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This is taken from Gatheral's book "The Volatility Surface", where he tries to go from equation 2.3 to equation 2.4.

We have the following PDE,

$$ \frac{\partial V}{\partial t}+\frac{1}{2}vS^2\frac{\partial ^2 V}{\partial S^2} + \rho\eta vS\frac{\partial ^2 V}{\partial S\partial v}+\frac{1}{2}\eta^2v\frac{\partial ^2 V}{\partial v^2} + rS\frac{\partial V}{\partial S}-rV-\lambda(v-\bar{v})\frac{\partial V}{\partial v} = 0 $$

By using a change of variables,

$$ x=\ln{\frac{Se^{r\tau}}{K}}, \tau = T-t $$

show that it reduces to

$$ -\frac{\partial C}{\partial \tau}+\frac{1}{2}v\frac{\partial ^2 C}{\partial x^2} -\frac{1}{2}v\frac{\partial C}{\partial x} +\frac{1}{2}\eta^2v\frac{\partial ^2 C}{\partial v^2} + \rho\eta v\frac{\partial ^2 C}{\partial x\partial v} - \lambda(v-\bar{v})\frac{\partial V}{\partial v} = 0 $$

My working are as follows...

We have $C(x,v,\tau)=V(Ke^{x-r\tau}, v, T-\tau)$. The partial derivatives are,

$$ \begin{aligned} \frac{\partial V}{\partial t} &= \frac{\partial V}{\partial S}\frac{\partial S}{\partial t} + \frac{\partial V}{\partial t}\\ &=\frac{\partial V}{\partial x}\frac{\partial x}{\partial S}\frac{\partial S}{\partial \tau}\frac{\partial \tau}{\partial t}+\frac{\partial V}{\partial \tau}\frac{\partial \tau}{\partial t}\\ &=\frac{\partial V}{\partial x}\frac{1}{S}(-rS)(-1)-\frac{\partial V}{\partial \tau}\\ &= r\frac{\partial V}{\partial x}-\frac{\partial V}{\partial \tau}\\ \frac{\partial V}{\partial S} &= \frac{\partial V}{\partial x}\frac{\partial x}{\partial S}=\frac{1}{S}\frac{\partial V}{\partial x} \\ \frac{\partial^2 V}{\partial S^2}&= \frac{1}{S}\frac{\partial }{\partial x}\frac{\partial x}{\partial S}\frac{\partial V}{\partial x}-\frac{1}{S^2}\frac{\partial V}{\partial x} \\ &=\frac{1}{S^2}\left(\frac{\partial^2 V}{\partial x^2}-\frac{\partial V}{\partial x} \right)\\ \frac{\partial V}{\partial S\partial v} &= \frac{\partial }{\partial S}\frac{\partial V}{\partial v}=\frac{\partial }{\partial x}\frac{\partial x}{\partial S}\frac{\partial V}{\partial v} = \frac{1}{S}\frac{\partial V}{\partial x\partial v} \end{aligned} $$

Substituting into the original PDE, I unfortunately get,

$$ r\frac{\partial C}{\partial x}-\frac{\partial C}{\partial \tau}+\frac{1}{2}v\frac{\partial ^2 C}{\partial x^2} -\frac{1}{2}v\frac{\partial C}{\partial x} +\frac{1}{2}\eta^2v\frac{\partial ^2 C}{\partial v^2} + \rho\eta v\frac{\partial ^2 C}{\partial x\partial v} +r\frac{\partial C}{\partial x}-rC- \lambda(v-\bar{v})\frac{\partial V}{\partial v} = 0 $$

I'm not sure how to get rid of the $r\frac{\partial C}{\partial x}$ and $rC$ terms.

I think perhaps I left of crucial steps involving his remark that "Further, suppose that we consider only the future value to expiration C of the European option price rather than its value today and define $\tau = T − t$."

Can someone help? Thanks!

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First note that you have a typo in the definition of the moneyness. It should be

\begin{equation} x = \ln \left( F_{t, T} / K \right) = \ln \left( S e^{r \tau} / K \right). \end{equation}

Following the remark that you cited, we then define

\begin{equation} e^{-r \tau} C(x, \nu, \tau) = V(S, \nu, t). \end{equation}

Note that $C$ is a function of $\tau$ and not $t$ - this seems strange in your notation. The corresponding partial derivatives are given by

\begin{eqnarray} \frac{\partial V}{\partial t} & = & e^{-r \tau} \left( r C - r \frac{\partial C}{\partial x} - \frac{\partial C}{\partial \tau} \right), \\ \frac{\partial V}{\partial S} & = & e^{-r \tau} \frac{1}{S} \frac{\partial C}{\partial x}, \\ \frac{\partial^2 V}{\partial S^2} & = & e^{-r \tau} \frac{1}{S^2} \left( \frac{\partial^2 C}{\partial x^2} - \frac{\partial C}{\partial x} \right). \end{eqnarray}

Substituting back yields Equation (2.4) in Gatheral's book. I would also recommend you clearly distinguish between $C$ and $V$ and don't have $V$ on both sides of your partial derivatives.

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  • $\begingroup$ thanks for pointing out my several typos, I have fixed them up. $\endgroup$ – Danny Jun 17 '17 at 14:00
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    $\begingroup$ oh so what the paragraph is saying is that $C$ is defined to be the undiscounted price, so we have to add the discounting term back before we are able to equate that to $V$. thank you very much! $\endgroup$ – Danny Jun 17 '17 at 14:05

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