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Define $\mathcal F(u) = \int_U \frac{1}{2} |\nabla u|^2 - \int_U fu - \int_{\partial U} gu : C^\infty(\bar U) \to \mathbb R$, where $f \in C^\infty(\bar U), g \in C^\infty(\partial U)$. Suppose $u_0 \in C^\infty(\bar U)$ satisfies $\mathcal F(u_0) = \min_{u\in C^\infty(\bar U) \\ \int_U u^2=1}\mathcal F(u)$, prove

$$ -\Delta u_0 = f \text{ in } U \text{ and } \frac{\partial u_0}{\partial \vec{n}}=g \text{ on } \partial U $$

Here is my solution, but the result seems not the same as the first equation above. I am not sure what goes wrong.

Suppose $v \in C^\infty(\bar U), t \in \mathbb R $ and $u + tv \neq 0$. Since $u_0$ minimizes $\mathcal F$, we have

$$ 0 = \left .\frac{d}{dt}\right |_{t=0} \mathcal F \left ( \frac{u_0 + tv}{\lVert u_0 + tv \lVert_{L^2}} \right ) = \int_U \nabla u_0\cdot \nabla v - \mathcal F(u_0) \int_U u_0v - \int_U fv - \int_{\partial U} gv$$

Note $\int_U \nabla u_0 \cdot \nabla v = \int_{\partial U} v\frac{\partial u_0}{\partial \vec{n}} - \int_U v\Delta u_0$.

Now, choose $v$ such that $v|_{\partial U} = 0$, then we have

$$0 = -\int_Uv \Delta u_0-\mathcal F(u_0) \int_U u_0v - \int_U fv$$

Then we have $\Delta u_0 + \mathcal F(u_0) u_0 + f = 0$ in $U$.

Next, for $v \in C^\infty(\bar U)$, plug $\Delta u_0 = -\mathcal F(u_0) u_0 - f$ into the two equations above, we have $\int_{\partial U}\left ( \frac{\partial u_0}{\partial \vec{n}} - g\right )v=0$, then we have $\frac{\partial u_0}{\partial \vec{n}}=g$ on $\partial U$.

I checked the computation twice and couldn't find the mistake.

Update: Note that $U$ is a bounded, connected open set in $\mathbb R^n$ with $C^\infty$ boundary $\partial U$.

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  • $\begingroup$ In the computation of the directional derivative of $\mathcal F$ why are you dividing the argument by $\vert\vert u_0+t v\vert\vert_{L^2}$? $\endgroup$ – Uskebasi Sep 21 '18 at 10:14
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    $\begingroup$ @Uskebasi to keep the function satisfies $\int_U u^2 = 1$. $\endgroup$ – miskcoo Sep 21 '18 at 11:46
  • $\begingroup$ Yes you are right, I missed that part. Are you sure about the given answer? $\endgroup$ – Uskebasi Sep 21 '18 at 13:36
  • $\begingroup$ So my method is correct? I am not sure about the given answer. This is an exercise, maybe my teacher missed that term? $\endgroup$ – miskcoo Sep 21 '18 at 13:56
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    $\begingroup$ I have't done the exact same computation you did for the partial derivative of $\mathcal{F}$ because they seemed rather messy. I have used the method of the Lagrange multipliers and the Euler-Larange equation I obtained is similar to yours. Moreover $-\Delta u_0 = f $ is the Euler-Lagrange equation for the unconstrained problem, i.e. without the request $\vert\vert u\vert\vert_{L^2}=1$. $\endgroup$ – Uskebasi Sep 21 '18 at 14:05

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