Waves and First Order Equations Peter Romeo Nyarko Supervisor : Dr. J.H.M.
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Transcript Waves and First Order Equations Peter Romeo Nyarko Supervisor : Dr. J.H.M.
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
Introduction
What is a wave?
Application of waves
Light and sound
Water waves
Traffic flow
Electromagnetic waves
Introduction
Wave equations
Linear wave equation
t c0 x 0
Non-Linear wave equation
t c x 0
Continuous Solution
Linear Wave Equation
Solution of the linear wave equation
f x c0t
Continuous Solution
Non-Linear Wave Equation
t c x 0
If we consider
x and
as functions of
t
d dx
dt t dt x
Since
d
dx
0,
c
dt
dt
remains constant
c is a constant on the characteristic curve and
therefore the curve is a straight line in the x, t plane
Continuous Solution
We consider the initial value problem
f x , t o, x
If one of the characteristics intersects
t 0, x Then
f
is a solution of our equation, and the equation of the characteristics is
x F t
where
c c f F
Continuous Solution
t
t3
tB
t1
0
B
Characteristic diagram for nonlinear waves
x
Continuous Solution
We check whether our solution satisfy the equation:
t c x 0 ,
t f ' t , x f ' x
x F t
0 F 1 F ' tt
1 1 F ' x
Continuous Solution
F f '
f '
t
, x
1 F ' t
1 F ' t
t c x 0
1
tB
F '
Continuous Solution
Breaking
1 x 0
f x
2 x 0
c1 c 1 , x 0
F x
c2 c 2 , x 0
c1 c2 Breaking occur immediately
2
2
1
1
x
Compression wave with overlap
c ' 0
1
, 2
Continuous Solution
There is a perfectly continuous solution for the special case of Burgers equation
if
c2 c1
x
c
t
c1
x
c
t
c2
x
c2 c2
t
x
,
c1
t
x
, c2 c1
t
x
,
c2
t
1
1
2
2
Rarefaction wave
x
Continuous Solution
Kinematic waves
We define density
Flow velocity v
q
x, t per unit length ,and flux q x, t per unit time , x2 x x1
x2
d
x, t dx q x1 , t q x2 , t
dt x1
,
Integrating over an arbitrary time interval,
t1, t2
x2
t2
x, t x, t dx q x , t q x , t dt
2
x1
1
t1
q
t x t x dxdt 0
1 1
t2 x2
This is equivalent to
1
2
Continuous Solution
Therefore the integrand
q
0.
t x
The relation between q and
Then
t c x 0,
The conservation law.
is assumed to be
c Q '
q Q
Shock Wave
We introduce discontinuities into our solution by a simple jump in
as far as our conservation equation is feasible
Assume q and
d
q x2,t q x1 , t
dt
x2
q
x1 x xs t , xs t x x2.
are continuous
xs t
and
d
x, t dx
dt
xs , t s xs , t s
xs t
x1
x, t dx
xs t
x1
x, t dx x, t dx
t
x2
xs t
t
Shock Structure
where
xs , t , x s , t
are the values of
x, t , x xs t
from below and above.
dxs
s
,
dt
x1 xs
and
where
s
is the shock velocity
x2 xs
q xs , t q xs , t xs , t xs , t s
Shock Waves
Let
s Shock velocity
q2 q1
s
s 2 1
Q 2 Q 1
2 1
Traffic Flow
(Example)
Consider a traffic flow of cars on a highway .
: the number of cars per unit length
u
: velocity
0 max
max
:The restriction on density.
is the value at which cars are bumper to bumper
From the continuity equation ,
t u x 0
Traffic Flow (Example)
This is a simple model of the linear relation
u umax
1
max
umax
The conservative form of the traffic flow model
t Q x 0
where
Q umax (1 / max )
o
max
Traffic Flow
(Example)
The characteristics speed is given by
Q ' umax (1 2 / max )
The shock speed for a jump from
l
to
r
Q l Q ( r )
s
l r
umax (1 l r / max )
Traffic Flow
(Example)
Consider the following initial data
Case
0 l r max
l x 0
x,0
r x 0
u 0
r max
t
l max
0
x
characteristics
Shock structure
We consider
Assume
At breaking
Then
where
q as a function of the density gradient xas well as the density
q Q v x
x become large and v the correction term becomes crucial
t c x vxx ,
c Q ' , c x ,
Assume the steady profile solution is given by
X , X x Ut
Shock structure
Then
c U
x
xx
Integrating once gives
Q U A x , Ais a constant
d
Q U A
x
Qualitatively we are interested in the possibility of a solution which
tends to a constant state.
Shock Structure
1
as
x
2
,
x 0
If such a solution exist with
Then U and Amust satisfy
as
x
as
x
Q 1 U A Q 2 U 2 A 0
Q 2 Q 1
U
2 1
The direction of increase of
between the two zero’s
2
depends on the sign of Q
U A
1
Shock Structure
If
c ' 0, Q U A 0
c ' 0, Q U A 0
with
with
2 1
2 1
and
as required
The breaking argument and the shock structure agree.
Let
Q 2
for a weak shock , with
Q U A 1 2
0
where
U 1 2 , A 12
Shock structure
2
x
d
1
log
1 2 2 1
1
As
1 , x exponentially and as 2
x exponentially.
Weak Solution
A function
x, t is called a weak solution of the conservation law
q
0
t x
if
q dxdt x,0 x,0 dx
t
x
0
holds for all test functions
C01 R [0, ) .
Weak solution
Consider a weak solution
x, t which is continuously differentiable in the two parts
R1 andR2 but with a simple jump discontinuity across the dividing boundary S
between
R1 and
R.2Then
Q
Q
dxdt
R t x dxdt
t
x
R2
1
l Q m ds 0 , l , m
S
,is normal to
S
Weak Solutions
The contribution from the boundary terms of R1 and R2 on the line
integral s
t
S
R2
R1
0
Weak solution ,discontinuous across S
x
Since the equations must hold for all test functions,
l Q m 0, on s
l
u ,
m
This satisfy
Q
0
t
x
Points of discontinuities and jumps satisfy the shock conditions
Weak Solutions
Non-uniqueness of weak solutions
1) Consider the Burgers’ equation, written in conservation form
1 2
0
t
x 2
Subject to the piecewise constant initial conditions
x, 0
if
x0
if
x0
Weak Solutions
1
2
2
dx
1
2
s1
dt
2
2)
Let
2
2/3
,
0
t
x 3
2
3/2
3/2
3
3
2
2
2 r
2
2
l
s2
, s1 s2
2
2
3 r l
3
3
Weak Solutions
Entropy conditions
A discontinuity propagating with speed
s given by :
Q 2 Q 1
s
2 1
Satisfy the entropy condition if
Q ' 2 s Q ' 1
where Q ' is the characteristics speed.
Weak Solutions
t
a) l r
2
1
0
x
Shock wave
Characteristics go into shock in (a) and go out of the shock in (b)
b) l r
1
2
0
0
Entropy violating shock
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