Mem. Differential Equations Math. Phys. 46 (2009), 157–160
J. Kapanadze
ON SOLVABILITY OF THE PROBLEM WITH A
DIRECTIONAL DERIVATIVE FOR THE EQUATION ∆3 v = 0
Abstract. We consider the problem of directional derivative for the equation ∆3 v = 0 when the direction of the derivative belongs to the tangent
plane. It is proved that if the boundary functions belong to a certain class,
then the problem has a solution.
!
"
∆3 v = 0
2000 Mathematics Subject Classification: 35J40.
Key words and phrases: Problem with a directional derivative, volume
and single-layer potentials, Dirichlet problem.
The investigation of the problem with a directional derivative in an ndimensional domain for n ≥ 3 has been initiated in [1], [2] and continued in
[3]–[8]. In this note we treat the problem for the equation ∆3 v = 0.
Let us introduce the necessary notation. Let Ω be a bounded smooth
domain belonging to the class C (5,α) , `x be a smooth direction at x ∈ ∂Ω,
`x ∈ C (4,α) , |`x | = 1. For simplicity of presentation, we will consider the
three-dimensional case.
The volume potential and the single-layer potential will be denoted as
follows (see [9]):
Z
Z
1
1
ψ(y)
µ(y)
V µ (x) =
dy; U ψ (x) =
dSy,
4π
|x − y|
4π
|x − y|
Ω
∂Ω
where ∂Ω is the boundary of Ω, µ ∈ L1 (Ω), ψ ∈ L1 (∂Ω). If G is the Green
function for the Dirichlet problem for Ω, then the solution to the Dirichlet
problem has the form
Z
∂G(x, y)
ϕ(y) dSy, x ∈ Ω, ϕ ∈ C(∂Ω),
v(x) = −
∂νy
∂Ω
where νx is the outer normal at x ∈ ∂Ω, |νx | = 1.
The Green potential is denoted by
Z
VGf (x) = G(x, y)f (y) dy, f ∈ L1 (Ω).
Ω
158
The density of a balayaged measure µ ∈ C(Ω) (see [9]) in the volume potential case is defined as the linear operator
Z
∂G(x, y)
µ(y) dy, y ∈ ∂Ω.
T µ(y) = µ0 (y) = −
∂νy
Ω
It is well-known ([9], p. 260) that
0
V µ (x) = U µ (x), x ∈ R3 \ Ω.
Denote
n
o
E = x : (νx , `x ) = cos(νxb`x ) = 0, x ∈ ∂Ω .
We always assume that E consists of a finite number of closed curves. Denote by Γ(x, y) the Newton kernel (4π|x − y|)−1 .
In the sequel we use the following auxiliary equality [10]
∂G(x, y)
∂G(x, y)
= cos(νxb`x )
, x ∈ ∂Ω \ E, y ∈ Ω.
∂`x
∂νx
In the proof of the main assertion the following lemma will be used.
(1)
Lemma. Let µ ∈ C α (Ω), Ω ∈ C (2,α) . Then the following equality holds
∂ 2 Viµ (x) ∂ 2 Veµ
−
= −µ(x) cos2 (νxb`x ), x ∈ ∂Ω \ E.
∂`2x
∂`2x
The outwards and inwards limits for V µ are considered.
(2)
Proof. Let x0 ∈ ∂Ω − E. Obviously
3
∂V µ (x) X ∂V µ (x) 0
αk , `x0 = `0 = (α01 , α02 , α03 ).
=
∂`x0
∂xk
k=1
It is easy to see that
∂ 2 V µ (x) X ∂ 2 V µ (x) 0 0
=
α α .
∂`20
∂xj ∂xk j k
Consider the difference
3 2 µ
µ
X
∂ 2 Viµ (x) ∂ 2 Veµ (x)
∂ Vi (x) ∂ 2 V` (x) 0 0
αj αk .
−
=
−
∂`20
∂`20
∂xj ∂xk
∂xj ∂xk
(3)
j,k=1
Let us now use the following relation ([11], p. 115)
∂ 2 Viµ (x) ∂ 2 Veµ (x)
−
= −µ(x)νj νk , ν = (ν1 , ν2 , ν3 ).
∂xj ∂xk
∂xj ∂xk
(4)
(4) along with (3) gives (ν0 = νx0 )
3
X
∂ 2 Viµ (x0 ) ∂ 2 Veµ (x0 )
−
=
−µ(x
)
αj αk νj νk = −µ(x) cos2 (ν0b`0 ).
0
∂`20
∂`20
j,k=1
The lemma is proved.
Let us proceed to the main assertion of this note.
159
Theorem. There exists a solution to the equation ∆3 v = 0 in the domain Ω belonging to the space C (4,α) (Ω) and satisfying the following boundary conditions
v = ϕ0 , ϕ0 ∈ C (5,α) (∂Ω), H0 = ϕ0 , ∆H0 = 0,
∂Ω
∂Ω
∂H0 (x)
∂v(x)
=
+ ϕ1 (x)(νx , `x ), ϕ1 ∈ C (4,α) , x ∈ ∂Ω \ E,
∂`x
∂`x
∂ 2 v(x)
∂ 2 H0 (x)
=
+ ϕ2 (x)(νx , `x )2 +
2
∂`x
∂`2x
∂ 2 Uiϕ1 (x) ∂ 2 Ueϕ1 (x)
+
−
, ϕ2 ∈ C (3,α) , x ∈ ∂Ω \ E.
∂`2x
∂`2x
(5)
Proof. The solution v ∈ C (4,α) has the form
Z
Z
Z
v(x) = H0 (x) − G(x, y)H1 (y) dy + G(x, y) G(y, z)H2 (z) dz dy,
Ω
Ω
Ω
where H0 , H1 and H2 are harmonic functions such that ∆v(x) = H1 (x),
∆2 v(x) = H2 (x), x ∈ ∂Ω. In order to find H1 , let us rewrite the solution as
follows
Z
Z
v(x) = H0 (x)− G(x, y)Φ1 (y) dy, Φ1 (y) = H1 (y)− G(x, y)H2 (z) dz.
Ω
Ω
Due to the second boundary condition, we get
∂H0 (x)
∂H0 (x)
+ cos(νxb`x )Φ0 (x) =
+ ϕ1 (x) cos(νxb`x ).
∂`x
∂`x
Therefore, Φ01 (x) = ϕ1 (x) x ∈ ∂Ω (Φ1 (x) = H1 (x) x ∈ ∂Ω).
The third boundary condition also gives
∂ 2 Uiϕ1 (x) ∂ 2 Ueϕ1 (x)
∂ 2 H0
2
+
Φ(x)
cos
(ν
b`
)
+
−
=
x
x
∂`2x
∂`2x
∂`2x
∂ 2 H0 (x)
∂ 2 Uiϕ1 (x) ∂ 2 Ueϕ1 (x)
2
=
+
ϕ
(x)
cos
(ν
b`
)
+
−
.
2
x
x
∂`2x
∂`2x
∂`2x
Hence, Φ(x) = H1 (x) = ϕ2 (x), x ∈ ∂Ω.
In order to define H2 , we apply again the second boundary condition
∂H0 (x)
∂v(x)
=
+ ϕ1 cos(νxb`x ) =
∂`x
∂`x
∂H0 (x)
=
+ H10 (x) cos(νxb`x ) + (VGh2 )0 cos(νxb`x ).
∂`x
Hence, ϕ1 (x) − H10 (x) = (VGH2 (x))0 x ∈ ∂Ω.
Therefore,
Z
Z
ϕ(x)dSx
VGH2 (x)
dx =
, ϕ = ϕ1 − H10 ∈ C (4,α) , x ∈ R3 − Ω.
|x − y|
|x − y|
Ω
∂Ω
160
Due to the equality
Z
Z
VGH2 (y)dy
V H2 (y)dy ϕ(y)dSy
,
=
, x ∈ R3 \ Ω v1 (x) = G
|x − y|
|x − y|
|x − y|
Ω
∂Ω
we get the Dirichlet problem in the domain Ω for the equation ∆3 v1 = 0
v1 (x) = U ϕ (x), x ∈ ∂Ω,
∂U`ϕ (x)
∂v1 (x)
=
, x ∈ ∂Ω,
∂νx
∂νx
∂ 2 U`ϕ (x)
∂ 2 v1 (x)
=
, x ∈ ∂Ω.
∂νx2
∂νx2
Now H2 can easily be found by solving the Dirichet problem (∆2 v1 =
H2 ).
References
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(Received 18.05.2006)
Author’s address:
M. Nodia Institute of Geophysics
1, Aleksidze St., 0193 Tbilisi
Georgia