= Solution
Use the <fully developed flow> branch of <Hartmann flow>. Translation invariance along the walls makes the velocity and induced <magnetic field> functions of $z$ alone. <Incompressible flow> and zero normal velocity at the walls give $u_z'=0$ and hence $u_z=0$. The zero <divergence> of the <magnetic field> makes $B_z$ constant, equal to the imposed $B_0$. There is no forcing in $y$; the homogeneous $y$-components obey the same coupled viscous-resistive equations as the $x$-components, with zero boundary data. Multiplying them by $u_y$ and $b_y/(\mu_0\rho)$, integrating by parts and adding gives
$$
\nu\int_{-L}^{L}(u_y')^2\,dz+\frac\eta{\mu_0\rho}\int_{-L}^{L}(b_y')^2\,dz=0.
$$
Thus $u_y=b_y=0$, establishing the asserted forms on this <fully developed flow> branch. This is a symmetry reduction of the steady channel model, rather than a claim that every possible flow in a channel is translation invariant.
The <current density> and <Lorentz force density> are
$$
\mathbf J=\frac1{\mu_0}\nabla\times\mathbf B=(0,b'/\mu_0,0),
\qquad \mathbf J\times\mathbf B=\frac1{\mu_0}(B_0b',0,-bb').
$$
The advective acceleration vanishes because $\mathbf u\cdot\nabla=u(z)\partial_x$ acts on fields independent of $x$. The $x$-component of the <magnetohydrodynamic momentum equation> therefore gives
$$
\boxed{0=G+\frac{B_0}{\mu_0\rho}b'+\nu u''.}
$$
The $z$-component must also balance: it requires $\partial_z(p+b^2/(2\mu_0))=0$. Hence a compatible <pressure> is $p(x,z)=p_*-\rho Gx-b(z)^2/(2\mu_0)$; the specified streamwise <pressure gradient> does not require the ordinary pressure to be uniform in $z$. This accounts for the <magnetic pressure> of the induced field.
The <resistive induction equation> with constant <magnetic diffusivity> gives $\partial_t\mathbf B=\nabla\times(\mathbf u\times\mathbf B)+\eta\nabla^2\mathbf B$. Since $\mathbf u\times\mathbf B=(0,-B_0u,0)$, its <curl> has $x$-component $B_0u'$. Steadiness consequently gives
$$
\boxed{0=B_0u'+\eta b''.}
$$
Finally the <no-slip boundary condition> gives $u(\pm L)=0$, and the <normal magnetic field boundary condition> gives $b(\pm L)=0$. \b[The coupled equations and all wall conditions follow directly from momentum balance and magnetic induction.]
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