One step in the derivation of Black-Scholes One step in the derivation of Black-Scholes
Assumptions:(1) ${\displaystyle \frac{\partial F}{\partial t}(t,x)+\frac{1}{2}\sigma^{2}x^{2}\frac{\partial^{2}F}{\partial x^{2}}(t,x)-rF(t,x)+rx\frac{\partial F}{\partial x}(t,x)=0}$
(2) $\tilde{F}(t,x)=e^{-rt}F(t,xe^{rt})$ .
Show that ${\displaystyle \frac{\partial\tilde{F}}{\partial t}(t,x)=-\frac{1}{2}\sigma^{2}x^{2}\frac{\partial^{2}\tilde{F}}{\partial x^{2}}(t,x)}$.
Although to derive Black-Scholes we might need PDE and
Ito's lemma, I think here we only need calculus. However I cannot
seem to get it right. I guess my problem has something to do with
the partial derivative of $F(t,xe^{rt})$. Thank you!
 A: First notice that (2) implies 
$$F(t,x) = e^{r t} \tilde F(t,x e^{-r t}).$$
Plug this directly into the Black-Scholes partial differential equation (1). 
You should find, for example, 
$$\frac{\partial F(t,x)}{\partial t} 
= r e^{r t} \tilde F(t,x e^{-r t})
+ e^{r t} \tilde F^{(1,0)}(t,x e^{-r t}) 
- r x \tilde F^{(0,1)} (t,x e^{-r t})$$
where $\tilde F^{(0,1)}(t,x e^{-r t})$ indicates a first derivative on the second argument, i.e., 
$$\tilde F^{(0,1)}(t,x e^{-r t}) 
= \frac{\partial \tilde F(t,X)}{\partial X}|_{X=x e^{-r t}}.$$
Notice that the partial derivative acts on the factor $e^{r t}$, the first argument of $\tilde F$, as well as its second argument.
If you can get this term, you should be able to work out the others.
Working out the rest of the terms we find Black-Scholes implies 
$$e^{r t} \tilde F^{(1,0)}(t,x e^{-r t}) 
= -\frac{1}{2} \sigma^2 x^2 e^{-r t} \tilde F^{(0,2)} (t,x e^{-r t}).$$
Change variables. 
Let $X = x e^{-r t}$. 
We immediately find
$$\frac{\partial\tilde F(t,X)}{\partial t}
= -\frac{1}{2} \sigma^2 X^2 \frac{\partial^2 \tilde F (t,X)}{\partial X^2},$$
which is the required equation. 
