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Quantitative residual error under strong monotonicity

Statement

Assume the Axiom of Choice.

Let (V,H,J) be the evolution triple of Banach evolution triples and weak time derivatives, 1<p<∞, and assume A:V→V∗ is strongly monotone with constant μ>0 and exponent p: ⟨A(x)−A(y),x−y⟩≥μ∥x−y∥Vp(x,y∈V). Suppose u,v∈Lp(0,T;V) have derivatives in Lp′(0,T;V∗) and residual difference R=(u′+A(u))−(v′+A(v))∈Lp′(0,T;V∗). Then for every t∈[0,T], ∥u(t)−v(t)∥H2+μ∫0t∥u−v∥Vp≤∥u(0)−v(0)∥H2+2Cμ,p∥R∥Lp′(0,t;V∗)p′, where both functions use their continuous H representatives and Cμ,p=1p′(2/(μp))1/(p−1). In particular, equal initial data and zero residual imply u=v.

Facts & Assumptions

Given: The Axiom of Choice, the evolution triple, 1<p<∞, strong monotonicity with constant μ, two functions in the stated evolution space, and their residual difference R.

[F1]

Functions in Lp(V) with derivative in Lp′(V∗) have continuous H representatives and satisfy the integrated Hilbert norm chain rule (The Banach evolution-space energy identity and H trace).

[F2]

Hölder's inequality applies to the dual pairing of Lp′(V∗) and Lp(V) (Holder's inequality for integrals, including the endpoint cases).

[F3]

For conjugate exponents p,p′>1, Young's inequality is ab≤ap/p+bp′/p′ for a,b≥0 (Young's inequality for conjugate real exponents).

[A1]

Strong monotonicity with exponent p and constant μ is the additional hypothesis of this theorem; it is not part of ordinary monotonicity in Coercive monotone p-growth diffusion fluxes.

[A2]

The Axiom of Choice licenses the energy chain rule through its absolutely continuous representative prerequisite (The Axiom of Choice).

Proof

1.1A1A2F1F2givenalgebra

Set e=u−v. Subtracting the residual definitions gives e′+A(u)−A(v)=R in Lp′(0,T;V∗). By [F1], testing this identity against e∈Lp(0,T;V) is integrable by [F2], and for every t, 12∥e(t)∥H2−12∥e(0)∥H2+∫0t⟨A(u)−A(v),e⟩=∫0t⟨R,e⟩.

2.1A1A2F1F2F3step 1.1givenalgebra∎

Strong monotonicity bounds the left integral below by μ∫0t∥e∥Vp. Young's inequality [F3], applied to (μp/2)1/pa and (μp/2)−1/pb, gives the scaled form ab≤μ2ap+1p′(2μp)1/(p−1)bp′ gives ∣⟨R,e⟩∣≤μ2∥e∥Vp+Cμ,p∥R∥V∗p′. Integrating and absorbing the first term yields 12∥e(t)∥H2+μ2∫0t∥e∥Vp≤12∥e(0)∥H2+Cμ,p∥R∥Lp′(0,t;V∗)p′. Multiplying by two proves the estimate. If e(0)=0 and R=0, the nonnegative left side vanishes for every t, hence e(t)=0 in H and e=0 in Lp(V).

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