research30 min studyLesson 812 in Mathematics

Rayleigh quotients and min–max principles

Translate trial functions and trial subspaces into bounds with the correct form domain and eigenvalue index.

On this page

What you’ll understand

  • Use the energy quotient on the correct Sobolev form domain.
  • Distinguish a one-function bound from a higher-dimensional min–max bound.
  • Remove constants when estimating the first positive closed-manifold eigenvalue.

Eigenvalues can be characterized without explicitly solving the differential equation. The Rayleigh quotient compares gradient energy with squared amplitude, and the min–max principle turns subspace dimension into an eigenvalue index. We retain the positive scalar Laplacian and one-based multiplicity list, including zero modes. The essential choice is the form domain: admissible trial functions must satisfy the variational boundary requirement even when they need not satisfy the stronger operator-domain condition.

Energy divided by mass

For a nonzero form-domain function, divide its integrated squared gradient by its squared square-integrable norm. Multiplying the trial function by any nonzero constant leaves this quotient unchanged. The infimum is the first eigenvalue, which is zero for a closed connected manifold because constants are admissible. To estimate its first positive eigenvalue, impose orthogonality to constants, equivalently zero mean. Forgetting that constraint gives a formally correct bound on zero instead of information about the spectral gap.

Dirichlet trials belong to H01H_0^1, while Neumann trials belong to H1H^1 and need not have zero normal derivative. The latter condition belongs to the operator domain and arises from the variational equation at a critical point. Requiring it of every trial can unnecessarily restrict the space. Conversely, allowing a trial with nonzero trace in a Dirichlet problem invalidates the intended bound because it is not admissible in that problem's energy domain.

R(u)=∫∣∇u∣2 dV∫∣u∣2 dV,λk=inf⁡V⊂Q;dim⁡V=k sup⁡0≠u∈VR(u)R(u)=\frac{\int|\nabla u|^2\,dV}{\int|u|^2\,dV},\qquad \lambda_k=\inf_{\substack{V\subset\mathcal Q\,;\,\dim V=k}}\ \sup_{0\ne u\in V}R(u)
The form domain Q is H-one on a closed manifold, H-zero-one for Dirichlet, and H-one for Neumann.↔ Scroll wide equations horizontally.

A subspace bounds an indexed eigenvalue

A single trial function gives an upper bound for the first eigenvalue in the relevant constrained space. For the kth eigenvalue, a k-dimensional trial subspace gives an upper bound only after maximizing the quotient over all its nonzero combinations. Checking several basis vectors separately is insufficient because mixed energy terms can increase that maximum. The subspace formulation is the conceptual source of the Rayleigh–Ritz method, where the maximum becomes a generalized matrix eigenvalue problem.

An orthonormal eigenbasis makes the principle transparent. The quotient of a finite combination is a weighted average of its participating eigenvalues, with weights proportional to squared coefficients. The span of the first k eigenfunctions has maximal quotient exactly the kth eigenvalue. Every k-dimensional competitor contains a nonzero vector orthogonal to the preceding k−1 eigenfunctions, whose quotient is at least that value. This proof counts multiplicity automatically because it counts dimensions, not distinct numerical levels.

u=∑jcjϕj⟹R(u)=∑jλj∣cj∣2∑j∣cj∣2u=\sum_jc_j\phi_j\quad\Longrightarrow\quad R(u)=\frac{\sum_j\lambda_j|c_j|^2}{\sum_j|c_j|^2}
Orthogonality diagonalizes both mass and energy; mixtures average eigenvalues rather than amplitudes.↔ Scroll wide equations horizontally.

Scaling and geometric comparisons

When a metric is multiplied by the square of a length factor, the energy numerator scales by length to the dimension minus two, while the denominator scales by length to the dimension. Their ratio therefore scales inversely with squared length. This gives a quick consistency test for proposed geometric bounds. A bound proportional to inverse volume without a dimension-dependent power cannot have the correct general scaling for Laplace eigenvalues.

Variational bounds are often useful when exact eigenfunctions are unknown. A function almost constant with opposite signs on two large regions, changing only across a thin connector, can have small gradient energy and substantial mass. After arranging zero mean, it bounds the first positive eigenvalue from above. This is a geometric construction of slow relaxation. It does not by itself give a matching lower bound or identify the true first eigenfunction; those require additional analysis.

A two-frequency Dirichlet trial

On the interval of length three take u(x)=sin⁡(πx/3)+2sin⁡(2πx/3)u(x)=\sin(\pi x/3)+2\sin(2\pi x/3). Compute its Rayleigh quotient and distinguish that one-function estimate from using the entire span of the two sine modes.

  1. The sine modes are orthogonal and have equal squared norm 3/23/2. Their energy eigenvalues are π2/9\pi^2/9 and 4π2/94\pi^2/9, respectively. The coefficient two contributes squared weight four.

  2. The quotient is ((1+16)/(1+4))π2/9=17π2/45((1+16)/(1+4))\pi^2/9=17\pi^2/45. It is a valid upper bound for the first Dirichlet eigenvalue, although the exact first eigenvalue is smaller.

  3. On the complete two-dimensional span, the quotient ranges between the two participating eigenvalues. Its maximum is 4π2/94\pi^2/9, so min–max bounds the second eigenvalue by that number and achieves equality in this explicit example.

The result

A selected mixture and a maximized trial subspace answer different variational questions, even when they use the same basis functions.

Explore the model

Interactive lab

Interactive lab · spectral

Compute the quotient of a finite sine mixture

Pause/Reset, then set both mode count to 2 and coefficient decay to 0. At zero time the plotted sum is sin(x)+sin(2x); predict its periodic Rayleigh quotient 5/2 by orthogonality. Reset and set both count to 2 and decay to 1; the quotient becomes 8/5 for sin(x)+sin(2x)/2. Compare their shapes and calculate the quotients separately.

Compute the quotient of a finite sine mixtureA finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Current readings: Sum at x = π/2: 0.867; Smallest wavelength: 2π / 5; Highest-mode coefficient: 0.2. All parameters are adjustable with the labeled controls below.01.573.144.716.28−2.56−1.2801.282.56x · radiansFinite sine sum
Finite sumHighest included mode
Sum at x = π/2
0.867
Smallest wavelength
2π / 5
Highest-mode coefficient
0.2
t = 0 s

A finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe panel shows the actual finite mixtures. It does not report a Rayleigh quotient or solve an unknown domain's eigenproblem; playback changes phase and is unnecessary for this comparison.

Check your understanding

What gives a valid upper bound for the kth eigenvalue in the one-based min–max principle?

Try it yourself

On a circle of length 4π4\pi, take the mean-zero function v(x)=cos⁡(x/2)+3sin⁡xv(x)=\cos(x/2)+3\sin x. Compute its Rayleigh quotient and the resulting upper bound for the first positive eigenvalue. Compare with the exact gap.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The two modes are orthogonal with equal norms, but their eigenvalues are one quarter and one.

Compare with the solution

The squared coefficient weights are one and nine, so the quotient is (1/4+9)/10=37/40(1/4+9)/10=37/40. The zero-mean constraint makes this an upper bound for the first positive eigenvalue, which is the second entry in the full multiplicity list. The exact gap is 1/41/4, achieved by the lowest nonzero frequency alone. The trial bound is valid but loose because most of its mass sits at the higher frequency.

Take it with you

  • Match the trial space to the variational boundary condition.
  • A kth-eigenvalue bound requires a whole k-dimensional subspace.
  • Closed-manifold gap estimates must exclude constants.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.