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Take the primitive state vector $\mathbf{U}=(\rho ,p,u_x,u_y,u_z,B_x,B_y,B_z{)}^T$, with $\rho >0$. In the absence of gravity, the ideal magnetohydrodynamic equations are first order: expanding the material derivatives, pressure gradient and Lorentz force makes every term linear in a spatial derivative of $\mathbf{U}$, with coefficients depending on $\mathbf{U}$. Dividing momentum balance by $\rho$ therefore gives a quasilinear partial differential equation system ${\partial }_t\mathbf{U}+A_i(\mathbf{U}){\partial }_i\mathbf{U}=0$ with eight-by-eight matrices. The solenoidal magnetic-field constraint is an additional constraint on initial data; the ideal magnetohydrodynamic induction equation preserves it because the divergence of a curl vanishes.

For a one-dimensional simple wave in magnetohydrodynamics, write $\mathbf{U}=\mathbf{U}(q(x,t))$ with a nonzero state-space tangent. Substitution gives ${\mathbf{U}}_q{q}_t+A_x{\mathbf{U}}_q{q}_x=0$. A nonconstant profile requires the tangent to be a right eigenvector of $A_x$:The chain rule then gives the wave-speed equation:This is the Inviscid Burgers equation, including the special case of constant $v$. Its characteristic curves are $x=a+v_0(a)t$, with $v=v_0(a)$. As long as the mapping remains invertible,Thus a region with $v_0^{\prime }<0$ steepens: faster characteristics catch slower ones, and the first gradient catastrophe occurs at $t_{\ast }=-1/{\min }_a{v}_0^{\prime }(a)$ when the minimum is negative. A genuinely nonlinear compressive simple wave therefore forms a shock wave. Rarefactive profiles can spread instead; steepening is not inevitable for every initial profile.

To continue past this time, use weak solutions of the conservative mass, momentum, fluid total-energy equation and ideal magnetohydrodynamic induction equation. The Rankine-Hugoniot conditions fix the jumps and shock velocity, while physical entropy production selects admissible magnetohydrodynamic shocks. Diffusive shock wave layers are replaced by moving discontinuities, so their microscopic structure need not be explicitly resolved. In particular, the smooth adiabatic pressure equation must be replaced by total-energy conservation when imposing the ideal magnetohydrodynamic shock conditions.

For the Alfvén waves, impose the one-dimensional solenoidal magnetic-field constraint, so $B_x$ is constant and $\delta B_x=0$. Put ${\mathbf{B}}_{\perp }=(B_y,B_z)$ and $c=v-u_x$. The transverse components of the linearized ideal magnetohydrodynamic equations giveThe longitudinal momentum equation additionally gives $\delta p+{\mathbf{B}}_{\perp }\cdot \delta {\mathbf{B}}_{\perp }/{\mu }_0=0$. The Alfvén eigenvectors have $\delta \rho =\delta p=\delta u_x=0$, hence their transverse polarization is perpendicular to ${\mathbf{B}}_{\perp }$. For $s=\pm 1$, their speed and explicit right Alfvén characteristic eigenvectors areFor ${\mathbf{B}}_{\perp }\ne 0$ one may choose $(h_y,h_z)=(B_z,-B_y)$, up to a nonzero scalar factor. If ${\mathbf{B}}_{\perp }=0$, either transverse polarization is allowed and this characteristic speed is degenerate. These expressions describe the two propagating Alfvén branches for $B_x\ne 0$; if $B_x=0$, they coalesce with advected, nonpropagating transverse disturbances.

Integrating along an Alfvén simple wave leaves $\rho ,p,u_x,B_x$ unchanged and gives $d{\mathbf{u}}_{\perp }=-sd{\mathbf{B}}_{\perp }/\sqrt{{\mu }_0\rho }$ and ${\mathbf{B}}_{\perp }\cdot d{\mathbf{B}}_{\perp }=0$. Consequently the finite-amplitude nonlinear Alfvén wave relations areThe constant-magnitude condition is essential: otherwise a varying magnetic pressure would drive longitudinal compression. An arbitrary smooth phase profile gives an explicit family, $f_3=B_{\perp }\cos \Theta (x-v_{s}t)$ and $f_4=B_{\perp }\sin \Theta (x-v_{s}t)$, with $f_1,f_2$ as above. Direct substitution in the transverse momentum and ideal magnetohydrodynamic induction equation gives $(u_x-v_s){\mathbf{u}}_{\perp }^{\prime }=B_x{\mathbf{B}}_{\perp }^{\prime }/({\mu }_0\rho )$ and $(u_x-v_s){\mathbf{B}}_{\perp }^{\prime }=B_x{\mathbf{u}}_{\perp }^{\prime }$, confirming the solution without relying on the eigenvector argument. Longitudinal momentum holds because $p+B_{\perp }^2/(2{\mu }_0)$ is constant. The speed $v_s$ is independent of amplitude along this family: these linearly degenerate characteristic fields translate without steepening.

Let $e$ be specific internal energy, $Q$ the heat supplied per unit volume per unit time, and $E=\rho (e+|\mathbf{u}{|}^2/2)$. For an inviscid perfect gas in a prescribed Newtonian gravitational potential $\Phi$, the fluid total-energy equation isThere is no gravitational term if no external body is present. For a time-independent $\Phi$, adding potential energy gives the equivalent conservative form ${\partial }_t(E+\rho \Phi )+\nabla \cdot [(E+p+\rho \Phi )\mathbf{u}]=Q$. Subtracting the kinetic-energy equation and using mass conservation givesFor a perfect gas with constant specific-heat ratio $\gamma$, $e=p/[(\gamma -1)\rho ]$. The isothermal equation of state therefore makes $e=c_s^2/(\gamma -1)$ constant. Hence the required heat supply and heat loss areCompression requires cooling; expansion requires heating. The isothermal sound speed $c_s$ differs from the adiabatic sound speed $\sqrt{\gamma }{c}_s$.

Define the isothermal Mach numbers by ${\mathcal{M}}_i=u_i/c_s$. Orient the normal to a stationary isothermal shock along the flow, so the mass flux $j={\rho }_1{u}_1={\rho }_2{u}_2>0$. Tangential velocity is continuous for this planar hydrodynamic shock. Normal momentum conservation givesFor a genuine discontinuity $u_1\ne u_2$, factorization gives $u_1{u}_2=c_s^2$. Thus the isothermal shock jump relations areAcross a thin shock, gravitational potential and tangential kinetic energy are unchanged. The specific enthalpy $h=\gamma c_s^2/(\gamma -1)$ is the same on both sides, so the energy removed per unit area per unit time isPositive net cooling requires $u_1>u_2$, equivalently ${\rho }_2>{\rho }_1$ and ${\mathcal{M}}_1>1>{\mathcal{M}}_2$. Cooling across an isothermal shock permits only compression shocks; the formal expansion discontinuity requires energy supplied by the surroundings. The continuous state $u_1=u_2$ is not a shock.

For steady radial flow, mass conservation gives $r^2\rho u_r=\text{constant}$. Differentiating it and eliminating the mass density gradient from radial momentum balance givesA regular sonic point has $|u_r|=c_s$, so the right-hand side must vanish there as well. ThereforeThis is the Bondi sonic point for accretion; the same critical radius appears in an isothermal Parker wind. A general steady solution need not pass through a sonic point, but a smooth transonic one must obey both critical conditions.

Put $x=r/r_s$ and $\mathcal{M}=|u_r|/c_s$ on a branch whose flow direction is fixed. The radial equation becomesIntegrating yields the integrated isothermal Bondi flow relation:For a transonic branch, $x=\mathcal{M}=1$ gives $C=-3/2$. Expansion about the critical point gives $(\mathcal{M}-1{)}^2=(x-1{)}^2+O((x-1{)}^3)$, with slope $-1$ for inflow supplied by gas at rest at infinity and slope $+1$ for a transonic outflow.

For Isothermal Bondi accretion, integrate radial momentum balance once more, using $u_r\to 0$ and $\rho \to {\rho }_{\infty }$ at infinity:At the Bondi sonic point, $GM/r_s=2c_s^2$ and $u_r^2=c_s^2$, so ${\rho }_s=e^{3/2}{\rho }_{\infty }$. The positive inward isothermal Bondi accretion rate is consequentlyThe critical mass density and rate are fixed by the regular transonic solution and the specified reservoir at infinity; an arbitrary subsonic solution does not share this accretion rate.