Stone M., Goldbart P. Mathematics for Physics: A Guided Tour for Graduate Students

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7.2 Making waves 245

from Rayleigh’s equation (7.51). Multiplying by ψ

∗

and integrating gives

0 =





∗



− k



ψ + k



∂

∂y



(ω − Uk)

|ψ|



. (7.58)

An integration by parts then gives



∗

dψ







dψ

+ k

|ψ|



∂

∂y



(U − ω/k)

|ψ|



. (7.59)

The lower limit makes no contribution, since ψ

∗

is zero there. On using (7.52) and taking

the imaginary part, we find



∗

dψ



= sgn





∂

∂y



∂U

∂y

−1

|ψ(y

)|, (7.60)



dψ



= sgn





∂

∂y



∂U

∂y

−1

|ψ(y

. (7.61)

This equation is most useful if the interaction with the flow does not substantially perturb

ψ(y) away from the still-water result ψ(y) = sinh(|k|y), and assuming this is so provides

a reasonable first approximation.

If we insert (7.61) into (7.56), where we approximate,





≈ g





− 2ig





γ ,

we find

γ =

2gk



dψ



= sgn



2gk



∂

∂y



∂U

∂y

−1

|ψ(y

. (7.62)

We see that either sign of γ is allowed by our analysis. Thus the resonant interaction

between the shear flow and wave appears to lead to either exponential growth or damping

of the wave. This is inevitable because our inviscid fluid contains no mechanism for

dissipation, and its motion is necessarily time-reversal invariant. Nonetheless, as in our

discussion of “friction without friction” in Section 5.2.2, only one sign of γ is actually

observed. This sign is determined by the initial conditions, but a rigorous explanation

246 7 The mathematics of real waves

of how this works mathematically is not easy, and is the subject of many papers. These

show that the correct sign is given by

γ =−π

2gk



∂

∂y



∂U

∂y

−1

|ψ(y

. (7.63)

Since our velocity profile has ∂

U /∂y

< 0, this means that the waves grow in amplitude.

We can also establish the correct sign for γ by computing the change of momentum

in the background flow due to the wave.

The crucial element is whether, in the neigh-

bourhood of the critical depth, more fluid is overtaking the wave than lagging behind it.

This is exactly what the quantity ∂

U /∂y

measures.

7.3 Nonlinear waves

Nonlinear effects become important when some dimensionless measure of the amplitude

of the disturbance, say P/P for a sound wave, or h/λ for a water wave, is no

longer 1.

7.3.1 Sound in air

The simplest nonlinear wave system is one-dimensional sound propagation in a gas. This

problem was studied by Riemann.

The one-dimensional motion of a fluid is determined by the mass conservation

equation

∂

ρ + ∂

(ρv) = 0, (7.64)

and Euler’s equation of motion

ρ(∂

v +v∂

v) =−∂

P. (7.65)

In a fluid with equation of state P = P(ρ), the speed of sound, c, is given by

dρ

. (7.66)

It will in general depend on P, the speed of propagation being usually higher when the

pressure is higher.

Riemann was able to simplify these equations by defining a new thermodynamic

variable π(P) as

π =



ρc

dP, (7.67)

G. E. Vekstein, Amer. J. Phys., 66 (1998) 886.

7.3 Nonlinear waves 247

where P

is the equilibrium pressure of the undisturbed air. The quantity π obeys

dπ

ρc

. (7.68)

In terms of π, Euler’s equation divided by ρ becomes

∂

v +v∂

v +c∂

π = 0, (7.69)

whilst the equation of mass conservation divided by ρ/c becomes

∂

π + v∂

π + c∂

v = 0. (7.70)

Adding and subtracting, we get Riemann’s equations

∂

(v +π) + (v + c)∂

(v +π) = 0,

∂

(v −π) + (v − c)∂

(v −π) = 0. (7.71)

These assert that the Riemann invariants v ± π are constant along the characteristic

curves

= v ±c. (7.72)

This tells us that signals travel at the speed v ± c. In other words, they travel, with

respect to the fluid, at the local speed of sound c. Using the Riemann equations, we can

propagate initial data v(x, t = 0), π(x, t = 0) into the future by using the method of

characteristics.

In Figure 7.9 the value of v + π is constant along the characteristic curve C

, which

is the solution of

= v +c (7.73)





Figure 7.9 Nonlinear characteristic curves.

248 7 The mathematics of real waves

passing through A. The value of v − π is constant along C

−

, which is the solution of

= v −c (7.74)

passing through B. Thus the values of π and v at the point P can be found if we know

the initial values of v + π at the point A and v − π at the point B. Having found v

and π at P we can invert π(P) to find the pressure P, and hence c, and so continue the

characteristics into the future, as indicated by the dotted lines. We need, of course, to

know v and c at every point along the characteristics C

and C

−

in order to construct

them, and this requires us to treat every point as a “P”. The values of the dynamical

quantities at P therefore depend on the initial data at all points lying between A and B.

This is the domain of dependence of P.

A sound wave caused by a localized excess of pressure will eventually break up

into two distinct pulses, one going forwards and one going backwards. Once these

pulses are sufficiently separated that they no longer interact with one another they are

simple waves. Consider a forward-going pulse propagating into undisturbed air. The

backward characteristics are coming from the undisturbed region where both π and v

are zero. Clearly π − v is zero everywhere on these characteristics, and so π = v.

Now π + v = 2v = 2π is constant on the forward characteristics, and so π and v are

individually constant along them. Since π is constant, so is c. With v also being constant,

this means that c + v is constant. In other words, for a simple wave, the characteristics

are straight lines.

This simple-wave simplification contains within it the seeds of its own destruction.

Suppose we have a positive pressure pulse in a fluid whose speed of sound increases

with the pressure. Figure 7.10 shows how, with this assumption, the straight-line char-

acteristics travel faster in the high-pressure region, and eventually catch up with and

intersect the slower-moving characteristics. When this happens the dynamical variables

will become multivalued. How do we deal with this?

Figure 7.10 Simple wave characteristics.

7.3 Nonlinear waves 249

(a) (b)

(d)(c)

Figure 7.11 A breaking nonlinear wave.

(d)

Figure 7.12 Formation of a shock.

7.3.2 Shocks

Let us untangle the multivaluedness by drawing another set of pictures. Suppose u obeys

the nonlinear “half” wave equation

(∂

+ u∂

)u = 0. (7.75)

The velocity of propagation of the wave is therefore u itself, so the parts of the wave

with large u will overtake those with smaller u, and the wave will “break”, as shown in

Figure 7.11.

Physics does not permit such multivalued solutions, and what usually happens is that

the assumptions underlying the model which gave rise to the nonlinear equation will no

longer be valid. New terms should be included in the equation which prevent the solution

becoming multivalued, and instead a steep “shock” will form (Figure 7.12).

Examples of an equation with such additional terms are Burgers’ equation

(∂

+ u∂

)u = ν∂

u, (7.76)

and the Korteweg–de Vries (KdV) equation (4.11), which, by a suitable rescaling of x

and t, we can write as

(∂

+ u∂

)u = δ∂

xxx

u. (7.77)

250 7 The mathematics of real waves

Burgers’ equation, for example, can be thought of as including the effects of thermal

conductivity, which was not included in the derivation of Riemann’s equations. In both

these modified equations, the right-hand side is negligible when u is varying slowly, but it

completely changes the character of the solution when the waves steepen and try to break.

Although these extra terms are essential for the stabilization of the shock, once we

know that such a discontinuous solution has formed, we can find many of its properties

– for example the propagation velocity – from general principles, without needing their

detailed form. All we need is to know what conservation laws are applicable.

Multiplying (∂

+ u∂

)u = 0byu

n−1

, we deduce that

∂





+ ∂



n + 1

n+1



= 0, (7.78)

and this implies that



∞

−∞

dx (7.79)

is time independent. There are infinitely many of these conservation laws, one for each n.

Suppose that the n-th conservation law continues to hold even in the presence of the

shock, and that the discontinuity is at X (t). Then





X (t)

−∞

dx +



∞

X (t)



= 0. (7.80)

This is equal to

−

(X )

X − u

(X )

X +



X (t)

−∞

∂

dx +



∞

X (t)

∂

dx = 0, (7.81)

where u

−

(X ) ≡ u

(X −) and u

(X ) ≡ u

(X +). Now, using (∂

+u∂

)u = 0inthe

regions away from the shock, where it is reliable, we can write this as

− u

−

)

X =−

n + 1



X (t)

−∞

∂

n+1

dx −

n + 1



∞

X (t)

∂

n+1



n + 1



n+1

− u

n+1

−

). (7.82)

The velocity at which the shock moves is therefore

X =



n + 1



n+1

− u

n+1

−

)

− u

−

)

. (7.83)

Since the shock can only move at one velocity, only one of the infinitely many

conservation laws can continue to hold in the modified theory!

7.3 Nonlinear waves 251

Example: Burgers’ equation. From

(∂

+ u∂

)u = ν∂

u, (7.84)

we deduce that

∂

u + ∂



− ν∂



= 0, (7.85)

so that Q



udx is conserved, but further investigation shows that no other

conservation law survives. The shock speed is therefore

X =

− u

−

)

− u

−

)

+ u

−

). (7.86)

Example: KdV equation. From

(∂

+ u∂

)u = δ∂

xxx

u, (7.87)

we deduce that

∂

u + ∂



− δ∂



= 0,

∂





+ ∂



− δu∂

u +

δ(∂



= 0

where the dots refer to an infinite sequence of (not exactly obvious) conservation laws.

Since more than one conservation law survives, the KdV equation cannot have shock-like

solutions. Instead, the steepening wave breaks up into a sequence of solitons.

Example: Hydraulic jump, or bore.

A stationary hydraulic jump is a place in a stream where the fluid abruptly increases in

depth from h

to h

, and simultaneously slows down from supercritical (faster than wave

speed) flow to subcritical (slower than wave speed) flow (Figure 7.13). Such jumps are

Figure 7.13 A hydraulic jump.

252 7 The mathematics of real waves

commonly seen near weirs, and white-water rapids.

A circular hydraulic jump is easily

created in your kitchen sink. The moving equivalent is the tidal bore.

The equations governing uniform (meaning that v is independent of the depth) flow

in channels are mass conservation

∂

h + ∂

{

}

= 0, (7.88)

and Euler’s equation

∂

v +v∂

v =−∂

{gh}. (7.89)

We could manipulate these into the Riemann form, and work from there, but it is more

direct to combine them to derive the momentum conservation law

∂

{hv}+∂





= 0. (7.90)

From Euler’s equation, assuming steady flow, ˙v = 0, we can also deduce Bernoulli’s

equation

+ gh = const, (7.91)

which is an energy conservation law. At the jump, mass and momentum must be

conserved:

= h

, (7.92)

and v

may be eliminated to find





+ h

). (7.93)

A change of frame reveals that v

is the speed at which a wall of water of height h =

− h

) would propagate into stationary water of depth h

Bernoulli’s equation is inconsistent with the two equations we have used, and so

+ gh

=

+ gh

. (7.94)

This means that energy is being dissipated: for strong jumps, the fluid downstream

is turbulent. For weaker jumps, the energy is radiated away in a train of waves – the

so-called “undular bore”.

The breaking crest of Frost’s “white wave” is probably as much an example of a hydraulic jump as of a

smooth downstream wake.

7.3 Nonlinear waves 253

Example: Shock wave in air. At a shock wave in air we have conservation of mass

= ρ

, (7.95)

and momentum

+ P

= ρ

+ P

. (7.96)

In this case, however, Bernoulli’s equation does hold,

so we also have

+ h

. (7.97)

Here, h is the specific enthalpy (U + PV per unit mass). Entropy, though, is not con-

served, so we cannot use PV

= const. across the shock. From mass and momentum

conservation alone we find





− P

− ρ

. (7.98)

For an ideal gas with c

= γ , we can use energy conservation to eliminate the

densities, and find

= c



1 +

γ + 1

2γ

− P

. (7.99)

Here, c

is the speed of sound in the undisturbed gas.

7.3.3 Weak solutions

We want to make mathematically precise the sense in which a function u with a

discontinuity can be a solution to the differential equation

∂





+ ∂



n + 1

n+1



= 0, (7.100)

even though the equation is surely meaningless if the functions to which the derivatives

are being applied are not in fact differentiable.

Recall that enthalpy is conserved in a throttling process even in the presence of dissipation. Bernoulli’s

equation for a gas is the generalization of this thermodynamic result to include the kinetic energy of the gas.

The difference between the shock wave in air, where Bernoulli holds, and the hydraulic jump, where it does

not, is that the enthalpy of the gas keeps track of the lost mechanical energy, which has been absorbed by

the internal degrees of freedom. The Bernoulli equation for channel flow keeps track only of the mechanical

energy of the mean flow.

254 7 The mathematics of real waves

We could play around with distributions like the Heaviside step function or the Dirac

delta, but this is unsafe for nonlinear equations, because the product of two distributions

is generally not meaningful. What we do is introduce a new concept. We say that u is a

weak solution to (7.100)if



dx dt



∂

ϕ +

n + 1

n+1

∂



= 0, (7.101)

for all test functions ϕ in some suitable space T . This equation has formally been

obtained from (7.100) by multiplying it by ϕ(x, t), integrating over all space-time, and

then integrating by parts to move the derivatives off u, and onto the smooth function ϕ.

If u is assumed smooth then all these manipulations are legitimate and the new equation

(7.101) contains no new information. A conventional solution to (7.100) is therefore also

a weak solution. The new formulation (7.101), however, admits solutions in which u has

shocks.

Let us see what is required of a weak solution if we assume that u is everywhere

smooth except for a single jump from u

−

(t) to u

(t) at the point X (t). Let D

be the

regions to the left and right of the jump, as shown in Figure 7.14. Then the weak-solution

condition (7.101) becomes

0 =



−

dx dt



∂

ϕ +

n + 1

n+1

∂





dx dt



∂

ϕ +

n + 1

n+1

∂



(7.102)

Let

n =





1 +|

X |

−



1 +|

X |



(7.103)

X(t)





Figure 7.14 The geometry of the domains to the right and left of a jump.