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I am only familiar with some of the basics of Sobolev spaces and variational calculus, so please keep that in mind with your answer. I have read other posts regarding the, "inverse problem of calculus of variation" which I think aptly describes my question. Upon some further digging it seems that some people are proving (or have proved, rather) existence of functionals which have an extremum at some PDE's solution. By this I mean take your PDE of interest to be represented as some $n$-th order operator $\mathcal{D}^n[u] = 0$, and take the energy functional to be $\mathcal{F}[u]$ where $\mathcal{F} : H_0^1(\Omega) \rightarrow \mathbb{R}$, then \begin{gather} \forall u, \phi \in H_0^1(\Omega) \; \exists \mathcal{F} \; | \; \mathcal{D}^n[u] = \left. \frac{d \mathcal{F}}{d t}(u + \phi t) \right|_{t = 0} = 0 \end{gather} This is my understanding of what I think that people have shown. In the above I take $H_0^1$ to be the Sobolev space, and the functional derivative to be the "first variation" I believe it is called.

OK so now for my question:

Is there a systematic (algorithmic) way to determine an energy functional that is extremized by the solution of a PDE?

Any qualitative explanations are welcome! I would note that the reason I am interested in this is formulating a general finite element algorithm, and thus I need a weak formulation of the PDE (not sure if a weak formulation of the PDE is the same thing as the energy functional formulation, so bonus point question is to educate me the distinction between weak formulation and energy functional formulation).

Thanks!

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    $\begingroup$ Though I am no expert on the topic, it would seem that you would want to multiply your PDE by the undetermined variable itself (and derivatives of said quantity) and integrate. This is how many of the known energy functionals are constructed, for example the Dirichlet energy comes from multiplying the Laplace equation $- \Delta u = 0$ by $u$ and integrating over the volume, using integration by parts and the boundary conditions to make the boundary terms vanish. However, I'm not sure if this works for all equations. $\endgroup$
    – Matthew Cassell
    Dec 2, 2021 at 12:18
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    $\begingroup$ Emerson, I think this MathOverflow Q&A answers to your question. $\endgroup$
    – Daniele Tampieri
    Dec 18, 2021 at 18:11
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    $\begingroup$ Also this Q&A relies on the previous MathOverflow answers and solves your question. $\endgroup$
    – Daniele Tampieri
    Dec 18, 2021 at 19:06

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