Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


share|cite|improve this question
A few questions: Is there a relationship between $r$, $\delta$ and $\sigma$? Are all constants positive? One bigger than the other, etc? What's the domain for $V$? What's the behavior you're expecting for $S$ as the price $V$ goes to infinity? – Pragabhava Jul 11 '13 at 5:21

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




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

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

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

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

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




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

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.