1

Riemann problems for hyperbolic systems

A Dissertation Submitted in partial fulfillment

FOR THE DEGREE OF

MMASTER OF SCIENCE IN MATHEMATICS

UNDER THE ACADEMIC AUTONOMY NATIONAL INSTITUTE OF TECHNOLOGY ROURKELA

By

T.SWATI

Under the Guidance of

DR. RAJA SEKHAR TUNGALA

DEPARTMENT OF MATHEMATICS NATIONAL INSTITUTE OF TECHNOLOGY

ROURKELA – 769008 ODISHA

2

CERTIFICATE

Dr. RAJA SEKHAR TUNGALA Assistant Professor Department of Mathematics NIT, Rourkela-ODISHA

This is to certify that the dissertation entitled“Riemann problems for hyperbolic

systems” being submitted by T.Swati to the Department of mathematics, National

Institute of Technology, Rourkela, Odisha, for the award of the degree of Master of

Science in mathematics is a record of bonafide research work carried out by them

under my supervision and guidance. I am satisfied that the dissertation report has

reached the standard fulfilling the requirements of the regulations relating to the

nature of the degree.

Rourkela-769 008

Date:

Dr. Raja SekharTungala

Supervisor

3

DECLARATION

I hereby certify that the work which is being presented in thesis entitled “ RiemannProblems for

hyperbolic systems” in partial fulfillment of the requirement for the award of the Degree of Master

of Science, submitted in the Department of Mathematics, National Institute of Technology, Rourkela

is an authentic record of my work carried out under the supervision

of Dr. Raja SekharTungala.

The matter embodied in this has not been submitted by me for the award of any other degree.

(T.Swati)

This is to certify that the above statement made by the candidate is carried to the best of the

Knowledge.

DR. RAJA SEKHAR TUNGALA

Assistant Professor, Department of mathematics

National Institute of Technology

Rourkela

Odisha

4

Acknowledgements

I deem it a privilege and honour to have worked in association under Dr. Raja sekhar Tungala, Assistant Professor, Department of mathematics, National Institute of Technology, Rourkela. I express my deep sense of gratitude and indebtedness to him for suggesting me the problem and guiding me throughout the dissertation. Words and inadequate to express my feelings of thankfulness for all the parental care and affection he has shown while my work was in progress.

I thank all faculty members of the Department of mathematics, who have always inspire

me to work hard and helped me learn new concepts during our stay at NIT, Rourkela.

I would like to thanks our parents for their unconditional love and support. They have

helped me in every situation throughout our life, I am grateful for their support.

I would like to accord our sincere gratitude to Mr Bibekananda bira for his valuable suggestions,

guidance in carrying out my project and his sincere help. I also thankful to C Bada and

Abhimanyu maharana the department assistant and peon respectively for their cooperation

and help.

Finally, I would like to thank all our friends for their support and the great almighty to

shower his blessing on us and making dreams and aspirations.

T.Swati

5

Abstract

In this report, we defined hyperbolic system and given some examples. We study the behaviour of

hyperbolic system. Later, we revised the exact solution of the Riemann Problem for the non- linear

PDE, which in hyperbolic system of the general form of conservation laws which governs one-

dimensional isentropic magnetogasdynamics. Lastly, we find the solution using phase plane analysis

and interactions of elementary waves between the same families as well as different families.

6

INTRODUCTION

The Riemann problem is defined as the initial value problem for the system with two valued

piecewise constant initial data. The Riemann problem is a fundamental tool for studying the

interaction between waves. It has played a central role both in the theoretical analysis of

systems of hyperbolic conservation laws and in the development and implementation of

practical numerical solutions of such systems.

Basically, the Riemann problem gives the micro-wave structural of the flow.

One can think of the propagation of the flow as a set of small scale Riemann problem between

the wave arising from these Riemann problems.

7

TABLE OF CONTENTS

CHAPTER 1: Introduction to hyperbolic system

1.1 Definition and Examples…………………………………………………. 8

1.2

1.3

1.4

1.5

Hyperbolic system for conservation laws………………………………...

Cauchy problem……….……………………………............…………….

Riemann problem………………………………………………………….

Weak solution……………………………………………………………...

9

10 10 15

CHAPTER 2: Riemann Problem for isentropic magnetogasdynamics

2.1

2.1.1

2.1.2

2.2

2.3

Shock and Rarefaction waves………… …………………………………

Shocks…………………………….……………………………………

Rarefaction waves………………………………………………………

Riemann problem……………………………………………………….

Interaction of elementary waves………………………………….…….

19

22

29

39

43

2.3.1 Interaction of waves from different families……………………….…... 44

2.3.2 Interaction of waves from same family……………………………..

REFERENCES

48

8

Chapter-1

Introduction to Hyperbolic Systems

1.1Definitions and Examples: The general form of system of conservation laws in several space variables

� � . ufxt

uj

d

j j

01

���

��� �

�

� �1.1

Here � be an open subset of pR , ;:f jpR� where , ,u : R p ���� 0

� � � � 02121 ��� ,tRx.........,,xx ,Xu.........,,uuu ddp .

The set � is called, the set of states and the functions, � �pjjj ..,f..........ff 1� are called flux

functions, the system (1.1) is written in conservation form, the conservation of the p real quantities,u,.........,uu 21 .We have a simplest differential equation model for a fluid flow:

.02

2

����

����

���

��� u

xtu

This equation is called inviscid Burger’s equation, which is also known as one- dimensional conservation law.

� � � � 0���

���

��� ug

yuf

xtu

,

which is a two dimensional equation. From this equation, we get following system of two dimensional equations:

� � � �

� � � � 0

0

2122122

2112111

��

��

��

���

��

��

��

���

y,uug

x,uuf

tu

y,uug

x,uuf

tu

Let D be an arbitrary domain of pR and let � �Td...,n,.........nn 1� be the outward unit normal to the

boundary D� of D. Then, it follow from (1.1) that,

� � . ds nufxu dt j

d

j Dj

D

01

���� � ��

� �

This is conservation law in integral form. This equation has a physical meaning that the Variation of

xu dD� is equal to the losses through the boundary D� .

9

1.2 Hyperbolic System of Conservation Laws:

For all ,......,dj 1� , let pki, ku

uijfujA ��

�

��

������

�

�

������

�

� ����

��

����

�� 1 be an Jacobian matrix of ;����

��ujf

equation (1.1)is called a hyperbolic system .

If for any Ωu� and ,,wdRdw,........,ww 01 ��� ���

����

�

the matrix ����

���

���

�� �

�� ujA

d

j jwu,wA1

has p real eigenvalues with Independent eigenvectors

, u,wk ru,wkλ u,wkru,wA

, i.e.u,wp ,....,ru,w,ru,wr

121

����

���

���

���

���

���

���

��

����

���

���

���

���

��

�

����

��u,wkr are right eigenvectors.

� � � � � � � � p.k, u,w lu,wλ u,wAu,wl kkk ��� 1

� �u,wlk are left eigenvectors.

If � �u,wA has p real eigenvalues and p corresponding linear independent eigenvectors, and if

� �wuk ,� real distinct eigenvalues, then the system is called strictly hyperbolic.

Example:

02

)12

����

����

���

��� u

xtu

Let � � ���

����

��

2

2uuf , then � 11�����

�����

� uufA

Here the eigenvalue is 1 and eigenvector is u. Example:

( )( ) 0=+ ;0= )2 vpxt

uxu

tυ

� � � �

� � 0

2

22

11

��

����

�

�

����

�

�

��

��

��

��

�

��

���

� ��

vpλuf

vf

uf

vf

vpu

f,v,uu

'

10

It is hyperbolic system. If the eigenvalues � �wu,x� are all distinct. The system (1.1) is called strictly

hyperbolic.

1.3 Cauchy Problem: Let

� �� � ,0ut �� xuf � � � � 0, �� stssx

be the partial differential equation with initial data of the curve. We have the surface which contains

the curve is called Cauchy problem. � � ,0:tx,u ���dR ,for t>0 and 0u is the function of x

alone and which have initial value �dRu :0

!"#

$

�0 ,0 ,

0 xuxu

ur

l

.

Where lu and ru are constants, then the Cauchy problem is called Riemann problem.

1.4 Riemann Problem: The conservation laws is given,

� �� � .0���

��� uf

xtu

Let lu and ru be two states of ;RΩ p% we have for piecewise smooth continuous function

t)u(x,t)(x,:u solutions of (1.1) that connection lu and ru :with initial condition

!"#

$

�00

0 , xu, xu

ur

l

is called Riemann problem.

3)Example:

The equation of gas dynamics in Eulerian coordinate:

In Eulerian coordinates, the Euler equations for a compressible inviscid fluid in the conservation form.

( )

( ) ( ) ,.31 ,0=++t

0=+t

3

1=

3

1=

iδpuuρx

uρ

uρx

ρ

ijijj

i

jj

� �� � � � .0 ''

'2

$&�

�

vpvp

vp

�

�

11

( ) ( )( ) 0=++t

3

1=j

jupeρ

xeρ

fluid theofdensity =ρ , � � velocity theu,u,uu 321� , energy internal specific pressure,p �� ' ,

energy totalspecific the2

2ue �� '

� �

� � � �

� � � �� �

� �

� � � �

� � � �

� � � �� �

� � � �

� � � �

� � 0

0

0

10

0

0

01

11

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011

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puρ

u γ γpρ

u

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puρ

px

uux

γppt

upγ

pxγ

pt

upργ

ρpxργ

ρpt

upρex

ρet

xp

ρtρuu

tu

pρux

ρut

ux

ρρx

utρ

ρuxt

ρ

.upρex

ρet

,pρux

ρut

,ρuxt

ρ

xt

12

� �

� � � �

� �

� �

� �

000

010

0

-u p 0

1 -u 0

0 -

01 0 00 1 00 0 1

-

0

10

0

0-,

. ,-, are seigenvalue

0-

01

0

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1 0

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λ u - γ ρ

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ρ u -λ

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))(

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�

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px

xxp

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xxpp

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xxx

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p

p

ργpu

uxx

x

.1

1

1 xassume

01

0.

0.

)- multiple(

01

(1.4) 0

(1.3) 01

(1.2) 0

0

p 0

1 0

0

2) (1,0,0). isr eigenvecto ing Correspond with eigenvalue The

00

1 assume

2

32

3

32

32

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21

3

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1

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0]

0 .

0 x

equation, from

1.7 0 x

1.6 01

1.5 0

0

0

1 0

0

)3

.

0

(1.2)equation in putting

32

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21

3

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1

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15

� �

.hyperbolicstrictly a isit and

1,1,),1,1,(,1,0,0 are rseigenvecto its and

)1,1,(,

)1,1,(x isr eigenvecto The

0

01

0

,equation from

1

1

1

01

321

1

1

1

1

21

2

32

3

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()

)()

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()

)()

()

)()

()

()

)(

()

)(

()

))

)(

))(

()

)

))(

))(

1.5 Weak solution:

Characteristics curve in one-dimensional case: Let RR:f be a 1C function. The conservation laws, with initial data:

� � � �1.8 0, tR, x,0tu

����

��� uf

x

� � � � R, x,0, 0 �� xuxu

Here u be a smooth solution, which follows the above equations

16

� � 1C tx,u �

Let u be smooth solution of Equation (1.1), then the non-conservation form � � 0u t �*� xuuf .

We take (u)fa(u) *�

From above equation, we have non-conservation from

0)(

tu

���

���

xuua

.

The characteristics curve of above condition; it will be define as the solution is integral curve of the differential equation

� �� � .,dtdx txua� (1.9)

Theorem (1): Assume that u is a smooth of (1.1) the characteristic curve are straight lines along, which u is constant. Proof: Consider a characteristic curve passing through the point � �,0,x0 a solution of the ordinary

differential equation is using the Method of characteristics,

� �

� �

� � Cxx

ufdtdxso,

duuf

dxdt'

��

*�

��

00 valueinitialwith

01

Along a curve, u is constant.

� �� � � �� � � �� �

� � .0 ,.

,,,

����

���

��*�

��

��

���

�

xuuf

tuei

dtdxttx

xuttx

tuttxu

dtd

By above equation is using by chain rule,

� �� � 0, so, �ttxudtd

Hence the characteristic curves are straight lines, whose constant slopes depends on the initial value

� �

� � � �� � � � .

curve) intergalin (

0xtuftxCtuftx

ufdtdx

�*��*�

*�

17

Example:

4) � � 0 �� xt uuau , � � � �xuxu 00, �

Solution: � �

� � � �

� �

� �� � � �� � � � � �� �� � � �

function.smooth is ,.

00 data, initial toaccording

are curves sticcharacteri e th

let 0

0

0

000

0

ueiatxu,ttx u

atxuxu,xu,ttxu

xattx

uaufuuau xt

� ���

��

�*��

Non-smooth Solution: � � � � 00 $** ** uf and uf are two cases for convex and concave

respectively.

Existences of non-smooth solution: We consider convex case � � 0uf i.e, ** .

Let R,,xx �21 such that 12 xx

if � �xu0 is decreasing function, then � � � �2010 u xu x .

Since, � � ,uf 0 ** then � � � �)u( )xu( 2010 xff * *

� � � �1011, xutxu � , implies that � � � �1020 xuxu $ .

So that characteristics intersect after finite time and form non smooth solution. Example: 5) The Burgers’ equation (inviscid equation) is ,uuu xt 0�� with initial condition

� �+!

+"

#

��

$�

1 xif 0,1x0 if x,-1

0 xif , 10,xu

.

Solution:

02

2

����

����

��

xt

uu

By solving characteristic curve we get.

18

� � � �� �

� � � � 000

0

0

xxtutxxx,tt ux(t)

xufttx

����

�*�

In these means, the characteristics curve passes through the point � �.00,x Then we have

� � � �

� �

++!

++"

#

$,

��

��

�

++!

++"

#

,

��

�

�

���

�

�

,�� �

����

110

11

11 1

1

11

1 that,know we

1101

0

0

00

000

00

0

,t x, if

x , if t-t

x-tx , if

x,tu

f xx, i

x, if t-t

x-tt, if xx

x

x if, xx, if xtx

if xt, x,txxx

At t=1, the characteristic intersect

� �!"#

$

�1011

1 if x, if x,

x,u

Now, it is discontinuities may develop after a finite time if f is nonlinear, when 0u is smooth in

fig. (1).

fig. (1)

19

Chapter-2

Riemann Problem for isentropic magnetogasdynamics

2.1 Shock and rarefaction waves: When flow of an isentropic, inviscid and perfectly conducting compressible fluid is subjected to a transverse magnetic field, then conservation form can be written as

� � � �

� � � �2.1 002

0

22 R , x, t)

μBρu(p

tρu

t

ρut

ρt

� �����

���

���

���

Where ρ 0, , , 0,p , it may represent density, velocity, pressure, 0,B transversal magnetic

field and 0 - denote magnetic permeability, respectively; p and B are functions in which are γρk p 1� and ρkB 2� , where k1 and 2k are positive constants and ( is the adiabatic constant

which lies in the range 21 �$ ( for most of the gases. The independent variables are t and x.

� �

� �

xρργk

xp , ρk p

xk

xk

xuu

tu

xBB

xp

xuu

tu

xu

xu

tu

xBB

xp

xuu

tu

xu

xuu

tu

xBB

xuu

xp

tu

xBB

xuu

xu

xp

tuu

Bx

ut

xu

xu

ux

γγ

��

���

�

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111

221

1

2

2

22

thenwhere

02

02

02

02

0212t

02

up

0t

0t

))-

))())

-))

)))-

))

)))-

))

-))))

-))

)))

))

(

20

� �� �

.xρ

μk

xρργk

xuu

tu

,xρ

μk

xρργk

xuu

tuρ

,xρρ

μk

xρργk

xuρu

tuρ

ρkρk/μBBxρk

xB

ρ, kB

γ

γ

γ

xx

02

02

02

22

thatimpliesit

222

1

222

1

221

1

22

2

2

���

���

���

���

����

����

���

���

���

���

���

���

���

���

���

����

above equation can be written as

0u

2

02

0t

222

1

222

1

���

���

�

���

�

�

���

�

�

���

�

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�

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���

���

���

x

x

t ukk

u

u

xk

xk

xuu

tu

xu

xu

)

-)(

))

)-

))(

)))

(

�

for smooth solutions, system (2.1) can be written as

0 AUU xt �� (2.2)

where the matrix A is defined as A =

���

�

�

���

�

�

uρ

wu ρ

2 , and � �21

22 bcw �� is the magneto-acoustic

speed with � �� �21

ρpc *� is the local sound speed and � �� � 2

12

)μρ

ρB(b � , which is Alfven speed;

0�� xt AUU

� � ,bcw

, u

ρwu ρ

A

21

22

2

where, ��

���

�

�

���

�

��

21

� �� � � �� �

� �

010

01

0

2

212

21

����

�

�

���

�

�

��

���

�

���

�

�

���

�

�

�

�*�

λ u

ρwu ρ

λIA

)μρ

ρB(,bρpc

� �

� �� �� � � �� �/ 0

� �

� �

wuuwλ

wλuwuλ

uwλwλu

wλuw)λu(wλu

wλu

ρ

wρλu

u-λ ρ

w ρλ u

�� � ��

� �

� ��

�

�

�

����

�

�

���

�

�

2

2

1

22

22

22

2

0

00

0

0

0

�

The eigenvalues of A are wu �1� and wu ��2� . Thus, the system (2.2) is strictly hyperbolic

when w>0. Let � �tr1 ,r w) �

and � � , tr2 wr ) �

are the right eigenvectors corresponding to the

eigenvalues 1� and 2� respectively. We have

� � r. 11 ���

����

� ���

�

����

���

���

�1

wwuj

ui

))

�

� �� �

� �� � .

221

2

���

����

� **���

ρwρpρ

ρμρwρB

when � � 0,** ρp , the first characteristic field is genuinely nonlinear. Similarly, it can be

shown that the second characteristic field is nonlinear when � � . ρp 0,**

22

The waves associated with

1r and

2r characteristic field will be either shock or rarefaction waves.

2.2 Shock: Let lρ and ρ , the left and right hand states of either a shock or a rarefaction wave are

� � � � � �lll ,, ))) BBppuu lll ��� and � � � � � �))) BBppuu ��� ,, denotes respectively; the system

(2.1) are using in Rankine–Hugonist jump conditions, the given by

� � � �

� � �2.4 2

2.3 2

2��

���

����

�

-))2

))2

Bupu

u

where �. denote the jump across a discontinuity curve � �txx � and dtdxυ � is the shock speed.

Lemma 2.1:

Let 1S and 2S respectively denote 1- shock and 2-shock associated with 1� and 2� characteristic fields.

Let the states lU and U satisfy the Rankine-Hugoniot jump conditions (2.4) and (2.3). Then the shock

curves satisfy,

� �,ρρguu ll � (2.5)

Where � � ρρ

ρρμ

Bpμ

Bp,ρρgl

lll ��

�

����

� ���

����

� ��

22

22

such that for, ,21 $$ ( we have for 0 $* u,ρρ l and u 0 ** on 1S , whilst for lρρ $ we have

0 *u and 0$**u on 2S .

Proof: The 2 -elimination of

� �

�

� �

� ��

���

����

�

��

���

����

�

-))2

))2-

))2

2

2

22

22

Bupu

u

Bupu

ρuρυ

23

� �

� �

� �

� �

� �

� �

l

llll

l

llll

l

llll

ll

llll

ll

llllll

lllllll

lllll

lllll

lllll

ll

lllllll

lll

l

ll

BpBpuu

BpBpuu

BpBpuu

BpBpuuuu

BpBpuuuu

uupuuB

pBpuuuu

uuB

pBpuuuu

BupBupuuuu

Bupuu

Bupuu

Bupu

))))

--

))))

--

))))

--

))--

))

))--

))))))

)))))))--

))))

))))))--

))))

))-

)-

)))))

))-

)))

-)

))))

-)

))

)(22

)(22

)(22

)(22

2

)(22

2

)(22

2

)()(22

2

)(22

2

)(2

)(

2)(

2

22

22

222

2222

2222

22222222

2222

2222

2222

22

222222

222

22

2

22

2

���

����

� � �

���

����

� ��

���

����

� ��

���

����

� �� �

���

����

� �� �

� � ���

����

� �� �

� ���

����

� �� �

���

�

���

� ��� �

��

���

����

��

���

������

�

����

�

��

���

����

Let � � � �;,2lg )))3 � on the differentiating (2.5) with respect to ) we obtain

� �� �))

4

*�*

���

����

� ���

����

� �

���

����

� ����

�

����

� ���

����

� *�*

�

��

2-

22

2222

21

22

22

Ψu

ρρ

ρρμ

Bpμ

Bp

μ

Bpμ

Bpρρ

ρρμBBp

ρu

l

ll

ll

l

24

which is negative for l)) . We can already show that 3 and 3 * are positive for l)) , and

� � � � ;ρψρψ l 0�*� further, 2,1 $$ ( 0 **ψ , whilst 0$***ψ .

Let � � � �� � � � � �� �ρψρψρψρχ ** *� 22

so that � � 0�lρχ . Since � � � � � �ρψρψρχ *** �* 2 , it follows for � � 02,1 *$$ ρχ( . Hence, � � � �l)5)5

for l)) . Thus, for 21 $$ ( , if we differentiate again, we get

� �� � � � � �� �

.Son 04

21

23

2

** *

�**ρψ

ρψρψρψu

Similarly, for l)) $ and 2,1 $$ ( we have 0 dρdu

on again differentiating, then 2

2

Son 0d$

)du

.

Now, these shock curves are satisfied the Lax entropy conditions.

Lemma(2.2):

If p satisfies 0 *p and ,p 0,** then the Lax condition hold,

i.e., 1-shock satisfies

� � � � � �UUUl 211 , �2��2 $$$ � �2.6

Whilst the 2-shock satisfies

� � � � � � , 221 2��2� $$$ UUU ll � � 2.7

Proof:

Let us consider 1-shock curve to prove � �lU1�2 $ . On apply 1-shock, we know that ,l)) since 0 *p

and 0,**p , by Lagrange’s mean value theorem, there exist exists a � �))6 ,l� such that

� � � � � �� � � �a,b, cabcfbfaf � *�

� � � � � �� �

� � � � � �� � � �,,ρρ, ξ

ρρρpρpξp

ρρξpρpρpρ, bρp, af

ll

l

ll

l

�

�*

*� ���

25

further, since ,0p ,** we have 00 ,** * p,p and p* is an increasing function � � , ,, l 6)))6 $� l

1l

))

, � � � � 2ll cξpρp �*$* and thus,

� � � � 12 *$*$l

ll

l ρρ, ρ, ρ

ρρξpξpc

and it implies that,

� � � �

� � � �2.8 2 . ρρ

ρρ)p(pc

ξpρρξp

ll

ll

l

$

* *

Also, since ll ρ ρρ 2 � , we have ll ρ ρρ

�2

it implies that

� �� �

� �� � l

ll

l

l

ll

l

ll

ρμ

kρρ.

ρρμρρk

ρρ

ρμ

kρρμρρk

ρμ

k) ρ (ρμ

k

22

1

22

22

22

2222

22

2222

22

22

�

This implies by that

� �� �

� �� �

� �� �

� �� �

� �� �

� �� � l

l

l

l

ll

l

l

l

l

l

ll

l

l

ll

l

l

ll

l

μρB

ρρμBB

ρρ

ρρμBB

μρB

ρρμBB

ρρ

ρρμBB

ρρ

μkρ

μk

ρρμBB

ρρ

ρρμBB

222

222

2222

22222

22222

222

22

2222

�

l

ll μρ

Bb2

22

and therefore

26

� �� � .

2B

l

222

))

))- l

ll

Bb

$ (2.9)

From equation (2.8) and (2.9), we have

� �� �� �

� �� �� �

� �� �

� �� �

� �� �

� �� �

� �

� �� �

� �� �

���

����

� �

�

���

� �

���

����

� �

�

���

� �

��

���

� �

��

���

� �

��

���

� �

��

���

� �

$

�

$

�

$�

))-))) 11

2B)(p

112

1

2

2

2

2

2

2

22

222

22

2

22

222

222

222

2222

l

ll

ll

l

ll

ll

ll

ll

ll

ll

lll

ll

lll

ll

lll

ll

l

ll

ll

ll

l

ll

lll

Bpw

ρρρ

μBB)p(p

ρρw

μBB)p(p

ρρρρρρw

μBB)p(p

ρρρρw

μBB)p(p

ρρρρw

μBB)p(p

ρρρρw

ρρ

ρρμBB

ρρ

ρρ)p(pw

ρρ

ρρμBB

ρρ

ρρ)p(pbc

In (2.5), the above inequality holds that � � ,l

l

l wuu $

))

)and hence � � .1 lll wuU �$ �2

In same manner another condition, since 0,**p and )) $l on 1-shock, we have

� � � �ρp ρρppηp

l

l *$

�* for some � �,ρρη l� ,and hence

� � . ρρ

ρρ)p(pc l

l

l

2

(2.10)

Further, since ,2

���

��� �

lρρρ

27

� �� � ,

ρρ

ρρμBBb l

l

l

2

222

(2.11)

and hence from (2.10) and (2.11), we obtain

� �� � � � ,

ρρ

ρρ)p(p

ρρ

ρρμBBcb l

l

ll

l

l

�

�2

2222

(2.12)

It implies that

� �� �.2

22

l

lll ρρρ

ρμBBpp-w

���

����

� � $

From equation (2.3) and (2.5) imply that

� � .hence and 1 υU λυ,ρρuρu-ρu-wl

ll $�

$

Lastly, we show that � �U2�2 $ . In this way, the equation (2.12), which implying that,

For 1-shock curve, using (2.5) we have� �

l

ll

ρρρuuw

, which implies that )(2 U�2 $ .

Hence 1-shock satisfied Lax condition; as well as way satisfied by the lax condition for the

2-shock.

Now we will show that the density, pressure, velocity magnetic field vary across a shock.

Applying equation (2.1) for 1-shock the left and right states have to satisfies Lax conditions

(2.6). Let us define -u V 2� ; then since � �lU1�2 $ , it follows that ll wV $ implies that

0$lV , and hence luυ $ .

Similarly, by using second condition i.e., � � � �UU 21 �2� $$ , we get

� �� � � � .

2

22

ρρ

ρρ)p(p

ρρ

ρρμBBw l

l

ll

l

l

�

28

,so wV -ww, vυu-w $$�$$

hence wV $ . From (2.3) we have llVρρV � . Since ) and lρ are positive, both V and Vl

must have same sign; since ,Vl 0$ we have 0$V . For 1-shock, the gas speed on the both sides of

shock is greater than shock speed, and therefore the particles cross the from left to right.

In case of 2-shock, applying Lax conditions, (2.7) this implying ll wV $ since )(2 2� $U or

equivalently υwu �� , which follows that Vw$ , and hence 0 V . In case of 2-shock particles

cross from right to left.

Let the states ahead of, and behind the shock be designated the 1- state and 2-state, respectively.

Then, for 1-shock 2 ,1 �� rl , and

hence .ll W V WV 2

22

222 and $ ;

for 2-shock .ll W and VW, so V rl 2

22

22212 $ �� .

Thus, for both shocks we have

.ll WV WV 2

22

222 and $ .

To this conditions satisfied the equation (2.4) holds that

;μ

BVρpμ

BVρp22

222

222

212

111 �����

which implies that

μBwρp

μBVρp

μBVρp

μBwρp

2222

222

222

222

222

212

111

212

111 ��$�����$�� ;

so 22

222

2222

222

111 23

23 )ρ

μk(cρp)ρ

μk(cρp l ��$�� .

the above inequalities follows that 2l )) $ , and

therefore .B B pp 2121 and $$ and from (2.3) we have

2211 VρVρ � ;

29

Since 21 )) , it follows that .21 VV Since for 1-shock 01 $V and 21 )) $ , and it follows that

12 VV , implies that 21 uu . Similarly, for 2-shock 02 V and 21 )) $ , its implying that 21 VV

and so 21 uu $ .

2.3 Rarefaction waves:

The ���

���

txU which are of the piecewise smooth continuous solutions of (2.2) such that

� �

� � � �

� �

U,λU

UλtxU),λ

txU(

Uλtx,U

(x,t)

rnr

rnln

lnl

+++

!

+++

"

#

��

�

�U

(2.13)

If we take txη � , then the equation (2.2) is a system of ordinary differential equations and it can be

written as � � ,0,-.

����

�

�

���

�

� 77tr

uIA )8

where I is 22� identity matrix and the differentiating with respect to the variable 8 is denoted by

dot. If )0,0(,.

����

�

�

���

�

� 77tr

u) , then ) and u become constants. If � �0.0,.

����

�

�

���

�

� 77tr

u) , then there exist a

eigenvector of the matrix A corresponding to the eigenvalue8 . Since it has two real and distinct

eigenvalues 21 �� $ , so it has two families of the rarefaction waves 1R and 2R which are 1-Rarefaction

waves and 2-Rarefaction waves respectively; Let us consider 1-rarefaction waves, since

� � 0,-.

����

�

�

���

�

� 77tr

uIA )8

and with wu �1� we have

30

� �

� �

� �

� �

� � (2.14) .0

0 ddu

ddw

0 ddu

ddw

0w0

0 ww

w

0w-u-u w

w-u-u

01 0

0 1u w

u

01 0

0 1u w

u

1

2

.

2

2

2

2

���9

��

��

��

��

����

�

�

���

�

�

���

�

�

���

�

�

����

�

�

���

�

�

���

�

�

���

�

�

���

�

�

���

�

�

����

�

�

���

�

�

���

�

�

���

�

�

��

���

�

���

�

�

���

�

�

����

�

�

���

�

�

���

�

�

���

�

�

��

���

�

���

�

�

���

�

�

�

77

77

7

7

7

7

7

7

dyyywu

uw

uw

u

u

uwu

wu

88)

)

8)

8)

))

))

)

)

)

)

)

)

)

)

)

)

)

Where 19 1-Riemann invariant is (2.14) represents 1R curve. Similarly, 29 2-Riemann invariant of the

2-Rarefaction wave curve is represent R2 curve

� � .02 � �9 � dyyywu

(2.15)

Theorem 2.1

On � �21 Rly respectiveR the Riemann invariant )ly respective( 21 99 is constant.

Lemma2.3:Across1-rarefaction waves (respectively, 2-rarefaction waves), l)) � and u ul �

(respectively, l)) , and u ul , if and only if, characteristic speed increases from left hand state to

right hand state.

31

Proof: Since, we know that

� � 022

2

22

���

����

� *�

**�

��

μwB

wp

dρdw

bcw

w is an increasing function of ;) so it form for 1-rarefaction waves, � � � �lρwρw � or it can be written

as w-wl � . These inequalities are uul � and w-wl � show that � � � �UλU λ l 11 � . Similarly we can

prove � � � �UλUλ l 22 � for 2-rarefaction waves. Then the conversely for 1-rarefaction waves,

since � � � �UUl 11 �� � , we have

. uuw-w ll � (2.16)

In further, since 1-rarefaction wave region 19 is constant, then we get

� � � �dyyywdy

yywu-u

ρρl

�� �00

1 ,

the equation (2.16) shows that

� � � �dy,yywdy

yyww-w

ρρ

l

l

�� �00

which implies that l)) � and

� � � � 000

1 , � �� dyyywdy

yywu-u

ρρl

.

Hence l)) � and uul � .Similarly, it can be shown that the 2-rarefaction waves, l)) , and uul , .

Introducing a new parameter ,: where r

l)): � obtain from (2.5) the following formulas for shock

curves, for 1-shock curve 1 : and 2-shock curve 1$: .(Respectively, rarefaction curves in the term of parameterizations).

From equation (2.5),

32

� � � �

γγ

l

r

l

r

l

llllll

ρkpθppθ,

ρρ

ρρρρ

μBp

μBp,ρρg, ,ρρguu

1

22

by thereimplyingit

22 where

���

���

����

� ���

����

� �� �

ρk BBB

l

r2 thatimplies ��: , so that

ll

r

BB

)): r�� ,

� � � � � (2.17) 11111

11111

111

221

221

2

2

2

2

2

22

22

. θ

θBθAuu

θ)

BB()

pp(

uu

ρρρρ)

BB()

pp(

uu

ρρρρ

μB

μBpp

uu

ρρρρ

μBp

μBp

uu

γ

l

r

l

r

l

r

l

r

l

l

l

r

l

r

l

r

l

lll

l

r

l

lll

l

r

���

��� � �

���

��� ���

����

� � �

���

����

� ���

����

� � �

���

����

� ���

����

� � �

���

����

� ���

����

� � �

For 1-rarefaction waves 1),( $: since 1-Riemann invariant is constant, we have

γ

l

r

l

r θppθ,

ρρ

�� , ,θBB

l

r � so that ll

r

BB

)): r��

if we set dtdθt,θ �� , dtθ

BtγAtuu γ

l

r ���

��� ��� 1121 1

dtt

BtγAtuu γ

l

r 21

1 ���

(2.18)

Similarly, for 2-rarefaction wave � �1 : , we have

, ,r (::))

��l

r

l pp :�

l

r

BB

then as well as we set ll

r

BB

)): r��

33

dtt

BtγAtuu

dtθ

BtγAtuu

γ

l

r

γ

l

r

21

1121

1

1

���

���

��� ���

Thus, for 1-family, either shock or rarefaction wave, we have

� � � � � (2.19)

1 11111

1 2

1

2

1

θ ,if

θθBθA

,θifdt, t

BtγAt

uu

γ

γ

l

r

++

!

++

"

#

���

��� �

��

��

In the similar way,

� � � � �(2.2

1 2

1

1 11111

1

2

,θ ifdt,t

BtγAt

θif ,θ

θBθA

uu

γ

γ

l

r

++

!

++

"

#

,�

�

$���

��� �

�

0)

where 2

11

l

γl

uρkA

� and 2

22

2 l

l

μuρkB � ; is as to expression in above equations (2.19) and (2.20).

Theorem 3.1:

The 1R curve is convex and monotonic decreasing while 2R curve is concave and monotonic

increasing.

Proof: We know that, 1-rarefaction wave is

� � (3.1) ρρ dy ,if yywuu l

ρ

ρl

l

��� �

On differentiating with respect to , ) we have

0$�ρw-

dρdu

34

(3.2) 22

2

. ρw

ρw

dρud *

�

We know that ,cw 22 b�� since ρk,Bρkp γ21 �� in the following equations

Again, on differentiating with respect to ) ,we have

)))

'

22

2d wwd

u �

� �

� �

� �

� �

� �''22

112

2

''22

2

2

''2

2

2

''

22

2

22

2

kd

d

d

d

d

ccbbkd

u

ccbbwd

uw

ccbbwd

uw

ccbbwd

uw

ccbbwd

u

� ��

� �

� �

� �

*�* �

)-))(

)

))

))

)

)))

)))

(

We know that b and c are differentiating, we have

� �� �

21

12 that so and2

thereimpliesit

21

21

'11

2

22

'22

2

�

��

��

γ'

γγ

ρkγγ cc

ρkγγccργkcμ

kbbμ

kb

� �

� �w

ccbbw

bcccbbw

μρρkργkw

μρρk,bργkc

γ

γ

*�*�*

�

*�*�*

��

��

22

22

21

1

22

221

12

222

35

� �

� �

� �

� �0

2

3

023

211

21

2

2

11

22

2

2

2

2

11

22

2

2

2

11

22

2

2

112

22

21

12

2

�

�

�

�

�

���

��� ��

��

μBBpρ

ρkγγμ

ρk

dρud

wρ

ρkγγμ

ρk

dρud

γργkμ

ρkdρ

ud

ρkγγμρk

μρkργk

dρud

''

γ

γ

γ

γγ

for 21 �� ( hold

� �0

2

k3d

''2

11

22

2

2

2

�

�

�

-)

)((-)

)

(

BBp

k

du

and, therefore, u is convex with respect ) for1-rarefaction waves. Similarly, we can show for

2-rarefaction waves.

Now we prove that the shock curves are starlike with respect to � �ll u,) for p and B, these has a good

geometry in Riemann invariant coordinates whenever .0and0 ,** * p p

Theorem 3.2:

The 1-shock and 2-shock curve are starlike with respect to � �lu,l) when γρkp 1� and ρkB 2� for

values of ( lying in the range � �21 �� ( .

Proof:

We have to prove that any ray through the point � �lu,l) be intersected 1-shock curve in at atmost one

point for this is sufficient to prove the rays � �lu,l) through the two different points � � � �2211 ,,, uu )) on

the 1-shock curve, and whose slope are different. The slope of the line joining � �lu,l) with

� � � � ,,, 2211 uu )) isl

l

ρρ-uu 1

1 and l

l

ρρuu

2

2 .

For the 1-shock equation (2.5) are implies that

36

� � � �ρfρfρ-ρu-u

l

l21

2

�����

����

�,

where � � � � � � � �ll

l

ll

l

ρρρμρBBρ,f

ρρρρppρf

�

�

2

22

21

we prove that � � 0' $)lf and � � 0'2 $)f . When putting ρkB 2� in � �)2f and differentiating with

respect to ,) and we have

� � � �� � � �

� �

� � � �� �

� � � �� �� �

� � � �

� � ���

����

���

��

�

�

�

�

�

ρρμkρ f

ρμρρρkρ f

ρρρμρρρρρkρf

ρρρμρρρkρ f

ρρρμρ ρkρ k

ρρρμρBBρf

l

l

l

ll

ll

ll

l

ll

l

ll

l

112

2

2

2

22

22

2

22

2

22

2

2222

2

22

22

22

2

and its again differentiating with respect to ,) we have

� �

� � 02

12

2

22

2

2

22

2

$

�*

���

����

� �*

μρkρf

ρμkρf

It also proved that, when � �)1f in γρkp 1� and differentiating with respect to ,) and we obtain

� � � �

� � � �

� � � �ll

γl

γll

γl

γll

l

ρρρρ) ρ (ρkρf

ρρρρ ρk ρkρf

,ρρρρ

ppρf

�

�

�

11

111

1

on differentiating, we have

37

� � � �

� � � �� � � � � �� �

� �� �

� � � �� � � �� �� �� �

� � � �� � � �� �� �

� � � �� � � �� �� �

� � � �� � � �� �� �

� � � �� � � �� �� �

� �� �� �

� �� �� �

� � � � � �� �� �2

21

11

21

111

211

121

21

111

11

2

211

121

11

211

11

2

211

211

11

221

2

211

211

11

211

21

2

211

211

11

221

2

211

211

111

2

211

111

2

11111

1

1

212

22

22

22

22

22

22

2

ll

γl

γl

γl

γl

ll

γll

γl

γl

γγl

ll

γl

γll

γl

γl

γγl

ll

lγl

γlll

γγl

γll

ll

lγl

γlll

γγl

γl

γl

ll

lγl

γlll

γγl

γll

ll

lγl

γlll

γγl

γll

ll

llγl

γγll

ll

llγl

γγl

γll

ll

γl

γ

ρρρρρkρkρργkρργk

ρρf

ρρρρρkkρρργρkρρkρρkγkρρ

ρρf

ρρρρρkkρρρρkρρkργρkγkρρ

ρρf

ρρρρρρkρρkρρρkρρkργρkρρρρ

ρρf

ρρρρρρkρρkρρρkρρkργρkρργρkρρ

ρρf

ρρρρρρkρρkρρρkρρkργρkρρρρ

ρρf

ρρρρρρkρρkρρρkρρkργρ.kρρρρ

ρρf

ρρρρρρρρk ρkβγρ.kρρρρ

ρρf

ρρρρ

ρρρρρρ)ρk ρ(kρk ρk

ρρρρρρ

ρρf

ρρρρρ

) ρ (ρk

ρρf

� �

��

�

� �

��

�

��

��

�

�

��

�

�

��

�

�

��

�

�

��

�

��

�

�

�

�

��

�

��

��

���

���

����

Let � � � � � � ,ρkρkρργkρργkρg γl

γl

γl

γl

21

11

21

111 212 ��� � � � then � � 01 �lρg .

Then � � � �� � � � 11

11

12111 2112 �� � �� � γ

lγl

γl

γl

' ρkρkρργγkρργγkρg , � � 01 �l' ρg .

Since � � � �� � � �� � 221

111 1112 �� � γ

lγ

l'' ρργγγkργργγ kρg , if above condition are follows that the

values of ( in 21 �� ( , we have � � 0g ''1 $) . The above equation to be held in 1-shock and then )) $l

38

and � � � � 0gg '1

'1 �$ l)) , implying that � �)1g is a decreasing function of ) ; and this follows that

� � � �l)) 11 gg $ , it therefore 0'1 $f . Thus,

l

l

-u-u))

is a decreasing function of ) ; we hence 1-shock curve

is starlike with respect to � �lul) , as usually in same way 2-shock curve and is also a starlike with respect

to � �lul) .

Lemma 3.1

With � � 0 ρp' and ,0,''p the inequalities 102

1 $$dΠdΠ

and 101

2 $$dΠdΠ

hold along 1-shock and 2-

shock respectively, with

� �

� � (3.4) .

(3.3)

2

1

dyyywuΠ

dy, yywuΠ

ρ

ρ

�

�

�

��

Proof: From (3.3) and (3.4), we have

� � � �

� � � �ρρw

dρρdu

dρdΠ

ρρw

dρρdu

dρdΠ

�

��

2

1 and

We know that the above theorem as long a 1-shock curves.

� � 0$dρ

ρdu, it has that 02 $

dρdΠ

.

Further, as along 1-shock curvesdρ

dΠdρ

dΠ 21 $ , we have .12

1 $dΠdΠ

In order to prove that

,102

1 $$dΠdΠ

in sufficiently part dρ

dΠ1 <0. For a condition 1-shock curve, equation (2.5) it imply that

� �ρ

μBBp

ρρw

''

���

����

��

�

39

� � � �

,

222

11122

222

1222

2

2

22

2

22

22

2

22

���

����

� ���

����

� �

���

�

���

����

����

� ��

�

����

�� ��

�

����

� �

��

���

����

� ���

����

� �

���

����

� ����

�

����

� ���

����

�� ��

�

����

��

��

l

lll

l

''l

l

l

lll

ll

l

l'

''

'

ρρρρ

μBp

μBp

μBBp

ρμBp

μBp

ρρρρ

μBp

μBp

ρμBp

μBp

ρρρρ

μBBp

ρμ

BBp

ρρw

dρρdu

))

hence the above condition holds that

� � � � 0$�ρρw

dρρdu

, and these implies that 01 $ dρ

dΠ.

Similarly, we show that 12

10 $$dΠ

dΠ along 2-shock curves.

4. Riemann Problem:

The system (2.1) be the initial condition as

� �!"#

$

�0

00 ,

, ,,

xxifUxxifU

txUr

l , (4.1)

is called as Riemann problem. Where lU be the state to the left of 0xx � and rU be the state to the

right of 0xx � the constant states are separated by in both waves either a shock waves or rarefaction

wave. The Riemann invariant coordinates are

� � � �dyyywuΠdy

yywu Π

ρρ

�� ��� 21 and .

Lemma (4.1):

The mapping � � � �21,ΠΠρ,u is one to one and the Jacobian of this mapping is nonzero when 0 ) .

40

Proof: Since, � � � �dy

yywuΠdy

yywuΠ

ρρ

�� ��� 21 and

On differentiating with respect to ,) we get

,u

Π,ρw

ρΠ

,u

Π,ρw

ρΠ

1

1

22

11

���

���

���

���

Thus, the Jacobian of the mapping � � � �21,ΠΠρ,u

1

1

22

11

ρw

ρw

uΠ

ρΠ

uΠ

ρΠ

�

��

��

��

��

ρw

ρw

ρw 2��

This is one-one and onto.

We consider Riemann invariants as coordinate system. Let us will take a plane � �21,ΠΠ in that plane we

draw the curves 121 , ,S RS and 2 R which divide the plane can into four distinct regions. I, II, III, and IV.

Let lU are left state. Fixing lU and varying rU . Let us consider rU belong to any of the four region as

fig.4 (a). For 2RU� .

� � � � � �/ 0� � � � � �/ 0 � � � � � � 21 and 21

21

2121

2121

,n,URUSU T ,,nR,ΠΠ:,ΠΠUR.,,nS,ΠΠ:,ΠΠUS

nnnnn

nn

��������

�

In an above wave curves, the plane � �ρ,u divides into a four region. To solve the Riemann problem,

consider the wave curve � � 2 mUT for �mU � �lUT1 . And we have to verify that two curves � �mUT2 and

� �m*UT2 , where � �lm*m UT,UU 1� , so these a two curves are non-intersecting and the set of all such

curves to entire half space 0 ) in the plane � �21,ΠΠ in one-one fashion.

If �rU I, draw a vertical line rΠΠ 22 � in fig.4 (a). Which will be intersects S1 uniquely at a point1

mU .

The solution to Riemann problem is now obvious; we taking on constant state of 1mU by a 1-shock and

then from 1mU to the constant state rU by a 2-rarefaction wave.

41

Let �rU II region, draw a vertical line rΠΠ 22 � in fig.4 (a) which is intersect R1 uniquely at a point

2mU . The solution is going from lU and 2mU by 1R and to from 2m U to rU by 2R .

If �rU III region, we define the concept of inverse shock curve. The inverse curve denote by *2S

consists of those states � �21,ΠΠ which can be connected to the state � �rr ,ΠΠ 21 on the right by 2Sshock in fig.4 (a). These represented, from (2.5) by

Fig 4(a). Rarefaction curves (R1 and R2) and shock curves (S1 and S2) in the plane � �21,ΠΠ .

.22

22

���

����

� ���

����

� ���

r

rrrr ρρ

ρρμ

Bpμ

Bpuu The above curve intersect the 1R uniquely a point 3mU

.Therefore, rU can be connected with Ul by 1R are followed by 2S .

If �rU IV region (see fig.4(c)), (as from lemma 3.1) 11

2 dΠdΠ

on 1S . It also defines in 11

2 $dΠdΠ

on *2S .

This mean that, the 1S and *2S will intersecting uniquely at the point

4mU , therefore, the solution consists

of 1-shock and 2-shock. Thus we have shown that set � � � �/ 0lmm UT:UUT 12 � , covers the entire half

space 0, ) in the plane � �21ΠΠ in a one-one way.

When the vacuum state � �0�ρ it’s not satisfied the same condition.

Lemma (4.2):

If ,21 rl ΠΠ � the vacuum occurs.

42

Proof:

From fig.4 (a), ; and 2211 rmlm ΠΠΠ Π ��

if .0 then , 212121 � � � rlmmrl ΠΠΠΠΠ Π

But it will be � �dyyywΠΠ

mρ

mm �� 0

21 2 y

Which implying that that .ρm 0� Hence, vacuum occurs.

ρ,u).( planein curves waveFig4(b).

Theorem (4.1):

Assume that 00 , ''' ,pp and that we are given initial states lU and rU where ,ρl 0 0 )) for

the Riemann problem of system (2.1). Assume that rl ΠΠ 21 . Then there exists a solution of the

Riemann problem for system (2.1). Moreover, the solution is given by 1- wave following by a 2-wave satisfying 0, ) and the solution is unique in the class of constant states separated by shock waves

and rarefaction waves.

43

fig.4(c). 2-shock wave and 1-shock wave.

fig4(d). vacuum curve

5. Interaction of Elementary waves:

The interaction of elementary waves, obtaining from the Riemann problem (4.1), gives rise to new emerging elementary waves. And then two jump discontinuities at , and 21 ,xx it as follows:

� � � �5.1

2

21

1

0 x,if xU

xx,if xUxx,ifU

x,t U

r

*

l

+!

+"

#

�$$�$�$�

�

The choice of *U and rU in the terms of lU and an arbitrary . and 21 Rx x � With the initial data, we

have two Riemann problem locally. The first Riemann problem of the elementary wave may interact the second Riemann problem of the elementary wave, and the time of interaction at formed a new Riemann problem at one dimensional Euler equation. It may be found of the interaction of the elementary waves. Here we like ,212 RSR it means that a 2-rarefraction waves � �* touul ,R 2 of the first Riemann

problem interacts with 1-shock, , S1 of the second Riemann problem rl uu to .Then it interacts to new

Riemann problem mrl uuu via to 21RS . In different families are possible to interaction of elementary

waves and as well as the same family are respectively )SR,RR,RS,SS ( 12121212 and

),,,,,(S 222211111122 SRRSRSSRSSS .

44

5.1 Interaction of Elementary waves from different Families:

(a) Collision of two shocks (S2S1):

Let lU is connection to *U by the 2- shock 2S is a first Riemann problem and * U is connected to rU

by a 1-shock, 1S of the second Riemann problem. For a given ,Ul we consider *U and rU in such a

way that ,ρρ l* $ from (2.5) we have � �*ll* ,ρρguu � in other way that ,* r)) $ then we have

written .),ρg(ρuu r**r �

fig. 5.1(a). S2S1 collision

Since, speed of 1-shock of the second Riemann problem is negative, 2S and speed 2-shock of the first

Riemann problem is positive, S1 overtakes 2S . Then it shows that for any arbitrary state lU , the state

U r lies in the region IV (in fig.4 (b)). It in sufficient to prove that

� � � � � � . and for 0,,,g **** )))))))))) $$ � lll gg Let take in contrary that

� � � � � � .0,,,g ** �� )))))) ll gg If in this, then

� � � � � �� � � �,,,,2,,g l2

**l*2

*l2 )))))))))) ggg ���

Implying thereby that,

45

� � � � (5.2) 0,,2112

P112

P **l*

22*

**

22*

* �����

����

� ��

�

����

� � ���

�

����

� ��

�

����

� � ))))

))-))-ggBBPBBP l

ll

The above equation (5.2), is strictly positive, which is a contradiction. Hence,

� � � � � � 0,,,g **l �� )))))) lgg , i.e., the curve � �*1S U are lies below the curves � �lU1S , therefore,

U r lies in the region IV. Thus, and it follows interaction results is 2112S SSS interaction results, in

case of illustrate in fig.5.1 (a).

(b) Collision of a shock and rarefaction (S2R1):

Here � �l* USU 2� and � �*r URU 1� i.e., for a given lU , Let *U and rU such that )) $* from

equation (2.5), we have � � , and *r )) � � *ll* ,ρρguu from equation (2.15) we have

� �dy.yywuu

*

r

ρ

ρ*r ��� Since 2-shock of the Riemann problem is positive and 1- rarefaction wave of the

second Riemann problem is negative velocity, it follows that 1R overtakes 1S . Since, for any given lU ,

� � � � � � 0,ywyw *

*

� �� )))

)

)

)lgdy

ydy

y

l

for ,* l))) $$ and can be follows that the curve � �*UR1 lies below the curve � �lUR1 , hence rU lies

in the region III, subsequently 2112S SRR . The compute results this case in fig.5.1 (b).

fig. 5.1(b). S2R1 collision

46

(c) Collision of two rarefaction waves (R2R1):

We consider � �l* URU 2� and � �*r URU 1� . In any other way, for a given lU , Let *U and rU such that

*,l )) �� � and

*

* dyyywuu

l

l ���)

)*)) �r , then

� � . *

* dyyywuu

r

r ���)

)

Since, the trailing end of 2-

rarefaction wave has a positive velocity (bounded above) in � �tx, - plane and that 1-rarefaction wave

has a negative velocity (bounded above), interaction will take place. Since *l )) $ and

� � � � � � ,0 ***

� ��� dyyywdy

yywdy

yyw

l

)

)

)

)

)

)

It follows that the curve � � UR *l lies above the curve � �lUR1 ; hence rU lies in the region II and the

interaction results II and the interaction result is 2112R RRR . Then computed results, in fig.5.1 (c).

fig. 5.1(c) R2R1 collision

(d) Collision of a rarefaction wave and a shock (R2S1):

Here � �l* URU 2� and � �*r USU 1 � , i.e., for given lU , we choose *U and rU such that *l )) � ,

� �rl dy

yywuu

l

)))

)

$�� � ** and *

and � �.r**l ,ρρguu � Since, the second Riemann problem of 1-shock

speed is less than 2-rarefaction wave of first Riemann problem of the speed of trailing end in plane,-)(x,t and therefore 1S penetrates 2R . For any given lU . It show that �rU I, then it, to show

that

47

� � � � � � .0,, *

*

�� )))))

)

ggdyyyw

l

l

(5.3)

Since � �)) ,lg is a decreasing function with respect to the first variables l ) , then we will have

� � � � *l* for ,, )))))) $ gg l . Hence, the equalities (5.3), there imply that curves )(US *1 lies above

the curve � �lU S1 and rU lies in the region I. Thus the interaction result is 2112R RSS ; and its

computed results in fig. 5.1(d).

fig. 5.1(d) R2S1 collision.

5.2. Interaction of Elementary waves from same family:

(a) 2- shock wave overtakes another 2-shock wave (S2S2):

We consider the situation in which l U is connection to * U by a shock of the first Riemann problem

and *U is connected to rU by a 2- shock of the second Riemann problem. In other situation a given

left state lU , the intermediate state *U , and the right state rU are chose such that l)) $* and

� �*ll* .ρρguu � with Lax conditions satisfy

� � � � � � � � � � ,, ,, *2*22*21 UUUUUUU llll 2��2� $$$ (5.4)

and *r )) $ , and � �r**r ,ρρguu � with Lax stability conditions

� � � � � � � � � �rr UUUUUUU , ,, *2*2*2*2*1 2��2� $$$ (5.5)

48

where � �*2 ,UUl2 is the speed of shock connection lU to *U , and similarly, � �rUU ,*22 is the speed of

shock connecting *U to .rU From (5.4) and (5.5), we obtained � � � �,,, *2*2 UUUU lr 22 $ i.e., the

second Riemann problem of 2-shock overtakes the first Riemann problem of 2-shock at a finite time, then its give rise to new Riemann problem with data Ul and rU . To prove this problem. We must

have to determine the region in which rU lies respect to lU . Let be claim that rU vary lies in region III

so this have solution of the new Riemann problem consists of 1R and 2S . In any more way, to show that

to our claim: we have to prove that � � S2 *U lies in entirely in the region III; to prove this required to

show that for .* l))) $$

� � � � � � .0,,g-, ** )))))) ll gg We consider, on the contradiction that

� � � � � � .0,,g-, ** � )))))) ll gg for l))) $$ * . Then the follow that, if we take, then

� � � � � �� � � � ,,,,2,,g *2

**l*2

l2 )))))))))) ggg l ��� (5.6)

Implying there by that,

� � � � 0,,2112

p112 *l

22*

**

22

�����

����

� ��

�

����

� � ���

�

����

� ��

�

����

� � ))))

))-))- ll

ll

l

ll ggBBpBBpp

Proving that

0112

112

2222

, ρρμ

BBppρρμ

BBppl

l*l*

*l

ll �

���

�

���

����

����

� ��

�

����

� � ���

�

����

� ��

�

����

� � (5.7)

which is contradiction on as left hand of inequalities (5.7) is positive. Hence, 2122 R SS S ; and

computed results in above situation fig.5.2(a).

fig. 5.2(a) S2 overtakes S2.

49

(b) 1-shock wave overtakes another 1- shock wave (S1S1):

Let rU lies in a region I, so that 2111S RSS , is similarly to the previous case and its above situation

illustrate computer results.

fig. 5.2(b) S1 overtakes R1.

(C) 1-shock wave overtakes 1-Rarefaction wave (R1S1):

In case, the Riemann problem of the lU is connected to *U by 1- rarefaction wave and the second

Riemann of the *U is connected to rU by 1-shock. i.e., a given lU , Let *U and rU in such a way that

, * l)) � � �dyyywuu

l

l ���)

)*

* and � �., , *** rrr guu )))) �$ So we show that � �*1S U lies below of

the � � R1 lU for l))) �$* , in other way, for l))) �$* ,

� � � � � � 0,*

* � �� dyyywdy

yywg

ll )

)

)

)

)) (5.8)

Let us define, � � � � � � � �dyyywdy

yywg

ll

�� ��)

)

)

)

)))*

,F *1 so that � � 0F *1 �) . On differentiating � �)1F with

respect to ,) we obtain � � 0, F1 * ) implying that � � � �)) 1*1 FF $ . i.e., � � , 0F1 ) and hence � �*1S U lies

below the curve � �lU1R for l))) �$* . In another to show that it is sufficiently to � �lU1S lies above

the curve � �*1S U for ; l )) � the sufficiently part, the claim has

50

� � � � � � )))))))

)

�; � l* ,0,,*

dyyywgg

l

l ; (5.9)

Let us define � � � � � � � � ,,F*

*2 dyyywgg

l

ll � �)

)

)))))

So that � � � � 0FF 12 � ll )) .

Let us consider that � � � � � � ,for ,,, *** ))))))))) $$� lll ggg

implying that,

� � � � � � � � � �,,,,2,, 2***

2*

2 )))))))))) lll ggggg �

implying thereby that

� � � �;,,2 112

P112

P ***

2*

2

**

2*

2

*l ll

l

l

l ggBBPBBP ))))))-))-

����

����

� ��

�

����

� � ���

�

����

� ��

�

����

� �

or equivalently,

.0112

P112

P2

*

2*

2

**

22

*l ����

�

���

����

����

� �

��

����

� � ��

�

����

� ��

�

����

� �

ll

l BBPBBP

))-))- (5.10)

But the left hand side of inequality (5.10) is positive, which leaves us with a contradiction.

Hence, � � � � � � ,for ,g,g-,g **l* ))))))))) $$ ll implying that,

� � � � � � � � � � � � 0,g ,, 2**

**

� �� lll Fdyyywdy

yywgg

ll

))))))))

)

)

)

We define a new function,

� � � � � � � � .for ,,F **3

*

ll dyyywgg

l

)))))))))

)

�� � �

At some point � �11~,~ u) intersected in � �lU2S and � �,S *1 U for .~

1* l))) $$ Since, � � 0F3 ) and

� � ,0F *3 $) it is intermediate value property, there exists a ,~1) between *) and ,l) such that

� � ,0~F 13 �) by virtue of monotonicity. Thus, � �lU2S and � �*1S U is uniquely determined of the

51

intersection, and the computer results in fig. 5.2(b). We distinguished three cases to depending on the value of ,r)

(a) If ,~1r )) $ indeed 1-shock is weak as compared to 1- rarefaction wave, when �rU III and the

interaction results is .R 2111 SRS

(b) If 1r~ )) � , indeed two waves of first family interact, they annihilate each other, and give rise to

wave of second family, when rU lies on � �lU2S and the interaction result is 211R SS .

(c) If ,~1r )) and the interaction result is ,R 2111 SSS on ;IVUr � indeed , the 1-rarefaction of

the first Riemann problem is weak as compare to the 1-shock of second Riemann problem, which is stronger, overtakes and the trailing end of 1-rarefaction wave a reflected shock

� �rm UU ,S2 , and a connection new connection constants state mU on the left to rU on the right, is

produced. The transmitted wave, after interaction, is the 1-shock that joins state lU on the left and mU

on the right.

fig.5.2(c) R2 overtakes S2

(d) 1- Rarefaction wave overtakes 1-shock wave (S1R1):

Here for a given lU , we consider *U and , rU such that � �l* USU 1� and � �*1rU UR� , i.e., l)) *

from equation (2.5) we have � �*ll* ,ρρ-guu � and ,* l)) , from equation (2.5) we have

� �dyyyw uu

l

*

ρ

ρ*r ��� . In the plane (x,t) the speed of trailing end of, � �*1 U� is less than 1-shock speed

� �*1 ,UUl2 and therefore the 1-rarefaction wave from right overtakes 1-shock from left a finite time.

We show that the curve � �*1R U lies below the curve � � S1 lU for *l ))) $� ; for this we have to that

52

� � � � � �*l*l for 0,,g

*

))))))))

)

$� � dyyywg l . Let us a new function � �,G1 ) to show that

� � � � � � � �*l*l1 for 0,,gG

*

)))))))))

)

$� � � dyyywg l , then in this way it define the for � �)w

and � �,,))lg to hold that on � �)1G differentiating, we have that

� � � � 0,2

1112

p-p G

2

2

22

l

1 $��

���

����

����

� ��

�

����

� *�* ��

�

����

� �

�*))

))-)-)

l

l

l

g

BBpBB

Implying thereby that � � � �*11 GG )) , since � � ,0G *1 �) we have � � ,0G1 ) then we prove that,

� �*1R U lies below the curve � �lUlR for ,*l ))) $� then

� � � � � � .for 0,g *l*l

*

))))))

)

)

)

$� � �� dyyywdy

yywl

Since the left hand side of this inequalities, for

*l ))) $� , to be � �,Gl l) which is � �l)lG is positive, so the conclusion above. So, we show that to

� �*1R U and � �lU2S intersect into uniquely at some points � �22~,~ u) ; to show that, for this

� � � � � � 0,g,g*

l*l � � dyyyw)

)

)))) a uniquely root has 2~) such that l)) $2

~ . To show a new function,

we define � �)2G , there implying that � � � � � � � �dyyyw

� �*

,g,g ,0G l*l2

)

)

))))) ; and we know,

� � 0G 2 �) it takes negative as close to zero, then � � 0G2 l) . The curves are interest uniquely at the

� �*1R U and � �lU2S it follows that the intermediate value property, and in view of monotonically; here

three cases the value of r) on depending, we distinguish like,

(i) when 2r~)) , and IVUr � ; the interaction result is 2111 S RS S ; indeed in sufficient

case the both curves are interaction and then the 1-rarefaction wave is weak compared to the 1-shock is stronger, which is produced a new elementary wave.

(ii) when 2r~ )) � , and � �,USU lr 2� the interaction result is 211S SR i.e., interaction of the

first family of elementary waves. Gives rise to a second family of a new elementary wave. (iii) when 2r

~)) $ , and IIIUr � the interaction result is 2111S SRR .

53

Fig.5.1(d) R2 overtakes S2.

(e) 2-Rarefaction wave overtakes 2- shock wave (S2R2):

When � �lUS2*U � and � �*2rU UR� of the 22S R interaction takes place, in any another word, in a

given lU , we have consider *U and r U are in a such way that l)) $* , from (2.5), we have

� �** ,u ))ll gu � and l)) �* , from (2.15) we have � �dyyywu

r

���)

)*

*ru . We show that for

l))) �$* , � �lU2S lies above the curve � �*2 UR , i.e.,

� � � � � � � �ldyyyw )))))))

)

)

, 0,g,g *l*l

*

�; � (5.11)

We define, to show that � �) M1 , � � � � � � � �dyyyw

� �)

)

)))))*

,g,g M l*l1 . Since there implying by that

� �l2 US lies above � �*2 UR ,On differentiation � �) M1 , since � � ,0 M1 * ) we have � � � �*11 M M )) ,

since � � ,0 M *1 �) it follows that � � ,0 M1 ) we prove that the � �l2 UR lies above the curve � �*2 UR

for ;* l))) �$ to show that for this it is enough � � � � � � � � 0,g M*

*l1 � �� dyyywdy

yyw

l

l

)

)

)

)

))) for

))) �$ l* and the curve � �l2 UR lies above the curve � �l2 UR for ))) �$ l* , � � ; 0M1 l) the left

hand side of this inequalities is � �l)1M which to positive, we show that � �*2 UR of the intersect

uniquely � �l1 US at point � �33~,~ u) for 3l*

~))) $$ . We define

� � � � � � � � or , ,g-,M l**l2

*

)))))))))

)

�$ � � fdyyywg l so that � � ,0M2 $) and we consider a

54

constant ,K 0 such that � � 0M2 ) for all Kρ . Then there exists a 3~) such that � � 0~M 32 �) .

Thus, � �*2R U and � �lU1S are intersect uniquely at � �33~,~ u) as � �*2R U and � �lU1S in a terms of

monotone, and the computed results shown in fig.5.2(c). Here three cases are following,

(i) If ,~3r )) $ IVUr � the interaction result is 2122S SSR , indeed, the strength of 2R is

small compared to the elementary wave ,S 2 and 2S annihilates 2R in a finite time. The

strength of the reflected 1S is small compared to the incident waves 2S and 2R .

(ii) When 3r~)) � and )(S U 1r lU� the interaction result is 122 SRS indeed 1S is weaker

than 2R compared to the incident waves 2R and 2S .

(iii) If ,~3r )) the interaction results is 2122 SRS R ; indeed, 2R is stronger than 2S .

(f) 2-Shock waves overtakes 2-Rarefaction(R2S2):

For a given lU , we have *U and rU , here � �lUR2*U � and � �lUS1rU � such that *l )) � ,

� �dyyywu

l

l ���*

*u)

)

and � �rrr guu )))) , , *** �� . We prove that � �lU2R lies above the curve

� �*2S U for *l ))) $� .

� � � � � �*l* ,0,g

*

**

))))))

)

)

)

$�; � �� dyyywdy

yyw

ll

(5.12)

To show that, we have a new function

� � � � � � � �*l*1 for ,g

*

**

)))))))

)

)

)

�� �� �� dyyywdy

yywN

ll

; so that � � 0 1 �)N .

This , in view of the expression for � �)w and � �,,g * )) yields

� � � � 0,2

2111

*

22*

*2*

1 $��

���

����

����

� � ��

�

����

� ��

�

����

� *�*

�*))

-)))-)

g

BBppBBpN

There implying by that,. Hence this result � � � � 0NN *11 � )) we show that � �lU2S lies above the

curve � �*2S U for *l ))) $� ; to show, it is sufficient for this

55

� � � � � �*l* for 0,,

*

))))))))

)

$� � dyyywgg

l

l . If � � � � � �ll ggg )))))) ,,, ** then

� � � � � � � � � � � � 0N,,, 1**

**

� �� lll dyyywgdy

yywgg

ll

))))))))

)

)

)

. We consider, which is

contradiction that � � � � � �ll ggg )))))) ,,, ** � . Thus, it we have that � � � � � �)))))) ,,, ** ll gg � .

There implies by that, � � � � � � � � � �)))))))))) ,,,2- ,, 2***

2*

2lll ggggg � ; this expression, in

terms of � � � �lgg )))) ,,, ** and � �)) ,lg yields

� � � �ll

l ggBBppBBpp ))))))-))-

,,2112

112 **

*

2*

2

**

2*

2

* ����

����

� ��

�

����

� � ���

�

����

� ��

�

����

� � ;

or equivalently

0112

112

2

*

2*

2

**

2*

2

* ���

���

����

����

� ��

�

����

� � ���

�

����

� ��

�

����

� �

))-))-BBppBBpp l

l . (5.13)

Which is contraction, the above equation (5.13) is positive for *,l ))) $� hence,

� � � � � �ll ggg )))))) ,,, ** for *,l ))) $� we proved that, a point )~,~( 44 u) at � �*2S U and

� �lU1S intersect uniquely for *4~ ))) $$l . Here again we distinguish three cases depending on the

value of r ) .

(i) If , U,~r4r I� )) the interaction results is 2122R RSS , indeed, the elementary wave

2R is stronger compared to 2S , the strength of reflected 1S is small compared to the

incident waves 2S and 2R .

(ii) If � �lUS1r4r Uand ,~ �� )) the interaction result is 122R SS .

(iii) If IV�$ r4r Uand, ~ )) the interaction result is 2122R SSS ; indeed, 2S is stronger

than compared to the elementary wave 2R is weaker.

56

REFERENCES

[1] E. Godlewski, P.A. Raviart, Numerical Approximation of Hyperbolic Systems of Conservation

Laws, Springer-Verlag, New York, 1996.

[2] T. Raja Sekhar, V.D. Sharma, Riemann Problem and Elementary wave Interaction in Isentropic

Magnetogasdynamics, Nonlinear Analysis: Real World Application, 11(2010), 619-636

[3] YehudaPinchover,Jacob Rubinstein, An Introduction to Partial Differential Equations,

Cambridge University Press, UK, 2005.

[4] J.Kevorkian, Partial Differential Equations(Analytical Solution Techniques) Second Edition,

Springer-Verlag, New York, 2000.