Download 1. Introduction

Proceedings of the Institute of Mathematics and Mechanics,
National Academy of Sciences of Azerbaijan
Volume 42, Number 1, 2016, Pages 3–9
ON SOLVABILITY OF AN EXTERNAL PROBLEM WITH
IMPEDANCE BOUNDARY CONDITION FOR HELMHOLTZ
EQUATION BY INTEGRAL EQUATIONS METHOD
RAHIB J. HEYDAROV
Abstract. In the paper we study the solvability of an external problem
with an impedance boundary condition for the Helmholtz equation by
the method of weakly singular integral equations.
1. Introduction
It is known that in theory of acoustic waves, external boundary value problems for the Helmholtz equation is of great importance. Existence of the solutions
of Dirichlet and Neumann external boundary value problems for the Helmholtz
equation by the method of weakly singular integral equations was considered in
the papers [1,3-7]. In the monograph [2] the solvability of an external problem
with an impedance boundary condition for the Helmholtz equation by the method
of singular integral equations is given. It should be noted that the solution of a
singular integral equation is much complicated than the solution of a weakly singular integral equation (since a singular integral operator is not compact, while a
weakly singular integral operator is). Therefore, it is suitable to reduce an external problem with an impedance boundary condition for the Helmholtz equation
to a weakly singular integral equation, and this paper is devoted to this matter.
Let D ⊂ R3 be a bounded domain with boundary S ∈ Λα , where Λα is a
class of Lyapunov surfaces with an index 0 < α ≤ 1. Recall that an external
problem with an impedance boundary condition for the Helmholtz equation is to
find the function u twice continuously-differentiable on R3 \D and continuous on
S , prossessing a normal derivative in the sense of uniform convergence and satisfying the Helmholtz equation ∆u + k 2 u = 0 in R3 \D, the Sommerfeld radiation
condition
x
1
, grad u (x) − iku (x) = o
, |x| → ∞,
|x|
|x|
uniformly along all directions x/ |x|, and the boundary condition
∂u (x)
+ f (x) u (x) = g (x)
∂~n (x)
on S,
(1.1)
2010 Mathematics Subject Classification. 30E25; 31B10.
Key words and phrases. Helmholtz equation, Dirichlet’s boundary value problem, Neumann’s
boundary value problem, boundary value problem with impedance boundary conditions, singular
integral equation.
3
4
RAHIB J. HEYDAROV
where k is a wave number, moreover Imk ≥ 0, ~n (x) is a unit external normal at
the point x ∈ S, while f and g are the given continuous functions on S, and
Im kf ≥ 0 on S.
(1.2)
It should be pointed that, in particular, for f ≡ 0 we get Neumann’s external
boundary value problem for the Helmholtz equation, while for f ≡ const 6= 0 a
mixed boundary value problem for the Helmholtz equation.
2. Main result
Let ν (x, ϕ) be a simple layer acoustic potential, while w (x, ϕ) a double layer
acoustic potential, i.e.
Z
Z
∂Φk (x, y)
ϕ (y) dSy ,
ν (x, ϕ) = Φk (x, y) ϕ (y) dS, w (x, ϕ) =
∂~n (y)
S
S
eik|x−y| / (4π |x
R3 ,
where Φk (x, y) =
− y|), x, y ∈
x 6= y.
We will look for the solution of an external problem with an impedance boundary condition for the Helmholtz equation in the form
u (x) = ν (x, ϕ) + iηw (x, ν0 ) , x ∈ R3 \D,
where η is a real number, and if Imk > 0, then η = 0, if Imk = 0, then η 6= 0,
and ν0 (x, ϕ) is a simple layer potential for the Laplace equation, i.e.
Z
ν0 (x, ϕ) = ν (x, ϕ)|k=0 = Φ0 (x, y) ϕ (y) dSy .
S
It is known that (see [2]) the function u (x) satisfies the Helmholtz equation
and the Sommerfeld radiation condition at infinity, the functions Φ0 (x, y) and
ν0 (x, y) satisfy the Laplace equation. Applying Green’s second formula, we get
Z
Z
∂Φ0 (x, y)
∂ν0 (y, ϕ)
ν0 (y, ϕ)
dSy = Φ0 (x, y)
dSy , x ∈ R3 \D,
∂~n (y)
∂~n (y)
S
S
where the normal derivative on S is understood in the sense
∂ν0 (y, ϕ)
∂ν0 (y − h~n (y) , ϕ)
= lim
, y ∈ S.
h→0
∂~n (y)
∂~n (y)
h>0
Then, taking into account the limit value of the normal derivative of a simple
layer potential, we have:
Z
(Φk (x, y) − Φ0 (x, y))
w (x, ν0 ) =
ν0 (y, ϕ) dSy +
∂~n (y)
S


Z
Z
∂Φ0 (y, t)
+ Φ0 (x, y) 
ϕ (t) dSt  dSy +
∂~n (y)
S
S
Z
1
+
Φ0 (x, y) ϕ (y) dSy , x ∈ R3 \D,
2
S
ON SOLVABILITY OF AN EXTERNAL PROBLEM
5
and it means



Z
Z
∂ (Φk (x, y) − Φ0 (x, y)) 
Φ0 (y, t) ϕ (t) dSt  dSy +
u (x) = ν (x, ϕ) + iη 
∂~n (y)
S
S


Z
Z
∂Φ
(y,
t)
0
+ Φ0 (x, y) 
ϕ (t) dSt  dSy +
∂~n (y)
S
S

Z
1
Φ0 (x, y) ϕ (y) dSy  , x ∈ R3 \D.
+
2
(2.1)
S
It is easy to calculate
→ ~n (y)) (1 − ik |x − y|) eik|x−y| − 1
(−
yx,
∂ (Φk (x, y) − Φ0 (x, y))
=
,
∂~n (y)
4π |x − y|3
∂
∂ (Φk (x, y) − Φ0 (x, y))
K (x, y)
=
,
∂~n (y)
∂~n (y)
4π |x − y|5
where
Since
→ ~n (x)) ×
K (x, y) = (−
yx,
h
i
→ ~n (y)) 3 − 3ik |x − y| − k 2 |x − y|2 eik|x−y| − 3 +
× (−
xy,
h
i
−
+ (→
n (y) , ~n (x)) (1 − ik |x − y|) eik|x−y| − 1 |x − y|2 .
ik|x−y|
− 1 ≤ M |x − y|2 , ∀x, y ∈ S,
(1 − ik |x − y|) e
then
∂ (Φk (x, y) − Φ0 (x, y)) ≤ M |x − y|α ,
∂~n (y)
∂
∂ (Φk (x, y) − Φ0 (x, y)) M
∂~n (x)
≤ |x − y| .
∂~n (y)
Here and in what follows, M denotes positive constants different at different
inequalities.
Therefore,




Z
Z
∂  ∂ (Φk (x, y) − Φ0 (x, y)) 
Φ0 (y, t) ϕ (t) dSt  dSy  =
∂~n (x)
∂~n (y)
S
Z
=
∂
∂~n (x)
∂ (Φk (x, y) − Φ0 (x, y))
∂~n (y)
S


Z
Φ0 (y, t) ϕ (t) dSt  dSy , x ∈ S.

S
S
Furthermore, taking into account in equality (2.1) the limit value of the normal
derivative of a simple layer potential, we get
Z
∂u+ (x)
2 + iη
∂Φk (x, y)
=−
ϕ (x) +
ϕ (y) dSy +
∂~n (x)
4
∂~n (x)
S
6
RAHIB J. HEYDAROV

Z
+iη 
S


Z
∂ (Φk (x, y) − Φ0 (x, y)) 
Φ0 (y, t) ϕ (t) dSt  dSy +
∂~n (y)
S



Z
Z
∂Φ0 (x, y)  ∂Φ0 (y, t)
+
ϕ (t) dSt  dSy  .
(2.2)
∂~n (x)
∂~n (y)
∂
∂~n (x)
S
S
Finally, considering the limit value of double layer potential, we find
Z
+
u (x) = lim u(t) = Φk (x, y) ϕ (y) dSy +
t→x
t∈R3 \D
S



Z
Z
∂Φ
(x,
y)
0
 Φ0 (y, t) ϕ (t) dSt  dSy +
+iη 
∂~n (y)
S
S

Z
1
Φ0 (x, y) ϕ (y) dSy  , x ∈ S.
+
2
(2.3)
S
As a result, taking into account (2.2) and (2.3) in boundary condition (1.1),
we get a boundary integral equation (BIE)
Z
2 + iη
∂Φk (x, y)
−
ϕ (x) +
ϕ (y) dSy +
4
∂~n (x)
S


Z
 Φ0 (y, t) ϕ (t) dSt  dSy +
∂ (Φk (x, y) − Φ0 (x, y))
∂~n (y)
S
S



Z
Z
Z
∂Φ0 (x, y)  ∂Φ0 (y, t)


ϕ (t) dSt dSy + f (x)
Φk (x, y) ϕ (y) dSy +
+iη
∂~n (x)
∂~n (y)
S
S
S



Z
Z
Z
∂Φk (x, y) 
iη
+iη
Φ0 (y, t) ϕ (t) dSt  dSy +
Φ0 (x, y) ϕ (y) dSy  =
∂~n (y)
2
Z
+iη
∂
∂~n (x)
S
S
S
= g (x) , x ∈ S,
that may be rewritten in the operator form
ϕ + Aϕ = ψ,
(2.4)
where
ψ = −4 (2 + iη)−1 g,
A = −2 (2 + iη)−1 (2K + 2iη (T + G) + f (2L + 2iηF + iηL0 )) ,
Z
Z
(Lϕ) (x) = Φ0 (x, y) ϕ (y) dSy , (L0 ϕ) (x) = Φ0 (x, y) ϕ (y) dSy ,
S
S
Z
(Kϕ) (x) =
S
∂Φk (x, y)
ϕ (y) dSy ,
∂~n (y)
ON SOLVABILITY OF AN EXTERNAL PROBLEM
7


Z
∂Φk (x, y) 
Φ0 (y, t) ϕ (t) dSt  dSy ,
∂~n (y)
Z
(F ϕ) (x) =
S
S


Z
Z
∂Φ0 (x, y)  ∂Φ0 (y, t)
(Gϕ) (x) =
ϕ (t) dSt  dSy ,
∂~n (x)
∂~n (y)
S
S
Z
(T ϕ) (x) =
∂
∂~n (x)
∂ (Φk (x, y) − Φ0 (x, y))
∂~n (y)
×
S


Z
×
Φ0 (y, t) ϕ (t) dSt  dSy , x ∈ S.
S
Theorem 2.1. Integral equation (2.4) for an external problem with an impedance
boundary condition has a unique solution for all wave numbers with Imk ≥ 0 and
for any values of impedances f satisfying condition (1.2).
Proof. At first consider the case Imk > 0. Then η = 0, it means that equation
(2.4) takes the form
ϕ − Kϕ − f Lϕ = −2g.
(2.5)
In this case, the wave member k is not an eigen value of the Dirichlet internal
problem, and therefore equation (2.5) has a unique solution (see [2]).
Now consider the case Imk = 0 (therewith η 6= 0 and the number k 2 is real).
Since the operator A is compact, then by virtue of Riesz-Fredholm theory it
suffices to show that the homogeneous equation
ϕ + Aϕ = 0
(2.6)
has only a trivial solution ϕ = 0. Let ϕ ∈ C (S) be the solution of equation (2.6).
Since the function u is the solution of the homogeneous external problem with an
impedance boundary condition, then u = 0 in R3 \D (see [2]). By the theorem
for a step of simple and double layer potentials and by the theorem for a jump
of the normal derivative of simple and double layer potentials (see [2]), we get
u+ (x) − u− (x) = iην0 (x, ϕ) ,
∂u+ (x) ∂u− (x)
−
= −ϕ (x) , x ∈ S,
∂~n (x)
∂~n (x)
and this means
u− (x) = −iην0 (x, ϕ) ,
∂u− (x)
= ϕ (x) , x ∈ S,
∂~n (x)
where
u− (x) = lim u (t) ,
t→x
t∈D
∂u− (x)
∂u (x − h~n (x))
= lim
.
h→0
∂~n (x)
∂~n (x)
h>0
Hence
u− (x) = −iη
Z
Φ0 (x, y)
S
∂u− (y)
dSy , x ∈ S.
∂~n (y)
(2.7)
8
RAHIB J. HEYDAROV
Applying the Green first formula and considering (2.7), we find:
Z
Z
∂u− (x)
dSx =
iη ϕ (x) ν 0 (x, ϕ) dSx = u− (x)
∂~n (x)
S
S
=
Z |gradu (x)|2 − k 2 |u (x)|2 dx,
(2.8)
D
Z
Z
−iη
ϕ (x) ν0 (x, ϕ) dSx =
S
u− (x)
∂u− (x)
dSx =
∂~n (x)
S
=
Z |gradu (x)|2 − k 2 |u (x)|2 dx.
D
Hence


Z

Z
2iη
Φ0 (x, y) ϕ (y) dSy  −
ϕ (x) 
S
S


Z
−ϕ (x)  Φ0 (x, y) ϕ (y) dSy  dSx = 0,
S
this means
Z Z
Z Z
Φ0 (x, y) ϕ (x) ϕ (y) dSy dSx =
Φ0 (x, y) ϕ (x) ϕ (y) dSy dSx =
S S
S S


Z Z
=
Φ0 (x, y) ϕ (x) ϕ (y) dSy dSx .
S S
The latter means that the expression
Z Z
Φ0 (x, y) ϕ (x) ϕ (y) dSy dSx
S S
is valid. Having taken the imaginary part of equation (2.8), we get
Z Z
Φ0 (x, y) ϕ (x) ϕ (y) dSy dSx = 0.
S S
Hence we have ϕ = 0, and this completes the proof of the theorem.
ON SOLVABILITY OF AN EXTERNAL PROBLEM
9
References
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Helmholtzsche Schwingungsgleichung. Arch. Math., 16 (1965), 325-329. (German)
[2] D. Colton, R. Kress, Integral Equation Methods in Scattering Theory. Wiley, New
York, (1983); Mir, Moscow, (1987).
[3] V.D. Kupradze, Boundary Problems of the Theory of Vibrations and Integral Equations. Gosudarstv. Izdat. Tehn.-Teor. Ltt., Moscow-Leningrad, (1950). (Russian)
[4] R. Leis, Zur Dirichletschen Randwertaufgabe des Aussenraums der Schwingungsgleichung. Math. Zeit., 90 (1965), 205-211. (German)
[5] S. Muller, Zur Methode der Strahlungskapazittit von. H. Weyl. Math. Zeit., 56
(1952), 80-83. (German)
[6] I.N. Vekua, On Metaharmonic Functions. Trudy Tbilis. Mat. Inst., 12 (1943), 105174. (Russian)
[7] H. Weyl, Kapazitat von Strahlungsfeldern. Math. Zeit., 55 (1952), 187-198 (German).
Rahib J. Heydarov
Ganja State University, Ganja, Azerbaijan
E-mail address: [email protected]
Received: December 25, 2015; Accepted: February 10, 2016