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Exact viscous diffusion layer and material-circulation decay

Statement

Let $ρ_{0}, μ > 0$ , $ν = μ / ρ_{0}$ , and $U \in R$ . Use the constant-density incompressible model

$\partial_{t} u + (u \cdot \nabla) u + ρ_{0}^{- 1} \nabla p - ν Δ u = 0, div u = 0,$

with Cartesian derivatives and componentwise Laplacian; its unsteady Stokes system omits the convective term. Set $I = (0, \infty)$ and $Ω_{+} = R \times (0, \infty) \times R$ , with coordinates $(x, y, z)$ . Define $I_{0} = \int_{0}^{\infty} e^{- r^{2}} d r$ and $E (s) = I_{0}^{- 1} \int_{s}^{\infty} e^{- r^{2}} d r$ for $s \geq 0$ . On $I \times Ω_{+}$ define $u : I \times Ω_{+} \to R^{3}$ by $u (t, x, y, z) = (U E (y / (2 \sqrt{ν t})), 0, 0)$ and let $p$ be any constant scalar field. The field is smooth for $t > 0, y > 0$ and satisfies the pointwise unsteady Stokes equation $\partial_{t} u + ρ_{0}^{- 1} \nabla p - ν Δ u = 0$ and $div u = 0$ . It also solves the full constant-density Navier–Stokes system because its convective term vanishes. For each fixed $t > 0$ and $x, z \in R$ , its one-sided boundary limit is $\lim_{y ↓ 0} u (t, x, y, z) = (U, 0, 0)$ and its limit as $y \to \infty$ is $0$ ; for each fixed $x, z \in R$ and $y > 0$ , its limit as $t ↓ 0$ is $0$ . The corner $(t, y) = (0, 0)$ is excluded from the classical claim. For every $x, z \in R$ , $t > 0$ , $0 < ε < 1$ , and $y \geq 2 \sqrt{ν t} \sqrt{\log (1 / ε)}$ , the absolute value of the first velocity component satisfies $∣ U E (y / (2 \sqrt{ν t})) ∣ \leq ε ∣ U ∣$ .

Separately, let $T_{L} = R / (L Z)$ for $L > 0$ , set $T^{3} = T_{L_{x}} \times T_{L_{y}} \times T_{L_{z}}$ , and let $π : R^{3} \to T^{3}$ be the quotient map, where $L_{x}, L_{y}, L_{z} > 0$ . Equip this quotient with the flat smooth structure and metric induced by Euclidean translations, identifying each tangent space with $R^{3}$ through the derivative of $π$ . In this branch, the Navier–Stokes equations mean that the periodic lift satisfies the displayed constant-density system pointwise on $R^{3}$ . Choose an integer $m \geq 1$ , put $k = 2 π m / L_{y}$ , and let $A \in R$ . The periodic lift $\tilde{u} (t, x, y, z) = (A e^{- ν k^{2} t} \sin (k y), 0, 0)$ defines a smooth vector field $u$ on $T^{3}$ by $u (t, π (q)) = d π_{q} (\tilde{u} (t, q))$ ; with constant pressure, it is an exact Navier–Stokes solution for all $t \in R$ . Choose $c, b > 0$ with $c < L_{x} / 2$ , $b < L_{y} / 2$ , and $0 < α : = k b < π$ . Define the closed lifted loop and its quotient by

$\tilde{X} (t, s) = (c \sin (2 π s) + \frac{A}{ν k^{2}} (1 - e^{- ν k^{2} t}) \sin (k b \cos (2 π s)), b \cos (2 π s), 0), X (t, s) = π (\tilde{X} (t, s)), t \in R, 0 \leq s \leq 1.$

The displayed lift satisfies $\tilde{X} (t, 0) = \tilde{X} (t, 1)$ . The loop is material when $\partial_{t} \tilde{X} (t, s) = \tilde{u} (t, \tilde{X} (t, s))$ , as it does here. Define its flat-torus circulation by $Γ (t) = \int_{0}^{1} \tilde{u} (t, \tilde{X} (t, s)) \cdot \partial_{s} \tilde{X} (t, s) d s$ ; this value is unchanged if the lift is translated by a vector in $L_{x} Z \times L_{y} Z \times L_{z} Z$ . Let $C_{α} = \int_{0}^{2 π} \cos θ \sin (α \cos θ) d θ$ . Then $C_{α} > 0$ and $Γ (t) = A c C_{α} e^{- ν k^{2} t}$ , which is nonconstant when $A \neq 0$ . The corner-excluded wall-layer solution and periodic material-loop solution are distinct exact mathematical models; neither states an experiment or a general boundary-layer theorem. Under SI interpretation, $x, y, z, L_{x}, L_{y}, L_{z}, b, c$ are in metres, $t$ in seconds, $u, U, A$ in metres per second, $ρ_{0}$ in kilograms per cubic metre, $μ$ in pascal-seconds, $ν$ in square metres per second, and $k$ in inverse metres.

Statement references

Shreyas Mandre, Fluid Dynamics Lecture Notes, Chapter 3, §3.3.1

MIT 2.25 Advanced Fluid Mechanics, Solution to Stokes First Problem ATP

Constant-density equations and Euclidean circulation conventions: Formal parameterized fluid systems and observables.

Facts & Assumptions

Given: Use the constants, domains, velocities, initial and wall limits, loop orientation and regularity in the Statement. The equations are pointwise formal model equations; the first solution is classical only for $t > 0, y > 0$ .

[F1]

Formal parameterized fluid systems and observables: for constant density, $ν = μ / ρ_{0}$ and the incompressible Navier–Stokes equation is $u_{t} + (u \cdot \nabla) u + ρ_{0}^{- 1} \nabla p - ν Δ u = 0$ with $div u = 0$ ; Cartesian derivatives and componentwise Laplacian are used. This supplies the pointwise system for the half-space field and, through the periodic-lift convention in the Statement, for the torus field. The Euclidean material-loop definition there has domain in an open subset of $R^{3}$ ; it is not used for the torus loop, whose material condition and circulation are defined through lifts in the Statement.

[A1]

On the first domain the pressure is constant. On the torus, $A$ and $k$ are constant and the displayed field is periodic because $k L_{y} = 2 π m$ .

Proof

1.1givenF1A1algebra

The Gaussian integral $I_{0}$ is positive and finite: its integrand is positive and, for $r \geq 1$ , $e^{- r^{2}} \leq e^{- r}$ , whose tail is integrable. Differentiation gives $E^{'} (s) = - e^{- s^{2}} / I_{0}$ and $E^{''} (s) = - 2 s E^{'} (s)$ . For $s = y / (2 \sqrt{ν t})$ , $s_{t} = - s / (2 t)$ and $s_{y} = 1 / (2 \sqrt{ν t})$ , so $u_{t} = - U s E^{'} (s) / (2 t)$ and $ν u_{y y} = - U s E^{'} (s) / (2 t)$ . Thus $u_{t} = ν u_{y y}$ ; the field has zero divergence and zero convection because it has only an $x$ component and is independent of $x$ . The pressure gradient is zero, so [F1] gives the exact equation. Also $E (0) = 1$ , $E (s) \to 0$ as $s \to \infty$ , and for fixed $y > 0$ , $s \to \infty$ as $t ↓ 0$ .

1.2givenF1A1algebra

For the periodic lift, $\partial_{t} \tilde{u} = - ν k^{2} \tilde{u}$ , $Δ \tilde{u} = - k^{2} \tilde{u}$ , $div \tilde{u} = 0$ , and $(\tilde{u} \cdot \nabla) \tilde{u} = {\tilde{u}}_{x} \partial_{x} \tilde{u} = 0$ . A constant pressure therefore makes every term in [F1] sum to zero. Let $y_{0} (s) = b \cos (2 π s)$ and $x_{0} (s) = c \sin (2 π s)$ ; the displacement $d (y, t) = A (1 - e^{- ν k^{2} t}) \sin (k y) / (ν k^{2})$ satisfies $d_{t} (y, t) = A e^{- ν k^{2} t} \sin (k y)$ . Hence $\tilde{X} (t, s) = (x_{0} (s) + d (y_{0} (s), t), y_{0} (s), 0)$ is closed and $\partial_{t} \tilde{X} = \tilde{u} (t, \tilde{X})$ , so its quotient $X = π \circ \tilde{X}$ is a smooth material loop. Translating the lift by a lattice vector changes neither its derivative nor the periodic velocity, proving the stated lift invariance. Its circulation is $\int_{0}^{1} A e^{- ν k^{2} t} \sin (k y_{0} (s)) (x_{0}^{'} (s) + d_{y} (y_{0} (s), t) y_{0}^{'} (s)) d s$ . The second term is the integral of $\partial_{s} H (y_{0} (s), t)$ for a primitive $H_{y} (y, t) = A e^{- ν k^{2} t} \sin (k y) d_{y} (y, t)$ ; it vanishes because $y_{0} (0) = y_{0} (1)$ and $H$ is single-valued. Changing variables $θ = 2 π s$ in the remaining term gives $Γ (t) = A c e^{- ν k^{2} t} \int_{0}^{2 π} \cos θ \sin (k b \cos θ) d θ = A c C_{α} e^{- ν k^{2} t}$ .

2.1givenstep 1.1algebra

Substituting $r = s + z$ in the tail integral gives $E (s) = I_{0}^{- 1} \int_{0}^{\infty} e^{- (s + z)^{2}} d z \leq e^{- s^{2}} I_{0}^{- 1} \int_{0}^{\infty} e^{- z^{2}} d z = e^{- s^{2}}$ , since $(s + z)^{2} \geq s^{2} + z^{2}$ . If $y \geq 2 \sqrt{ν t} \sqrt{\log (1 / ε)}$ , then $s^{2} \geq \log (1 / ε)$ and hence $∣ u ∣ = ∣ U ∣ E (s) \leq ε ∣ U ∣$ .

3.1givenstep 1.2algebra∎

Since $0 < α < π$ , $\sin (α \cos θ)$ has the same sign as $\cos θ$ whenever $\cos θ \neq 0$ . Thus $\cos θ \sin (α \cos θ) > 0$ except at its zeros, so its integral $C_{α}$ is strictly positive. With $A \neq 0$ , $c > 0$ , $ν > 0$ , and $k \neq 0$ , step 1.2 has a nonzero coefficient multiplying the nonconstant exponential $e^{- ν k^{2} t}$ . This exact viscous solution therefore has a smooth material loop with nonconserved circulation; it does not assert Kelvin conservation for a viscous model.

Remarks

The sketch shows the initial material loop in the $(x, y)$ cross-section, with $x_{0} = c \sin θ$ and $y_{0} = b \cos θ$ . It is only a schematic of the exact parametrization in the Statement; the proof uses the parametrized curve, not the drawing.

\begin{tikzpicture}[scale=1.1]
  \draw[->] (-1.4,0) -- (1.5,0) node[right] {$x$};
  \draw[->] (0,-1.1) -- (0,1.2) node[above] {$y$};
  \draw[thick,domain=0:360,samples=80,variable=\t] plot ({0.9*sin(\t)},{0.65*cos(\t)});
  \node at (1.08,0.86) {$y_0=b\cos\theta$};
\end{tikzpicture}

Depends on

Formal parameterized fluid systems and observables