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In this section one can find exam papers of 2017. Users are strongly advised to post comments, hints, full solutions.

Exam paper #5 added.

Exam paper #6 added.

Attached are some hints and partial solutions (exam paper 2017-1 coincides with exam paper 2018-1 ).

Hello! I believe there's a mistake in the 5th exam paper's solution. Probably I am wrong, but at least I want to understand why.
In the 3rd exercise, part (a), case \(l=\pi\), there is the inequality \(-\cos(a)+\cos(b) \geq \frac{a^2-b^2}{8}\) for any \((a,b)\) in a suitable neighborhood of \((0,0)\).
I believe this is wrong.
Indeed, considering WLOG neighborhood = \(B_{\delta} ((0,0))\), we can swap \(a\) and \(b\).
Therefore we have \(-\cos(a)+\cos(b) \geq \frac{a^2-b^2}{8}\) and \(-\cos(b)+\cos(a) \geq \frac{b^2-a^2}{8}\), thus it must be an equality.
So there exists a constant \(k\) such that for any \(a\) with \(|a|\) small enough we have \(\cos(a) +\frac{a^2}{8} = k\), which is clearly false.

Moreover I believe that, for \(l=\pi, u_0\) is not a (WLM).

That's because (just outlining the main points) if \(u_0\) were a (WLM), then \(u_{\epsilon}(x) = \epsilon \cdot sin(x)\) would be another (WLM) for \(\epsilon\) small enough (\(F(u_{\epsilon}) = F(u_0)\) because of the parity of \(cos(\cdot)\)), but \(u_{\epsilon}\) doesn't satisfy (ELE) for any \(\epsilon \neq 0\) (but it should be easier to prove there is a small enough value of \(\epsilon\)).

LucaMac wrote:I believe there's a mistake in the 5th exam paper's solution.

Absolutely right That inequality holds true only when \(a^2\geq b^2\).

You also provided a nice proof that \(u_0\) is not a WLM.

A possible alternative approach is considering competitors of the form \(u_\varepsilon(x)=\varepsilon\sin x\pm\varepsilon^2 v(x)\), where \(v(x)\) is any function that is zero at the boundary and satisfies

\(\displaystyle\int_0^\pi\left(\cos x\cdot v'(x)-\sin x\cdot v(x)\right)\,dx\neq 0\).

It should not be difficult to show the existence of such a \(v(x)\) (but one has to "break the symmetry").

Massimo Gobbino wrote:where \(v(x)\) is any function that is zero at the boundary and satisfies

\(\displaystyle\int_0^\pi\left(\cos x\cdot v'(x)-\sin x\cdot v(x)\right)\,dx\neq 0\).

It should not be difficult to show the existence of such a \(v(x)\)

This is not true ... the integral is always 0 because it is a Null Lagrangian ...