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Positive matrix-potential wave system energy

Statement

Let N≥1, I:=[0,T) with 0<T≤∞, and let J⊆R be an open interval containing I. Let U,S∈C∞(J×R,RN). Write Sym⁡N(R) for the real symmetric N×N matrices and use the Euclidean dot product and norm on RN. Let V∈C∞(R,Sym⁡N(R)) be bounded, independent of t, and positive semidefinite at every x, meaning zTV(x)z≥0 for every z∈RN. Assume U and all its mixed derivatives of total order at most two, and S, are Schwartz in x locally uniformly for t∈J: for every compact K⊂J, every such derivative D and every integer m≥0, sup⁡t∈K,x∈R(1+∣x∣)m∣DU(t,x)∣<∞, and the analogous bounds hold for S. If Utt−Uxx+V(x)U=S, define E(t):=12∫R(∣Ut∣2+∣Ux∣2+UTV(x)U) dx, where ∣⋅∣ is the Euclidean norm and ∥v(t,⋅)∥22:=∫R∣v(t,x)∣2 dx for any vector-valued function v. Then E(t)≥0, E′(t)=∫RUt⋅S dx,E(t)≤E(0)+12∫0t∥S(τ,⋅)∥2 dτ(t∈I). In particular E is constant when S=0. If t,x have length units and U has unit [U], then V has unit m−2, S has unit [U]m−2, E has unit [U]2m−1, and both sides of the square-root estimate have unit [U]m−1/2. This abstract estimate supplies neither pointwise decay nor a reconstruction of coupled Einstein–Maxwell perturbations.

Facts & Assumptions

Given: The stated smooth matrix wave system and the spatial decay assumptions.

[F1]

Differentiation under the integral sign: if f(x,t) is integrable in x for each t in an open interval, has a measurable pointwise t-derivative, and that derivative has one integrable majorant locally in t, then ddt∫f(x,t) dx=∫∂tf(x,t) dx.

[F2]
[A1]

N≥1, I=[0,T) with T>0, J⊆R is open and contains I, U,S∈C∞(J×R,RN), and V is bounded, real symmetric, positive semidefinite and independent of t. The stated local Schwartz bounds justify differentiation under the integral and vanishing spatial boundary terms.

Proof

1.1givenA1algebra

For each component multiply Ui,tt−Ui,xx+∑jVijUj=Si by Ui,t and sum over i. The product rule gives ∂t12∣Ut∣2−∂x(Ut⋅Ux)+∂t12∣Ux∣2+UtTVU=Ut⋅S. Because V is symmetric and independent of t, UtTVU=∂t(12UTVU). Thus with e:=12(∣Ut∣2+∣Ux∣2+UTVU) the identity is ∂te−∂x(Ut⋅Ux)=Ut⋅S.

2.1step 1.1F1A1algebra

Integrate step 1.1 over [−R,R]. By the Schwartz assumptions, the boundary term [Ut⋅Ux]−RR tends to 0 as R→∞. The derivative of the energy density is bounded locally in t by an integrable multiple of (1+∣x∣)−2: its terms are products of derivatives of U through order two and the bounded coefficient V times such products. Thus [F1] applies to the energy density and gives E′(t)=∫R∂te dx. Taking the limit in the integrated identity from step 1.1 yields E′(t)=∫RUt⋅S dx. Positive semidefiniteness in [A1] also gives E≥12∥Ut∥22≥0.

3.1step 2.1F2A1algebra∎

By [F2] and step 2.1, E′≤∥Ut∥2∥S∥2≤2E∥S∥2. For ε>0, differentiation yields ddtE+ε=E′2E+ε≤∥S∥22EE+ε≤∥S∥22. Integrate from 0 to t and let ε↓0 to obtain the claimed estimate, including times at which E=0.

Remarks

The estimate is finite-time energy control, not decay. To use it for an actual coupled perturbation, one must first derive that system's exact variables, angular sectors, symmetric structure, and reconstruction.

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