Simplification of Differential Equation I am trying to understand the following simplification made in a text I am reading:
For:
\begin{align*}
dS_t
&=r S_t dt+\sqrt{V_t}S_tdW^S_t\\
dV_t
&=k(\theta - V_t)dt+\sigma \sqrt{V_t}dW_t^V\\
\end{align*}
$$
d\left<W^S,W^V \right>_t=\rho dt
$$
$$
\frac{\partial U}{\partial t}+\frac{1}{2}\frac{\partial^2 U}{\partial S^2} V S^2+\frac{\partial^2 U}{\partial S \partial V} V S \sigma \rho dt +  \frac{1}{2}\frac{\partial^2 U}{\partial V^2} V \sigma^2  -rU+rS\frac{\partial U}{\partial S}=-k(\theta - V ) {\frac{\partial U}{\partial V}}
$$
where U is the option price and V is the volatility.
Then, to simplify, we define:
$x=log(F_{t,T}/K)$ and I assume that this means  $=\log(\frac{Se^{r\tau}}{K})$
Where F is the time T forward price , K is the strike price.
We only consider the future value to expiration, C of the option price rather than its value today, and define $\tau=T-t$. Then the equation above simplifies to:
$$
-\frac{\partial C }{\partial \tau}+\frac{1}{2}VC_{11} -\frac{1}{2}VC_1 +\frac{1}{2}\sigma^2VC_{22}+\rho\sigma V C_{12}+k(\theta - V ) C_2=0
$$
Where subscript 1 = differentiate with respect to x, and 2 = differentiate with respect to V.
Could someone please explain how this simplification is possible? What I've done so far is substitute C for U in the first equation.
$$
\frac{\partial C}{\partial \tau}\frac{\partial \tau}{\partial t}+\frac{1}{2}\frac{\partial^2 C}{\partial S^2} V S^2+\frac{\partial^2 C}{\partial S \partial V} V S \sigma \rho dt +  \frac{1}{2}C_{22} V \sigma^2  -rU+rS\frac{\partial C}{\partial x}\frac{\partial x}{\partial S}=-k(\theta - V )C_2
$$
Where $\frac{\partial x}{\partial S}=\frac{1}{S}$. I am finding it impossible to cancel out the dt terms though.. any hints? Further, $\frac{\partial \tau}{\partial t}=-1$, so that explains the first expression..
 A: Discounted option price is a martingale under risk-neutral measure. Likewise, the future value of the option, 
C(t,T) = exp(r(T-t)) U(t,T)

is a martingale.
In the initial settings, U(t,T) is a function of S and V. But you can write it, and C(t,T), as a function of x and V, too. Then you will differentiate C(t,T), and the deterministic part, that is equal to 0 since C(t,T) is a martingale, should be exactly the equation you are trying to find.
In other words, you do not need to find the equation # 2 from the equation # 1, differentiating the future value as a function of x and V will yield the answer right away.
EDIT:
Since $dS_t =r S_t dt+\sqrt{V_t}S_tdW^S_t$, $$dx(t) = d\ log \frac{S_t \exp{r(T-t)}}{K} = \sqrt{V_t} dW_t^S - 0.5 V_t dt$$
Then, let us differentiate $C(t,T) = C\left(t,T,x(\cdot),V(\cdot)\right)$ (we use Ito lemma):
\begin{align*}
\frac{d}{dt} C(t,T)
&=\frac{\partial}{\partial t} C(t,T) dt + \frac{\partial}{\partial x} C(t,T) dx + \frac{\partial}{\partial V} C(t,T) dV \\
&+\frac{1}{2} \frac{\partial^2}{\partial x^2} C(t,T)<dx> + \frac{1}{2} \frac{\partial^2}{\partial V^2} C(t,T)<dV> +  \frac{\partial^2}{\partial x \ \partial V} C(t,T)<dx, dV>\\
&= \left(\frac{\partial C }{\partial t}+\frac{1}{2}VC_{11} -\frac{1}{2}VC_1 +\frac{1}{2}\sigma^2VC_{22}+\rho\sigma V C_{12}+k(\theta - V ) C_2\right) dt
+ \frac{\partial}{\partial x} C(t,T) \sqrt{V_t} dW_t^S + \frac{\partial}{\partial V} C(t,T) \sigma \sqrt{V_t}dW_t^V\\
\end{align*}
The term you multiply $dt$ by should be $0$ (due to no arbitrage constraint) and this is the equation that we need to obtain.
Let me know if anything is unclear.
