Consider Black Scholes problem $\frac{\partial V}{\partial t} + \frac{\sigma^2 S^2}{2}\frac{\partial^2V}{\partial S^2} + rS\frac{\partial V}{\partial S} -rV = 0$ with boundary condition $V(S,T)=f(S)$, where $V(S,t)$ is the solution to the PDE above.

Let $ρ(S,t)=\frac{∂V}{∂r}(S,t)$, then it can be shown that $\rho$ satisfies the following PDE: $\frac{∂ρ}{∂t} + \frac{σ^2 S^2}{2}\frac{∂^2ρ}{∂S^2} + rS\frac{∂ρ}{∂S} −rρ=V−S\frac{∂V}{∂S}$ with boundary condition $\rho(S,T)=0$ .

${\it QUESTION}$: given that the PDE above uniquely determines $ρ$ , and assuming that both $V$ and $S\frac{∂V}{∂S}$ are of the order $o(1/(T−t))$ as $t→T^−$ , show that $ρ(S,t)=(T−t)\Big(S\frac{∂V}{∂S}(S,t) −V(s,t)\Big) $,

I have tried to tackle the problem by following this:

Under sufficient regularity/growth/etc. conditions, an equation $Lu=v$ with specified initial and boundary conditions has a unique solution. So to verify that two solutions $f$ and $g$ of the same equation $Lu=v$ with identical initial and boundary conditions are the same, it is sufficient to verify that their difference satisfies $L(f−g)=0 $ with zero initial and boundary conditions. So it's enough to verify regularity conditions and check that the l.h.s. satisfies exactly the same equation and conditions as the r.h.s.

However I couldnt produce the required result.

I would be grateful if anyone could suggest how to approach this and similar problems.


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