I have a problem at hand that I'm trying to solve with the Euler-Lagrange equation, but I think I made a mistake somewhere.
Supposed I have two functions on two real variables, $u(x, t)$ and $v(x,t)$. I am trying to find a curve $x(t)$, with the boundary condition $x(0)=1$ and $x(1)=0$, such that the following area in the $uv$-plane is minimized.
\begin{align*}
A = \int v \, du = \int_{0}^1 v \left(\frac{\partial u}{\partial t} + \frac{\partial u}{\partial x} \dot{x} \right) \, dt
\end{align*}
For my problem, $u, v \geq 0$ for all $(x,t)$, and there are some "forbidden region" (something like a circle, but I'm trying to find its exact description) in the $uv$-plane containing the origin. I also have $u(1,0)=0$, $v(1,0)=v_0 > 0$, and $u(0,1)=u_0>0$, $v(0,1)=0$, so the boundary of the forbidden region should be the minimum curve. The Lagrangian is therefore
\begin{align*}
L(\dot{x}, x, t) = v \left(\frac{\partial u}{\partial t} + \frac{\partial u}{\partial x} \dot{x} \right)
\end{align*}
Then, the Euler-Lagrange equation is
\begin{align*}
\frac{\partial L}{\partial x} = \frac{d}{dt} \frac{\partial L}{\partial \dot{x}}
\end{align*}
For the RHS,
\begin{align*}
\frac{\partial L}{\partial \dot{x}} &= v \frac{\partial u}{\partial x}
\end{align*}
Then taking the full derivative,
\begin{align*}
\frac{d}{dt} \frac{\partial L}{\partial \dot{x}} &= \frac{d}{dt} \left(v \frac{\partial u}{\partial x} \right) \\
&= \frac{\partial}{\partial t} \left(v \frac{\partial u}{\partial x} \right) + \frac{\partial}{\partial x} \left(v \frac{\partial u}{\partial x} \right) \dot{x}
\end{align*}
For the LHS,
\begin{align*}
\frac{\partial L}{\partial x} = \frac{\partial}{\partial x} \left( v \frac{\partial u}{\partial t} \right) + \frac{\partial}{\partial x} \left( v \frac{\partial u}{\partial x} \right) \dot{x}
\end{align*}
This is already starting to feel wrong, as I am expecting a first-order differential equation. However, when I equate LHS=RHS, the term with $\dot{x}$ gets canceled off. In fact, upon further simplification, the second order terms get canceled too and I get
\begin{align*}
\frac{\partial v}{\partial x} \frac{\partial u}{\partial t} = \frac{\partial v}{\partial t} \frac{\partial u}{\partial x}
\end{align*}
which is certainly not true in general, since I haven't actually specified $u$ and $v$. So which step went wrong?
Best Answer
Denote $$ M(t,x)=v\frac{\partial u}{\partial t} ,\quad N(t,x)=v\frac{\partial u}{\partial x}. $$ You are to minimize $$ A=\int_0^1(M(t,x)+N(t,x)\dot x)\,dt $$ and get EL equation as $$ \frac{\partial M}{\partial x}=\frac{\partial N}{\partial t}.\tag{*} $$ There is no mistake. The Lagrangian $M+N\dot x$ is degenerate. You may see that $\dot x$ can be made as close to $+\infty$ or $-\infty$ on the interval as you wish, so (provided $N\ne 0$ at some point) the functional has no minimum. With one exception: it is constant (independent on $x$), and it happens with very specific $M$ and $N$ - those that satisfy the EL equation above. To see that: use Green's theorem to the curve integral $$ A=\int M\,dt+N\,dx $$ to claim it is path-independed when (*) holds.