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```Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 203, pp. 1–5.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
A NONLOCAL BOUNDARY PROBLEM FOR THE LAPLACE
OPERATOR IN A HALF DISK
GANI A. BESBAEV, ISABEK ORAZOV, MAKHMUD A. SADYBEKOV
Abstract. In the present work we investigate the nonlocal boundary problem
for the Laplace equation in a half disk. The difference of this problem is the
impossibility of direct applying of the Fourier method (separation of variables).
Because the corresponding spectral problem for the ordinary differential equation has the system of eigenfunctions not forming a basis. Based on these
eigenfunctions there is constructed a special system of functions that already
forms the basis. This is used for solving of the nonlocal boundary equation.
The existence and the uniqueness of the classical solution of the problem are
proved.
1. Formulation of the problem
¯ ∩ C 2 (D) satisfying equation
Our goal is to find a function u(r, θ) ∈ C 0 (D)
∆u = 0
(1.1)
in D, with the boundary conditions
u(1, θ) = f (θ),
u(r, 0) = 0,
0 ≤ θ ≤ π,
r ∈ [0, 1],
∂u
∂u
(r, 0) =
(r, π) + αu(r, π),
∂θ
∂θ
(1.2)
(1.3)
r ∈ (0, 1)
(1.4)
where D = {(r, θ) : 0 < r < 1, 0 < θ < π}; α > 0; f (θ) ∈ C 2 [0, π], f (0) = 0,
f 0 (0) = f 0 (π) + αf (π).
Problem (1.1)–(1.4) with α = 0 was considered in [3, 4] for the Laplace equation,
and in [5, 6] for the Helmholtz equation. The existence and the uniqueness of
the solution of the problem are proved by applying the method of separation of
variables and proving the basis of the special function systems of the SamarskiiIonkin type in Lp . In contrast to these papers in case of α 6= 0 it is impossible
to use directly the Fourier method of the separation of the variables. Because the
corresponding spectral problem for the ordinary differential equation has the system
of eigenfunctions not forming a basis.
2000 Mathematics Subject Classification. 33C10, 34B30, 35P10.
Key words and phrases. Laplace equation; basis; eigenfunctions;
nonlocal boundary value problem.
c
2014
Texas State University - San Marcos.
Submitted July 11, 2014. Published September 30, 2014.
1
2
G. A. BESBAEV, I. ORAZOV, M. A. SADYBEKOV
EJDE-2014/203
2. Uniqueness of the solution
Theorem 2.1. The solution of problem (1.1)–(1.4) is unique.
Proof. Suppose that there exist two functions u1 (r, θ) and u2 (r, θ) satisfying the
conditions of the problem (1.1) - (1.4). We show that the function u(r, θ) =
u1 (r, θ) − u2 (r, θ) is equal to 0.
Consider the function
U (r, θ) = u(r, θ) + u(r, π − θ)
in D1 = {(r, θ) : 0 < r < 1, 0 < θ < π/2}. It is easy to see that
∆U = 0;
∂U
(r, π/2) = 0;
∂θ
∂U
(r, 0) = αU (r, 0) for 0 < r < 1;
∂θ
U (1, θ) = 0 for 0 ≤ θ ≤ π/2.
Since α > 0, it follows that U = 0 in D¯1 by the maximum principle and the
Zaremba-Giraud principle [1, p. 26] for the Laplace equation. This means that
u(r, θ) = −u(r, π − θ), in particular u(r, 0) = u(r, π) = 0 at r ∈ [0, 1]. The equality
¯ follows from the uniqueness of the solution of the Dirichlet problem
u(r, θ) = 0 in D
for the Laplace equation. The proof of the theorem is complete.
3. Forming the basis
If solutions to (1.1) satisfying the conditions (1.3), (1.4) are sought in the form
u(r, θ) = R(r)ϕ(θ),
√
then R(r) = r
problem
λ
, Re
√
λ ≥ 0, and for the function ϕ(θ) we have the spectral
−ϕ00 (θ) = λϕ(θ),
0 < θ < π;
(3.1)
ϕ(0) = 0, ϕ (0) = ϕ (π) + αϕ(π).
This problem has two groups of eigenvalues. All the eigenvalues are simple and
the corresponding system of eigenfunctions does not form the basis in L2 (0, π) .
However, in  a special system of functions is built based of these eigenfunctions
which forms the basis. This fact was applied for the solution of the nonlocal initialboundary problem for the heat equation. In  one family of problems simulating
the determination of the temperature and density of heat sources from given values
of the initial and final temperature is similarly considered.
Let us present the necessary facts from . Problem (3.1) has two groups of
(1)
(2)
eigenvalues λk = (2k)2 , k = 1, 2, . . . , λk = (2βk )2 , k = 0, 1, 2, . . . . Herein
βk are roots of the equation tgβ = α/2β, β > 0, they satisfy the inequalities
k < βk < k + 1/2, k = 0, 1, 2, . . . , and two-side estimates are carried out for
δk = βk − k where k is large enough,
α
1
α
1
1−
< δk <
1+
.
(3.2)
2k
2k
2k
2k
The eigenfunctions of the problem (3.1) have the form
0
(1)
ϕk (θ) = sin(2kθ),
0
(2)
k = 1, 2, . . . ; ϕk (x) = sin(2βk θ), k = 0, 1, 2, . . . .
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A NONLOCAL BOUNDARY PROBLEM
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This system is almost normed but does not form even an ordinary basis in L2 (0, π).
The additional system constructed from the previous one
(2)
ϕ0 (θ) = (2β0 )−1 ϕ0 (θ),
(1)
ϕ2k (θ) = ϕk (θ),
(2)
(1)
ϕ2k−1 (θ) = (ϕk (θ) − ϕk (θ))(2δk )−1 ,
k = 1, 2, . . .
is a Riesz basis in L2 (0, π). Biorthogonal to it, is the system
(2)
ψ0 (θ) = 2β0 ψ0 (θ),
(2)
(1)
ψ2k (θ) = ψk (θ) + ψk (θ),
(2)
ψ2k−1 (θ) = 2δk ψk (θ), k = 1, 2, . . . .
This system is constructed from the eigenfunctions
(1)
(1)
ψk (θ) = Ck cos(2kθ + γk ),
(2)
ψk (θ)
=
(2)
Ck
k = 1, 2, . . . ,
cos(βk (1 − 2θ)),
k = 0, 1, 2, . . . .
(j)
of the problem conjugated to (3.1). The constants Ck are taken from the biorthog(j)
(j) onal relations ϕk , ψk = 1, j = 1, 2.
If the function f (θ) is in C 2 [0, π] and satisfies the boundary conditions of problem
(3.1), then its Fourier series by the system ϕk (θ) converges uniformly. We can
calculate that
(2)
(1)
ϕ000 (θ) = −λ0 (θ), ϕ002k (θ) = −λk ϕ2k (θ),
(2)
(2)
ϕ002k−1 (θ) = −λk ϕ2k−1 (θ) −
(1)
λk − λk
ϕ2k (θ).
2δk
(3.3)
4. Construction of the formal solution to the problem
Considering section 3, we can write any solution of (1.1)–(1.4) in the form of a
biorthogonal series
∞
X
Rk (r)ϕk (θ),
(4.1)
u(r, θ) =
k=0
Rπ
where Rk (r) = (u(r, ·) and ψk (·)) ≡ 0 u(r, θ)ψk (θ)dθ. Functions (4.1) satisfy the
boundary conditions (1.3) and (1.4).
Substituting (4.1) in (1.1) and the boundary conditions (1.2), taking into account
(3.3), for finding unknown functions Rk (r) we obtain the following problems
(2)
r2 R000 (r) + rR00 (r) − λ0 R0 (r) = 0,
(2)
00
0
r2 R2k−1
(r) + rR2k−1
(r) − λk R2k−1 (r) = 0,
(2)
(1)
(4.2)
λk − λk
R2k−1 (r),
2δk
with the boundary conditions Rk (1) = fk , where fk are the Fourier coefficients of
the expansion of the function f (θ) into the biorthogonal series by ϕk (θ).
(1)
00
0
r2 R2k
(r) + rR2k
(r) − λk R2k (r) =
4
G. A. BESBAEV, I. ORAZOV, M. A. SADYBEKOV
EJDE-2014/203
The regular solution of (4.2) exists, is unique and can be written in the explicit
form
q
q
(2)
(2)
λ0
R0 (r) = f0 r
, R2k−1 (r) = f2k−1 r
q
(2)
λk
−r
2δk
Substituting (4.3) in (4.1), we obtain a formal solution
q
R2k (r) = f2k r
(1)
λk
+ f2k−1
r
λk
q
,
(4.3)
(1)
λk
.
∞
u(r, θ) = f0
X
r2β0
r2k 2δk
sin(2β0 θ) +
[r sin(2(k + δk )θ) − sin(2kθ)]
f2k−1
2β0
2δk
k=1
+
∞
X
f2k r
2k
(4.4)
sin(2kθ).
k=1
5. Main Theorem
Our main result reads as follows.
Theorem 5.1. If f (θ) ∈ C 2 [0, π], f (0) = 0, f 0 (0) = f 0 (π) + αf (π), then there
¯ ∩ C 2 (D) of problem (1.1)–(1.4).
exists a unique classical solution u(r, θ) ∈ C 0 (D)
Proof. The uniqueness of the classical solution of the problem follows from Theorem
2.1. The formal solution of the problem is shown in the form of (4.4). To make
sure that these functions are really the desired solutions we need to verify the
applicability of the superposition principle. For it we need to show the convergence
of the series, the possibility of termwise differentiation, and to prove the continuity
of these functions on the boundary of the half-disk.
The possibility of differentiating the series (4.4) any number of times at r < 1 is
an obvious consequence of the convergence of power series and two-sided estimates
(3.2) for δk . Let us justify the uniform convergence of the series (4.1) at r ≤ 1.
For this we use the sign of the uniform convergence of Weierstrass. By direct
calculation it is easy to see that the series (4.4) is majorized by the series C1 (|f0 | +
|f1 | + |f2 | + . . . ). This series converges  due to the requirements of the theorem
imposed on f (θ). Since all the terms of the series (4.4) are continuous functions,
¯ The proof is
then the function u(r, θ) is continuous in the boundary domain D.
complete.
6. Conjugated problem: existence and uniqueness of the solution
Let us now formulate a problem conjugated to (1.1)-(1.4). We look for a function
¯ ∩ C 2 (D) satisfying the equation
v(r, θ) ∈ C 0 (D)
∆v = 0
(6.1)
in D with the boundary conditions
v(1, θ) = g(θ),
v(r, 0) = v(r, π),
0 ≤ θ ≤ π,
(6.2)
r ∈ [0, 1],
(6.3)
∂v
(r, π) + αv(r, π) = 0, r ∈ (0, 1),
∂θ
where g(θ) ∈ C 2 [0, π], g(0) = g(π), g 0 (π) + αg(π) = 0.
(6.4)
EJDE-2014/203
A NONLOCAL BOUNDARY PROBLEM
5
We can easily verify the conjugacy of the problems (1.1)–(1.4) and (6.1)–(6.4)
by direct calculation. The uniqueness of the solution of problem (6.1)–(6.4) follows
from the maximum principle and the Zaremba-Giraud principle [1, p. 26] for the
Laplace equation. The existence of the solution and its representation in the form
of a biorthogonal series can be proved similar to Theorem 5.1. Let us show this
result without the proof.
Theorem 6.1. If g(θ) ∈ C 2 [0, π],g(0) = g(π), g 0 (π) + αg(π) = 0, then there exists
¯ ∩ C 2 (D) of problem (6.1)-(6.4).
a unique classical solution v(r, θ) ∈ C 0 (D)
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Gani A. Besbaev
Faculty of Information technology, Auezov South Kazakhstan state University,
Shymkent, Kazakhstan