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PDE change of variables Black-Scholes equation

  1. Oct 11, 2013 #1
    1. The problem statement, all variables and given/known data
    By changing variables from [itex](S,t,V)[/itex] to [itex](x,\tau,u)[/itex] where
    [tex]\tau = T - t,[/tex]
    [tex]x = \ln\left(\frac{S}{K}\right) + \left(r - \frac{\sigma^2}{2}\right)(T-t),[/tex]
    [tex]u=e^{r\tau}V,[/tex]
    where [itex]r, \sigma, \tau, K[/itex] are constants, show that the Black-Scholes equation
    [tex]\frac{\partial V}{\partial t} + \frac{\sigma^2}{2}S^2 \frac{\partial^2 V}{\partial S^2} + rS\frac{\partial V}{\partial S} - rV = 0[/tex]
    reduces to the diffusion equation
    [tex]\frac{\partial u}{\partial \tau} - \frac{\sigma^2}{2}\frac{\partial^2 u}{\partial x^2}=0.[/tex]

    2. Relevant equations
    Chain rule.


    3. The attempt at a solution
    I know that [tex]\frac{\partial}{\partial t} = \frac{\partial x}{\partial t}\frac{\partial}{\partial x} + \frac{\partial \tau}{\partial t}\frac{\partial}{\partial \tau}[/tex] and similarly for [itex]\partial / \partial S[/itex]. I don't know what to do with the second-order derivative though. Since [itex]\partial / \partial S[/itex] turns out to be [itex]\frac{1}{S}\frac{\partial}{\partial x}[/itex], I reckoned that [tex]\frac{\partial^2 V}{\partial S^2} = \frac{1}{S}\frac{\partial}{\partial x} \left( \frac 1 S \frac{\partial V}{\partial x} \right),[/tex] but that seems to just make things more complicated.
     
  2. jcsd
  3. Oct 11, 2013 #2

    pasmith

    User Avatar
    Homework Helper

    Use
    [tex]
    \frac{\partial^2 V}{\partial x^2} = S\frac{\partial }{\partial S}\left(S\frac{\partial V}{\partial S}\right)
    = S\frac{\partial V}{\partial S} + S^2 \frac{\partial^2 V}{\partial S^2}[/tex]
    to eliminate [itex]S^2 \frac{\partial^2 V}{\partial S^2}[/itex] to leave you with terms in [itex]\partial V/\partial t[/itex], [itex]S\partial V/\partial S[/itex], [itex]V[/itex] and [itex]{\partial^2 V}/{\partial x^2}[/itex]. Then you can tidy up the first-order derivatives.
     
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