Black Scholes Merton PDE with a time variant boundary condition

Question

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

Answer 1

Case $1$: $r\neq0$

Let $S(V,t)=S_c(V,t)+\dfrac{c}{r}$ ,

Then $\dfrac{\partial S(V,t)}{\partial t}=\dfrac{\partial S_c(V,t)}{\partial t}$

$\dfrac{\partial S(V,t)}{\partial V}=\dfrac{\partial S_c(V,t)}{\partial V}$

$\dfrac{\partial^2S(V,t)}{\partial V^2}=\dfrac{\partial^2S_c(V,t)}{\partial V^2}$

$\therefore r\left(S_c(V,t)+\dfrac{c}{r}\right)=c-\dfrac{\partial S_c(V,t)}{\partial t}+\delta V\dfrac{\partial S_c(V,t)}{\partial V}+\dfrac{1}{2}\sigma_h^2V^2\dfrac{\partial^2S_c(V,t)}{\partial V^2}$

$rS_c(V,t)+c=c-\dfrac{\partial S_c(V,t)}{\partial t}+\delta V\dfrac{\partial S_c(V,t)}{\partial V}+\dfrac{1}{2}\sigma_h^2V^2\dfrac{\partial^2S_c(V,t)}{\partial V^2}$

$\dfrac{\partial S_c(V,t)}{\partial t}=\dfrac{1}{2}\sigma_h^2V^2\dfrac{\partial^2S_c(V,t)}{\partial V^2}+\delta V\dfrac{\partial S_c(V,t)}{\partial V}-rS_c(V,t)$

Of course we use separation of variables (similar to Black-Scholes PDE with non-standard boundary conditions):

Let $S_c(V,t)=P(V)Q(t)$ ,

Then $P(V)Q'(t)=\dfrac{1}{2}\sigma_h^2V^2P''(V)Q(t)+\delta VP'(V)Q(t)-rP(V)Q(t)$

$P(V)Q'(t)=\dfrac{\sigma_h^2Q(t)}{2}\biggl(V^2P''(V)+\dfrac{2\delta}{\sigma_h^2}VP'(V)-\dfrac{2r}{\sigma_h^2}P(V)\biggr)$

$\dfrac{2Q'(t)}{\sigma_h^2Q(t)}=\dfrac{V^2P''(V)+\dfrac{2\delta}{\sigma_h^2}VP'(V)-\dfrac{2r}{\sigma_h^2}P(V)}{P(V)}=-s^2-\dfrac{1}{4}\biggl(\dfrac{2\delta}{\sigma_h^2}-1\biggr)^2-\dfrac{2r}{\sigma_h^2}$

$\begin{cases}\dfrac{Q'(t)}{Q(t)}=-\dfrac{\sigma_h^2s^2}{2}-\dfrac{\sigma_h^2}{8}\biggl(\dfrac{2\delta}{\sigma_h^2}-1\biggr)^2-r\\V^2P''(V)+\dfrac{2\delta}{\sigma_h^2}VP'(V)+\biggl(s^2+\dfrac{1}{4}\biggl(\dfrac{2\delta}{\sigma_h^2}-1\biggr)^2\biggr)P(V)=0\end{cases}$

$\begin{cases}Q(t)=c_3(s^2)e^{-t\Bigl(\frac{\sigma_h^2s^2}{2}+\frac{\sigma_h^2}{8}\bigl(\frac{2\delta}{\sigma_h^2}-1\bigr)^2+r\Bigr)}\\P(V)=\begin{cases}c_1(s^2)V^{\frac{1}{2}-\frac{\delta}{\sigma_h^2}}\sin((\ln V-\ln V_s)s)+c_2(s^2)V^{\frac{1}{2}-\frac{\delta}{\sigma_h^2}}\cos((\ln V-\ln V_s)s)&\text{when}~s\neq0\\c_1V^{\frac{1}{2}-\frac{\delta}{\sigma_h^2}}\ln V+c_2V^{\frac{1}{2}-\frac{\delta}{\sigma_h^2}}&\text{when}~s=0\end{cases}\end{cases}$

$\therefore S(V,t)=\int_0^\infty C_1(s^2)V^{\frac{1}{2}-\frac{\delta}{\sigma_h^2}}e^{-t\Bigl(\frac{\sigma_h^2s^2}{2}+\frac{\sigma_h^2}{8}\bigl(\frac{2\delta}{\sigma_h^2}-1\bigr)^2+r\Bigr)}\sin((\ln V-\ln V_s)s)~ds+\int_0^\infty C_2(s^2)V^{\frac{1}{2}-\frac{\delta}{\sigma_h^2}}e^{-t\Bigl(\frac{\sigma_h^2s^2}{2}+\frac{\sigma_h^2}{8}\bigl(\frac{2\delta}{\sigma_h^2}-1\bigr)^2+r\Bigr)}\cos((\ln V-\ln V_s)s)~ds+\dfrac{c}{r}$

$S(V_s,t)=f(t)$ :

$\int_0^\infty C_2(s^2)V_s^{\frac{1}{2}-\frac{\delta}{\sigma_h^2}}e^{-t\Bigl(\frac{\sigma_h^2s^2}{2}+\frac{\sigma_h^2}{8}\bigl(\frac{2\delta}{\sigma_h^2}-1\bigr)^2+r\Bigr)}~ds+\dfrac{c}{r}=f(t)$

$V_s^{\frac{1}{2}-\frac{\delta}{\sigma_h^2}}e^{-\Bigl(\frac{\sigma_h^2}{8}\bigl(\frac{2\delta}{\sigma_h^2}-1\bigr)^2+r\Bigr)t}\int_0^\infty C_2(s^2)e^{-\frac{\sigma_h^2ts^2}{2}}~ds=f(t)-\dfrac{c}{r}$

$\int_0^\infty\dfrac{C_2(s^2)e^{-\frac{\sigma_h^2ts^2}{2}}}{2s}d(s^2)=\left(f(t)-\dfrac{c}{r}\right)V_s^{\frac{\delta}{\sigma_h^2}-\frac{1}{2}}e^{\Bigl(\frac{\sigma_h^2}{8}\bigl(\frac{2\delta}{\sigma_h^2}-1\bigr)^2+r\Bigr)t}$

$\int_0^\infty\dfrac{C_2(s)e^{-\frac{\sigma_h^2ts}{2}}}{2\sqrt{s}}ds=\left(f(t)-\dfrac{c}{r}\right)V_s^{\frac{\delta}{\sigma_h^2}-\frac{1}{2}}e^{\Bigl(\frac{\sigma_h^2}{8}\bigl(\frac{2\delta}{\sigma_h^2}-1\bigr)^2+r\Bigr)t}$

$\mathcal{L}_{s\to\frac{\sigma_h^2t}{2}}\left\{\dfrac{C_2(s)}{2\sqrt{s}}\right\}=\left(f(t)-\dfrac{c}{r}\right)V_s^{\frac{\delta}{\sigma_h^2}-\frac{1}{2}}e^{\Bigl(\frac{\sigma_h^2}{8}\bigl(\frac{2\delta}{\sigma_h^2}-1\bigr)^2+r\Bigr)t}$

$C_2(s)=2V_s^{\frac{\delta}{\sigma_h^2}-\frac{1}{2}}\sqrt{s}\mathcal{L}^{-1}_{\frac{\sigma_h^2t}{2}\to s}\left\{f(t)e^{\Bigl(\frac{\sigma_h^2}{8}\bigl(\frac{2\delta}{\sigma_h^2}-1\bigr)^2+r\Bigr)t}\right\}-\dfrac{2c}{r}V_s^{\frac{\delta}{\sigma_h^2}-\frac{1}{2}}\sqrt{s}\delta\biggl(s+\dfrac{1}{4}\biggl(\dfrac{2\delta}{\sigma_h^2}-1\biggr)^2+\dfrac{2r}{\sigma_h^2}\biggr)$

Answer 2

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.