partial differential equations
Consider the PDE on a bounded smooth domain $\Omega \subset R^n$, $\triangle u(x)= 0$ on $\Omega$, $\frac{\partial u(x)}{\partial n}|_{\partial \Omega}=g(x)$. Prove that if it admits a smooth solution u, we have $\int_{\partial \Omega} g d \sigma=0$
I know that $\triangle u=div(\triangledown u)$. Can someone give me a hint to do this?
Weak solutions of the equation are critical points of the energy functional $$ E(u)=\int_\Omega\Bigl(\frac12|\nabla u|^2+\frac1{3}|u|^3-f\,u\bigr)dx. $$ Observe that $E(u)$ is well defined on the Banach space $X=H^1_0(\Omega)\cap L^3(\Omega)$.
To show the existence of a solution it is enough to show that $E$ attains its minimum on $X$. This is done by showing that $E$ is strictly convex and coercive (i.e. $E(u)\to\infty$ if $\|u\|_X\to\infty$).
Uniqueness follows (by a different method) because the function $g(u)=u\,|u|$ is increasing.
There are many books on variational methods. One of them is Introduction à la théorie des points critiques by O. Kavian (in french.) A more general case of your question is solved on page 139.
The following is a formal explanation; I will not go into the details of the correct spaces,...
The weak formulation of the problem is $$ \int_\Omega\bigl(\nabla u\cdot\nabla u+u\,|u|\,v-f\,v\bigr)dx=0\quad v\text{ a test function.} $$ The energy functional is a functional $E:X\to\mathbb{R}$ whose derivative with respect to $u$ is given precisely by the weak formulation. What does ths mean? The differential at $u$ is defined as an element $E'(u)$ of de dual of $X$ such that $$ E(u+v)=E(u)+\langle E'(u),v\rangle+o(\|v\|). $$ One way to compute it is $$ \langle E'(u),v\rangle=\lim_{t\to0}\frac{E(u+t\,v)-E(v)}{t}. $$ Using this it is easy to show that $E(u)$ is in fact the energy functional associated to the equation. Observe that $|u|^3/3$ is a primitive of $u\,|u|$.
The existence and uniqueness of the pure Neumann boundary value problem for smooth data can be proved using the Green representation formula, explicitly.
First notice that for $-\Delta \Phi = \delta(x)$, by Green's second identity and convolution formula: for $x\in \Omega$ $$ \int_{\Omega} \Big(u(y) \Delta \Phi(y-x) - \Phi(y-x)\Delta u(y)\Big) dy = \int_{\partial \Omega}\left(u(y)\frac{\partial \Phi}{\partial n}(y-x) - \Phi(y-x)\frac{\partial u}{\partial n}(y) \right) dS(y), $$ we have $$ u(x) = -\int_{\Omega} \Phi(y-x)\Delta u(y) \, dy + \int_{\partial \Omega}\left(\color{red}{\Phi(y-x)}\frac{\partial u}{\partial n}(y) - u(y)\color{blue}{\frac{\partial \Phi}{\partial n}(y-x)} \right) dS(y). $$ For Dirichlet problem, a correction function is constructed so that red term vanishes.
For Neumann problems, Green function is constructed so that $u(x)$ on each point is unique up to a constant. Some people prefer to set: $$ \left\{\begin{aligned} -\Delta \Phi &= \delta(x) \quad \text{in }\Omega, \\ \frac{\partial \Phi}{\partial n} &= \frac{1}{|\partial \Omega|} \quad \text{on } \partial\Omega, \end{aligned}\right. $$ where absolute value just means the Lebesgue measure of codimension $1$, i.e., surface area, so that the constant difference is the average of $u$ on $\partial \Omega$. I myself prefer to set something like: $$ \left\{\begin{aligned} -\Delta \Phi &= \delta(x) - \frac{1}{|\Omega|}\quad \text{in }\Omega, \\ \frac{\partial \Phi}{\partial n} &=0 \quad \text{on } \partial\Omega, \end{aligned}\right.\tag{$\star$} $$ so that you have: $$ u(x) = \frac{1}{|\Omega|}\int_{\Omega}u(y)dy -\int_{\Omega} \Phi(y-x)\Delta u(y) \, dy + \int_{\partial \Omega} \Phi(y-x)\frac{\partial u}{\partial n}(y)dS(y). $$ This means the solution is also unique up to a constant, that constant is the average of $u$ in $\Omega$, in other words, once we fix this average, $u$ is pinned down to just one function. The existence of the Green function is another story, which relies on functional analysis, e.g. see here.
The unique of $u$ up to a constant can be either proved by maximum principle or energy method, but for Helmholtz decomposition does not address the uniqueness, we don't have to address this issue here.
Now for your equation, we know that the compatibility condition is satisfied because: $$ \int_{\Omega}\Delta u = \int_{\Omega}\nabla \cdot u = \int_{\partial \Omega} \frac{\partial u}{\partial n} dS = \int_{\partial \Omega} E\cdot n\,dS. $$ Consider $u$ is smooth and $\int_{\Omega} u = 0$, then $$ u(x) = -\int_{\Omega} \Phi(y-x) (\nabla_y \cdot E(y)) \, dy + \int_{\partial \Omega} \Phi(y-x) (E(y)\cdot n) dS(y). $$
Here are another two proofs for smooth vector field. The Neumann boundary approach is good for less regular vector field (say the one has component in some Sobolev space), for smooth vector field, using vector potentials would be more preferable (at least for me).
The first one is using a pointwisely hold formula if $\Omega$ is convex with respect to a point $x_0 \in \mathbb{R}^3$: $$ E = \nabla \phi + A, $$ where $$ A = -(x - x_0)\times \int^1_0 t \nabla \times E\big(x_0 + t(x- x_0)\big)\,dt , $$ and $$ \phi = (x - x_0)\cdot \int^1_0 tE\big(x_0 + t(x- x_0)\big). $$ Though this does not bear the exact same form, but the formula gives you more physical intuition. For reference please see here.
The second one is using the Dirichlet problem's green function. Now we want to solve a vector Poisson equation: $$ \left\{\begin{aligned} -\Delta F &= E\quad \text{in }\Omega, \\ F &=0 \quad \text{on } \partial\Omega, \end{aligned}\right. $$ Then for the Dirichlet Green function: $$ \left\{\begin{aligned} -\Delta \Phi &= \delta(x) \quad \text{in }\Omega, \\ \Phi &=0 \quad \text{on } \partial\Omega. \end{aligned}\right. $$ Use the Green representation formula component-wise, we have $$ F_i(x) = -\int_{\Omega} \Phi(y-x)\Delta F_i(y) \, dy - \int_{\partial \Omega} F_i(y) \frac{\partial \Phi}{\partial n}(y-x) dS(y), $$ and this is $$ F_i(x) = \int_{\Omega} \Phi(y-x)E_i(y) \, dy. $$ Now use the vector Laplacian's identity for smooth vector fields $$ E(x) = -\Delta F(x) = \nabla \times \nabla \times F - \nabla \nabla \cdot F \\ =\nabla_x \times \left(\nabla_x \times \int_{\Omega} \Phi(y-x)E (y) \, dy\right) - \nabla _x\left(\nabla_x \cdot \int_{\Omega} \Phi(y-x)E(y) \, dy\right), $$ where the subscript $x$ means taking gradient/div/curl w.r.t. $x$. Further moving the derivative into the integral: $$ E(x) = \nabla_x \times \left(\int_{\Omega} \nabla_x \times \Big(\Phi(y-x)E (y)\Big) \, dy\right) - \nabla_x\left(\int_{\Omega} \nabla_x \cdot \Big(\Phi(y-x)E(y)\Big) \, dy\right) \\ = \nabla_x \times \left(\int_{\Omega} \nabla_x \Phi(y-x)\times E (y) \, dy\right) - \nabla_x\left(\int_{\Omega} \nabla_x \Phi(y-x)\cdot E(y) \, dy\right) \\ = - \nabla_x \times \left(\int_{\Omega} \nabla_y \Phi(y-x)\times E (y) \, dy\right) + \nabla_x\left(\int_{\Omega} \nabla_y \Phi(y-x)\cdot E(y) \, dy\right) \\ = \nabla_x \times \left(\int_{\Omega} \Big( \Phi(y-x) \nabla_y \times E (y) - \nabla_y \times (\Phi(y-x)E(y))\Big) \, dy\right) \\ + \nabla_x\left(\int_{\Omega} \Big( -\Phi(y-x) \nabla_y \cdot E (y) + \nabla_y \cdot (\Phi(y-x)E(y))\Big) \, dy\right) \\ = \nabla_x \times \left(\int_{\Omega} \Phi(y-x) \nabla_y \times E (y)\,dy - \int_{\partial \Omega} \Phi(y-x)n\times E(y)\,dS(y) \right) \\ + \nabla_x\left(-\int_{\Omega} \Phi(y-x) \nabla_y \cdot E (y) \, dy +\int_{\partial \Omega}\Phi(y-x)E(y)\cdot n\,dS(y) \right). $$ So that: in $\Omega$ $$ E = \nabla \phi + \nabla \times A, $$ where $$ \phi(x) = -\int_{\Omega} \Phi(y-x) \nabla_y \cdot E (y) \, dy +\int_{\partial \Omega}\Phi(y-x)E(y)\cdot n\,dS(y), \\ A(x) = \int_{\Omega} \Phi(y-x) \nabla_y \times E (y)\,dy - \int_{\partial \Omega} \Phi(y-x)n\times E(y)\,dS(y). $$
Best Answer
We have
$\nabla \cdot \nabla u = \triangle u = 0 \tag 1$
on $\Omega$. Thus the divergence theorem yields
$\displaystyle \int_{\partial \Omega} \nabla u \cdot \mathbf n \; dS = \int_\Omega \nabla \cdot \nabla u; dV = \int_\Omega 0 \; dV = 0, \tag 2$
where $\mathbf n$ is the outward-pointing unit normal vector field on $\partial \Omega$ and $dS$ is the volume element on $\partial \Omega$; but on $\partial \Omega$ we have
$\nabla u \cdot \mathbf n = \dfrac{\partial u}{\partial n} = g = 0, \tag 3$
whence
$\displaystyle \int_{\partial \Omega} g \; dS = 0. \tag 4$