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PDE and finding a general solution

  1. Apr 10, 2012 #1
    Hi everyone,

    I am doing a sheet on Asian Options and The Black Scholes equation.

    I have the PDE,

    [itex]\frac{∂v}{∂τ}[/itex]=[itex]\frac{1}{2}[/itex][itex]σ^{2}[/itex][itex]\frac{X^{2}}{S^{2}}[/itex][itex]\frac{∂^{2}v}{∂ε^{2}}[/itex] + ([itex]\frac{1}{T}[/itex] + (r-D)X)[itex]\frac{∂v}{∂ε}[/itex]

    I have to seek a solultion of the form v=[itex]α_{1}[/itex](τ)ε + [itex]α_{0}[/itex](τ) and determine the general solution for [itex]α_{1}[/itex](τ) and [itex]α_{0}[/itex](τ).

    We are given that ε=[itex]\frac{I}{TS}[/itex] - [itex]\frac{X}{S}[/itex], τ=T-t and V(S, I, t)=[itex]e^{-Dτ}[/itex]Sv(ε, τ)

    Can anybody help me with this problem?
    Thanks
     
    Last edited: Apr 10, 2012
  2. jcsd
  3. Apr 10, 2012 #2
    Hi meghibbert17 !

    several notations are not clear enough. For exemple three different typographies for "v".
    What exactly is the list of variables and the list of constants ?
    In the first equation, are you sure that dv/dr = 0 ?
     
  4. Apr 10, 2012 #3
    Hello,

    Sorry, I am new to this and it does look rather messy!

    in the first equation it is not dv/dr but dv/d(tau).

    All the v's in the equation are the same as the v=[itex]α_{1}[/itex](τ)ε + [itex]α_{0}[/itex](τ) which we are seeking a solution for and then V(S, I, t) = e−DτSv(ε, τ).

    Incase it is also not clear, its tau = T-t
    Is that any clearer? Thankyou
     
  5. Apr 10, 2012 #4
    Sorry, I cannot understand what are the constants and what are the variables and the functions.
    Moreover, I see that dV/d(tau)=0 in the first equation. And V is a function of (tau) in the given relationship V(S,I,t)=exp(-D*tau)*S*V(epsilon,tau). This is in contradiction.
    All this is too messy for me. I hope that someone else could help you.
     
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