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The Rayleigh quotient on an interval

Example

Assume the Axiom of Choice, the ultrafilter lemma, DC and HB (The Axiom of Choice, The ultrafilter extension principle (UL/BPI), The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain, The real dominated-extension principle as an additional hypothesis over ZF), inherited from The first Dirichlet eigenfunction by constrained minimisation; the explicit computation below consumes only Countable Choice, through The sharp Dirichlet Poincare inequality on an interval. On Ω=(0,1) the constrained minimisation of The first Dirichlet eigenfunction by constrained minimisation is explicit: a minimiser of E(u)=∫01u′2 on the L2-unit sphere S⊆H01(0,1) is u0(x)=2 sin⁡(πx), the minimum is λ1=π2, and the weak eigenvalue equation is −u0′′=π2u0 with u0(0)=u0(1)=0.

Facts & Assumptions

Given: The interval (0,1), the energy E(u)=∫01u′2 dx on H01(0,1;R), the unit sphere S={u∈H01(0,1):∥u∥L2=1} (The notation Hk and the reserved zero-boundary symbol, Integer-order Sobolev spaces and their norms, Zero-boundary Sobolev space as a norm closure), and u0(x)=2 sin⁡(πx).

[F1]

The sharp Dirichlet Poincare inequality on an interval: with L=1 and ϕ(x)=sin⁡(πx), every u∈H01(0,1) satisfies ∥u∥L2≤π−1∥u′∥L2, the function ϕ attains equality, and ∫01ϕ′v′ dx=π2∫01ϕv dx for every v∈H01(0,1).

[F3]

The second fundamental theorem: if G is differentiable on [a,b] with G′=f and f is integrable, then ∫abf=G(b)−G(a): for every ψ∈Cc∞(0,1), choose 0<a<b<1 with ψ=0 near a,b; applying the fundamental theorem to u0ψ on [a,b] gives ∫01u0ψ′=−∫01u0′ψ. Thus the classical derivative u0′ is also the weak derivative.

[F4]

Explicit compactly supported smooth cutoffs: in dimension one there is χ∈Cc∞(R) with 0≤χ≤1, χ=1 on [−1,1], and χ=0 outside [−2,2]; for every R>0, the dilate χR(x)=χ(x/R) has derivative R−1χ′(x/R). The construction requires no choice.

[F5]

Double-angle and quadratic power-reduction identities, The second fundamental theorem: if G is differentiable on [a,b] with G′=f and f is integrable, then ∫abf=G(b)−G(a), The derivatives of sine and cosine are cosine and minus sine, Integrable functions on [a,b] form a set closed under sums and scalar multiples, and ∫ab(λf+μg)=λ∫abf+μ∫abg, Pi is the first positive zero of sine: sin⁡2t=(1−cos⁡2t)/2 and cos⁡2t=(1+cos⁡2t)/2; sin⁡(π)=sin⁡(2π)=0; and the fundamental theorem of calculus applied to sin⁡(2πx)/(2π) gives ∫01cos⁡(2πx) dx=0, since the integral is linear.

[F6]

The first Dirichlet eigenfunction by constrained minimisation: on a nonempty bounded open set, E attains its infimum λ1 on S, and every minimiser u0 satisfies ∫u0′h′=λ1∫u0h for all h∈H01 with λ1=E(u0).

Verification

technique · direct

Given: The interval, the energy and the function u0 above.

1.1givenF2F3F4F5

The function u0 is smooth on [0,1], with classical derivative u0′(x)=2πcos⁡(πx) by [F2]. For each test function ψ∈Cc∞(0,1), integration of (u0ψ)′ over an interior interval containing its support gives ∫01u0ψ′=−∫01u0′ψ [F3], so u0′ is its weak derivative; both u0 and u0′ are bounded, hence u0∈H1(0,1). Take χ from [F4] and put M=∥χ′∥L∞(R)<∞. For integers m≥5, define ξm(x)=(1−χ(mx))(1−χ(m(1−x))). Then ξm∈Cc∞(0,1), ξm=1 on [2/m,1−2/m], ∥ξm′∥∞≤2Mm, and both 1−ξm and ξm′ are supported in Bm=(0,2/m)∪(1−2/m,1). On Bm, ∣u0(x)∣≤22π/m, while ∣u0′(x)∣≤2π everywhere; since ∣Bm∣≤4/m, these bounds give ∥(1−ξm)u0∥L2→0 and ∥(1−ξm)u0′−ξm′u0∥L2→0. Hence ξmu0→u0 in H1(0,1), and the closure definition of H01 gives u0∈H01(0,1). Finally u0(0)=u0(1)=0 because sin⁡0=sin⁡π=0 [F5].

2.1step 1.1F5

Normalisation and energy: by the power-reduction identities and the vanishing of ∫01cos⁡(2πx) dx [F5], ∫01sin⁡2(πx) dx=12∫01(1−cos⁡(2πx)) dx=12 and ∫01cos⁡2(πx) dx=12; hence ∥u0∥L22=2⋅12=1, so u0∈S, and E(u0)=∫01u0′2=2π2∫01cos⁡2(πx) dx=π2.

3.1step 1.1step 2.1F1

Minimality: for every v∈S the sharp inequality [F1] gives 1=∥v∥L2≤π−1∥v′∥L2, that is E(v)=∥v′∥L22≥π2; since u0∈S with E(u0)=π2 by step 2.1, the infimum over S is the minimum λ1=π2, attained at u0.

4.1step 1.1step 3.1F1F2F5

Weak eigenvalue equation: the weak identity of [F1] for ϕ=sin⁡(πx) scales by 2 to ∫01u0′h′=π2∫01u0h for every h∈H01(0,1), and by [F2] u0′′=−π2u0 classically with u0(0)=u0(1)=0; thus −u0′′=π2u0 holds in the weak sense, with λ1=π2 as the eigenvalue.

5.1step 1.1step 2.1step 3.1step 4.1F1F6∎

Steps 1.1-4.1 exhibit the minimiser, the minimum and the eigenvalue equation explicitly, so the constrained minimisation of [F6] on (0,1) has u0(x)=2sin⁡(πx) as a minimiser with λ1=π2, in agreement with the general statement; the only choice principle consumed by this computation is Countable Choice through the sharp interval inequality [F1].

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