Advection Diffusion Equation on Semi-Infinite Domain Regarding the BVP
$$u_t(x,t) - v\, u_x(x,t) = k\, u_{xx}(x,t),\qquad x\geq0$$
with BC $u_x(0, t)=0$ for $t\geq 0$, and parameters $v,k>0$, I have some questions.
Does an expression for the Green's function in some relatively nice form exist? If so, what is it?
There's the obvious change of variables to convert the above to the heat equation, but with non-standard boundary conditions. I could not think of any nice image singularity solution. We also have the problem of the domain being semi-infinite, making it difficult for there to be any series solutions.
If such a solution exists, how would I derive such a Green's function?
As another note, I would also be happy with a fundamental solution for the BC $u(0,t)=0$. A related question is how does one go about solving the heat equation with moving boundary conditions.
 A: I am going to outline a solution with a few differences:
1) I am going to use the boundary condition $u(0,t) = u_0$ for all $t \ge 0$.
2) I am going to assume that $u(x,0)=0$ for all $x \gt 0$. 
3) I am going to use Laplace transforms in $t$ to help with the solution.
That all said, define the Laplace transform of the solution
$$\hat{u}(x,s) = \int_0^{\infty} dt \, u(x,t) \, e^{-s t}$$
The PDE above then becomes an ODE in $x$:
$$k \hat{u}'' + v \hat{u}'-s \hat{u}=0$$
$$\hat{u}(0,s) = \frac{u_0}{s}$$
$$\lim_{x \to \infty} \hat{u}(x,s) = 0$$
The general solution to the ODE may be written as
$$\hat{u}(x,s) = A(s) e^{r_+ x} + B(s) e^{r_- x}$$
where
$$r_{\pm} = -\frac{v}{2 k} \pm \frac{\sqrt{v^2+4 k s}}{2 k}$$
Our boundary conditions imply that $A(s)=0$ and $B(s)=u_0/s$.  The solution to the equation is then an inverse Laplace transform:
$$u(x,t) = \frac{u_0}{i 2 \pi} e^{-v x/(2 k)} \int_{c-i \infty}^{c+i \infty} \frac{ds}{s} e^{-x\sqrt{v^2+4 k s} /(2 k)} \, e^{s t}$$
for some $c \gt 0$.  Our job is now to evaluate this integral, which we will do via the residue theorem as follows.
Consider the following integral in the complex plane:
$$\oint_C \frac{dz}{z} e^{-x\sqrt{v^2+4 k z} /(2 k)} \, e^{z t}$$
where $C$ is the following contour:

Note the point is the branch point.  I will assert without proof that the integral vanishes along the sections $C_2$, $C_4$, and $C_6$ of $C$.  This leaves $C_1$ (the ILT), $C_3$, and $C_5$.  There is a pole within the contour at $z=0$, so by the residue theorem, we have
$$\frac{1}{i 2 \pi}\left [\int_{C_1} + \int_{C_3} + \int_{C_5}  \right ]\frac{dz}{z} e^{-x\sqrt{v^2+4 k z} /(2 k)} \, e^{z t}  = e^{-v x/(2 k)}$$
Along $C_3$, note that there is a branch point at $z=-v^2/(4 k)$.  Thus we parametrize $z=-v^2/(4 k) + e^{i \pi} y$; the integral over $C_3$ becomes
$$e^{-v^2 t/(4 k)} \int_{\infty}^0 \frac{dy}{y+\frac{v^2}{4 k}} \frac{e^{-i (x/(\sqrt{k})) \sqrt{y}}}{\sqrt{y}} e^{-t y}$$
Similarly, along $C_5$, let $z=-v^2/(4 k) + e^{-i \pi} y$; the integral over $C_5$ becomes
$$e^{-v^2 t/(4 k)} \int_0^{\infty} \frac{dy}{y+\frac{v^2}{4 k}} e^{i (x/(\sqrt{k})) \sqrt{y}}\, e^{-t y}$$
Putting this all together as above, we get an expression for the ILT:
$$\frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} \frac{ds}{s} e^{-(x/(2 k)) \sqrt{v^2+4 k s}} \, e^{s t} = e^{-x v/(2 k)} - \frac{e^{-v^2 t/(4 k)}}{\pi} \int_0^{\infty} \frac{dy}{y+\frac{v^2}{4 k}} \sin{\frac{x \sqrt{y}}{\sqrt{k}}} \, e^{-t y}$$
ADDENDUM
I now address the evaluation of the integral on the RHS.  Making the substitution $y=u^2$, we get
$$\frac{e^{-a^2 t}}{\pi} \int_{-\infty}^{\infty} du \frac{u \sin{b u}}{a^2+u^2} e^{-t u^2}$$
where $a^2=v^2/(4 k)$ and $b=x/\sqrt{k}$.  This integral is a challenge to say the least, with Mathematica unable to evaluate analytically.  That said, a minor rewrite of the integrand reveals a strategy for evaluation:
$$\frac{e^{-a^2 t}}{\pi} \int_{-\infty}^{\infty} du  \frac{u^2}{a^2+u^2} \frac{\sin{b u}}{u} e^{-t u^2}$$
which in turn may be written as
$$\frac{e^{-a^2 t}}{\pi} \int_{-\infty}^{\infty} du \frac{\sin{b u}}{u} e^{-t u^2} - \frac{a^2 \,e^{-a^2 t}}{\pi} \int_{-\infty}^{\infty} \frac{du }{a^2+u^2} \frac{\sin{b u}}{u} e^{-t u^2}$$
The former integral may be evaluated any number of ways, e.g., Parseval's Theorem.  I get
$$\int_{-\infty}^{\infty} du \frac{\sin{b u}}{u} e^{-t u^2} = \pi \, \text{erf}\left(\frac{b}{2 \sqrt{t}}\right)$$
The latter integral is over a product of three functions with known Fourier transforms.  To evaluate this integral using Parseval's Theorem, we need to take the Fourier transform of a product of two of the functions by using the convolution theorem.  As an illustration, let 
$$f(u) = \frac{1}{a^2+u^2} \frac{\sin{b u}}{u}$$
Then by the convolution theorem, the Fourier transform $\hat{f}(k)$ is
$$\hat{f}(k) = \frac{\pi}{2 a} \int_{-b}^b dk' e^{-a |k-k'|}$$
Believe it or not, Mathematica wouldn't even evaluate this one.  That said, it is not too difficult: just consider the cases $k \lt -b$, $k \gt b$, and $|k| \le b$.  The result is
$$\hat{f}(k) = \begin{cases}\frac{\pi}{a^2} \left (1-e^{-a b} \cosh{a k} \right ) & |k| \le b\\ \frac{\pi}{a^2} \sinh{a b} \, e^{-a |k|} & |k| \gt b \end{cases}$$
The integral we seek is then
$$\int_{-\infty}^{\infty} \frac{du }{a^2+u^2} \frac{\sin{b u}}{u} e^{-t u^2} = \frac{1}{2 \pi} \sqrt{\frac{\pi}{t}} \int_{-\infty}^{\infty} dk \, \hat{f}(k) \, e^{-k^2/(4 t)}$$
which is expressible in terms of error functions.
A: Let $u(x,t)=X(x)T(t)$ ,
Then $X(x)T'(t)-vX'(x)T(t)=kX''(x)T(t)$
$X(x)T'(t)=kX''(x)T(t)+vX'(x)T(t)$
$X(x)T'(t)=(kX''(x)+vX'(x))T(t)$
$\dfrac{T'(t)}{T(t)}=\dfrac{kX''(x)+vX'(x)}{X(x)}=-\dfrac{4k^2s^2+v^2}{4k}$
$\begin{cases}\dfrac{T'(t)}{T(t)}=-\dfrac{4k^2s^2+v^2}{4k}\\kX''(x)+vX'(x)+\dfrac{4k^2s^2+v^2}{4k}X(x)=0\end{cases}$
$\begin{cases}T(t)=c_3(s)e^{-\frac{t(4k^2s^2+v^2)}{4k}}\\X(x)=\begin{cases}c_1(s)e^{-\frac{vx}{2k}}\sin xs+c_2(s)e^{-\frac{vx}{2k}}\cos xs&\text{when}~s\neq0\\c_1xe^{-\frac{vx}{2k}}+c_2e^{-\frac{vx}{2k}}&\text{when}~s=0\end{cases}\end{cases}$
$\therefore u(x,t)=\int_0^\infty C_1(s)e^{-\frac{2vx+t(4ks^2+v^2)}{4k}}\sin xs~ds+\int_0^\infty C_2(s)e^{-\frac{2vx+t(4ks^2+v^2)}{4k}}\cos xs~ds$
$u_x(x,t)=\int_0^\infty C_1(s)e^{-\frac{2vx+t(4ks^2+v^2)}{4k}}\left(-\dfrac{v}{2k}\sin xs+s\cos xs\right)ds+\int_0^\infty C_2(s)e^{-\frac{2vx+t(4ks^2+v^2)}{4k}}\left(-\dfrac{v}{2k}\cos xs-s\sin xs\right)ds$
If the B.C. is $u_x(0,t)=0$ :
$\int_0^\infty sC_1(s)e^{-\frac{t(4ks^2+v^2)}{4k}}~ds-\int_0^\infty\dfrac{v}{2k}C_2(s)e^{-\frac{t(4ks^2+v^2)}{4k}}~ds=0$
$\int_0^\infty\left(sC_1(s)-\dfrac{v}{2k}C_2(s)\right)e^{-\frac{t(4ks^2+v^2)}{4k}}~ds=0$
$sC_1(s)-\dfrac{v}{2k}C_2(s)=0$
$C_2(s)=\dfrac{2ks}{v}C_1(s)$
$\therefore u(x,t)=\int_0^\infty C_1(s)e^{-\frac{2vx+t(4ks^2+v^2)}{4k}}\sin xs~ds+\int_0^\infty\dfrac{2ks}{v}C_1(s)e^{-\frac{2vx+t(4ks^2+v^2)}{4k}}\cos xs~ds=\int_0^\infty C_1(s)e^{-\frac{2vx+t(4ks^2+v^2)}{4k}}\left(\sin xs+\dfrac{2ks}{v}\cos xs\right)ds$
If the B.C. is $u(0,t)=0$ :
$\int_0^\infty C_2(s)e^{-\frac{t(4ks^2+v^2)}{4k}}~ds=0$
$C_2(s)=0$
$\therefore u(x,t)=\int_0^\infty C_1(s)e^{-\frac{2vx+t(4ks^2+v^2)}{4k}}\sin xs~ds$
