I am looking at solutions for a diffusion equation, where we have:
\begin{equation} \frac{1}{4}u_{xx}=u_t , \ \ \ \ \ \ \, 0<x<\infty, t>0 \\ u_x(0,t)=0 \\ u(x,0)=\begin{cases} 6 \ \ \ \ 0 \leq x\leq 3\\ 0 \ \ \ \ x>3 \end{cases} \end{equation}
Clearly, here the domain of x is infinite. A method which I am trying to understand is using the "Reflection method" also called the "Odd extension method". Here we form an extension, $u^\bullet$ of $u$ as such
\begin{equation} u^\bullet(x,t)= \begin{cases} u(-x,t) \ \ \ \ x<0 \\ u(x,t) \ \ \ \ x>0 \end{cases} \end{equation}
Then the method just refers to the "reflection lemma" which yields the following new form of the original PDE:
\begin{equation} \begin{cases} u^\bullet_t-\frac{1}{4}u^\bullet_{xx}=\big(u_t-\frac{1}{4}u_{xx}\big)^\bullet-2\cdot\frac{1}{4}u_x(0,t)\delta(x)=0\\ u^\bullet(x,0)=\begin{cases} 6 \ \ \ for |x|<3 \overset{def}{=}g(x)\\ 0 \ \ \ for |x|>3 \end{cases} \end{cases} \end{equation}
which can be subjected to Laplace transform, and then get a solution in terms of the error function (which is really laborious).
But I have no literature of this reflection method, and no example which clearly illustrates how to derive that new form of the original PDE.
Does anyone have an idea how that form came about?
Thanks