### Transformation from the Black-Scholes differential equation to the diffusion equation - and back

• vonjd

10 years ago

I know the derivation of the Black-Scholes differential equation and I understand (most of) the solution of the diffusion equation. What I am missing is the transformation from the Black-Scholes differential equation to the diffusion equation (with all the conditions) and back to the original problem.

All the transformations I have seen so far are not very clear or technically demanding (at least by my standards).

My question:
Could you provide me references for a very easily understood, step-by-step solution?

• 10 years ago

One starts with the Black-Scholes equation $$\frac{\partial C}{\partial t}+\frac{1}{2}\sigma^2S^2\frac{\partial^2 C}{\partial S^2}+ rS\frac{\partial C}{\partial S}-rC=0,\qquad\qquad\qquad\qquad\qquad(1)$$ supplemented with the terminal and boundary conditions (in the case of a European call) $$C(S,T)=\max(S-K,0),\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(2)$$ $$C(0,t)=0,\qquad C(S,t)\sim S\ \mbox{ as } S\to\infty.\qquad\qquad\qquad\qquad\qquad\qquad$$ The option value $C(S,t)$ is defined over the domain $0< S < \infty$, $0\leq t\leq T$.

Step 1. The equation can be rewritten in the equivalent form $$\frac{\partial C}{\partial t}+\frac{1}{2}\sigma^2\left(S\frac{\partial }{\partial S}\right)^2C+\left(r-\frac{1}{2}\sigma^2\right)S\frac{\partial C}{\partial S}-rC=0.$$ The change of independent variables $$S=e^y,\qquad t=T-\tau$$ results in $$S\frac{\partial }{\partial S}\to\frac{\partial}{\partial y},\qquad \frac{\partial}{\partial t}\to - \frac{\partial}{\partial \tau},$$ so one gets the constant coefficient equation $$\frac{\partial C}{\partial \tau}-\frac{1}{2}\sigma^2\frac{\partial^2 C}{\partial y^2}-\left(r-\frac{1}{2}\sigma^2\right)\frac{\partial C}{\partial y}+rC=0.\qquad\qquad\qquad(3)$$

Step 2. If we replace $C(y,\tau)$ in equation (3) with $u=e^{r\tau}C$, we will obtain that $$\frac{\partial u}{\partial \tau}-\frac{1}{2}\sigma^2\frac{\partial^2 u}{\partial y^2}-\left(r-\frac{1}{2}\sigma^2\right)\frac{\partial u}{\partial y}=0.$$

Step 3. Finally, the substitution $x=y+(r-\sigma^2/2)\tau$ allows us to eliminate the first order term and to reduce the preceding equation to the form $$\frac{\partial u}{\partial \tau}=\frac{1}{2}\sigma^2\frac{\partial^2 u}{\partial x^2}$$ which is the standard heat equation. The function $u(x,\tau)$ is defined for $-\infty < x < \infty$, $0\leq\tau\leq T$. The terminal condition (2) turns into the initial condition $$u(x,0)=u_0(x)=\max(e^{\frac{1}{2}(a+1)x}-e^{\frac{1}{2}(a-1)x},0),$$ where $a=2r/\sigma^2$. The solution of the heat equation is given by the well-known formula $$u(x,\tau)=\frac{1}{\sigma\sqrt{2\pi \tau}}\int_{-\infty}^{\infty} u_0(s)\exp\left(-\frac{(x-s)^2}{2\sigma^2 \tau}\right)ds.$$

Now, if we evaluate the integral with our specific function $u_0$ and return to the old variables $(x,\tau,u)\to(S,t,C)$, we will arrive at the usual Black–Merton-Scholes formula for the value of a European call. The details of the calculation can be found e.g. in The Mathematics of Financial Derivatives by Wilmott, Howison, and Dewynne (see Section 5.4). Sorry to resurrect an old question, but I'm having some difficulties doing the substitution in step 3. Can anyone give a little extra detail? I managed to understand, through the help of this page http://planetmath.org/AnalyticSolutionOfBlackScholesPDE. I might edit the answer above to elaborate later. for step 3, there are two standard ways to do it. One is as above, the other is multiplying by an exponential. I go into both in detail in my book "concepts" May I ask what happened to equation 1 if equation 2 is not an option but some other derivative, defined as: G(S_t,t)= s_b-S_t, S_t