Assume I am given a functional of the form: $$ I_0[u]:= - \int \nabla u \cdot \nabla u dx $$ then, I know that by the Euler-Lagrange equations, I have: $$ \frac{\delta I_0}{\delta u }= 2\Delta u $$
But, what if I have higher-order derivatives inside the functional? For example, what is the variational derivative $\frac{\delta I_2}{\delta u }$ of the functional: $$ I_1 [u]:=\int \Delta u dx $$ or $$ I_2 [u]:=\int (\Delta u)^2 dx $$ ?
Hope I made myself clear.
Thank you all in advance
The Euler-Lagrange equations implicitly assume that you have a functional which is of the form $I[y,y';x]=\int f(y,y',x)dx$. To do this, we use a Taylor expansion, incrementing $y$ by a small amount $\delta y$: $$\int_{a}^{b}f(y+\delta y,y'+\delta y',x)dx\\=\int_{a}^{b}f(y,y',x)dx+\int_{a}^{b}\left(\delta y \frac{\partial f}{\partial y}+\delta y' \frac{\partial f}{\partial y'}\right)dx\\=\int_{a}^{b}f(y,y',x)dx+\int_{a}^{b}\delta y\left(\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'}\right)dx$$ Where we neglected any quadratic and higher order terms. Clearly, if $f$ depends on second (order higher) order derivatives, you would need to do a different Taylor series expansion (of a similar kind). I'll leave that as an exercise. For example, if $f$ depends on $y''$ as well, Euler-Lagrange becomes $$\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'}+\frac{d^{2}}{dx^{2}}\frac{\partial f}{\partial y''}=0$$