Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 318 (Magnetohydrodynamics)

Use the fully developed flow branch of Hartmann flow. Translation invariance along the walls makes the velocity and induced magnetic field functions of $z$ alone. Incompressible flow and zero normal velocity at the walls give $u_z^{\prime }=0$ and hence $u_z=0$. The zero divergence of the magnetic field makes $B_z$ constant, equal to the imposed $B_0$. There is no forcing in $y$; the homogeneous $y$-components obey the same coupled viscous-resistive equations as the $x$-components, with zero boundary data. Multiplying them by $u_y$ and $b_y/({\mu }_0\rho )$, integrating by parts and adding givesThus $u_y=b_y=0$, establishing the asserted forms on this fully developed flow branch. This is a symmetry reduction of the steady channel model, rather than a claim that every possible flow in a channel is translation invariant.

The current density and Lorentz force density areThe advective acceleration vanishes because $\mathbf{u}\cdot \nabla =u(z){\partial }_x$ acts on fields independent of $x$. The $x$-component of the magnetohydrodynamic momentum equation therefore givesThe $z$-component must also balance: it requires ${\partial }_z(p+b^2/(2{\mu }_0))=0$. Hence a compatible pressure is $p(x,z)=p_{\ast }-\rho Gx-b(z{)}^2/(2{\mu }_0)$; the specified streamwise pressure gradient does not require the ordinary pressure to be uniform in $z$. This accounts for the magnetic pressure of the induced field.

The resistive induction equation with constant magnetic diffusivity gives ${\partial }_t\mathbf{B}=\nabla \times (\mathbf{u}\times \mathbf{B})+\eta {\nabla }^2\mathbf{B}$. Since $\mathbf{u}\times \mathbf{B}=(0,-B_{0}u,0)$, its curl has $x$-component $B_0{u}^{\prime }$. Steadiness consequently givesFinally the no-slip boundary condition gives $u(\pm L)=0$, and the normal magnetic field boundary condition gives $b(\pm L)=0$. The coupled equations and all wall conditions follow directly from momentum balance and magnetic induction.

Integrating the resistive induction equation once introduces a constant $C$:Because $b(L)=b(-L)=0$, integrating this relation again over the channel gives $C=B_{0}Q/(2L)$, where $Q$ is the volumetric flow rate per unit span. Put $m^2=B_0^2/({\mu }_0\rho \nu \eta )$ and $H=mL$, taking $B_0\ge 0$ without loss of generality. Reversing the imposed field reverses $b$ but leaves $u$ and $Q$ unchanged. Eliminating $b^{\prime }$ from the magnetohydrodynamic momentum equation givesIts two zero wall values remove the odd homogeneous solution. Write $u=U[1-\cosh (mz)/\cosh H]$. Integrating the first relation with the two magnetic wall values yieldsTherefore $U=(G{L}^2/\nu H^2)H\coth H$. Substitution and integration produce the velocity and induced field:Here $H=B_{0}L/\sqrt{{\mu }_0\rho \nu \eta }$ is the Hartmann number. The hyperbolic cosine makes $u$ even, while the hyperbolic sine makes $b$ odd. Direct differentiation verifies both coupled equations and all four wall values.

Integrating the velocity gives the fluxThe apparent singularity at $H=0$ is removable. The small-field limit is plane Poiseuille flow:The cubic coefficient of $b$ is its first-order response in $B_0$; $b$ itself vanishes at zero imposed field. For a fixed positive $G$, the flux decreases monotonically as $H>0$ increases. One exact way to see the sketch's monotonicity is the partial-fraction expansion of the hyperbolic cotangent:whose derivative is strictly negative for $H>0$. The flux has a horizontal tangent at $H=0$ and approaches zero asThe flux is even if signed $H$ is used.

For $H\gg 1$, put $\zeta =z/L$. Away from the walls,The velocity has a nearly uniform core, with thin Hartmann layers of thickness $\delta =L/H$ enforcing the no-slip boundary condition. Near the upper wall, with $s=H(1-\zeta )$ fixed, $u/U_c\sim 1-e^{-s}$ and $b/B_s\sim -1+s/H+e^{-s}$; the lower wall follows by even/odd symmetry. Thus $b$ is negative for $0<z<L$, positive for $-L<z<0$, and returns rapidly to zero at both walls. The required sketches show a decreasing flux, a flat velocity core, and an odd induced field with magnetic wall layers.

During the positive half-period, write $Y=(A,B{)}^T$ and $\dot{Y}=K_{+}Y$, with $K_{+}=\left(\begin{array}{cc}0& D^2\\ i& 0\end{array}\right)$. Then $K_{+}^2=i{D}^{2}I={\sigma }^{2}I$. Its two eigenvalues are $\pm \sigma$, and corresponding eigenvectors are $(\sigma ,i{)}^T$ and $(-\sigma ,i{)}^T$. Since $D>0$, they are independent. Thus the general half-period solution isUsing the even and odd terms in the matrix exponential, $e^{T{K}_{+}}=I\cosh (\sigma T)+(K_{+}/\sigma )\sinh (\sigma T)$. Because $D^2/\sigma =-i\sigma$, the fundamental matrix of a linear differential equation givesThe lower-left denominator is $\sigma$, as printed in the PDF. In the negative half-period $K_{-}=\left(\begin{array}{cc}0& -D^2\\ i& 0\end{array}\right)$ and $K_{-}^2=({\sigma }^{\ast }{)}^{2}I$. The analogous matrix exponential has the stated $M^{-}$, replacing $\sigma$ by its complex conjugate ${\sigma }^{\ast }=-i\sigma$. Both matrices satisfywith ${\sigma }^{\ast }$ in the second determinant. Equivalently, their generators have zero trace, so the determinant of each matrix exponential is one.

At each complete period the state is multiplied by the monodromy matrix of a periodic linear system $N=M^{-}{M}^{+}$, so $Y(2nT)=N^{n}Y(0)$. The determinant is therefore $\det N=1$. Let $c_{\pm }=\cosh ({\sigma }^{(\ast )}T)$ and $s_{\pm }=\sinh ({\sigma }^{(\ast )}T)$. Multiplying the two matrices and taking the trace givesThe ratio sum vanishes because the two ratios are $-i$ and $i$. With $x=\sqrt{2}DT$, the hyperbolic-function identity for $|\cosh (a+ia){|}^2$ consequently yieldsFor $x>0$, $F>2$: differentiating gives $F^{\prime }=\sinh x-\sin x>0$, since this derivative has derivative $\cosh x-\cos x>0$ and vanishes initially. The characteristic polynomial is ${\lambda }^2-F\lambda +1$. Hence the two Floquet multipliers are positive, distinct and reciprocal, with dominant multiplier and growth rateThe finite-time propagation within each half-period is bounded independently of the number of cycles, so it does not change this asymptotic Floquet growth rate. The expression is the maximal rate, attained for generic nonzero initial data. The exceptional initial state in the reciprocal multiplier's eigenspace decays with rate $-\gamma$; the zero state stays zero. Thus a vanishing mean alpha effect does not preclude dynamo action in this periodically switched Parker dynamo wave model.

As $x\to \infty$, $F=\frac{1}{2}{e}^x[1+O(e^{-x})]$, $\Lambda =F[1+O(F^{-2})]$, and $\log \Lambda =x-\log 2+o(1)$. ThereforeIn particular, at $T=1$ the rate is asymptotic to $(\sqrt{2}D-\log 2)/2$, as required. The growth calculation uses the product $M^{-}{M}^{+}$, rather than averaging the two generators: the two generators do not commute.

Under the first-order smoothing approximation, neglect the fluctuating product in the resistive induction equation while retaining its contribution to the mean electromotive force. Since the test field ${\mathbf{B}}_0$ is constant and the velocity is a solenoidal vector field, the fluctuation equation isLet $a=\eta k^2>0$, $d=a^2+{\Omega }^2$, and ${\mathbf{C}}_j=i({\mathbf{B}}_0\cdot {\mathbf{k}}_j){\hat{\mathbf{u}}}_j$. For each Fourier mode, the diffusion equation becomes a scalar linear relaxation equation applied to each vector component. Take its long-time periodic response, with $\Omega \ne 0$. For the first mode,while for the second,Thus the complex response coefficients areThey giveAn arbitrary initial fluctuation also contains a homogeneous diffusive transient. In the forced modes this is $\mathrm{Re}{\sum }_j{\mathbf{r}}_j{e}^{i{\mathbf{k}}_j\cdot \mathbf{x}-at}$; more general initial modes also diffuse. The displayed harmonic expression is the periodic particular solution after those transients, not the most general solution at finite time. The constants $a,d$ and the factors of $i$ are unchanged by taking the final real part.

To justify the helicity relation, write ${\hat{\mathbf{u}}}_j={\mathbf{a}}_j+i{\mathbf{c}}_j$ with real vectors. Solenoidality gives ${\mathbf{k}}_j\cdot {\mathbf{a}}_j={\mathbf{k}}_j\cdot {\mathbf{c}}_j=0$. Their cross product is parallel to ${\mathbf{k}}_j$, andThis is the helicity vector of a solenoidal Fourier mode. In these conventions the spatially averaged kinetic helicity density of $\mathrm{Re}({\hat{\mathbf{u}}}_j{e}^{i{\mathbf{k}}_j\cdot \mathbf{x}})$ is $-H_j{k}^2$; the sign is set by the specified cross-product order.

For equal wavevectors, spatial averaging of two real harmonic fields givesTime averaging gives another factor $1/2$, because $⟨{\cos }^2\Omega t{⟩}_t=⟨{\sin }^2\Omega t{⟩}_t=1/2$ and the mixed temporal average is zero. Consequently each diagonal mode contributes to the mean-field electromotive forceDistinct modes with ${\mathbf{k}}_1\ne \pm {\mathbf{k}}_2$ have no spatially averaged cross term. The printed condition also permits the opposite-wavevector case. The possible cross terms still cancel after time averaging: the out-of-phase response coefficients are ${\mathbf{p}}_2=-\Omega {\mathbf{C}}_2/d$ and ${\mathbf{q}}_1=\Omega {\mathbf{C}}_1/d$, while integration by parts gives $⟨{\mathbf{u}}_1\times ({\mathbf{B}}_0\cdot \nabla ){\mathbf{u}}_2{⟩}_x=⟨{\mathbf{u}}_2\times ({\mathbf{B}}_0\cdot \nabla ){\mathbf{u}}_1{⟩}_x$ for the real spatial fields. This handles all the distinct wavevectors allowed by the question, assuming the usual periodic-cell or whole-space spatial average.

Thus the alpha tensor isBoth averaging factors matter. The alpha tensor is real and symmetric, and this contribution vanishes when the mode kinetic helicity densities vanish.

For the two perpendicular wavevectors and common $H_1=H_2=H$, put ${\alpha }_0=\eta H{k}^4/[2({\Omega }^2+{\eta }^2{k}^4)]$. ThenMean-field dynamo action is possible when ${\alpha }_0\ne 0$ and sufficiently long mean-field wavelengths are allowed. The missing $zz$ entry does not prevent an alpha-squared dynamo. Indeed a slowly varying mean magnetic field transverse to $z$, with dependence $e^{iKz+\lambda t}$, obeysThe two growth rates are $-\eta K^2\pm {\alpha }_{0}K$, so the larger is positive if $0<|K|<|{\alpha }_0|/\eta$. This supplies a brief constructive reason; scale separation also requires $|K|\ll k$. A spatially uniform test field itself has zero curl of its mean-field electromotive force and does not grow. If $H=0$, or if boundary conditions exclude all unstable long wavelengths, this particular alpha effect does not yield growth. The conclusion concerns the stated first-order smoothing approximation; no uncomputed turbulent-diffusion correction is assumed.