Euler lagrange equation is a constant

Question

I'm working through exercises which require me to find the Euler-Lagrange equation for different functionals. I've just come across a case where the Euler Lagrange equation simplifies to $$1=0.$$ Please could someone explain what can be concluded about the set of extremals to problems where this is the case.

Answer 1

Notation $$L[u]=\int(f(u)u_{x}+u)dx=\int F(u,u_{x})dx $$

$$\frac{d}{dx}\left( \frac{\partial F}{\partial u_{x}}\right)-\frac{\partial F}{\partial u}=0 $$

As $F(u,u_{x})=f(u)u_{x}+u$ we have

$$\frac{\partial F}{\partial u_{x}}=f(u) $$

$$ \frac{\partial F}{\partial u}=\frac{df}{du}\frac{du}{dx}+1$$

so $$ $$

Hence $$\frac{d}{dx}(f(u)+1)-\frac{df(u)}{du}\frac{du}{dx}=0 $$

$$\frac{df}{du}\frac{du}{dx}-\frac{df}{du}\frac{du}{dx}=0 \rightarrow 0=0$$

Answer 2

This is breeze can no man do calculus of variations no more? Kmt

$$ L[u]=\int(f(u)u_{x}+u)dx $$

$$ \frac{d}{dx}\left( \frac{\partial F}{\partial u_{x}}\right)-\frac{\partial F}{\partial u}=0 $$

$$ f'(u)u_x +1 -f'(u)u_x=0 $$

$$ 1=0 $$

so for any $u$, the E-L equations is 1=0, thus the set of extremals is empty - hence the functional has no stationary value - consider the graph of $e^x$. Infimum, not minimum. (Do you even math?)