Prove that a Lagrangian that Induces a Linear Elliptic Euler-Lagrange PDE has Unique Form

Question

I am asking if existence, taken as an assumption for a solution $L$ to the linear operator equation $$\mathcal{E}L = F$$ with conditions on $F$, implies further conditions on $F$ and uniqueness of a generic form for the terms in $L$ not annihilated by $\mathcal{E}$.

The variational problem giving rise to the operator equation is to find a stationary solution $u$ that minimizes the action functional $$I[u]=\int_{\Omega}{L(Du,u,x)dx}$$

This induces the corresponding Euler-Lagrange equation $$\frac{\partial L}{\partial u}-\sum_{i=1}^n{\frac{\partial}{\partial x_i}\left(\frac{\partial L}{\partial(\partial_i u)} \right)}=0$$

and the corresponding operator $$\mathcal{E}=\partial_u-\nabla\cdot \partial_{\partial_i u}$$

This can be adjusted to accommodate any number of higher order derivatives (see here). A nice lemma would justify only allowing dependence up to $Du$, not $D^2u$ and higher.

Suppose $L$ induces Euler-Lagrange equations that are linear-elliptic. In the operator formulation this is $\mathcal{E}L=F=0$ where $$F(Du,u,x)=\sum_{i=1}^{n}{\sum_{j=1}^n{ a_{ij}(x)\partial_{ij}u }}+\sum_{k=1}^n b_k(x)\partial_{k}u+c(x)u-f(x) $$

with smoothness and other conditions on $a,b,c,f$ to ensure ellipticity. Can we prove that

1. the divergence of the $k$th column of the $a_{i,j}$ matrix is such that $$ b_k=\nabla\cdot a_{.,k} $$ or more generally
2. the (Lagrangian) function $L$ has the form $$ L(Du,u,x)=\frac{c(x)u^2}{2}-f(x)u-\text{Tr}\left(A(x)(\nabla u \otimes \nabla u) \right) + L_0 $$ where
   • $c,f$ are some smooth functions,
   • $A_{i,j}$ is some $n\times n$ matrix of smooth functions and
   • $L_0$ is such that $L_0$ $\mathcal{E}L_0=0$?

The converse is easy enough to verify. (Note, the $a_{i,j}$ are $(1+\delta_{i,j})A_{i,j}$, since applying $\mathcal{E}$ introduces a $2$ on the diagonal where there is none off diagonal). Try as I might I can't find any way to add other terms into $L$ that produce linear elliptic terms in the Euler-Lagrange equation (see here). But I also can't see all the way through the logic for why this should not be possible.

It seems 'obvious' that $F$ restricts $L$ to a quadratic polynomial in $Du,u$ with smooth function coefficients, and that each possible term is either captured in (2) or forced to have coefficient zero from incompatibility with $F$. I'm looking for confirmation (or disconfirmation) of this logic.

Edit. Could someone at least elucidate what kind of problem this is and what relevant theory is helpful understanding it? I understand that the variational problem in $u$ has its own peculiarities, and that $\mathcal{E}L=0$ is merely a related PDE. But what kind of problem is $\mathcal{E}L=F$ where, instead of solving for $u$, we are solving for $L$? What is known about this problem in general? Is there a Green's Function approach that makes sense to find $L$ constructively? Or at least one that shows existence and maybe demonstrates some properties?

Answer 1

Premises to the answer

The question you are asking is not an easy one as it is a version of the inverse problem of the calculus of variation (see [1] and [3] for some bibliographical references). In the form of your operator equation, the problem is hardly solvable, but if you leave the classic formulation of the problem and adopt a generalised one, you can get a necessary and sufficient condition on $F$ in order for your problem to have a solution. The generalised inverse problem of the calculus of variation asks for a solution of the following first variation equation $$\DeclareMathOperator{\dm}{d\!} \frac{\delta\mathfrak{L}}{\delta u }[u](\varphi) = \mathfrak{F}[u](\varphi) \label{2}\tag{FV} $$ where

• $\mathfrak{L}$ is a functional on a given Banach (or more general) space) $E$: precisely, in our starting problem $$ \mathfrak{L}(u)=I(u)=\int_{\Omega}{L(Du,u,x)\dm x} $$
• $\mathfrak{F}$ is an operator mapping the domain of $\mathfrak{L}$ to the dual $E^\ast$ of $E$. In our starting problem, $$ \mathfrak{F}[u](\varphi)\triangleq\langle \mathfrak{F}[u],\varphi\rangle= \int_{\Omega}{F(Du,u,x)}{\varphi(x)}{\dm x} $$
• the left side of \eqref{2} is the functional derivative of $\mathfrak{L}$, i.e. the operator $E\to E^\ast$ defined as $$ \frac{\delta\mathfrak{L}}{\delta u }[u](\varphi)\triangleq D\mathfrak{L}[u](\varphi) \triangleq \left.\frac{\dm\,}{\dm \epsilon} \mathfrak{L}(u+\epsilon \varphi)\right|_{\epsilon=0} $$ The functional derivative can be nonlinear: however in most applications, including the one we are dealing, it is linear and thus it is at least a Gâteaux derivative.

This problem is known to be solvable if and only if $\mathfrak{F}$ is a potential operator (and a Gâteaux derivative, see [1], pp. 5-6, §11 pp. 92-93, and [3], §3 pp. 152-153). Being a potential operator means that, for any real curve in $E$ i.e. any function $\varepsilon\mapsto u(\varepsilon)\in E$ (here we assume $\varepsilon\in [0,1]$ without restriction to generality), the value of the functional $$ \int\limits_0^1 \mathfrak{F}[u(\epsilon)]\left( \frac{\dm\,}{\dm\epsilon} u(\varepsilon)\right)\dm\epsilon = \int\limits_0^1 \left\langle\mathfrak{F}[u(\epsilon)], \frac{\dm\,}{\dm\epsilon} u(\varepsilon)\right\rangle\dm\epsilon $$ depends only on the functions $u(\varepsilon)|_{\varepsilon=0}\triangleq u_0$ and $u(\varepsilon)|_{\varepsilon=1}\triangleq u_1$. This in turn can be shown to be equivalent to the fact that the functional derivative of $\mathfrak{F}$ is symmetric (see [1], pp. 5-6, §11 pp. 92-93, [2], §5.2 pp. 55-58, and [3], §3 pp. 152-153), i.e. $$ \left\langle\frac{\delta\mathfrak{F}}{\delta u }[u](u_0),u_1\right\rangle=\left\langle\frac{\delta\mathfrak{F}}{\delta u }[u](u_1),u_0\right\rangle\label{3}\tag{Sym} $$ and this condition implies that $$ \mathfrak{L}(u) \triangleq \mathfrak{L}(u_0) + \int\limits_0^1 \left\langle\mathfrak{F}[u(\epsilon)], \frac{\dm\,}{\dm\epsilon} u(\varepsilon)\right\rangle\dm\epsilon\label{5}\tag{Lag} $$ where $u(\epsilon) \triangleq u_0 + \varepsilon (u-u_0)$.
This is the condition needed in order to answer the question.

Answer to the question

Can we prove that

1. the divergence of the $k$th column of the $a_{i,j}$ matrix is such that $$ b_k=\nabla\cdot a_{.,k} $$ or more generally
2. the (Lagrangian) function $L$ has the form $$ L(Du,u,x)=\frac{c(x)u^2}{2}-f(x)u-\text{Tr}\left(A(x)(\nabla u \otimes \nabla u) \right) + L_0 $$ where
   • $c,f$ are some smooth functions,
   • $A_{i,j}$ is some $n\times n$ matrix of smooth functions and
   • $L_0$ is such that $L_0$ $\mathcal{E}L_0=0$?

The condition $$\DeclareMathOperator{\divg}{\nabla\cdot} b_k=\nabla\cdot a_{.,k} \iff \mathbf b(x) = \divg\mathbf{A}(x) $$ is necessary and sufficient for the linear PDE to be self-joint or, equivalently, symmetric: therefore we have that $$ \begin{split} F(Du,u,x) &=\sum_{i=1}^{n}{\sum_{j=1}^n{ a_{ij}(x)\partial_{ij}u }}+\sum_{k=1}^n b_k(x)\partial_{k}u+c(x)u-f(x) \\ & = \divg \big(\mathbf{A}(x)\nabla u\big)+c(x)u-f(x) \end{split} $$ and this is equivalent to say that $$ \begin{split} \mathfrak{F}[u](\varphi) = \langle \mathfrak{F}[u],\varphi\rangle &= \int_{\Omega}{F(Du,u,x)}{\varphi(x)}{\dm x}\\ &= -\int_{\Omega} \big\langle\mathbf{A}(x)\nabla u, \nabla \varphi\big\rangle {\dm x} + \int_{\Omega} [c(x)u-f(x)] \varphi {\dm x}. \end{split} $$ Thus $$ \begin{split} \left\langle\frac{\delta\mathfrak{F}}{\delta u }[u](u_0),u_1\right\rangle =\left\langle\frac{\delta\mathfrak{F}}{\delta u }(u_0),u_1\right\rangle &= -\int_{\Omega} \big\langle\mathbf{A}(x)\nabla u_0, \nabla u_1 \big\rangle {\dm x} + \int_{\Omega} c(x)u_0u_1 {\dm x}\\ & = -\int_{\Omega} \big\langle\mathbf{A}(x)\nabla u_1, \nabla u_0 \big\rangle {\dm x} + \int_{\Omega} c(x)u_1u_0 {\dm x} \\ &= \left\langle\frac{\delta\mathfrak{F}}{\delta u }(u_1),u_0\right\rangle =\left\langle\frac{\delta\mathfrak{F}}{\delta u }[u](u_1),u_0\right\rangle \end{split} $$ thus \eqref{3} is verified (I dropped the dependence on the generic $u\in E$ as $\mathfrak{F}$ is a linear functional of this argument: the condition is not restricted to linear operators). In turn the symmetry/self-adjointness of $F$ is equivalent to \eqref{3} which is equivalent to the fact that $F$ is the Euler-Lagrange operator associated to a given Lagrangian function $L$ solving the operator equation $\mathcal{E} L =F$. For the second part of the question, formula \eqref{5} gives the exact form of the Lagrangian function $L$, which happens to be the same one you asked about, i.e. $$ L(Du,u,x)=\frac{c(x)u^2}{2}-f(x)u-\text{Tr}\left(\mathbf A(x)(\nabla u \otimes \nabla u) \right) + L_0 $$ where the involved quantities have the same meaning as above.

References

[1] Vladimir Mikhailovich Filippov, Variational principles for nonpotential operators. With an appendix by the author and V. M. Savchin, Translated 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, Zbl 0682.35006.

[2] Mikhail Mordukhovich Vaĭnberg, Variational methods for the study of nonlinear operators. With a chapter on Newton’s method by L.V. Kantorovich and G.P. Akilov, translated and supplemented by Amiel Feinstein, Holden-Day Series in Mathematical Physics. San Francisco-London- Amsterdam: Holden-Day, Inc. pp. x+323 (1964), MR0176364, ZBL0122.35501.

[3] Enzo Tonti, "Extended variational formulation", Vestnik Rossiĭskogo Universiteta Druzhby Narodov, Seriya Matematika 2, No. 2, 148-162 (1995). Zbl 0965.35036.