Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 331 / 3

Choose a length $\ell$, velocity $U_s$, time $\ell /U_s$, and pressure $\rho U_s^2$, giving Reynolds number $Re=U_s\ell /\nu$. Write perturbation velocity as $(u,v,w)$ and let $\mathcal{L}={\partial }_t+U{\partial }_x-R{e}^{-1}{\nabla }^2$. Linearizing the incompressible Navier-Stokes equations givesTaking the divergence gives ${\nabla }^{2}p=-2U^{\prime }{v}_x$. Applying ${\nabla }^2$ to the wall-normal momentum equation and usingthen cancels the pressure derivatives and yields $\mathcal{L}{\nabla }^{2}v-U^{\prime \prime }{v}_x=0$. Define wall-normal vorticity $\eta =u_z-w_x$. Applying ${\partial }_z$ to the streamwise equation minus ${\partial }_x$ to the spanwise equation gives $\mathcal{L}\eta =-U^{\prime }{v}_z$.

For the assumed normal modes, set $D=d/dy$ and ${\kappa }^2={\alpha }^2+{\beta }^2$. The two equations become the Orr-Sommerfeld equation and Squire equation:At rigid no-slip walls the conditions are $\hat{v}=D\hat{v}=\hat{\eta }=0$ for nonzero horizontal wavenumber.

An Orr-Sommerfeld mode has $\hat{v}\ne 0$ satisfying the Orr-Sommerfeld equation, with accompanying vorticity satisfying the forced Squire equation. A Squire mode has $\hat{v}=0$ and $\hat{\eta }\ne 0$, so its Squire equation is homogeneous.

For a Squire mode multiply that homogeneous equation by $\overline{\eta }$ and integrate over the channel. Under the no-slip condition $\eta =0$ at the walls, integration by parts givesSince $U,\alpha$ are real, taking the real part provesThe numerator is strictly positive for a nontrivial finite-channel mode, and $Re>0$. Thus Squire modes are exponentially damped in the convention $e^{-i\omega t}$. This does not rule out transient growth from non-normal modes in the coupled system.

With $U=U_0$ and $\beta =0$, the Orr-Sommerfeld equation factors into constant-coefficient operators:For distinct nonzero characteristic roots, the general solution isEither square-root choice for $\gamma$ gives the same solution space. The no-slip boundary condition and impermeability impose $\hat{v}=D\hat{v}=0$ at each wall.

The displayed exponential basis needs the usual repeated-root qualification. If ${\gamma }^2={\alpha }^2\ne 0$, replace it by $(a_1+a_{2}y)e^{\alpha y}+(a_3+a_{4}y)e^{-\alpha y}$. If just one root pair is zero, that pair contributes $1,y$; if both pairs are zero, the basis is $1,y,y^2,y^3$. These are the complete limiting cases, not four independent copies of a repeated exponential.

Take $k^{\ast }>0$ and positive shear scale $S^{\ast }$, with kinematic viscosity ${\nu }^{\ast }$. ChooseThe velocity scale is $S^{\ast }/k^{\ast }$. A reversed shear can be treated by reversing the corresponding coordinate orientation and mode convention. The dimensional Orr-Sommerfeld equation, with $U^{\ast \prime \prime }=0$, isSubstitution of the scales and cancellation gives

In two dimensions at nondimensional streamwise wavenumber one, incompressibility gives $i\hat{u}+D\hat{v}=0$, so $\hat{u}=iD\hat{v}$. The spanwise vorticity is the $z$ component of the curl:ConsequentlyThe symbol $\hat{\omega }$ here denotes a vorticity amplitude, not the temporal frequency used elsewhere. In dimensional variables, ${\hat{\zeta }}^{\ast }=-i(D_{\ast }^2-(k^{\ast }{)}^2){\hat{v}}^{\ast }/k^{\ast }$, with the same sign convention.

Put $W=(D^2-1)\hat{v}$ and ${\delta }^3=R{e}^{-1}$. The constant-shear Orr-Sommerfeld equation givesLet $a=e^{i\pi /6}$ and $Z=a(y-c-i{\delta }^3)/\delta$. Since $d/dy=(a/\delta )d/dZ$ and $a^3=i$, the equation becomesThis is the Airy ordinary differential equation, yielding the Airy reduction of the Orr-Sommerfeld equation in constant shear. It is distinct from the evolution PDE represented by the existing Airy equation article.

Write $W(s)=a_1\mathrm{Ai}(Z(s))+a_2\mathrm{Bi}(Z(s))$, using the independent Airy functions. To solve $(D^2-1)\hat{v}=W$, the variation of parameters formula, equivalently a direct convolution, gives the velocity reconstruction from the Orr-Sommerfeld vorticity:The reference point $y_0$ is arbitrary. Twice differentiating the integral produces $W(y)$ because $\sinh 0=0$ and its derivative at zero is one, verifying the formula. The four constants are independent: applying $D^2-1$ to an identically zero combination first forces $a_1=a_2=0$, then the complementary solutions force $C_{+}=C_{-}=0$. The boundary conditions would be $\hat{v}(y_j)=D\hat{v}(y_j)=0$, $j=1,2$. Their compatibility selects the eigenvalues; they need not be imposed here.