given the functional:
$$ F(\phi)=\frac{1}{2} \langle L\phi,\phi \rangle – \frac{1}{2} \langle \phi,f \rangle – \frac{1}{2} \langle f,\phi \rangle $$
where the differential equation with operator $L$ and function $\phi$:
$$ L\phi=f $$
and $ \langle F,G \rangle $ denotes the inner product of two functions.
how is its first variation calculated?
$$ \delta F=\frac{1}{2} \langle L\delta\phi,\phi \rangle + \frac{1}{2} \langle L\phi,\delta\phi \rangle – \frac{1}{2} \langle \delta\phi,f \rangle – \frac{1}{2} \langle f,\delta\phi \rangle $$
Best Answer
Assuming that $L$ is a linear operator, we have $$ \frac{d}{dt} \langle L (\phi+t\eta), \phi+t\eta \rangle|_{t=0}= \\ \frac{d}{dt} \langle L \phi, \phi \rangle+t(\langle L \phi, \eta \rangle+ \langle L \eta, \phi \rangle) +t^2 \langle L \eta, \eta \rangle |_{t=0}= \\ \langle L \phi, \eta \rangle+ \langle L \eta, \phi \rangle $$ At this point, you are already at your whit`s end - if you knew your operator (e.g. $L=- \Delta$), you could probably perform an integration by parts. Adapting your notation, this means that the first term of $\delta F$ is (with the factor of $\frac{1}{2}$ in front) $$ \frac{1}{2}(\langle L \phi, \delta \phi \rangle+ \langle L \delta\phi, \phi \rangle) $$ for the linear terms, we have $$ \frac{d}{dt} \langle \phi+t\eta, f \rangle|_{t=0}=\frac{d}{dt} \langle \phi,f \rangle + t \langle \eta, f \rangle|_{t=0}=\langle\eta,f\rangle $$ Switching back to your notation, we have with the factor of $\frac{1}{2}$ in front $$ \frac{1}{2}\langle\delta \phi,f\rangle $$ The term $\langle f, \phi \rangle$ is taken care of in the same manner. Adding all terms give you the desired variation $\delta F$. However, I wanna note that I computed a Gateaux derivative of a functional (also called "variation" sometimes) and if you knew how $F$ depended on $\phi, \nabla \phi$ explicitly, i.e. $F=F(\phi, \nabla \phi)$, you would arrive at the weak form of the Euler-Lagrange equations. On another note, you still have to take care of the domain of your differential operator $L$, which is often NOT the whole $L^2$ space. I omitted this part, as it would require more informations on $L$.