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Simple Waves

Reference: Landau & Lifshitz, Fluid Mechanics

1 Nonlinear waves

1.1 Small amplitude waves

When we considered sound waves we considered small ampli-tude perturbations to the fluid equations. In 1D

(1)2x2

------------1

cs2

-----2t2

------------– 0=

Astrophysical Gas Dynamics: Simple Waves 1/33

The solution to this equation, of course, is:

(2)

where and are arbitrary functions. The perturbation to the

velocity is given by:

(3)

Note that for the wave travelling in each direction

(4)

f1 x cst– f2 x cst+ +=

f1 f2

vxt

-----------cs

2

-----x

---------–cs

2

----- f1

x cst– f2 x cst+ + –= =

vxcs---- f1 x cst– f2 x cst+ – =

vxcs----=


1.2 Simple waves

The above relationship suggests looking for a full nonlinear so-lution in which

(5)

Such a solution of the Euler equations is called a simple wave.

Take the 1D Euler equations

(6)

v v =

t------

v x

--------------+ 0=

vt----- v

vx-----

1---px------+ + 0=


Since and then and the aboveequations can be written:

(7)

1.2.1 Aside - Profile velocity

We would like to work out the velocity of a point on the densityprofile, at a fixed density, as shown in the figure. The aim of the

following is to determine .

v v = p p = p p v =

t------

dd v

x------+ 0=

vt----- v

1---dpdv------+

xddv

+ 0=

xt-----


To determine the vlocity of a point with constant density, weconsider a transformation of independent variables from to

defined by:

(8)

x

t1 t2

x

x t t

x t =

t t=


the reverse transformation being:

(9)x x t =

t t=


In the new variables

(10)

t

t

x

t tx

t

x tt

+=

x

t tx

t

x+ 0= =

tx

t

x

x

t

----------–=


Similarly, in order to consider the profle velocity at a given ve-locity value, we consider a transformation

(11)

from to and the rate of change of at a fixed is giv-en by

(12)

v v x t =

t t=

x t v t x t

txv

tv

xv------–=


1.3 Continued solution for simple wave:

Given the above expression for a simple wave and the continuityequation we have

(13)

The momentum equation implies that

(14)

tx

dd v v

ddv

+= =

txv

v1---vddp

+=


Now the crucial point in the development of this solution is thatsince then corresponds to and

(15)

v v = v constant= constant=

txv t

x

=


Hence,

(16)

v1---vddp

+ v ddv

+=

cs2

-----

vdd

ddv

=

ddv 2 cs

2

2------=

ddv cs

----=


Thus as a function of is given by:

(17)

We take a polytropic equation of state

(18)

and take to be a reference density corresponding to .

Hence,

(19)

v

vcs---- d=

p K s =

0 v 0=

K p00–= c0

2 K0 1–=

pp0-----

0------ = cs

2 c02 0------ 1–

= cs c00------

1–2

-----------=


Hence the relationship between and for a polytropic gas is:

(20)

Solving for the sound speed:

(21)

V

vcs----d c0

0------

3–2

-----------d

0------ = =

2c0 1–-----------

0------

1–2

-----------1–

2c0 1–-----------

csc0----- 1–= =

cs c0 1–

2-----------v=


1.3.1 Wave velocity

(22)

Since then

(23)

where is an arbitrary function of v. Using the solution for then

(24)

This is one of the most useful forms of the solution for a simplewave.

tx

vv

1---vddp

+ vcs

2

-----

vdd

+ v cs= = =

cs cs cs v = =

x t v cs f v +=

f v cs

x t 1+

2-----------v c0 f v +=


NB. Waves travelling to the right represented by the + sign in theabove expressions; waves travelling to the left by the - sign.

1.3.2 Velocity and the formation of shocks

From the above

(25)

and this shows the nonlinear effect on the profile velocity. When

then , i.e. the standard relation for a sound

wave. However, when nonlinear effects have a marked

effect on the profile.

tx

v

1+2

-----------v c0=

v c0«t

x

vc0=

v c0


The increase of the profile velocity with increasing positive vand the decrease with decreasing negative means that the pro-file steepens as shown in the figure. The end result is a series ofever steepening profiles which eventually become unphysical.

vt1

x

v


When the profile becomes infinitely steep a shock forms. In thisregion the assumptions embodied in the Euler equations breakdown i.e. it is no longer possible to neglect viscosity. (Moreabout this later.)

v

x

Infinitely steepprofile

Unphysicalt1 t2 t3 t4


At the time of formation of the shock:

(26)

Moreover, the shock represents a point of inflection in the profileso that

(27)

If we use the solution:

(28)

vxt

0=

2x

v2--------

t

0=

x t c0 1+

2-----------v+

f v +=


then the conditions for the formation of a shock are:

(29)

and

(30)

Both of these conditions have to be satisfied simultaneously.

1.3.3 Fluid outside the region at rest

This is a special case.

vx 1+

2-----------t f v + 0= = t 2f v

1+--------------–=

f v 0=


In this case the condition for a shock to form is simply

(31)

since the velocity profile beyond this point is simply . This

gievs the condition

v

x

vxt

0 at v 0= =

v 0=

t2f 0 1+

--------------–=


2 Piston moving out of a pipe as an example of a simple wave

2.1Formation of solution

x

v–

c0tat2

2--------–

x 0= c0u at–=


The piston is initially at and moves towards the left withacceleration a. A sound wave travels towards the right with ve-locity since the gas is at rest to the right of the piston.

We can solve for this flow as a simple wave. Take the generalsolution:

(32)

The + sign applies to a wave travelling to the right so that

(33)

(Note: the wave is travelling to the right although the piston istravelling to the left.)

x 0=

c0

x t c0 12--- 1+ v+ f v +=

x t c012--- 1+ v+ f v +=


The arbitrary function may be calculated using the boundarycondition that the gas remains in contact with the piston, i.e.

(34)

Write the solution in the form:

(35)

v at at x–at2

2--------–= =

f v x t c0 1+

2-----------v+–=

f at– at2

2--------– t c0

1+ 2

----------------at––=

c0a----- at–

2a------ at– 2+=


Therefore

(36)

and the solution for x as a function of x and t is given by the so-lution of the quadratic:

(37)

f v c0a-----v

2a------v2+=

x t c0 1+

2----------------v+

c0a-----v

2a------v2+ +=

2a------v2

c0a-----

1+2

-----------t+ v c0t x–+ + 0=

v–c0

----- 1+

2----------- at

c0

----- 1+

2-----------at+

2 2a

------ x c0t– ++=


The other boundary condition to incorporate is the condition thata sound wave moves away from the piston. i.e. at

. We can readily see that the - sign in the above expres-

sion gives the correct boundary condition so that

(38)

v 0=x c0t=

v–c0

----- 1+

2----------- at

c0

----- 1+

2-----------at+

2 2a

------ x c0t– +–+=


Does a shock form in this solution?

(39)

The second derivative is always non zero so that if a shock formsit does so when i.e. when

(40)

f v c0a-----v

2a------v2+=

f v c0a-----

a---v+=

f v a---=

v 0=

t2f 0 1+

--------------–2

1+-----------

c0a-----– 0= =


Negative values of t are outside the domain of validity of our so-lution so that a shock does not form.

2.2Maximum Velocity

An intriguing aspect of the above solution is that a velocity isreached at which a cavity forms behind the piston. Consider oursolution for the sound speed in a simple wave:

(41)cs c0 1–

2-----------v c0

1–2

-----------v+= =


Now in order that the density remain positive, the sound speedmust also remain positive and

(42)

Once the gas (and hence the piston) reaches this velocity a vac-uum forms.

c0 1–

2-----------v+ 0

v2c0 1–-----------–


xc0tat2

2--------– Vacuum

v–

v–2c0 1–-----------=


3 Piston moving into gas

3.1 Solution

Piston accelerates from zero velocity to a velocity and a

position at time t.

u at=

V

xxpiston

12---at2=

u at=

x12---at2=


Again we use the general simple wave for a wave moving to theright

(43)

and we use the boundary condition for gas at the surfaceof the piston:

(44)

x t c0 1+

2-----------v+ f v +=

v at=

12---at2 t c0

1+2

-----------at+ f at +=

f at c0a----- at –

2a------ at 2–=

f v c0a-----v–

2a------v2–=


3.2 Formation of shock:

We have

(45)

so that the only place where a shock can form is at . Thetime of formation of the shock in this case is:

(46)

Note that the velocity of the piston is not supersonic with respectto the undisturbed gas when the shock forms:

(47)

f v c0a-----–

a---v–= and f v

a---– 0=

v 0=

t2f 0 1+

--------------–2

1+-----------

c0a-----= =

u at2

1+-----------c0 c0 for 1= =


3.3 Velocity of gas

The expression for the velocity profile is, in this case (exercise):

(48)

One can show directly from this expression that the velocity pro-

file becomes infinitely steep at when .

vc0

----- 1+

2-----------at–

–c0

----- 1+

2-----------at–

2 2a

------ x c0t– –+=

x c0t= t2

1+-----------

c0a-----=


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