ns/stress — zero-mean waves, nonzero flux

A wave around a ring: w_r = A·cos(nθ), w_θ = B·cos(nθ + φ). Each component averages to zero around the ring. Their product does not: ⟨w_rw_θ⟩ = ½AB·cos φ. In phase, outward motion carries a swirl surplus and inward motion a deficit — both move angular momentum outward. Flip the sign of both components and nothing changes; flip one and the flux reverses. That product is a Reynolds stress, and its radial divergence is a force on the mean flow.

angular frequency · n

phase lag · φ

amplitudes

A

B

w_r (radial) w_θ (azimuthal) w_r·w_θ and its mean

The background's leftover azimuthal residual R_θ(r) is written as a stress divergence, R_θ = −(∂_r + 2/r)T_rθ, and regularity at the axis fixes T_rθ(r) = −r⁻²∫₀^r s²R_θ(s)ds (§3.2). For T to vanish beyond the annulus — so that no wave is needed at infinity — the weighted integral ∫₀^∞ r²R_θ dr must be exactly zero. Shape a residual out of two bumps and watch what a nonzero moment does to the stress: it leaves an r⁻² tail that never ends. The five cumulative integrals of (4.15) exist to kill such tails.

residual R_θ = bump(r₁) − β·bump(r₂)

r₁

r₂

β

residual R_θ(r) stress T_rθ(r) = −r⁻²∫₀^rs²R_θ r²R_θ, whose total integral is the moment

Two pulse families, σ = ±, each with a fixed direction of momentum flux per unit squared amplitude. Because squared amplitudes are positive, the stresses they can jointly realise form a cone — Figure 4 of the paper — and the profiles are built so the required stress T = (T_rθ, T_rz) lies strictly inside it. The cone is not free: it is determined by the background shear through the frame (N, K) and the constant c₀² = (v_s − 2)/2 of §7.1. Drag the target.

shear direction · t_s = −b_s/a

shear strength · v_s

c₀² = (v_s − 2)/2. Near v_s = 2 the pulses barely grow but the cone is the whole half-plane; strong shear narrows it toward −N.

pulse tilt · u*

The realised cone at the pulse midpoint (s = u*) sits inside the admissible cone of (7.1) and opens toward it as u* grows — which is exactly how the paper chooses u*.

admissible cone (4.23)/(7.1) cone realised by the two families at s = u* target T and its decomposition c₁v₊ + c₂v₋

§2.2 · §3.2–3.3 · §4.3 · Lemma 4.5 · Propositions 7.5, 7.6 · Lemma A.8 of the paper

Averaging the nonlinear term

Add a divergence-free increment w with pressure π to a background (u_B, p_B). The residual changes exactly by

R(u_B + w, p_B + π) = R(u_B, p_B) + L_{u_B}(w, π) + ∇·(w ⊗ w)

The linear term is the pulse equation of the pulse page. The quadratic term is the one that matters here: if w oscillates fast with zero mean, its angular average vanishes but the average of w ⊗ w does not, and ∇·⟨w ⊗ w⟩ is a smooth force acting on the mean flow. In cylindrical components the two entries that transport tangential momentum radially are ⟨w_rw_θ⟩ and ⟨w_rw_z⟩; their radial divergences −(∂_r + 2/r) and −(∂_r + 1/r) are what the background's azimuthal and axial residuals are written as (§3.2). Proposition 7.5 chooses positive squared amplitudes y so that the leading covariance equals T; the higher-order terms are smaller by positive powers of q.

Why the moments

Integrating a residual from the axis outward produces a stress; for that stress to be zero beyond the annulus, the total weighted integrals ∫r²R_θ and ∫rR_z must vanish, otherwise stresses proportional to r⁻² and r⁻¹ survive to infinity (§4.2, Lemma A.8). The paper keeps five cumulative radial integrals — M, I, J, S and the pressure increment C_p of (4.15) — and matches them whenever two profile pieces are joined (Lemma 4.4), because the stress at a radius depends on the profile at every smaller radius through these integrals. Each correction stage of §9 solves the same five moment equations again (8.25). The second tab is the one-dimensional version of that bookkeeping.

The cone

In the frame of §7.1 — N along the background's tangential shear g₀ = F₀(−a, b_s), K its quarter turn — the pointwise admissible condition on the leading stress is

T·N < 0, |c₀ · (T·K)/(T·N)| < 1, c₀² = (v_s − 2)/2 (7.1)

which is the same as the inequalities of (4.23): T_θ + t_sT_z > 0 and (v_s − 2)(T_z − t_sT_θ)² < 2(T_θ + t_sT_z)². The two pulse families have transverse amplitude t = x(e_r − sK) + yN with y/x = c₀√(1 + s²) (Lemma 7.4), so their radial–tangential covariance per unit x² is ½(−sK + c₀√(1 + s²)N), with s = ±u* at the midpoint of the pulse. Both directions point into the half-plane T·N < 0 (because c₀ < 0), on opposite sides of −N, and the cone they span opens toward the admissible one as u* grows — since |c₀|·|s|/(|c₀|√(1 + s²)) = |s|/√(1 + s²) → 1. That is how u* is fixed in §7.1: large enough that u*/√(1 + u*²) exceeds the supremum of |c₀(T·K)/(T·N)| over the closed annulus, with the strict margin κ of Theorem 4.6(iii) making the supremum less than one.

Once the target has positive coefficients, so does anything close to it, so a signed change of the stress can be produced by a signed change of the amplitudes about their fixed positive values: the linearised operator ℒ of Proposition 7.6 supplies the stress increments the correction cycle asks for, of either sign, without ever needing a negative squared amplitude.

What the canvases compute

a ring wave — the two components and their product sampled around the ring, averaged numerically, checked against ½AB cos φ.
stress from residual — the integral T_rθ(r) = −r⁻²∫₀^rs²R_θ(s)ds by the trapezoid rule, the total moment, and the size of the tail at the right edge of the plot.
the cone — N, K, c₀ from (t_s, v_s); the admissible cone from (7.1); the two family directions at s = ±u*; the 2×2 solve for (c₁, c₂); and the smallest u* that would put your target inside the realised cone.