Rayleigh quotient $Q=(\frac{||\triangledown w||}{||w||})^2$ in using the eigenfunction $\sin(x)$ on the segment $(0,\pi)$

Question

I would like to well understanding the Rayleigh quotient $Q=(\frac{\|\nabla w\|}{\|w\|})^2$. Does anyone could explain to me why we divide the norm of the gradient $\| \nabla w \|$ by $\| w \|$, and we finally square that quotient? I tried to understand myself the introduction of that quotient in using the eigenfunction $\sin(x)$ and $k \sin (x)$, $c \not = 0$ on the segment $(0,\pi)$, but it is not evident to understand.

Answer 1

If $w$ is an eigenvector of the Laplacian (with homogeneous Dirichlet or Neumann boundary conditions) on a bounded domain, then the Dirichlet energy is $$\int \langle \nabla w, \nabla w\rangle = \lambda \int \langle w, w\rangle$$ where $\lambda$ is $w$'s eigenvector. Therefore $$\frac{\|\nabla w\|^2}{\|w\|^2} = \lambda$$ and if $w$ is not an eigenvector, $Q(w)$ is a convex combination of the eigenvalues depending on how $w$ is expressed in the eigenvector basis. So you should think of $Q$ not as a quotient of norms, which is then squared, but rather as quotient of inner products, where the denominator is there to normalize the eigenvectors.

Answer 2

Here's the intuition. If we have a Hilbert space with a norm $n$ and a positive quadratic form $q$, then we have two sets of "unit spheres": the set of vectors with $n(v,v) = 1$ and the set of vectors with $q(v,v) = 1$. Since $q$ is positive, the $q$-unit sphere is an ellipsoid. The Rayleigh quotient $q/n$ measures how far the two subspaces are from each other, and critical points of the Rayleigh quotient correspond to eigenspaces. (The ellipsoid and the unit sphere are parallel at an eigenspace.)

So given an ellipsoid and a (say) $2$-plane, think about the maximum stretch factor for vectors in the plane. Unless the plane is exactly the span of the first two eigenspaces, it must "bleed into" one of the higher eigenspaces, and such vectors have higher Rayleigh quotient.

What helped me with this was just drawing ellipsoids and their intersection with planes and staring at the picture until it made sense. At that point, you should be able to follow the proof.