\quad\quad设有界区域 \Omega \subset \mathbb{R}^n ，f \in C^1\left(\Omega \times \mathbb{R} \times \mathbb{R}^n \right)，泛函 E(u) 定义为E(u)=\int_{\Omega} f(x, u(x), \nabla u(x)) \mathrm{d} x, 且满足\quad\quad(i) 存在 a \in L^1(\Omega)，b,c>0，p>1，使得对任意 x,u,\xi，f(x,u,\xi) \geqslant a(x)+b|u|^p+c|\xi|^p；\quad\quad(ii) 存在 u_0 \in W^{1,p}(\Omega) 使得 E(u_0) < \infty；\quad\quad(iii) 存在常数 a',a''>0，使得对任意 u,\xi，\begin{aligned}\left| f_u(x, u, \xi) \right|& \leqslant a^{\prime}\left(1+|u|^{p-1}+|\xi|^{p}\right), \\ \left|\nabla_{ \xi} f(x, u, \xi)\right| & \leqslant a^{\prime \prime}\left(1+|u|^{p}+|\xi|^{p-1}\right), \end{aligned} \quad \quad \mathrm{ a.e.} \ x, 那么对于 u \in W^{1,p}(\Omega)，记 E(u) 的 Gâteaux 导数为 E'(u)，则 E'(u)=0 等价于 \frac{\partial }{\partial u}f(x, u, \nabla u)-\sum_{k=1}^{n} \frac{\partial}{\partial x_{k}}\left(\frac{\partial }{\partial \xi_{k}}f(x, u, \nabla u)\right) =\frac{\partial f}{\partial u} - \mathrm{div}\left(\frac{\partial f}{\partial \nabla u}\right)=0. \quad\quad如果 E(u) 是凸的，则 E'(u)=0 的解也是 E(u) 极小化问题的解. E'(u)=0 即为 Euler-Lagrange 方程. 更一般地，如果泛函 E(u) 定义为E(u)=\int_{\Omega} f(x, u, \nabla u, D^2 u, \cdots, D^i u, \cdots) \mathrm{d} x, 则 Euler-Lagrange 方程的形式为 \dfrac{\partial f}{\partial u} + \sum_i \sum_{\mu_1,\cdots,\mu_i} (-1)^i \dfrac{\partial ^i}{\partial x_{\mu_i}\cdots\partial x_{\mu_1}}\left(\dfrac{\partial f}{\partial u_{\mu_1 \cdots \mu_i}}\right) = 0 . \quad\quad 接下来给出几个由泛函求 Euler-Lagrange 方程的例子.
\quad\quad例 1\quad\quad E(u)=\displaystyle \int_{\Omega}\nabla u^{\mathrm{T}}A \nabla u \mathrm{d} x + \lambda \int_{\Omega} (u-f)^2 \mathrm{d} x.\quad\quad由 E'(u)=0 即 2\lambda (u - f) - \mathrm{div} \left(2A \nabla u\right) = 0 得到 Euler-Lagrange 方程为 -\mathrm{div} \left(A \nabla u \right) + \lambda u = \lambda f.
\quad\quad例 2\quad\quad E(u)=\dfrac{1}{p(x)} \displaystyle \int_{\Omega}\left|\nabla u\right|^{p(x)} \mathrm{d} x + \dfrac{\lambda}{2} \int_{\Omega} (u-f)^2 \mathrm{d} x. \quad\quad由 E'(u)=0 即 \lambda (u - f) - \mathrm{div} \left(|\nabla u|^{p(x)-1} \dfrac{\nabla u}{|\nabla u|}\right) = 0 得到 Euler-Lagrange 方程为 -\mathrm{div} \left(|\nabla u|^{p(x)-2} \nabla u\right) + \lambda u = \lambda f.
\quad\quad例 3\quad\quad E(u)=\displaystyle \int_{\Omega}|\nabla u|-\dfrac{1}{K} \log (1 + K \nabla u) \mathrm{d} x + \dfrac{\lambda}{2} \int_{\Omega} (u-f)^2 \mathrm{d} x.\quad\quad由 E'(u)=0 即 \lambda (u - f) - \mathrm{div} \left(\dfrac{\nabla u}{|\nabla u|}-\dfrac{1}{1+K|\nabla u|}\dfrac{\nabla u}{|\nabla u|}\right) = 0 得到 Euler-Lagrange 方程为 -\mathrm{div} \left(\dfrac{K\nabla u}{1+K|\nabla u|} \right) + \lambda u = \lambda f.
\quad\quad例 4\quad\quad E(u)=\displaystyle \int_{\Omega}\sqrt{1+|\nabla u|} \mathrm{d} x + \dfrac{\lambda}{2} \int_{\Omega} (u-f)^2 \mathrm{d} x.\quad\quad由 E'(u)=0 即 \lambda (u - f) - \mathrm{div} \left(\dfrac{|\nabla u|}{\sqrt{1+|\nabla u|}}\dfrac{\nabla u}{|\nabla u|}\right) = 0 得到 Euler-Lagrange 方程为 -\mathrm{div} \left(\dfrac{\nabla u}{\sqrt{1+|\nabla u|}} \right) + \lambda u = \lambda f.
\quad\quad例 5\quad\quad E(u)=\displaystyle \int_{\Omega}\dfrac{|\nabla u|^2}{1+|\nabla u|} \mathrm{d} x + \dfrac{\lambda}{2} \int_{\Omega} (u-f)^2 \mathrm{d} x.\quad\quad由 E'(u)=0 即 \lambda (u - f) - \mathrm{div} \left(\dfrac{|\nabla u|^2+2|\nabla u|}{\left(1+|\nabla u|\right)^2}\dfrac{\nabla u}{|\nabla u|}\right) = 0 得到 Euler-Lagrange 方程为 -\mathrm{div} \left(\dfrac{2+|\nabla u|}{\left(1+|\nabla u|\right)^2} \nabla u \right) + \lambda u = \lambda f.
\quad\quad例 6\quad\quad E(u)=\displaystyle \int_{\Omega}f(|\Delta u|) \mathrm{d} x，Euler-Lagrange 方程为 \Delta\left(f'(|\Delta u|)\dfrac{\Delta u}{|\Delta u|}\right)=0.
\quad\quad例 7\quad\quad E(u)=\displaystyle \int_{\Omega}\left|u_{x x}\right|+\left|u_{y y}\right| \mathrm{d} x \mathrm{d} y+\dfrac{\lambda}{2} \int_{\Omega}(u-f)^{2} \mathrm{d} x \mathrm{d} y.\quad\quad由 E'(u)=0 即 \lambda (u - f) +\left(\dfrac{u_{x x}}{\left|u_{x x}\right|}\right)_{x x}+\left(\dfrac{u_{y y}}{\left|u_{y y}\right|}\right)_{y y} = 0 得到 Euler-Lagrange 方程为 \left(\dfrac{u_{x x}}{\left|u_{x x}\right|}\right)_{x x}+\left(\dfrac{u_{y y}}{\left|u_{y y}\right|}\right)_{y y} + \lambda u = \lambda f.
\quad\quad例 8\quad\quad E(u)=\displaystyle \int_{\Omega}\left|D^2u\right| \mathrm{d} x \mathrm{d} y+\dfrac{\lambda}{2} \int_{\Omega}(u-f)^{2} \mathrm{d} x \mathrm{d} y.\quad\quad由 E'(u)=0 即 \lambda (u - f) +\left(\dfrac{u_{x x}}{\left|D^2u\right|}\right)_{x x}+\left(\dfrac{u_{x y}}{\left|D^2u\right|}\right)_{y x}+\left(\dfrac{u_{y x}}{\left|D^2u\right|}\right)_{x y}+\left(\dfrac{u_{y y}}{\left|D^2u\right|}\right)_{y y} = 0 得到 Euler-Lagrange 方程为 \left(\dfrac{u_{x x}}{\left|D^2u\right|}\right)_{x x}+\left(\dfrac{u_{x y}}{\left|D^2u\right|}\right)_{y x}+\left(\dfrac{u_{y x}}{\left|D^2u\right|}\right)_{x y}+\left(\dfrac{u_{y y}}{\left|D^2u\right|}\right)_{y y} + \lambda u = \lambda f.