$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$

Now we consider functionals, defined on functions of several variables.

Let us consider functional
\begin{equation}
\Phi[u]= \iiint_\Omega L(x, u,\nabla u)\,dx
\label{eq-10.3.4}
\end{equation}
where $\Omega$ is $n$-dimensional domain and $L$ is some function of $n+2$ variables. Let us consider $u+\delta u$ where $\delta u $ is a "small" function. We do not formalize this notion, just $\varepsilon \phi$ with fixed $\phi$ and $\varepsilon\to 0$ is considered to be small. We call $\delta u$ *variation* of $u$ and important is that we change a function as a whole object. Let us consider
\begin{multline}
\Phi[u+\delta u]-\Phi[u]=
\iiint_\Omega \Bigl(L(x,u+\delta u,\nabla u +\nabla \delta u)-L(x, u,\nabla u)
\Bigr)\,dx\\
\approx \iiint_\Omega \Bigl(\frac{\partial L}{\partial u}\delta u +\sum_{1\le j\le n} \frac{\partial L}{\partial u_{x_j}}\delta u_{x_j}
\Bigr)\,dx\qquad
\label{eq-10.3.5}
\end{multline}
where we calculated the linear part of expression in the parenthesis; if $\delta u=\varepsilon \phi$ and all functions are sufficiently smooth then $\approx$ would mean "equal modulo $o(\varepsilon)$ as $\varepsilon\to 0$".

**Definition 1.**

- Function $L$ we call
*Lagrangian*. - The right-hand expression of (\ref{eq-10.3.5}) which is a linear functional with respect to $\delta u$ we call
*variation of functional $\Phi$*and denote by $\delta \Phi$.

**Assumption 1.**
All functions are sufficiently smooth.

Under this assumption, we can integrate the right-hand expression of (\ref{eq-10.3.5}) by parts: \begin{multline} \delta \Phi:= \iiint_\Omega \Bigl(\frac{\partial L}{\partial u}\delta u +\sum_{1\le j\le n} \frac{\partial L}{\partial u_{x_j}}\delta u_{x_j} \Bigr)\,dx\\ = \iiint_\Omega\Bigl(\frac{\partial L}{\partial u} - \sum_{1\le j\le n} \frac{\partial\ }{\partial x_j} \frac{\partial L}{\partial u_{x_j}} \Bigr)\delta u \,dx - \iint_{\partial \Omega} \Bigl(\sum_{1\le j\le n} \frac{\partial L}{\partial u_{x_j}}\nu_j \Bigr)\delta u \,d\sigma\qquad \label{eq-10.3.6} \end{multline} where $d\sigma$ is an area element and $\nu$ is a unit interior normal to $\partial \Omega$.

**Definition 2.**
If $\delta \Phi=0$ for all *admissible variations* $\delta u$ we call $u$ a *stationary point* or *extremal* of functional $\Phi$.

**Remark 1.**

- We consider $u$ as a point in the functional space.
- In this definition we did not specify which variations are admissible. Let us consider as admissible all variations which are $0$ at the boundary: \begin{equation} \delta u |_{\partial\Omega}=0. \label{eq-10.3.7} \end{equation} We will consider different admissible variations later.

In this framework \begin{equation} \delta \Phi= \iiint_\Omega\Bigl(\frac{\partial L}{\partial u} - \sum_{1\le j\le n} \frac{\partial\ }{\partial x_j} \frac{\partial L}{\partial u_{x_j}} \Bigr)\delta u \,dx . \label{eq-10.3.8} \end{equation}

**Lemma 1.**
Let $f$ be a continuous function in $\Omega$.
If $\iiint_\Omega f(x)\phi(x)\,dx=0$ for all $\phi$ such that
$\phi|_{\partial \Omega}=0$ then $f=0$ in $\Omega$.

*Proof.* Indeed, let us assume that $f(\bar{x})> 0$ at some point
$\bar{x}\in \Omega$ (case $f(\bar{x})< 0$ is analyzed in the same way). Then $f(x)>0$ in some vicinity $\mathcal{V}$ of $\bar{x}$. Consider function $\phi(x)$ which is $0$ outside of $\mathcal{V}$, $\phi\ge 0$ in $\mathcal{V}$ and $\phi(\bar{x})>0$. Then $f(x)\phi(x)$ has the same properties and $\iiint_{\Omega} f(x)\phi(x)\, dx>0$. Contradiction!

As a corollary we arrive to

**Theorem 2.**
Let us consider a functional (\ref{eq-10.3.4}) and consider as admissible all $\delta u$ satisfying (\ref{eq-10.3.7}). Then $u$ is a stationary point of $\Phi$ if and only if it satisfies *Euler-Lagrange equation*
\begin{equation}
\frac{\delta \Phi}{\delta u}:=
\frac{\partial L}{\partial u} - \sum_{1\le j\le n} \frac{\partial\ }{\partial x_j} \left(\frac{\partial L}{\partial u_{x_j}}\right) =0.
\label{eq-10.3.9}
\end{equation}

**Remark 2.**

- This is a second order PDE.
- We do not have statements similar to Remark 10.1.4 [2]-[4], see also Remark 10.2.2

**Definition 3.**
If $\Phi[u]\ge \Phi[u+\delta u]$ for all small admissible variations $\delta u$ we call $u$ a *local maximum* of functional $\Phi$. If
$\Phi[u]\le \Phi[u+\delta u]$ for all small admissible variations $\delta u$ we call $u$ a *local minimum* of functional $\Phi$.

Here again we do not specify what is *small admissible variation*.

**Theorem 3.**
If $u$ is a *local extremum* (that means either local minimum or maximum) of $\Phi$ and variation exits, then $u$ is a stationary point.

*Proof.*
Consider case of minimum. Let $\delta u =\varepsilon \phi$. Then
$\Phi [u+\delta u]- \Phi [u]=\varepsilon (\delta \Phi)(\phi) +o(\varepsilon)$. If $\pm \delta \Phi> 0$ then choosing $\mp \varepsilon <0$ we make
$\varepsilon (\delta \Phi)(\phi)\le -2\epsilon\_0 \varepsilon$ with some $\epsilon\_0>0$. Meanwhile for sufficiently small $\varepsilon$ "$o(\varepsilon)$" is much smaller and $\Phi [u+\delta u]- \Phi [u]\le -\epsilon_0 \varepsilon<0$ and $u$ is not a local minimum.

**Remark 3.**
We consider neither sufficient conditions of extremums nor *second variations* (similar to second differentials). In some cases they will be obvious.

**Example 2.**

- Consider a surface $\Sigma = \{(x,y,z):\, (x,y)\in \Omega , z=u(x,y)\}$ which has $(x,y)$-projection $\Omega$. Then the surface area of $\Sigma$ is
\begin{equation}
A(\Sigma)= \iint_{\Omega} \bigl(1+u_x^2+u_y^2\bigr)^{\frac{1}{2}}\,dxdy.
\label{eq-10.3.10}
\end{equation}
We are interested in such surface of minimal area (a.k.a.
*minimal surface*) under restriction $u=g$ at points $\partial \Omega$. It is a famous*minimal surface problem*(under the assumption that it projects nicely on $(x,y)$-plane (which is not necessarily the case). One can formulate it: find the shape of the soap film on the wire.

Then Euler-Lagrange equation is \begin{equation} -\frac{\partial\ }{\partial x} \Bigl(u_x\bigl(1+u_x^2+u_y^2\bigr)^{-\frac{1}{2}}\Bigr)- \frac{\partial\ }{\partial y} \Bigl(u_y\bigl(1+u_x^2+u_y^2\bigr)^{-\frac{1}{2}}\Bigr)=0. \label{eq-10.3.11} \end{equation} 2. Assuming that $u_x, u_y \ll 1$ one can approximate $A(\Sigma)-A(\Omega)$ by \begin{equation} \frac{1}{2}\iint_{\Omega} \bigl(u_x^2+u_y^2\bigr)\,dxdy \label{eq-10.3.12} \end{equation} and for this functional Euler-Lagrange equation is \begin{equation} -\Delta u=0. \label{eq-10.3.13} \end{equation} 3. Both [1] and [2] could be generalized to higher dimensions.

**Remark 4.**
Both variational problem and equation come with the boundary condition $u|_{\partial\Omega}=g$. In the next section we analyze the case when such condition is done in the original variational problem only on the part of the boundary.