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Let us considère the following Laplace-Neumann problem $-\Delta u=0$ with homogenuous boundary condition of type neumann, i:e $\frac{{\partial u}}{{\partial n}} = 0$. The variational formulation is given as follows $a(u,v)=l(v)$, where $$a(u,v) = \int_\Omega {\nabla u\nabla v} dx = 0$$ and $l(v)=0$. Let us cinsidère the space
$$\left\{ V={u \in H^1(\Omega), \int_\Omega v(x)ds=0} \right\}$$ $l$ is continuous ( is zero), and $a$ is continuous and corecive. By lax-Milgram theoremn there exists a unique solution of the variational problem $$a(u,v)=l(v)$$. We can notice that the constant are alos solutions of the problem, but the constants don't satisfy the zero average condition, I don't understand that.

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In your case the problem is $-\Delta u = 0$ subject to $\frac{\partial u}{\partial n} = 0,$ which only has trivial solutions. Indeed if $u$ is such a solution with zero average $\int_{\Omega} u \,\mathrm{d}x = 0,$ since $a(u,u) = \int_{\Omega} |\nabla u|^2\,\mathrm{d}x = 0$ we conclude that $u$ is a.e. constant and hence zero as it has zero average.

So the solution $u \equiv 0$ is the unique solution subject to the zero average condition. The other constant solutions $u \equiv k \neq 0$ are also solutions, but they don't satisfy the zero average condition. I assume this is what is meant here.

For a non-homogenous problem $-\Delta u = f$, we get $\ell(v) = \int_{\Omega} fv \,\mathrm{d}x$ is non-trivial and hence we get non-trivial solutions. However if $u$ is a solution, then so is $u+k$ for any constant $k.$ The zero average condition removes this extra degree of freedom.

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