Use the fully developed flow branch of Hartmann flow. Translation invariance along the walls makes the velocity and induced magnetic field functions of alone. Incompressible flow and zero normal velocity at the walls give and hence . The zero divergence of the magnetic field makes constant, equal to the imposed . There is no forcing in ; the homogeneous -components obey the same coupled viscous-resistive equations as the -components, with zero boundary data. Multiplying them by and , integrating by parts and adding gives
Thus , 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
The advective acceleration vanishes because acts on fields independent of . The -component of the magnetohydrodynamic momentum equation therefore gives
The -component must also balance: it requires . Hence a compatible pressure is ; the specified streamwise pressure gradient does not require the ordinary pressure to be uniform in . This accounts for the magnetic pressure of the induced field.
The resistive induction equation with constant magnetic diffusivity gives . Since , its curl has -component . Steadiness consequently gives
Finally the no-slip boundary condition gives , and the normal magnetic field boundary condition gives . The coupled equations and all wall conditions follow directly from momentum balance and magnetic induction.
Integrating the resistive induction equation once introduces a constant :
Because , integrating this relation again over the channel gives , where is the volumetric flow rate per unit span. Put and , taking without loss of generality. Reversing the imposed field reverses but leaves and unchanged. Eliminating from the magnetohydrodynamic momentum equation gives
Its two zero wall values remove the odd homogeneous solution. Write . Integrating the first relation with the two magnetic wall values yields
Therefore . Substitution and integration produce the velocity and induced field:
Here is the Hartmann number. The hyperbolic cosine makes even, while the hyperbolic sine makes odd. Direct differentiation verifies both coupled equations and all four wall values.
Integrating the velocity gives the flux
The apparent singularity at is removable. The small-field limit is plane Poiseuille flow:
The cubic coefficient of is its first-order response in ; itself vanishes at zero imposed field. For a fixed positive , the flux decreases monotonically as increases. One exact way to see the sketch's monotonicity is the partial-fraction expansion of the hyperbolic cotangent:
whose derivative is strictly negative for . The flux has a horizontal tangent at and approaches zero as
The flux is even if signed is used.
For , put . Away from the walls,
The velocity has a nearly uniform core, with thin Hartmann layers of thickness enforcing the no-slip boundary condition. Near the upper wall, with fixed, and ; the lower wall follows by even/odd symmetry. Thus is negative for , positive for , and returns rapidly to zero at both walls. The required sketches show a decreasing flux, a flat velocity core, and an odd induced field with magnetic wall layers.
Figure 1.
Hartmann-flow flux versus magnetic field strength, and velocity and induced-field profiles at Hartmann number 20
. Details.

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