= Solution
The <MHD induction equation> can be written in material form as
$$
\frac{D\mathbf B}{Dt}
=(\mathbf B\cdot\nabla)\mathbf u-\mathbf B\,\nabla\cdot\mathbf u,
\qquad
\frac D{Dt}=\partial_t+u_x\partial_x+u_z\partial_z.
$$
Its $y$ component immediately gives
$$
\boxed{\frac{DB_y}{Dt}=\mathbf B\cdot\nabla u_y-B_y\nabla\cdot\mathbf u.}
$$
For the <Cartesian magnetic flux function>, the $x,z$ components of the <MHD induction equation>, or the $y$ component of the <magnetic vector potential> equation, give
$$
\partial_t\psi+u_x\partial_x\psi+u_z\partial_z\psi=C(t).
$$
The additive function of time in $\psi$ has no effect on $\mathbf B$. Choose this gauge so that $C=0$. Then \b[the flux label is materially conserved]:
$$
\boxed{\frac{D\psi}{Dt}=0.}
$$
This is <magnetic flux freezing> in the two-dimensional geometry.
To prove the absence of axial motion, construct an invariant solution with $u_y=0$. Along each <Lagrangian trajectory>, $\psi$ is fixed, and the assumed divergence gives
$$
\frac{DB_y}{Dt}=-g(\psi,t)B_y,\qquad
B_y(\mathbf x,t)=
f(\psi(\mathbf x,t))
\exp\left[-\int_0^t g(\psi(\mathbf x,t),s)\,ds\right].
$$
Thus $B_y$ remains a function of $\psi$ and time alone. Its <gradient> stays parallel to $\nabla\psi$, so the $y$ component of the <Lorentz force> remains zero. The axial momentum equation is
$$
\rho\frac{Du_y}{Dt}
=\frac1{\mu_0}(B_x\partial_x+B_z\partial_z)B_y=0,
$$
because $(B_x\partial_x+B_z\partial_z)\psi=0$. With $u_y=0$ initially, it remains zero. Hence \b[neither axial force nor axial motion is generated]. This <flux-surface preservation during magnetic-tube expansion> argument assumes the smooth ideal evolution for which the initial-value problem is unique; the particular in-plane rising motion need not be calculated.
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