I build a model with the following governing equation. I don't know if there is an analytic solution existed for this problem. Any suggestions or comments are appreciated.
$$ \frac{\partial c}{\partial t}-\frac{\partial ^{2}c}{\partial x^2}+\frac{\partial c}{\partial x} -f(x)\frac{\partial ^{2}c}{\partial z^2}+f(x)\frac{\partial c}{\partial z} =0 $$ where $f(x)$ is a step function (sum of two heaviside functions). The boundary condition at $x=0$ is first type B.C. and all other boundaries are second type with zero flux. Initial condition is zero everywhere.
Hint:
Let $c(x,z,t)=X(x)Z(z)T(t)$ ,
Then $X(x)Z(z)T'(t)-X''(x)Z(z)T(t)+X'(x)Z(z)T(t)-f(x)X(x)Z''(z)T(t)+f(x)X(x)Z'(z)T(t)=0$
$X(x)Z(z)T'(t)=(X''(x)-X'(x))Z(z)T(t)+(Z''(z)-Z'(z))f(x)X(x)T(t)$
$\dfrac{T'(t)}{T(t)}=\dfrac{X''(x)-X'(x)}{X(x)}+\dfrac{(Z''(z)-Z'(z))f(x)}{Z(z)}=-\dfrac{4(g(u))^2+1}{4}$
$\begin{cases}T'(t)=-\dfrac{4(g(u))^2+1}{4}T(t)\\\dfrac{X''(x)-X'(x)}{f(x)X(x)}+\dfrac{Z''(z)-Z'(z)}{Z(z)}=-\dfrac{4(g(u))^2+1}{4f(x)}\end{cases}$
$\begin{cases}T'(t)=-\dfrac{4(g(u))^2+1}{4}T(t)\\\dfrac{X''(x)-X'(x)}{f(x)X(x)}+\dfrac{4(g(u))^2+1}{4f(x)}=-\dfrac{Z''(z)-Z'(z)}{Z(z)}=\dfrac{4(h(v))^2+1}{4}\end{cases}$
$\begin{cases}T'(t)=-\dfrac{4(g(u))^2+1}{4}T(t)\\\begin{cases}Z''(z)-Z'(z)+\dfrac{4(h(v))^2+1}{4}Z(z)=0\\X''(x)-X'(x)+\dfrac{4(g(u))^2+1-(4(h(v))^2+1)f(x)}{4}X(x)=0\end{cases}\end{cases}$