Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 57 / 3 / b

Set $\xi =\tanh (z/H)$, so $h=1-{\xi }^2$ and $d/dz=(1-{\xi }^2)H^{-1}d/d\xi$. The vertical equation becomesFor $|\xi |<1$, division by the common factor gives the Legendre differential equation with eigenvalue $k^2{H}^2$. Requiring boundedness at both surfaces $\xi =\pm 1$ selectsThe second independent solution is unbounded at an endpoint. These are the bounded vertical modes of a sech-squared magnetized disk. The division by $H$ is outside the square root, as in the original PDF; the converted TeX places it incorrectly. The magnetic perturbation involves $F_n^{\prime }$, which tends to zero at either surface.

The constant $n=0$ function has $k=0$. The specified magnetic ansatz contains $1/(ik)$ and must not be applied to it. A uniform horizontal velocity with no magnetic perturbation is an epicyclic motion, not a growing magnetic mode; the zero-wavenumber formal degeneracy does not provide growing MRI at arbitrary field strength. Nontrivial magnetically coupled vertical modes have $n\ge 1$.

For $A>0$, the magnetorotational instability grows when $A<3{\Omega }^2$. Indeed, the quadratic in $s^2$ then has a negative constant term, and one positive root. For $A\ge 3{\Omega }^2$, the two roots are nonpositive, since their sum is negative, their product nonnegative, and their discriminant is ${\Omega }^4+16{\Omega }^{2}A>0$. The smallest nonzero vertical eigenvalue is $k_1^2=2/H^2$, so all vertical modes are stable ifThe strict inequality requested gives stability, and equality is marginal for $n=1$. A sufficiently strong field increases magnetic tension. The unstable MRI needs a sufficiently long vertical wavelength to exchange angular momentum without excessive restoring tension; bounded vertical structure imposes a smallest nonzero effective wavenumber. Beyond the finite-thickness magnetorotational instability criterion, no allowed magnetic mode is long enough to remain unstable.