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Sine modes decay under Dirichlet heat flow

Example

Assume Countable Choice. For every integer k≥1 and every T>0, the function uk(x,t):=e−k2tsin⁡(kx) is a classical solution of the heat equation ut=Δu on (0,π)×(0,T] with Dirichlet data uk(0,t)=uk(π,t)=0 and initial data uk(x,0)=sin⁡(kx); it is smooth up to t=0 in this one-dimensional setting. Its L2(0,π) norm decays at the rate of the k-th Dirichlet eigenvalue, ∥uk(⋅,t)∥2=e−k2tπ/2(t≥0), so higher modes decay faster, and the nodal set of uk(⋅,t) does not depend on t.

Facts & Assumptions

Given: Countable Choice, an integer k≥1, T>0, and the function uk(x,t)=e−k2tsin⁡(kx) on [0,π]×[0,T].

[A1]

Countable Choice is the ambient hypothesis, inherited through the L2 dictionary in step 3.1 (The Axiom of Countable Choice (ACω)).

[F1]

The heat operator is ∂t−Δ, with Δ=∂x2 in one space dimension (The heat operator, the heat equation, and the Cauchy problem, The Laplacian of a C2 function and of a C2 vector field).

[F3]

The L2(0,π) inner product is ⟨f,g⟩=∫0πfg on the quotient space of The space Lp(μ) as the quotient by null functions (L2 with the integral pairing is a Hilbert space), and ∫0πsin⁡(kx)2dx=π/2 (L2 normalisation of the sine modes on an interval).

Verification

Given: Countable Choice, k≥1, T>0, and uk(x,t)=e−k2tsin⁡(kx).

1.1F1F2given

The function uk is smooth on the closed rectangle (a product of a smooth exponential and a smooth sine), and [F2] gives ∂tuk=−k2e−k2tsin⁡(kx)=−k2uk together with ∂x2uk=−k2e−k2tsin⁡(kx)=−k2uk; hence uk,t=∂x2uk=Δuk on the open rectangle by [F1].

1.2F2given

The boundary values are uk(0,t)=e−k2tsin⁡0=0 and uk(π,t)=e−k2tsin⁡(kπ)=0 for every t, while uk(x,0)=e0sin⁡(kx)=sin⁡(kx); the nodal set at time t is {x∈(0,π):sin⁡(kx)=0}={mπ/k:1≤m≤k−1}, independent of t because the positive factor e−k2t never vanishes.

2.1A1F3given

By [F3] the squared norm of step 1.1's function is ∥uk(⋅,t)∥22=e−2k2t∫0πsin⁡(kx)2dx=e−2k2tπ/2, so ∥uk(⋅,t)∥2=e−k2tπ/2 for every t≥0.

3.1step 2.1given

Since k↦e−k2t is strictly decreasing in k≥1 for every fixed t>0, higher modes decay faster at each positive time, with the ratio e−(k2−l2)t between the k-th and l-th modes for k>l.

4.1step 1.1step 1.2step 2.1step 3.1given∎

Steps 1.1, 1.2, 2.1 and 3.1 verify that uk is a classical solution smooth up to t=0 with Dirichlet data, decay rate e−k2tπ/2, faster decay for higher modes, and a time-independent nodal set.

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