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If $H$ is continuous, piecewise twice differentiable, and has derivative jump $[H^{\prime }]=H^{\prime }(y+)-H^{\prime }(y-)$ at $y$, then its distributional second derivative is $H_{\mathrm{reg}}^{\prime \prime }+[H^{\prime }]{\delta }_y$. If the regular part is integrable and the boundary terms vanish, two integrations by parts give the displayed asymptotic expansion by the Riemann-Lebesgue lemma. A jump $[H]$ in $H$ itself instead produces $[H]e^{-iqy}/(iq)$ at leading order. This distinguishes the different decay rates of the streamfunction and streamwise velocity in linear inviscid damping in unbounded Couette flow.

With the streamfunction convention $(u,v)=({\psi }_y,-{\psi }_x)$, disturbance vorticity is $\omega =-\Delta \psi$. The basic Couette flow has constant vorticity $-1$. Linearizing the two-dimensional Euler equations for an inviscid fluid in vorticity form givesThe second term vanishes, soThis derivation uses incompressibility and an inviscid homogeneous fluid, without introducing a viscous decay term.

For the initial streamwise Fourier mode, write $\psi =e^{i\alpha x}a(y,t)$. Its transported Laplacian has amplitudeThe decaying Green function on the line isThis is the negative of the one-dimensional modified Helmholtz Green function for $-{\partial }_y^2+{\alpha }^2$: it is continuous and its first derivative with respect to $y$ has jump one at $y=s$. Consequently $({\partial }_y^2-{\alpha }^2)G=\delta (y-s)$, and inversion yieldsUnder the usual decay of the initial profile and its derivatives, the initial integral returns $\varphi$. There is no nonzero homogeneous solution of $a^{\prime \prime }-{\alpha }^{2}a=0$ decaying at both infinities.

For the large-time calculation, put $q=\alpha t$ and $H(s)=G(y,s)F(s)$, keeping the observation point $y$ fixed. Assume sufficient localization to justify the following integrations by parts; for example, $F$ may be a Schwartz function. The function $H$ is continuous, but its derivative in $s$ has jumpsince $G_s(y,y+)-G_s(y,y-)=1$. The Fourier decay from a Green-function derivative jump givesHere the regular second derivative is computed separately on either side of $s=y$. If it is integrable, its Fourier transform tends to zero by the Riemann-Lebesgue lemma. HenceThe cusp in the Green function, rather than a stationary point of the linear phase, supplies this leading contribution.

To obtain the streamwise velocity, differentiate the integral itself. This avoids an unjustified differentiation of a pointwise remainder. The kernel $G_y$ has a jump of $-1$ as a function of $s$ at $s=y$, so one integration by parts givesThe transverse velocity follows exactly from $v=-i\alpha \psi$, while the Laplacian equation gives vorticity without an asymptotic approximation:Thus the generic orders are $u=O(t^{-1})$, $v=O(t^{-2})$, and $\omega =O(1)$, at each fixed $y$. If $F(y)=0$, the leading coefficients vanish and faster decay can occur. For real physical disturbances, take the real part of these complex Fourier modes. The vorticity generally keeps oscillating rather than approaching a pointwise limit, even while the velocity decays. This is linear inviscid damping in unbounded Couette flow caused by phase mixing.

For arbitrary initial vorticity, the characteristics satisfy $\dot{x}=y$, $\dot{y}=0$, and $\dot{\omega }=0$. Therefore the exact solution isPutting ${\omega }_0=-F(y)e^{i\alpha x}$ recovers the preceding nondecaying mode. In general the shear is area preserving, so its transport preserves every finite Lp norm of vorticity. Local pointwise behavior for arbitrary ${\omega }_0$ depends on its spatial profile: an initially localized profile can move away from a fixed point. There is no universal pointwise nonzero limit or $t^{-2}$ decay of vorticity for arbitrary initial data.

The stated smoothness and decay of $\varphi$ alone do not guarantee the claimed algebraic rates. The intended integration by parts calculation requires control of derivatives and tails; convergence of the displayed integral alone is insufficient. Here is a counterexample to the literal unrestricted rate, with $\alpha =1$. Choose a nonnegative smooth function $r$ supported in $(-1/4,1/4)$ with $\int r=1$, and a sufficiently rapidly increasing sequence $t_n$. SetChoose $\sum e^n/t_n<\infty$. Then $F$ is smooth and integrable, $\varphi$ is smooth and tends to zero at both infinities, and every evolution integral converges absolutely. At the observation point $(x,y)=(0,0)$, all supports have $s>0$, soAt $t=t_n$, the $n$th summand is $-1/(2t_n)$. The Fourier transform of $r$ decays faster than any inverse power. Inductively choose $t_n$ so large that all earlier summands at $t_n$ together have modulus less than $1/(8t_n)$, and choose subsequent times so that ${\sum }_{m>n}1/t_m<1/(4t_n)$. The later summands then also have modulus less than $1/(8t_n)$. Thus $|\psi (0,0,t_n)|\ge 1/(4t_n)$, contradicting $O(t^{-2})$. This construction meets the printed assumptions; what it lacks is uniform derivative localization. The preceding boxed decay formulas are the correct intended result under the explicitly stated sufficient localization hypothesis.