The Stefan problem
The Stefan problem in its classical statement is a mathematical model of the process of propagation of heat in a medium with di erent phase states, e.g., in a medium with liquid and solid phases.The process of propagation of heat in each phase is described by the parabolic equations. In this work we...
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Інститут прикладної математики і механіки НАН України
2008
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| Цитувати: | The Stefan problem / M.A. Borodin // Нелинейные граничные задачи. — 2008. — Т. 18. — С. 218-229. — Бібліогр.: 2 назв. — англ. |
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Digital Library of Periodicals of National Academy of Sciences of Ukraine| _version_ | 1860245312163020800 |
|---|---|
| author | Borodin, M.A. |
| author_facet | Borodin, M.A. |
| citation_txt | The Stefan problem / M.A. Borodin // Нелинейные граничные задачи. — 2008. — Т. 18. — С. 218-229. — Бібліогр.: 2 назв. — англ. |
| collection | DSpace DC |
| description | The Stefan problem in its classical statement is a mathematical model of the process of propagation of heat in a medium with di erent phase states, e.g., in a medium with liquid and solid phases.The process of propagation of heat in each phase is described by the parabolic equations. In this work we prove the existence of the global classical solution in a two-phase multidimensional Stefan problem. We apply a method which consists of the following. First, we construct approximating problems, then we prove some uniform estimates and pass to the limit.
|
| first_indexed | 2025-12-07T18:35:56Z |
| format | Article |
| fulltext |
218 Нелинейные граничные задачи 18, 218-229 (2008)
c©2008. M. A. Borodin
THE STEFAN PROBLEM
The Stefan problem in its classical statement is a mathematical model of
the process of propagation of heat in a medium with different phase states, e.g.,
in a medium with liquid and solid phases.The process of propagation of heat in
each phase is described by the parabolic equations. In this work we prove the
existence of the global classical solution in a two-phase multidimensional Stefan
problem. We apply a method which consists of the following. First, we construct
approximating problems, then we prove some uniform estimates and pass to the
limit.
Keywords and phrases: free boundary problem, global classical solution,
the Stefan problem
MSC (2000): 35R35
1. Statement of the problem.
The Stefan problem in its classical statement is a mathematical
model of the process of propagation of heat in a medium with different
phase states, e.g., in a medium with liquid and solid phases. As
a result of melting or crystallization, the domains occupied by the
liquid and solid phases undergo certain changes. This unknown interface
is called a free boundary. The process of propagation of heat in each
phase is described by the heat equation.
Let D ∈ R
3 - the bounded domain , which boundary consists
of two C2+α surfaces ∂D1 and ∂D2 such, that ∂D1 contains inside
∂D2 and limits domain D, DT = D × (0, T ). The problem is to
find a function u(x, t) and domains ΩT , GT , satisfying to following
conditions:
∆u− a(u)
∂u
∂t
= 0 in ΩT ∪GT , (1.1)
ΩT ={(x, t) ∈ DT : 0 < u(x, t) < 1}, GT ={(x, t) ∈ DT : u(x, t) > 1},
where a(u) - step function a1 > 0 в ΩT and a2 > 0 in GT ; on the
known boundary ∂DT
u(x, t) = 0 on ∂D1 × (0, T ), u(x, t) = ϕ(x, t) > 1 on ∂D2 × (0, T );
(1.2)
The Stefan problem 219
on the unknown (free) boundary γT = DT ∩ ∂ΩT = DT ∩ ∂GT
u−(x, t) = u+(x, t) = 1,
3
∑
i=1
(∂u−
∂xi
− ∂u+
∂xi
)
cos(n, xi)+λ cos(n, t) = 0;
(1.3)
where λ is a positive constant, n is the normal to the surface γT
directed to the side of increase of the function u(x, t); u+(x, t), u−(x, t)
are the boundary values on the surface γT taken from the domains
GT , ΩT respectively. The function u(x, t) is interpreted as the tempe-
rature of the medium, γt is the interface between the liquid and solid
phases, and u(x, t) = 1 is the temperature of melting. The initial
conditions are
u(x, 0) = ψ(x) > 0 in D, ψ(x)
∣
∣
∣
∂D1
= 0,
ψ(x)
∣
∣
∣
∂D2
= ϕ(x, 0) > 1,
Ω0 = {x ∈ D : 0 < ψ(x) < 1},
G0 = {x ∈ D : ψ(x) > 1}, γ0 = D ∩ ∂Ω0.
(1.4)
2.Construction of the approximating problem.
Assume that problem has a classical solution. Multiply the
equation (1.1) by a smooth function η(x, t) which vanishes on ∂DT
and integrate by parts:
∫
DT
(
∇u(x, t)∇η(x, t) + a(u)∂u
∂t
η(x, t) + λχ(u)∂η
∂t
)
dxdt+
+λ
∫
D
χ(ψ)η(x, 0)dx = 0,
(2.1)
where
χ(u) =
{
1, if u(x, t) < 1,
0, if u(x, t) > 1
.
For ∀ε > 0 we introduce a function χε(τ) ∈ C∞(R1):
χε(τ) = 1 ∀τ ≤ 1 − ε, χε(τ) = 0 ∀τ ≥ 1, χ′
ε(τ) ≤ 0, χ(n)
ε (x) ≤ c
εn
,
Let
aε(x) = a1 + χε(x)(a2 − a1).
Define the function {uε(x, t)} as solutions of the following problem:
220 M. A. Borodin
∆uε(x, t) − aε(u
ε(x, t))
∂uε(x, t)
∂t
= −λ∂χε(u
ε(x, t))
∂t
in DT (2.2)
uε(x, t) = 0 on ∂D1 × [0, T ), uε(x, t) = ϕ(x, t) on ∂D2 × [0, T );
(2.3)
uε(x, 0) = ψ(x) in D, ψ(x) = 0 on ∂D1, ψ(x) = ϕ(x, 0) > 1 on ∂D2.
(2.4)
Then, as is well known [1], takes place the statement
Theorem 2.1. Let
ψ(x) ∈ C l+α(D), ϕ(x, t) ∈ H l+α, l+α
2 (DT ), 0 < α < 1,
and assume that the corresponding consistency conditions hold at
t = 0, x ∈ ∂D. Then this problem is solvable and
∥
∥
∥
uε(x, t)
∥
∥
∥
Hl+α,
l+α
2 (DT )
≤ c
M(ε)
, (2.5)
where positive constant c do not depend on ε,M(ε) → 0, if ε→ 0.
The equation (2.2) we shall transform to a form
∆uε(x, t) − ∂
∂t
uε(x,t)
∫
0
[aε(τ) − λχ′
ε(τ)]dτ = 0.
We divide the cylinder DT by the planes t = kτ, k = 1, 2, ...N,Nτ =
T, integrate equation (2.2) with respect to the variable t, from (k−1)τ
to kτ .
kτ
∫
(k−1)τ
∆uε(x, τ)dτ −
uε
k
(x)
∫
uε
k−1
(x)
[aε(τ) − λχ′
ε(τ)]dτ = 0,
where uεk(x) ≡ uε(x, kτ). After simple transformations we obtain
4uεk(x) − βεk(x)
∂uεk(x)
∂t
= f εk(x), (2.6)
uεk(0) = 0, uεk(l) = ϕ(kτ) = ϕk, uε0(x) = ψ(x). (2.7)
где
∂uεk(x)
∂t
=
uεk(x) − uεk−1(x)
τ
,
The Stefan problem 221
βεk(x) =
1
∫
0
{kε[uεk−1 + τ(uεk − uεk−1)] − λχ′
ε[u
ε
k−1 + τ(uεk − uεk−1)]}dτ,
f εk(x) =
1
τ
kτ
∫
(k−1)τ
[∆uε(x, kτ) − ∆uε(x, t]dt.
3. The fundamental solutions and its properties.
For studying of the problem (2.4), (2.5) we need in the integral
representation of the solution. Let KR(x0) be ball with center at the
point x0 and the radius R and
Γn−k+1(|x− x0|) =
iτ
2πan
∮
∂L
(sinh
√
zR)−1 sinh
√
z(R− |x− x0|)dz
4π|x− x0|(1 − zτ
an
)(1 − zτ
an−1
)(1 − zτ
ak
)
,
(3.1)
where
L = {z = ξ + iη : Rez > − π2
R2
, |z| < %},
(a1
τ
, 0
)
,
(a2
τ
, 0
)
, ...
(an
τ
, 0
)
∈ L, ai > 0, i = 1, 2, ...n.
Property 1.
Let |x− x0| 6= 0, then
∆Γn−k+1 − ak
Γn−k+1 − Γn−k
τ
= 0, ∀k = 1...(n− 1),
∆Γ1 − an
Γ1
τ
= 0, Γ1(|x− x0|) =
sinh
√
an
τ
(R − |x− x0|)
4π|x− x0| sinh
√
an
τ
. (3.2)
Property 2. There is the estimation
∫
KR(x0)
Γn(|x− x0|)
τ
dx ≤
∫
KR(x0)
Γn−1(|x− x0|)
τ
dx ≤ ...
≤
∫
KR(x0)
Γ1(|x− x0|)
τ
dx ≤ 1
an
(
1 −
√
an
τ
R
sinh
√
an
τ
R
)
≤ 1. (3.3)
Property 3. Let Kδ(x0) denote the ball with its center at the point
x0 and radius δ. Then
lim
δ→0
∮
∂Kδ(x0
∂Γm−k+1
∂n
ds =
{
1, if k = m,
0, if k 6= m,
222 M. A. Borodin
where n is the inner normal. All reduced above property can be
received by immediate calculations.
Property 4. Let {vk(x) ∈ C2(D)} and they satisfy the following
equations:
∆vk −
akvk − ak−1vk−1
τ
= −fk − fk−1
τ
,
then there is an integral representation
vm(x0) =
∫
KR(x0)
(a0v0 − f0)Γm(|x− x0|
τ
dx−
m
∑
k=1
∫
∂KR(x0)
vk
∂Γm−k+1
∂n
ds+
+
m
∑
k=1
∫
KR(x0)
fk
Γm−k+1 − Γm−k
τ
dx. (3.4)
This integral representation follows from the previous properties of
the fundamental solutions and Green’s for elliptic equations.
Property 5. The functions {Γm−k+1(|x − x0|) − Γm−k(|x − x0|)}
change the sign on the interval 0 < |x− x0| < R no more than once
and
∣
∣
∣
∣
∂Γk−1(r)
∂r
∣
∣
∣
∣
r=R
≤
∣
∣
∣
∣
∂Γk(r)
∂r
∣
∣
∣
∣
r=R
. (3.5)
This property follows from property1 and from principle of the
maximum.
Property 6. We will denote by rk,k−1 the points, where the functions
Γk(r) − Γk−1(r), k = 1, 2...n, are equal to zero. Then we have the
inequality
rk,k−1 ≤ rk+1,k, and
r2,1 =
√
τ
ln an−1 − ln an√
an−1 −
√
an
+ o(
√
τ ) τ → 0, if an−1 6= an, (3.6)
r2,1 =
√
τ
2
an
, if an−1 = an.
Proof. The functions Γk(r) − Γk−1(r) satisfy the equation
∆(Γk+1(|x−x0|)−Γk(|x−x0|))−am−k
Γk+1(|x− x0|) − Γk(|x− x0|)
τ
=
= −am−k+1
Γk(|x− x0|) − Γk−1(|x− x0|)
τ
. (3.7)
The Stefan problem 223
We construct an integral representation at the center of the sphere
Krk,k+1
(x0)
Γk+1(0) − Γk(0) =
=
∫
Krk,k+1
(x0)
am−k+1
Γk(|x−x0|)−Γk−1(|x−x0|)
τ
Γ1(|x− x0|)dx =
= −
∫
∂Krk,k+1
(x0)
(Γk+1(|x− x0|) − Γk(|x− x0|)
∂Γ1
∂n
ds,
Γ1(|x− x0|) =
sinh
√
am−k
τ
(rk+1,k − |x− x0|)
4π|x− x0| sinh
√
am−k
τ
rk+1,k
.
Taking into account that the function Γk+1(r)−Γk(r) is equal to zero
at the points r = 0, rk+1,k = 0, we obtain
0 =
∫
Krk,k+1
(x0)
am−k+1
Γk(|x− x0|) − Γk−1(|x− x0|)
τ
Γ1(|x− x0|)dx.
If rk,k−1 > rk+1,k, then this equality is impossible.
Formula (3.1) implies
4π|x− x0|Γ1(|x− x0|) = e−|x−x0|
√
am
τ +O
(
e−R
√
am
τ
)
,
4π|x−x0|Γ2(|x−x0|) =
am−1
am−1 − am
(
e−|x−x0|
√
am
τ − e−|x−x0|
√
am−1
τ
)
+
+O
(
e−R
√
amin
τ
)
, if am−1 6= am,
4π|x− x0|Γ2(|x− x0|) =
|x− x0|
2
√
amτ
e−|x−x0|
√
am
τ +O
(
e−R
√
am
τ
)
if am−1 = am. From here we obtain
r2,1 =
√
τ
ln am−1 − ln am√
am−1 −
√
am
+ o(τ) τ → 0.
In particular, if am−1 = am, then
r2,1 =
√
τ
2√
am
.
224 M. A. Borodin
The function Γ2(r)−Γ1(r) changes the sign once. Therefore, as follows
from the equations (3.7), the functions Γk(r) − Γk−1(r) change the
sign once too. It means that the inequalities (3.6) hold.
Property 7. We have following estimate
∣
∣
∣
∣
∂ΓN
∂n
∣
∣
∣
∣
≤M1
{
1
qNR
exp{−M2
R√
τ
} + exp
{
− Tπ2
R2amax
}}
, (3.8)
where q ≥ 2, and positive constants M1,M2 do not depend onN, τ, R.
Proof. Let us estimate integral
∂ΓN
∂n
∣
∣
∣
∣
∣
∂KR(x0)
=
−iτ
2πaN
∮
L
√
z
2πR sinh(
√
zR)
· dz
(1 − zτ
a1
)(1 − zτ
a2
)...(1 − zτ
aN
)
,
(3.9)
where ∂L is the boundary of the domain
L = {z : % = |z| <
(1+q) max
1≤k≤N
ak
τ
,
Rez = b0 > − π2
R2 , b0 < 0, % >
max
1≤k≤N
ak
τ
}.
Let us represent the integral (3.9) as a sum of two terms: I1 and I2,
where I1 denotes the integral along the part of the curve ∂L which is
an arch of a circle, and I2 denotes the integral along the part of the
contour which lies inside the straight line Rez = b0. Let us estimate
the integral I1. The estimates
∣
∣
∣
∣
(1 − zτ
a1
)(1 − zτ
a2
)...(1 − zτ
aN
)
∣
∣
∣
∣
≥
∣
∣
∣
∣
|z|τ
amax
− 1
∣
∣
∣
∣
N
≥ qN ,
∣
∣sinh(
√
zR)
∣
∣ ≥ sinh[
√
|z|R cos(argz/2)] = sinh[
√
|z|R cosϕ],
where ϕ→ π
4
, if τ → 0, imply
|I1| ≤ c1
1
qNR
exp{−c2
R√
τ
},
where the constants c1 and c2 do not depend on τ . Let us now esti-
mate the integral I2. As Rez = b0, we obtain
∣
∣
∣
∣
(1 − zτ
a1
)(1 − zτ
a2
)...(1 − zτ
aN
)
∣
∣
∣
∣
≥
(
1 + τ
|b0|
amax
)
T
τ
.
The Stefan problem 225
Assume b0 = − π2
2R2 . Then
∣
∣
∣
∣
√
|z|R
sinh
√
zR
∣
∣
∣
∣
≤ c4. Thus we obtain
|I2| ≤ c5 exp
{
− Tπ2
R2amax
}
.
The constants c3, c4, c5 do not depend on τ .
4. Uniform estimates. Passage to the limit.
Theorem 4.1. Let
ψ(x) ∈ C l+α(D), ϕ(x, t) ∈ H l+α, l+α
2 (DT ), 0 < α < 1, l ≥ 3,
and assume that the corresponding consistency conditions hold at t =
0, x ∈ ∂D. Then exists the constant c, which not depend ε, that the
following estimate holds:
max
x∈D
∣
∣
∣
∣
∂uε(x, t)
∂t
∣
∣
∣
∣
+ max
DT \{ωε(x,t)∩ωε(x,0)}
∣
∣
∣
∣
∂uε(x, t)
∂x
∣
∣
∣
∣
≤ c, (4.1)
where ωε(x, t) = {(x, t) ∈ DT : 1 < uε(x, t) < 1 + ε}.
Let x0 be an arbitrary point in D. Denote by KR(x0) the sphere
with the center at the point x0 of radius R = τσ, KR(x0) ⊂ D. Let’s
transform the equation (2.6) to the following form
∆
(
∂uε
k
(x)
∂t
)
− βε
k
(x0)
τ
∂uε
k
(x)
∂t
+
βε
k−1
(x0)
τ
∂uε
k−1
(x)
∂t
=
=
∂[fε
k
(x)−fε
k−1
(x)]
∂t
+ +
βε
k
(x)−βε
k
(x0)
τ
∂uε
k
(x)
∂t
− βε
k−1
(x)−βε
k−1
(x0)
τ
∂uε
k−1
(x)
∂t
.
In order to obtain the estimate for the functions
∂uε
k
(x)
∂t
we will use
the integral representation (Property 4). It will give
∂uεn(x0)
∂t
= −
∫
KR(x0)
∆uε0(x)
Γn(|x− x0|)
τ
dx−
226 M. A. Borodin
n
∑
k=1
∫
∂KR(x0)
∂uε
k
∂t
∂Γn−k+1
∂n
ds−
−
n
∑
k=1
∫
KR(x0)
∂(fε
k
(x)−fε
k−1
(x))
∂t
Γn−k+1(|x− x0|)dx−
−
n
∑
k=1
∫
KR(x0)
{
βε
k
(x)−βε
k
(x0)
τ
∂uε
k
(x)
∂t
− βε
k−1
(x)−βε
k−1
(x0)
τ
∂uε
k−1
(x)
∂t
}
×
×Γn−k+1(|x− x0|)dx = I1 + I2 + I3 + I4.
Let’s estimate every term. From (3.4) follows
|I1| ≤
∫
KR(x0)
1
τ
|∆uε0(x)|Γn(|x− x0|)dx ≤ max
x∈D
|∆uε0(x)|
1
an
.
Let assume 0 < σ < 1
2
. Then from (3.8) follows
|I2| ≤
Tc1
τM(ε)
{
1
qNR
exp{−c2
R√
τ
} + exp
{
− Tπ2
R2amax
}}
≤
≤ Tc1
τM(ε)
{
1
qNτσ
exp{−c2
τσ√
τ
} + exp
{
− Tπ2
τ 2σamax
}}
≤ c(σ)
M(ε)
τ,
where c(σ) do not depend ε.
From (2.5), (3.4) follows
|I3| ≤
n
∑
k=1
∫
KR(x0)
|f εk(x)|
∣
∣
∣
Γn−k+1(|x−x0|)−Γn−k(|x−x0|)
τ
∣
∣
∣
dx ≤
≤
∫
Kr2,1
(x0)
max
x∈D,1≤k≤N
|f εk(x)|Γ1(|x−x0|)
τ
dx+
+
∫
KR\Kr2,1
(x0)
max
x∈D,1≤k≤N
|f εk(x)| 1
τ |x−x0|dx ≤ c
(
τ
M(ε)
+ τ2σ
M(ε)
)
,
where constant c do not depend ε, τ.
From (3.5), (3.6) follows
|I4| ≤
n
∑
k=1
∫
KR(x0)
|βεk(x) − βεk(x0)|×
×|∂u
ε
k
(x)
∂t
|
∣
∣
∣
Γn−k+1(|x−x0|)−Γn−k(|x−x0|)
τ
∣
∣
∣
dx+
The Stefan problem 227
+
n
∑
k=1
∫
KR\Kr2,1
(x0)
|βεk(x) − βεk(x0)|×
×|∂u
ε
k
(x)
∂t
|
∣
∣
∣
Γn−k+1(|x−x0|)−Γn−k(|x−x0|)
τ
∣
∣
∣
dx ≤
≤
∫
Kr2,1
(x0)
c1|x−x0|
ε2M(ε)
Γ1(|x−x0|)
τ
dx+
+
∫
KR\Kr2,1
(x0)
c1|x−x0|
ε2M(ε)
1
τ |x−x0|dx ≤ c
ε2M(ε)
(τ 1/2 + τ 3σ−1).
In result we shall receive
∣
∣
∣
∣
∂uεn(x0)
∂t
∣
∣
∣
∣
≤ c1 max
x∈D
|∆uε0(x)| + c2
τ 3σ−1
ε2M(ε)
,
where constants c1, c2 do not depend from τ, ε. Let us assume that
τ 3σ−1 < ε2M(ε), 1/3 < σ < 1/2. Then
∣
∣
∣
∣
∂uε(x0, t)
∂t
∣
∣
∣
∣
≤ ∂uεk(x0)
∂t
+
∣
∣
∣
∣
∣
∣
∣
1
τ
kτ
∫
(k−1)τ
{
∂uε(x, t)
∂t
− ∂uε(x, η)
∂t
}
dη
∣
∣
∣
∣
∣
∣
∣
≤
≤ c1 max
x∈D
|∆uε0(x)| + c2
τ 3σ−1
ε2M(ε)
+ c3
τ
M(ε)
≤ c4,
where the constant c4 do not depend from ε.
Near of boundary of domain D the equation (2.6) become the
linear equation with the constant coefficients. Therefore the appro-
priate estimate can be easily received.
We differentiate the equation (2.6) with respect to one of the
variables xi.
∂2
∂x2
(
∂uεk(x)
∂x
)
− bε(u
ε
k(x))u
ε′
k (x) − bε(u
ε
k−1(x))u
ε′
k−1(x)
τ
=
∂f εk(x)
∂x
.
We use the property 4. It gives
uε′n (x0) =
∫
KR(x0)
βε0u
ε′
0 (x)Γn(|x−x0|)
h
dx−
−
n
∑
k=1
{
∫
∂KR(x0)
uε′k (x)∂Γn−k+1
∂n
ds−
∫
KR(x0)
f ε′k (x)Γn−k+1dx
}
−
−
n
∑
k=1
∫
KR(x0)
(βεk(x0) − βεk(x))u
ε′
k (x)Γm−k+1(|x−x0|)−Γm−k(|x−x0|)
h
dx
(4.3)
228 M. A. Borodin
From this integral representation, applying the same reasoning as
above, we obtain the second part of the estimate (4.1). The estimation
(4.2) can be proved similarly.
Let’s return to integral representation (4.3) and we will estimate
from below the first term. We will assume, that ∂ψ(x)
∂x
6= 0 everywhere
in D. For definiteness we will assume ∂ψ(x)
∂x
> 0. Then
∫
KR(x0)
βε(ψ(x0))
∂u0(x)
∂x
Γn(|x−x0|)
τ
dx ≥
≥ amin minD(ψ′(x)
∫
KR(x0)
Γn(|x−x0|)
τ
dx =
= amin minD(ψ′(x)
{
∫
∂KR(x0)
n−1
∑
k=1
1
βε
k
(x0)
∂Γn−k+1(|x−x0|)
∂n
dx+
+
∫
KR(x0
Γ1(|x−x0|)
τ
dx.
}
= I1 + I2.
From (3.8) follows
|I1| ≤
Tc
τM(ε)
{
1
qNτσ
exp{−c2
τσ√
τ
} + exp
{
− Tπ2
τ 2σamax
}}
≤ c(σ)
M(ε)
τ,
where c, c(σ) do not depend ε. From (3.3) follows
|I2| =
∫
KR(x0)
Γ1(|x− x0|)
τ
dx =
1
βεn(x0)
(
1 −
√
βε
n(x0)
τ
τσ
sinh
√
βε
n(x0)
τ
τσ
)
Similarly to the previous theorem it is possible to prove that all other
terms in the received integral representation (4.3) have limits equal
to zero when σ → 0. From here follows
Theorem 4.2. Let
ψ(x) ∈ C2+α(D), ϕ(x, t) ∈ H2+α, 2+α
2 (DT ), min
D
|∇ψ(x)| > 0,
(0 < α < 1) and assume that the corresponding consistency conditions
hold at t = 0, x ∈ ∂D. Then exists the constant c, which not depend
ε, that the following estimate holds:
|∇uε(x, t)| ≥ c > 0 ∀(x, t) ∈ DT \ ωε(x, t).
The Stefan problem 229
Let the function η(x, t) ∈ C2,1(DT ) be equal to zero on (∂D ×
(0, T ))∪ (D× (t = T )). We multiply (2.2) by η(x, t), integrate it over
DT . After simple transformations we obtain
∫
DT
(
∇uε(x, t)∇η(x, t) + a(u)∂u
ε
∂t
η(x, t) + λχε(u
ε)∂η
∂t
)
dxdt+
+λ
∫
D
χε(ψ)η(x, 0)dx = 0,
The possibility of passage to the limit follows from the statements
proved above. As a result we obtain
Theorem 4.3. Let the following conditions be satisfied:
ψ(x) ∈ C2+α(D), ϕ(x, t) ∈ H2+α,1+α/2(DT ), min
D
|∇ψ(x)| > 0,
(0 < α < 1) and we will assume that corresponding consistency
conditions hold at t = 0, x ∈ ∂D. Then ∀T > 0 there exists a solution
of the problem (1.1)-(1.4) and
u(x, t) ∈ C(DT ) ∩
(
H2+α,1+α/2(ΩT \ γ0) ×H2+α,1+α/2(GT \ γ0)
)
,
the free boundary is given by the graph of a function from H 2+α,1+α/2
class.
In this work existence of classical solution is proved at more
natural limitations, than in work [2].
1. Ladyshenskaja O. A., Solonnikov V. A. and Uraltceva N. N. Linear and
Quasi-linear Parabolic Equations, Nauka, Moscow, 1967.
2. Borodin M.A. Existence of the clobal classical solution for a two-phase Stefan
problem // SIAM J. Math. Anal., vol. 30, No. 6(1999), pp. 1264-1281.
Donetsk Nationality university,
Dept. of math. physics, st. Universitetskaya 24,
Donetsk 83055, Ukraine
Received 6.10.08
|
| id | nasplib_isofts_kiev_ua-123456789-124263 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 0236-0497 |
| language | English |
| last_indexed | 2025-12-07T18:35:56Z |
| publishDate | 2008 |
| publisher | Інститут прикладної математики і механіки НАН України |
| record_format | dspace |
| spelling | Borodin, M.A. 2017-09-23T09:45:02Z 2017-09-23T09:45:02Z 2008 The Stefan problem / M.A. Borodin // Нелинейные граничные задачи. — 2008. — Т. 18. — С. 218-229. — Бібліогр.: 2 назв. — англ. 0236-0497 MSC (2000): 35R35 https://nasplib.isofts.kiev.ua/handle/123456789/124263 The Stefan problem in its classical statement is a mathematical model of the process of propagation of heat in a medium with di erent phase states, e.g., in a medium with liquid and solid phases.The process of propagation of heat in each phase is described by the parabolic equations. In this work we prove the existence of the global classical solution in a two-phase multidimensional Stefan problem. We apply a method which consists of the following. First, we construct approximating problems, then we prove some uniform estimates and pass to the limit. en Інститут прикладної математики і механіки НАН України The Stefan problem Article published earlier |
| spellingShingle | The Stefan problem Borodin, M.A. |
| title | The Stefan problem |
| title_full | The Stefan problem |
| title_fullStr | The Stefan problem |
| title_full_unstemmed | The Stefan problem |
| title_short | The Stefan problem |
| title_sort | stefan problem |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/124263 |
| work_keys_str_mv | AT borodinma thestefanproblem AT borodinma stefanproblem |