Conversion of the Black-Scholes Equation to the Diffusion Equation

We first bring the equation into the standard form of the diffusion equation, and then solve it using the Green's function for the diffusion equation on the initial condition at t=t^*.

The first difference we notice from the canonical equation is that the coefficients depend on x. However, the equation is homogeneous or invariant under the scaling of $x \to ax$. The standard way to simplify this and eliminate the explicit coordinate dependence is to define a new variable $u=\ln(x/c)$, where we have scaled x by c to make it dimensionless. Then under $x \to ax$, $u \to u + \ln(a)$. Since the equation is invariant under this, it cannot have any explicit dependence on u in the coefficients. Changing variables to u using $\frac{\partial u}{\partial x}=\frac{1}{x}$, and defining $\tilde{w}(u,t)=w(x,t)$, the derivatives become

$$\frac{\partial \tilde{w}}{\partial x} \frac{\partial \tilde{w}}{\partial u} \frac{\partial u}{\partial x}= \frac{1}{x}\frac{\partial \tilde{w}}{\partial u},$$ = (7)

$$\frac{\partial^2 \tilde{w}}{\partial x^2} \frac{1}{x^2} \left(\frac{\partial^2 \tilde{w}}{\partial u^2} - \frac{\partial \tilde{w}}{\partial u}\right).$$ = (8)

The 1/x and 1/x² factors cancel those in the Eq. 6 giving an equation with constant coefficients

$$\frac{\partial \tilde{w}}{\partial t}= r \tilde{w} - (r -v^2... ...u} -\frac{1}{2} v^2 \frac{\partial^2 \tilde{w}}{\partial u^2}.$$ (9)

Now we observe that even if $\tilde{w}(u,t)$ is independent of u, it still grows as e^rt from the $r\tilde{w}$ term. Factoring this out at the start will remove the $r\tilde{w}$ term. We normalize this behavior where the boundary condition is at t=t^* by writing the solution as

$$\tilde{w}(u,t) = e^{-r(t^*-t)} y(u,t).$$ (10)

Substituting this into Eq. 9 eliminates the $r\tilde{w}$ term giving

$$\frac{\partial y}{\partial t}= - (r -v^2/2) \frac{\partial y}{\partial u} -\frac{1}{2} v^2 \frac{\partial^2 y}{\partial u^2}.$$ (11)

We next scale towards a canonical form. First we scale u to get a common coefficient for the u derivatives, and then absorb that coefficient into a rescaling for t. The new variables are

$$u'=u \frac{(r-v^2/2)}{v^2/2},$$ (12)

and

$$t'= \frac{(r-v^2/2)^2}{v^2/2}(t^*-t).$$ (13)

With $\hat{y}(u',t')=y(u,t)$ the equation has become

$$\frac{\partial \hat{y}}{\partial t'}= \frac{\partial \hat{y}}{\partial u'} + \frac{\partial^2 \hat{y}}{\partial u'^2}.$$ (14)

Now, even with a constant gradient in u' there is an increase with time in $\hat{y}$. This is because the $\hat{y}$ is moving at constant velocity bringing a larger value to a fixed u' point. This term can then be eliminated by going to a comoving frame, or changing the u' spatial coordinate to z = u' + t' where the velocity is 1. Thus with

$$\tilde{y}(z,t')=\tilde{y}(u'+t',t')=\hat{y}(u,t)$$ (15)

we finally get the canonical form of the diffusion equation with unit diffusion coefficient

$$\frac{\partial \tilde{y}}{\partial t'}= \frac{\partial^2 \tilde{y}}{\partial z^2}.$$ (16)

The boundary conditions at t=t^* are now at t'=0 where z=u'. $x \geq c$ translates into $u \geq 0$. We now use the case where $(r-v^2/2) \geq 0$ so the condition on u translates into $u^\prime \geq 0$. The boundary conditions are then

 

$$\tilde{y}(z,0) x-c = c (e^u -1) \quad {\rm for} ~ z\geq 0,~ {\rm and}$$ = (17)

$$0 \quad {\rm for} ~ z\leq 0.$$ = (18)