Proving separable solution to a heat-type equation contains a 1st ord diff eq and 2nd ord diff eq?

Question

I have a heat type equation $$\frac{d}{dt}V + \frac{1}{2} \sigma^{2} S^{2} \frac{d^{2}}{dS^{2}}V + (r-D) S \frac{d}{ds} V - rV = 0$$

Am asked to prove the solution is separable $$V=A(t) B(s)$$ and that A(t) is 1st order diff eq and B(S) 2nd order diff eq.

I did $$\frac{d}{dt}V=B(s) \frac{d}{dt}A(t)$$ and $$\frac{d}{dS}V=A(t) \frac{d}{dS}B(S)$$ and $$\frac{d^2}{dS^2}V=A(t) \frac{d^2}{dS^2}B(S)$$

Plugged it in, got

$$\frac{\frac{d}{dt} A(t)}{A(t)} + \frac{1}{2} \sigma^{2} S^{2} \frac{\frac{d^{2}}{dS^{2}} B(S)}{B(S)} + (r-D) S \frac{\frac{d}{dS} B(S)}{B(S)} - r = 0$$

but don't know where to go from here?

Answer 1

For simpler notation, denote $F_x(x,t)$ the partial of $F$ wrt $x$. Your PDE is then $$V_t + \frac{\sigma^2 S^2}{2} V_{SS} + (r-D)s V_S - rV = 0,$$ which looks like the Black-Scholes equation BTW.

Under the assumption $V(t,S) = A(t) B(S)$ you have $V_t = A_t B, V_S = A B_S$ and $V_{SS} = AB_{SS}$, substituting you get $$ A_t B + \frac{\sigma^2 S^2}{2} AB_{SS} + (r-D)s AB_S - rAB = 0 $$ Dividing both sides by $V = AB$ you get $$ \frac{A_t}{A} + \frac{\sigma^2 S^2}{2} \frac{B_{SS}}{B} + (r-D)s \frac{B_S}{B} - r = 0 $$ which is equivalent to $$ \frac{\sigma^2 S^2}{2} \frac{B_{SS}}{B} + (r-D)s \frac{B_S}{B} = r - \frac{A_t}{A} $$ but now LHS only depends on $S$ and all RHS only depends on $t$, for all values of $S,T$, which is only possible if both LHS and RHS are equal to the same constant, say $c$.

So you get two independent ODEs out of your PDE: $$ \begin{split} \frac{\sigma^2 S^2}{2} \frac{B_{SS}}{B} + (r-D)s \frac{B_S}{B} &= c\\ r - \frac{A_t}{A} &= c \end{split} $$

Answer 2

We have \begin{eqnarray*} -\frac{\frac{d}{dt} A(t)}{A(t)} = \frac{1}{2} \sigma^{2} S^{2} \frac{\frac{d^{2}}{dS^{2}} B(S)}{B(S)} + (r-D) S \frac{\frac{d}{dS} B(S)}{B(S)} - r. \end{eqnarray*} The RHS is a function of $t$ only and the LHS is a function of $S$ only, so they are both constant. So we have \begin{eqnarray*} -\frac{\frac{d}{dt} A(t)}{A(t)} = k_i \\ \frac{1}{2} \sigma^{2} S^{2} \frac{\frac{d^{2}}{dS^{2}} B(S)}{B(S)} + (r-D) S \frac{\frac{d}{dS} B(S)}{B(S)} - r= k_i \\ \end{eqnarray*} where $k_i$ is a constant and a general solution will be of the form \begin{eqnarray*} V=\sum_{k_i} A_{k_i} (t) B_{k_i}(S) . \end{eqnarray*}