# Similarity reductions in Black Scholes PDE

Suppose that $$V(S,I,t)$$ satisfies the equality

$$\frac{\partial V}{\partial t}+\frac{1}{2}\sigma^2S^2 \frac{\partial^2V}{\partial S^2}+S\frac{\partial V}{\partial I}+rS\frac{\partial V}{\partial S}-rV=0$$

Here $$I=\int _0^t S dt$$.

Now, let $$R=S/I$$ and $$V(S,R,t)=I\,W(R,t)$$. Is it true that $$W$$ satisfies the following equality?

$$\frac{\partial W}{\partial t}+\frac{1}{2}\sigma^2R^2\frac{\partial^2 W}{\partial R^2}+R(r-R)\frac{\partial W}{\partial R}-(r-R)W=0$$

• For arbitrary $f$? Commented Sep 7, 2021 at 20:46
• Sorry about that .f(S,t) should be replaced by S.I have edited
– abc
Commented Sep 7, 2021 at 21:33

Yes, but it looks like two things ought to be mentioned. First, the equality for $$V$$ needs to be clarified. I think it is a PDE for a function $$V(S,I,t)$$ of three independent variables. If that is so, the sentence "Here $$I = \int_0^t Sdt$$" doesn't fit, because it assumes $$S$$ and $$I$$ are functions of $$t$$, and the equality for $$V$$ then only holds along a thin curve instead of a region of $$(S,I.t)$$ space, which I think is what is intended.
The second point is that we ought to write $$V(S,I,T) = IW(R,t), \qquad R = S/I$$ rather than $$V(S,R,t)$$. Then the result follows from ordinary chain rule. For example $$V_I = W+IW_R\frac{-S}{I^2} = W-RW_R$$ and so forth.
If you really intend for $$S$$ to be a function of $$t$$ in the $$V$$ equation, you would still need it to hold in a region of $$(S,I,t)$$ space in order to use the chain rule. But if that is the case then you could specialize the $$W$$ PDE to the resulting curve in $$(R,t)$$ space. I'm not clear what the advantage of that would be since the PDEs have to hold in the entire region containing the curve anyway.
• Thanks!.So you mean we need $\frac{\partial I}{\partial t}=0$