I have a problem concerning the transformation of a PDE. (It is actually a linear complementarity problem for American option, however my problem only concerns the underlying mathematics/PDE).
For a function $V(S,t)$, $t \in [0,T]$ and $S \geq 0$, I consider the problem (with $r,\delta > 0$) $$ (V - V(S,T))\left(V_t + \frac{\sigma^2}2 S^2V_{SS}+(r-\delta)SV_S -rV\right) = 0, \\ V - V(S,T) \geq 0 \quad \text{and} \quad -\left(V_t + \frac{\sigma^2}2 S^2V_{SS}+(r-\delta)SV_S -rV\right) \geq 0 $$ with boundary and final conditions given as follows: $$ V(0,t) = K, \quad \lim_{S \to \infty}V(S,t) = 0, \quad V(S,T) = (K - S)^+. $$
Now, my problem: One can make the transformation $$ x = \log(K/S), \quad \tau = \frac{\sigma^2}2(T-t) $$ and define with $k_\delta = \frac{2(r-\delta)}{\sigma^2}$, $$ u(x,\tau) = \exp\left(\frac12 (k_\delta-1)x+\frac14(k_\delta-1)^2\tau+k_0\tau)\right)\frac{V(S,t)}K. $$ Then the above PDE (when set equal to 0) is transformed into the nice equation $u_\tau = u_{xx}$. However, then in my lecture notes, the function $f$ is defined as follows $$ f(x,\tau) = \exp\left(\frac12 (k_\delta-1)x+\frac14(k_\delta-1)^2\tau+k_0\tau)\right)\frac{V(Ke^x,T)}K $$ and it is claimed that the above term $(V - V(S,T))$ turns into $(u - f)$. Obviously $V$ turns into $u$, but how can $V(S,T)$ (a function not depending on time, not on $t$ or $\tau$) turn into $f$, which clearly depends on time (in the form of $\tau$)? Additionally, it is claimed that the second boundary condition is now given by $$ \lim_{x \to -\infty}(u(x,\tau) - f(x,\tau)) = 0, $$ where I am also confused about the use of the function $f$.
Can someone explain to me, what the idea behind the function $f$ is or should it maybe done differently? How can I understand the transformation of the transformed boundary condition?
Thanks in advance!