Supersonic Flow Can't See Upstream — Characteristics and Sound Waves in the Euler Equations

A supersonic fighter arrives before the sound of its own engine. The roar catches up only after the jet has passed overhead. The split is already in the characteristic speeds of the Euler equations. A small disturbance breaks into three signals, and their tracks in the $(x,t)$ plane are the characteristic curves (the path a signal leaves behind).

Set the Mach slider to 0.4 and the blue left-going wave reaches upstream of the source. Push it past 1 and that track leaves the picture.

That is the moment the slope of the blue line changes sign. Watch whether all three signals run only to the right, and whether the left side goes empty.

The compressible equations are signal propagation#

One-dimensional inviscid compressible flow is written as three conservation laws.

Here $\rho$ is density, $u$ is velocity, $P$ is pressure, and $e_{\text{tot}}$ is the total specific energy. On the surface these look like equations for a fluid that "flows." But the essence of these equations is not transport — it's signal propagation. All the dynamics of the flow decompose into a handful of signals, each running at a fixed speed. To see what those speeds are, we just need to linearize the equations for a moment.

A small disturbance builds a wave equation#

Lay a very small disturbance on top of a background state $(\rho_0, u_0, P_0)$. Set $\rho = \rho_0 + \rho_1$ and $u = u_0 + u_1$, use $P = K\rho^\gamma$, and keep only first-order terms. Define the derivative that moves with the background, $D_t \equiv \partial_t + u_0\,\partial_x$ (the co-moving derivative). Combining the two disturbance equations yields a single wave equation.

Here $c_0$ is the speed of sound.

$\gamma$ is the specific-heat ratio. Plugging $\rho_1 \sim e^{i(kx-\omega t)}$ into this equation drops out the dispersion relation.

$\omega$ is the angular frequency and $k$ is the wavenumber. The disturbance rides along with the background flow $u_0$ and spreads left and right at the sound speed $c_0$. In other words, a disturbance at one point splits into two speeds: $u_0 + c_0$ and $u_0 - c_0$.

Characteristics that split three ways#

The two sound waves aren't the whole story. The third signal is the fluid itself. Suppose we dye the fluid. If at $t=0$ we color the left side blue and the right side red, the color boundary $\phi$ is simply carried along with the flow.

This signal moves at speed $u_0$. This is the entropy (or contact) characteristic. To summarize, a one-dimensional flow has three characteristic speeds.

The lines in the $(x,t)$ plane that satisfy $\dfrac{dx}{dt} = \lambda$ are called characteristic curves. The $\lambda_\pm$ are the sound waves, and $\lambda_0$ is the transport of material. Along each acoustic characteristic the Riemann invariant $u \pm \dfrac{2c}{\gamma-1}$ stays constant, and along the material characteristic the entropy is conserved.

At $M = u_0/c_0$ of 0.4, $\lambda_-$ is still negative, so the left-going sound wave reaches upstream. Once $M$ passes 1, $\lambda_- = u_0 - c_0$ turns positive and all three signals run only downstream.

The same split in more than one dimension picks up a repeated root and tangential pieces, laid out in the Euler eigensystem for an arbitrary face normal. In one dimension that repeat folds into the single entropy wave.

Above Mach 1, the cone of influence tips over#

The same story is sharper on a spacetime diagram. The wedge between the three characteristic curves $C_-,\,C_0,\,C_+$ emanating from the source is the domain of influence. It's the set of spacetime events the source can send a signal to.

When $M<1$, this wedge opens on both sides of the vertical line. A signal can travel either upstream or downstream. When $M>1$, the whole wedge tips over to the right. Even the $C_-$ characteristic leans to the right, so no upstream event is inside the domain of influence. This is why, in supersonic flow, "upstream can't see downstream." Numerically, this is exactly the reason you must not impose a boundary condition at a supersonic outlet.

In the diagram below, drag the round probe around.

Place the probe inside the wedge and it turns green (reachable); place it outside and it turns red (unreachable). Raise $u_0$ above 1, and no matter where you put the probe to the left of the source, it stays red. Upstream is completely cut off from the signal.

Checking the characteristic speeds in Python#

Let's confirm that the characteristic speeds really are $u \pm c$ and $u$. Writing the 1D Euler equations in primitive variables $W=(\rho,u,P)$ gives the quasilinear form $\partial_t W + A(W)\,\partial_x W = 0$, and the eigenvalues of the coefficient matrix $A$ are exactly the characteristic speeds.

import numpy as np
 
def characteristic_speeds(rho, u, p, gamma=1.4):
    """Jacobian A(W) of the 1D Euler primitive variables, W=(rho,u,p).
    Eigenvalues of A = the three characteristic speeds."""
    c = np.sqrt(gamma * p / rho)              # speed of sound
    A = np.array([[u,      rho,        0.0  ],
                  [0.0,    u,          1/rho],
                  [0.0,    rho*c**2,   u    ]])
    lam = np.sort(np.linalg.eigvals(A).real)
    return c, lam
 
rho, u, p = 1.225, 220.0, 1.0e5               # air, u = 220 m/s
c, lam = characteristic_speeds(rho, u, p)
print(f"c0         = {c:8.2f} m/s")
print(f"eig(A)     = {np.round(lam, 2)}")
print(f"u-c,u,u+c  = {np.round([u-c, u, u+c], 2)}")
print(f"Mach       = {u/c:5.3f}  -> {'supersonic' if u > c else 'subsonic'}")

The output is as follows.

c0         =   338.06 m/s
eig(A)     = [-118.06  220.    558.06]
u-c,u,u+c  = [-118.06  220.    558.06]
Mach       =  0.651  -> subsonic

The eigenvalues match $u-c,\,u,\,u+c$ exactly. This time, let's actually advance a disturbance in time and measure the speed of its leading front. The acoustic Riemann invariants $w_\pm = u_1 \pm (c_0/\rho_0)\,\rho_1$ are scalars carried at $u_0 \pm c_0$ respectively, so we can transport them separately with upwinding.

N, L = 400, 1.0
x = np.linspace(0, L, N, endpoint=False)
dx = L / N
rho0, c0 = 1.0, 1.0
u0 = 0.6 * c0                                  # background Mach 0.6
 
bump = np.exp(-((x - 0.5) ** 2) / (2 * 0.01 ** 2))
wp = bump.copy()                              # right-going invariant (u0 + c0)
wm = bump.copy()                              # left-going  invariant (u0 - c0)
 
def upwind_step(w, a, dt):
    if a > 0:
        dwdx = (w - np.roll(w, 1)) / dx
    else:
        dwdx = (np.roll(w, -1) - w) / dx
    return w - a * dt * dwdx
 
CFL = 0.4
dt = CFL * dx / (abs(u0) + c0)
steps = int(0.25 / dt)
for _ in range(steps):                        # time-advance loop
    wp = upwind_step(wp, u0 + c0, dt)
    wm = upwind_step(wm, u0 - c0, dt)
 
T = steps * dt
def front_speed(w):
    return (x[np.argmax(w)] - 0.5) / T
print(f"u0+c0: target {u0 + c0:+.3f}, measured {front_speed(wp):+.3f}")
print(f"u0-c0: target {u0 - c0:+.3f}, measured {front_speed(wm):+.3f}")

u0+c0: target +1.600, measured +1.600
u0-c0: target -0.400, measured -0.400

The two fronts move at $u_0+c_0$ and $u_0-c_0$ respectively. The right-going invariant heads downstream, the left-going invariant heads upstream. Raise $u_0$ above $c_0$ and the measured value on the second line flips positive too.

The air numbers make the gap obvious. With $\rho=1.225$, $u=220$ m/s, $P=10^5$, and $\gamma=1.4$, the sound speed is 338.06 m/s. The printed eigenvalues are $-118.06$, $220$, and $558.06$, identical to $u-c$, $u$, and $u+c$. The span $558.06-(-118.06)$ is 676.12 m/s, twice 338.06. The middle value, 220, only carries material. It does not know the sound speed. Mach 0.651 is below 1, so $\lambda_-$ stays negative and one signal still travels upstream.

The second run sets the sound speed to 1. With $u_0=0.6$ the targets are $+1.600$ and $-0.400$, and the measured fronts hit those numbers. $-0.400$ is just $0.6-1$. The minus sign is the upstream reach. Once $u_0$ exceeds $c_0$, that line turns positive, the same state as the spacetime wedge tipped to the right.

In the Riemann invariant $u \pm \dfrac{2c}{\gamma-1}$, $\gamma=1.4$ gives $\gamma-1=0.4$ and $2/0.4=5$. What stays constant along an acoustic characteristic is $u\pm 5c$. The scalars the code carries, $w_\pm = u_1 \pm (c_0/\rho_0)\,\rho_1$, are that invariant cut down to a small disturbance. With $\rho_0=c_0=1$ the coefficient is 1.

Fixing pressure at a supersonic outlet sends the wave back#

The moment the domain-of-influence wedge crosses the vertical is $M=1$. Above that, no characteristic runs upstream. At a supersonic outlet all three signs point outward, so there is nothing left to prescribe. Freezing the pressure invents an incoming characteristic that was not there, and a reflected wave walks back into the domain.

A subsonic outlet is different. Only $\lambda_-$ points inward, so one condition is enough. That incoming wave is what NSCBC scales by a reflection factor σ at the outlet. Set the pressure before counting signs, and even the Mach 0.651 air mixes an upstream signal into the outlet treatment.