= 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 $-rB_x+irA_x=irP_x$, so the complex evolution is
$$
P_t=\mathcal L_rP,\qquad \mathcal L_r=\partial_x^2+ir\partial_x-1.
$$
Both 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 gives
$$
\mathcal L_rP=e^{-irx/2}\left[F''+\left(\frac{r^2}{4}-1\right)F\right].
$$
The <eigenfunctions> satisfying the <Dirichlet boundary condition> are therefore $e^{-irx/2}\sin(n\pi x/\ell)$, with temporal <eigenvalues>
$$
\sigma_n=\frac{r^2}{4}-1-\frac{n^2\pi^2}{\ell^2},\qquad n=1,2,\ldots.
$$
The first nonzero stationary mode appears when $\sigma_1=0$, giving
$$
\boxed{r_0=2\sqrt{1+\pi^2/\ell^2},\qquad P_0=C e^{-ir_0x/2}\sin(\pi x/\ell).}
$$
The 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 $P_D=C_0F(y)e^{-ihy}$, so
$$
\boxed{A_D=C_0\cos(\pi y/\ell)\cos(hy),\qquad B_D=-C_0\cos(\pi y/\ell)\sin(hy).}
$$
Here $A_D$ is an <even function> and $B_D$ an <odd function> of $y$, and the cosine envelope vanishes at both boundaries. The quadrupole choice is $P_Q=iP_D$, which gives
$$
\boxed{A_Q=C_0\cos(\pi y/\ell)\sin(hy),\qquad B_Q=C_0\cos(\pi y/\ell)\cos(hy).}
$$
Now $A_Q$ 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.
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