Conversion of BS PDE to Heat Equation

In its simplest form, the Black-Scholes equation is a PDE of the form

$V_t = rV - rx V_x - \frac{1}{2} \sigma^2 x^2 V_{xx}$ where $r, \sigma$ are positive constants and subscripts correspond to partial differentiation. A solution is a smooth function $V = V(t,x)$ satisfying the BS PDE. Normally, solutions for which $x > 0$ and $0 \le t \le T$ are sought. For a particular option pricing problem, solutions must also satisfy the boundary condition $V(T,x) = f(x)$ for some given payoff function $f$ .

It is possible to convert the variable-coefficient PDE into a constant-coefficient PDE by change of variables. Consider the global differeomorphism $y = \ln x$ and let the function $W$ be defined by $W(t,y) = V(t,x)$ . Then

$V_x = W_y y_x = W_y x^{-1}$

$V_{xx} = W_{yy} y_x^2 + W_y y_{xx} = W_{yy} x^{-2} + W_y (-x^{-2})$ .

and so

$W_t = rW - r W_y - \frac{1}{2} \sigma^2 (W_{yy} - W_y) = r W - (r - \frac{1}{2} \sigma^2) W_y - \frac{1}{2} \sigma^2 W_{yy}$

is a constant-coefficient PDE. Exponential integrating factors can be included to reduce the equation to the heat equation.

Let $W(t,y) = U(t,y) e^{at} e^{by}$ . Then

$W_t = (U_t + aU) e^{at} e^{by}$

$W_y = (U_y + bU) e^{at} e^{by}$

$W_{yy} = (U_{yy} + 2bU_y + b^2 U) e^{at} e^{by}$

and so

$U_t + aU = rU - (r-\frac{1}{2} \sigma^2)(U_y + bU) - \frac{1}{2} \sigma^2 (U_{yy} + 2bU_y + b^2 U)$

Choosing

$a = r - (r - \frac{1}{2} \sigma^2) b - \frac{1}{2} \sigma^2 b^2 = r + \frac{1}{2} \sigma^2 b^2$

$0 = -(r - \frac{1}{2} \sigma^2) - \frac{1}{2} \sigma^2 2b$

these added degrees of freedom may be used to zero coefficients multiplied by $U$ and $U_y$ to give

$U_t = -\frac{1}{2} \sigma^2 U_{yy}$

Finally, let $\tau = T-t$ , and $S(\tau,y) = U(t,y)$ . Then $U_t = - S_\tau$ and the standard form of the heat equation

$S_\tau = \frac{1}{2} \sigma^2 S_{yy}$

(where $\tau \in [0,T]$ and $y$ is unconstrained) is obtained. The boundary condition becomes $S(y,0) = e^{by} f(e^y)$ .

Finding the Fundamental Solution

[ignore this incorrect development]

For the PDE $S_\tau = \frac{1}{2} \sigma^2 S_{yy}$ and $S(y,0) = \delta(y - y_0)$ . Trying $S(\tau,y) = A(\tau) B(y)$ ,

$A'(\tau) B(y) = \frac{1}{2} \sigma^2 A(\tau) B''(y)$ and so $A'(\tau)/ A(\tau) = \frac{1}{2} \sigma^2 B''(y) / B(y)$ , the LHS is a function of $\tau$ and the RHS is a function of $y$ . This can only be simultaneously satisfied when the equation is some constant $k$ . Hence $A(\tau) = a e^{k\tau}$ and $B(y) = b e^{k^{1/2} (\sigma^2/2)^{-1/2} y}$ so $S(\tau,y) = ab e^{k\tau + k^{1/2} (\sigma^2/2)^{-1/2} y}$ . Using the boundary conditon $S(0,y_0) = 1 = ab e^{k^{1/2} (\sigma^2/2)^{-1/2} y_0}$ which requires

[to be continued]

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