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QUESTION

Find the most general solution of the Black-Scholes equation that
has the special form $V(S,t)=A(t)B(S)$. (Hint: separate the
variables).


ANSWER

$V(S,t)=A(t)B(t)$

Substituting this into Black-Scholes:

$$\frac{\partial V}{\partial
t}+\frac{1}{2}\sigma^2S^2\frac{\partial^2V}{\partial
S^2}+rS\frac{\partial V}{\partial S}-rv=0$$

$\frac{\partial V}{\partial t}=\frac{dA}{dt}B(S)\\ \frac{\partial
V}{\partial S}=A(t)\frac{dB}{dS}\\ \frac{\partial^2V}{\partial
S^2}=A(t)\frac{d^2B}{dS^2}$.

Therefore

$$B\frac{dA}{dt}+\frac{1}{2}\sigma^2S^2A\frac{d^2B}{dS^2}+rSA\frac{dB}{dS}-rAB=0$$

Goal: seperate all the $t$ dependence to one side and all the $S$
dependence to the other side.

To do this divide through by $AB$ and rearrenge:

$$\underbrace{\frac{1}{A}\frac{dA}{dt}-r}_{\textrm{only $t$
dependence (or
const.)}}=\underbrace{-\frac{1}{2}\frac{\sigma^2S^2}{B}
\frac{d^2B}{dS^2}-\frac{rS}{B}\frac{dB}{dS}}_{\textrm{only $S$
dependence}}$$

The only way this can happen is if LHS=RHS=const.

Let that constant be $C$.

Therefore

\begin{eqnarray}
\frac{1}{A}\frac{dA}{dt}-r&=&C\\
-\frac{1}{2}\frac{\sigma^2S^2}{B}\frac{d^2B}{dS^2}-\frac{rS}{B}\frac{dB}{dS}&=&C
\end{eqnarray}

Solve(1):

\begin{eqnarray*}
\frac{1}{A}\frac{dA}{dt}&=&r+c\\
\Rightarrow\int\frac{dA}{A}&=&\int(r+c)\,dt\\ \Rightarrow \ln
A&=&(r+c)t+\textrm{const.}\\ \Rightarrow A&=&A(0)e^{(r+c)t}
\end{eqnarray*}

Solve(2):

$$\frac{\sigma^2}{2}S^2\frac{d^2B}{dS^2}+rS\frac{dB}{dS}+CB=0$$
Euler-equation: try solutions $A=S^n$ with $n$ to be found by
substitution.

$$\frac{\sigma^2}{2}S^2n(n-1)S^{n-2}+rSnS^{n-1}+CS^n=0$$

Dividing by $S^n$

\begin{eqnarray*}
\frac{\sigma^2}{2}n(n-1)+rn+C&=&0\\
\frac{\sigma^2}{2}n^2+\left(r-\frac{\sigma^2}{2}\right)n+c&=&0
\end{eqnarray*}

$$n=\frac{\left(\frac{\sigma^2}{2}-r\right)\pm\sqrt{\left(r-\frac{\sigma^2}{2}\right)^2-4\frac{\sigma^2}{2}C}}{\sigma^2}$$

Call these values $n^+,\ n^-$. Then the general solution is

$$B(S)=\alpha r^{n^+}+\beta r^{n^-}$$

$\alpha,\ \beta$ const. Therefore the most general solution to
Black-Scholes is

\begin{eqnarray*}
V=A(t)B(S)=A(0)e^{(r+c)t}\left[\alpha r^{n^+}+\beta
r^{n^-}\right]\\ &=&e^{(r+c)t}\left(\gamma r^{n^+}+\delta
r^{n^-}\right)
\end{eqnarray*}

$\gamma,\ \delta=$ arbitrary constants to be determined by
boundary data. ($C$ is determined in the same way.)




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