Suppose I have the energy functional given by $$\Pi(u) : = \frac{1}{2} \int_{\Omega} k\nabla u \cdot \nabla u dx - \int_{\Omega} fu dx + \int_{\partial \Omega} \sigma_0 u^2 ds,$$ where $f \in L^2(\Omega)$ and $\sigma_0 \in C^{\infty}(\partial \Omega)$ and $\partial \Omega$ denotes the boundary of $\Omega$. Also note that $\sigma_0 > 0$.

How do I show this energy functional attains a minimum in $H^1(U)$?

According to Evan's, we need to show that $\Pi(u)$ is coercive and convex, but I haven't had any luck.

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    $\begingroup$ Are you assuming $\sigma_0 > 0$ or something similar? If yes, it would be important to add to the question, if not, this functional will not in general attain a minimum. $\endgroup$ – Lukas Geyer May 29 '17 at 13:54
  • $\begingroup$ @LukasGeyer Yes, I am assuming $\sigma_0 > 0$. $\endgroup$ – user412674 May 30 '17 at 0:23
  • $\begingroup$ @user412674: It is not fair to turn the question into a totally different one by some edit after a complete answer has been given. (For later reference: in the original version of the question, the last term read $- \int_{\Gamma_N} \sigma_0 u ds$, i.e., the stated problem was a Poisson equation with pure Neumann boundary conditions) $\endgroup$ – gerw May 30 '17 at 8:14
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    $\begingroup$ @gerw Forgive me, I made a mistake with the initial statement of the question. $\endgroup$ – user412674 May 31 '17 at 4:42
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    $\begingroup$ @gerw Are you not able to look at the edited question? I'll be happy to award you the bounty $\endgroup$ – user412674 May 31 '17 at 22:59

Note: This answers the first version of the question, in which $$\Pi(u) : = \frac{1}{2} \int_{\Omega} k\nabla u \cdot \nabla u dx - \int_{\Omega} fu dx - \int_{\Gamma_N} \sigma_0 u ds.$$

This functional is not coercive - try to plug in constant functions from $H^1(U)$. In fact, if a certain compatibility condition (which one?) on $(f,\sigma_0)$ is violated, $\Pi$ is not even bounded from below.

If this compatibility condition is satisfied, you have several possibilities to obtain the solvability:

  • Discuss everything in the quotient space $H^1(U) / \mathbb R$ (where $\mathbb R$ denote the constant functions).
  • Discuss everything in the subspace $\{ v \in H^1(U) \mid \int_U v \, \mathrm dx = 0\}$.
  • Use a Fredholm argument: The necessary and sufficient optimality condition of your problem can be written as $A u = F$, with $A : H^1(U) \to H^1(U)^*$ and $F \in H^1(U)^*$. Denote by $E : H^1(U) \to H^1(U)^*$ the operator $(Eu)(v) = \int_U u \, v \, \mathrm dx$. Then, $A + E$ is invertible (by Lax-Milgram theorem). Due to Rellich-Kondrachov, $E$ is compact, hence $A$ is a Fredholm operator of index $0$. Now, it remains to study the kernel and cokernel of $A$.

Edit: Answer to the new question: Assuming some regularity of the boundary $\partial\Omega$ (Lipschitz is enough), one can show the existence of $c > 0$ such that $$\| u \|_{H^1(\Omega)}^2 \ge c \, \Big( \frac12 \int_\Omega |\nabla u|^2 \, \mathrm{d}x + \int_{\partial\Omega} \sigma_0 \, u^2 \, \mathrm{d}x\Big),$$ e.g., by a contradiction argument or, alternatively, by proving the inequality first for $u \in C^\infty(\overline\Omega)$. This yields the desired coercivity.

  • $\begingroup$ Plugging in a constant function, call it $c$, then $$I(c) = -c \int_{\Omega} f(x)dx + c^2 \int_{\partial \Omega} \sigma_0 ds.$$ If we assume $\sigma_0 \geq c > 0$ and $f,\sigma_0$ are smooth. Does this simplify the problem? I'm not sure why this wouldn't be bounded below. We could even suppose that $\Omega$ has smooth boundary. $\endgroup$ – user412674 May 28 '17 at 0:05
  • $\begingroup$ Potentially dumb question, but would Fredholm theory still apply despite the square in the boundary term? Its my understanding that Fredholm theory is generally for linear operators, no? $\endgroup$ – Matt May 28 '17 at 0:15
  • $\begingroup$ @user412674: Under which conditions is a parabola $x \mapsto a \, x + x^2$ bounded from below (on all of $\mathbb{R}$)? $\endgroup$ – gerw May 28 '17 at 6:44
  • $\begingroup$ @Matt: Of course, Fredholm theory is for linear operators. The minimization problem is quadratic, hence, its optimality conditions $I'(u) = 0$ are linear in $u$. $\endgroup$ – gerw May 28 '17 at 6:45

To show that the functional is coercive, perhaps try a weak convergence argument? See the proof of the Poincare inequality in Evan's Chapter 6.

  • $\begingroup$ @gerw Would you approach the problem this way? $\endgroup$ – Phillip Wiggins Jun 1 '17 at 7:00

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