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analytic solution of BlackScholes PDE
Here we present an analytical solution for the BlackScholes 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 $0<x<\infty,\>0\leq t\leq T$, 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 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:
$rf=re^{{rt}}\,u=re^{{rt}}\,u+e^{{rt}}\frac{\partial u}{\partial t}+rxe^{{rt}}% \frac{\partial u}{\partial x}+\frac{1}{2}\sigma^{2}x^{2}e^{{rt}}\frac{\partial% ^{2}u}{\partial x^{2}}\,;$ 
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:
$y=\log x\,,\quad s=Tt\,.$ 
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 BlackScholes formula itself), so that $\log x$ describes a Brownian motion, possibly with a drift. Then $\log x$ should satisfy some sort of diffusion equation (wellknown 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 distribution 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 transformation 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 $\infty<z<\infty$ 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
$f(t,x)=e^{{rt}}\,u\Bigl(Tt,\>\log x+(r\tfrac{1}{2}\sigma^{2})\tau\Bigr)\,.$ 
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 $f$:
$f(t,x)=\frac{e^{{r\tau}}}{\sqrt{2\pi\sigma^{2}\tau}}\int_{{\infty}}^{\infty}% \psi(e^{\zeta})\,\exp\Bigl(\frac{\bigl(\log x+(r\frac{1}{2}\sigma^{2})\tau% \zeta\bigr)^{2}}{2\sigma^{2}\tau}\Bigr)\,d\zeta\,,$ 
($\tau=Tt$) which agrees with the result derived using probabilistic methods.
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