How to prove this is a self-adjoint operator? I have this operator from $H^1_0$ to $H^1_0$ defined by: $$Au(t)=\int_0^1 G(t,s) f(s,u(s))\mathsf ds$$ where $$G(t,s)=\begin{cases} t(1-s), &t\leq s\\s(1-t), &s\leq t.\end{cases}$$
I want to see if $A'(0)$ is self-adjoint, if my calculus are right we have 
$$A'(u)[v](t)= \int_0^1G(t,s) f_{u}(s,u(s)) v(s)\mathsf ds $$
Have you an idea please ?
 A: Assume that $f$ is continuously differentiable on $\mathbb R^2$.
Also assume that $f_u(s, 0) := \frac{\partial f}{\partial u}(s,0) = c$ for every $s\in[0,1]$.
Notice that $G(t,s) = G(s,t)$ for every $s,t\in[0,1]$.
Thus, for $v,w \in H^1_0$ we have
$$\begin{align}
\int_0^1 A'(0)[v](t) w(t) \;\mathrm d t
&= \int_0^1 \left( \int_0^1 G(t,s)f_u(s,0)v(s) \;\mathrm d s\right) \; w(t) \;\mathrm d t \\
&= \int_0^1 \int_0^1 G(t,s) c v(s) w(t) \;\mathrm d s \;\mathrm d t \\
&= \int_0^1 \int_0^1 G(s,t) c w(t) v(s)  \;\mathrm d t \;\mathrm d s \\
&= \int_0^1 \left( \int_0^1 G(s,t) f_u(t,0) w(t) \;\mathrm d t \right) \; v(s) \;\mathrm d s \\
&= \int_0^1 v(s) A'(0)[w](s) \;\mathrm d s.
\end{align}$$
Further, we have
$$\begin{align}
\int_0^1 (A'(0)[v])'(t) w'(t) \;\mathrm d t
&= \int_0^1 \frac{\mathrm d}{\mathrm d t}\left( \int_0^1 G(t,s)f_u(s,0)v(s) \;\mathrm d s\right) \; w'(t) \;\mathrm d t \\
&= \int_0^1 \left( \int_0^1 \frac{\partial G}{\partial t}(t,s)f_u(s,0)v(s) \;\mathrm d s\right) \; w'(t) \;\mathrm d t \\
&= \int_0^1 \underbrace{\int_0^1 \frac{\partial G}{\partial t}(t,s) v(s) \;\mathrm d s}_{=:\phi(t)} \; cw'(t) \;\mathrm d t.
\end{align}$$
From
$$\phi'(t) 
= \frac{\mathrm d}{\mathrm dt}\left(\int_0^t (-s) v(s) \;\mathrm d s + \int_t^1 (1-s) v(s) \;\mathrm d s \right)
= (-t) v(t) - (1-t) v(t) = -v(t) 
$$ 
follows
$$\begin{align}
\int_0^1 (A'(0)[v])'(t) w'(t) \;\mathrm d t
&= -\int_0^1 c v(t) w'(t) \;\mathrm d t \\
&= \underbrace{-\left. cv(t)w(t) \right|_0^1}_{=0} + \int_0^1 c v'(t) w(t) \;\mathrm d t \\
&= \int_0^1 v'(t) (A'(0)[w])'(t) \;\mathrm d t.
\end{align}$$
That is, $A'(0)$ is self adjoint.
