Okay this is probably going to be an extremely easy/straightforward question but I thought I should post it here just to double check. Suppose I have a payoff $\Phi = (S_{T}-K)^{+}$. Now let's say I now have an equation: $u = s\partial_{s}\Phi - \Phi$, this means that given a payoff $\Phi$ as given above then, substituting this payoff into the equation and assuming $S_T = S_{0}\exp((r-1/2)T+\sigma\sqrt{T}Z_i))$ then I should get:
$u = max(S_{T},0) - max(S_{T}-K,0)$, right?
And from this equation, the possible solutions should be:
If $S_{T} > K$, $u = K$, if $S_{T} < K$ and $S_{T} > 0$, $u = S_{T}$, and if $S_{T} < K$ and $S_{T} < 0$ then $u = 0$.
Is all of this correct? I know this is really trivial but I just thought I should check...
I assume you mean $s\Phi'(s) - \Phi(s),$ to be evaluated at $s = S_T$. In that case, the specific stochastic nature of $S_T$ is irrelevant.
We have
$$\Phi(s) = \begin{cases} s - K, &\mbox{if } s > K \\ 0, &\mbox{if } s \leqslant K\end{cases} \\ \Phi'(s) = \begin{cases} 1, &\mbox{if } s > K \\ 0, &\mbox{if } s < K\end{cases}$$
Hence,
$$s\Phi'(s) - \Phi(s) = \begin{cases} K, &\mbox{if } s > K \\ 0, &\mbox{if } s < K\end{cases} $$
Note that the derivative $\Phi'(K)$ does not exist, but the left-hand and right-hand derivatives are $\Phi_L'(K) = 0$ and $\Phi_R'(K) = 1$.
Thus,
$$S_T\Phi'(S_T) - \Phi(S_T) = \begin{cases} K, &\mbox{if } S_T > K \\ 0, &\mbox{if } S_T < K\end{cases} $$