I have an integral operator and I wonder how I can characterize the critical point. I'll give a simplified example so maybe people can comment on and I can maybe generalize in another question.
Disclaimer : I am not an expert in this subject and my knowledge of calculus of variation is quite elementary.
Suppose we have the following integral operator
$$ T[f](y) = \int_{\Omega} K(x,y) \frac{f^2(x)}{2} d\mu(x) $$
Where $f, f^2 \in L^2(\Omega,\mu)$, $K \in L^2(\Omega^2,\mu\times \mu)$. I was trying to characterize the critical point of this non linear operator. Therefore a calculation of the variational derivative gives me the differential $DT_f$ defined by
$$ DT_f[h](y) = \int_{\Omega}K(x,y)f(x)h(x)d\mu(x) $$
To find such critical point I need to find $f$ such that for every $y \in \Omega$ an $h \in L^2(\Omega,\mu)$ yields
$$ DT_f[h](y) = 0 \iff \int_{\Omega}K(x,y)f(x)h(x)d\mu(x) = 0 \iff \int_{\Omega} f(x) h(x) K(x,y) d\mu(x) = \int_{\Omega} f(x)h(x)w_y(x) d\mu(x) = \int_{\Omega} f(x)h(x) d\lambda_y(x) = 0 $$
where $w_y(x) = K(x,y)$ and $d \lambda_y = w_y d\mu$. The last equality of the chain above is equivalent to
$$ \left\langle f, h\right\rangle_{L^2(\Omega,\lambda_y)} = 0 \;\;\; \forall y \in \Omega, h \in L^2(\Omega,\lambda_y) $$
- Is my derivation correct?
- If not where is it wrong? Or can you reccomend further readings on this subject.
