I need to solve a PDE in 2D representing a (time-independent) nonlinear diffusion process. The unknown function is $\phi(x,y)$ and its gradients create fluxes $\vec J$ through a nonlinear relation: $$\vec J = (\phi+h)^\alpha\frac{\vec\nabla\phi}{\sqrt{|\vec \nabla\phi|}} \ .$$ The PDE to be solved (representing steady-state flow) is $$ \operatorname{div}\vec J = \partial_x J_x+\partial_y J_y =0 \ . $$ $h(x,y)$ is a given function and the values of $\phi$ are known on the boundary of the domain (Dirichlet problem).

For various reasons, both numerical and analytic, I want to represent the problem as an energy-minimization problem. That is, to find a functional $\mathcal{L}(\phi, \vec\nabla\phi)$ whose minimizer is a solution to the PDE (or yet in other words: a functional whose Euler-Lagrange equation is the PDE).

Is there a general way to do this, or to prove that it is impossible? Without the $(\phi+h)^\alpha$ term I can easily guess the functional, but I couldn't make it work with it.


2 Answers 2


As already said by Kosh, you are trying to solve an inverse problem in the calculus of variation: in its classical formulation, given a system of PDE, the problem consists in finding a functional whose Euler-Lagrange equations are exactly the given PDE(s).
To my knowledge, a necessary and sufficient condition for a system of PDE (jointly with its boundary/Cauchy conditions) to be the Euler-Lagrange equation(s) of a Lagrangian functional (meaning functional which is the integral of a function whose arguments, apart from a "independent" variable $\mathbf{x}$, are the unknown function $\phi$ and its lower order derivatives) is still unknown. However, it is possible to approach the problem from a more general yet concrete point of view: consider a general operator $$ \phi\mapsto\mathsf{N}(\phi)(\mathbf x)-f(\mathbf{x}),\label{op}\tag{OP} $$ where

  • $\mathsf N$ is a linear or nonlinear operator (possibly your differential operator, an integral operator, a system of PDEs etc.),
  • $\phi\in\operatorname{Dom}(\mathsf N)$, the domain of $\mathsf N$ in an appropriate function space, and
  • $f\in\operatorname{Range}(\mathsf N)$ the range of $\mathsf N$ again in an appropriate function space.

The kernel of this operator is characterized by the following equation

$$ \mathsf N(\phi)(\mathbf x)=f(\mathbf{x}).\label{1}\tag{1} $$ If there exists a general functional $F$ such that its functional derivative vanish on the set of solution(s) of \eqref{1}, i.e. $$ \left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon}F(\phi+\varepsilon\psi)\right|_{\varepsilon =0}\!\!\!=0\;\;\,\forall\psi \in \operatorname{Dom}(\mathsf N)\iff \phi(\mathbf x)\text{ is a solution of \eqref{1},}\label{gvf}\tag{GVF} $$ the problem \eqref{1} is said to admit an extended variational formulation and the operator \eqref{op} is said to be a potential operator (or conservative operator according to [2] §3 p. 152).
Enzo Tonti (see [1], chapter III §11.1 p. 94 and chapter IV §§17.1-17.4, pp. 168-162 and the recent paper [2] or the Author's web site, where a brief survey of the result is given) proved that, under fairly general conditions, a such a functional $F$ always exists. Precisely, Tonti's theorem says that every linear / nonlinear problem \eqref{1}, provided some natural hypotheses are assumed, admits an extended variational formulation. The proof is constructive in that it shows that it is possible to explicitly construct a compact self-adjoint invertible linear operator $\mathsf K$ such that the functional $$ F(\phi)=\int\limits_\Omega \mathsf N(\phi)\mathsf K\big(\mathsf N(\phi)-f\big)\mathrm{d}\mathbf{x}\label{2}\tag{2} $$ satisfies \eqref{gvf}: moreover, if $\mathsf{K}$ is also a positive definite operator (i.e. $\langle v, \mathsf{K}v\rangle>0$ for all functions $v\in \operatorname{Dom}(\mathsf K)$ such that $v\not\equiv 0$), the the solutions of \eqref{1} are the minimum points of the functional \eqref{2} ([2] §4 theorem 2 pp. 155-156). Thus applying the techniques described in [1] and [2] you can surely find variational formulation for your problem, even if it cannot be expressed as the Euler-Lagrange equation(s) of an appropriate energy functional.


  • When $\phi, f$ belong to function spaces on sufficiently regular domains $\Omega$, then the Green's operator associated to the Green's function for a any symmetric linear PDE on $\Omega$ can be chosen as the needed $\mathsf{K}$ operator. When $\Omega$ is not regular, there are nevertheless other possible choices: some of them are described in §5, pp. 156-161 of [2].

  • The analytical approach to the inverse problem of the calculus of variations is perhaps comprehensively described in the wonderful monograph [1], which is however up to date up to 1989: on the other hand [2] surveys nicely the theory of potential operator up to the more recent development due to the Author.

  • In remark 17.1 of [1] chapter IV, §17.2 pp. 171-172, Filippov gives a brief but very interesting survey on the conditions of potentiality for a nonlinear differential operator (ordinary or partial): perhaps, if something more tailored to the form of a given equation respect to \eqref{2} is needed, it would be worth to have a look at the references cited there.


[1] Filippov, Vladimir Mikhailovich, Variational principles for nonpotential operators. With an appendix by the author and V. M. Savchin, Transl. from the Russian by J. R. Schulenberger. Transl. ed. by Ben Silver, Translations of Mathematical Monographs, 77. Providence, RI: American Mathematical Society (AMS). pp. xiii+239 (1989), ISBN: 0-8218-4529-2. MR1013998, ZBL0682.35006.

[2] Tonti, Enzo, "Extended variational formulation", Vestnik Rossiĭskogo Universiteta Druzhby Narodov, Seriya Matematika 2, No. 2, 148-162 (1995). ZBL0965.35036.


In general, the problem can be complicated because some nonlinearities arise as Lagrange multipliers. Therefore, an energy functional (whose Euler-Lagrange equations coincide with the nonlinear PDE) can exist provided that some constraints to the energy functional are imposed.

Anyhow, the general problem is called The Inverse Problem of the Calculus of Variations. There is a quite recent book on that. You can take a look at it.



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