Let $\Omega$ be an open bounded domain with a boundary $\partial\Omega$. Consider the following Neumann eigenvalue problem for Laplacian: find $(\phi_n,\lambda_n)\in H^1(\Omega)\times \mathbb{R}$ \begin{align*} -\Delta \phi_n& = \lambda_n \phi_n\quad \mbox{in }\Omega,\\ \frac{\partial \phi_n}{\partial \nu} & = 0 \quad \mbox{on }\partial\Omega. \end{align*}

Now by spectral theory (see the notes at here https://faculty.math.illinois.edu/~laugesen/595Lectures.pdf), it is known that the sequence of eigenfunctions $\phi_n$ (ordered nondecreasingly by the eigenvalues, with multiplicity counted) can be taken to be a complete orthonormal basis in $L^2(\Omega)$, and also forms a complete orthogonal basis in $H^1(\Omega)$. Thus any function $u\in H^1(\Omega)$ can be expanded into \begin{equation*} u = \sum_{n=1}^\infty (u,\phi_n)_{L^2(\Omega)}\phi_n\quad \mbox{in } L^2(\Omega), \end{equation*} and this expansion holds also in $H^1(\Omega)$, since $u\in H^1(\Omega)$ by assumption.

I am puzzled over the fact that all the eigenfunctions $\phi_n$ have zero Neumann boundary condition, so any $n$-term truncation $u_n$ $$ u_n = \sum_{i=1}^n(u,\phi_n)_{L^2(\Omega)}\phi_n $$ has a zero Neumann boundary condition. However, a function $u$ in $H^1(\Omega)$ may not have a zero Neumann boundary condition. How shall one understand the convergence in $H^1(\Omega)$ ?

  • $\begingroup$ Thanks for the comment. I would also expect so but it is a result proved on page 37 of the note (in the link), and I do not find any problem with the proof. $\endgroup$ – user118240 Jan 15 '18 at 13:11
  • $\begingroup$ you can also think of the simpler case with the Dirichelt boundary condition eigenfunctions; so they are zero on the boundary. Yet any $L^2(\Omega)$ function can be written as an infinite sum. $\endgroup$ – Math604 Jan 18 '18 at 15:04
  • $\begingroup$ Thank you for your comments. I completely agree with it: just like all the sines $\{\sin i\pi x\}_{i=1}^\infty$ are complete in $L^2(0,1)$. $\endgroup$ – user118240 Jan 19 '18 at 17:12

The boundary has measure=0, and so there is no contradiction, because the convergence takes place in $L^2$ (or in $H^1$). The convergence is not pointwise. It is probably best to convince yourself of this for an interval $\Omega=(0,L)$ in $\mathbb R$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.