How to show that a delta function solution to a PDE (Fokker-Planck) is really a solution I would like to show that a given solution really is a solution to a PDE. 
The discussion of this is from a book "Quanum Noise" by Gardiner and Zoller (around page 125).
The partial differential equation is (I've taken $\hbar$  to be 1):
$$\frac{\partial P(\alpha, \alpha^{*}, t) }{ \partial t} = i \left( -\omega \frac{\partial}{\partial \alpha} \alpha  + \omega \frac{\partial}{\partial \alpha^{*}} \alpha^{*} -\lambda \frac{\partial}{\partial \alpha}   + \lambda^{*} \frac{\partial}{\partial \alpha^{*}}   \right) P(\alpha, \alpha^{*}, t) $$
It is "shown" that a solution to it is 
$$P(\alpha, \alpha^{*}, t) = \delta^{(2)}[\alpha - \beta(t)] = \delta(Re(\alpha) - Re(\beta(t))) \delta(Im(\alpha) - Im(\beta(t)))$$
where $\alpha$ satisfies:
$$\dot \beta = i \left( \omega \beta + \lambda(t) \right)$$
I would like to be able to show that the function above for $P(\alpha, \alpha^{*}, t)$ written in terms of the $\delta$ functions really satisfies the PDE by directly substituting it in, but I can't quite do it. I've tried various things, and in particular some properties of the delta function like 
$$\int \alpha \frac{\partial}{\partial \alpha} \delta(\alpha) d\alpha = - \int \delta(\alpha) d\alpha$$
but can't get RHS to agree with the LHS.
Does anyone know how one would do this? 
Thanks!
 A: Lets check by direct computation. Let $\phi$ be some test function and consider $\partial_t P$ in the sense of distributions.
\begin{align*}
(\partial_t P, \phi) = & - (P , \partial_t \phi)
\end{align*}
Lets check the RHS first, since $P = \delta^{(2)} ( \alpha - \beta (t) )$, we have that
$$ - ( P, \partial_t \phi ) = - (\Im \partial_t \phi)( \alpha) (\Re \partial_t \phi)(\alpha) $$
where $\Im$ is the imaginary part, and $\Re$ is the real part. Now lets check the LHS by plugging in the PDE.
$$(\partial_t P, \phi) = (- i\omega \partial_\alpha \alpha P, \phi) + (i \omega \partial_{\alpha^*} \alpha^* P, \phi ) - ( i \lambda \partial_\alpha P , \phi) + ( i \lambda^* \partial_{\alpha^*} P, \phi)$$
We can simplify things via product rule 
\begin{align*}
 (\partial_t P, \phi) =& (-i \omega  \alpha \partial_{\alpha}P, \phi) + (i \omega \alpha^* \partial_{ \alpha^*} P, \phi) - ( i \lambda \partial_\alpha P , \phi) + ( i \lambda^* \partial_{\alpha^*} P, \phi) \\
 =& -([i \omega \alpha + i \lambda] \partial_a P , \phi ) + ([ i \omega \alpha^* + i \lambda ] \partial_{\alpha^*} P, \phi) \\
=& -( \partial_t \alpha \partial_\alpha P, \phi ) + ( \partial_t \alpha^* \partial_ {\alpha^*}P, \phi )\\
\end{align*}
Note that
$$P (\alpha, \alpha^*,t) = \delta \left ( \frac{\alpha + \alpha^*}{2} - \frac{ \beta(t) + \beta^*(t)}{2} \right ) \delta \left ( \frac{\alpha - \alpha^*}{2i} - \frac{ \beta(t) - \beta^*(t)}{2i} \right ) $$
Now it's just a derivative calculation. Note that $( \delta' ,\phi) = -\phi'(0)$  (Sorry about switching between $\alpha$ and $\beta$ :s)
