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Bounded-domain Stokes residual estimate

Statement

Assume ${A C}_{ω}$ . Let $n \geq 2$ and let $Ω \subset R^{n}$ be a bounded $C^{1}$ domain, meaning a bounded connected open set whose boundary is locally a $C^{1}$ graph and which lies on one side of that graph after a rigid change of coordinates. Let $μ > 0$ , $w \in C^{2} (\overline{Ω}; R^{n})$ , $q \in C^{1} (\overline{Ω})$ , and $f \in C (\overline{Ω}; R^{n})$ ; $C^{k} (\overline{Ω})$ means that derivatives through order $k$ extend continuously to the closure. For ordinary Euclidean coordinates $x_{1}, \dots, x_{n}$ , set $\partial_{j} = \partial / \partial x_{j}$ , define $\nabla r = (\partial_{1} r, \dots, \partial_{n} r)$ for scalar $r$ , define $\nabla \cdot v = \sum_{i = 1}^{n} \partial_{i} v_{i}$ and $(\nabla v)_{i j} = \partial_{j} v_{i}$ for vector $v$ , define $(v \cdot \nabla) v = \sum_{j = 1}^{n} v_{j} \partial_{j} v$ , and define the vector Laplacian by $(Δ v)_{i} = \sum_{j = 1}^{n} \partial_{j}^{2} v_{i}$ . For $a = (a_{i})_{i = 1}^{n} \in R^{n}$ set $∣ a ∣ = (\sum_{i = 1}^{n} a_{i}^{2})^{1 / 2}$ , so $∣ \nabla v ∣^{2} = \sum_{i, j = 1}^{n} ∣ \partial_{j} v_{i} ∣^{2}$ . For vector fields $g, v$ , write $∥ g ∥_{2} = (\int_{Ω} ∣ g ∣^{2} d x)^{1 / 2}$ and $∥ \nabla v ∥_{2} = (\int_{Ω} \sum_{i, j = 1}^{n} ∣ \partial_{j} v_{i} ∣^{2} d x)^{1 / 2}$ . Suppose pointwise that $\nabla \cdot w = 0$ , $w ∣_{\partial Ω} = 0$ , and $- μ Δ w + \nabla q = f$ . Also assume that a stated constant $C_{Ω} > 0$ satisfies the zero-trace Poincare inequality $∥ v ∥_{2} \leq C_{Ω} ∥ \nabla v ∥_{2}$ for every $v \in C^{1} (\overline{Ω}; R^{n})$ with $v ∣_{\partial Ω} = 0$ . Then

$∥ \nabla w ∥_{2} \leq \frac{C_{Ω}}{μ} ∥ f ∥_{2}, ∥ w ∥_{2} \leq \frac{C_{Ω}^{2}}{μ} ∥ f ∥_{2} .$

For $L_{x}, L_{z}, h > 0$ , set $T_{L} = R / (L Z)$ for $L > 0$ and $Q = T_{L_{x}} \times (0, h) \times T_{L_{z}}$ , the periodic channel cell. A field on $\overline{Q}$ is identified with its lift to $R \times [0, h] \times R$ periodic with periods $L_{x}, L_{z}$ in the first and third coordinates. Integrals on $Q$ use Lebesgue measure on one fundamental cell in each periodic coordinate; $∥ v ∥_{L^{2} (Q)}^{2} = \int_{Q} ∣ v ∣^{2}$ and $∥ \nabla v ∥_{L^{2} (Q)}^{2} = \int_{Q} \sum_{i, j = 1}^{3} ∣ \partial_{j} v_{i} ∣^{2}$ . Every $v \in C^{1} (\overline{Q}; R^{3})$ periodic in the two torus variables and zero on $y = 0, h$ satisfies $∥ v ∥_{L^{2} (Q)} \leq h ∥ \nabla v ∥_{L^{2} (Q)}$ . This channel inequality is a separate direct estimate on the periodic cell; it does not treat that cell as a bounded $C^{1}$ Euclidean domain.

In dimension three, let $ρ_{0} > 0$ , $ν = μ / ρ_{0}$ , and suppose already-existing time-independent fields $u_{N S}, u_{S} \in C^{2} (\overline{Ω}; R^{3})$ and $p_{N S}, p_{S} \in C^{1} (\overline{Ω})$ are divergence-free, have the same boundary velocity, and satisfy the constant-density incompressible Navier-Stokes and Stokes equations $ρ_{0} (u_{N S} \cdot \nabla) u_{N S} + \nabla p_{N S} - μ Δ u_{N S} = 0$ and $\nabla p_{S} - μ Δ u_{S} = 0$ . If $f = - ρ_{0} (u_{N S} \cdot \nabla) u_{N S}$ , then the preceding estimate applies to $w = u_{N S} - u_{S}$ and $q = p_{N S} - p_{S}$ whenever the stated Poincare bound holds. This is an a-posteriori residual estimate for already-given fields; it asserts no solution existence, no exterior-domain bound, and no error bound from Reynolds number alone. The constant $C_{Ω} = h$ established for the periodic channel is not transferred to an exterior domain.

Statement references

Guermond, Part I, Chapter 3, energy method and Poincare-Steklov context

Tong, Chapter 2, Couette and Stokes-flow context; not the residual estimate

Facts & Assumptions

Given: The fields, domain, residual equation, zero trace, and Poincare bound in [A1]. For the comparison clause, use the exact formal Navier-Stokes/Stokes conventions in [F1].

[A1]

Assume ${A C}_{ω}$ . Let $n \geq 2$ , let $Ω \subset R^{n}$ be a bounded connected $C^{1}$ domain, $μ > 0$ , $w \in C^{2} (\overline{Ω}; R^{n})$ , $q \in C^{1} (\overline{Ω})$ , $f \in C (\overline{Ω}; R^{n})$ , $\nabla \cdot w = 0$ , $w ∣_{\partial Ω} = 0$ , $- μ Δ w + \nabla q = f$ , and $∥ v ∥_{2} \leq C_{Ω} ∥ \nabla v ∥_{2}$ for every $C^{1}$ zero-trace vector field.

[F1]

Formal parameterized fluid systems and observables on $f d - x 01$ defines, for constants $ρ_{0} > 0$ and $μ > 0$ with $ν = μ / ρ_{0}$ , the pointwise constant-density incompressible equation $u_{t} + (u \cdot \nabla) u + ρ_{0}^{- 1} \nabla p - ν Δ u = 0$ , $\nabla \cdot u = 0$ . Its steady Stokes system is $ρ_{0}^{- 1} \nabla p - ν Δ u = 0$ , $\nabla \cdot u = 0$ , and omits both the time derivative and convection.

[F2]

Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation: for $a, b \in R^{n}$ with the Euclidean inner product and norm, $∣ a \cdot b ∣ \leq ∣ a ∣ ∣ b ∣$ .

[F3]

Divergence on a bounded C1 Euclidean domain: assuming ${A C}_{ω}$ , for $n \geq 2$ , bounded $C^{1}$ Euclidean $Ω$ , and $G \in C^{1} (\overline{Ω}; R^{n})$ , $\int_{Ω} \nabla \cdot G d x = \int_{\partial Ω} G \cdot ν d S$ , with outward normal on every boundary component.

[F4]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): ${A C}_{ω}$ states that every sequence $(X_{j})_{j \in N}$ of nonempty sets has a function $c$ with $c (j) \in X_{j}$ for every $j$ . This exact countable-choice assumption is used only to invoke [F3]; the proof assumes no full Axiom of Choice.

Proof

1.1givenA1F3F4constructalgebra

For each component $i$ , set $G_{i} = w_{i} \nabla w_{i} \in C^{1} (\overline{Ω}; R^{n})$ . The product rule gives $\nabla \cdot G_{i} = ∣ \nabla w_{i} ∣^{2} + w_{i} Δ w_{i}$ . Since $w_{i} = 0$ on the boundary, $G_{i} \cdot ν = w_{i} \partial_{ν} w_{i} = 0$ there. Apply [F3] to each $G_{i}$ and sum over $i$ to obtain $- \int_{Ω} w \cdot Δ w d x = \int_{Ω} ∣ \nabla w ∣^{2} d x$ .

1.2givenA1F3F4constructalgebra

The field $G = q w$ is $C^{1}$ on $\overline{Ω}$ . Its divergence is $\nabla q \cdot w + q \nabla \cdot w$ , and its boundary flux is $q w \cdot ν = 0$ . Applying [F3] and using $\nabla \cdot w = 0$ yields $\int_{Ω} \nabla q \cdot w d x = 0$ .

1.3givenA1F2algebra

Put $A = \int_{Ω} ∣ f ∣^{2} d x$ , $B = \int_{Ω} ∣ w ∣^{2} d x$ , and $C = \int_{Ω} ∣ f ∣ ∣ w ∣ d x$ . By [F2], $∣ \int_{Ω} f \cdot w d x ∣ \leq C$ . For every $t \in R$ , $0 \leq \int_{Ω} (∣ f ∣ - t ∣ w ∣)^{2} d x = A - 2 t C + t^{2} B$ ; if $B > 0$ , choosing $t = C / B$ gives $C^{2} \leq A B$ , while if $B = 0$ , continuity forces $w = 0$ throughout $Ω$ and hence $C = 0$ . Thus $∣ \int_{Ω} f \cdot w d x ∣ \leq ∥ f ∥_{2} ∥ w ∥_{2}$ .

1.4givenF2algebra

On the periodic channel, fix $(x, z, y)$ and set $g (s) = \partial_{y} v (x, s, z)$ , so $v (x, y, z) = \int_{0}^{y} g (s) d s$ because $v$ vanishes at $y = 0$ . By bilinearity and repeated integration, $∣ v (x, y, z) ∣^{2} = \int_{0}^{y} \int_{0}^{y} g (s) \cdot g (t) d s d t$ . By [F2] and $2 a b \leq a^{2} + b^{2}$ , this is at most $y \int_{0}^{y} ∣ g (s) ∣^{2} d s \leq h \int_{0}^{h} ∣ g (s) ∣^{2} d s$ . Integrating over $0 < y < h$ and the two periodic coordinates gives $∥ v ∥_{L^{2} (Q)}^{2} \leq h^{2} ∥ \partial_{y} v ∥_{L^{2} (Q)}^{2} \leq h^{2} ∥ \nabla v ∥_{L^{2} (Q)}^{2}$ ; periodic faces are identified and no $C^{1}$ Euclidean-domain theorem is invoked on this cell.

1.5givenA1F1algebra

For already-existing stationary fields in the comparison clause, [F1] gives $- μ Δ u_{N S} + \nabla p_{N S} = - ρ_{0} (u_{N S} \cdot \nabla) u_{N S}$ and $- μ Δ u_{S} + \nabla p_{S} = 0$ . Subtraction shows $w = u_{N S} - u_{S}$ has zero trace and satisfies the residual equation with $q = p_{N S} - p_{S}$ and the stated $f$ ; it does not assert either field exists.

2.1givenA1step 1.1step 1.2algebra

Take the Euclidean inner product of $- μ Δ w + \nabla q = f$ with $w$ , integrate over $Ω$ , and substitute steps 1.1 and 1.2. The result is $μ ∥ \nabla w ∥_{2}^{2} = \int_{Ω} f \cdot w d x$ .

3.1givenA1step 2.1step 1.3algebra

Let $r = ∥ \nabla w ∥_{2}$ . Steps 2.1 and 1.3 and the assumed Poincare inequality give $μ r^{2} \leq C_{Ω} ∥ f ∥_{2} r$ . If $r = 0$ the gradient bound is immediate; if $r > 0$ , division by $μ r$ gives $r \leq C_{Ω} ∥ f ∥_{2} / μ$ .

4.1givenA1step 3.1algebra

Apply the Poincare inequality once more to $w$ and then use step 3.1 to obtain $∥ w ∥_{2} \leq C_{Ω} r \leq C_{Ω}^{2} ∥ f ∥_{2} / μ$ . This proves both residual estimates.

5.1givenA1step 1.5step 3.1step 4.1∎

The residual fields in step 1.5 satisfy the hypotheses of the first estimate, so steps 3.1 and 4.1 give the stated a-posteriori bounds whenever the explicit Poincare inequality holds.

Bounded-domain Stokes residual estimate

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources