= Solution
For the parallel flow $\mathbf u=(u(y,t),0)$ with no pressure gradient, the <Navier-Stokes equations> reduce to the <diffusion equation>
$$
\partial_tu=\nu\,\partial_y^2u.
$$
Substitution of
$$
u=\operatorname{Re}\!\left[U_0f(y)e^{i\omega t}\right]
$$
gives
$$
\nu f''=i\omega f.
$$
The boundary velocities require $f(L_0)=1$ and $f(-L_0)=-1$. The problem is antisymmetric under $y\mapsto-y$, so $f$ is odd. With
$$
k=\sqrt{\frac{i\omega}{\nu}}
=(1+i)\sqrt{\frac{\omega}{2\nu}}
=\frac{(1+i)\Delta}{L_0},
$$
the unique odd solution is
$$
\boxed{
f(y)=\frac{\sinh[(1+i)\Delta\widehat y]}
{\sinh[(1+i)\Delta]},
\qquad
\widehat y=\frac y{L_0},
\qquad
\Delta=\left(\frac{\omega L_0^2}{2\nu}\right)^{1/2}
}.
$$
Back to article page
