analytic solution of BlackScholes PDE
Here we present an analytical solution for the BlackScholes partial differential equation,
$rf={\displaystyle \frac{\partial f}{\partial t}}+rx{\displaystyle \frac{\partial f}{\partial x}}+{\displaystyle \frac{1}{2}}{\sigma}^{2}{x}^{2}{\displaystyle \frac{{\partial}^{2}f}{\partial {x}^{2}}},f=f(t,x),$  (1) 
over the domain $$, with terminal condition $f(T,x)=\psi (x)$, by reducing this parabolic PDE to the heat equation of physics.
We begin by making the substitution:
$$u={e}^{rt}f,$$ 
which is motivated by the fact that it is the portfolio value discounted by the interest rate $r$ (see the derivation of the BlackScholes formula) that is a martingale^{}. Using the product rule^{} on $f={e}^{rt}u$, we derive the PDE that the function $u$ must satisfy:
$$rf=r{e}^{rt}u=r{e}^{rt}u+{e}^{rt}\frac{\partial u}{\partial t}+rx{e}^{rt}\frac{\partial u}{\partial x}+\frac{1}{2}{\sigma}^{2}{x}^{2}{e}^{rt}\frac{{\partial}^{2}u}{\partial {x}^{2}};$$ 
or simply,
$0={\displaystyle \frac{\partial u}{\partial t}}+rx{\displaystyle \frac{\partial u}{\partial x}}+{\displaystyle \frac{1}{2}}{\sigma}^{2}{x}^{2}{\displaystyle \frac{{\partial}^{2}u}{\partial {x}^{2}}}.$  (2) 
Next, we make the substitutions:
$$y=\mathrm{log}x,s=Tt.$$ 
These changes of variables can be motivated by observing that:

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The underlying process described by the variable $x$ is a geometric Brownian motion (as explained in the derivation of the BlackScholes formula itself), so that $\mathrm{log}x$ describes a Brownian motion^{}, possibly with a drift. Then $\mathrm{log}x$ should satisfy some sort of diffusion equation (wellknown in physics).

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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 $\partial u/\partial t$ is positive in equation (2). (Compare with the standard heat equation, $0=\partial u/\partial t+\partial u/\partial x$, which describes a temperature evolving forwards in time.) So to get to the heat equation, we have to use a substitution to reverse time.
Since
$$\frac{\partial u}{\partial s}=\frac{\partial u}{\partial t},\frac{\partial u}{\partial x}=\frac{\partial u}{\partial y}\frac{dy}{dx}=\frac{1}{x}\frac{\partial u}{\partial y},$$ 
and
$$\frac{{\partial}^{2}u}{\partial {x}^{2}}=\frac{\partial}{\partial x}\left(\frac{1}{x}\frac{\partial u}{\partial y}\right)=\frac{1}{{x}^{2}}\frac{\partial u}{\partial y}+\frac{1}{{x}^{2}}\frac{{\partial}^{2}u}{\partial {y}^{2}},$$ 
substituting in equation (2), we find:
$0={\displaystyle \frac{\partial u}{\partial s}}+(r\frac{1}{2}{\sigma}^{2}){\displaystyle \frac{\partial u}{\partial y}}+{\displaystyle \frac{1}{2}}{\sigma}^{2}{\displaystyle \frac{{\partial}^{2}u}{\partial {y}^{2}}}.$  (3) 
The first partial derivative^{} with respect to $y$ does not cancel (unless $r=\frac{1}{2}{\sigma}^{2}$) 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:
$$z=y+(r\frac{1}{2}{\sigma}^{2})\tau ,\tau =s.$$ 
Under the new coordinate system $(z,\tau )$, we have the relations amongst vector fields:
$$\frac{\partial}{\partial z}=\frac{\partial}{\partial y},\frac{\partial}{\partial \tau}=(r\frac{1}{2}{\sigma}^{2})\frac{\partial}{\partial y}+\frac{\partial}{\partial s},$$ 
leading to the following of equation (3):
$0={\displaystyle \frac{\partial u}{\partial \tau}}(r\frac{1}{2}{\sigma}^{2}){\displaystyle \frac{\partial u}{\partial z}}+(r\frac{1}{2}{\sigma}^{2}){\displaystyle \frac{\partial u}{\partial z}}+{\displaystyle \frac{1}{2}}{\sigma}^{2}{\displaystyle \frac{{\partial}^{2}u}{\partial {z}^{2}}};$ 
or:
$\frac{\partial u}{\partial \tau}}={\displaystyle \frac{1}{2}}{\sigma}^{2}{\displaystyle \frac{{\partial}^{2}u}{\partial {z}^{2}}},u=u(\tau ,z),$  (4) 
which is one form of the diffusion equation. The domain is on $$ and $0\le \tau \le T$; the initial condition is to be:
$u(0,z)={e}^{rT}\psi ({e}^{z}):={u}_{0}(z).$ 
The original function $f$ can be recovered by
$$f(t,x)={e}^{rt}u(Tt,\mathrm{log}x+(r\frac{1}{2}{\sigma}^{2})\tau ).$$ 
The fundamental solution of the PDE (4) is known to be:
$${G}_{\tau}(z)=\frac{1}{\sqrt{2\pi {\sigma}^{2}\tau}}\mathrm{exp}\left(\frac{z}{2{\sigma}^{2}\tau}\right)$$ 
(derived using the Fourier transform^{}); and the solution $u$ with initial condition ${u}_{0}$ is given by the convolution:
$$u(\tau ,z)={u}_{0}*{G}_{\tau}(z)=\frac{{e}^{rT}}{\sqrt{2\pi {\sigma}^{2}\tau}}{\int}_{\mathrm{\infty}}^{\mathrm{\infty}}\psi ({e}^{\zeta})\mathrm{exp}\left(\frac{{(z\zeta )}^{2}}{2{\sigma}^{2}\tau}\right)\mathit{d}\zeta .$$ 
In terms of the original function $f$:
$$f(t,x)=\frac{{e}^{r\tau}}{\sqrt{2\pi {\sigma}^{2}\tau}}{\int}_{\mathrm{\infty}}^{\mathrm{\infty}}\psi ({e}^{\zeta})\mathrm{exp}\left(\frac{{\left(\mathrm{log}x+(r\frac{1}{2}{\sigma}^{2})\tau \zeta \right)}^{2}}{2{\sigma}^{2}\tau}\right)\mathit{d}\zeta ,$$ 
($\tau =Tt$) which agrees with the result derived using probabilistic methods (http://planetmath.org/BlackScholesFormula).
Title  analytic solution of BlackScholes PDE 

Canonical name  AnalyticSolutionOfBlackScholesPDE 
Date of creation  20130322 16:31:34 
Last modified on  20130322 16:31:34 
Owner  stevecheng (10074) 
Last modified by  stevecheng (10074) 
Numerical id  6 
Author  stevecheng (10074) 
Entry type  Derivation 
Classification  msc 60H10 
Classification  msc 91B28 
Related topic  ExampleOfSolvingTheHeatEquation 
Related topic  BlackScholesPDE 
Related topic  BlackScholesFormula 