Axisymmetric mixed free boundary value problem
Mixed Free Boundary Value Problem for Laplace equation in axisymmetrical case is considered. We take into consideration mean and Gauss curvatures of the free boundary. The problem of this type arise on investigating thermal equillibrium of two phases. We take into account capillary forces acting in...
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2007
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| Zitieren: | Axisymmetric mixed free boundary value problem / E. Shcherbakov // Нелинейные граничные задачи. — 2007. — Т. 17. — С. 130-150. — Бібліогр.: 12 назв. — англ. |
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| citation_txt | Axisymmetric mixed free boundary value problem / E. Shcherbakov // Нелинейные граничные задачи. — 2007. — Т. 17. — С. 130-150. — Бібліогр.: 12 назв. — англ. |
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| description | Mixed Free Boundary Value Problem for Laplace equation in axisymmetrical case is considered. We take into consideration mean and Gauss curvatures of the free boundary. The problem of this type arise on investigating thermal equillibrium of two phases. We take into account capillary forces acting in intermediate layer separating different phases. Plane model of the equillibrium without capillary forces was considered in the paper([1]). We consider variational problem whose solution is generalized solution of the boundary value problem.We prove regularity of solution, analyticity of free boundary and investigate its properties.
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130 Нелинейные граничные задачи 17, 130-150 (2007)
c©2007. E. Shcherbakov
AXISYMMETRICAL MIXED
FREE BOUNDARY VALUE PROBLEM
Mixed Free Boundary Value Problem for Laplace equation in axisymmetrical
case is considered. We take into consideration mean and Gauss curvatures of the
free boundary. The problem of this type arise on investigating thermal equillibrium
of two phases. We take into account capillary forces acting in intermediate layer
separating different phases. Plane model of the equillibrium without capillary
forces was considered in the paper([1]). We consider variational problem whose
solution is generalized solution of the boundary value problem. We prove regularity
of solution, analyticity of free boundary and investigate its properties.
Keywords and phrases: Mixed boundary value problem, mean curvature,
Gausss curvature, variational problem, free boundary
MSC (2000): 35R35
1. Boundary Value Problem.
All the domains considered in the paper are meridional sections
of axisymmetrical domains.We are searching for a domain Ω, lying
in the half stripΠ = {(x, y) |x ∈ (−1, 1) , y ∈ (−∞, 0)}which is sym-
metrical with regard to the axis y. The boundary of Ω consists of the
set Γ = {(±1, y) , y ∈ (−∞, 0)} and unknown Σ with the endponts
(±1, 0). We suppose the curve Σto be twice differentiable everywhwere
except of the set ΛΣ of the points lying on the y-axis where it possibly
posesses singularity. We are searching also for a function u = u (x, y),
u (x, y) = u (−x, y), a solution of the following equation
1
x
·
∂
∂x
(
x ·
∂u (x, y)
∂x
)
+
∂2u (x, y)
∂y2
= 0 (x, y) ∈ Ω, x > 0 (1.1)
We suppose that it satisfies the following boundary conditions
u (x, y) = 1, (x, y) ∈ Σr, (1.2)
∂u
∂ν
(±1, y) + α · u (±1, y) = 0, y ∈ (−∞, 0) , (1.3)
Axisymetric mixed free boundary value problem 131
u (∞) = 0 (1.4)
l 1
2
· |∇u|2 + κ · H(x, y) + θ · K(x, y) = λ,
(x, y) ∈ Σr
⋂
(Π − ΛΣ) , x 6= ±1.
(1.5)
Here Σr is the set of regular points of the free arcΣ for the Dirichlet
boundary value problem (see[2]). The functions H (x, y), K (x, y) re-
present values of mean and Gauss curvature of the axisymmetrical
surface S at a point (x, y) of its meridional section Ω. The letter
νdenotes external normal to the set Γ. The numbers α, κ, θ are
nonnegative real numbers. When function u satisfies Laplace equation,
κ = θ = 0, we have plane boundary value problem studied in (see[1]).
In the sequel we shall call to the problem (1.1)-(1.5) as the main
boundary value problem.
2. Variational Problem.
We use the variational method for procuring solution of described
free boundary value problem. Let
I(u, Ω) :=
∫ ∫
Ω
|∇u(x, y)|2 |x| dx dy − λ
∫ ∫
B
|x| dx dy
+ κ
∫
Σ
|x| ds + 2θ
∫
Σ
f(ẋ) ds + α
∫
Γ
u2 ds
(2.1)
Here B is complement Π \ Ω̄ to closed domain Ω̄ in the half strip Π,
z = z (s) = y (s) + ix (s) is the natural parametric representation of
curve Σ, ẋ (s)- derivative
dx
ds
, and f is the function of the following
type:
2f(t) = −
√
1 − t2 ·
t∫
0
(arcsin σ + σ
√
1 − σ2 −
π
2
)×
×(1 − σ2)−
3
2 dσ + E0
√
1 − t2.
(2.2)
Let us consider the functional I over the setDof the admissible pairs
(u, Ω) defined as follows. The boundary of the domain Ω consists
132 E. Shcherbakov
of rectifiable curves Σ = δΩ
⋂
Π, symmetrical with regard toy axis
and of the set Γ of the above mentioned type. It is clear that the
curve Σ connecting the points (±1, 0) define admissible domains. We
denote as Ξthe set of admissible curves Σ.We denote as W the class
of admissible functions u,u (x, y) = u (−x, y). These functions are
continuous in the domainΩ̄ \ΛΣ. They assume boundary value equal
to one on the curve Σ, and vanish at the infinite point. We will
suppose that the functions u (x, y) posess on the setΩ generalized
derivatives of the first order such that weighted Dirichlet integral is
bounded ∫ ∫
Ω
|∇u(x, y)|2 · |x| dx dy < ∞
Variational problem: Find minimum of the functionalI = I (u, Ω) on
the set D.
3. Symmetrization of functions and domains.
Let Ω be an admissible domain, Ωh =Ω
⋂
[(x, y) , −1 ≤ x ≤ 1,
y > −h, h > 0] and
I0 (u, Ωh) = I0
′
(u, Ωh) + I0
′′
(u, Ωh) (3.1)
I0
′
(u, Ωh) =
∫ ∫
Ωh
|∇u(x, y)|2 |x| dx dy (3.2)
I0
′′
(u, Ωh) = α ·
∫ ∫
Γh
|u(x, y)|2 ds (3.3)
h ≤ ∞. In this section we are going to study the behaviour of the
functional I0 under symmetrization of the solution of the boundary
value problem (1.1)-(1.4). We start with the following principle of
maximum (see[1]).
Lemma 3.1. Let Σ be analytic curve symmetrical with regard to
axis y and u = u (x, y) - a solution of the boundary value problem
(1.1)-(1.4). Then the functionuattains its maximum value on the
curveΣand this value is equal to one.
Axisymetric mixed free boundary value problem 133
Proof. For the function u given we can choose a number h sufficiently
large and such that the following inequality takes place |u(x,−h)| <
1
2
· sup {u (z) , z ∈ Ω}. The function u cannot assume its maximum
value inside of the domain Ω. It cannot also have the maximum value
on the line Γ. It means that the maximum value can be achieved only
on the curve Σ. It is clear that this value is equal to one. The lemma
is proved.
Let us now consider symmetrization of the function u ∈ W with
regard to the axis x. We can extend the function under consideration
as unity into the complement to the domain Ω to the domain Π. Let us
put t = x2 , y = y ν (t, y) = u
(√
t, y
)
, if ≥ 0,and u
(√
|t|, y
)
, ift <
0. Besides we put ν equal to one at the points where the function
uassumes the same value. Thus we we have that the function ν is
defined everywhere in the half-strip Π. Let
G = {(t, y, z) : (t, y) ∈ Π, z < ν (t, y)}
We have from the lemma (3.1) that
G ⊂ {(t, y, z) : (t, y) ⊂ Π, z < 1}
We denote as G (y0) a section of the domain G by the plane {(t, y, z) :
y = y0}. It is clear that δG (y0) coincides with the graph of the
functionν0 (t) := ν (t, y0). Let us now put
ρ1 (y0) = inf {ν0 (t) : |t| ≤ 1},
ρ2 (y0) = sup {ν0 (t) : |t| ≤ 1}.
We see that the intersection of δG (y0)with the line
l (y0, ρ) = {(t, y0, ρ) : |t| ≤ 1, ρ1 ≤ ρ ≤ ρ2}
consists of the even number of the points tk (ρ), 1 ≤ k ≤ 2n. Let us
denote as 2 · T (ρ) the measure of the intersection of G (y0) with the
line l0 (y0, ρ),
2 · T (ρ) = t2 (ρ) − t1 (ρ) + · · · + t2n (ρ) − t2n−1 (ρ)
It is clear that this measure is equal to 2 when ρ < ρ1 (y0) The
function T (ρ)thus defined is an increasing function.
Definition 3.1. Let ν• = ν• (t, y) be the function corresponding to
the function ν in the following way
134 E. Shcherbakov
• ν• (−t, y) = ν• (t, y) , t ∈ [0, 1],
• ν• (T (ρ)) = ρ, ρ1 (y) ≤ ρ ≤ ρ2 (y) , y ≤ y0.
We call the function u• = u• (x, y) = ν• (x2, y) , |x| ≤ 1 the sym-
metrization of the function u with regard to the axis x. The line Σ
is the level line for the function u which means that symmetrization
of the function leads to the Steiner symmetrization of the domain Ω
with regard to the axis y. We denote as Ω• the symmetrization of
the domain Ωand we select the notation u• for the symmetrization
of the function u. Now we are going to define the symmetrization of
the function u with regard to the axis x. Once again we extend this
function into the half-strip Π by the rule u (x, y) = u (x,−y) ,−∞ <
y ≤ 0. Let us define as G
′
the domain G with its reflection in the
plane (x, y). We denote as G
′
(x0)the intersection of the domain G
with the plane {x = x0}. It is clear that the boundary δG
′
(x0)is
the graph of the function u0 (x0, y) , |x0| < 1.We can transform this
graph in the same way as it was done with the graph of G (y0) of
the function ν0.As a result we get the function u•• (t, y) whose graph
will be symmetric with regard to the plane (x, z). As in preceding
case the symmetrization of the function of this type leads us to the
symmetrization of the domain Ω
⋃
{(x, y) |x| ≤ 1, y > 0, (x,−y) ∈ Ω}
- in this case with regard to the axisx. Let us study now the behaviour
of the functionalI0 under symmetrizations of the solutions of the
boundary value(1.1)-(1.4) just defined.
Theorem 3.1. Let u (x, y)be a solution of the problem (1.1)-(1.4)in
the domain Ω and u• = u• (x, y)-its symmetrization with regard to
some of its axis. Then
I0 (u•, Ω•) ≤ I0 (u, Ω) (3.4)
Proof. For the beginning we consider symmetrization with regard to
the axis y. After substitution of the variables u, ν by the variables u•,
ν• under the sign of integral I0 we arrive at the following expression
I
′
0 (u•, Ω•) = 4 ·
1∫
0
1∫
0
t
(
ν•
t
2
)
· dt · dy +
1∫
0
1∫
0
(
ν•
y
2
)
· dt · dy (3.5)
Axisymetric mixed free boundary value problem 135
Now the function T = T (ρ) is monotone one for each y0 ∈ (−∞, 0).
Hence the following representation takes place
1∫
0
(
ν•
y
)2
(t, y0) · dt =
ρ2(y0)∫
ρ1(y0)
(
δT
δy
)2
∣∣∣ δT
δρ
∣∣∣
(ρ, y0) · dρ (3.6)
This representation is a plane consequence of the second property
from the defintion 3.1(see[4]). The function u is a real analytic func-
tion. This means that the graph of the function ν0 (t) is the union of
the monotone curves. For each of this curves we can do calculations
which leads us to the formula (3.6). Thus we get as a result the
following formula
1∫
0
(νy)
2 (t, y0) dt =
ρ2(y0)∫
ρ1(y0)
2n(ρ)∑
i=1
(
δti
δy
)2
∣∣∣ δti
δρ
∣∣∣
(ρ, y0) dρ (3.7)
Now we use the Shwartz inequality to get the following result
2·n(ρ)∑
i=1
(−1)i ·
δti
δy
2
≤
2n(ρ)∑
i=1
(
δti
δy
)2
∣∣∣ δti
δρ
∣∣∣
·
2n(ρ)∑
ı=1
(−1)i ·
δti
δρ
=
=
2n(ρ)∑
i=1
(
δti
δy
)2
∣∣∣ δti
δρ
∣∣∣
∣∣∣∣
δT
δρ
∣∣∣∣
From this inequality we easily get
ρ2(y0)∫
ρ1(y0)
(
δT
δy
)2
∣∣∣ δT
δρ
∣∣∣
(ρ, y0) dρ =
ρ2(y0)∫
ρ1(y0)
(
δti
δy
)2
∣∣∣ δti
δρ
∣∣∣
(ρ, y0) dρ (3.8)
Comparing the expressions (3.6)-(3.8) we get the following result
0∫
−∞
1∫
0
(
νy
•2
)
(t, y) dt · dy ≤
0∫
−∞
1∫
0
(
νy
•2
)
(t, y) dt · dy (3.9)
136 E. Shcherbakov
We prove now that there also takes place the following inequality
0∫
−∞
1∫
0
(
νt
•2
)
(t, y) · t dt · dy ≤
0∫
−∞
1∫
0
(
νt
•2
)
(t, y) · t dt · dy (3.10)
To this end let us consider the following quiet evident inequality
2n(ρ)∑
i=1
(−1)i ti
2
≤
2n(ρ)∑
i=1
ti∣∣∣ δti
δρ
∣∣∣
·
2n(ρ)∑
i=1
(−1)i δti
δρ
=
=
2n(ρ)∑
i=1
ti∣∣∣ δti
δρ
∣∣∣
·
∣∣∣∣
δT
δρ
∣∣∣∣ .
We get from it the result
T∣∣∣ δT
δρ
∣∣∣
≤
2n(ρ)∑
i=1
ti∣∣∣ δti
δρ
∣∣∣
.
It leads us to the inequality
0∫
−∞
1∫
0
t (νt
•)2 (t, y) dt · dy ≤
0∫
−∞
ρ2(y0)∫
ρ1(y0)
T∣∣∣ δT
δρ
∣∣∣
dρ dy ≤
0∫
−∞
ρ2(y0)∫
ρ1(y0)
2n(ρ)∑
i=1
ti∣∣∣ δti
δρ
∣∣∣
dρ dy =
0∫
−∞
1∫
0
t
(
νt
2
)
(t, y) dt · dy.
It is in fact the inequality (3.10). Now uniting the inequalities (3.9)-
(3.10)we arrive at the result
Í0 (u•, Ω•) ≤ Í0 (u, Ω) (3.11)
It is now left to prove that
∫
Γ
u•2 ds ≤
∫
Γ
u2 ds (3.12)
Axisymetric mixed free boundary value problem 137
The function T (ρ) is a monototone one. Hence for each fixed y0,−∞ ≤
y0 ≤ 0, the function u• achieves its minimal value on on the line
{x = ±1}. It means that on the setΓ the values of the function u•
do not exceed the values of the function u. These arguments lead
us to the conclusion that the inequality(3.12) is valid. From the
inequalities(3.11)-(3.12)it now follows that under the symmetrization
of the function uwith regard to the axis y the inequality (3.4)takes
place. The same result also takes place for the symmetrization with
regard to the axis x. This time it will be even more easy to prove
the assertion as we will have now in the expression (3.12) equality
instead of the inequality. The theorem is proved.
4. Dirichlet Principle.
We will show now that solutions of mixed boundary value prob-
lem(1.1)-(1.4) possess extremal property usually called as Dirichlet
Principle. Let Ω be an admisssible domain with the curve Σ monotone
in each quadrant from lower half-plane. We supppose also that this
curve consists of the regular points and is symmetric with regard
to the axis y. The aim of this section consists of the proof of the
Dirichlet Principle for the solutions of the problem (1.1)–(1.4). This
problem is singular because of infinity of the domain. In the paper
[1] the principle was proved in the plane case using the method of
exhausting the infinite domain with finite ones. Dealing with our case
in the same way we prove the folowing result.
Theorem 4.1. Let Ω be an admissible domain whose boundary arc
Σ was just described.Let u be admissible function so that (u, Ω) ∈ D.
Let u0 = u0 (x, y) be a solution of the problem (1.1)–(1.4) in the
domainΩ. Then
I0 (u0, Ω) ≤ I0 (u, Ω) . (4.1)
Proof. Let Eh be the class of functions defined in Ωh with finite
integral I0 assuming in the mean the boundary value equal to unity
on the arc Σ and equal to zero on the line y = −h. Let us consider
the extremal sequence νn (h) for the functional I0(u, Ωh) defined by
the condition
lim
n→∞
I0 (νn (h) , Ωh)
138 E. Shcherbakov
inf {I0 (u, Ωh) , u ∈ Eh} = d
From parallelogram equality we get
I0 (νn (h) , Ωh) + I0 (νm (h) , Ωh)
2
= I0
(
νn (h) + νm (h)
2
, Ωh
)
+
+I0
(
νn (h) − νm (h)
2
, Ωh
)
≥ d + I0
(
νn (h) − νm (h)
2
, Ωh
) (4.2)
In accordance with the definition of the constantdwe get from (4.1)
that the sequences Í0 (νn (h) , Ωh) ,
´́
I0 (νn (h) , Ωh) are fundamental
ones. As for the domains of the considered type the inequality of
Friedrichs takes place (see [6]) than the sequence νn (h)is fundamental
in the functional space W 1,2 (Ωh). We denote by νh = νh (x, y) the
limit of this sequence. On the dislocation of the boundary of the
domain the functions of the bounded set from the space W 1,2 (Ωh)
behave themselves in equicontinuous way. It is clear that this function
satisfies the equation (1.3). Passing to the limit under the sign of the
integral we get also that the following condition takes place
lim
n→∞
0∫
−h
(νn)2 (±1, y) dy =
0∫
−h
(νh)
2 (±1, y) dy.
Let hn := −n, and Ωn,νn - sequences corresponding to hn. Now we
are going to construct a solution ν = ν (x, y) of the problem (1.1)–
(1.4)considering the sequence νnin the domain Ω =
⋃
Ωn. Let y0 be
the maximal distance from the points of the curve Σ to the axis x.
Let us consider the function ν∞ = ν∞ (x, y) defined as follows
v∞ (x, y) =
= −
∑∞
m=0
2α
λm
2·J0(λm)·[λm
2+α2]
· eλm·(y+y0) · J0 (λm · x)
(4.3)
Here J0 = J0 (t)-Bessel function of zero order and λm-solution of the
equation
λm · J ′
0 (λm) + α · J0 (λm) = 0
In the half-strip Π the function v∞ (x, y)is the unique solution of the
problem (1.1)–(1.4) The difference between the functions v∞ (x, y)−
Axisymetric mixed free boundary value problem 139
vn (x, y) ≥ 0, |x| ≤ 1, y = y0 Taking the condiion (1.3) into account
we get that the functionvn (x, y) satisfies the following condition
νn (x, y) ≤ ν∞ (x, y) , (x, y) ∈ Π (y0) (4.4)
Let us consider the difference νn−νm, m < n in the domain Ωh. From
the condition (4.4)for the points (x, y) ∈ (y = −h) we get that the
limit of this difference for mtending to infinity is equal to zero. The
difference under consideration is equal to zero on the arc Σ and on the
set Γ satisfies to the condition(1.3). It means that the sequenceνnis
fundamental in the sense of the uniform convergence in each domain
Ω (n0) = Ω
⋂
{y > −n0} , n0 ∈ N
Let ν = ν (x, y) be the limit of the sequence {νn}. It is clear that
it satisfies the equation (1.1). The class of the functions admissible
for the domain Ωh does not diminish when nincreases. It means that
the functionν = ν (x, y) satisfies the condition (1.2)at the points of
the curve Σ. The inequality (4.4) means that this function satisfies
also the condition (1.4)at the infinity. It is easy to prove that it
also satisfies the condition (1.3)on the set Γ. The above said means
that the function ν = ν (x, y) satisfies (1.1)–(1.4). It is an admissible
function for the the variational problem for the functional I0 (u, Ω).
In the usual way we prove that this functional achieves its minimal
value on this function [2]. The theorem is proved.
5. Solution of boundary value problem.
We are going to prove here that using the solutions of the problem
(1.1)–(1.4) we are able to construct a solution of the main problem.
To begin with we recall that the numberE0 can be selected arbitrarily
large. This permits us to prove the following result.
Lemma 5.1. For arbitrary non negative values of λ,κ, θthere exists
a numberE0such that the values of the functional I (u, Ω) are non
negative on the set of admissible pairs (u, Ω) ∈ D.
Proof. Let Q (u, Ω) be the following expression
Q (u, Ω) := −λ ·
∫ ∫
B
|x| · dx · dy + θ ·
∫
Σ
f (ẋ) · ds (5.1)
140 E. Shcherbakov
In accordance with the formula (2.2) we get for the values of E0
sufficiently large the following inequality
Q ≥ −λ · θ − θ · ρ ·
1∫
0
(
arcsin σ + σ ·
√
1 − σ2
)
×
× (1 − σ2)
− 3
2 dσ + θ · E0 · ρ ≥ 0
(5.2)
Here ρ is the maximal distance from the points of the arc Σto the
axis x. It is clear that the functional I (u, Ω)−Q (u, Ω) assumes non-
negative values over the set of the admissible functions. Now we are
ready to prove the main result.
Theorem 5.1. Let the number c0 > 0, satisfies condition κ−c0θ > 0
Then for all nonnegative α, λ there exists an admissible pair (ue, Ωe) ∈
D, such that
I (ue, Ωe) = inf {I (u, Ω) , (u, Ω) ∈ D}
The curve Σe corresponding to the domain Ω is piece-wise analytic
and it is nondecreasing for the points with x > 0 The function ue is
a solution of the main problem.
Proof. Let{en} be a minmizing sequence en = (un, Ωn) for the variati-
onal problem from the section 2. Without restriction we may assume
that the curves Σn = δΩn
⋂
Π are piece-wise analytic ones consisting
of the regular points(see [7]). The functional Q behaves monotonically
under symmetrization of the functions and domains defined before.
The functional I behaves itself in the same way. It means that we
may assume monotone behaviour of the curves Σnin each quadrant.
Using Dirichlet principle we can also assume that the functions un
are the solutions of the problem (1.1)–(1.4) in the domain Ωn. In
accordance with the lemma 5.1 we can asssume that for the numbers
E0 sufficiently large the functional I is non negative over the set
of admissible functions. It means that the set of ordinates of the
points of intersection of the curves Σn with y - axis is bounded.
It follows from the Helly theorem that the sequence of the curves
Σn converges to the limit curve Σe. Besides we get convergence of
the domains Ωn to the domain Ωe as to the kernel(see [9]). The
Axisymetric mixed free boundary value problem 141
functions un are the solutions of the equation(1.4). It means that we
can assume that they are traces of the harmonic functions defined
over the domainsΩ•
n obtained by their rotation about y-axis. From
this it follows that the sequence of the functions unis compact in the
sense of the uniform convergence inside of the domain Ωe. We leave
for the convergent subsequence the notation of the proper sequence.
Hence the sequence{un}is convergent inside of the domain Ωe. Let
us denoteue := lim
n→∞
un (x, y). The functions unare limited by the
function ν∞. It means that the limit function ueassumes at the infinity
the value equal to zero. Let us now consider the behaviour of the
function ue on the set Σe
⋃
Γe. It was said earlier that the functions
from the limited set of the Sobolev space are equicontinuous in the
mean with regard to the shift of the boundary. The arcs of the set
Σe−{x = 0} satisfy Lipshitz condition. It means that the function ue
assumes in the mean the value equal to one on the arc Σ(see[6],[10]).
The points of the set Σe\{x = 0} are regular ones. Hence the function
ue assumes on Σe the boundary value equal to one with possibl
eexeption of the points lying on the y-axis. We can assume that the
functions un are extended across analytic set Γ. Passing to the limit
on Γ we get that the limit function is a solution of equations(1.1)–
(1.4). We will show now that the function ue satisfies almost every-
where on Σe the boundary condition (1.5). The functional I (u, Ω)
is semicontinuous from below on the set D (see [4],[7]), whence it
follows that it attains its minimal value on the pair (ue, Ωe) ∈ D.
Let z0 = y0 + i · x0, 0 < x0 < 1-be any point on the curveΣesuch
that tangential line exists at this point. Let z• = z + ǫ · F be a
transformation defined by the infinitely differentiable function F with
support lying in the disk B (z0, δ). For the values ǫ > 0 sufficiently
small the mapping z• (z) is the topological one. The Bjorke theorem
(see [10]) permits us to consider the function F as the extension
on B (z0, δ) of the finite infinitely differentiable function given on
Σδ
e = Σe ∩B (z0, δ). The necessary condition for the element (ue, Ωe)
142 E. Shcherbakov
to be extremal can be written in the form
0 =
|Σ(ǫ)|∫
0
[(
2 · i · λ · x · ¯̇z + 4 · i · u2
ezx −
κ
i
)
F
]
ds
−
|Σ(ǫ)|∫
0
[
(κ · x · ¯̇z − 2 · i · θ · fẋ · +ẏ · ¯̇z + 2 · θ · f · ¯̇z)
dF
ds
]
ds
(5.3)
Here
z = y + i · x, uez = 2−1 ·
(
δue
δy
− i
δue
δx
)
and |σ (ǫ)| is the length of the curve Σǫ
e = Σe
⋂
{|z − z0| < ǫ} . The
condition (see(5.2)), is the condition of existence of generalized deri-
vative for the function κ ·x · ¯̇z−2 · i ·θ · fẋ · ẏ · ¯̇z+2 ·θ · f · ¯̇z ∈ L2 ([0, |Σǫ|]).
Let Φ1 = Φ1 (s) , Φ2 = Φ2 (s) are the following functions
Φ1 (s) = κ · x · ẏ − 2θ · fẋ · ẋ · ẏ + 2θ · f · ẏ, (5.4)
Φ2 (s) = −κ · x · ẋ − 2θ · fẋ · ẏ
2 − 2θ · f · ẋ (5.5)
From above said it follows that they are absolutely continuous. The
following equations are the consequences of the conditions
G (ẏ, Φ1, Φ2, y) = 0, G = ẏ ·Φ1 −
√
1 − ẏ2 ·Φ2 − 2 · θ · f − κ · y (5.6)
H (ẋ, Φ1, Φ2) = 0, H = ẋ ·Φ1+
√
1 − ẋ2 ·Φ2+2 ·θ · fẋ ·
√
1 − ẋ2, (5.7)
It is easy to show that these equations are solvable in ẋ, ẏ for the
values of the parameterE0 sufficiently large. We get also that ẋ =
Ψ1 (Φ1, Φ2), ẏ = Ψ (Φ1Φ2) for some differentiable functions Ψ1, Ψ2.
The functions Φ1, Φ2, y are absolutely continuous as the functions of
the parameter s of natural parametrization of the curve Σe. It means
that the functions ẋ, ẏ are differentiable almost everywhere on Σǫ
e.
Taking into account the above-mentioned results we get from (5.3)
the equality
−i · κ · ż · x · k (z) + κ · ẋ · ¯̇z + 2 · ı θ · ẍ · ¯̇z = (5.8)
−2 · i · λ · x · ¯̇z − 4 · i ·
(
δue
δz
)2
· ż +
κ
i
(5.9)
Axisymetric mixed free boundary value problem 143
Here k (z)-curvature of the curve Σǫ
e. In the axially symmetric case
we have
2 · H (z) = k (z) +
(
δue
δz
)
· (x · |∇ue|)
−1
(5.10)
K (z) = −
ẍ
x
. (5.11)
Using (5.6)-(5.8) we get that the function ue satisfies almost every-
where on the curve Σe boundary condition (1.5). Using (5.6)–(5.8)
we get that the function ue satisfies almost everywhere on the curve
Σe boundary condition (1.5).
Using a‘priory estimates for |∇ue| from the paper [2] we get that
the curve Σe
⋂
{x > 0} is a Liapunov curve. From Shauder estimates
it follows that this curve is infinitely differentiable one. In the usual
way (see [4], [7]) we prove that the curve Σ consists of the analytic
arcs.
The theorem is proved.
6. Free boundary.
In this part we investigate the contact of free boundary with the
lines {x = ±1} and {y = 0}. Let us begin with the set
{
Σ̄ ∩ {x = 1}
}
.
We shall prove for the first the following lemma which is the simple
generalization of the result proved in ([1]).
Lemma 6.1 Let b be the length of the segment Σ̄ ∩ {x = 1} and
{un}-the sequence of the functions from extremal sequence {un, Ωn}.
Then
lim
n→∞
0∫
−b
un (1, y)dy = b (6.1)
Proof. Let us consider the parts Σn = δΩn ∩Π lying in the half space
{y > −b + ǫ0}. Then from assumption of the lemma it follows that
they are contained in the rectangular
Πǫ = (−b + ǫ0 < y < 0 ) × (1 − ǫ, 1) , 0 < ǫ < 1, 0 < ǫ0 < b
144 E. Shcherbakov
Let us compare the functions un, n > n (ǫ, ǫ0), in the rectangular
Πǫwith the functionuǫwhich is a solution of (1.1) and satisfies the
following conditions
uǫ (x, b − ǫ) = 0 = uǫ (x, 0) , 1 − ǫ < x < 1 (6.2)
uǫ (1 − ǫ, y) = 1,−b + ǫ0 < y < 0 (6.3)
duε
dx
+ α · uε (1, y) = 0,−b + ε0 < y < 0 (6.4)
The function uεcan be represented in the following form
uε (x, y) =
∑
m≥0
Am · Vm (x, ε) · sinm
π
−b + ε0
· y (6.5)
Vm (x, y) = −
α
m
· Y 1
m (x, ε) + Y2
m (x, ε) (6.6)
y′′ + x−1 · y′−m2 · y = 0 (6.7)
Y 1
m (1, ε) = 0,
(
Y 1
m
)′
(1, ε) = 1 (6.8)
Y 2
m (1, ε) = 1,
(
Y 2
m
)′
(1, ε) = 0 (6.9)
The functions Y k
m, k = 1, 2, are constructed as linear combinations of
the functions l0 (mx) , K0 (mx), constituting the fundamental system
for equation (6.6) (see [11]). It is can be easily verified that
Y 1
m (x) =
−K0 (m) · l0 (m · x) + l0 (m) · K0 (m · x)
m · l0 (m) · K ′
0 (m) − l′0 (m) · K0 (m)
(6.10)
Y 2
m (x) =
−K ′
0 (m) · l0 (m · x) − l0 (m) · K0 (m · x)
l0 (m) · K ′
0 (m) − l′0 (m) · K0 (m)
(6.11)
Using integral representations for the functions l0 (mx) , K0 (mx)
(see [11]) we obtain the following estimate for the functions Vm
Vm (1 − ε, ε0) >
m
4
· eε·m
2 (6.12)
The functions uε from (6.5) evidently satisfy the condition (6.2). Let
us take coefficients Am in the form
Am =
4
π · m · Vm (1 − ε, ε0)
,
Axisymetric mixed free boundary value problem 145
Than we get that the function uεsatisfies also the condition (6.3).
The estimate (6.12) is not exact but it permits us to differentiate the
series (6.5)by terms. This means that the function uε satisfies the
condition (1.1). Thus we have that the functionuε is a solution of the
problem (1.1),(6.2)–(6.4). Now we evidently have
0∫
−b+ǫ0
uǫ (1, y) dy =
m0∫
−b+ǫ0
0
∑
m=0
1
4 · π · Vm (1 − ǫ, ǫ0)
· sin
π
−b + ǫ0
y · dy+
+
∑
m≥m0
4 (−b + ǫ0)
π · m2 · Vm (1 − ǫ, ǫ0)
· [1 − (−1)m]
For a given positive number ǫ0 we can choose the numbers m0,
n (ǫ0) sufficiently large and a positive number ǫ sufficiently small such
that for n > n (ǫ0) we get the inequality
0 ≤ −
0∫
−b+ǫ0
un (1, y) · dy + b ≤ 2 · ǫ0
Tending n to infinity we get the result we need. The lemma is proved.
Theorem 6.1. Let (u, Ω) be a solution of variational problem from
the point 2. Let us assume that the inequalityλ < α takes place.Then
for the arc Σeof the boundary Ωeconnecting the points (−1, 0) , (1, 0)we
have
Σ̄e ∩ {x = ±1} = {(−1, 0) , (1, 0)} (6.13)
Proof. Let us suppose that the theorem is not correct. In this case the
number b from the precedent lemma is positive. Using Green theorem
we get
∫ ∫
Ωe
|x| · |∇u|2 dx dy + α ·
∫
Γ
u2
e dy = α ·
∫
Γ
ue dy (6.14)
Let us consider the pair (u•, Ω•) for which u• (x, y) = u (x, y − b)and
Ω•is the domain obtained by the dislocation by the number b of the
146 E. Shcherbakov
domain Ω in the positive direction of the axis y. The pair (u•, Ω•),
evidently, belongs to the set D. We denote as Σ• the part of the
boundary of Ω• lying in the closure of the half-strip Π and connecting
the points (−1, 0) , (1, 0) . Taking (6.13) into account we get
I0 (u•, Ω•) = I0 (u, Ω) + 2 · b · (λ − α) −
b2
2
(6.15)
∫
Σ•
|x| · ds <
∫
Σ
|x| · ds (6.16)
|l•|∫
0
f (ẋ) · ds =
|l|∫
0
f (ẋ) · ds (6.17)
Here l,l•are the lengths of the curves Σ,Σ•respectively. From the
conditions (6.13)–(6.16) it follows that
I (u•, Ω•) < I (u, Ω) .
The theorem is proved.
We will show now that the curve Σe = ∂Ω
⋂
Π̄ for extremal pair
(ue, Ωe) does not have points on the line {y = 0} for the exception of
its end points.
Lemma 6.2. Let u0 = u0 (x, y) be a solution of the problem (1.1)-
(1.4) in the half strip Π. Then there exist a number М such that
for a sufficiently small neighbourhood of the point (1, 0) the following
inequality takes place
u0
y (x, y) ≥ M · ln
[
1 +
1
α · (1 − x)
]
(6.18)
Proof. Without any difficulties we get for the function u0 the following
representation
u0 (x, y) = 1 +
2 · α
π
·
0∫
−∞
I0 (µ · x) · sin µ · y
µ · [µ · I1 (µ) + α · I0 (µ)]
rmdµ (6.19)
Axisymetric mixed free boundary value problem 147
Here I0, I1-Bessel functions of purely imaginary argument. The pos-
sibility of such representation follows from asymptotic estimate (see
[12])
In (µ) =
eµ
√
2π · µ
·
[
1 + O
(
µ
−1
2
)]
, µ → ∞ (6.20)
It follows from (6.19) that
|u0
y(x, 0)| >
2α
π
·
N1∫
0
I0 (µ · x)
[µ · I ′
0 (µ) + α · I0 (µ)]
dµ
+ 3−1 ·
(1−x)−1)∫
N1
eµ·(x−1)
µ + α
dµ
(6.21)
The number N1 in the condition (6.21) is selected in such way that
for the function O
(
µ
−1
2
)
from (6.20) takes place inequality
∣∣∣O
(
µ
−1
2
)∣∣∣ <
1
2
.
From inequality (6.21) we now get
|u0
y(x, 0)| >
2α
π
·
N1∫
0
I0(µx)
[µI1(µ) + αI0(µ)]
dµ
+3−1e−1 · ln
[
1
1 − x
+ α
]
− 3−1e−1 · ln
[
1
N1
+ α
] (6.22)
The inequality (6.22) implies the result we need. The lemma is proved
Lemma 6.3. Let (u, Ω) be a solution of variational problem from
section 2 such that Σ ∩ (−1, 1) 6= ∅. Then the function uy tends to
infinity logarithmically when x tends to unit.
Proof. Let us extend the function u as unity to the half strip Π. Let us
consider the function w := u− u0 in the half strip Π. We can extend
the function w as an odd function across the axis x. From assumption
of the lemma it follows that there exists a positive number such that
148 E. Shcherbakov
Σ ∩ (0, 1) = [c, 1) , c < 1. We consider now the reduction of the
extended function w to the set
Πc = {(x, y) |0 < c < 1 ,−∞ < y < ∞} .
Let us consider sine transform
ws (c, µ) :=
0∫
−∞
w (c, y) · sin µ · ydy
of the function w (c, y). We can easily verify that for the function w
in the strip Πc takes place the following representation.
w (x, y) = 2
π
·
0∫
−∞
ws (c, µ)×
×µ·I0[2µ·(1−x)]+α·I1[2µ·(1−x)]
µ·I0[2µ·(1−c)]+α·I1[2µ·(1−c)]
· sin µ · ydµ
(6.23)
From the representation (6.23) it follows that that the function
wy (x, 0) is bounded in the neighbourhood of the point (1, 0). It
means that the function uy (x, o) jointly with the function u0
y (x, 0)
logarithmically tends to the infinity. The lemma is proved.
Theorem 6.2. Let (ue, Ωe) be a solution of variational problem from
section 2. Let us suppose that for the numbers λ, α takes place inequali-
ty λ2 < α. Then the number c, [c, 1) = Σe ∩ (0, 1), is equal to unit.
Proof. From the theorem 5.1 it follows that the curve Σe consists of
the analytic curves. Let us consider variation δI under local variation
δx̃ of the boundary
δI (Ωe, ue, δ~x, δu) =
∫
Σe
[
λ − |∇ue|
2] · δ~x · ~ν · ds+
+κ · δ
∫
Σe
|x| · ds + θ · δ
∫
Σe
f (ẋ) · ds
(6.24)
We can select as δx̃ the function whose graph for x > 0 represents a
step with its basis of the length
∣∣∣ln
−1
2 (1 − x0)
∣∣∣
Axisymetric mixed free boundary value problem 149
centered at the point x0 and of the same height. In this case we haveh
δ
∫
Σe
|x| ds + δ
∫
Σe
f (ẋ) ds = O
(∣∣∣ln
−1
2 (1 − x0)
∣∣∣
)
, x0 → 1 (6.25)
From lemma 6.3 we get
∫
Σe
[
λ2 − |∇ue|
2]·δx̃·~ν · ds
x
+κ·δ
∫
Σe
|x|· ds+θ·δ
∫
Σe
f (ẋ)· ds < 0 (6.26)
It means that
δI (Ωe, ue,δx̃,δu) < 0
for the points x0 sufficiently near to one. We denote by the letter Ω̃
the domain obtained from Ωe with the help of displacement δx̃. Let
ũ be a solution of boundary problem (1.1)-(1.4) in the domain Ω̃. On
the basis of inequality (6.26) and Dirichlet principle we now get
I
(
ũ, Ω̃
)
< I (ue, Ωe) (6.27)
Let us symmetrize the function ũ and domain Ω̃ in order to axis y.
We denote as Ũ the symmetrization of the function ũ and through
Ω• the symmetrization of the domain Ω̃ in regard to the axis y. Then
x0∫
−∞
U2 (1, y) · dy ≤
0∫
−∞
ũ2 (1, y) · dy (6.28)
Consider now the function Ú (x, y) = U (x, y + h) and domain Ώ
obtained from domain Ω• by dislocation in the negative direction
of the axis ydefined by the numberh. The pair
(
ú, Ώ
)
is admissible
for the extremal problem from the part 2. Using inequalities (6.27)-
(6.28)we arrive at the condition
I
(
Ú , Ώ
)
< I (ue, Ωe) + 2 · (α − λ) · ln
−1
2 |(1 − x0)| (6.29)
The function
[
λ − |∇ue|
2] · δx̃ is negative and is of the order
ln− 1
2 |(1 − x0)| when x0 tends to 1. The length of the supporter of δx̃
150 E. Shcherbakov
is equal to ln− 1
2 |(1 − x0)|. It means that for the points x0 sufficiently
near the unity we get on the basis (6.29) the inequality
I
(
Ú , Ώ
)
< I (ue, Ωe) (6.30)
As it was already said the pair
(
Ú Ώ
)
is admissible. Thus the inequali-
ty (6.30)means that the assumption c < 1 leads us to the contradiction.
The theorem is proved.
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of Thermal Equilibrium with a Free Boundary // Amer. Math. Soc. Transl.,
(2)126, (1985), 77-92 .
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Order, Springer Verlag, Berlin, Heidelberg, 1983.
3. Polya G.and Szego G. Isoperimetric Inequalities in Mathematical Physics,
Princeton, Univ.Press,1951.
4. Garabedian P.R., Lewy H. and Shiffer M. Axially Symmetric Сavitational
Flow // Annals of Mathematics, 56(1952), 560-602.
5. Garabedian P.R. Partial Differential Equations, New York-London-Sydney,
J.Willey, Inc., 1964.
6. Michlin S.G. Linear Partial Differential Equations, Moscow, Vischaya Shkola,
1977(in Russian).
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with Surface Tension and Wedging Forces, // Zeitschrift fur Analysis und ihre
Anwendungen,17(4), (1998), 937-961.
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Theory of Functions, Moscow, Izdatelstvo Fiziko-matematicheskoi Literatury,
1972 (in Russian).
9. Suvorov G.D. Families of Topological Mappings, Novosibirsk,1965 (in Russian).
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Kuban State University
149 Stavropolskaya st.,
350040, Krasnodar, Russian Federation
Received 1.03.07
|
| id | nasplib_isofts_kiev_ua-123456789-10114 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 0236-0497 |
| language | English |
| last_indexed | 2025-12-02T02:07:46Z |
| publishDate | 2007 |
| publisher | Інститут прикладної математики і механіки НАН України |
| record_format | dspace |
| spelling | Shcherbakov, E. 2010-07-23T14:25:06Z 2010-07-23T14:25:06Z 2007 Axisymmetric mixed free boundary value problem / E. Shcherbakov // Нелинейные граничные задачи. — 2007. — Т. 17. — С. 130-150. — Бібліогр.: 12 назв. — англ. 0236-0497 https://nasplib.isofts.kiev.ua/handle/123456789/10114 Mixed Free Boundary Value Problem for Laplace equation in axisymmetrical case is considered. We take into consideration mean and Gauss curvatures of the free boundary. The problem of this type arise on investigating thermal equillibrium of two phases. We take into account capillary forces acting in intermediate layer separating different phases. Plane model of the equillibrium without capillary forces was considered in the paper([1]). We consider variational problem whose solution is generalized solution of the boundary value problem.We prove regularity of solution, analyticity of free boundary and investigate its properties. en Інститут прикладної математики і механіки НАН України Axisymmetric mixed free boundary value problem Article published earlier |
| spellingShingle | Axisymmetric mixed free boundary value problem Shcherbakov, E. |
| title | Axisymmetric mixed free boundary value problem |
| title_full | Axisymmetric mixed free boundary value problem |
| title_fullStr | Axisymmetric mixed free boundary value problem |
| title_full_unstemmed | Axisymmetric mixed free boundary value problem |
| title_short | Axisymmetric mixed free boundary value problem |
| title_sort | axisymmetric mixed free boundary value problem |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/10114 |
| work_keys_str_mv | AT shcherbakove axisymmetricmixedfreeboundaryvalueproblem |