}
\begin{cases}
x=\frac{r+s}{2}\\
t=\frac{r-s}{2}
\end{cases}
so,
$$\frac{\partial^2 v}{\partial x^2}=\frac{\partial^2 v}{\partial s^2}.(\frac{\partial s}{\partial x})^2+2\frac{\partial^2 v}{\partial s \partial r}.\frac{\partial s}{\partial x}.\frac{\partial r}{\partial x} +\frac{\partial^2 v}{\partial r^2}.(\frac{\partial r}{\partial x})^2+\frac{\partial v}{\partial s}.\frac{\partial^2 s}{\partial x^2}+\frac{\partial v}{\partial r}.\frac{\partial^2 r}{\partial x^2}$$
$$
v_{xx}=v_{ss}+2v_{sr}+v_{rr}
$$
and
$$\frac{\partial^2 v}{\partial t^2}=\frac{\partial^2 v}{\partial s^2}.(\frac{\partial s}{\partial t})^2+2\frac{\partial^2 v}{\partial s \partial r}.\frac{\partial s}{\partial t}.\frac{\partial r}{\partial t} +\frac{\partial^2 v}{\partial r^2}.(\frac{\partial r}{\partial t})^2+\frac{\partial v}{\partial s}.\frac{\partial^2 s}{\partial t^2}+\frac{\partial v}{\partial r}.\frac{\partial^2 r}{\partial t^2}$$
$$
v_{tt}=v_{ss}-2v_{sr}+v_{rr}
$$
Now we have:
$$
v_{ss}-2v_{sr}+v_{rr}=v_{ss}+2v_{sr}+v_{rr}+(\frac{-2}{\pi}(\frac{r+s}{2})+2)\sin(\frac{r-s}{2})
$$
$$
4v_{sr}=(\frac{r+s}{\pi}-2)\sin(\frac{r-s}{2})
$$
Now this is easy to solve :)
