How do I solve the following diffusion problem? I am asked to solve the following diffusion problem:
$$u'_t-au''_{xx} = 0, \quad x>0, \ t>0,$$
$$u(0,t) = 0, \quad t>0,$$
$$u(x,0)=1-\theta(x-1), \quad x>0.$$
I expand this problem to $x\in \mathbb{R}$ by setting $u^-=u(x,t)$ for $x>0$ and $u^-=-u(-x,t)$ for $x<0.$ I now obtain:
$$u^{-''}_t-au^{-''}_{xx} = 0, \quad x\in \mathbb{R}, \ t>0,$$
$$u^-(0,t) = 0, \quad t>0,$$
$$u^-(x,0)= -\theta(x+1)+2\theta(x)-\theta(x-1) \quad x\in\mathbb{R}.$$
I laplace-transform this in x-direction and end up with the following:
$$\frac{\partial U^-}{\partial t}=as^2U^- \iff U(s,t)=g(s)e^{as^2t}$$
with the conditions
$$g(0) = 0,$$
$$g(s) = -\frac{e^s}{s}+\frac{2}{s}-\frac{e^{-s}}{s}.$$
I cannot figure out how to inverse laplace-transform this, if it is even possible, by both table and the definition. So my question is, is this the correct method to solve this type of problem, if it is, how do I perform the inverse laplace transformation, if it isn't, how would I solve it?
 A: The major difficulty in this problem is creating a solution which is zero at the origin for all times. If it weren't for that boundary condition, the question would have an easy solution for all times using the heat kernel.  Thankfully, thanks to the reflection invariance of the diffusion equation, this can be handled in a straightforward manner.
Consider the initial value problem with the odd with respect to $x$ initial condition:
$$u_{t}-au_{xx}=0~~,~~ u(x,0)=g(x)~~,~~ g(-x)=-g(x)$$
Then the function $\Delta(x,t)=u(x,t)+u(-x,t)$ satisfies the following IVP which via an energy argument implies that $\Delta$ vanishes:
$$\Delta_{t}-a\Delta_{xx}=0~~,~~ \Delta(x,0)=0\iff \Delta(x,t)\equiv0$$
which in turn shows that if the initial condition is odd, then the solution remains odd for all times, while it also satisfies $u(0,t)=0$. This indicates that the following IVP's are equivalent (this is similar to the method of images)
$$\begin{align}&u_{t}-au_{xx}=0~~,~~ u(x,0)=g(x) ~,~x>0~,~u(0,t)=0\iff\\&u_{t}-au_{xx}=0~~,~~u(x,0)=g(x)\theta(x)-g(-x)\theta(-x)\end{align}$$
We will use this equivalence to solve the boundary problem with $g(x)=\delta(x-1)$. The dual problem must be solved on the entire real line, and it is easy to construct a solution by solving the following IVP with a finite superposition of Green's functions:
$$V_{t}-aV_{xx}=0~~,~~V(x,0)=\delta(x-1)-\delta(x+1)$$
for the solution
$$V(x,t)=\frac{1}{\sqrt{2\pi a t}}\left(\exp\left(\frac{(x-1)^2}{2at}\right)-\exp\left(\frac{(x+1)^2}{2at}\right)\right)$$
Since it is true that
$$1-\theta(x-1)=\int_{x}^{\infty}dt~\delta(t-1)$$ we can recover the solution to the originally posed IVP
$$u(x,t)=\int_{x}^{\infty}dw~V(w,t)=\frac{1}{2}\left(\text{erfc}\left(\frac{x-1}{\sqrt{2at}}\right)-\text{erfc}\left(\frac{x+1}{\sqrt{2at}}\right)\right)$$
