I want to obtain the PDE for the Black-76 model. I believe it has to be the following PDE:
$$\left(\frac{\partial V}{\partial t}+\frac{1}{2}\sigma^{2}F^{2}\frac{\partial^{2} V}{\partial F^{2}}\right)dt-rV=0.$$
I know that the PDE of Black-Scholes model is given by: $$\left(\frac{\partial V}{\partial t}+\frac{1}{2}\sigma^{2}S^{2}\frac{\partial^{2} V}{\partial S^{2}}+rS\frac{\partial V}{\partial S}\right)dt-rV=0.$$
Here you start with geometric Brownian motion process, $dS_{t}=\mu S_{t}dt+\sigma S_{t}dW_{t}$. Further you consider a trading strategy under which one holds one option and continuously trades in the stock in order to hold some $\Delta$ shares.
What I have done so far:
Now I start with the process: $dF=\sigma F_{t} dW_{t}$. By Ito's lemma I have
\begin{align} dV &= \frac{\partial V}{\partial F}dF+\frac{\partial V}{\partial t}dt+\frac{1}{2}\frac{\partial^{2} V}{\partial F^{2}}dF^{2} \\ &= \frac{\partial V}{\partial F}\sigma F dW+\frac{\partial V}{\partial t}dt+\frac{1}{2}\frac{\partial^{2} V}{\partial F^{2}}(\sigma F dW)^{2} \\ &= \frac{\partial V}{\partial F}\sigma F dW+\frac{\partial V}{\partial t}dt+\frac{1}{2}\frac{\partial^{2} V}{\partial F^{2}}\sigma^{2} F^{2} dt \\ &= \left(\frac{\partial V}{\partial t}+\frac{1}{2}\frac{\partial^{2} V}{\partial F^{2}}\sigma^{2} F^{2}\right)dt+\frac{\partial V}{\partial F}\sigma F dW. \end{align}
Now I make use of the same hedging strategy as in the Black Scholes case so I have: $$\Pi=V-\Delta F \Rightarrow d\Pi=dV-\Delta dF.$$
So I have: \begin{align} d\Pi &= dV-\Delta dF \\ &= \left(\frac{\partial V}{\partial t}+\frac{1}{2}\frac{\partial^{2} V}{\partial F^{2}}\sigma^{2} F^{2}\right)dt+\frac{\partial V}{\partial F}\sigma F dW-\Delta dF \\ &= \left(\frac{\partial V}{\partial t}+\frac{1}{2}\frac{\partial^{2} V}{\partial F^{2}}\sigma^{2} F^{2}\right)dt+\frac{\partial V}{\partial F}\sigma F dW-\Delta (\sigma F dW) \end{align} Thus: $$\Delta=\frac{\partial V}{\partial F}$$ Also $d\Pi=r\Pi dt$, so:
$$r\Pi dt = \left(\frac{\partial V}{\partial t}+\frac{1}{2}\frac{\partial^{2} V}{\partial F^{2}}\sigma^{2} F^{2}\right)dt.$$
Integrating gives:
$$\frac{\partial V}{\partial t}+\frac{1}{2}\frac{\partial^{2} V}{\partial F^{2}}\sigma^{2} F^{2}-r\Pi = 0$$
Now $$\Pi=V-\Delta F=V-\frac{\partial V}{\partial F}F,$$ so we have
$$\frac{\partial V}{\partial t}+\frac{1}{2}\frac{\partial^{2} V}{\partial F^{2}}\sigma^{2} F^{2}-rV+\frac{\partial V}{\partial F}rF = 0$$
This is the Black-Scholes PDE not Black PDE. What am I doing wrong? How do I get the correct PDE?