Use Laplace Transform to find Green's-like function I am trying to solve the PDE
$$\frac{\partial^2u}{\partial x^2}=\frac{\partial u}{\partial t}\hspace{10pt}-\infty<x<\infty, \hspace{5pt} t>0$$
subject to the initial conditions
$$u(x,0)=f(x), \hspace{5pt}u(x,t)\text{ bounded}$$
and get an answer in the form
$$u(x,t)=\int_{-\infty}^\infty K(|x-\xi|,t)f(\xi)d\xi.$$
I have already shown that the laplace transform of $$\frac{e^{-a^2/4t}}{\sqrt{\pi t}}$$
is
$$s^{-1/2}e^{-|a|\sqrt{s}}$$
and then I transform the PDE with respect to $t$ to get
$$y''(x)-sy(x)+f(x)=0$$
which I'm given has particular solution
$$y(x,s)=\frac{e^{-x\sqrt{s}}}{2\sqrt{s}}\int_0^xe^{\xi \sqrt{s}}{f(\xi)d\xi}-\frac{e^{x\sqrt{s}}}{2\sqrt{s}}\int_0^xe^{-\xi \sqrt{s}}{f(\xi)d\xi}$$
and I found the bounded homogenous solution as $Ae^{-|x|\sqrt{s}}$, so the general solution is of the form $$g(x,s)=Ae^{-|x|\sqrt{s}}+By(x,s).$$
This is where I'm stuck. Firstly, I can't use my known Laplace transform on $y(x,s)$ since it depends on the sign of $x-\xi$ (whereas the known Laplace transform only depends on the absolute value). Even if I can inverse Laplace transform $y(x)$, I can't see how I would get the homogenous solution to be inside an integral with $f(\xi)$. There is also the problem of the integral bounds being from $-\infty$ to $\infty$ in the solution and only from $0$ to $x$ in the particular solution, but I think this might be solved if I can find a way to include the homogenous solution.
I'm hoping for a hint on how to proceed, or some intuition about what is going on in this question.
 A: I suppose that your particular solution can be adjusted to meet all requirements.
$$y(x,s)=\frac{e^{-x\sqrt{s}}}{2\sqrt{s}}\int_0^xe^{\xi \sqrt{s}}{f(\xi)d\xi}-\frac{e^{x\sqrt{s}}}{2\sqrt{s}}\int_0^xe^{-\xi \sqrt{s}}{f(\xi)d\xi}$$ is not perfect due to two reasons:

*

*It does not include the value of function $f(x)$ for negative $x$ (though, $x<0$ should also contribute, because the function is finite, but defined for all $x\in R$)

*at $x=0$ $\,\,y(o,s)=0\,$ for all $s\,$ - what can not be true.

Therefore, let's take
$$y(x,s)=\frac{e^{-x\sqrt{s}}}{2\sqrt{s}}\int_{-\infty}^xe^{\xi \sqrt{s}}{f(\xi)d\xi}+\frac{e^{x\sqrt{s}}}{2\sqrt{s}}\int_x^\infty e^{-\xi \sqrt{s}}{f(\xi)d\xi}$$
You can check that this is also a particular solution of the equation and it $\neq0$ at $x=0$.
Now you can find a general solution, making inverse transformation or using the found Laplace transform of the function $\frac{e^{-a^2/4t}}{\sqrt{\pi t}}$. For both terms you get a right (negative) sign.
$$y(x,t)=\frac{1}{2\sqrt{\pi t}}\int_{-\infty}^xe^{-\frac{|\xi-x|^2}{4t}}f(\xi)d\xi+\frac{1}{2\sqrt{\pi t}}\int_{x}^\infty e^{-\frac{|x-\xi|^2}{4t}}f(\xi)d\xi=\frac{1}{2\sqrt{\pi t}}\int_{-\infty}^\infty e^{-\frac{|\xi-x|^2}{4t}}f(\xi)d\xi$$
At $t\to0$ $K(|x-\xi|,t)\to\delta (x-\xi)$; integrating you get $y(x,t=0)=f(x)$ as required.
