Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 74 / 3

Let $K^2=|\mathbf{k}{|}^2=k_x^2+k_z^2+{\Omega }^2{t}^2{k}_x^2$. For the complex wave before taking its real part,since ${\dot{k}}_y=-\Omega k_x$. The asserted velocity equation therefore follows directly from $\dot{\mathbf{v}}+\Omega v_y\hat{\mathbf{x}}=-i\mathbf{k}q$. To verify the pressure amplitude, solenoidality is $\mathbf{k}\cdot \mathbf{v}=0$. Its derivative requiresand hence $q=2i\Omega k_x{v}_y/K^2$, as claimed.

The transverse amplitude equations areBoth coefficients are real. Differentiating $v_y{v}_z^{\ast }-v_z{v}_y^{\ast }$ makes the terms proportional to $b|v_y{|}^2$ cancel, while the remaining terms give $a(v_y{v}_z^{\ast }-v_z{v}_y^{\ast })$. Therefore the helicity invariant of an inviscid shearing wave obeysHere $H$ is real because the difference in its definition is purely imaginary. Its sign is preserved, and shear reduces its magnitude away from $t=0$.

Write the total velocity as $\mathbf{U}=\Omega y\hat{\mathbf{x}}+\mathbf{u}$ and the magnetic field as $B_0\hat{\mathbf{x}}+\mathbf{b}$. For solenoidal fields the resistive induction equation is $({\partial }_t+\mathbf{U}\cdot \nabla )\mathbf{B}=\mathbf{B}\cdot \nabla \mathbf{U}+\eta {\nabla }^2\mathbf{B}$. The first-order smoothing approximation drops the fluctuating terms $\mathbf{b}\cdot \nabla \mathbf{u}-\mathbf{u}\cdot \nabla \mathbf{b}$, retaining shear exactly. The uniform field is not stretched by the background shear because ${\partial }_x(\Omega y)=0$. The other retained stretching terms are $\mathbf{b}\cdot \nabla (\Omega y\hat{\mathbf{x}})=\Omega b_y\hat{\mathbf{x}}$ and $B_0{\partial }_x\mathbf{u}$. HenceThe same phase cancellation as in part (i), together with ${\nabla }^2{e}^{i\mathbf{k}\cdot \mathbf{x}}=-K^2{e}^{i\mathbf{k}\cdot \mathbf{x}}$, proves that a response can be taken as $\mathrm{Re}(\mathbf{c}(t)e^{i\mathbf{k}(t)\cdot \mathbf{x}})$, whereIf $\mathbf{k}\cdot \mathbf{c}=0$ initially, its derivative is $-\eta K^2\mathbf{k}\cdot \mathbf{c}$, so magnetic solenoidality is preserved.

For $j=y,z$ the response equation is ${\dot{c}}_j+\eta K^2{c}_j=i{k}_x{B}_0{v}_j$. Its integrating-factor solution isThe velocity amplitudes and wavevector vary on the shear timescale. In the fast-diffusion limit $\eta (k_x^2+k_y^2)\gg |\Omega |$, and after the free transient, the integral is concentrated within time $(\eta K^2{)}^{-1}$ of $t$. Replacing the slowly varying factors there by their current values gives the quasistatic magnetic response of a helical shearing wave:Complex conjugation introduces a minus sign in $c_j^{\ast }$. With precisely the EMF normalization specified in the question,A literal spatial average of the product of the two real Fourier waves would be half this quantity; no extra factor of $1/2$ belongs in the question's defined $\mathcal{E}$.

Use $A=k_x^2+k_z^2$ and part (i) to obtain $\mathcal{E}(t)\simeq -B_0{H}_{0}A/[\eta (A+{\Omega }^2{k}_x^2{t}^2{)}^2]$. For the time integral, take the driven response without a free magnetic transient and suppose the fast-diffusion approximation holds through the contributing interval, in particular at $t=0$; $\eta k_x^2\gg |\Omega |$ is a sufficient uniform condition. For $\Omega k_x\ne 0$, substitute $z=|\Omega k_x|t/\sqrt{A}$. Since ${\int }_{-\infty }^{\infty }(1+z^2{)}^{-2}dz=\pi /2$, this givesThe integral identity follows immediately from $z=\tan \chi$, reducing it to ${\int }_{-\pi /2}^{\pi /2}{\cos }^2\chi d\chi$. If $k_x=0$, the forcing and $H$ vanish and the driven EMF is zero. If $\Omega =0$ but $H_0\ne 0$, the leading EMF is constant rather than time-integrable; the finite integral relies on nonzero shear.