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Backward Euler MINI step existence and discrete energy bound

Statement

Assume Countable Choice, the mini-element hypotheses of The mixed mini-element backward Euler Navier–Stokes scheme, f∈L2(0,T;L2(D)d), and an initial value uh0∈Zh. On every finite positive time partition there is at least one mixed solution (uhn,phn)n=1N of the backward Euler scheme. Every such solution satisfies, for 0≤m≤N, the work-form inequality 12∥uhm∥22+12∑n=1m∥uhn−uhn−1∥22+ν∑n=1mτn∥∇uhn∥22≤12∥uh0∥22+∑n=1mτn(fhn,uhn)L2. Consequently it satisfies the mesh-independent estimate 12∥uhm∥22+12∑n=1m∥uhn−uhn−1∥22+ν2∑n=1mτn∥∇uhn∥22≤12∥uh0∥22+CP(D)22ν∥f∥L2(0,T;L2)2. The constant is independent of mesh size and time-step sizes.

Facts & Assumptions

Given: Countable Choice, the MINI spaces, ν>0, a finite partition, f∈L2(0,T;L2), and uh0∈Zh.

[A1]

Countable Choice is inherited from the MINI/Stokes stability supplier; no arbitrary-index choice is used inside a single finite partition. (The Axiom of Countable Choice (ACω), Uniform Fortin stability of the mini element)

[F1]

The velocity mass form is real Hilbert; aν(z,z)=2ν∥ε(z)∥22≥ν∥∇z∥22 on H01. (Stokes velocity and pressure spaces and weak forms, The homogeneous Dirichlet strain identity)

[F2]

c(z;z,z)=0 for every z∈V, and c is continuous on the finite-dimensional MINI space. (The skew convective form for incompressible flow, Continuity, energy cancellation, and Sobolev bounds for skew convection)

[F3]

A continuous finite-dimensional map with positive outward radial pairing on a sphere has a zero in the ball. (An outward-pointing finite-dimensional map has a zero)

[F4]

The stable MINI pair has a positive discrete inf-sup constant, so the pressure map is injective and the velocity constraint map is onto. (Uniform Fortin stability of the mini element, Weak and pointwise divergence-free discrete velocities)

[F5]

The identity 2(x−y,x)=∥x∥2−∥y∥2+∥x−y∥2 holds in L2, and Poincare gives ∥z∥2≤CP(D)∥∇z∥2. (The Hilbert-space discrete energy identity, The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction)

[F6]

Holder and Young absorb ∥fhn∥2∥uhn∥2 into ν∥∇uhn∥22/2 with remainder CP(D)2∥fhn∥22/(2ν). (Holder's inequality for integrals, including the endpoint cases, Young's inequality for conjugate real exponents)

[F7]

The norm of a Bochner integral is at most the integral of the norm. (Bochner integral norm inequality)

Proof

1.1A1F2givenconstruct

Fix a time cell and the previous value uhn−1∈Zh. On Zh, define Fn(z)∈Zh∗ by ⟨Fn(z),wh⟩=τn−1(z−uhn−1,wh)L2+aν(z,wh)+c(z;z,wh)−(fhn,wh)L2. The map is continuous by [F2] and finite dimensionality. If Zh={0} its only velocity is zero; otherwise represent Fn in coordinates of a gradient-orthonormal basis of Zh.

2.1F1F2F3givenstep 1.1algebra

By [F1]–[F2], ⟨Fn(z),z⟩≥ν∥∇z∥22−CP(D)(τn−1∥uhn−1∥2+∥fhn∥2)∥∇z∥2. Choose R so the right side is strictly positive whenever ∥∇z∥2=R. The outward-zero result [F3] supplies uhn∈Zh with Fn(uhn)=0.

3.1F4step 2.1algebra

Let Gn(vh) be the velocity residual (time term plus viscosity and convection minus forcing). Step 2.1 gives Gn=0 on Zh=ker⁡B, where Bvh(qh)=b(vh,qh). Since [F4] makes B onto Qh∗, B∗ is injective and its range is contained in (ker⁡B)∘. Rank-nullity gives both spaces dimension dim⁡Qh, so they are equal. Thus −Gn=B∗phn for a unique phn∈Qh, which gives the stated mixed equation with the correct residual sign. Repeating the finite step for n=1,…,N gives a mixed solution.

4.1F1F2F5givenstep 3.1

Test the mixed equation with uhn. Its pressure term is zero since uhn∈Zh, and its convective term is zero by [F2]. Apply [F5] to the time term and [F1] to the viscosity term, obtaining 12(∥uhn∥22−∥uhn−1∥22+∥uhn−uhn−1∥22)+ντn∥∇uhn∥22≤τn(fhn,uhn)L2. The right side is retained with its sign; it follows directly from the tested equation, skew cancellation, and the lower Korn bound.

5.1F5F6F7step 4.1algebra∎

Before applying Young, retain the actual work term in step 4.1 and sum over n=1,…,m; telescoping gives the first displayed inequality. For the coarser estimate, Cauchy–Schwarz, Poincare and Young give τn(fhn,uhn)≤ν2τn∥∇uhn∥22+CP(D)22ντn∥fhn∥22. For each interval, [F7] and Holder [F6] give τn∥fhn∥22=τn−1∥∫tn−1tnf(t) dt∥22≤τn−1(∫tn−1tn∥f(t)∥2 dt)2≤∫tn−1tn∥f(t)∥22 dt. Summing proves ∑nτn∥fhn∥22≤∥f∥L2L22. Substitution into the work-form inequality proves the second displayed estimate.

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