$ \Delta p=0$ and $(\nabla p)\cdot \nu\big|_{\partial\Omega}=0 $ in $H^1(\Omega)$ implies $p$ is a constant

Question

In the proof of a theorem regarding the characterization of some space in Temam's Navier Stokes Equations (p11 (1.37)), the following argument is made.

Suppose $\Omega$ is a Lipschitz open bounded set in $\mathbb{R}^n$, and $p\in H^1(\Omega)$ is such that $$ \Delta p=0, \quad(\nabla p)\cdot \nu\big|_{\partial\Omega}=0 $$ where $\nu$ is the unit outer normal. Then $p$ is a constant.

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I don't have much experience solving PDE in terms of distributions. Would anybody come up with a reference or giving an argument regarding why the above is true?

Answer 1

For any $f∈ H^1$, $$ 0 = ∫_\Omega fΔ p = ∫_{\partial \Omega}f ∇ p· \nu - ∫_\Omega∇ f ∇ p $$ so $∫_\Omega ∇ f ∇ p = 0$. With $f=p$, $‖∇ p‖_{L^2} = 0$. By Poincaré, $$\left\| p - \frac{1}{|\Omega|}∫_\Omega p\right\|_{L^2} \leq C‖∇ p‖_{L^2} = 0$$ so $p$ equals the constant $\frac{1}{|\Omega|}∫_\Omega p$ (almost everywhere).

Answer 2

For the sake of a reference, this appears in an exercise of Evans's PDE (problem 10 in chapter 6 with smoother assumption of $\partial\Omega$). Also, one should assume that $\Omega$ is connected in the statement.