General Boundary Value Problem for the Third Order Linear Differential Equation of Composite Type
The boundary value problem is considered for the linear two-dimensional integro-di®erential loaded third order equation of composite type with nonlocal terms in the boundary conditions. The principal part of the equation is a derivative of the two-dimensional Laplace equation with respect to the var...
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irk-123456789-1067132016-10-04T03:02:20Z General Boundary Value Problem for the Third Order Linear Differential Equation of Composite Type Gharehgheshlaghi, A.D. Aliyev, N. The boundary value problem is considered for the linear two-dimensional integro-di®erential loaded third order equation of composite type with nonlocal terms in the boundary conditions. The principal part of the equation is a derivative of the two-dimensional Laplace equation with respect to the variable x₂. Taking into account the ill-posedness of boundary value problems for hyperbolic di®erential equations, the principal part of the boundary conditions is chosen in a special form dictated by the obtained necessary conditions. Рассмотрена граничная задача для линейного двумерного интегро-дифференциального нагруженного уравнения композитного типа третьего порядка с нелокальными граничными условиями. Основная часть уравнения - производная по переменной от двумерного уравнения Лапласа относительно переменной x₂. Учитывая некорректность граничной задачи для гиперболических уравнений, главная часть граничных условий выбрана в специальной форме, продиктованной полученными необходимыми условиями. 2012 Article General Boundary Value Problem for the Third Order Linear Differential Equation of Composite Type / A.D. Gharehgheshlaghi, N. Aliyev // Журнал математической физики, анализа, геометрии. — 2012. — Т. 8, № 2. — С. 119-134. — Бібліогр.: 24 назв. — англ. 1812-9471 http://dspace.nbuv.gov.ua/handle/123456789/106713 en Журнал математической физики, анализа, геометрии Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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The boundary value problem is considered for the linear two-dimensional integro-di®erential loaded third order equation of composite type with nonlocal terms in the boundary conditions. The principal part of the equation is a derivative of the two-dimensional Laplace equation with respect to the variable x₂. Taking into account the ill-posedness of boundary value problems for hyperbolic di®erential equations, the principal part of the boundary conditions is chosen in a special form dictated by the obtained necessary conditions. |
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Gharehgheshlaghi, A.D. Aliyev, N. General Boundary Value Problem for the Third Order Linear Differential Equation of Composite Type Журнал математической физики, анализа, геометрии |
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Gharehgheshlaghi, A.D. Aliyev, N. |
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Gharehgheshlaghi, A.D. |
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General Boundary Value Problem for the Third Order Linear Differential Equation of Composite Type |
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General Boundary Value Problem for the Third Order Linear Differential Equation of Composite Type |
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General Boundary Value Problem for the Third Order Linear Differential Equation of Composite Type |
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General Boundary Value Problem for the Third Order Linear Differential Equation of Composite Type |
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General Boundary Value Problem for the Third Order Linear Differential Equation of Composite Type |
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general boundary value problem for the third order linear differential equation of composite type |
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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General Boundary Value Problem for the Third Order Linear Differential Equation of Composite Type / A.D. Gharehgheshlaghi, N. Aliyev // Журнал математической физики, анализа, геометрии. — 2012. — Т. 8, № 2. — С. 119-134. — Бібліогр.: 24 назв. — англ. |
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Журнал математической физики, анализа, геометрии |
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Journal of Mathematical Physics, Analysis, Geometry
2012, vol. 8, No. 2, pp. 119–134
General Boundary Value Problem for the Third Order
Linear Differential Equation of Composite Type
A. Delshad Gharehgheshlaghi
Institute of Mathematics and Mechanics of NAS of Azerbaijan
E-mail: ali-delshad@rambler.ru
N. Aliyev
Baku State University, Baku, Azerbaijan
Received September 14, 2010, revised October 11, 2011
The boundary value problem is considered for the linear two-dimensional
integro-differential loaded third order equation of composite type with non-
local terms in the boundary conditions. The principal part of the equation
is a derivative of the two-dimensional Laplace equation with respect to the
variable x2. Taking into account the ill-posedness of boundary value prob-
lems for hyperbolic differential equations, the principal part of the boundary
conditions is chosen in a special form dictated by the obtained necessary con-
ditions.
Key words: composite type equations, nonlocal boundary conditions,
global boundary conditions, necessary conditions, regularization, Fredholm
property.
Mathematics Subject Classification 2000: 35J25, 35J40.
1. Introduction
It is well known that in the case of ordinary differential operators, the La-
grange formula is the main tool [1] to solve the boundary value problems. But
when the operator is generated by the boundary value problem for partial equa-
tions, Green’s second formula [2, 3] becomes basic. For each concrete case [3–6],
some potentials (with unknown densities) that are the solutions of the considered
problems are derived proceeding from boundary conditions.
The form of the kernel of the potential is determined by Green’s formula men-
tioned above. The study of the properties of the constructed potentials enables
to define the unknown densities of some integral equations. By studying the pro-
perties of simple and double layers, it was possible to investigate the solutions of
c© A. Delshad Gharehgheshlaghi and N. Aliyev, 2012
A. Delshad Gharehgheshlaghi and N. Aliyev
Dirichlet and Neumann problems. Despite of being known, limit theorems both
for normal derivatives of double layer potentials and tangential derivatives of
simple and double layers [2] for some reasons were not often applied to boundary
value problems.
To solve the boundary value problems with oblique derivatives, a jump for-
mula obtained in [6–8] is usually used.
Notice that the problem under consideration is new. In the classic papers,
boundary value problems are mainly considered for even-order elliptic equations.
On the other hand, the mathematical model of the nuclear reactor in some cases
is described by the integro-differential equation of first order with the boundary
condition given on a part of the boundary [3]. As A.V. Bitsadze noted at one
of the Steklov seminars, in connection with the Tricomi problems, the whole
boundary should be present in the boundary condition. Therefore, the boundary
condition given in [3] is not appropriate. If we replace the boundary condition
given in [3] by the Dirichlet condition (given on the whole boundary), the obtained
problem will have no solution.
The boundary conditions given here (nonlocal conditions) correspond to the
first-order derivative.
The applied method is a continuation and generalization of the potential
theory. The solution is sought in the form of Green’s discrete second formula.
Therefore, in our case we always have jumps.
If the potential of the simple layer is continuous, then we have jumps because
of the double layer potential. If the double layer is continuous, then we have
a jump because of the simple layer potential. In the case of the problem with
non-local conditions, both potentials have jumps.
In the classic papers, in the case of inclined derivatives, if the inclination
is tangential, then we do not have a jump, i.e., we get the Fredholm integral
equation of first kind.
In our case such difficulties do not arise. Therefore, we can consider a problem
with oblique derivatives of arbitrary form.
Notice that in our case the whole boundary is a support for each boundary
condition.
Finally, we consider an example corresponding to the stated problem. Here
the problem is discretized in a certain sense, the system of 39 algebraic equations
with 39 variables is solved, and the errors of the obtained solution are shown.
120 Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 2
General Boundary Value Problem for the Third Order Linear Differential Equation
2. Problem Statement
Let D be a bounded convex in the direction of x2 plane domain with a
Lyapunov-type boundary Γ [3]. When the domain D is orthogonally projected on
the axis x1 (parallel to x2), the boundary Γ is divided into the parts Γ1 and Γ2.
The equations of these lines are denoted by x2 = γk (x1) , k = 1, 2; x1 ∈ [a1, b1].
Consider the boundary value problem
lu ≡ ∂3u (x)
∂x3
2
+
∂3u (x)
∂x2
1∂x2
+
2∑
k=0
a2k (x)
∂2u (x)
∂xk
1∂x2−k
2
+
2∑
k=1
a1k (x)
∂u (x)
∂xk
+ a0 (x) u (x)
+
2∑
m=0
2∑
n=1
∫ b1
a1
K2mn (x, η1)
∂2u (η)
∂ηm
1 ∂η2−m
2
∣∣∣∣
η2=γn(η1)
dη1
+
2∑
m=1
2∑
n=1
∫ b1
a1
K1mn (x, η1)
∂u (η)
∂ηm
∣∣∣∣
η2=γn(η1)
dη1
+
2∑
n=1
∫ b1
a1
K0n (x, η1) u (η1, γn (η1)) dη1 = f (x) , x ∈ D ⊂ R2, (1)
lku ≡ ∂2u (x)
∂x2
2
∣∣∣∣
x2=γk(x1)
−
2∑
p=1
2∑
j=1
αkjp (x1)
∂u (x)
∂xj
∣∣∣∣
x2=γp(x1)
−
2∑
p=1
αkp (x1)u (x1, γp (x1))−
2∑
p=1
2∑
j=1
∫ b1
a1
αkjp (x1, η1)
∂u (η)
∂ηj
∣∣∣∣
η2=γp(η1)
dη1
−
2∑
p=1
∫ b1
a1
αkp (x1, η1) u (η1, γp (η1)) dη1 = fk (x1) , k = 1, 2; x1 ∈ [a1, b1] ,
(2)
l3u ≡ ∂2u (x)
∂x2
1
∣∣∣∣
x2=γ2(x1)
−
2∑
p=1
2∑
j=1
α3jp (x1)
∂u (x)
∂xj
∣∣∣∣
x2=γp(x1)
−
2∑
p=1
α3p (x1)u (x1, γp (x1)) +
2∑
p=1
2∑
j=1
∫ b1
a1
α3jp (x1, η1)
∂u (η)
∂ηj
∣∣∣∣
η2=γp(η1)
dη1
−
2∑
p=1
∫ b1
a1
α3p (x1, η1) u (η1, γp (η1)) dη1 = f3 (x1) , x1 ∈ [a1, b1] , (3)
Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 2 121
A. Delshad Gharehgheshlaghi and N. Aliyev
where all data of equation (1) and boundary conditions (2), (3) are assumed to
be continuous functions.
If we consider the sufficiently smooth data of boundary value problem (1)–(3),
then this problem reduces to the Fredholm integral equation of the second kind
with respect to the function u (x). Otherwise, we get the system of the Fredholm
integral equations of the second kind with respect to the unknown function u (x)
and its derivatives. The kernels of these equations or of the obtained system have
no singularities.
3. Fundamental Solutions and Their Basic Properties
Applying the Fourier transformations [2, 3] for the principal part of equation
(1), namely, for the first two terms, we get the fundamental solution in the form
U (x− ξ) =
1
4π2
∫
R2
ei(α,x−ξ)
α2
(
α2
1 + α2
2
)dα, (4)
where x− ξ = (x1 − ξ1, x2 − ξ2), and (α, x− ξ) = α1 (x1 − ξ1) + α2 (x2 − ξ2), R2
is a real plane.
Then, by means of Hormander’s ladder method [9], for the integral (4) we
obtain
U (x− ξ) =
x2 − ξ2
2π
[
ln
√
|x1 − ξ1|2 + (x2 − ξ2)
2 − 1
]
+
|x1 − ξ1|
2π
arctg
x2 − ξ2
|x1 − ξ1| .
(5)
Differentiating (4) or (5), one can easily get
∂3U (x− ξ)
∂x3
2
+
∂3U (x− ξ)
∂x2
1∂x2
= δ (x− ξ) , (6)
where
∂U (x− ξ)
∂x1
=
e (x1 − ξ1)
π
arctg
x2 − ξ2
|x1 − ξ1| , (7)
∂U (x− ξ)
∂x2
=
1
2π
ln
√
|x1 − ξ1|2 + (x2 − ξ2)
2 , (8)
∂2U (x− ξ)
∂x2
1
= e (x2 − ξ2) δ (x1 − ξ1)− 1
2π
x2 − ξ2
|x1 − ξ1|2 + (x2 − ξ2)
2 , (9)
∆xU (x− ξ) = e (x2 − ξ2) δ (x1 − ξ1) . (10)
Here e (t) is the symmetric Heaviside function, δ (x− ξ) = δ (x1 − ξ1) δ (x2 − ξ2)
is a two-dimensional Dirac delta function [2, 3].
122 Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 2
General Boundary Value Problem for the Third Order Linear Differential Equation
4. Basic Relations
Using fundamental solution (5), its property (6) and considering equations (1),
we get Green’s second formula [2–6]. From these formulas we get representations
for any solution of equation (1) and expressions for the boundary values of this
solution
∫
Γ
∂2u (x)
∂x2
2
U (x− ξ) cos (ν, x2) dx +
∫
Γ
∂2u (x)
∂x2
1
U (x− ξ) cos (ν, x2) dx
−
∫
Γ
∂u (x)
∂x2
∂U (x− ξ)
∂x2
cos (ν, x2) dx−
∫
Γ
∂u (x)
∂x1
∂U (x− ξ)
∂x2
cos (ν, x1) dx
+
∫
Γ
u (x)
∂2U (x− ξ)
∂x1∂x2
cos (ν, x1) dx +
∫
Γ
u (x)
∂2U (x− ξ)
∂x2
2
cos (ν, x2) dx
+
∫
D
l0u · U (x− ξ) dx−
∫
D
f (x) U (x− ξ) dx =
u (ξ) , ξ ∈ D,
1
2
u (ξ) , ξ ∈ Γ,
(11)
where
l0u ≡
2∑
k=0
a2k (x)
∂2u (x)
∂xk
1∂x2−k
2
+
2∑
k=1
a1k (x)
∂u (x)
∂xk
+ a0 (x) u (x)
+
2∑
m=0
2∑
n=1
∫ b1
a1
K2mn (x, η1)
∂2u (η)
∂ηm
1 ∂η2−m
2
∣∣∣∣
η2=γn(η1)
dη1
+
2∑
m=1
2∑
n=1
∫ b1
a1
K1mn (x, η1)
∂u (η)
∂ηm
∣∣∣∣
η2=γn(η1)
dη1+
2∑
n=1
∫ b1
a1
K0n (x, η1) u (η1, γn (η1)) dη1 .
(12)
Then applying the schemes used in [10–14], we obtain the remaining basic
relations that give representations for the derivative of the unknown function
and the boundary values of these derivatives
−
∫
Γ
∂2u (x)
∂x2
2
∂U (x− ξ)
∂x2
cos (ν, x2) dx−
∫
Γ
∂2u (x)
∂x1∂x2
∂U (x− ξ)
∂x2
cos (ν, x1) dx
+
∫
Γ
∂u (x)
∂x2
∂2U (x− ξ)
∂x2
2
cos (ν, x2) dx +
∫
Γ
∂u (x)
∂x2
∂2U (x− ξ)
∂x1∂x2
cos (ν, x1) dx
−
∫
D
l0u · ∂U (x− ξ)
∂x2
dx +
∫
D
f (x)
∂U (x− ξ)
∂x2
dx =
∂u (ξ)
∂ξ2
, ξ ∈ D,
1
2
∂u (ξ)
∂ξ2
, ξ ∈ Γ,
(13)
Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 2 123
A. Delshad Gharehgheshlaghi and N. Aliyev
−
∫
Γ
∂2u (x)
∂x2
2
∂U (x− ξ)
∂x1
cos (ν, x2) dx−
∫
Γ
∂2u (x)
∂x2
1
∂U (x− ξ)
∂x1
cos (ν, x2) dx
+
∫
Γ
∂u (x)
∂x2
∂2U (x− ξ)
∂x1∂x2
cos (ν, x2) dx +
∫
Γ
∂u (x)
∂x1
∂2U (x− ξ)
∂x1∂x2
cos (ν, x1) dx
−
∫
Γ
∂u (x)
∂x2
∂2U (x− ξ)
∂x2
2
cos (ν, x1) dx−
∫
Γ
∂u (x)
∂x1
∂2U (x− ξ)
∂x2
2
cos (ν, x1) dx
−
∫
D
l0u · ∂U (x− ξ)
∂x1
dx +
∫
D
f (x)
∂U (x− ξ)
∂x1
dx =
∂u (ξ)
∂ξ1
, ξ ∈ D,
1
2
∂u (ξ)
∂ξ1
, ξ ∈ Γ,
(14)
∫
Γ
∂2u (x)
∂x2
2
∂2U (x− ξ)
∂x2
1
cos (ν, x2) dx +
∫
Γ
∂2u (x)
∂x2
1
∂2U (x− ξ)
∂x2
1
cos (ν, x2) dx
−
∫
Γ
∂2u (x)
∂x2
2
∂2U (x− ξ)
∂x1∂x2
cos (ν, x1) dx +
∫
Γ
∂2u (x)
∂x1∂x2
∂2U (x− ξ)
∂x1∂x2
cos (ν, x2) dx
−
∫
Γ
∂2u (x)
∂x1∂x2
∂2U (x− ξ)
∂x2
2
cos (ν, x1) dx +
∫
Γ
∂2u (x)
∂x2
1
∂2U (x− ξ)
∂x2
2
cos (ν, x2) dx
+
∫
D
l0u · ∂2U (x− ξ)
∂x2
1
dx−
∫
D
f (x)
∂2U (x− ξ)
∂x2
1
dx =
∂2u (ξ)
∂ξ2
1
, ξ ∈ D,
1
2
∂2u (ξ)
∂ξ2
1
, ξ ∈ Γ.
(15)
Notice that in the remaining two expressions (see also [9–11]) the derivatives
higher than third order in the domain D (both for u (x) and for U (x− ξ)) and
the derivatives higher than second order on the boundary Γ do not appear in the
integrand, i.e.,
∫
Γ
∂2u (x)
∂x2
2
∂U2 (x− ξ)
∂x2
2
cos (ν, x2) dx +
∫
Γ
∂2u (x)
∂x1∂x2
∂2U (x− ξ)
∂x2
2
cos (ν, x1) dx
−
∫
Γ
∂2u (x)
∂x1∂x2
∂2U (x− ξ)
∂x1∂x2
cos (ν, x2) dx +
∫
Γ
∂2u (x)
∂x2
2
∂2U (x− ξ)
∂x1∂x2
cos (ν, x1) dx
+
∫
D
l0u · ∂2U (x− ξ)
∂x2
2
dx−
∫
D
f (x)
∂2U (x− ξ)
∂x2
2
dx =
∂2u (ξ)
∂ξ2
2
, ξ ∈ D,
1
2
∂2u (ξ)
∂ξ2
2
, ξ ∈ Γ,
(16)
124 Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 2
General Boundary Value Problem for the Third Order Linear Differential Equation
∫
Γ
∂2u (x)
∂x2
2
∂2U (x− ξ)
∂x1∂x2
cos (ν, x2) dx +
∫
Γ
∂2u (x)
∂x1∂x2
∂2U (x− ξ)
∂x1∂x2
cos (ν, x1) dx
−
∫
Γ
∂2u (x)
∂x2
2
∂2U (x− ξ)
∂x2
2
cos (ν, x1) dx +
∫
Γ
∂2u (x)
∂x1∂x2
∂2U (x− ξ)
∂x2
2
cos (ν, x2) dx
+
∫
D
l0u · ∂2U (x− ξ)
∂x1∂x2
dx−
∫
D
f (x)
∂2U (x− ξ)
∂x1∂x2
dx =
∂2u (ξ)
∂ξ1∂ξ2
, ξ ∈ D,
1
2
∂2u (ξ)
∂ξ1∂ξ2
, ξ ∈ Γ.
(17)
Thus we establish the following
Theorem 1. If D ⊂ R2 is a bounded convex domain with Lyapunov boundary
Γ, the data of equation (1), a2k (x) , k = 0, 2, x ∈ D; a1k (x) , k = 1, 2, x ∈
D; a0 (x) , x ∈ D; K2mn (x, η1) , m = 0, 2, n = 1, 2, x ∈ D, η1 ∈ (a1, b1) ;
K1mn (x, η1) , m = 1, 2, n = 1, 2, x ∈ D, η1 ∈ (a1, b1) ; Kon (x, η1) , x ∈ D,
η1 ∈ (a1, b1), f (x) , x ∈ D are continuous functions, then every solution of
equation (1) determined in the domain D satisfies basic relations (11), (13)–(17).
5. Necessary conditions
Considering the second expressions of basic relations (11), (13)–(17) and pass-
ing from the integrals over the boundary Γ to those over its parts Γk (k = 1, 2) ,
one obtains
u (ξ1, γk (ξ1)) = 2
∫
Γ
∂2u(x)
∂x2
2
U(x1 − ξ1, x2 − γk(ξ1)) cos(ν, x2)dx
+2
∫
Γ
∂2u(x)
∂x2
1
U(x1 − ξ1, x2 − γk(ξ1)) cos(ν, x2)dx
− 2
∫
Γ
∂u(x)
∂x2
∂U(x−ξ)
∂x2
∣∣∣∣
ξ2=γk(ξ1)
cos(ν, x2)dx− 2
∫
Γ
∂u(x)
∂x1
∂U(x−ξ)
∂x2
∣∣∣∣
ξ2=γk(ξ1)
cos(ν, x1)dx
+2
∫
Γ
u(x)∂2u(x−ξ)
∂x1∂x2
∣∣∣∣
ξ2=γk(ξ1)
cos(ν, x1)dx + 2
∫
Γ
u(x)∂2U(x−ξ)
∂x2
2
∣∣∣∣
ξ2=γk(ξ1)
cos(ν, x2)dx
+2
∫
D
l0uU(x1 − ξ1, x2 − γk(ξ1))dx
−2
∫
D
f(x)U(x1 − ξ1, x2 − γk(ξ1))dx, k = 1, 2, ξ1 ∈ [a1, b1],
(18)
Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 2 125
A. Delshad Gharehgheshlaghi and N. Aliyev
∂u(ξ)
∂ξ1
∣∣∣
ξ2=γk(ξ1)
= − 2
∫
Γ
∂2u(x)
∂x2
2
∂U(x−ξ)
∂x1
∣∣∣∣
ξ2=γk(ξ1)
cos(ν, x2)dx
−2
∫
Γ
∂2u(x)
∂x2
1
∂U(x−ξ)
∂x1
∣∣∣∣
ξ2=γk(ξ1)
cos(ν, x2)dx + 2
∫
Γ
∂u(x)
∂x2
∂2U(x−ξ)
∂x1∂x2
∣∣∣∣
ξ2=γk(ξ1)
cos(ν, x2)dx
+2
∫
Γ
∂u(x)
∂x1
∂2U(x−ξ)
∂x1∂x2
∣∣∣∣
ξ2=γk(ξ1)
cos(ν, x1)dx− 2
∫
Γ
∂u(x)
∂x2
∂2U(x−ξ)
∂x2
2
∣∣∣∣
ξ2=γk(ξ1)
cos(ν, x1)dx
− 2
∫
Γ
∂u(x)
∂x1
∂2U(x−ξ)
∂x2
2
∣∣∣∣
ξ2=γk(ξ1)
cos(ν, x1)dx− 2
∫
D
l0u
∂U(x−ξ)
∂x1
dx
+2
∫
D
f(x)∂U(x−ξ)
∂x1
dx, k = 1, 2, ξ1 ∈ [a1, b1]
(19)
and
∂u(ξ)
∂ξ2
∣∣∣
ξ2=γk(ξ1)
= −2
∫
Γ
∂2u(x)
∂x2
2
∂U(x−ξ)
∂x2
∣∣∣∣
ξ2=γk(ξ1)
cos(ν, x2)dx
−2
∫
Γ
∂2u(x)
∂x1∂x2
∂U(x−ξ)
∂x2
∣∣∣∣
ξ2=γk(ξ1)
cos(ν, x1)dx +
∫
Γ
∂u(x)
∂x2
∂U2(x−ξ)
∂x2
2
∣∣∣∣
ξ2=γk(ξ1)
cos(ν, x2)dx
+ 2
∫
Γ
∂u(x)
∂x2
∂U2(x−ξ)
∂x1∂x2
∣∣∣∣
ξ2=γk(ξ1)
cos(ν, x1)dx− 2
∫
D
l0u
∂U(x−ξ)
∂x2
∣∣∣∣
ξ2=γk(ξ1)
dx
+ 2
∫
D
f(x)∂U(x−ξ)
∂x2
∣∣∣∣
ξ2=γk(ξ1)
dx, k = 1, 2, ξ1 ∈ [a1, b1],
(20)
where ξ2 = γk (ξ1) , k = 1, 2, are the equations of Γk, and the “dots” denote the
sums of nonsingular terms.
As it is seen from the fundamental solution (5), for the boundary values of
the second derivative we have
∂2U (x− ξ)
∂x2
2
∣∣∣∣x2=γp(x1)
ξ2=γp(ξ1)
=
1
2π
γ′p (σp (x1, ξ1))
(x1 − ξ1)
[
1 + γ′2p (σp)
] , p = 1, 2; (21)
∂2U (x− ξ)
∂x1∂x2
∣∣∣∣x2=γp(x1)
ξ2=γp(ξ1)
=
1
2π
1
(x1 − ξ1)
[
1 + γ′2p (σp)
] , p = 1, 2; (22)
∂2U (x− ξ)
∂x2
1
∣∣∣∣x2=γp(x1)
ξ2=γp(ξ1)
= − 1
2π
γ′p (σp)
(x1 − ξ1)
[
1 + γ′2p (σp)
] , p = 1, 2; (23)
∂2U (x− ξ)
∂x2
1
∣∣∣∣x2=γp(x1)
ξ2=γq(ξ1)
= δ (x1 − ξ1) e (γp (x1)− γq (ξ1)) , p, q = 1, 2; p 6= q,
(24)
where σp (x1, ξ1) is located between x1 and ξ1.
126 Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 2
General Boundary Value Problem for the Third Order Linear Differential Equation
∂u2(ξ)
∂ξ2
2
∣∣∣
ξ2=γk(ξ1)
= −2
b1∫
a1
∂2u(x)
∂x2
2
∣∣∣
x2=γ1(x1)
∂2U(x−ξ)
∂x2
2
∣∣∣
x2=γ1(x1)
ξ2=γk(ξ1)
dx1
+2
b1∫
a1
∂2u(x)
∂x2
2
∣∣∣
x2=γ2(x1)
∂2U(x−ξ)
∂x2
2
∣∣∣
x2=γ2(x1)
ξ2=γk(ξ1)
dx1
+2
b1∫
a1
∂2u(x)
∂x1∂x2
∣∣∣
x2=γ1(x1)
∂2U(x−ξ)
∂x2
2
∣∣∣
x2=γ1(x1)
ξ2=γk(ξ1)
γ′1(x1)dx1
−2
b1∫
a1
∂2u(x)
∂x1∂x2
∣∣∣
x2=γ2(x1)
∂2U(x−ξ)
∂x2
2
∣∣∣
x2=γ2(x1)
ξ2=γk(ξ1)
γ′2(x1)dx1
+2
b1∫
a1
∂2u(x)
∂x1∂x2
∣∣∣
x2=γ1(x1)
∂2U(x−ξ)
∂x1∂x2
∣∣∣
x2=γ1(x1)
ξ2=γk(ξ1)
dx1
−2
b1∫
a1
∂2u(x)
∂x1∂x2
∣∣∣
x2=γ2(x1)
∂2U(x−ξ)
∂x1∂x2
∣∣∣
x2=γ2(x1)
ξ2=γk(ξ1)
dx1
+2
b1∫
a1
∂2u(x)
∂x2
2
∣∣∣
x2=γ1(x1)
∂2U(x−ξ)
∂x1∂x2
∣∣∣
x2=γ1(x1)
ξ2=γk(ξ1)
γ′1(x1)dx1
−2
b1∫
a1
∂2u(x)
∂x2
2
∣∣∣
x2=γ2(x1)
∂2U(x−ξ)
∂x1∂x2
∣∣∣
x2=γ2(x1)
ξ2=γk(ξ1)
γ′2(x1)dx1
+2
∫
D
l0u
∂2U(x−ξ)
∂x2
2
dx + 2
∫
D
f(x)∂2U(x−ξ)
∂x2
2
dx
= (−1)k−1
π
b1∫
a1
∂2u(x)
∂x1∂x2
∣∣∣
x2=γk(x1)
dx1
x1−ξ1
+ · · · , k = 1, 2.
(25)
∂2u(ξ)
∂ξ2
1
∣∣∣
ξ2=γk(ξ1)
= −2
b1∫
a1
∂2u(x)
∂x2
2
∣∣∣
x2=γ1(x1)
∂2U(x−ξ)
∂x2
1
∣∣∣
x2=γ1(x1)
ξ2=γk(ξ1)
dx1
+2
b1∫
a1
∂2u(x)
∂x2
2
∣∣∣
x2=γ2(x1)
∂2U(x−ξ)
∂x2
1
∣∣∣
x2=γ2(x1)
ξ2=γk(ξ1)
dx1
−2
b1∫
a1
∂2u(x)
∂x2
1
∣∣∣
x2=γ1(x1)
∂2U(x−ξ)
∂x2
1
∣∣∣
x2=γ1(x1)
ξ2=γk(ξ1)
dx1
+2
b1∫
a1
∂2u(x)
∂x2
1
∣∣∣
x2=γ2(x1)
∂2U(x−ξ)
∂x2
1
∣∣∣
x2=γ2(x1)
ξ2=γk(ξ1)
dx1
−2
b1∫
a1
∂2u(x)
∂x2
2
∣∣∣
x2=γ1(x1)
∂2U(x−ξ)
∂x1∂x2
∣∣∣
x2=γ1(x1)
ξ2=γk(ξ1)
γ′1(x1)dx1
+2
b1∫
a1
∂2u(x)
∂x2
2
∣∣∣
x2=γ2(x1)
∂2U(x−ξ)
∂x1∂x2
∣∣∣
x2=γ2(x1)
ξ2=γk(ξ1)
γ′2(x1)dx1
−2
b1∫
a1
∂2u(x)
∂x1∂x2
∣∣∣
x2=γ1(x1)
∂2U(x−ξ)
∂x1∂x2
∣∣∣
x2=γ1(x1)
ξ2=γk(ξ1)
dx1
Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 2 127
A. Delshad Gharehgheshlaghi and N. Aliyev
+2
b1∫
a1
∂2u(x)
∂x1∂x2
∣∣∣
x2=γ2(x1)
∂2U(x−ξ)
∂x1∂x2
∣∣∣
x2=γ2(x1)
ξ2=γk(ξ1)
dx1
−2
b1∫
a1
∂2u(x)
∂x1∂x2
∣∣∣
x2=γ1(x1)
∂2U(x−ξ)
∂x1∂x2
∣∣∣
x2=γ1(x1)
ξ2=γk(ξ1)
γ′1(x1)dx1
+2
b1∫
a1
∂2u(x)
∂x1∂x2
∣∣∣
x2=γ2(x1)
∂2U(x−ξ)
∂x2
2
∣∣∣
x2=γ2(x1)
ξ2=γk(ξ1)
γ′2(x1)dx1
−2
b1∫
a1
∂2u(x)
∂x2
1
∣∣∣
x2=γ1(x1)
∂2U(x−ξ)
∂x2
2
∣∣∣
x2=γ1(x1)
ξ2=γk(ξ1)
dx1
+2
b1∫
a1
∂2u(x)
∂x2
1
∣∣∣
x2=γ2(x1)
∂2U(x−ξ)
∂x2
2
∣∣∣
x2=γ2(x1)
ξ2=γk(ξ1)
dx1
+2
∫
D
l0u
∂2U(x−ξ)
∂x2
1
dx− 2
∫
D
f(x)∂2U(x−ξ)
∂x2
1
dx, k = 1, 2, ξ1 ∈ [a1, b1].
(26)
Then the remaining necessary conditions take the form
∂2u(ξ)
∂ξ2
1
∣∣∣
ξ2=γ1(ξ1)
− ∂2u(ξ)
∂ξ2
1
∣∣∣
ξ2=γ2(ξ1)
− ∂2u(ξ)
∂ξ2
2
∣∣∣
ξ2=γ2(ξ1)
= − 1
π
∫ b1
a1
∂2u(x)
∂x1∂x2
∣∣∣
x2=γ1(x1)
dx1
x1−ξ1
+ . . .
(27)
∂2u(ξ)
∂ξ2
1
∣∣∣
ξ2=γ2(ξ1)
− ∂2u(ξ)
∂ξ2
1
∣∣∣
ξ2=γ1(ξ1)
− ∂2u(ξ)
∂ξ2
2
∣∣∣
ξ2=γ1(ξ1)
= 1
π
∫ b1
a1
∂2u(x)
∂x1∂x2
∣∣∣
x2=γ2(x1)
dx1
x1−ξ1
+ . . .
(28)
∂2u(ξ)
∂ξ1∂ξ2
∣∣∣
ξ2=γk(ξ1)
= −2
b1∫
a1
∂2u(x)
∂x2
2
∣∣∣
x2=γ1(x1)
∂2U(x−ξ)
∂x1∂x2
∣∣∣
x2=γ1(x1)
ξ2=γk(ξ1)
dx1
+2
b1∫
a1
∂2u(x)
∂x2
2
∣∣∣
x2=γ2(x1)
∂2U(x−ξ)
∂x1∂x2
∣∣∣
x2=γ2(x1)
ξ2=γk(ξ1)
dx1
+2
b1∫
a1
∂2u(x)
∂x1∂x2
∣∣∣
x2=γ1(x1)
∂2U(x−ξ)
∂x1∂x2
∣∣∣
x2=γ1(x1)
ξ2=γk(ξ1)
γ′1(x1)dx1
−2
b1∫
a1
∂2u(x)
∂x1∂x2
∣∣∣
x2=γ2(x1)
∂2U(x−ξ)
∂x1∂x2
∣∣∣
x2=γ2(x1)
ξ2=γk(ξ1)
γ′2(x1)dx1
−2
b1∫
a1
∂2u(x)
∂x2
2
∣∣∣
x2=γ1(x1)
∂2U(x−ξ)
∂x2
2
∣∣∣
x2=γ1(x1)
ξ2=γk(ξ1)
γ′1(x1)dx1
128 Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 2
General Boundary Value Problem for the Third Order Linear Differential Equation
+2
b1∫
a1
∂2u(x)
∂x2
2
∣∣∣
x2=γ2(x1)
∂2U(x−ξ)
∂x2
2
∣∣∣
x2=γ2(x1)
ξ2=γk(ξ1)
γ′2(x1)dx1
−2
b1∫
a1
∂2u(x)
∂x1∂x2
∣∣∣
x2=γ1(x1)
∂2U(x−ξ)
∂x2
2
∣∣∣
x2=γ1(x1)
ξ2=γk(ξ1)
dx1
+2
b1∫
a1
∂2u(x)
∂x1∂x2
∣∣∣
x2=γ2(x1)
∂2U(x−ξ)
∂x2
2
∣∣∣
x2=γ2(x1)
ξ2=γk(ξ1)
dx1
+2
∫
D
l0u
∂2U(x−ξ)
∂x1∂x2
dx− 2
∫
D
f(x)∂2U(x−ξ)
∂x1∂x2
dx
= (−1)k
π
b1∫
a1
∂2u(x)
∂x2
2
∣∣∣
x2=γk(x1)
dx1
x1−ξ1
+ · · · , k = 1, 2, ξ1 ∈ [a1, b1].
(29)
Thus we obtained the following statements.
Theorem 2. Under the conditions of Theorem 1 every solution of equation
(1) satisfies the necessary regular conditions (18)–(20).
Theorem 3. Under the conditions of Theorem 1 every solution of equation
(1) satisfies the necessary singular conditions (25)–(29).
6. Fredholm Property
Considering the necessary singular conditions (29) and taking into account
the boundary conditions (2), we get
∂2u (ξ)
∂ξ1∂ξ2
∣∣∣∣
ξ2=γk(ξ1)
=
(−1)k
π
∫ b1
a1
dx1
x1 − ξ1
fk (x1) +
2∑
p=1
2∑
j=1
αkjp (x1)
∂u (x)
∂xj
∣∣∣∣
x2=γp(x1)
+
2∑
p=1
αkp (x1)u (x1, γp (x1)) +
2∑
p=1
2∑
j=1
∫ b1
a1
αkjp (x1, η1)
∂u (η)
∂ηj
∣∣∣∣
η2=γp(η1)
+
2∑
p=1
∫ b1
a1
αkp (x1, η1) u (η1, γp (η1)) dη1
. (30)
The first term in the right-hand side is easily regularized if
fk (x) ∈ C(1) [a1, b1] , fk (a1) = fk (b1) = 0, k = 1, 2, (31)
as shown in [15].
Concerning the second and the third terms in the right-hand side of (30), they
are regularized by using (18)–(20).
Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 2 129
A. Delshad Gharehgheshlaghi and N. Aliyev
After substituting (18)–(20) into (30), it suffices to replace the regular inte-
grals in (18)–(20) by the singular integrals from (30). For the last two terms in
the right-hand side of (30) it suffices to replace the integrals contained in it.
Finally, consider the necessary condition (27). Its right-hand side contains
singular integrals, and we substitute
∂2u (x)
∂x1∂x2
∣∣∣∣
x2=γ1(x1)
by its regular expression
obtained by means of (30). Then in the left-hand side of expression (27), instead
of
∂2u (ξ)
∂ξ2
1
∣∣∣∣
ξ2=γ2(ξ1)
and
∂2u (ξ)
∂ξ2
2
∣∣∣∣
ξ2=γ2(ξ1)
we use their expressions from the bound-
ary conditions (2) and (3). Thus we get the regular relation for
∂2u (ξ)
∂ξ2
1
∣∣∣∣
ξ2=γ1(ξ1)
as well.
Thus we proved.
Theorem 4. Under the conditions of Theorem 1, (31) and if αkjp (x1), k =
1, 3, j = 1, 2, p = 1, 2; αkp (x1), k = 1, 3, p = 1, 2, x1 ∈ [a1, b1] ; αkjp (x1, η1),
k = 1, 3, j = 1, 2, p = 1, 2; x1 ∈ [a1, b1] , η1 ∈ [a1, b1] ; αkp (x1, η1), k = 1, 3,
p = 1, 2; x1 ∈ [a1, b1] , η1 ∈ [a1, b1] and f3 (x1), x1 ∈ [a1, b1] are continuous
functions, then for the boundary values of u (x) and its derivatives up to the
second order we get a normal system of the second order integral equations whose
Fredholm kernel formulas do not contain singularities (i.e., the singularity in the
trace formula is weak).
If all boundary values up to the second order are determined by the above
mentioned system of integral equations, then after substitution these boundary
values into the left-hand side of (11), (13)–(17), we obtain for the unknown
function u (x) and its derivatives up to the second order, inclusively for ξ ∈ D,
the system of the Fredholm integral equations of the second kind with regular
kernels.
Theorem is proved.
Theorem 5. Under the conditions of Theorem 4, the stated boundary value
problem is reduced to the system of the Fredholm integral equations of the second
kind whose Fredholm kernel does not contain singular integrals.
A boundary value problem for the second order equation of composite type
was studied in [16–21].
Various special cases of boundary value problems for the composite type equa-
tions of third order were considered in [22–24].
130 Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 2
General Boundary Value Problem for the Third Order Linear Differential Equation
7. Example
Consider the domain D in the upper half-plane enclosed by the abscissa axis
(x2 = 0) and the parabola (x2 = 1− x2
1),
D =
{
x = (x1, x2), x2 ∈ (0, 1− x2
1), x1 ∈ (−1, 1)
}
.
Consider the boundary value problem with the solution u(x) = x1(x2
2 − x2
1)
∂3u(x)
∂x3
2
+
∂3u(x)
∂x2
1∂x2
= 0, x ∈ D, (32)
under the boundary conditions
∂2u(x)
∂x2
2
∣∣∣∣
x2=0
= 2x1, (33)
∂2u(x)
∂x2
2
∣∣∣∣
x2=1−x2
1
= 2x1, (34)
and
∂2u(x)
∂x2
1
∣∣∣∣
x2=1−x2
1
= −6x1. (35)
Notice that for discretization we use the following pattern. If discretization is
performed at the point m,n, then it is necessary to know the values of this
function on the four layers
(m,n− 1); (m− 1, n), (m,n), (m + 1, n);
(m− 1, n + 1), (m, n + 1), (m + 1, n + 1); (m,n + 2).
(36)
Then equation (32) yields the following discrete equations at the point (m,n):
ym+1,n+1 − ym+1,n + ym,n+2 − 5ym,n+1 + 5ym,n
− ym,n−1 + ym−1,n+1 − ym−1,n = 0,
(37)
where, m = −3, 3 , n = 1; m = −2, 2 , n = 2.
Similarly, the equations obtained from the boundary condition (33) are in the
form (notice that here h = 1
4 , x1 = 1
4m)
ym,n+1 − 2ym,n + ym,n−1 =
1
32
m, (38)
where, m = −4, 4 , n = 1.
Equations (38) are obtained from the boundary condition (34) with m =
−3, n = 2;m = −2, 2, n = 3;m = 3, n = 2.
Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 2 131
A. Delshad Gharehgheshlaghi and N. Aliyev
Finally, the algebraic equations are obtained from the boundary condition
(35)
ym+1,n − 2ym,n + ym−1,n = − 3
32
m. (39)
Here
m = −3, n = 0, 2;m = 3, n = 0, 2;m = −1, 1, n = 4;m = −2, 2, n = 3
are obtained. So, we get a system of 39 variables and 39 linear algebraic equations.
Since, u(x) = x1(x2
2 − x2
1), then,
ym,n = y(mh, nh) = mh(n2h2 −m2h2) = mh3(n2 −m2) =
m
64
(n2 −m2). (40)
In the considered example we give the exact values at the knot points of the
domain, the approximate values of the solution of the system of linear algebraic
equations, and finally their absolute error in the table below.
It should be noted that here e = |y(Exact)− y(App)|.
Table
ym,n Exact App Err
y−4,0 1 0.1107 0.8893
y−3,0 0.4218 0.1141 0.3077
y−2,0 0.125 -0.0076 0.1326
y−1,0 0.0126 -0.1461 0.1617
y0,0 0 -0.0511 0.0511
y1,0 -0.0156 -0.0497 0.0341
y2,0 -0.125 -0.1422 0.0172
y3,0 -0.4218 -0.1166 0.3052
y4,0 -1 0.0072 1.0072
y−4,1 0.9375 0.0684 0.8691
y−3,1 0.375 0.0677 0.3073
y−2,1 0.0937 0.0045 0.0892
y−1,1 0 -0.0852 0.0852
y0,1 0 -0.0271 0.0271
y1,1 0 -0.0002 0.0002
y2,1 -0.0937 -0.0023 0.9347
y3,1 -0.375 -0.0357 0.3393
y4,1 -0.9375 0.0399 0.9774
y−4,2 0.75 0.0256 0.7244
y−3,2 0.2343 0.0113 0.223
y−2,2 0 0.0099 0.0099
y−1,2 -0.0468 0.0084 0.0552
y0,2 0 0.1650 0.1550
y1,2 0.0468 0.0852 0.3828
132 Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 2
General Boundary Value Problem for the Third Order Linear Differential Equation
y2,2 0 0.0367 0.0367
y3,2 -0.2343 0.0488 0.2831
y4,2 -0.75 0.0922 0.8422
y−3,3 0 0.1964 0.1964
y−2,3 -0.1562 0.1192 0.2754
y−1,3 -0.125 0.1045 0.2295
y0,3 0 0.1597 0.1597
y1,3 0.125 0.2773 0.1523
y2,3 0.1562 -0.0343 0.1905
y3,3 0 -0.0225 0.0225
y−2,4 -0.375 0.0831 0.4581
y1,4 -0.2343 0.2823 0.5176
y0,4 0 -0.4307 0.4307
y1,4 0.2343 -0.2748 0.5088
y2,4 0.375 0.0061 0.3689
As it is seen from the table, for the given test the errors are as given in the
table. It should also be noted that the maximal error is 1.0072 and the minimal
error is 0.0002. Subsequently this problem will be reduced to the integral equation
and solved approximately, and the error will be found. Finally, the errors obtained
in this way will be compared.
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