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?

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    $\begingroup$ What is $\frac{\partial F}{\partial y}$ equal to? $\endgroup$ – jacob1729 Mar 6 at 15:24
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    $\begingroup$ @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 ? $\endgroup$ – Jim Β Mar 6 at 15:34
  • $\begingroup$ @JimΒ I made my comment before that post. Do you see why that is a contradiction then? $\endgroup$ – jacob1729 Mar 6 at 16:13
  • $\begingroup$ @jacob1729 yeah I do . Thank you and the OP of the answer below for your input. $\endgroup$ – Jim Β Mar 6 at 16:14
  • $\begingroup$ Related Math.SE question: math.stackexchange.com/q/3136864/11127 $\endgroup$ – Qmechanic Mar 7 at 2:10

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


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