# analytic solution of Black-Scholes PDE

Here we present an analytical solution for the Black-Scholes partial differential equation,

 $\displaystyle rf=\frac{\partial f}{\partial t}+rx\,\frac{\partial f}{\partial x% }+\frac{1}{2}\sigma^{2}x^{2}\,\frac{\partial^{2}f}{\partial x^{2}}\,,\quad f=f% (t,x)\,,$ (1)

over the domain $00\leq t\leq T$, 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:

 $rf=re^{rt}\,u=re^{rt}\,u+e^{rt}\frac{\partial u}{\partial t}+rxe^{rt}\frac{% \partial u}{\partial x}+\frac{1}{2}\sigma^{2}x^{2}e^{rt}\frac{\partial^{2}u}{% \partial x^{2}}\,;$

or simply,

 $\displaystyle 0=\frac{\partial u}{\partial t}+rx\frac{\partial u}{\partial x}+% \frac{1}{2}\sigma^{2}x^{2}\frac{\partial^{2}u}{\partial x^{2}}\,.$ (2)

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 $\partial u/\partial t$ 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

 $\frac{\partial u}{\partial s}=-\frac{\partial u}{\partial t}\,,\quad\frac{% \partial u}{\partial x}=\frac{\partial u}{\partial y}\,\frac{dy}{dx}=\frac{1}{% x}\,\frac{\partial u}{\partial y}\,,$

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:

 $\displaystyle 0=-\frac{\partial u}{\partial s}+(r-\tfrac{1}{2}\sigma^{2})\frac% {\partial u}{\partial y}+\frac{1}{2}\sigma^{2}\,\frac{\partial^{2}u}{\partial y% ^{2}}\,.$ (3)

The first partial derivative with respect to $y$ does not cancel (unless $r=\tfrac{1}{2}\sigma^{2}$) 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:

 $z=y+(r-\tfrac{1}{2}\sigma^{2})\tau\,,\quad\tau=s\,.$

Under the new coordinate system $(z,\tau)$, we have the relations amongst vector fields:

 $\frac{\partial}{\partial z}=\frac{\partial}{\partial y}\,,\quad\frac{\partial}% {\partial\tau}=-(r-\tfrac{1}{2}\sigma^{2})\frac{\partial}{\partial y}+\frac{% \partial}{\partial s}\,,$

leading to the following of equation (3):

 $\displaystyle 0=-\frac{\partial u}{\partial\tau}-(r-\tfrac{1}{2}\sigma^{2})% \frac{\partial u}{\partial z}+(r-\tfrac{1}{2}\sigma^{2})\frac{\partial u}{% \partial z}+\frac{1}{2}\sigma^{2}\frac{\partial^{2}u}{\partial z^{2}}\,;$

or:

 $\displaystyle\frac{\partial u}{\partial\tau}=\frac{1}{2}\sigma^{2}\frac{% \partial^{2}u}{\partial z^{2}}\,,\quad u=u(\tau,z)\,,$ (4)

which is one form of the diffusion equation. The domain is on $-\infty and $0\leq\tau\leq T$; the initial condition is to be:

 $\displaystyle u(0,z)=e^{-rT}\,\psi(e^{z}):=u_{0}(z)\,.$

The original function $f$ can be recovered by

 $f(t,x)=e^{rt}\,u\Bigl{(}T-t,\>\log x+(r-\tfrac{1}{2}\sigma^{2})\tau\Bigr{)}\,.$

The fundamental solution of the PDE (4) 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 $f$:

 $f(t,x)=\frac{e^{-r\tau}}{\sqrt{2\pi\sigma^{2}\tau}}\int_{-\infty}^{\infty}\psi% (e^{\zeta})\,\exp\Bigl{(}-\frac{\bigl{(}\log x+(r-\frac{1}{2}\sigma^{2})\tau-% \zeta\bigr{)}^{2}}{2\sigma^{2}\tau}\Bigr{)}\,d\zeta\,,$

($\tau=T-t$) which agrees with the result derived using probabilistic methods (http://planetmath.org/BlackScholesFormula).

Title analytic solution of Black-Scholes PDE AnalyticSolutionOfBlackScholesPDE 2013-03-22 16:31:34 2013-03-22 16:31:34 stevecheng (10074) stevecheng (10074) 6 stevecheng (10074) Derivation msc 60H10 msc 91B28 ExampleOfSolvingTheHeatEquation BlackScholesPDE BlackScholesFormula