# Euler lagrange equation is a constant

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.

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It would be extremely helpful if you post the functionals. If you did all the calculations right means that the functional has no stationary value, for a contradiction you found. –  Jorge May 18 '13 at 17:10
Please post the integral –  Occupy Gezi May 18 '13 at 17:18
The functional concerned is: $$L[u]=\int(f(u)u_x+u) dx$$, where $f$ may be any function. It is always the case that 1=0. Does this indeed mean that no stationary point exists? –  user45503 May 18 '13 at 17:20

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

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

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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?)

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There's no need for that attitude. –  Javier May 19 '13 at 19:57
Well, the commentary about "can no one do calculus of variations any more?" and the "Do you even math?" would be pretty good candidates. There's no reason to talk like that to someone who is having trouble with an exercise. I would be very surprised if you never made a single mistake in your math ever. –  Javier May 19 '13 at 20:22