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Asian-European Journal of Mathematics
Vol. 7, No. 2 (2014) 1450030 (
9
pages)
c
World Scientific Publishing Company
DOI:
10.1142/S1793557114500302
On a nonlocal problem for mixed parabolic–hyperbolic type
equation with nonsmooth line of type changing
E. T. Karimov
Institute of Mathematics
National University of Uzbekistan named after Ulugbek
Tashkent, Uzbekistan
erkinjon@gmail.com
N. A. Rakhmatullaeva
Tashkent State Technical University named after A. R. Beruni
Tashkent, Uzbekistan
rakhmatullaeva@mail.ru
Communicated by M. O. Perestyuk
Received July 18, 2013
Accepted March 12, 2014
Published June 16, 2014
In this paper, we investigate a boundary problem with nonlocal conditions for mixed
parabolic–hyperbolic type equation with three lines of type changing. Considered domain
contains a rectangle as a parabolic part and three domains bounded by smooth curves
and type-changing lines as a hyperbolic part of the mixed domain. Applying method of
energy integrals we prove the uniqueness of the solution for the considered problem. The
proof of the existence will be done by reducing the original problem into the system of
the second kind Volterra integral equations.
Keywords: Parabolic–hyperbolic equation; Volterra integral equation; nonlocal boundary
problem; Green’s function.
AMS Subject Classification: 35M10
1. Introduction
Theory of mixed type equations is one of the main parts of the general theory of
partial differential equations. First, fundamental works on this theory were done
by Tricomi [
24
], Gellerstedt [
8
], Frankl [
7
], Morawetz [
13
], Protter [
15
], Lavrent’ev
and Bitsadze [
12
], and etc.
Due to many applications in gas and aerodynamics, mechanics, this direction
was very rapidly developed [
17
]. Nowadays, this theory has many branches due to
the usage of various methods of mathematical and functional analysis, topological
methods and the method of Fractional Calculus [
9
,
16
,
25
].
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Omitting huge amount of works, related to the study of mixed type equations, we
only note some recent works on local and nonlocal boundary problems for parabolic–
hyperbolic equations [
1
–
3
,
10
,
22
].
Regarding the investigations of mixed parabolic–hyperbolic equations with
nonsmooth lines of type changing we note works [
4
,
6
,
11
,
14
,
23
].
In this paper, we study nonlocal problem for parabolic–hyperbolic equation with
three lines of type changing in a special mixed domain, hyperbolic parts of which
bounded by smooth curves and type-changing lines. Under the certain assumptions
on these curves, we find conditions to parameters, participated in nonlocal condi-
tions and we require definite regularity from the given function f (x, y) in order to
prove the unique solvability of the stated problem.
2. Formulation of a Problem
Consider an equation
Lu = f ( x, y)
(1)
in a domain Ω = Ω
0
∪ Ω
i
∪ AB ∪ BC ∪ AD, where
Lu =
u
xx
− u
y
,
(x, y)
∈ Ω
0
,
u
xx
− u
yy
,
(x, y)
∈ Ω
i
(i = 1, 3).
Smooth curves γ
1
: y =
−γ
1
(x), γ
1
(0) = γ
1
(1) = 0, γ
2
: x =
−γ
2
(y), γ
2
(0) =
γ
2
(1) = 0, γ
3
: x = γ
3
(y), γ
3
(0) = γ
3
(1) = 1 are located inside of the appropriate
characteristic triangles. Moreover, regarding the curves γ
i
(t) (i = 1, 3) we suppose
that they are twice continuously differentiable and t
± γ
i
(t) (0
≤ t ≤ 1 , i = 1 , 3) are
monotonically increase.
We formulate the following problem for Eq. (1).
Fig. 1.
Domain Ω.
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On a nonlocal problem for mixed parabolic–hyperbolic type equation
Problem. To find a regular solution of Eq. (1), satisfying conditions
[u
x
− u
y
](θ
1
(t)) = σ
1
[u
x
+ u
y
](θ
∗
1
(t)),
0
≤ t ≤ 1 ,
(2)
[ u
x
− u
y
](θ
2
(t)) = σ
2
[u
x
+ u
y
](θ
∗
2
(t)),
0
≤ t ≤ 1 ,
(3)
[ u
x
+ u
y
](θ
3
(t)) = σ
3
[u
x
− u
y
](θ
∗
3
(t)),
0
≤ t ≤ 1 ,
(4)
u( A) = u( B) = 0 .
(5)
Here σ
1
, σ
2
, σ
3
are arbitrary real numbers, θ
1
(t), θ
2
(t), θ
3
(t)[θ
∗
1
(t), θ
∗
2
(t), θ
∗
3
(t)] are
affixes of points of intersection of the curves γ
i
(t) (i = 1, 3) and characteristics
x
− y = t, y − x = t, x + y = 1 + t [ x + y = t, x − y = t, y − x = 1 + t]
of Eq. (1), respectively.
We mean as a regular solution of the problem in the domain Ω function u(x, y)
∈
W =
{u : u(x, y) ∈ C(Ω) ∩ C
1
(Ω)
∩ C
2,1
x,y
(Ω
0
)
∩ C
2
(Ω
i
), i = 1, 3
}, satisfying Eq. (1)
in domains Ω
j
(j = 0, 3).
3. Main Result
Theorem. If conditions σ
2
, σ
3
∈ [−1, 1], f(x, y) ∈ C
1
(Ω) are fulfilled, then the
problem has unique regular solution.
Proof. First, we deduce main functional relations.
Solution of the problem in Ω
i
(i = 1, 3) can be represented by the D’Alembert’s
formula [
18
]:
u( ξ, η) =
1
2
τ
−
1
(ξ) + τ
−
1
(η)
−
η
ξ
ν
−
1
(t)dt
−
η
ξ
dξ
1
η
ξ
1
f
1
(ξ
1
, η
1
) dη
1
,
(6)
u( ξ, η) =
1
2
τ
−
2
(ξ) + τ
−
2
(
−η) −
−η
ξ
ν
−
2
(t)dt
−
−η
ξ
dξ
1
−η
ξ
1
f
1
(ξ
1
, η
1
) dη
1
,
(7)
u( ξ, η) =
1
2
τ
+
3
( ξ
− 1) + τ
+
3
(1
− η) −
1−η
ξ−1
ν
+
3
( t) dt
−
1−η
ξ−1
dξ
1
1 −η
ξ
1
f
1
(ξ
1
, η
1
) dη
1
,
(8)
where ξ = x + y, η = x
− y, 4 f
1
(ξ, η) = f (
ξ+ η
2
,
ξ−η
2
), τ
±
1
(x) = u(x,
±0), τ
±
2
(y) =
u(
±0 , y), τ
±
3
(1
± 0 , y), ν
±
1
(y) = u
y
(x,
±0), ν
±
2
(y) = u
x
(
±0, y), ν
±
3
(y) = u
x
(1
± 0, y).
By virtue of conditions to γ
i
(i = 1, 3), equation of the curve represented as
ξ = ρ( η) and η = υ( ξ) such that ρ( υ( ξ)) = ξ.
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Since
θ
1
x
− γ
1
(x) + t
2
;
x
− γ
1
(x)
− t
2
;
θ
∗
1
x + γ
1
(x) + t
2
;
−
x + γ
1
(x) + t
2
;
θ
2
y
− γ
2
(y)
− t
2
;
y
− γ
2
(y) + t
2
;
θ
∗
2
−
y + γ
2
(y)
− t
2
;
y + γ
2
(y) + t
2
;
θ
3
−
y + γ
3
(y)
− 1 − t
2
;
y + γ
3
(y) + 1 + t
2
;
θ
∗
3
y
− γ
3
(y) + 1
− t
2
;
y
− γ
3
(y)
− 1 + t
2
,
from (2) and (6), (3) and (7), (4) and (8), we deduce main functional relations on
the line of type changing of Eq. (1), respectively:
(1
− σ
1
)τ
−
1
(x)
− (1 + σ
1
)ν
−
1
(x) = A
1
(x),
0 < x < 1,
(9)
(1 + σ
2
)τ
−
2
(y) + (
−1 + σ
2
)ν
−
2
(y) = A
2
(y),
0 < y < 1,
(10)
(1 + σ
3
)τ
+
3
(y) + (
−1 + σ
3
)ν
+
3
(y) = A
3
(y),
0 < y < 1,
(11)
where
A
1
(x) = 2σ
1
υ(x)
x
f
1
(x; η
1
)dη
1
+ 2
x
ρ( x)
f
1
(ξ
1
; x)dξ
1
,
A
2
(y) =
−2 σ
2
υ( y)
y
f
1
(x; η
1
)dη
1
+ 2
y
ρ( y)
f
1
(ξ
1
; x)dξ
1
,
A
3
(y) = 2σ
3
υ(y)
y
f
1
(x; η
1
)dη
1
− 2
y
ρ( y)
f
1
(ξ
1
; x)dξ
1
.
3.1.
The uniqueness of the solution
In case, when
|σ
i
| = 1 by formulas (9)–(11) we can find τ
±
i
(t) or ν
±
i
(t) (i = 1, 3),
therefore the problem will be divided into four problems, which can be solved
directly.
Consider the case, when
|σ
i
| = 1.
In order to prove the uniqueness we use well-known method called as “abc-
method” [
18
]. We note that using this method Rassias and Karimov proved the
uniqueness theorem for parabolic equations with two [
20
] and three [
21
] lines of
degeneration. In the work by Rassias [
19
] bi-parabolic elliptic bi-hyperbolic Tricomi
problem has been studied by similar method.
We multiply equation u
xx
− u
y
= 0 by the function u(x, y) and rewrite it as
(u
· u
x
)
x
− u
2
x
−
1
2
( u
2
)
y
= 0.
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On a nonlocal problem for mixed parabolic–hyperbolic type equation
Integrating this equality along the domain Ω
ε
0
=
{( x, y) : ε < x < 1 − ε, ε < y <
1
− ε, ε > 0} we obtain
Ω
ε
0
(u
· u
x
)
x
− u
2
x
−
1
2
( u
2
)
y
dxdy = 0.
Using Green’s formula [
18
] we deduce
A
ε
B
ε
1
2
u
2
(x, +0)dx +
B
ε
C
ε
u(1
− 0 , y) u
x
(1
− 0, y)dy +
C
ε
D
ε
1
2
u
2
(x, 1
− 0) dx
+
D
ε
A
ε
u(+0 , y) u
x
(+0, y)dy
−
Ω
ε
0
u
2
x
dxdy = 0.
Finally, we pass to the limit as ε
→ 0 and considering introduced notations, in the
domain Ω
0
we have the following equality
Ω
0
u
2
x
(x, y)dxdy +
1
0
τ
+
2
( y) ν
+
2
( y) dy
−
1
0
τ
−
3
(y)ν
−
3
(y)dy
+
1
2
1
0
u
2
(x, 1)dx
−
1
2
1
0
[ τ
+
1
( x)]
2
dx = 0.
(12)
In order to prove the uniqueness, first we prove that u(x,
±0) = τ
1
±
(x) = 0.
Equation u
xx
− u
y
= 0 can be reduced to τ
+
1
( x) = ν
+
1
( x) and substituting it into
the integral I
1
=
1
0
τ
+
1
(x)ν
+
1
(x)dx, taking condition (5) into account we have
I
1
=
1
0
τ
+
1
( x) τ
+
1
(x)dx =
−
1
0
((τ
+
1
(x))
)
2
dx.
(13)
Now considering (9) we obtain
I
1
=
1
− σ
1
1 + σ
1
1
0
τ
+
1
( x) τ
+
1
(x)dx =
1
− σ
1
2(1 + σ
1
)
(τ
+
1
(x))
2
1
0
= 0.
(14)
From (13) and (14) it follows that τ
±
1
(x)
≡ 0.
Now we prove that
I
2
=
1
0
τ
+
2
( y) ν
+
2
( y) dy
≥ 0 , I
3
=
1
0
τ
+
3
( y) ν
+
3
( y) dy
≤ 0 .
Taking (10) and (11) into account we get respectively the following:
I
2
=
1
0
τ
+
2
( y) ν
+
2
( y) dy =
1 + σ
2
1
− σ
2
1
0
τ
+
2
( y) τ
+
2
(y)dy =
1 + σ
2
2(1
− σ
2
)
τ
+
2
2
(1),
I
3
=
1
0
τ
+
3
( y) ν
+
3
( y) dy =
1 + σ
3
1
− σ
3
1
0
τ
−
3
(y)τ
−
3
(y)dy =
−
1 + σ
3
2(1
− σ
3
)
τ
−
3
2
(1) .
If
1+σ
2
1−σ
2
> 0,
1+σ
3
1−σ
3
> 0, then I
2
≥ 0, I
3
≤ 0.
Considering that I
2
≥ 0, I
3
≤ 0 and τ
+
1
= 0 from the equality (12) we state that
u( x, y) = 0 in Ω
0
. Since u(x, y)
∈ C(Ω), we can conclude, u( x, y) = 0 in Ω.
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3.2.
The existence of the solution
Now we prove the existence of the solution for the problem.
We pass to the limit in Ω
0
at y
→ +0 and from the equation u
xx
− u
y
= f (x, y),
considering (9), we get
τ
+
1
(x)
−
1
− σ
1
1 + σ
1
τ
+
1
(x) = f
∗
(x),
(15)
where f
∗
(x) = f (x, 0)
−
1
1+σ
1
A
1
(x). From the condition (5) we have
τ
+
1
(0) = 0 ,
τ
+
1
(1) = 0 .
(16)
Solution of Eq. (15) together with conditions (16) can be represented as [
5
]
τ
+
1
( x) =
1 + σ
1
1
− σ
1
×
x
0
(e
1 −σ1
1+σ1
(x−t)
− 1)f
∗
(t)dt
−
e
1−σ1
1+σ1
x
− 1
e
1−σ1
1+σ1
− 1
1
0
(e
1−σ1
1+σ1
(1−t)
− 1)f
∗
(t)dt
.
Solution of the first boundary problem for Eq. (1) in the domain Ω
0
has the
form [
5
]
u( x, y) =
1
0
τ
+
1
( x
1
)G(x, y; x
1
, 0) dx
1
+
y
0
τ
+
2
(y
1
)G
x
1
(x, y; 0, y
1
)dy
1
−
y
0
τ
−
3
(y
1
)G
x
1
(x, y; 1, y
1
)dy
1
−
1
0
dx
1
y
0
f (x
1
, y
1
)G(x, y; x
1
, y
1
)dy
1
,
(17)
where
G( x, y; x
1
, y
1
) =
1
2
π(y
− y
1
)
∞
n= −∞
[e
−
(x−x1+2n)2
4(y−y1 )
− e
−
(x+x1+2n)2
4(y−y1 )
]
is the Green’s function of the first boundary problem for the heat equation.
Differentiating (17) once by x we have
u
x
(x, y) =
1
0
τ
+
1
( x
1
)G
x
(x, y; x
1
, 0)dx
1
+
y
0
τ
+
2
(y
1
)G
x
1
x
(x, y; 0, y
1
)dy
1
−
y
0
τ
−
3
(y
1
)G
x
1
x
(x, y; 1, y
1
)dy
1
−
1
0
dx
1
y
0
f (x
1
, y
1
)G
x
(x, y; x
1
, y
1
)dy
1
.
Considering
G
x
1
x
(x, y; x
1
, y
1
) = N
y
1
(x, y; x
1
, y
1
),
G
x
(x, y; x
1
, y
1
) =
−N
x
1
(x, y; x
1
, y
1
)
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On a nonlocal problem for mixed parabolic–hyperbolic type equation
we obtain
u
x
(x, y) =
−
1
0
τ
+
1
( x
1
)N
x
1
(x, y; x
1
, 0) dx
1
+
y
0
τ
+
2
(y
1
)N
y
1
(x, y; 0, y
1
)dy
1
−
y
0
τ
−
3
(y
1
)N
y
1
(x, y; 1, y
1
)dy
1
+
1
0
dx
1
y
0
f ( x
1
, y
1
)N
x
1
(x, y; x
1
, y
1
)dy
1
,
(18)
where
N ( x, y; x
1
, y
1
) =
1
2
π(y
− y
1
)
∞
n= −∞
[e
−
(x−x1+2n)2
4(y−y1)
+ e
−
(x+x1+2n)2
4(y−y1 )
].
(19)
In (18), we use formula of integration by parts and taking (16), (19) into account,
one can find that
u
x
(x, y) =
−
1
0
(τ
+
1
(x
1
))
N ( x, y; x
1
, 0)dx
1
−
y
0
(τ
+
2
(y
1
))
N ( x, y; 0 , y
1
)dy
1
+
y
0
(τ
−
3
(y
1
))
N ( x, y; 1 , y
1
)dy
1
+
y
0
f (1, y
1
)N (x, y; 1, y
1
)dy
1
−
y
0
f (0, y
1
)N (x, y; 0, y
1
)dy
1
−
1
0
dx
1
y
0
df (x
1
, y
1
)
dx
1
N (x, y; x
1
, y
1
)dy
1
.
(20)
Passing to the limit as x
→ +0 and x → 1 −0, according to the notation u
x
(+0, y) =
ν
+
2
( y), u
x
(1
− 0, y) = ν
−
3
(y), and considering (10), (11) we obtain the following:
1 + σ
2
−1 + σ
2
τ
+
2
(y) +
y
0
τ
+
2
( y
1
)N (0, y; 0, y
1
)dy
1
−
y
0
τ
−
3
(y
1
)N (0, y; 1, y
1
)dy
1
= E
1
(y),
(21)
1 + σ
3
−1 + σ
3
τ
−
3
(y)
−
y
0
τ
−
3
(y
1
)N (1, y; 1, y
1
)dy
1
+
y
0
τ
+
2
( y
1
)N (1, y; 0, y
1
)dy
1
= E
2
(y),
(22)
where
E
1
(y) =
−
y
0
[f (1, y
1
)N (0, y; 1, y
1
)
− f(0, y
1
)N (0, y; 0, y
1
)]dy
1
+
A
2
(y)
−1 + σ
2
+
1
0
τ
+
1
(x
1
)N (0, y; x
1
, 0) dx
1
+
1
0
dx
1
y
0
df (x
1
, y
1
)
dx
1
N (0, y; x
1
, y
1
)dy
1
,
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WSPC/246-AEJM
1450030
E. T. Karimov
& N. A. Rakhmatullaeva
E
2
(y) =
−
y
0
[f (1, y
1
)N (1, y; 1, y
1
)
− f(0, y
1
)N (1, y; 0, y
1
)]dy
1
+
A
3
(y)
−1 + σ
3
+
1
0
τ
+
1
(x
1
)N (1, y; x
1
, 0) dx
1
+
1
0
dx
1
y
0
df (x
1
, y
1
)
dx
1
N (1, y; x
1
, y
1
)dy
1
.
Since (21) is the second kind Volterra integral equation regarding the function
τ
+
2
(y), we can formally write its solution via resolvent kernel. Substituting the
found solution τ
+
2
(y) into (22), we will get second kind Volterra integral equation
regarding the function τ
−
3
(y), which is uniquely solvable due to continuity of the
kernel and continuous differentiability of the right-hand side of the integral equation.
After the finding of functions τ
±
i
(t)(i = 1, 3), we find unknown functions ν
±
i
(t)(i =
1, 3) by formulas (9)–(11).
Since we have all functions required to write solution of the problem, it can be
represented in the domain Ω
0
by the formula (17) and in domains Ω
i
(i = 1, 3) by
formulas (6)–(8), respectively.
Theorem is proved.
4. Open Problems
(1) If we replace parabolic part of the mixed equation with fractional case, i.e.
u
xx
−
C
D
α
0y
u, the solvability question of the problem is open. Here
C
D
α
0y
g =
1
Γ(1
− α)
y
0
(y
− t)
−α
g
(t)dt
is the Caputo fractional differential operator of the order α (0 < α < 1) [
16
].
(2) Consideration of nonlocal conditions, used in [
11
] instead of conditions (2)–(4) is
as well interesting, but one need to find certain nonlocal conditions, connecting
some part of one curve with part of another curve. Successful formulation could
generalize several local problems.
Acknowledgments
Authors would like to thank Professor A. S. Berdyshev for his useful remarks and
anonymous referees for their valuable suggestions, which made paper more readable.
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1450030-8
2
nd
Reading
June 18, 2014
14:51
WSPC/246-AEJM
1450030
On a nonlocal problem for mixed parabolic–hyperbolic type equation
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