Let $I$ be the functional defined by $$I[u]=\int_{a}^b L(x,u(x),u(x)')\,dx$$ where $a$ and $b$ are constants and $u'=du/dx$. The Beltrami identity states that if $\partial L/\partial x=0$, then the Euler-Lagrange equation $$ \frac{\partial L}{\partial u}-\frac{d}{dx}\left(\frac{\partial L}{\partial u'}\right)=0$$ can be integrated once, giving rise to $$ L-u'\frac{\partial L}{\partial u'}=\text{constant}.$$
Now assume that $I$ is a functional of two functions that depends on two variables, that is $$ I[u,v]=\int_a^b\int_c^d L(x,y,u(x,y),v(x,y),u_{,x}(x,y),u_{,y}(x,y),v_{,x}(x,y),v_{,y}(x,y))\,dxdy,$$ where $a$, $b$, $c$ and $d$ are constants and $u_{,x}=\partial u/\partial x$, etc. If $\partial L/\partial x=0$ (or $\partial L/\partial y=0$). Is there any identity analogous to the Beltrami identity for the Euler-Lagrange equation?
Well, for a Lagrangian density ${\cal L}(x,u(x),\partial u(x))$ in $n$ variables $\{x^1,\ldots,x^n\}$ with no explicit dependence $$\frac{\partial {\cal L}}{\partial x^k}~=~0$$ along the $k$'th coordinate, there is a corresponding continuity equation $$ \sum_{j=1}^n\frac{d T^{j}{}_k}{dx^j}~=~0,$$ where the canonical stress-energy-momentum tensor is $$ T^{j}{}_{\ell}~:=~\frac{\partial {\cal L}}{\partial u_{,j}} u_{,\ell}-\delta^j_{\ell} {\cal L}. $$ One can in principle create a conserved quantity by integrating $T^{j}{}_k$ over (an appropriate region of) the hyperplane $\{x^j={\rm const}\}$, cf. Noether's theorem. Note that the choice of hyperplane is not unique.