Shearing wave on an affine flow

Core idea

Suppose $U(t,x)=A(t)x$ , with $\operatorname{tr}A=0$ , is a smooth unforced Navier–Stokes solution with pressure $p_U$ . Let $k(t)\in\mathbb R^d\setminus\{0\}$ and $a(t)\in\mathbb C^d$ solve

k'=-A^T k,\qquad a'=-Aa+2\frac{k(k\cdot Aa)}{|k|^2}-\nu|k|^2a,

with $k(0)\cdot a(0)=0$ . Then the shearing wave

u=U+\operatorname{Re}(a(t)e^{ik(t)\cdot x})

is an exact solution with pressure

p=p_U+\operatorname{Re}\left(\frac{2i(k\cdot Aa)}{|k|^2}e^{ik\cdot x}\right).

Why the nonlinear wave term vanishes

The wavevector equation transports the phase by $U$ . Differentiating $k\cdot a$ with the displayed ODEs gives zero, preserving transversality. Since the amplitude is spatially constant and perpendicular to $k$ , every term in $(w\cdot\nabla)w$ , for $w=\operatorname{Re}(ae^{ik\cdot x})$ , vanishes. The remaining linear and viscous terms give exactly the amplitude and pressure formulas.

A simple shear and scope

For $U=Sx_1e_2$ , one has $k_1(t)=k_1(0)-Stk_2(0)$ , with the other components constant. General affine backgrounds use the same displayed evolution. These waves on all of Euclidean space are generally not finite-energy fields. Spatial localization creates additional terms that need correction; the exact plane-wave identity alone does not supply a localized solution.

References

Nishant K. Singh and S. Sridhar, Plane shearing waves of arbitrary form: exact solutions of the Navier–Stokes equations.