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Here we present an analytical solution for the Black-Scholes partial differential equation,
over the domain $0 < x < \infty, \: 0 \leq t \leq T$ , with terminal condition $f(T,x) = \psi(x)$ , by reducing this parabolic PDE to the heat equation of physics.
We begin by making the substitution:$$ u = e^{-rt} \, f\,,$$ which is motivated by the fact that it is the portfolio value discounted by the interest rate $r$ (see the derivation of the Black-Scholes formula) that is a martingale. Using the product rule on $f = e^{rt} \, u$ , we derive the PDE that the function $u$ must satisfy:$$ rf = re^{rt} \, u = re^{rt} \, u + e^{rt} \pd{u}{t} + rx e^{rt} \pd{u}{x} + \frac12 \sigma^2 x^2 e^{rt} \pdd{u}{x^2}\,;$$ or simply,
Next, we make the substitutions:$$ y = \log x\,, \quad s = T-t\,.$$ These changes of variables can be motivated by observing that:
- The underlying process described by the variable $x$ is a geometric Brownian motion (as explained in the derivation of the Black-Scholes formula itself), so that $\log x$ describes a Brownian motion, possibly with a drift. Then $\log x$ should satisfy some sort of diffusion equation (well-known in physics).
- The evolution of the system is backwards from the terminal state of the system. Indeed, the boundary condition is given as a terminal state, and the coefficient of $\ipd{u}{t}$ is positive in equation (
![[*]](http://images.planetmath.org:8080/cache/objects/8706/js//usr/share/latex2html/icons/crossref.png "[*]") ). (Compare with the standard heat equation, $0 = -\ipd{u}{t} + \ipd{u}{x}$ , which describes a temperature distribution evolving forwards in time.) So to get to the heat equation, we have to use a substitution to reverse time.
Since$$ \pd{u}{s} = -\pd{u}{t}\,, \quad \pd{u}{x} = \pd{u}{y} \, \od{y}{x} = \frac{1}{x} \, \pd{u}{y}\,,$$ and$$ \pdd{u}{x^2} = \pd{}{x} \left( \frac{1}{x} \pd{u}{y} \right) = -\frac{1}{x^2} \, \pd{u}{y} + \frac{1}{x^2} \pdd{u}{y^2}\,,$$ substituting in equation (), we find:
The first partial derivative with respect to $y$ does not cancel (unless $r = \tfrac12 \sigma^2$ ) because we have not take into account the drift of the Brownian motion. To cancel the drift (which is linear in time), we make the substitutions:$$ z = y + (r - \tfrac12 \sigma^2) \tau\,, \quad \tau = s\,.$$ Under the new coordinate system $(z,\tau)$ , we have the relations amongst vector fields:$$ \pd{}{z} = \pd{}{y}\,, \quad \pd{}{\tau} = -(r-\tfrac12 \sigma^2) \pd{}{y} + \pd{}{s}\,,$$ leading to the following transformation of equation (): \begin{align*} 0 = -\pd{u}{\tau} -(r-\tfrac12 \sigma^2) \pd{u}{z} + (r-\tfrac12 \sigma^2) \pd{u}{z} + \frac12 \sigma^2 \pdd{u}{z^2}\,; \end{align*}or:
which is one form of the diffusion equation. The domain is on $-\infty < z < \infty$ and $0 \leq \tau \leq T$ ; the initial condition is to be: \begin{align*} u(0, z) = e^{-rT} \, \psi(e^{z} ) := u_0(z)\,. \end{align*}The original function $f$ can be recovered by$$ f(t, x) = e^{rt} \, u\Bigl(T-t,\: \log x + (r-\tfrac12 \sigma^2) \tau \Bigr)\,.$$
The fundamental solution of the PDE () is known to be:$$ G_\tau(z) = \frac{1}{\sqrt{2\pi \sigma^2 \tau}} \exp\Bigl( -\frac{z}{2\sigma^2 \tau} \Bigr)$$ (derived using the Fourier transform); and the solution $u$ with initial condition $u_0$ is given by the convolution:$$ u(\tau, z) = u_0 * G_\tau(z) = \frac{e^{-rT}}{\sqrt{2\pi \sigma^2 \tau}} \, \int_{-\infty}^\infty \psi(e^\zeta) \, \exp \Bigl( - \frac{(z-\zeta)^2}{2\sigma^2 \tau} \Bigr) \, d\zeta\,.$$ In terms of the original function $f$ :$$ f(t, x) = \frac{e^{-r\tau}}{\sqrt{2\pi
\sigma^2 \tau}} \int_{-\infty}^\infty \psi (e^\zeta) \, \exp \Bigl( - \frac{\bigl(\log x + (r-\frac12 \sigma^2)\tau -\zeta\bigr)^2} {2\sigma^2 \tau} \Bigr) \, d\zeta\,,$$ ($\tau = T -t$ ) which agrees with the result derived using probabilistic methods.
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![[parent]](http://images.planetmath.org:8080/images/uparrow.png)
