'
2015
9
Sept. 2015
Communication on Applied Mathematics and Computation
29 Æ 3 Vol. 29 No. 3
DOI 10.3969/j.issn.1006-6330.2015.03.007
Two-dimensional Riemann problems in angular
region for linear hyperbolic systems
with real constant coefficients
CHEN Jianjun, SHENG Wancheng
(College of Sciences, Shanghai University, Shanghai 200444, China)
Abstract The explicit solutions to the two-dimensional Riemann problems in
the angular region with angle α (0 6 α 6 π2 ) for linear hyperbolic systems with
real constant coefficients are given.
Key words two-dimensional Riemann problem; hyperbolic system; explicit solution
2010 Mathematics Subject Classification 35L65; 35L60
Chinese Library Classification O175.27
1B<E>GI?:H2.Q7L;0 Riemann D
RTU VWS
($$& 200444)
MK ("% #!+ Riemann )
α * (0 6 α 6 ).
46/ Riemann !+
2010 =J385 35L65; 35L60
PA385 O175.27 CF-O9 A CN,5 1006-6330(2015)03-0313-09
π
2
0
Introduction
We are concerned with a 2 × 2 system of linear partial differential equations with real
constant coefficients
n
∂U X ∂U
+
Ai
= 0,
∂t
∂xi
i=1
(1)
Received 2012-12-20;
Revised 2013-03-21
Supported by the National Basic Research Program of China (10971130); the Shanghai Leading Academic Discipline Project (J50101); the First-Class Discipline of Universities in Shanghai; the Shanghai
Municipal Education Commission of Scientific Research Innovation Project (11ZZ84)
Corresponding author SHENG Wancheng, research interests are nonlinear partial differential equations and its applications. E-mail: mathwcsheng@shu.edu.cn
314
Communication on Applied Mathematics and Computation
(t, x) = (t, x1 , x2 , · · · , xn ) ∈ R+ × Rn ,
Vol. 29
U = (u1 , u2 , · · · , un ) ∈ Rn ,
where Ai (1 6 i 6 n) are real constant 2 × 2 matrices, and n is a positive integer.
The Riemann problem is the most fundamental problem for hyperbolic systems. A lot
of work has been conducted on 2 × 2 hyperbolic systems[1-4] . In [1], the authors proved
that each homogeneous linear first-order hyperbolic system of two unknowns in N space
dimensions with constant coefficients can be reduced to one of three canonical systems.
In addition, when the Riemann data was given as a piecewise constant vector function,
they also gave the explicit solution to the Riemann problem. In [2], in the case that the
Riemann data is a piecewise smooth vector function of the polar angle θ, Li and Sheng gave
the explicit solution to the general multi-dimensional Riemann problem for the canonical
form of 2 × 2 hyperbolic systems with real constant coefficients. Moreover, Li and Sheng[5]
gave the explicit solution to the general Riemann problem, in which the Riemann data is a
piecewise smooth function of the polar angle θ, for the linearized system of a two-dimensional
isentropic flow in gasdynamics.
The system is called to be hyperbolic, if for any α = (α1 , α2 , · · · , αn ) ∈ S n−1 in the
n
P
time direction, the matrix
αi Ai is diagonalizable with real eigenvalues.
i=1
Theorem 1 (see [1-2, 4]) If the system is hyperbolic in the time direction, then (1)
can be reduced to the canonical form
∂U
∂U
+A
= 0,
∂t
∂x
or
∂U
∂U
∂U
+A
+B
= 0,
∂t
∂x
∂y
or
∂U
= 0,
∂t
A=
(2)
∀(t, x, y) ∈ R+ × R2 ,
∀(t, x) ∈ R+ × Rn
with
1
∀(t, x) ∈ R+ × R,
1
0
0
−1
!
,
B=
0 1
1 0
(3)
(4)
!
.
Preliminaries
For the canonical system, (2) and (4) can be easily solved. Now, we mainly focus on
seeking the solution to the two-dimensional Riemann problem for the canonical system
∂U
∂U
∂U
+A
+B
= 0,
∂t
∂x
∂y
∀(t, x, y) ∈ R+ × R2
(5)
with the following Riemann data:
U (0, x, y) =
u10 (θ)
u20 (θ)
!
H(y)H(x sin α − y cos α),
∀x, y ∈ R,
(6)
No. 3
CHEN Jianjun, et al.: Two-dimensional Riemann problems for hyperbolic systems 315
where H is the Heaviside function and u10 (θ), u20 (θ) are bounded, piecewise continuous and
smooth functions on [0, 2π], α ∈ [0, π2 ], and θ is a polar angle satisfying
(
x = r cos θ
y = r sin θ
0 6 θ 6 2π,
,
0 6 r 6 +∞.
(7)
Thus, the system (5) and the Riemann data (6) may be transformed to the following form
by (7):
and

∂u
∂u
1
∂u
∂u
1
∂u

 1 + cos θ 1 − sin θ 1 + sin θ 2 + cos θ 2 = 0,
∂t
∂r
r
∂θ
∂r
r
∂θ
∂u
∂u
1
∂u
∂u
1
∂u

2
1
1
2
2

+ sin θ
+ cos θ
− cos θ
+ sin θ
= 0,
∂t
∂r
r
∂θ
∂r
r
∂θ
t=0:
u1
u2
!
u10 (θ)H(sin θ)H(sin(α − θ))
=
u20 (θ)H(sin θ)H(sin(α − θ))
0 6 θ 6 2π,
06α6
!
,
(8)
(9)
π
.
2
Moreover, the system (5) satisfies the wave equation
∂2U
−
∂t2
∂2U
∂2U
+
∂x2
∂y 2
= 0.
(10)
Thus, the Riemann problem (5) and (6) can be transformed to the following Cauchy problems, respectively:
and
 2
∂ 2 u1 ∂ u1 ∂ 2 u1


−
+
= 0,


∂t2
∂x2
∂y 2



u1 |t=0 = u10 (θ)H(sin θ)H(sin(α − θ)),
∂u1 1



= (f (θ)H(sin(θ))H(sin(α − θ)) − u20 (0)δ(θ)


∂t
r
t=0


− u10 (α) sin αδ(α − θ) + u20 (α) cos αδ(α − θ)),
 2
∂ u2 ∂ 2 u2
∂ 2 u2 

−
+
= 0,


∂t2
∂x2
∂y 2



u |
2 t=0 = u20 (θ)H(sin θ)H(sin(α − θ)),
 ∂u2 
1


 ∂t t=0 = r (g(θ)H(sin(θ))H(sin(α − θ)) − u10 (0)δ(θ)



+ u10 (α) cos αδ(α − θ) + u20 (α) sin αδ(α − θ)),
where 0 6 r 6 +∞, 0 6 θ 6 2π, 0 6 α 6
π
2,
f (θ) = sin θu′10 (θ) − cos θu′20 (θ),
(11)
(12)
and
g(θ) = − cos θu′10 (θ) − sin θu′20 (θ).
(13)
316
Communication on Applied Mathematics and Computation
Vol. 29
2 Explicit solution to two-dimensional Riemann problem
for canonical linear system
For (11) and (12), by using the Poisson formula,
ZZ
ZZ
1 ∂
1
U0 dξdη
Ut dξdη
q
q
U (x, y, t) =
+
,
2πa ∂t P
2πa P
2
2
2
2
t2 −(ξ−x) −(η−y)
t2 −(ξ−x) −(η−y)
M
M
t
where
PM
t
t
: (ξ − x)2 + (η − y)2 6 t2 , we get the solution as follows (see [2]).
When r > t,

Z
1
1 θ+ϕ


(u
(θ
+
ϕ)
+
u
(θ
−
ϕ))
+
f (σ)dσ,
u
(t,
r,
θ)
=

10
10
1

2
2 θ−ϕ
Z

1
1 θ+ϕ


 u2 (t, r, θ) = (u20 (θ + ϕ) + u20 (θ − ϕ)) +
g(σ)dσ,
2
2 θ−ϕ
(14)
where ϕ = sin−1 rt ; while, when r 6 t,

√
Z
Z

t t2 −r2 2π
u10 (σ)
1 2π
−r cos(σ−θ)

u1 (t, r, θ) =
dσ+
f (σ)cos−1 q
dσ,

2
2
2

2π
2π 0

0 t −r sin (σ−θ)
t2 −r2 sin2 (σ−θ)
(15)
√
Z
Z 2π

2 −r2 2π

t
t
u
(σ)
1
−r
cos(σ−θ)

20
−1
u2 (t, r, θ) =
dσ+
dσ.
g(σ) cos p

2

2
2
2π
2π 0
0 t −r sin (σ−θ)
t2 −r2 sin2 (σ−θ)
Theorem 2
The solution to the multi-dimensional Riemann problem (11) and (12)
is decomposed into the six domains (Di )16i66 in R2 × R+ as follows:
(i) in D1 = {(α, t, x, y) | t > 0, x sin α − y cos α − t > 0, y > t},

1


 u1 = (u10 (θ − ϕ) + u10 (θ + ϕ)) +

2

1


 u2 = (u20 (θ − ϕ) + u20 (θ + ϕ)) +
2
1
2
1
2
Z
θ+ϕ
f (σ)dσ,
θ−ϕ
Z θ+ϕ
(16)
g(σ)dσ;
θ−ϕ
(ii) in D2 = {(α, t, x, y) | t > 0, x sin α−y cos α−t < 0, x sin α−y cos α+t > 0, cos(θ−α) > 0},

Z
1
1 α


u
=
(u
(θ
−
ϕ)
−
u
(α)
sin(α))
+
f (σ)dσ + u20 (α) cos α ,

10
 1 2 10
2 θ−ϕ
Z
(17)

1 α
1


g(σ)dσ + u10 (α) cos α ;
 u2 = (u20 (θ − ϕ) + u20 (α) sin(α)) +
2
2 θ−ϕ
(iii) in D3 = {(α, t, x, y) | t > 0, x sin α − y cos α + t < 0, cos(θ − α) > 0, y < −t},
u1 = 0,
u2 = 0;
(18)
No. 3
CHEN Jianjun, et al.: Two-dimensional Riemann problems for hyperbolic systems 317
(iv) in D4 = {(α, t, x, y) | t > 0, x sin α − y cos α − t > 0, −t < y < t},

Z

1
1 θ+ϕ


u
=
(u
(θ
+
ϕ)
−
u
(0))
+
f (σ)dσ,
20
 1 2 10
2 0
Z

1
1 θ+ϕ


 u2 = (u20 (θ + ϕ) − u10 (0)) +
g(σ)dσ;
2
2 0
(v) in D5 = {(α, t, x, y) | t > 0, x sin α − y cos α − t < 0, y < t, cos(θ − α) > 0},
Z

1 α
1


f (σ)dσ + u20 (α) cos α ,
 u1 = − 2 (u20 (0) + u10 (α) sin(α)) + 2
0
Z α

1
1

 u2 = (u20 (α) sin(α) − u10 (0)) +
g(σ)dσ + u10 (α) cos α ;
2
2 0
(19)
(20)
(vi) in D6 = {(α, t, x, y) | t > 0, x sin α − y sin α > 0, x2 + y 2 < t2 },
√

Z
Z α
t t2 − r 2 α
1
u10 (σ)
−r cos(σ − θ)



u1 =
dσ +
f (σ)cos−1 √
dσ

2
2
2
2

2π
2π
t − r2 sin2 (σ − θ)

0 t − r sin (σ − θ)
0




1 −r cos θ
−r cos(α − θ)


−
u20 (0)cos−1 p
+ u10 (α) sin αcos−1 q


2
2π

t2 − r2 sin θ

t2 − r2 sin2 (α − θ)






−r cos(α − θ)

−1

q
−
u
(α)
cos
α
cos
,
20




t2 − r2 sin2 (α − θ)

√
Z
Z α
(21)
u20 (σ)
−r cos(σ − θ)
t t2 − r 2 α
1

−1

q
u
=
dσ
+
g(σ)cos
dσ

2

2
2
2

2π
2π 0
0 t − r sin (σ − θ)

t2 − r2 sin2 (σ − θ)






−r cos θ
1 −r cos(α − θ)


−
u10 (0)cos−1 p
− u20 (α) sin αcos−1 q


2
2π

t2 − r2 sin θ

t2 − r2 sin2 (α − θ)






−r cos(α − θ)

−1

q
−
u
(α)
cos
αcos
.

10


t2 − r2 sin2 (α − θ)
Proof For (11) and (12), we make use of (14) or (15) in each regular domain of solution
(see Fig. 1), i.e.,
Fig. 1
Six domains for Riemann problem
318
Communication on Applied Mathematics and Computation
Vol. 29
(i) in D1 , we have x sin α − y cos α − t > 0, y > t, then
ϕ < θ < α − ϕ,
and thus we get (16);
(ii) in D2 , we have x sin α − y cos α − t < 0, x sin α − y cos α + t > 0, cos(θ − α) > 0, then
α − ϕ < θ < α + ϕ,
θ<ϕ
0 < θ − ϕ < α < θ + ϕ,
and thus we get (17);
(iii) in D3 , we have x sin α − y cos α + t < 0, cos(θ − α) > 0, y < −t, then
α < θ − ϕ < θ + ϕ < 2π,
and thus we get (18);
(iv) in D4 , we have x sin α − y cos α − t > 0, −t < y < t, then
θ − ϕ < 0 < θ + ϕ < α,
and thus we get (19);
(v) in D5 , we have x sin α − y cos α − t < 0, y < t, cos(θ − α) > 0, then
θ − ϕ < 0 < α < θ + ϕ,
and thus we get (20);
(vi) in D6 , we have r < t, 0 < θ < α, we use (15), then
√
Z
t t2 − r 2 α
u10 (σ)
u1 =
dσ
2
2
2
2π
0 t − r sin (σ − θ)
Z α
−r cos(α − σ)
1
+
f (σ)H(sin σ)H(sin(α − σ))cos−1 q
dσ
2π 0
t2 − r2 sin2 (α − σ)
Z α
1
−
(u20 (0)δ(σ) + u10 (α) sin αδ(α − σ)
2π 0
−r cos(σ − θ)
− u20 (α) cos αδ(α − σ)) cos−1 q
dσ,
t2 − r2 sin2 (α − σ)
√
Z
t t2 − r2 α u20 (σ) 2
u2 =
sin (σ − θ)dσ
2π
t2 − r 2
Z α 0
1
−r cos(α − σ)
+
g(σ)H(sin σ)H(sin(α − σ))cos−1 q
dσ
2π 0
t2 − r2 sin2 (α − σ)
Z α
1
+
(u20 (α) sin αδ(α − σ) − u10 (0)δ(σ)
2π 0
−r cos(α − σ)
+ u10 (α) cos αδ(α − σ)) cos−1 q
dσ.
t2 − r2 sin2 (α − σ)
No. 3
CHEN Jianjun, et al.: Two-dimensional Riemann problems for hyperbolic systems 319
Then, we can easily get (21).
Thus, we obtain the solution to the Riemann problem (11) and (12).
From Theorem 2, we can easily obtain the results in [2] as follows.
Corollary 1
If α =
π
2
in Theorem 2, then
(i) in D1 = {(t, x, y) | t > 0, x > t, y > t},

1



 u1 = 2 (u10 (θ − ϕ) + u10 (θ + ϕ)) +

1


 u2 = (u20 (θ − ϕ) + u20 (θ + ϕ)) +
2
(ii) in D2 = {(t, x, y) | t > 0, −t < x 6 0, y >
1
2
1
2
Z
θ+ϕ
f (σ)dσ,
θ−ϕ
Z θ+ϕ
(22)
g(σ)dσ;
θ−ϕ
√
t2 − x2 or 0 < x < t, y > t},

π 1


u10 (θ − ϕ) − u10
+
 u1 =

2
2
π 
1


u20 (θ − ϕ) + u20
+
 u2 =
2
2
π
2
1
2
Z
1
2
Z
f (σ)dσ,
θ−ϕ
π
2
(23)
g(σ)dσ;
θ−ϕ
√
(iii) in D3 = {(t, x, y) | t > 0, x < −t or x > 0, y < −t or − t 6 x 6 0, y < − t2 − x2 },
u1 = 0,
(iv) in D4 = {(t, x, y) | t > 0, x >
u2 = 0;
(24)
p
t2 − y 2 , −t < y 6 0 or x > t, 0 6 y < t},


1


 u1 = 2 (u10 (θ + ϕ) − u20 (0)) +

1


 u2 = (u20 (θ + ϕ) − u10 (0)) +
2
1
2
θ+ϕ
Z
f (σ)dσ,
0
1
2
Z
(25)
θ+ϕ
g(σ)dσ;
0
√
(v) in D5 = {(t, x, y) | t > 0, 0 < x < t, t2 − x2 < y < t},

π 
1


+
 u1 = − 2 u20 (0) + u10 2

1 π 

− u10 (0) +
 u2 = − u20
2
2
1
2
1
2
π
2
Z
f (σ)dσ,
0
Z
0
π
2
(26)
g(σ)dσ;
320
Communication on Applied Mathematics and Computation
(vi) in D6 = {(t, x, y) | t > 0, x2 + y 2 < t2 },

√
Z

t t2 − r 2 α
u10 (σ)


u
=
dσ
1

2
2
2

2π

0 t − r sin (σ − θ)


Z α


1
−r cos(σ − θ)


+
f (σ)cos−1 q
dσ



2π
2
2 sin2 (σ − θ)
0

t
−
r




π


1 −r cos θ
−r cos(α − θ)

−1
−1

p
q
−
u
(0)cos
+
u
cos
,

20
10

2π
2

t2 − r2 sin2 θ

t2 − r2 sin2 (α − θ)
√
Z

u20 (σ)
t t2 − r 2 α


u
=
dσ

2

2 − r2 sin2 (σ − θ)

2π
t

0

Z α



1
−r cos(σ − θ)


g(σ)cos−1 q
+
dσ


2π

2 − r2 sin2 (σ − θ)
0

t






1 π −1
−r cos θ −r cos(α − θ)


.
+
u20
cos q
− u10 (0)cos−1 p


2π
2

t2 − r2 sin2 θ
t2 − r2 sin2 (α − θ)
Vol. 29
(27)
Also from Theorem 2, we can easily obtain the results in [1] as follows.
Corollary 2
then
If u10 (θ) ≡ u10 , u20 (θ) ≡ u20 in Theorem 2, where u10 , u20 are constants,
(i) in D1 = {(α, t, x, y) | t > 0, x sin α − y cos α − t > 0, y > t},
u1 = u10 ,
u2 = u20 ;
(28)
(ii) in D2 = {(α, t, x, y) | t > 0, x sin α− y cos α− t < 0, x sin α− y cos α+ t > 0, cos(θ − α)
> 0},
1 − sin θ
cos θ
cos θ
1 − sin θ
u10 +
u20 , u2 =
u10 +
u20 ;
2
2
2
2
(iii) in D3 = {(α, t, x, y) | t > 0, x sin α − y cos α + t < 0, cos(θ − α) > 0, y < −t},
u1 =
u1 = 0,
u2 = 0;
(29)
(30)
(iv) in D4 = {(α, t, x, y) | t > 0, x sin α − y cos α − t > 0, −t < y < t},
u10 − u20
u20 − u10
, u2 =
;
2
2
(v) in D5 = {(α, t, x, y) | t > 0, x sin α − y cos α − t < 0, y < t, cos(θ − α) > 0},
u1 =
u1 =
cos θ − 1
sin θ
u20 −
u10 ,
2
2
u2 =
cos θ − 1
sin θ
u10 +
u20 ;
2
2
(31)
(32)
(vi) in D6 = {(α, t, x, y) | t > 0, x sin α − y sin α > 0, x2 + y 2 < t2 },
u1 =
−u10 sin α + u20 cos α
−x cos α − y sin α
arccos q
2π
t2 − (x sin α − y cos α)2
−
u20
−x
u10
t2 cos α + y(x sin α − y cos α)
q
arccos p
+
arccos p
,
2π
2π
2
t2 − y 2
t2 − y 2 t2 − (x sin α − y cos α)
(33)
CHEN Jianjun, et al.: Two-dimensional Riemann problems for hyperbolic systems 321
No. 3
u10 cos α + u20 sin α
x cos α + y sin α
arccos q
2π
2
t2 − (x sin α − y cos α)
u2 =
−
u10
u20
−x
t2 cos α + y(x sin α − y cos α)
q
arccos p
+
arccos p
.
2π
2π
2
t2 − y 2
t2 − y 2 t2 − (x sin α − y cos α)
(34)
Proof We only carry out the calculations for D6 because the others can be similarly
obtained.
Because of
Z
α
0
we get
t√t2 − r2
σ=α
u10 (σ)
tan(σ − θ) ,
dσ = u10 arctan
2π
t2 − r2 sin2 (σ − θ)
σ=0
t√t2 − r2
t√ t2 − r 2
u10 u1 =
arctan
tan(α − θ) + arctan
tan θ
2π
2π
2π
−r cos θ
−r cos(α − θ)
1 −1
−1
+ u10 sin αcos q
u20 cos p
−
2π
t2 − r2 sin2 θ
t2 − r2 sin2 (α − θ)
−r cos(α − θ)
− u20 cos αcos−1 q
.
t2 − r2 sin2 (α − θ)
Let
√
t t2 − r 2
tan θ1 =
tan(α − θ),
2π
√
t t2 − r 2
tan θ2 =
tan θ.
2π
Then,
cos(θ1 + θ2 ) =
s
1+
t2 tan α2 (t2 − r2 )
t2 cos α + y(x sin α − y cos α)
q
,
=p
2
2
(t + xy tan α − y )
2
2
2
2
t − y t − (x sin α − y cos α)
and thus, we obtain (33). Equation (34) can be obtained in the same way.
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