Past exam of the advanced fluid dynamics course of the University of Oxford / 2017 / 1 / d

Use the original scanned boundary condition. The scan fixes the normal-field amplitude at the corner $x=z=0$. The supplied TeX drops the condition $x=0$ and would instead prescribe a constant normal field on the entire bottom. Those are different boundary-value problems; the solution below uses the legible original scan.

Choose a positive integer $n$, put $k=n\pi /a$, and seek a separated decaying mode. A suitable flux function isThis is a decaying linear force-free magnetic arcade. The first and third curl components then satisfy the force-free equation automatically; its second component requires $-({\partial }_x^2+{\partial }_z^2)A={\lambda }^{2}A$. ThusDirect differentiation gives $\nabla \cdot \mathbf{B}=0$ and $\nabla \times \mathbf{B}=\lambda \mathbf{B}$. Both horizontal components vanish at $x=0,a$, and $B_z(0,0)=B_0$. The magnetic field on the bottom is $B_0\cos kx$, not a uniform $B_0$.

A strictly decaying real normal mode requires $q>0$, soFor the fundamental, single-arcade choice $n=1$, this becomes $|\lambda |<\pi /a$. Equality gives no vertical exponential decay, while a larger magnitude gives an oscillatory rather than decaying vertical dependence for that mode. The data do not explicitly fix the horizontal mode number. If arbitrarily fine higher-mode arcades are permitted, every finite real $\lambda$ admits some integer $n$ with $n\pi /a>|\lambda |$. The fundamental restriction must therefore not be advertised as a necessary bound for every possible field satisfying only the stated pointwise and wall data.

For a fundamental field line, $A$ is constant along its projection in the $(x,z)$ plane. If $0<s<1$ denotes its normalized flux-function value, its arch isThe maximum height is $q^{-1}\log (1/s)$ at $x=a/2$. In the horizontal projection,Thus the vertical projections are arches and the horizontal projections are straight sheared segments joining their paired footpoints. Positive $\lambda$ gives positive horizontal slope, negative $\lambda$ reverses the shear, and $\lambda =0$ gives an unsheared potential arcade. Higher $n$ repeats such cells across the width.

The transcription's alternative uniform-bottom condition would be inconsistent with exponential decay and zero side flux for $B_0\ne 0$: solenoidality gives $d{\int }_0^a{B}_{z}dx/dz=-B_x(a,z)+B_x(0,z)=0$, whereas a uniform nonzero bottom would give a conserved flux $a{B}_0$ that cannot decay to zero. This integral check is an independent reason not to silently solve that mistranscribed problem.