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

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?

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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).

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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? –  Mattheus Jan 15 '14 at 18:57
I managed to understand, through the help of this page planetmath.org/AnalyticSolutionOfBlackScholesPDE. I might edit the answer above to elaborate later. –  Mattheus Jan 15 '14 at 20:17
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" –  Mark Joshi May 22 at 7:01

I'd like to give an alternative derivation not involving the clever (mystifying?) transformation to the heat equation and thus present a more general technique for solving constant coefficeint advection-diffusion PDEs. All we need is the Fourier transform: \begin{align*} \mathcal{F}[f] & = \int_{-\infty}^\infty e^{-i \omega y} f(y) dy, \end{align*} where $\mathcal{F}[f] = \mathcal{F}[f](\omega)$.

We'll use the following well-known facts: \begin{align*} \mathcal{F}\left[ \frac{1}{s \sqrt{2\pi}} \exp\left( -\frac{1}{2}\left(\frac{y - m}{s}\right)^2\right) \right] & = \exp(-i \omega m - s^2\omega^2/2), \\ \mathcal{F}\left[\frac{ \partial^n f}{\partial y^n}\right] & = (i \omega)^n \mathcal{F}[f], \\ \mathcal{F}[cf] & = c \mathcal{F}[f], \\ \mathcal{F}[f \ast g] & = \mathcal{F}[f]\mathcal{F}[g], \end{align*} where the convolution $(f \ast g)(y) = \int_{-\infty}^\infty f(z)g(y-z)dz$.

Using @olaker's notation, start with the constant coefficient PDE ((3) above) $$\frac{\partial C}{\partial \tau} = \frac{\sigma^2}{2}\frac{\partial^2 C}{\partial y^2} + \left(r-\frac{\sigma^2}{2}\right)\frac{\partial C}{\partial y} - rC.$$ Step 1. Take the Fourier transform of each term term above and solve the resulting (very simple) ODE: \begin{align*} \frac{\partial \hat{C}}{\partial \tau} & = -\frac{\sigma^2 \omega^2}{2}\hat{C} + i \omega\left(r-\frac{\sigma^2}{2}\right)\hat{C} - r\hat{C}, \\ \hat{C} & = \hat{C}_0 e^{-r\tau} \exp \left(-\frac{\sigma^2 \omega^2}{2}\tau + i \omega\left(r-\frac{\sigma^2}{2}\right)\tau\right). \end{align*}

Step 2. Letting $m = \left(\frac{\sigma^2}{2} - r\right)\tau$ and $s = \sigma \sqrt{\tau}$ from the Fourier transform notation, note $$\exp \left(-\frac{\sigma^2 \omega^2}{2}\tau + i \omega\left(r-\frac{\sigma^2}{2}\right)\tau\right) = \mathcal{F}\left[ \frac{1}{\sigma \sqrt{2\pi\tau}} \exp\left( -\frac{1}{2}\left(\frac{y - \left(\frac{\sigma^2}{2} - r\right)\tau}{\sigma \sqrt{\tau}}\right)^2\right) \right],$$ so \begin{align*} \hat{C} & = \hat{C}_0 e^{-r\tau}\mathcal{F}\left[ \frac{1}{\sigma \sqrt{2\pi\tau}} \exp\left( -\frac{1}{2}\left(\frac{y - \left(\frac{\sigma^2}{2} - r\right)\tau}{\sigma \sqrt{\tau}}\right)^2\right) \right] \\ & = \frac{1}{\sigma \sqrt{2\pi\tau}} e^{-r\tau}\mathcal{F}\left[C_0 \ast \exp\left( -\frac{1}{2}\left(\frac{y - \left(\frac{\sigma^2}{2} - r\right)\tau}{\sigma \sqrt{\tau}}\right)^2\right) \right]. \end{align*}

Step 3. Take inverse transform: \begin{align*} C(y, \tau) & = \frac{1}{\sigma \sqrt{2\pi\tau}} e^{-r\tau} \int_{-\infty}^\infty C_0(z) \exp\left( -\frac{1}{2}\left(\frac{y - z - \left(\frac{\sigma^2}{2} - r\right)\tau}{\sigma \sqrt{\tau}}\right)^2\right) dz \\ C(y, \tau) & = \frac{1}{\sigma \sqrt{2\pi\tau}} e^{-r\tau} \int_{-\infty}^\infty C_0(z) \exp\left( -\frac{1}{2}\left(\frac{z - \left(y + \left(r - \frac{\sigma^2}{2}\right)\tau\right)}{\sigma \sqrt{\tau}}\right)^2\right) dz \end{align*}

Step 4. Finally, change variables back $y \to S$, where we had $y = \log S$. Before we do this, note $S$ is really the initial'' stock price in the usual sense, i.e. $S = S_0$, but to be consistent we'll stick with $S$ as the initial (known) stock price. We'll also transform the $z$ variable, suggestively calling it $S_T$ by $S_T = e^z$. \begin{align*} C(S, \tau) & = \frac{1}{\sigma \sqrt{2\pi\tau}} e^{-r\tau} \int_0^\infty C_0(S_T) \frac{1}{S_T} \exp\left( -\frac{1}{2}\left(\frac{\log S_T - \left(\log S + \left(r - \frac{\sigma^2}{2}\right)\tau\right)}{\sigma \sqrt{\tau}}\right)^2\right) dS_T. \end{align*} Just note \begin{align*} f(S_T) := \frac{1}{S_T \sigma \sqrt{2\pi\tau}} \exp\left( -\frac{1}{2}\left(\frac{\log S_T - \left(\log S + \left(r - \frac{\sigma^2}{2}\right)\tau\right)}{\sigma \sqrt{\tau}}\right)^2\right) \end{align*} is the probability density function for a log $\mathcal{N}\left(\log S + \left(r - \frac{\sigma^2}{2}\right)\tau, \sigma^2 \tau\right)$ random variable, and under Black-Scholes, this is indeed the distribution of $S_T$ under $\mathbb{Q}$. Hence \begin{align*} C(S,\tau) = e^{-r\tau}\mathbb{E}_{\mathbb{Q}}[C_0(S_T)|\mathcal{F}_t]. \end{align*}

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