Use Green's Functions to solve PDE $u_{tt}-u_{xx}=g(t)\sin(x)$ with IVP and BC Use Green's Functions to solve $$u_{tt}-u_{xx}=g(t)\sin(x)$$ for $$0<x<\pi,t>0$$ with initial conditions $$u(x,0)=u_t(x,0)=0$$ for $$0\leq x \leq \pi$$
and boundary conditions $$u(0,t)=u(\pi,t)=0, t \geq 0$$
I know that the answer is $$u(x,t)=\sin(x) \int_0^t g(t-\tau)\sin(\tau)d\tau$$ but I have no idea where to start. :(
If anyone could offer intuition and steps I could follow for these types of problems I would really appreciate the help. Is there a standard algorithm of sorts used for Green's functions solutions? They are really confusing to me
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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The $\ds{\sin\pars{x}}$ factor in the RHS suggests the definition
  $\ds{\mrm{u}\pars{x,t} \equiv \sin\pars{x}\mrm{U}\pars{t}}$ with
  $\ds{\mrm{U}\pars{0} = \mrm{U}_{t}\pars{0} = 0}$. $\ds{\mrm{U}\pars{t}}$
  satisfies $\ds{\totald[2]{\mrm{U}\pars{t}}{t} + \mrm{U}\pars{t} = \mrm{g}\pars{t}}$.

Then,
\begin{align}
\mrm{U}\pars{t} & = \int_{0}^{\infty}\mrm{G}\pars{t,t'}\mrm{g}\pars{t'}\,\dd t'
\quad\mbox{where}\quad
\left\{\begin{array}{rcl}
\ds{\pars{\partiald[2]{}{t} + 1}\mrm{G}\pars{t,t'}} & \ds{=} &
\ds{\delta\pars{t - t'}}
\\[2mm]
\ds{\mrm{G}\pars{0,t'} = \mrm{G}_{t}\!\pars{0,t'}} & \ds{=} & \ds{0}
\end{array}\right.
\end{align}

$\ds{\mrm{G}\pars{t,t'}}$ satisfies
  $\ds{\left.\pars{\partiald[2]{}{t} + 1}\mrm{G}\pars{t,t'}
\right\vert_{\ t\ \not=\ t'} = 0}$ and
  $\ds{\left.
\lim_{\epsilon \to 0^{+}}\partiald{\mrm{G}\pars{t,t'}}{t}
\right\vert_{\ t=\ t' - \epsilon}^{\ t=\ t' + \epsilon} = 1}$.

Then,
\begin{align}
&\mrm{G}\pars{t,t'} =
\left\{\begin{array}{lcl}
\ds{A\sin\pars{t} + B\cos\pars{t}} & \ds{t < t'}
\\[2mm]
\ds{C\sin\pars{t - t'}} & \ds{t > t'}
\end{array}\right.
\\[5mm] &
\mrm{U}\pars{0} = \mrm{U}_{t}\pars{0} = 0 \implies A = B = 0.
\end{align}

Continuity at $\ds{t = t'}$ is already satisfied by the 'form'
  $\ds{C\sin\pars{t - t'}}$, when $\ds{t > t'}$, and the 'jump' at
  $\ds{t = t'}$ yields
  $\ds{C\cos\pars{t - t} - 0 = 1 \implies C = 1}$.

Then,
$$
\mrm{U}\pars{t} = \int_{0}^{t}\sin\pars{t - t'}\mrm{g}\pars{t'}\,\dd t'  = \int_{0}^{t}\sin\pars{t'}\mrm{g}\pars{t - t'}\,\dd t'
$$
$$
\bbx{\mrm{u}\pars{x,t} =
\sin\pars{x}\int_{0}^{t}\mrm{g}\pars{t - t'}\sin\pars{t'}\,\dd t'}
$$
