= Solution
Take $B_0=B_z>0$, $v_a=B_0/\sqrt{\mu_0\rho_0}$ and $\zeta=k(z-v_at)$. The transverse <ideal magnetohydrodynamic induction equation> and <magnetohydrodynamic momentum equation> are satisfied by $u_x=-B_x/\sqrt{\mu_0\rho_0}$, $u_y=u_z=0$, with constant <mass density>. However, the longitudinal <magnetohydrodynamic momentum equation> also requires
$$
0=-\partial_z\left(p+\frac{B_x^2}{2\mu_0}\right).
$$
Here $B_x^2=a^2B_0^2\cos^2\zeta$ is not constant. Balancing its <gradient> would require
$$
p=p_*(t)-\frac{a^2B_0^2}{2\mu_0}\cos^2\zeta.
$$
The adiabatic <pressure> equation instead gives $\partial_t p=0$, since this <velocity> is divergence-free and has no component along the spatial variation. No choice of $p_*(t)$ removes the spatially varying time derivative. Thus \b[a nontrivial finite-amplitude linearly polarized sinusoidal <Alfvén wave> is not an exact pure traveling wave in a compressible ideal fluid.] The <magnetic-pressure obstruction to a linearly polarized Alfvén wave> is quadratic in $a$, so the wave is a valid linear approximation and drives a compressive response at higher order. In an incompressible model <pressure> is a constraint variable; it is not governed by the same compressible adiabatic evolution law.
Allowing longitudinal flow and varying <mass density> does not rescue a regular one-dimensional periodic traveling solution with precisely this transverse <magnetic field>. To see this, let $U=u_z-v_a$ and use the traveling coordinate. Integrated continuity, induction and transverse momentum give
$$
\rho U=m,\qquad UB_x-B_0u_x=A,\qquad
mu_x-\frac{B_0B_x}{\mu_0}=D,
$$
where $m,A,D$ are constants. Eliminating $u_x$ gives
$$
\left(\frac{m^2}{\rho}-\frac{B_0^2}{\mu_0}\right)B_x=mA+B_0D.
$$
At any zero of $B_x$, finite <mass density> and <velocity> require the right side to vanish. Wherever $B_x\ne0$, the coefficient then vanishes and fixes $\rho=\mu_0m^2/B_0^2$, constant; continuity fixes $U$, and adiabatic <entropy> transport fixes $p$. The longitudinal <magnetohydrodynamic momentum equation> again contradicts the varying $B_x^2$. The alternative $m=0$ makes transverse momentum force $B_x$ constant. Trivial cases such as $a=0$ or $k=0$ are excluded from the nontrivial wave conclusion.
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