Electronic Journal of Differential Equations, Vol. 2013 (2013), No. 138, pp. 1–5.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
EXISTENCE OF SOLUTIONS TO NON-LOCAL PROBLEMS FOR
PARABOLIC-HYPERBOLIC EQUATIONS WITH THREE LINES
OF TYPE CHANGING
ERKINJON T. KARIMOV, AKMAL I. SOTVOLDIYEV
Abstract. In this work, we study a boundary problem with non-local conditions, by relating values of the unknown function with various characteristics.
The parabolic-hyperbolic equation with three lines of type changing is equivalently reduced to a system of Volterra integral equations of the second kind.
1. Introduction
Consider an equation
uxx − uy = 0,
uxx − uyy ,
(x, y) ∈ Ω0 ,
(1.1)
(x, y) ∈ Ωi i = 1, 2, 3
in the domain Ω = Ω0 ∪ Ω1 ∪ Ω2 ∪ Ω3 ∪ AB ∪ AA0 ∪ BB0 ; see Figure 1.
↑
A0
1
B0
Z
Z
DZ
Ω2
Z
Z
Ω0
Z
Z
Z
Z
E
Z
-1/2
Ω3
B Z A
Z
1
0 Z
Ω1 Z
Z
-1/2
Z
C
3/2
→
x
Figure 1. Domain Ω
2000 Mathematics Subject Classification. 35M10.
Key words and phrases. Parabolic-hyperbolic equation; non-local condition;
Volterra integral equation.
c
2013
Texas State University - San Marcos.
Submitted May 14, 2013. Published June 20, 2013.
1
2
E. T. KARIMOV, A. I. SOTVOLDIEV
EJDE-2013/138
Problem AS. Find a regular solution of equation (1.1) in the domain Ω, satisfying
the following conditions:
a1 (t)u(−t, t) + a2 (t)u(t, −t) = a3 (t),
06t6
1
,
2
1
6 t 6 1,
2
1
c1 (t)(ux + uy )(t − 1, t) + c2 (t)(ux − uy )(2 − t, t) = c3 (t),
< t < 1.
2
b1 (t)u(t, t − 1) + b2 (t)u(2 − t, 1 − t) = b3 (t),
(1.2)
(1.3)
(1.4)
Here ai (t), bi (t), ci (t) (i = 1, 2, 3) are given functions, such that
a1 (0) + a2 (0) 6= 0,
b21 (t) + b22 (t) > 0,
b1 (1) + b2 (1) 6= 0,
c21 (t) + c22 (t) > 0,
a21 (t) + a22 (t) > 0,
a21 + b22 > 0,
a22 + b21 > 0.
Note that problem AS is a generalization of the following problems:
Case A a1 ≡ 0.
(1) a2 , b1 , b2 , c1 , c2 6= 0,
(2) b1 ≡ 0, a2 , b2 , c1 , c2 6= 0,
(3) c2 ≡ 0, a2 , b1 , b2 , c1 6= 0,
(4) b1 ≡ 0, c2 ≡ 0, a2 , b2 , c1 6= 0;
Case B a2 ≡ 0.
(1) a1 , b1 , b2 , c1 , c2 6= 0,
(2) b2 ≡ 0, a1 , b1 , c1 , c2 6= 0,
(3) c1 ≡ 0, a1 , b1 , b2 , c2 6= 0,
(4) b2 ≡ 0, c1 ≡ 0, a1 , b1 , c2 6= 0;
Case C b1 ≡ 0.
(1) a1 , a2 , b2 , c1 , c2 6= 0,
(2) c2 ≡ 0, a1 , a2 , b2 , c1 6= 0;
Case D b2 ≡ 0.
(1) a1 , a2 , b1 , c1 , c2 6= 0,
(2) c1 ≡ 0, a1 , a2 , b1 , c2 6= 0;
Case E c1 ≡ 0. a1 , a2 , b1 , b2 , c2 6= 0;
Case F c2 ≡ 0. a1 , a2 , b1 , b2 , c1 6= 0.
Also note that cases A4 and B4 were studied in [9]. Other cases were not
investigated, and the main result of this paper is true for these particular cases.
Boundary problems for parabolic-hyperbolic equations with two lines of type
changing were investigated in [1, 6, 7, 8], and with three lines of type changing in
[2, 3]. The main point in this present work is the non-local condition, which relates
values of the unknown function with various characteristics. It makes very difficult
the reduction of the considered problem to a system of integral equations, we need
a special algorithm for solving this problem.
2. Main results
In the domain Ω1 solution of the Cauchy problem with initial data u(x, 0) =
τ1 (x), uy (x, 0) = ν1 (x) can be represented, as in [4], by
Z x+y
2u(x, y) = τ1 (x + y) + τ1 (x − y) +
ν1 (z)dz.
(2.1)
x−y
EJDE-2013/138
EXISTENCE OF SOLUTIONS
3
Assuming that in condition (1.2),
u(−t, t) = ϕ1 (t),
06t6
1
,
2
(2.2)
from (2.1), we find that
τ10 (t) = ν1 (t) +
2[a ( t ) − a ( t )ϕ ( t )] 0
1 2
1 2
3 2
,
a2 ( 2t )
0 < t < 1.
(2.3)
In condition (1.3) introduce
u(2 − t, 1 − t) = ϕ2 (t),
1
6 t 6 1,
2
(2.4)
and from (2.1), we obtain
τ10 (t) = −ν1 (t) +
2[b ( t+1 ) − b ( t+1 )ϕ ( t+ )] 0
2 2
2 2
3 2
,
b1 ( t+1
)
2
0 < t < 1.
From (2.3) and (2.5), it follows that
a ( t ) − a ( t )ϕ ( t ) 0 b ( t+1 ) − b ( t+1 )ϕ ( t+1 ) 0
1 2
1 2
2 2
2 2
3 2
3 2
τ10 (t) =
+
,
t
t+1
a2 ( 2 )
b1 ( 2 )
(2.5)
0 < t < 1. (2.6)
The solution of the Cauchy problem in the domain Ω2 , with given data u(0, y) =
τ2 (y), ux (0, y) = ν2 (y), is written as follows [4],
Z y+x
2u(x, y) = τ2 (y + x) + τ2 (y − x) +
ν2 (z)dz.
(2.7)
y−x
Considering (2.2) from (2.7) we obtain
t
τ20 (t) = ν2 (t) + ϕ01 ( ), 0 < t < 1.
2
In condition (1.4) introduce another designation
(ux − uy )(2 − t, t) = ϕ3 (t),
(2.8)
1
< t < 1.
2
(2.9)
Then from (2.7) we obtain
t+1
t+1
c3 ( t+1
2 ) − c2 ( 2 )ϕ3 ( 2 )
= τ20 (t) + ν2 (t), 0 < t < 1.
t+1
c1 ( 2 )
(2.10)
From (2.8) and (2.10) we deduce
t+1
t+1
c3 ( t+1
t
2 ) − c2 ( 2 )ϕ3 ( 2 )
2τ20 (t) = ϕ01 ( ) +
,
2
c1 ( t+1
2 )
0 < t < 1.
(2.11)
The solution of the Cauchy problem with data u(1, y) = τ3 (y), ux (1, y) = ν3 (y)
in the domain Ω3 has a form [4]
Z y+x−1
2u(x, y) = τ3 (y + x − 1) + τ2 (y − x + 1) +
ν3 (z)dz.
(2.12)
y−x+1
Using (2.4) and (2.9) from (2.12), after some evaluations one can get
2τ30 (t) = −ϕ02 (
2−t
t+1
) − ϕ3 (
),
2
2
0 < t < 1.
(2.13)
4
E. T. KARIMOV, A. I. SOTVOLDIEV
EJDE-2013/138
Further, from the equation (1.1) we pass to the limit at y → +0 and considering
(2.3) we find
2 a ( t ) − a ( t )ϕ ( t ) 0
3 2
1 2
1 2
00
0
τ1 (t) − τ1 (t) = −
.
(2.14)
t
a2 ( 2 )
The solution of (2.14) with the conditions
τ1 (0) =
a3 (0)
,
a1 (0) + a2 (0)
τ1 (1) =
b3 (1)
,
b1 (1) + b2 (1)
which is deduced from (1.2) and (1.3), can be represented as [5]
i
h
a3 (0)
b3 (1)
a3 (0)
τ1 (x) =
+x
−
a1 (0) + a2 (0)
b1 (1) + b2 (1) a1 (0) + a2 (0)
Z 1
h
i
b3 (1)
a3 (0)
+
G(x, t)
−
dt
b1 (1) + b2 (1) a1 (0) + a2 (0)
0
Z 1
2[a ( t ) − a ( t )ϕ ( t )] 0
1 2
1 2
3 2
−
dt, 0 6 x 6 1,
G(x, t)
a2 ( 2t )
0
(2.15)
(2.16)
where G(x, t) is Green’s function of problem (2.14)-(2.15).
Continuing to assume that the function ϕ1 is known, using the formula (2.6)
we represent function ϕ2 via ϕ1 . Then using the solution of the first boundary
problem for equation (1.1) in the domain Ω0 (see [5]) and functional relations
between functions τj and νj (j = 2, 3), we obtain
Z y
Z y
τ 0 2 (y) =
τ 0 3 (η)N (0, y, 1, η)dη −
τ 0 2 (η)N (0, y, 0, η)dη + F1 (y),
(2.17)
Z0 y
Z0 y
0
0
τ 3 (y) =
τ 3 (η)N (1, y, 1, η)dη −
τ 0 2 (η)N (1, y, 0, η)dη + F2 (y),
0
0
where
Z
1
a3 (0)
N (0, y, 0, 0)
a
(0)
+ a2 (0)
1
0
y
b3 (1)
N (0, y, 1, 0) + ϕ0 1 ( ),
+
b1 (1) + b2 (1)
2
Z 1
a3 (0)
F2 (y) =
τ1 (ξ)Gx (1, y, ξ, 0)dξ −
N (1, y, 0, 0)
a1 (0) + a2 (0)
0
b3 (1)
y+1
+
N (1, y, 1, 0) − ϕ3 (
),
b1 (1) + b2 (1)
2
F1 (y) =
and
τ1 (ξ)Gx (o, y, ξ, 0)dξ −
∞ h
i
X
(x−ξ+2n)2
(x+ξ+2n)2
1
G(x, y, ξ, η) = p
e− 4(y−η) − e− 4(y−η)
2 π(y − η) n=−∞
is the Green’s function of the first boundary problem; see [5],
∞ h
i
X
(x−ξ+2n)2
(x+ξ+2n)2
1
N (x, y, ξ, η) = p
e− 4(y−η) + e− 4(y−η) .
2 π(y − η) n=−∞
From the first equation in (2.17), we represent function ϕ3 via ϕ1 and further,
from the second equation of (2.17), we find the function ϕ1 .
After the finding function ϕ1 , using appropriate formulas, we find functions ϕ2 ,
ϕ3 , τi , νi , (i = 1, 2, 3). Solution of the problem AS can be established in the domain
EJDE-2013/138
EXISTENCE OF SOLUTIONS
5
Ω0 as a solution of the first boundary problem, and in the domains Ωi (i = 1, 2, 3)
as a solution of the Cauchy problem.
Theorem 2.1. If the functions ai , bi , ci are continuously differentiable on a segment, and have continuous second-order derivatives on an interval, then problem
AS has a unique regular solution.
References
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Congress. Ankara, Turkey, 13-18 August, 2007, pp. 727-734.
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[8] Nakusheva, V. A.; First boundary problem for mixed type equation in a characteristic polygon,
Dokl. AMAN, 2012, Vol. 14, No 1, pp. 58-65.
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three lines of type changing, PhD thesis, Tashkent, 2011.
Erkinjon T. Karimov
Institute of Mathematics, National University of Uzbekistan “Mirzo Ulugbek”, Durmon
yuli str., 29, Tashkent, 100125, Uzbekistan
E-mail address: erkinjon@gmail.com, erkinjon.karimov@hotmail.com
Akmal I. Sotvoldiyev
Institute of Mathematics, National University of Uzbekistan “Mirzo Ulugbek”, Durmon
yuli str., 29, Tashkent, 100125, Uzbekistan
E-mail address: akmal.sotvoldiyev@mail.ru