3
$\begingroup$

In the derivation below, I cannot figure out how to solve for "Step 3". Can anyone help me walk through the steps in detail?

Derivation:

enter image description here

$\endgroup$
  • 3
    $\begingroup$ To convert the PDE for $u(y,\tau)$ (Step 2) in that for $u^*(x,\tau)$ (Step 3), write the change of variable $u(y,\tau) = \tilde{u}(x(y,\tau),\tau)$ and apply the chain rule to express the partial derivatives ([Hint] $ \frac{\partial u}{\partial \tau} = \frac{\partial \tilde{u}}{\partial x} \frac{\partial x}{\partial \tau} + \frac{\partial \tilde{u}}{\partial t}$). This is more of maths question than a quantitative finance one. I am therefore voting to close. $\endgroup$ – Quantuple Feb 15 '18 at 9:09
  • $\begingroup$ Please don't close this - I still think it's relevant to quant finance because it's using Black Scholes, and a common change of variables for Black Scholes. This is also a change of variables done in Paul Wilmott's book for quant finance $\endgroup$ – Zak Fischer Feb 15 '18 at 12:20
  • $\begingroup$ Can you elaborate on your hint? I understand how to calculate $\frac{dx}{d \tau}$ from your hint. But besides that I'm not sure how to leverage it. And in particular how does this lead to step three which has $\frac{\partial^2 u}{\partial x^2}$? I really appreciate any help anyone can provide. I've looked up this problem a lot online and many people say "it can be done" but I have not found anywhere where they actually show their work and steps so I'm stumped! :( $\endgroup$ – Zak Fischer Feb 15 '18 at 12:22
  • 1
    $\begingroup$ Use the hint to express $\partial u/\partial \tau$ as a function of the derivatives of $u^*$ (sorry I see that I've used both $u^*$ and $\tilde{u}$ notations, these are one and the same function). Now apply the same reasoning for the other partial derivatives i.e. $\partial u/\partial y$ and $\partial^2 u/\partial y^2$ (Hint: $\partial x/\partial y =1$). Plug that in the equation obtained at Step 2 and note that some terms cancel out producing the desired equation. $\endgroup$ – Quantuple Feb 15 '18 at 13:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.