Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 74 / 1

Adding $i$ times the equation for $B$ to the equation for $A$ combines the two real amplitudes into $P=A+iB$. With the fluctuating coupling absent, the derivative terms satisfy $-r{B}_x+ir{A}_x=ir{P}_x$, so the complex evolution isBoth real amplitudes satisfy the zero Dirichlet boundary condition, hence $P(0)=P(\ell )=0$. A unit-modulus change of dependent variable removes the first derivative: put $P=e^{-irx/2}F$. Direct differentiation givesThe eigenfunctions satisfying the Dirichlet boundary condition are therefore $e^{-irx/2}\sin (n\pi x/\ell )$, with temporal eigenvaluesThe first nonzero stationary mode appears when ${\sigma }_1=0$, givingThe complex constant $C$ supplies the phase freedom of this unperturbed mean-field dynamo model.

To display the dipole and quadrupole parity in a mean-field dynamo, center the coordinate at the midpoint: $y=x-\ell /2$, $F(y)=\cos (\pi y/\ell )$, $h=r_0/2$. Choose a real amplitude $C_0$. The dipole choice is $P_D=C_{0}F(y)e^{-ihy}$, soHere $A_D$ is an even function and $B_D$ an odd function of $y$, and the cosine envelope vanishes at both boundaries. The quadrupole choice is $P_Q=i{P}_D$, which givesNow $A_Q$ is an odd function and $B_Q$ an even function. Both choices have the same stationary threshold because the unperturbed complex equation is invariant under a constant phase rotation.

Write $P=P_0+d{P}_1+O(d^2)$ and $r=r_0+d{r}_1+O(d^2)$, with each coefficient satisfying the zero Dirichlet boundary condition. The steady complex equation, including the fluctuating coupling, is $0={\mathcal{L}}_{r}P-dB\sin (2\pi x/\ell )$. Since $B_0=(P_0-P_0^{\ast })/(2i)$, its order-$d$ terms giveThis is a regular perturbation equation for the threshold shift. The right projection uses the complex inner product, not a bilinear product without conjugation.

The linear operator ${\mathcal{L}}_{r_0}$ is self-adjoint on the domain $H^2(0,\ell )\cap H_0^1(0,\ell )$ in complex L2 space. For functions $v,w$ in that domain, two integration by parts steps giveThe boundary term vanishes. In particular $i{r}_0{\partial }_x$ is formally self-adjoint: both complex conjugation of $i$ and integration by parts contribute a minus sign. Taking $v=P_0$, $w=P_1$ and using ${\mathcal{L}}_{r_0}{P}_0=0$ provesThis is the Fredholm solvability condition for a self-adjoint operator. Define $s(x)=\sin (2\pi x/\ell )$ and $N={\int }_0^{\ell }|P_0{|}^{2}dx$. The gauged sine mode hassince the real envelope contributes an integral of its own derivative, whose endpoint term is zero. In particular the denominator in the solvability condition is nonzero. ThereforeFor either parity choice this expression is real. Indeed $|P_0{|}^2$, $A_0^2$ and $B_0^2$ are even about the midpoint, while $s$ is odd. Hence $\int |P_0{|}^{2}s=\int B_0^{2}s=0$, and the last numerator reduces to $\int A_0{B}_{0}s$.

No integral evaluation is needed to compare the two shifts. The phase change $P_Q=i{P}_D$ gives $A_Q=-B_D$, $B_Q=A_D$, and leaves $N$ and $I$ unchanged. Thus $A_Q{B}_Q=-A_D{B}_D$, andEquivalently, the term involving $|P_0{|}^2$ vanishes by parity, while $(P_Q^{\ast }{)}^2=-(P_D^{\ast }{)}^2$ reverses the remaining perturbation numerator. This is parity splitting of a mean-field dynamo threshold: the perturbation breaks the phase degeneracy and lowers one onset threshold by the amount it raises the other.