 Solution

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.