Euler-Lagrange equation for a functional

Question

What does it mean that the equation: $$ \text{div}_{x,y}(y^a\nabla_{x,y}u)=0,\quad \text{in }\mathbb{R}^n\times(0,\infty),$$ is the Euler-Lagrange equation for the functional: $$ J(u)=\int_{\mathbb{R}^n\times (0,\infty)}y^a|\nabla_{x,y}u(x,y)|^2\,dx\,dy?$$ Morover there exist a Banach space on which is defined $J$? I find this terminology in "An extension problem related to the fractional Laplacian" by Caffarelli-Silvestre at page 2.

Answer 1

It means that the functional derivative of $J(u)$ is zero (i.e. $J(u)$ has $u$ as a stationary point) if the function $u$ solves the given divergence form PDE, i.e. $$\DeclareMathOperator{\divg}{\mathrm{div}} \DeclareMathOperator{\grad}{\nabla} \bigg{[}\frac{\mathrm{d}}{\mathrm{d}\epsilon}J(u+\epsilon \varphi)\bigg{]}_{\epsilon = 0} =\lim_{\epsilon \to 0}\frac{J(u+\epsilon \varphi)-J(u)}{\epsilon} = 0\iff\divg_{x,y} \big(y^a \grad_{x,y} u\big)=0 $$ for all test function $\varphi\in C^1_c(\Bbb R^n\times [0,+\infty)\big)$, where $$ \bigg{[}\frac{\mathrm{d}}{\mathrm{d}\epsilon}J(u+\epsilon \varphi)\bigg{]}_{\epsilon = 0} = 2\int\limits_{\Bbb R^n \times [0,+\infty)} \!\!\!\!\! y^a\langle\grad_{x,y}u(x,y), \grad_{x,y}\varphi(x,y)\rangle\,\mathrm{d}x\,\mathrm{d}y\label{1}\tag{FD} $$ The given PDE is equivalent to the vanishing of the functional (as a matter of fact, Gâteaux or Frechet) derivative \eqref{1}, provided $u$ is $C^2$-smooth, has a sufficiently smooth boundary value on $y=0$, as can be seen by applying Du Boys-Reymond's lemma: this is classically known and due to Euler and Lagrange, and thus PDEs that come as necessary conditions for the vanishing of the functional derivative of a given (integral) functional are called "Euler-Lagrange equations" of the given functional.

Edit following the comments

As a follow-up of the discussion started on comments on how to prove the uniqueness of the solution of the following Dirichlet problem $$ \begin{cases} \divg_{x,y} \big(y^a \grad_{x,y} u\big)=0\\ u|_{y=0}=u_0\in \mathscr{S}(\Bbb R^n) \end{cases}\label{2}\tag{1} $$ where $\mathscr{S}(\Bbb R^n)$ is the ordinary Schwarz space, it may be noted that the functional $J(u)$ could be defined naturally on the following weighted Sobolev space $$ \begin{split} H^1_{y^a}\big(\Bbb R^n\times [0,+\infty)\big) & \equiv W^{1,2}_{y^a} \big(\Bbb R^n\times [0,+\infty)\big) \\ & \triangleq\Big\{u\in L^2\big(\Bbb R^n\times [0,+\infty)\big) : \|u_{0}\|_{L^2(\Bbb R^n)} + J(u) < +\infty\Big\} \\ \end{split} $$ (here we have put $u_0=u_{y=0}$ in order to simplify the notation) which is naturally Hilbert when equipped with the following inner product $$ \langle u , v \rangle_{H^1_{y^a}}= \langle u_0 , v_0 \rangle_{L^2(\Bbb R^n)} +\int\limits_{\Bbb R^n \times [0,+\infty)} \!\!\!\!\! y^a\langle\grad_{x,y}u(x,y), \grad_{x,y}v(x,y)\rangle\,\mathrm{d}x\,\mathrm{d}y $$ (proving the completeness of this space does not seem to be too difficult). Using this Hilbert space for studying problem \eqref{2} seems to be quite natural, since

• $\mathscr{S}(\Bbb R^n)\subsetneq L^2(\Bbb R^n)$ so it is possible to efficiently deal with the boundary condition, and
• $C_c^\infty(\Bbb R^n)\times C_c^\infty\big( \mathbb{R}^n\times[0,\infty)\big)$ is dense in $H^1_{y^a}$ as this spaces explicitly deals with boundary conditions and thus
• with the Hilbert space structure and the density of it is possible to use well known methods to prove existence and uniqueness of solution $u$.