Derivative of $f(u,v) = \int_\Omega (v-u)^+.$

Question

Here $\Omega$ is a bounded domain.

Define $f:H^1 \times H^1 \to \mathbb{R}$ by $$f(u,v) = \int_\Omega (v-u)^+.$$ (Note the $H^1$, not $L^2$).

I want to find an expression for the (directional or Frechet) derivative $f'(u,v)(h_1, h_2)$.

Define $I(w) = \int_\Omega w$, then $I'(w)h = \int_\Omega h$. Now I think we need to differentiate $g(x) : = (x)^+$ and somehow combine, but I don't know the details. Could someone help please?

Answer 1

We have the functions $f(u,v) = ∫_Ω (v-u)^+, I(w) := ∫_Ω w, g(w):= (w)^+, h(u,v) := v-u$ with the relationship $$ f(u,v) = I\circ g\circ h(u,v)$$ We use the notation $dF(x_0)[ε_0]$ to mean the derivative of $F$ at $x_0$ evaluated with increment $ε_0$. By chain rule the derivative at $u,v$ is $$ df(u,v)[(ε,\delta)] = dI(g\circ h(u,v))[dg(h(u,v))[dh(u,v)[(ε,\delta)]] ] $$ We have the individual derivatives $$dI(w)[h] = I(h), \quad dh(u,v)[(ε,\delta)] = h(ε,\delta)$$ by linearity, and $dg(w)[ε] = (\mathbb 1_{w\geq 0}) ε $. Combining,

$$ df(u,v)[(\epsilon,\delta)] = ∫_Ω \mathbb 1_{v(x) - u(x) ≥ 0} (ε(x)- δ(x) ) dx $$