Is there any way to transform the basic call option payoff $V(s,0) = \max(s-K,0)$ such that $g(V(s,0))\neq 0$ $\forall s $, where $g()$ is the transform function of the payoff. This is to use in a numerical method where the form $V(s,t) = f(s,t)g(V(s,0))$ is assumed, where $f(s,t)$ is to be approximated, hence $g(V(s,0)) \neq 0$ otherwise it kills off $f(s,t)$ for all OTM options.

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    $\begingroup$ I don't really understand what your purpose is, but you can always do: $V(s)=\max(s-K,0)=\max(s,K)-K$. Defining $v(s)=\max(s,K)$ we always have $v(s)>0$ for a strictly positive strike so you can solve for $v(s)$ (the value of $K$ is readily available through discounting). I understand function $g$ is some kind of "pricing function" but you might want to clarify your notation. $\endgroup$ Apr 5, 2018 at 16:31
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    $\begingroup$ @DaneelOlivaw Yes. Daniel previously provided a very complete answer to a similar question on a non-zero boundary conditions (quant.stackexchange.com/questions/33244/…). Solving a PDE for different boundary conditions follows the same process as solving for the original boundary conditions. $\endgroup$ Apr 5, 2018 at 17:14
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    $\begingroup$ @DaneelOlivaw i'm trying out a numerical method where I assume the form as $V(s,t)=f(s,t)V(s,0)$, where f(s,t) is to approximated, so the issue is when the the payoff = 0 it always kills off the approximated function $\endgroup$
    – Sam Palmer
    Apr 5, 2018 at 17:23
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    $\begingroup$ @SamPalmer interesting, maybe you should add those clarifications to your question and clearly define $g$. $\endgroup$ Apr 5, 2018 at 17:42
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    $\begingroup$ If that can be solved, $U(s,t)=f(s,t)\max(s,K)$ and $V(s,t)=U(s,t)+Ke^{-rt}$. I am not 100% sure you can freely separate the $Ke^{-rt}$ part and solve only for $U(s,t)$ with your numerical method but from no-arbitrage that should be right. $\endgroup$ Apr 5, 2018 at 18:56


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