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I have the following equation for $u:D\subseteq \mathbb{R}^d\rightarrow \mathbb{R}$ $$-\Delta \Delta u=0$$ Based on this I have to find the functional where this u would be a extreme point. Since we dealt with variantional calculus the solution should look something like $$\int L(\nabla u, u,x)dx$$ But I have no other Approach then just guessing and checking.

I suppose that I need something that does $$\frac{\partial L}{\partial \nabla u}=\frac{\partial }{\partial x_i}\sum_{j=1}^d\frac{\partial^2}{\partial x_j^2}u$$ But this doesnt really help.

Enoo_58
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    With what you have that would be impossible. Logically at best, you can get an extra derivative out of the gradient by applying Euler Lagrange, but that only gets you half way. What you need is $L(x,u,\nabla u, \Delta u)$ or $L(x,u,\nabla u, \Delta u, \nabla \Delta u)$, not $L(x,u,\nabla u)$ (more generally you would want to use Hessian of $u$ but this may not be necessary in this case). Then you will need to figure out how to apply Euler Lagrange to higher order derivatives. – Ninad Munshi Nov 06 '22 at 13:33
  • For some reason we didnt do these cases could you maybe give a solution for $L(x,u,\nabla u, \Delta u)$ because I dont really know how to hanlde them. Am I supposed to take $\frac{\partial L}{\partial \Delta u}$ at some point? – Enoo_58 Nov 06 '22 at 13:40
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    The action $$S = \int_{\Omega} \lvert \Delta u \rvert^{2} dx$$ is the action you are after, assuming you have boundary conditons that ensure any evaluated terms vanish. See here and here for more. – Matthew Cassell Nov 07 '22 at 02:42
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    You can use calculus of variations to derive Euler Lagrange. Follow any textbook you want and do the work yourself to extend the example process to higher order derivatives. – Ninad Munshi Nov 07 '22 at 03:09

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