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In the following we consider $\Omega\subset\mathbb{R}^n\ (n\geq2)$ to be open, bounded and with Lipschitz boundary. Consider the following regular variational integral: \begin{equation*} I[u]=\int_{\Omega}f(x, u, Du)\ \mathrm{d}x, \end{equation*}for all $u$ in the admissible class $W^{1, n}\cap L^{\infty}(\Omega;\mathbb{R}^N)$. Here $N$ is at least $1$, the integral is at least Caratheodory, coercive and $|f(x, z, \zeta)|\leq K|\zeta|^n$ for some $K>0$ and all $(x, z, \zeta)\in\Omega\times\mathbb{R}^N\times\mathbb{R}^{Nn}$.

There are clearly a wide number of problems framed in the above way. I was wondering if there is an example of a variational problem framed in the above way that has a non-constant continuous minimiser, $v$ and a non-constant critical point (not necessarily, but preferably continuous) $u$, that is distinct from $v$. The admissible class can be made smaller if it suits the example. I am looking for explicit examples and not existence theorems.

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  • $\begingroup$ Consider $$ I(u)= \frac{1}{2} \int_\Omega | \nabla u|^2 + \frac{1}{4} \int_\Omega u^4 - \frac{\lambda}{2} \int_\Omega u^2$$ where $ \lambda>\lambda_1 $ (first eigenvalue of $ -\Delta $). This should have two non-zero minimizers and $0$ should also be a critical point... $\endgroup$
    – Math604
    Commented Sep 27, 2015 at 20:06
  • $\begingroup$ Sorry, in the question I meant that both the critical point and the minimiser's are non-constant...Also am I correct in assuming that $\Omega\subset\mathbb{R}^2$ in the above example? $\endgroup$
    – Nirav
    Commented Sep 27, 2015 at 22:06
  • $\begingroup$ any dimension is fine.. $\endgroup$
    – Math604
    Commented Sep 27, 2015 at 22:57

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