Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 65 / 3

Interpret the two-dimensional dependence as independence of $z$, allowing the velocity to have a vertical component $U=u_z(x,y)$. Write $u_h=(u_x,u_y)$ and $B_h=(A_y,-A_x)$. The resistive induction equation, with a gauge fixing the potential's zero mean, givesThe first is a homogeneous advection-diffusion equation. Multiply it by $A$, average over the periodic box and use integration by parts. Advection contributes zero because $\nabla \cdot u_h=0$; there are no surface terms. ThusMultiplying by $e^{2\eta k^{2}t}$ makes its derivative nonpositive. With the initial norm convention $⟨A^2(0)⟩=A_0^2$, this provesThe Poincaré inequality uses the zero-mean assumption; a spatially constant flux potential would otherwise have no gradient. Zero mean remains preserved by this incompressible periodic equation.

Multiplying the vertical induction equation by $B$ and averaging eliminates advection and givesIntegrate the derivatives of $A$ by parts. Mixed derivatives of $U$ cancel, leavingThe pointwise cross-product bound and the Cauchy-Schwarz inequality bound its absolute value by $Q\sqrt{⟨A^2⟩⟨|\nabla B{|}^2⟩}$, where $Q=\Vert \nabla U{\Vert }_{\infty }$. HenceFor $J=⟨|\nabla B{|}^2⟩$ and $X=⟨A^2⟩$, completing the square givesApply part (i) and integrate, obtaining the periodic two-coordinate anti-dynamo energy boundThe printed bound has only one power of diffusivity and is false as written. Both powers above are needed: one comes from maximizing against diffusion and the other from integrating the decay time of $A$.

For an explicit counterexample in a nondimensional $2\pi$-periodic box, take $u=(0,0,\sin y)$, $A(x,y,0)=\cos x$, $B(x,y,0)=0$, $k=1$. All means vanish and the velocity is solenoidal. The exact solution isIndeed $B_h\cdot \nabla U=e^{-\eta t}\sin x\cos y$ and this mode's Laplacian is $-2B$. Here $A_0^2=1/2$, $Q=1$, and at $t=(\log 2)/\eta$,For $\eta =0.01$ this is $156.25$, exceeding the printed $1/(8\eta )=12.5$. The corrected bound is valid and finite.

Keep part of the diffusion rather than maximizing all of it. The elementary square inequality givesWriting $E=⟨B^2⟩$ and $\lambda =\eta k^2>0$, part (ii), the Poincaré inequality and part (i) implyMultiply by $e^{\lambda t}$ and integrate from zero:Thus stretching by a vertical velocity can transiently generate vertical magnetic energy, but cannot maintain it in this zero-mean periodic, $z$-independent setting. For a strictly planar velocity, $Q=0$ and the passive vertical component decays directly. The stated conclusion depends on positive magnetic diffusivity and the spectral gap supplied by zero mean.