I have been having some issues with proving a result seen in Jim Gatheral's book The Volatility Surfrace. Specifically, in chapter 2 section "The Solution for European Options" he states the following:
$-\frac{\partial C}{\partial \tau} + \frac{1}{2} v C_{11} - \frac{1}{2} \eta^2 v C_{22} + \rho \eta v C_{12} - \lambda (v - \bar{v})C_{2} = 0 \quad (2.4)$
Where:
$\tau = T-t$
$x := log(F_{t,T}/K)$
$F_{t,T} = S_{0}e^{\int_t^T \mu_t dt}$
And the subscripts 1 and 2 represent differentiation with respect to $x$ and $v$ respectively. $\mu_t$ represents the risk-neutral drift of the stock price process.
We are then told that the solution to (2.4) has the form:
$C(x,v,\tau) = K [e^{x}P_{1}(x,v,\tau) - P_{0}(x,v,\tau)] \quad (2.5)$
Finally we are told that substitution of (2.5) into (2.4) gives:
$-\frac{\partial P_{j}}{\partial \tau} + \frac{1}{2} v \frac{\partial^2 P_{j}}{\partial x^2} - \big (\frac{1}{2} - j) v \frac{\partial P_{j}}{\partial x} + \frac{1}{2} \eta^2 v \frac{\partial^2 P_{j}}{\partial v^2} + \rho \eta v \frac{\partial^2 P_{j}}{\partial x \partial v} + (a-b_j v)\frac{\partial P_{j}}{\partial v} = 0 \quad (2.6)$
Where:
$ a = \lambda \bar{v} $
$ b_{j} = \lambda - j \rho \eta $
It seems like in order to substitute (2.5) into (2.4) to achieve (2.6) we simply need to take the required derivatives of (2.5). However, when I do so I am unable to achieve the answer seen in (2.6).
I was wondering if anyone had some working to show how the result in (2.6) can be achieved?
Are you missing the term $-\frac{1}{2} v C_1$ in $(2.4)$? The correct equations: $$ \tag{2.7} - \frac{\partial C}{\partial \tau} + \frac{1}{2} v C_{11} - \frac{1}{2} v C_1 + \frac{1}{2} \eta^2 v C_{22} + \rho \eta v C_{12} - \lambda (v - \bar{v})C_2 = 0 $$
By the chain rule, we obtain the followings:
\begin{align} \frac{\partial C}{\partial \tau} &= Ke^x\frac{\partial P_1}{\partial \tau} - K\frac{\partial P_0}{\partial \tau}, \\ C_1 &= Ke^xP_1 + Ke^x\frac{\partial P_1}{\partial x} - K \frac{\partial P_0}{\partial x}, \\ C_{11} &= \frac{\partial C_1}{\partial x} = Ke^xP_1 + 2Ke^x\frac{\partial P_1}{\partial x} + Ke^x\frac{\partial^2 P_1}{\partial x^2} - K\frac{\partial^2 P_0}{\partial x^2}, \\ C_2 &= Ke^x\frac{\partial P_1}{\partial v} - K\frac{\partial P_0}{\partial v}, \\ C_{22} &= \frac{\partial C_2}{\partial v} = Ke^x\frac{\partial^2 P_1}{\partial v^2} - K\frac{\partial^2 P_0}{\partial v^2}, \\ C_{12} &= \frac{\partial C_1}{\partial v} = Ke^x\frac{\partial P_1}{\partial v} + Ke^x\frac{\partial^2 P_1}{\partial x \partial v}- K\frac{\partial^2 P_0}{\partial x \partial v}. \end{align}
By dividing these equations into the cases $j=0$ and $j=1$ and substituting them into $(2.7)$, we obtain the two equations (2.6).