Waves and First Order Equations

Peter Romeo Nyarko

Supervisor : Dr. J.H.M. ten Thije Boonkkamp

23rd September, 2009

Outline

• Introduction• Continuous Solution• Shock Wave• Shock Structure• Weak Solution• Summary and Conclusions

IntroductionWhat is a wave?

Application of wavesLight and soundWater wavesTraffic flowElectromagnetic waves

Wave equations

Introduction

Linear wave equation

0 0t xc Non-Linear wave equation

0t xc

Continuous Solution

Solution of the linear wave equation

0f x c t

Linear Wave Equation

Continuous Solution

If we consider and as functions of x

Since remains constant is a constant on the characteristic curve and

therefore the curve is a straight line in the plane

,x t

c

t

0d

dt

dx

cdt

,

Non-Linear Wave Equation

0t xc

d dxt xdt dt

Continuous Solution

We consider the initial value problem

, ,f x t o

0,t x If one of the characteristics intersects Then

f

is a solution of our equation, and the equation of the characteristics is

x F t

c c f F where

x

x

t

0

Bt

1t

3t

Characteristic diagram for nonlinear waves

B

Continuous Solution

Continuous Solution

We check whether our solution satisfy the equation:

x F t

0t xc

' , 't t x xf f

0 1 '

1 1 '

t

x

F F t

F

,

Continuous Solution

' ',

1 ' 1 't x

F f f

F t F t

0t xc

1

'Bt F

Continuous SolutionBreaking

1

2

0

0

xf x

x

1 1

2 2

, 0

, 0

c c xF x

c c x

1 2c c Breaking occur immediately ' 0c 2 1 ,

2

1

2

1

xCompression wave with overlap

Continuous SolutionThere is a perfectly continuous solution for the special case of Burgers equation

if 2 1c cx

ct

2 2

xc c

t

1 1

2 1

2 2

,

,

,

xc c

tx x

c c ct t

xc c

t

Rarefaction wave

2

1 1

2

x

Kinematic waves

We define density per unit length ,and flux per unit time ,

,x t ,q x t

,q

v

2

1

1 2, , ,x

x

dx t dx q x t q x t

dt

2 1x x x

Continuous Solution

2 2

1 1

2 1 1 2, , , ,x t

x t

x t x t dx q x t q x t dt

Flow velocity

Integrating over an arbitrary time interval, 1 2,t t

This is equivalent to

2 2

1 1

0t x

t x

qdxdt

t x

Continuous Solution

0.q

t x

The conservation law.

The relation between and is assumed to beq q Q

0,t xc Then 'c Q

Therefore the integrand

Shock WaveWe introduce discontinuities into our solution by a simple jump in and as far as our conservation equation is feasible

q

q 1 2, .s sx x x t x t x x Assume and are continuous

1

2

2, 1, , ,s

s

x t x

x x t

d dq x t q x t x t dx x t dx

dt dt

1

2

, , , ,s

s

x t x

s s t t

x x t

x t s x t s x t dx x t dx

Shock Structure

where , , ,s sx t x t are the values of , , sx t x x t

from below and above.

,sdxdt

s

1 sx x 2 sx x and

, , , ,s s s sx x x xq t q t t t s

where is the shock velocitys

Shock Waves

Let s Shock velocity

2 1 2 1

2 1

2 1

q q

Q Q

s

s

Traffic Flow (Example) Consider a traffic flow of cars on a highway . : the number of cars per unit length

: velocity

u

:The restriction on density. max0

max is the value at which cars are bumper to bumper

0t xu

From the continuity equation ,

maxmax

1u u

maxu

maxo

This is a simple model of the linear relation

0t xQ

max max(1 / )Q u

where

The conservative form of the traffic flow model

Traffic Flow (Example)

max max' (1 2 / )Q u

max max

( )

(1 / )

l r

l r

l r

Q Q

u

s

The characteristics speed is given by

The shock speed for a jump from to l r

Traffic Flow (Example)

Traffic Flow (Example)

Consider the following initial data

0

,00

l

r

xx

x

x

t

0characteristics

max0 l r

0u

maxr

maxl

Case

Shock structureWe consider as a function of the density gradient as well as the density

Assume

At breaking become large and the correction term becomes crucial

,t x xxc v

'c Q , ,xc

x

v

q

xq Q v x

Then

where

Assume the steady profile solution is given by

,X X x Ut

Shock structureThen

x xxc U

xQ U A

x d

Q U A

Integrating once gives

, Ais a constant

Qualitatively we are interested in the possibility of a solution which tends to a constant state.

Shock Structure

1 x 2 x 0x x

1 2 2 0Q U A Q U A

2 1

2 1

Q QU

as , as

If such a solution exist with as

Then and must satisfy

The direction of increase of depends on the sign of Q U A between the two zero’s

2

1

AU

Shock Structure ' 0,c 0Q U A 2 1 If with and

' 0,c 0Q U A with 2 1 as required

The breaking argument and the shock structure agree.

2Q

1 2Q U A

1 2U 1 2A

Let for a weak shock , with 0

where

,

Shock structure

1

2

1 2 12

1log

x d

x 1

x

2 As , exponentially and as

exponentially.

Weak Solution

A function is called a weak solution of the conservation law ,x t

0q

t x

0

,0 ,0t xq dxdt x x dx

if

holds for all test functions 10 [0, ) .C R

Weak solutionConsider a weak solution which is continuously differentiable in the two parts

and but with a simple jump discontinuity across the dividing boundary

between and . Then

,x t

1R 2R

1R 2R

1 2R R

Q Qdxdt dxdt

t x t x

0S

l Q m ds ,l m S,is normal to ,

S

Weak Solutions

Since the equations must hold for all test functions,

s1R 2R

0,l Q m

,l

um

on s

0

Q

t x

This satisfy

Points of discontinuities and jumps satisfy the shock conditions

The contribution from the boundary terms of and on the line integral

S

t

x

2R 1R

0Weak solution ,discontinuous across S

Weak SolutionsNon-uniqueness of weak solutions

1) Consider the Burgers’ equation, written in conservation form

Subject to the piecewise constant initial conditions

210

2t x

0

,0

0

if x

x

if x

Weak Solutions

2 2

1

1122

dxs

dt

2/320

3t x

2 , 2) Let

3/2 3/2 3 3 2 2

2 2 2

2 2 2,

3 3 3r l

r l

s

1 2s s

Weak Solutions

Entropy conditions

A discontinuity propagating with speed given by :

s

2 1

2 1

Q Qs

Satisfy the entropy condition if

2 1' 'Q s Q

'Q where is the characteristics speed.

t

x0

2

1

Weak Solutions

Entropy violating shock00

1

2

rl

rl

Shock wave

a)

b)

Characteristics go into shock in (a) and go out of the shock in (b)

x

Summary and Conclusion

1) Explicit solution for linear wave equations.

2) Study of characteristics for nonlinear equations.

3) Weak solutions are not unique.

THANK YOU