Документ взят из кэша поисковой машины. Адрес оригинального документа : http://www.mso.anu.edu.au/~geoff/AGD/Simple_Waves.pdf Дата изменения: Mon Feb 22 13:48:00 2016 Дата индексирования: Sun Apr 10 03:26:32 2016 Кодировка: Поисковые слова: universe

Simple Waves
Reference: Landau & Lifshitz, Fluid Mechanics 1 Nonlinear waves
1.1 Small amplitude waves

When we considered sound waves we considered small amplitude perturbations to the fluid equations. In 1D 2 1 2 ------------ ­ ----- ------------ = 0 2 x 2 cs t 2
(1)

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The solution to this equation, of course, is: = f 1 x ­ c s t + f 2 x + c s t
(2)

where f 1 and f 2 are arbitrary functions. The perturbation to the velocity is given by:
2 2 v x c s cs ----------- = ­ ----- --------- = ­ ----t x cs v x = ---- f f1 x ­ cs t + f2 x + cs t 1 x ­ cs t ­ f2 x + cs t

(3)

Note that for the wave travelling in each direction cs v x = ----
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(4)
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1.2 Simple waves

The above relationship suggests looking for a full nonlinear solution in which v = v
(5)

Such a solution of the Euler equations is called a simple wave. Take the 1D Euler equations ----- + t v v ----- + v ----t x v ------------- = 0 x 1 p + -- ----- = 0 - x

(6)

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Since v = v and p = p then p = p v and the above equations can be written: d ----- + v ----- = 0 t d x v 1 dp dv ----- + v + -- -----=0 t dv d x
1.2.1 Aside - Profile velocity

(7)

We would like to work out the velocity of a point on the density profile, at a fixed density, as shown in the figure. The aim of the following is to determine x . ----t

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t1 t2 x x To determine the vlocity of a point with constant density, we consider a transformation of independent variables from x t to t defined by: = x t t = t
(8)

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the reverse transformation being: x = x t t = t
(9)

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In the new variables t = x + t x tt t xt t = x + = 0 x tt t x t x x = ­ --------- t x t

(10)

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Similarly, in order to consider the profle velocity at a given velocity value, we consider a transformation v = v x t t = t
(11)

from x t to v t and the rate of change of x at a fixed t is given by v t x = ­ ----t v v x

(12)

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1.3 Continued solution for simple wave:

Given the above expression for a simple wave and the continuity equation we have x = d v = v + dv d d t The momentum equation implies that 1 x = v + -- dp dv t v
(14) (13)

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Now the crucial point in the development of this solution is that since v = v then v = constant corresponds to = constant and x = x t v t
(15)

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Hence, 1 dp dv v + -= v+ dv d
2 c s d dv ----= dv d 2 2 cs dv = ---- d 2

(16)

cs dv = ---d

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Thus v as a function of is given by: cs v = ---- d



(17)

We take a polytropic equation of state p = K s
(18)

and take 0 to be a reference density corresponding to v = 0 . Hence,
­ K = p0 0 2 c0 = K 0 ­ 1 ­ (19) 1 --------- ----- 2 cs = c0 0
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p ----- ----- = p0 0

2 = c 2 ----- ­ 1 cs 0 - 0

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Hence the relationship between V and for a polytropic gas is:
­3 cs ---------- ----- 2 d ----v = ---- d = c 0 0 0





­1 2 c 0 --------- ----- 2 ­ 1 = --------- ­1 0

(20)

2 c0 cs = ---------- ---- ­ 1 - ­ 1 c0

Solving for the sound speed: ­1 c s = c 0 ---------- v 2
(21)

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1.3.1 Wave velocity





x t

2 c s d 1 dp = v + -= v + ----= v cs dv dv v

(22)

Since c s = c s = c s v then x = t v cs + f v
(23)

where f v is an arbitrary function of v. Using the solution for c s then +1 x = t ----------- v c 0 + f v 2
(24)

This is one of the most useful forms of the solution for a simple wave.
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NB. Waves travelling to the right represented by the + sign in the above expressions; waves travelling to the left by the - sign.
1.3.2 Velocity and the formation of shocks

From the above +1 x = ----------- v c 0 v 2 t
(25)

and this shows the nonlinear effect on the profile velocity. When x = c , i.e. the standard relation for a sound v « c 0 then 0 t v wave. However, when v c 0 nonlinear effects have a marked effect on the profile.

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v t1 x The increase of the profile velocity with increasing positive v and the decrease with decreasing negative v means that the profile steepens as shown in the figure. The end result is a series of ever steepening profiles which eventually become unphysical.

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v t1

t2 t3

Infinitely steep profile Unphysical

t4 x

When the profile becomes infinitely steep a shock forms. In this region the assumptions embodied in the Euler equations break down i.e. it is no longer possible to neglect viscosity. (More about this later.)
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At the time of formation of the shock: x = 0 v t
(26)

Moreover, the shock represents a point of inflection in the profile so that 2x = 0 -------v2 t If we use the solution: c + + 1 v + f v x=t 0 ---------- 2
(28) (27)

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then the conditions for the formation of a shock are: x + 1 2f v = ----------- t + f v = 0 t = ­ ------------v 2 +1 and f v = 0
1.3.3 Fluid outside the region at rest

(29)

(30)

Both of these conditions have to be satisfied simultaneously. This is a special case.

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v x In this case the condition for a shock to form is simply x = 0 at v = 0 v t 2f 0 gievs the condition t = ­ -------------+1
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(31)

since the velocity profile beyond this point is simply v = 0 . This


2 Piston moving out of a pipe as an example of a simple wave
2.1Formation of solution

­v ­ ------2 u = ­a t x=0 at 2 c0 t x c0

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The piston is initially at x = 0 and moves towards the left with acceleration a. A sound wave travels towards the right with velocity c 0 since the gas is at rest to the right of the piston. We can solve for this flow as a simple wave. Take the general solution: 1 x = t c 0 + -- + 1 v + f v 2 The + sign applies to a wave travelling to the right so that 1 x = t c 0 + -- + 1 v + f v 2
(33) (32)

(Note: the wave is travelling to the right although the piston is travelling to the left.)
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The arbitrary function may be calculated using the boundary condition that the gas remains in contact with the piston, i.e. at 2 v = ­ a t at x = ­ ------2 Write the solution in the form: +1 f v = x ­ t c 0 + ----------- v 2 at 2 + 1 f ­ a t = ­ ------- ­ t c 0 ­ --------------- at 2 2 c0 = ---- ­ at + ----- ­ at 2 a 2a
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(34)

(35)


Therefore c0 2 f v = ---- v + ----- v a 2a c0 + 1 2 x = t c 0 + --------------- v + ---- v + ----- v 2 a 2a

(36)

and the solution for x as a function of x and t is given by the solution of the quadratic: 2 c0 + 1 ----- v + ---- + ----------- t v + c 0 t ­ x = 0 a 2a 2 c0 + 1 c0 + 1 2 2 a ­ v = ---- + ----------- at ---- + ----------- at + ----- x ­ c 0 t 2 2
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(37)

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The other boundary condition to incorporate is the condition that a sound wave moves away from the piston. i.e. v = 0 at x = c 0 t . We can readily see that the - sign in the above expression gives the correct boundary condition so that c0 + 1 2 2 a c0 + 1 ­ v = ---- + ----------- at ­ ---- + ----------- at + ----- x ­ c 0 t (38) 2 2

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Does a shock form in this solution? c0 2 f v = ---- v + ----- v a 2a c0 f v = ---- + -- v -aa v = -f a

(39)

The second derivative is always non zero so that if a shock forms it does so when v = 0 i.e. when 2f 0 2 t = ­ -------------- = ­ ----------+1 +1
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c0 ---- 0 a

(40)

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Negative values of t are outside the domain of validity of our solution so that a shock does not form.
2.2Maximum Velocity

An intriguing aspect of the above solution is that a velocity is reached at which a cavity forms behind the piston. Consider our solution for the sound speed in a simple wave: ­1 ­1 c s = c 0 ---------- v = c 0 + ---------- v 2 2
(41)

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Now in order that the density remain positive, the sound speed must also remain positive and ­1 c 0 + ---------- v 0 2 2 c0 ­ v ---------­1

(42)

Once the gas (and hence the piston) reaches this velocity a vacuum forms.

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2 c0 ­ v = ---------­1 ­v at 2 c0 t x
Vacuum

­ ------2

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3 Piston moving into gas V u = at 12 x piston = -- at 2
3.1 Solution

x

Piston accelerates from zero velocity to a velocity u = a t and a 12 position x = -- at at time t. 2

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Again we use the general simple wave for a wave moving to the right +1 x = t c 0 + ----------- v + f v 2
(43)

and we use the boundary condition v = a t for gas at the surface of the piston: 12 +1 -- at = t c 0 + ----------- at + f a t 2 2 c0 f a t = ­ ---- at ­ a c0 f v = ­ ---- v ­ ----a 2a
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----- at 2 2a v2

(44)

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3.2 Formation of shock:

We have c0 f v = ­ ---- ­ -- v -aa v = ­ -- 0 and f a
(45)

so that the only place where a shock can form is at v = 0 . The time of formation of the shock in this case is: 2f 0 2 t = ­ -------------- = ----------+1 +1 c0 ---a
(46)

Note that the velocity of the piston is not supersonic with respect to the undisturbed gas when the shock forms: 2 u = a t = ----------- c 0 c 0 for 1 +1
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(47)
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3.3 Velocity of gas

The expression for the velocity profile is, in this case (exercise): c0 + 1 c0 + 1 2 2 a v = ­ ---- ­ ----------- at + ---- ­ ----------- at ­ ----- x ­ c 0 t (48) 2 2 One can show directly from this expression that the velocity pro2 c0 file becomes infinitely steep at x = c 0 t when t = ----------- ---- . +1a

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