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Let $f(t) \in \mathbb{R}^+$, $g(t) \in \mathbb{R}^+$ be bounded and integrable functions, $A>0$ and $0 < \eta \leq 1$. If $h(t) = A\cdot f(t) - g(t)$, what needs to be the value of $A$ to satisfy
\begin{equation} \int_0^T \bigl(h^+ - \frac{h^-}{\eta}\bigr) dt =0 \end{equation} where $h^+$ and $h^-$ are the absolute positive and negative parts of $h(t)$, respectively.

Background: In solar energy storage, the size of both the storage and the solar field are not determined a priori. If $g(t)$ is the demand, $f(t)$ is the solar irradiation, $A$ is the size of the field and $\eta$ is the round-trip efficiency of the storage, an answer to the above question determines the size of the solar field and indirectly the size of the storage for arbitrary demand and supply curves.

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  • $\begingroup$ Despite you do not know a priori the demand $g(t)$ nor the irradiation $f(t)$, you may try to guess something more on their structure: otherwise the question is not well posed since, strictly speaking, $g$ and $f$ could be non-integrable. For example, considering that the solar irradiation during the day brings only a finite energy to a given unit surface, you may assume a finite energy i.e. $L^2$ condition to hold for it, i.e. $$ \int_0^{T_\mathrm{day}}\!\! |f(t)|^2\mathrm{d} t\le\infty $$ And similar conditions may also hold for $g$. $\endgroup$
    – Daniele Tampieri
    Commented Oct 8, 2019 at 15:17
  • $\begingroup$ True, thanks, I edited the question to make it clearer. $\endgroup$
    – Michael Wild
    Commented Oct 8, 2019 at 15:45
  • $\begingroup$ What sort of answer are you hoping for here? It seems unlikely that there is an analytic answer, but I suspect you could easily solve this numerically. Starting with $A$ sufficiently large to make the integral positive and then proceeding by bisection seems like it ought to work. $\endgroup$
    – Aaron Bergman
    Commented Oct 9, 2019 at 2:29
  • $\begingroup$ To be honest I am hoping that there is an obvious solution to this problem that I simply missed. I actually wrote a small program that numerically solves this problem, it is indeed very easy to find a solution. If you look at it graphically it's obvious how to get there. However the final application that this problem is a part of is quite a lot more involved, that's why an analytical solution would be very beneficial. $\endgroup$
    – Michael Wild
    Commented Oct 9, 2019 at 9:14

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