 Solution

Start from the constant-coefficient mean-field dynamo equationHere $\eta$ may include turbulent as well as microscopic diffusion. Use a local Cartesian model, neglect meridional circulation and curvature, and assume independence of $y$. Represent the poloidal magnetic field by a vector potential along $y$ and the toroidal magnetic field by its $y$ component:The shear contributes $(\overline{B}\cdot \nabla )U=\Omega A_x\hat{y}$. The poloidal part of $\alpha \nabla \times (B\hat{y})$ gives $A_t=\alpha B+\eta {\nabla }^{2}A$, while the toroidal part of $\alpha \nabla \times {\overline{B}}_{\mathrm{pol}}$ is $-\alpha {\nabla }^{2}A$. ThereforeRetaining one common transverse dependence $e^{i\ell z}$ changes ${\nabla }^2$ to ${\partial }_x^2-{\ell }^2$, producing the two requested equations. The first coupling is the alpha effect regenerating poloidal field from toroidal field; the shear is differential-rotation winding in an Alpha-Omega dynamo. Retaining the alpha contribution in the second equation also allows alpha-squared dynamo feedback, rather than making the pure Parker shear-dominated approximation.