interesting boundary problem of PDE Consider the following PDE
\begin{equation}
\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
\end{equation}
 with boundary condition 
\begin{equation}
V(S,T)=f(S)
\end{equation} 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
\begin{equation}
\frac{∂ρ}{∂t} + \frac{σ^2 S^2}{2}\frac{∂^2ρ}{∂S^2} + rS\frac{∂ρ}{∂S} −rρ=V−S\frac{∂V}{∂S}\end{equation}
 with boundary condition 
\begin{equation} \rho(S,T)=0. \end{equation} 
${\it QUESTION}$: 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 $t→T^−$ , is it possible to show that
\begin{equation}
ρ(S,t)=(T−t)\Big(S\frac{∂V}{∂S}(S,t) −V(s,t)\Big).
\end{equation}
Any ideas where to start?
 A: I give a derivation of the result from scratch below. However, this is long and technical. A much more simple way to proceed if we already know that the boundary condition for $\rho$ reads $\rho(S,T) = 0$, as we do here, and don't care about deriving it is to just substitute $\rho = (T-t)\rho_0$ where $\rho_0 = \left(S\frac{dV}{dS}-V\right)$ into the equation for $\rho$ to get
$$(T-t)\left(\frac{d\rho_0}{dt} + \frac{\sigma^2S^2}{2}\frac{d^2\rho_0}{dS^2} + rS \frac{d\rho_0}{dS} - r\rho_0\right) = 0$$
Now if $V$ is a solution to the original equation then one can show that $S\frac{dV}{dS}$ is also a solution and since the equation is linear so is any linear combination of these two solutions and the claim follows.

We will derive a formal integral-solution to the equation and use this to derive the desired result from scratch. Starting with
$$\frac{dV}{dt} + \frac{\sigma^2 S^2}{2}\frac{d^2 V}{dS^2} + rS \frac{dV}{dS} - rV = 0$$
we define $x=\log S$ to get it on a simpler form
$$\frac{dV}{dt} + \frac{\sigma^2}{2}\frac{d^2 V}{dx^2} + \left(r-\frac{\sigma^2}{2}\right) \frac{dV}{dx} - rV = 0$$
We can now simplify the equation above by taking the Fourier transform w.r.t. $S$ giving us
$$\frac{d\hat{V}}{dt} = \hat{V}\left(\frac{\sigma^2\omega^2}{2} - i\omega\left(r - \frac{\sigma^2}{2}\right) + r\right)$$
The solution of this ODE reads
$$\hat{V} = A(\omega) e^{\left(\frac{\sigma^2\omega^2}{2} - i\omega\left(r - \frac{\sigma^2}{2}\right) + r\right)t}$$
Applying the boundary condition to determine $A$ and taking the inverse Fourier transform gives us an expression for the solution
$$V(S,t) = \int_{-\infty}^\infty \hat{f}(\omega) e^{\left(\frac{\sigma^2\omega^2}{2} - i\omega\left(r - \frac{\sigma^2}{2}\right) + r\right)(t-T) + i\omega \log S}d\omega$$
where $\hat{f}(\omega)$ is the Fourier transform of $f(S) = V(S,T)$. With the solution in hand we can go ahead and calculate the quantities we are interested in:
$$\rho(S,t) = \int_{-\infty}^\infty \hat{f}(\omega)(t-T)(1-i\omega) e^{\left(\frac{\sigma^2\omega^2}{2} - i\omega\left(r - \frac{\sigma^2}{2}\right) + r\right)(t-T) + i\omega \log S}d\omega$$
$$S\frac{dV(S,t)}{dS} - V(S,t) = \int_{-\infty}^\infty \hat{f}(\omega)(-1 + i\omega) e^{\left(\frac{\sigma^2\omega^2}{2} - i\omega\left(r - \frac{\sigma^2}{2}\right) + r\right)(t-T) + i\omega \log S}d\omega$$
and it follows that
$$\rho(S,t) = (T-t)\left(S\frac{dV}{dS} - V\right)$$
