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4.3.2 Solution of the normalized diffusion equation $ \spadesuit$

[ SLIDE Fourier-Laplace fwd - bck || Solution binary options || same VIDEO as previous section: modem - LAN - DSL]


Readers interested in solving the diffusion equation (4.3.1#eq.5) analytically are likely to be familiar with the Fourier-Laplace transform that is commonly used to solve initial and boundary value problems. Others may skip the whole derivation and verify that the final result (4.3.2#eq.11) indeed satisfies the Black-Scholes equation (3.4#eq.4).
Start by transforming (4.3.1#eq.5) with a Laplace transform in time

$\displaystyle \int_0^\infty d\tau \exp(i\omega \tau)\left[ \pd{u}{\tau} -\pd{^2 u}{x^2} \right] = 0, \qquad \Im m(\omega)>0$ (4.3.2#eq.1)


Note that the condition $ \Im m(\omega)>0$ is important to ensure causality. Integrate the first term by parts and substitute a Dirac function $ \delta(x-\xi)$ for the initial condition $ u(x,0)$

\begin{displaymath}\begin{split}\left. u\exp(i\omega \tau) \right\vert _0^\infty... ... u(x,\omega) -\pd{^2 u(x,\omega)}{x^2} \right] &= 0 \end{split}\end{displaymath} (4.3.2#eq.2)


The notation $ u(x,\omega)$ refers to the Laplace transform of $ u(x,\tau)$ . Spatial derivatives are dealt with a Fourier transform

\begin{displaymath}\begin{split}\int_{-\infty}^\infty dx \exp(-ikx)\left[ -\delt... ...(-ik\xi) +i\omega u(k,\omega) -k^2 u(k,\omega) &= 0 \end{split}\end{displaymath} (4.3.2#eq.3)


and yields an explicit solution in Fourier-Laplace space

$\displaystyle u(k,\omega) = \frac{-i\exp(-ik\xi)}{\omega +ik^2}$ (4.3.2#eq.4)


The pole in the complex plane for $ \omega=-ik^2$ needs to be taken into account when inverting the Laplace transform in a causal manner

\begin{displaymath}\begin{split}u(k,\tau) &=\int_{-\infty+iC}^{+\infty+iC} \frac... ...} 2\pi i\exp(-k^2\tau) =\exp(-ik\xi) \exp(-k^2\tau) \end{split}\end{displaymath} (4.3.2#eq.5)


where the residue theorem has been used to evaluate the complex integral, closing the contour in the lower half plane where the phase factor $ \exp(-i\omega\tau)$ decays exponentially. Invert the Fourier transformation

\begin{displaymath}\begin{split}u(x,\tau) &= \int_{-\infty}^\infty \frac{dk}{2\p... ...t_{-\infty}^\infty dk \exp ik(x-\xi) \exp(-k^2\tau) \end{split}\end{displaymath} (4.3.2#eq.6)


and use the formula (3.323.2) from Gradshteyn & Ryzhik [9]

$\displaystyle \int_{-\infty}^{\infty} dx \exp(-p^2x^2)\exp(\pm qx) = \frac{\sqrt{\pi}}{p}\exp\left[\frac{q^2}{4p^2}\right] \qquad p>0$ (4.3.2#eq.7)


here with $ p=\sqrt{\tau}$ and $ q=i(x-\xi)$ to write down a solution of the diffusion equation

$\displaystyle u(x,\tau)=\frac{1}{2\sqrt{\pi \tau}}\exp\left[-\frac{(x-\xi)^2}{4\tau}\right]$ (4.3.2#eq.8)


This shows that a Dirac function $ \delta(x-\xi)$ assumed as initial condition in (4.3.2#eq.2) spreads out into a Gaussian as time evolves forward. A superposition of a whole series of Dirac functions can now be used to decompose any arbitrary initial condition $ u_0(\xi)$

$\displaystyle u(x,0) =\int_{-\infty}^\infty u_0(\xi)\delta(\xi-x)d\xi =\int_{-\infty}^\infty u_0(\xi)\delta(x-\xi)d\xi$ (4.3.2#eq.9)


and after evolving each Dirac functions separately using (4.3.2#eq.8), can again be superposed at a later time when $ \tau\ge 0$ :

$\displaystyle u(x,\tau)=\frac{1}{2\sqrt{\pi \tau}}\int_{-\infty}^\infty d\xi u_0(\xi)\exp\left[-\frac{(\xi-x)^2}{4\tau}\right]$ (4.3.2#eq.10)


Transforming back into financial variables (4.3.1#eq.1), some algebra finally yields a general formula for the price of a binary option with a terminal payoff $ \Lambda(S)$

$\displaystyle V(S,t)=\frac{\exp[-r(T-t)]}{\sigma\sqrt{2\pi(T-t)}} \int_0^\infty... ...) -(r-D-\frac{\sigma^2}{2})(T-t)\right)^2} {2\sigma^2(T-t)} \right]\frac{dX}{X}$ (4.3.2#eq.11)


SYLLABUS  Previous: 4.3.1 Transformation to log-normal  Up: 4.3 Methods for European  Next: 4.3.3 Black-Scholes formula