Show that the Euler–Lagrange equation for the functional:
$$I(y) = \int_{0}^{1}y dx$$
subject to y(0) = y(1) = 0 has no solutions. Explain why no extremum for I exists.
When forming the E-L equation I get 1=0. How would I go about doing this question?
On
So you've answered your own question. You obtain 1=0 which obviously doesn't have any solutions. There is no reason why this should have an extremum either. Here is a proof (ish) Take y to be any function from [0,1] to the real line satisfying the condition. Then if we add a function delta to y (with the right boundary conditions) it will change I up to first order in the integral of delta so if y where a minimum then adding a delta with a small negative integral would make I smaller and therefore it can't be a minimum. The same happens with a maximum.
If you consider the function $f_n$, where $f_n(x) = n, x \in [\frac{1}{n}, 1 -\frac{1}{n}]$, $f_n(x) = n^2x, x\in [0, \frac{1}{n}],$ and $f_n(x) = n^2(1-x), x\in [1 -\frac{1}{n}, 1],$ then clearly $I(f_n)$ can be as large as you want. $I(-f_n)$ can also be as large as you want.