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I have the PDE $$ -\frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2S^2\frac{\partial^2 V}{\partial S^2} + rS\frac{\partial V}{\partial S} - rV = 0$$ with initial and boundary conditions:

  1. $V(0,S)=max(E-S,0) $
  2. $V(t,S^*)=E-S^*(t) $
  3. $V(t,\infty)=0 $

$S^*$ finding from this condition - $\frac{\partial V(t,S)}{\partial S}\bigg|_{S=S^*}=-1$, and for $S\le S^*$ must be performed $\frac{\partial V(t,S)}{\partial t}=0$. Where $\sigma=0.2$, $r=0.08$, $E=100$.

I have to find $V(t,S)$ with $t=0.25$ and $S=100$.

So, I think that this PDE I can solve using finite difference method. Now I did approximation for PDE using forward and backward explicity methods. But I don't know how I can solve boundary condition. What I need to learn for solving this PDE? Or you can solve it, and tell me how you did it. :) Thanx!

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  • $\begingroup$ Let me try to understand. $S^*$ is an unknown point that can be moving? $\endgroup$
    – Pragabhava
    Nov 6, 2012 at 4:56
  • $\begingroup$ I think you're right. It's PDE with free boundary. $\endgroup$
    – spoluer
    Nov 6, 2012 at 5:21
  • $\begingroup$ Can you elaborate on the deduction of the model, or provide some reference? $\endgroup$
    – Pragabhava
    Nov 6, 2012 at 5:27
  • $\begingroup$ Option Pricing Using The Explicit Finite Difference Method. And C++ in Finance. Last link I used, because I need C++ algorithm for solve this PDE. $\endgroup$
    – spoluer
    Nov 6, 2012 at 6:40
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    $\begingroup$ From a quick read, shouldn't the last term $-rS$ be $-rV$? $\endgroup$
    – Pragabhava
    Nov 6, 2012 at 6:56

2 Answers 2

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I try to calculate ODE in answer from @Pragabhava, but I think I have a mistake in my solution.
For $S=S^*$, $V'(S)=-1$, then $V(S)=-S$ and $V''(S)=0$. To find a general solution I use Liouville's formula.
$\begin{vmatrix} -S & V \\ -1 & V' \\ \end{vmatrix} = ce^{-\int P(S)dS}$
$\int{P(S)dS=\int{\frac{rS}{\frac{1}{2}\sigma^2S^2}}dS}=2\frac{r}{\sigma^2}\int{\frac{dS}{S}}=2\frac{r}{\sigma^2}ln|S|$
$-SV'+V=ce^{-2\frac{r}{\sigma^2}ln|S|}=\frac{c}{S^{2\frac{r}{\sigma^2}}}$
Then I divided by $V^2(S)=S^2$,
$\frac{SV'-V}{S^2}=-\frac{c}{S^{2(\frac{r}{\sigma^2}+1)}}$;
$\big(\frac{V}{S}\big)'=-\frac{c}{S^{2(\frac{r}{\sigma^2}+1)}}$;
$\frac{V}{S}=-c\int\frac{dS}{S^{2(\frac{r}{\sigma^2}+1)}}=c\frac{1}{(2\frac{r}{\sigma^2}+1)S^{(2\frac{r}{\sigma^2}+1)}}$;
$V(S)=\frac{c}{(2\frac{r}{\sigma^2}+1)S^{2(\frac{r}{\sigma^2}+1)}}$
From this common solution I determine $S$:
$-S=\frac{c}{(2\frac{r}{\sigma^2}+1)S^{2(\frac{r}{\sigma^2}+1)}}$;
$S^{2\frac{r}{\sigma^2}+3}=-\frac{c}{2\frac{r}{\sigma^2}+1}$;
$S=\sqrt[2\frac{r}{\sigma^2}+3]{-\frac{c}{2\frac{r}{\sigma^2}+1}}$
Am I right?

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  • $\begingroup$ Why dont you use the ansat $V(S) = S^\alpha$ and see what happens with $\alpha$? $\endgroup$
    – Pragabhava
    Nov 8, 2012 at 4:41
  • $\begingroup$ Sorry, I don't understand. What does you mean? $\endgroup$
    – spoluer
    Nov 8, 2012 at 6:41
  • $\begingroup$ If $V(S) = S^\alpha$, then substituting in the time independent equation leads to $$\left(\frac{1}{2}\sigma^2\alpha + r \right)(\alpha - 1) = 0.$$ Either $\alpha = 1$ or $\alpha = -\frac{2 r}{\sigma^2}$. $\endgroup$
    – Pragabhava
    Nov 8, 2012 at 6:57
  • $\begingroup$ Ok, I understand you. But how can you find $S^*(t)$ using this solution? $\endgroup$
    – spoluer
    Nov 8, 2012 at 7:12
  • $\begingroup$ I'm not entirely sure, but I would use $V(S) = A S + B S^{-\frac{2 r}{\sigma^2}}$ and evaluate the condition $V(0,S)$, then determine $S^*$ with the condition $\frac{\partial V(S^*)}{\partial S} = 1$, and then do either separation of variables, Laplace transform, or finite differences, whatever works best. $\endgroup$
    – Pragabhava
    Nov 8, 2012 at 7:17
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Although I may have no idea about substitution of the initial and boundary conditions, I have idea of the form of the solution of this PDE, as I able to solve the similar type question.

I think you only interesting about the solution of $t,S\geq0$ :

Let $V(t,S)=F(t)G(S)$

Then $-F'(t)G(S)+\dfrac{1}{2}\sigma^2S^2F(t)G''(S)+rSF(t)G'(S)-rF(t)G(S)=0$

$F'(t)G(S)=\dfrac{\sigma^2F(t)}{2}\left(S^2G''(S)+\dfrac{2r}{\sigma^2}SG'(S)-\dfrac{2r}{\sigma^2}G(S)\right)$

$\dfrac{2F'(t)}{\sigma^2F(t)}=\dfrac{S^2G''(S)+\dfrac{2r}{\sigma^2}SG'(S)-\dfrac{2r}{\sigma^2}G(S)}{G(S)}=-\xi^2-\dfrac{1}{4}\left(\dfrac{2r}{\sigma^2}-1\right)^2-\dfrac{2r}{\sigma^2}$

$\begin{cases}\dfrac{F'(t)}{F(t)}=-\dfrac{\sigma^2\xi^2}{2}-\dfrac{\sigma^2}{8}\left(\dfrac{2r}{\sigma^2}-1\right)^2-r\\S^2G''(S)+\dfrac{2r}{\sigma^2}SG'(S)+\biggl(\xi^2+\dfrac{1}{4}\left(\dfrac{2r}{\sigma^2}-1\right)^2\biggr)G(S)=0\end{cases}$

$\begin{cases}F(t)=c_3(\xi)e^{-t\Bigl(\frac{\sigma^2\xi^2}{2}+\frac{\sigma^2}{8}\bigl(\frac{2r}{\sigma^2}-1\bigr)^2+r\Bigr)}\\G(S)=\begin{cases}c_1(\xi)S^{\frac{1}{2}-\frac{r}{\sigma^2}}\sin(\xi\ln S)+c_2(\xi)S^{\frac{1}{2}-\frac{r}{\sigma^2}}\cos(\xi\ln S)&\text{when}~\xi\neq0\\c_1S^{\frac{1}{2}-\frac{r}{\sigma^2}}\ln S+c_2S^{\frac{1}{2}-\frac{r}{\sigma^2}}\text{when}~\xi=0\end{cases}\end{cases}$

$\therefore V(t,S)=C_1e^{-\Bigl(\frac{\sigma^2}{8}\bigl(\frac{2r}{\sigma^2}-1\bigr)^2+r\Bigr)t}S^{\frac{1}{2}-\frac{r}{\sigma^2}}\ln S+C_2e^{-\Bigl(\frac{\sigma^2}{8}\bigl(\frac{2r}{\sigma^2}-1\bigr)^2+r\Bigr)t}S^{\frac{1}{2}-\frac{r}{\sigma^2}}+\int_0^\infty C_3(\xi)e^{-t\Bigl(\frac{\sigma^2\xi^2}{2}+\frac{\sigma^2}{8}\bigl(\frac{2r}{\sigma^2}-1\bigr)^2+r\Bigr)}S^{\frac{1}{2}-\frac{r}{\sigma^2}}\sin(\xi\ln S)~d\xi+\int_0^\infty C_4(\xi)e^{-t\Bigl(\frac{\sigma^2\xi^2}{2}+\frac{\sigma^2}{8}\bigl(\frac{2r}{\sigma^2}-1\bigr)^2+r\Bigr)}S^{\frac{1}{2}-\frac{r}{\sigma^2}}\cos(\xi\ln S)~d\xi$

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