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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
1
011
0
01
0
0
0
0
0
0
2
2
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λIA
puρ
u γ γpρ
u
u ρ
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
0
1 0
0
0100010001
0
10
0
1
2
3
2
3
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1
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xpx
x
x
xxxu
xxx
u p ρ
u
u ρ
xIAufor
upupu
pu
pu
puu
λuγp
ρ
λuu
λ u - γ ρ
-λ u
ρ u -λ
λ_
p u ρ
u
u ρ
()
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)�
)�
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(
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(
13
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)
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))(
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�
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px
xxp
xxp
xxpp
xpxp
p
xxp
xppx
xxp
xxp
xxx
p
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
2
1
3
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14
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� �
� �
0
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
32
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21
3
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1
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xxpp
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xxp
xxx
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x
)()
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))(
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()
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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
32
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� ��
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� �
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pppp
ppso
xp
xp
px
pxp
pxp
pxp
xxp
px
x xp
xassume
x xp
()
)()
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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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��
���
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���
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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
���
���
�
���
�
�
���
�
�
���
�
���
�
���
���
���
���
���
���
���
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.
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