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Peter Romeo Nyarko Supervisor : Dr. J.H.M. ten Thije Boonkkamp 23 rd September, 2009. Waves and First Order Equations. Outline. Introduction Continuous Solution Shock Wave Shock Structure Weak Solution Summary and Conclusions. Introduction. What is a wave?. Application of waves - PowerPoint PPT Presentation
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Waves and First Order Equations
Peter Romeo NyarkoSupervisor : 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
0ddt dx c
dt,
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 wavesB
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 fF t F t
0t xc
1'Bt F
Continuous SolutionBreaking
1
2
00
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 cxct
2 2xc ct
1 1
2 1
2 2
,
,
,
xc ct
x xc c ct t
xc ct
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
,qv
2
1
1 2, , ,x
x
d x t dx q x t q x tdt
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
q dxdtt x
Continuous Solution0.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 dxdt dt
1
2
, , , ,s
s
x t x
s s t tx 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.
,sdxdts
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 WavesLet 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 Structure1 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
1 logx 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
0qt 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,
s 1R 2R
0,l Q m
,lum
on s 0
Qt 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
21 02t x
0
,00
if xx
if x
Weak Solutions
2 2
1
1122
dxsdt
2/32 03t x
2 , 2) Let
3/2 3/2 3 3 2 2
2 2 2
2 2 2 ,3 3 3
r l
r l
s
1 2s s
Weak SolutionsEntropy 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 Conclusion1) Explicit solution for linear wave equations.
2) Study of characteristics for nonlinear equations.
3) Weak solutions are not unique.
THANK YOU