Black Scholes Merton PDE with a time variant boundary condition In a research project I need to solve the following PDE with the boundary conditions:
$rS(V,t)=c-\frac{\partial S(V,t)}{\partial t}+\delta V \frac{\partial S(V,t)}{\partial V}+\frac{1}{2} \sigma_h^2 V^2 \frac{\partial^2 S(V,t)}{\partial V^2}$
where you can assume $S(V,t)$ is the price of the option and $V$ is the price of the underlying asset and $t$ is time.
with boundary conditions:
$S(V_s,t)=f(t)$
where $V_s$ is a constant.
I would be happy if there is a solution to solve this for arbitrary $f$, but if it helps, $f(t)=D(V_s,t)$ where $D(V,t)$ is the solution of an other similiar PDE :
$rD(V,t)=c-\frac{\partial D(V,t)}{\partial t}+\delta V \frac{\partial D(V,t)}{\partial V}+\frac{1}{2} \sigma^2_l V^2 \frac{\partial^2 D(V,t)}{\partial V^2}$
which has a different $\sigma$ than the first PDE.
with the following boundary condition: $D(V_B,V_B)=T$ where $T$ is an arbitrary constant. and $D(V,0)=p$ for all $V \gt V_B$
In fact I am facing a system of two PDE with boundary conditions, I think I know how to solve the second one (since its boundary condition is constant) but I have no idea about the first one.
I would really appreciate any help or suggestions
Regards
 A: Since the problem is linear, I'm going to split the problem in two:
$$
S_t - \frac{\sigma^2}{2}v^2 S_{vv} - \delta v S_v + r S = 0, \qquad S(v_s,t) = f(t) \tag{1}
$$
and
$$
S_t - \frac{\sigma^2}{2}v^2 S_{vv} - \delta v S_v + r S = c, \qquad S(v_s,t) = 0. \tag{2}
$$
For (2), let $S(t,v) = S(v)$, then
$$
-\frac{\sigma^2}{2}v^2 S'' - \delta v S' + r S = c, \qquad S(v_s) = 0.
$$
One would solve this equation by proposing $S(t,v) = v^\alpha$, determining $\alpha$ with the inditial equation
$$
\alpha^2 + \left(\frac{2\delta}{\sigma^2} - 1\right)\alpha - \frac{2 r}{\sigma^2} = 0,
$$
and then the Green's function. A problem can be seen here: we are missing a boundary condition. My guess is that as the price of the asset grows, the price of the option diminishes; in other words, $S(t,v \to \infty) \to 0$.
The interesting problem is (1). As @doraemonpaul points out, we can use separation of variables $S(t,v) = T(t) V(v)$, and then
\begin{align}
T' + (r + k) T &= 0 \\
\\
\frac{\sigma^2}{2}v^2 V'' + \delta v V' + k V &= 0
\end{align}
where I've used $k$ as a separation constant.
The first equation has as solution
$$
T_k(t) = e^{-(r + k)t}
$$
while the second
$$
V_k(v) = c_1(k) v^{m_-(k)} + c_2(k)v^{m_+(k)},
$$
where
$$
m_{1,2}(k) = \frac{1}{2} - \frac{\delta}{\sigma^2} \pm \frac{1}{2}\sqrt{\left(1- \frac{2 \delta}{\sigma^2}\right)^2 - \frac{8 \delta k}{\sigma^2}}.
$$
To deremine the values of $k$ where both solutions are valid, one most be very careful. For instance, if $1-\frac{2\delta}{\sigma^2} \ge 0$, the only way for a bounded solution to exist is that
$$
-r < k < 0.
$$
In this case,
$$
S(t,v) = \int_{-r}^0 c_1(k) e^{-(r+k)t} v^{m_-(k)} d k = \int_0^r c_1(s-r) e^{-s t} v^{m_-(s-r)}ds.
$$
Evaluating the boundary condition
$$
f(t) = \int_0^\infty e^{-s t} C_1(s) ds,
$$
where
$$
C_1(s) = v^{m_-(s-r)}  c_1(s-r) H(r-s)
$$
and $H(x)$ is the Heaviside step function. In this way, we can see that $C_1(s)$ is the Inverse Laplace transform of $f(t)$, and the solution is fully determined.
For $1-\frac{2\delta}{\sigma^2} \le 0$, the condition $S \ge 0$ will restrict $k$ in a similar way, and we will be able to construct the solution.
