Reaction Diffusion Equation with a time-dependent Neumann b.c. on a semi-infinite interval I am trying to solve the following equation:
\begin{align}
\frac{1}{2 D}\frac{\partial m(x,t)}{\partial t} &= \frac{\partial^2 m(x,t)}{\partial x^2} - \frac{m(x,t)}{\lambda_{sf}^{2}}\\
m(x,0) &= 0\\
\frac{\partial m(0,t)}{\partial x}&= f(t)\\
 m(\infty,t) &= 0 
\end{align}
I have tried a method suggested here. However, I cannot work out the BCs for $\phi(x)$. I have tried using $\phi_x(0) = f(t) $ and $\phi(\infty) =0 $ but I don't think $\phi_x(x)$ can be a function of $t$ so that BC doesn't make sense.
I am sure I am missing something really obvious I just can't think of it. I have pages of latex notes I have not uploaded but can if someone would like to see them.
*** EDIT ***
The solution should tend to an equilibrium solution as $t \rightarrow \infty$.
Also I know $f(t)$ is approximately $-Ae^{-Bx} + D$. Although I do not know the exact formula as another (more complicated) PDE dictates the rate of change of the solution at the boundary which has not been solved either.
 A: Typically you would have boundary conditions specifying  both $m(0, t)$ and $\dfrac{\partial m}{\partial x}(0,t)$, as well as the initial condition $m(x,0)$.  You are missing the boundary condition on $m(0,t)$, so your problem is underdetermined. 
A: \begin{align}
\frac{1}{2 D}\partial_t m(x,t) &= \partial^2_x m(x,t) - \lambda_{sf}^{-2}m(x,t)\,,\\
m(x,0) &= 0\,,\\
\partial_x m(0,t)&= f(t)\,, \\
m(\infty,t) &= 0 \,.\\
\end{align}
To clean up the notation, I write $\lambda_{sf}^{-2}=a$, take $t=t'/(2D)$, and drop primes,
\begin{align}
\partial_t m(x,t) &= \partial^2_x m(x,t) - a m(x,t)\,,\\
m(x,0) &= 0\,,\\
\partial_x m(0,t)&= f(t) \,,\\
m(\infty,t) &= 0 \,.\\
\end{align}
A Laplace transformation $\hat{m}(x,s)=\mathcal{L}\left\{m(x,t)\right\}=\int_{0}^{\infty}{\rm d}t\, e^{st}m(x,t)$ takes the above PDE to
\begin{align}
s\hat{m}(x,s) -m(x,0)&=  \partial_x^2 \hat{m}(x,s)-a \hat{m}(x,s)\nonumber\\
\partial_x^2 \hat{m}(x,s) &=  n^2\hat{m}(x,s)\,,
\end{align}
where we used that $m(x,0)=0$ and where $n^2=a+s$. Clearly, this ODE is solved by $\hat{m}(x,s)=Ae^{nx}+Be^{-nx}$.
In terms of Laplace-transformed variables, the boundary conditions  read
\begin{align}
\partial_x \hat{m}(0,s)&=\hat{f}(s)\,,\\
\hat{f}(\infty,s)&=0\,.
\end{align}
Enforcing the boundary condition at $x=\infty$ yields $A=0$. The Neumann condition yields $B=-\hat{f}(s)/(ne^0)$; hence,
\begin{align}
\hat{m}(x,s)=-\hat{f}(s)\frac{e^{-x\sqrt{a+s}}}{\sqrt{a+s}}\equiv -\hat{f}(s)\hat{g}(x,s)\,,
\end{align}
with $\hat{g}(x,s)=e^{-x\sqrt{a+s}}/\sqrt{a+s}$.
Finding $m(x,t)$ amounts to performing the following inverse Laplace transform
\begin{align}
m(x,t)&=-\mathcal{L}^{-1}\left\{\hat{f}(s)\hat{g}(x,s)\right\}
=-\int_0^t f(t-\tau)g(x,\tau){\rm d}\tau\,,
\end{align}
where I used that multiplying two functions in Laplace space corresponds to convolution in real-time (see wikipedia).
We now use that the Laplace transform of $h(x,t)=e^{-x^2/(4t)}/\sqrt{t\pi}$ is $\hat{h}(s)=\mathcal{L}\left\{h(t)\right\}=e^{\sqrt{s}x}/\sqrt{s}$ (see here) and  $\hat{g}(x,s)=\hat{h}(-x,s+a)$ to find
\begin{align}
\mathcal{L}\left\{h(-x,t)\right\}&=\hat{h}(-x,s)\\
\mathcal{L}\left\{e^{-at}h(-x,t)\right\}&=\hat{h}(-x,s+a)\\
\mathcal{L}\left\{e^{-at}h(-x,t)\right\}&=\hat{g}(x,s)\\
e^{-at}h(-x,t)&=\mathcal{L}^{-1}\left\{\hat{g}(x,s)\right\}
\end{align}
from which we conlude, with  $h(x,t)=h(-x,t)$, that
\begin{equation}
g(x,t)=\frac{1}{\sqrt{t\pi}}\exp{\left(-\frac{x^2}{4t}-at\right)}\,.
\end{equation}
Finally, we find
\begin{align}
m(x,t)
=-\int_0^t f(t-\tau)\frac{1}{\sqrt{\tau\pi}}\exp{\left(-\frac{x^2}{4\tau}-a\tau\right)}{\rm d}\tau\,,
\end{align}
or, in terms of the original variables,
\begin{align}
m(x,t)&
=-\int_0^{2Dt} f[2D(t-\tau)]\sqrt{\frac{2D}{\tau\pi}}\exp{\left(-\frac{x^2}{8D\tau}-\frac{2D\tau}{\lambda_{sf}^{2}}\right)}{\rm d}\tau\,.
\end{align}
