I have been given the following PDE to solve: $$ u_{tt}-u_{xx}=g(t)\sin x \;\;\; (t,x)\in (0,\infty)\times(0,\pi)\\u(0,x)=u_t(0,x) \;\;\;x\in(0,\pi)\\ u(t,0) = u(t,\pi) =0 \;\;\; t>0 $$
This shouldn't be too hard. I have to solve it by separating variables so I write $u(t,x)=v(t)w(x)$ and substituting this into the pde it's clear that $w(x)=\sin x$ and then I have the following ODE for $v$: $$ v'' +v = g(t) \;\;\;\;\;: \;\;\;v(0)=v'(0)=0 $$
But I don't know how to solve this for a general $g(t)$. The solution they have given us says that: $$ v(t) = \int_{0}^{t}g(t-\tau)\sin \tau \;\;d\tau $$ I know this is something to do with Fourier series but I'm not sure why this applies in this case?
Any help welcome :)
Okay I think I have solved it but its not simple:
By looking at the ode we guess a solution of the form:
$$ v(t) = A(t)\cos(t)+B(t)\sin(t)$$ and then we differentiate twice. However, because we picked such a general form, we have a degree of freedom still to play with so we set: $$ A'\cos t +B'\sin t = 0 $$ With this equation and with the ODE we derive the following system of simultaneous equations: $$ B' \cos t - A' \sin t =g\\ A' \cos t + B\ \sin t =0 $$
We rearrange and solve for $A', B'$ and then integrate to get:
$$ A(t) = -\int_{0}^{t} g(s)\sin(s) \;ds\\ B(t) = \int_{0}^{t}g(s)\cos(s) \;\;ds $$
Now we substitute this into our formula for $v$ and apply the compound angle formula for $\sin$ and then change variables $\tau = t-s$ and this gives us the solution:
$$ v(t) = \int_{0}^{t}g(t-\tau)\sin(\tau) \;\;d\tau $$