# Proving that the Euler-Lagrange Equation has no solution [closed]

I'm trying to show that the Euler - Lagrange equation for the functional $$I(y)=\int_{a}^{b} y\:dx$$

subject to $$y(0)=y(1)=0$$ has no solutions.

The Euler - Lagrange equation states that: $$\frac{d}{dx}\frac{\partial F}{\partial y'}-\frac{\partial F}{\partial y}=0$$

For this specific problem, $$F=y$$. Hence, $$\frac{\partial F}{\partial y'}=0$$. So: $$-\frac{\partial F}{\partial y}=0$$

I don't understand why there's no solution when one could say that $$y=c$$, where $$c$$ is a constant. I understand that $$I(y)$$ does not have an extremum but how do I prove that the Euler Lagrange equation has no solution?

## closed as off-topic by JMac, GiorgioP, Kyle Kanos, ZeroTheHero, user191954 Mar 9 at 14:33

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• What is $\frac{\partial F}{\partial y}$ equal to? – jacob1729 Mar 6 at 15:24
• @jacob1729 I mean, $F=y$, so the derivative would just be $1$ and that would lead to $-1=0$ ( as posted below). Does that prove that the Euler - Lagrange equation has no solution ? – Jim Β Mar 6 at 15:34
• @JimΒ I made my comment before that post. Do you see why that is a contradiction then? – jacob1729 Mar 6 at 16:13
• @jacob1729 yeah I do . Thank you and the OP of the answer below for your input. – Jim Β Mar 6 at 16:14
• Related Math.SE question: math.stackexchange.com/q/3136864/11127 – Qmechanic Mar 7 at 2:10

$$\partial F/\partial y$$ is not an unknown, it is equal to 1, so your equation is $$-1=0$$.