Electronic Journal of Differential Equations, Vol. 2006(2006), No. 21, pp. 1–16.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
EXISTENCE AND UNIQUENESS FOR ONE-PHASE STEFAN
PROBLEMS OF NON-CLASSICAL HEAT EQUATIONS WITH
TEMPERATURE BOUNDARY CONDITION AT A FIXED FACE
ADRIANA C. BRIOZZO, DOMINGO A. TARZIA
Abstract. We prove the existence and uniqueness, local in time, of a solution
for a one-phase Stefan problem of a non-classical heat equation for a semiinfinite material with temperature boundary condition at the fixed face. We
use the Friedman-Rubinstein integral representation method and the Banach
contraction theorem in order to solve an equivalent system of two Volterra
integral equations.
1. Introduction
The one-phase Stefan problem for a semi-infinite material for the classical heat
equation requires the determination of the temperature distribution u of the liquid
phase (melting problem) or of the solid phase (solidification problem), and the
evolution of the free boundary x = s(t). Phase-change problems appear frequently
in industrial processes and other problems of technological interest [2, 3, 6, 8, 9, 10,
11, 12, 18, 29]. A large bibliography on the subject was given in [25].
Non-classical heat conduction problem for a semi-infinite material was studied
in [4, 7, 17, 27, 28], e.g. problems of the type
ut − uxx = −F (ux (0, t)),
u(0, t) = 0,
u(x, 0) = h(x),
x > 0, t > 0,
t>0
(1.1)
x>0
where h(x), x > 0, and F (V ), V ∈ R, are continuous functions. The function F ,
henceforth referred as control function, is assumed to satisfy the condition
(H1) F (0) = 0.
As observed in [27, 28], the heat flux w(x, t) = ux (x, t) for problem (1.1) satisfies
a classical heat conduction problem with a nonlinear convective condition at x = 0,
2000 Mathematics Subject Classification. 35R35, 80A22, 35C05, 35K20, 35K55, 45G15, 35C15.
Key words and phrases. Stefan problem; non-classical heat equation; free boundary problem;
similarity solution; nonlinear heat sources; Volterra integral equations.
c
2006
Texas State University - San Marcos.
Submitted November 1, 2005. Published February 9, 2006.
1
2
A. C. BRIOZZO, D. A. TARZIA
EJDE-2006/21
which can be written in the form
wt − wxx = 0,
x > 0, t > 0,
wx (0, t) = F (w(0, t)),
0
w(x, 0) = h (x) ≥ 0,
t > 0,
(1.2)
x > 0.
The literature concerning problem (1.2) has increased rapidly since the publication of the papers [19, 21, 22]. Related problems have been also studied; see
for example [1, 14, 16]. In [26], a one-phase Stefan problem for a non-classical
heat equation for a semi-infinite material was presented. There the free boundary
problem consists in determining the temperature u = u(x, t) and the free boundary
x = s(t) with a control function F which depends on the evolution of the heat flux
at the extremum x = 0 is given by the conditions
ut − uxx = −F (ux (0, t)),
0 < x < s(t), 0 < t < T,
u(0, t) = f (t) > 0,
u(s(t), t) = 0,
0 < t < T,
ux (s(t), t) = −ṡ(t),
u(x, 0) = h(x) ≥ 0,
0 < t < T,
0 6 x 6 b = s(0)
(1.3)
(b > 0).
The goal in this paper is to prove the existence and uniqueness, local in time, of
a solution to the one-phase Stefan problem (1.3) for a non-classical heat equation
with temperature boundary condition at the fixed face x = 0. First, we prove that
problem (1.3) is equivalent to a system of two Volterra integral equations (2.4)-(2.5)
following the Friedman-Rubinstein’s method given in [13, 23]. Then, we prove that
the problem (2.4)-(2.5) has a unique local solution by using the Banach contraction
theorem.
2. Existence and Uniqueness of Solutions
We have the following equivalence for the existence of solutions to the nonclassical free boundary problem (1.3).
Theorem 2.1. The solution of the free-boundary problem (1.3) is
Z
u(x, t) =
b
Z
t
G(x, t; ξ, 0)h(ξ)dξ +
Gξ (x, t; 0, τ )f (τ ) dτ
Z t
ZZ
+
G(x, t; s(τ ), τ )v(τ ) dτ −
G(x, t; ξ, τ )F (V (τ ))dξ dτ,
0
0
0
(2.1)
D(t)
Z
s(t) = b −
t
v(τ ) dτ ,
(2.2)
0
where D(t) = {(x, τ ) : 0 < x < s(τ ), 0 < τ < t}, with f ∈ C 1 [0, T ), h ∈ C 1 [0, b],
h(b) = 0, h(0) = f (0), F is a Lipschitz function over C 0 [0, T ], and the functions
v ∈ C 0 [0, T ], V ∈ C 0 [0, T ] defined by
v(t) = ux (s(t), t) ,
V (t) = ux (0, t)
(2.3)
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ONE-PHASE STEFAN PROBLEMS
3
must satisfy the following system of Volterra integral equations
b
Z
t
Z
0
N (s(t), t; 0, τ )f˙(τ ) dτ
N (s(t), t; ξ, 0)h (ξ)dξ − 2
v(t) = 2
0
0
t
Z
+2
(2.4)
Gx (s(t), t; s(τ ), τ )v(τ ) dτ
0
t
Z
[N (s(t), t; s(τ ), τ ) − N (s(t), t; 0, τ )]F (V (τ )) dτ.
+2
0
b
Z
N (0, t; ξ, 0)h0 (ξ)dξ
V (t) =
0
Z
t
N (0, t; 0, τ )f˙(τ ) dτ +
−
0
Z
Z
t
Gx (0, t; s(τ ), τ )v(τ ) dτ
(2.5)
0
t
[N (0, t; s(τ ), τ ) − N (0, t; 0, τ )]F (V (τ )) dτ,
+
0
where G, N are the Green and Neumann functions and K is the fundamental
solution of the heat equation, defined respectively by
G(x, t, ξ, τ ) = K(x, t, ξ, τ ) − K(−x, t, ξ, τ ),
N (x, t, ξ, τ ) = K(x, t, ξ, τ ) + K(−x, t, ξ, τ ),

2
 √ 1
t>τ
exp − (x−ξ)
4(t−τ
)
K(x, t, ξ, τ ) = 2 π(t−τ )
0
t≤τ,
where s(t) is given by (2.2),
Proof. Let u(x, t) be the solution to (1.3). We integrate, on the domain Dt,ε =
{(ξ, τ ) : 0 < ξ < s(τ ), ε < τ < t − ε}, the Green identity
(Guξ − uGξ )ξ − (Gu)τ = GF (uξ (0, τ )) .
(2.6)
Now we let ε → 0, to obtain the following integral representation for u(x, t),
Z
u(x, t) =
b
Z
t
G(x, t; ξ, 0)h(ξ)dξ +
Gξ (x, t; 0, τ )f (τ ) dτ
Z t
ZZ
+
G(x, t; s(τ ), τ ) uξ (s(τ ), τ ) dτ −
G(x, t; ξ, τ )F (uξ (0, τ ))dξ dτ .
0
0
0
D(t)
From the definition of v(t) and V (t) by (2.3), we obtain (2.1) and (2.2). If we
differentiate u(x, t) in variable x and we let x → 0+ and x → s(t), by using the
jump relations, we obtain the integral equations for v and V given by (2.4) and
(2.5).
Conversely, the function u(x, t) defined by (2.1) where v and V are the solutions
of (2.4)and (2.5), satisfy the conditions (1.3) (i),(ii),(iv) and (v). In order to prove
condition (1.3) (iii) we define ψ(t) = u(s(t), t). Taking into account that u satisfy
the conditions (1.3) (i),(ii),(iv) and (v), if we integrate the Green identity (2.6) over
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A. C. BRIOZZO, D. A. TARZIA
EJDE-2006/21
the domain Dt,ε , (ε > 0) and we let ε → 0 we obtain that
Z b
Z t
u(x, t) =
G(x, t; ξ, 0)h(ξ)dξ +
G(x, t; s(τ ), τ )v(τ ) dτ
0
0
Z t
+
ψ(τ )[Gx (x, t; s(τ ), τ ) − G(x, t; s(τ ), τ )v(τ )] dτ
0
Z t
ZZ
+
Gξ (x, t; 0, τ )f (τ ) dτ −
G(x, t; ξ, τ )F (V (τ ))dξ dτ.
0
D(t)
Then, if we compare this last expression with (2.1), we deduce that
Z t
M (x, t) =
ψ(τ )[Gx (x, t; s(τ ), τ ) − G(x, t; s(τ ), τ )v(τ )] dτ ≡ 0
(2.7)
0
for 0 < x < s(t), 0 < t < σ. We let x → s(t) in (2.7) and by using the jump
relations we have that ψ satisfy the integral equation
Z t
1
ψ(t) +
ψ(τ )[Gx (s(t), t; s(τ ), τ ) − G(s(t), t; s(τ ), τ )v(τ )] dτ = 0 .
2
0
Then we deduce that
t
|ψ(τ )|
√
dτ
t−τ
0
Z t
Z τ
dτ
|ψ(η)|
√
√
dη
≤ C2
τ −η
t−τ 0
0
Z t
Z t
dτ
= C2
|ψ(η)|dη
[(t
−
τ
)(τ
− η)]1/2
0
η
Z t
= πC 2
|ψ(η)|dη
Z
|ψ(t)| ≤ C
0
where C = C(t); therefore by using the Gronwall inequality we have that ψ(t) = 0
over [0, σ].
Next, we use the Banach fixed point theorem in order to prove the local existence
and uniqueness of solution v, V ∈ C 0 [0, σ] to the system of two Volterra integral
equations (2.4)-(2.5) where σ is a positive small number. Consider the Banach
space
v
CM,σ = w
~=
: v, V : [0, σ] → R, continuous, with kwk
~ σ≤M
V
with
kwk
~ σ := kvkσ + kV kσ := max |v(t)| + max |V (t)|
t∈[0,σ]
t∈[0,σ]
We define A : CM,σ −→ CM,σ , such that
~e = A(w(t))
w(t)
~
=
A1 (v(t), V (t))
A2 (v(t), V (t))
where
Z
t
[N (s(t), t, s(τ ), τ ) − N (s(t), t, 0, τ )]F (V (τ )) dτ
A1 (v(t), V (t)) = F0 (v(t)) + 2
0
(2.8)
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ONE-PHASE STEFAN PROBLEMS
5
with
b
Z
N (s(t), t, ξ, 0)h0 (ξ)dξ − 2
F0 (v(t)) = 2
t
Z
N (s(t), t, 0, τ )f˙(τ ) dτ
0
0
t
Z
Gx (s(t), t, s(τ ), τ )v(τ ) dτ
+2
0
and
b
Z
N (0, t, ξ, 0)h0 (ξ)dξ −
A2 (v(t), V (t)) =
Z
t
N (0, t, 0, τ )f˙(τ ) dτ
0
0
t
Z
(2.9)
Gx (0, t, s(τ ), τ )v(τ ) dτ
+
0
t
Z
[N (0, t, s(τ ), τ ) − N (0, t, 0, τ )]F (V (τ )) dτ.
+
0
Lemma 2.2. If v ∈ C 0 [0, σ], maxt∈[0,σ] |v(t)| ≤ M and 2M σ ≤ b then s(t) defined
by (2.2) satisfies
|s(t) − s(τ )| ≤ M |t − τ |
|s(t) − b| ≤
b
,
2
∀t, τ ∈ [0, σ].
To prove the following Lemmas we need the inequality
exp
−x2 nα n/2
/(t − τ )n/2 ≤
,
α(t − τ )
2ex2
α, x > 0, t > τ, n ∈ N.
(2.10)
Lemma 2.3. Let σ ≤ 1, M ≥ 1, f ∈ C 1 [0, T ), h ∈ C 1 [0, b], F a Lipschitz function
over C 0 [0, T ]. Under the hypothesis of Lemma 2.2, we have the following properties:
t
Z
|N (s(t), t, 0, τ )||f˙(τ )| dτ ≤ kf˙kt C1 (b)t
(2.11)
√
|Gx (s(t), t, s(τ ), τ )||v(τ )| dτ ≤ M 2 C2 (b) t
(2.12)
0
Z
t
0
b
Z
|N (s(t), t, ξ, 0)||h0 (ξ)|dξ ≤ kh0 k
(2.13)
0
Z
t
√
|N (s(t), t, s(τ ), τ ) − N (s(t), t, 0, τ )||F (V (τ ))| dτ ≤ C4 (L)M t
(2.14)
0
Z
b
|N (0, t, ξ, 0)||h0 (ξ)|dξ ≤ kh0 k
(2.15)
2kf˙kσ √
|N (0, t, 0, τ )||f˙(τ )| dτ ≤ √
t
π
(2.16)
|Gx (0, t, s(τ ), τ )||v(τ )| dτ ≤ C3 (b)M t
(2.17)
0
t
Z
Z
0
t
0
Z
0
t
√
|N (0, t, s(τ ), τ ) − N (0, t, 0, τ )||F (V (τ ))| dτ ≤ C4 (L)M t
(2.18)
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A. C. BRIOZZO, D. A. TARZIA
EJDE-2006/21
where L is the Lipschitz constant of F and
8 1/2 1
1
3b
2 3/2
) √ , C2 (b) = √ + √ (
)
2
eb
π
2 π 4 π 3eb2
3b 24
4L
C3 (b) = √ ( 2 )3/2 , C4 (L) = √ .
8 π eb
π
C1 (b) = (
Proof. To prove (2.11), we have
|N (s(t), t, 0, τ )| = |K(s(t), t, 0, τ ) + K(−s(t), t, 0, τ )| = 2K(s(t), t, 0, τ )
−s2 (t) (t − τ )−1/2
√
4(t − τ )
π
2
−b
(t − τ )−1/2
√
≤ exp
16(t − τ )
π
8 1/2 1
≤ ( 2 ) √ = C1 (b)
eb
π
= exp
then (2.11) holds. To prove (2.12), we have
|Gx (s(t), t, s(τ ), τ )| = Kx (s(t), t, s(τ ), τ ) + Kx (−s(t), t, s(τ ), τ )
(t − τ )−3/2 −(s(t) − s(τ ))2 √
=
(s(t) − s(τ )) exp
4(t − τ )
4 π
−(s(t) + s(τ ))2 − (s(t) + s(τ )) exp
4(t − τ )
(t − τ )−3/2 −9b2 √
≤
M (t − τ ) + 3b exp
4(t − τ )
4 π
1 2
3/2
≤ √ M (t − τ )−1/2 + 3b
.
3eb2
4 π
Then
Z
0
t
√
2 3/2 M ) t
|Gx (s(t), t, s(τ ), τ )||v(τ )| dτ ≤ √ 2M t + 3b(
3eb2
4 π
√
3b
2 3/2 1
√ (
≤ M2 t √ +
)
2 π M 4 π 3eb2
√
≤ M 2 C2 (b) t,
which implies (2.12). To prove (2.13), we have
Z b
Z ∞
|N (s(t), t, ξ, 0)||h0 (ξ)|dξ ≤ kh0 k
|N (s(t), t, ξ, 0)|dξ ≤ kh0 k
0
0
because
Z
∞
|N (s(t), t, ξ, 0)|dξ ≤ 1.
0
To prove (2.14), by taking into account that
2
|N (s(t), t, s(τ ), τ ) − N (s(t), t, 0, τ )| ≤ p
π(t − τ )
(2.19)
EJDE-2006/21
ONE-PHASE STEFAN PROBLEMS
7
we obtain
Z
t
Z
t
2
p
dτ
π(t − τ )
0
√
= C4 (L)M t.
|N (s(t), t, s(τ ), τ ) − N (s(t), t, 0, τ )||F (V (τ ))| dτ ≤ LM
0
The inequality (2.15) is prove in the same way as (2.13). To prove (2.16), we have
Z
t
|N (0, t, 0, τ )||f˙(τ )| dτ ≤ kf˙kσ
Z
t
|N (0, t, 0, τ )| dτ
0
0
= kf˙kσ
Z
0
t
1
p
π(t − τ )
dτ
kf˙kσ √
= √ 2 t.
π
To prove (2.17), we have
−(s(τ ))2 (t − τ )−3/2
√
s(τ ) exp
4(t − τ )
4 π
−b2 3b
≤ √ (t − τ )−3/2 exp
16(t − τ )
8 π
3b 24 3/2
≤ √ ( 2) .
8 π eb
|Gx (0, t, s(τ ), τ )| =
To prove (2.18), as in (2.14), we prove that
2
|N (0, t, s(τ ), τ ) − N (0, t, 0, τ )| ≤ p
π(t − τ )
and therefore (2.18) holds.
Lemma 2.4. Let s1 , s2 be the functions corresponding to v1 , v2 in C 0 [0, σ], respectively, with maxt∈[0,σ] |vi (t)| ≤ M , i = 1, 2, Then we have
|s2 (t) − s1 (t)| ≤ tkv2 − v1 kt ,
|si (t) − si (τ )| ≤ M |t − τ |,
b
3b
≤ si (t) ≤ ,
2
2
i = 1, 2,
∀t ∈ [0, σ], i = 1, 2.
(2.20)
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A. C. BRIOZZO, D. A. TARZIA
EJDE-2006/21
Lemma 2.5. Let f ∈ C 1 [0, T ), h ∈ C 1 [0, b], F a Lipschitz function in C 0 [0, T ].
We have
√
|F0 (v2 (t)) − F0 (v1 (t))| ≤ E(b, h, f ) tkv2 − v1 kt ;
Z
(2.21)
t
|N (s2 (t), t, s2 (τ ), τ ) − N (s2 (t), t, 0, τ )||F (V2 (τ )) − F (V1 (τ ))| dτ
√
≤ C4 (L) tkV2 − V1 kt ;
Z t
|N (s2 (t), t, 0, τ ) − N (s1 (t), t, 0, τ )||F (V1 (τ ))| dτ
0
0
(2.22)
(2.23)
≤ C5 (b, L, M )tkv2 − v1 kt ;
Z
t
|N (s2 (t), t, s2 (τ ), τ ) − N (s1 (t), t, s1 (τ ), τ )||F (V1 (τ ))| dτ
√
≤ [C6 (L, M ) t + C7 (b, L, M )t]kv2 − v1 kt ;
Z t
|Gx (0, t, s2 (τ ), τ )||v2 (τ ) − v1 (τ )| dτ ≤ C8 (b)tkv2 − v1 kt ;
0
Z t
|Gx (0, t, s2 (τ ), τ )v2 (τ ) − Gx (0, t, s1 (τ ), τ )v1 (τ )| dτ
0
0
≤ (C8 (b)t + C9 (b, M )t2 )kv2 − v1 kt ;
Z t
[N (0, t, s2 (τ ), τ ) − N (0, t, 0, τ )]F (V2 (τ ))
0
− [N (0, t, s1 (τ ), τ ) − N (0, t, 0, τ )]F (V1 (τ )) dτ
√
≤ C4 (L) tkV2 − V1 kt + C5 (b, L, M )t2 kv2 − v1 kt ,
(2.24)
(2.25)
(2.26)
(2.27)
where the constants are defined by
4L
3b 24
C4 (L) = √ , C5 (b, L, M ) = LM √ ( 2 )3/2 ,
π
8 π eb
3
LM
6
3bLM 2
C6 (L, M ) = √ , C7 (b, L, M ) = ( 2 )3/2 √ ,
eb
π
2 π
2
3
24
40 5 9b
1
24
M
C8 (b) = √ ( 2 )3/2 , C9 (b, M ) = [( 2 ) 2 √ + √ ( 2 )3/2 ] .
eb
2
4 π eb
16 π 2 π eb
Proof. The proof of (2.21) can be found in [13]. To prove (2.22), we have
2
|N (s2 (t), t, s2 (τ ), τ ) − N (s2 (t), t, 0, τ )| ≤ p
.
π(t − τ )
Then
Z
t
|N (s2 (t), t, s2 (τ ), τ ) − N (s2 (t), t, 0, τ )||F (V2 (τ )) − F (V1 (τ ))| dτ
0
4L √
≤√
tkV2 − V1 kt
π
(2.28)
EJDE-2006/21
ONE-PHASE STEFAN PROBLEMS
9
To prove (2.23), we use the mean value theorem: There exists c = c(t, τ ) between
s1 (t) and s2 (t) such that
|N (s2 (t), t, 0, τ ) − N (s1 (t), t, 0, τ )||F (V1 (τ ))|
= |Nx (c, t, 0, τ )||s2 (τ ) − s1 (τ )||F (V1 (τ ))|
c2 (t − τ )−3/2
√
LM τ |v2 (τ ) − v1 (τ )|
4(t − τ )
2 π
3b
b2
≤ √ exp −
(t − τ )−3/2 LM τ |v2 (τ ) − v1 (τ )|
16(t − τ )
4 π
3b 24
≤ √ ( 2 )3/2 LM τ |v2 (τ ) − v1 (τ )| .
4 π eb
≤ |c| exp −
Then
Z
t
|N (s2 (t), t, 0, τ ) − N (s1 (t), t, 0, τ )||F (V1 (τ ))| dτ
0
3b 24 3
≤ √ ( 2 ) 2 LM tkv2 − v1 kt = C5 (b, L, M )tkv2 − v1 kt .
8 π eb
To prove (2.24), we have
N (s2 (t), t, s2 (τ ), τ ) − N (s1 (t), t, s1 (τ ), τ )
= K(s2 (t), t, s2 (τ ), τ ) − K(s1 (t), t, s1 (τ ), τ )
+ K(−s2 (t), t, s2 (τ ), τ ) − K(−s1 (t), t, s1 (τ ), τ ) .
As in [24], for each (t, τ ), 0 < τ < t, we define
ft,τ (x) = exp
−x2 .
4(t − τ )
Then we have
K(s2 (t), t, s2 (τ ), τ ) − K(s1 (t), t, s1 (τ ), τ )
(s2 (t) − s2 (τ ))2 (t − τ )−1/2 (s1 (t) − s1 (τ ))2 √
exp −
− exp −
4(t − τ )
4(t − τ )
2 π
−1/2 (t − τ )
√
=
ft,τ (s2 (t) − s2 (τ )) − ft,τ (s1 (t) − s1 (τ ))
2 π
=
and
K(−s2 (t), t, s2 (τ ), τ ) − K(−s1 (t), t, s1 (τ ), τ )
(t − τ )−1/2 (s2 (t) + s2 (τ ))2 (s1 (t) + s1 (τ ))2 √
exp −
− exp −
4(t − τ )
4(t − τ )
2 π
−1/2 (t − τ )
√
=
ft,τ (s2 (t) + s2 (τ )) − ft,τ (s1 (t) + s1 (τ ))
2 π
=
10
A. C. BRIOZZO, D. A. TARZIA
EJDE-2006/21
By the mean value theorem there exists c = c(t, τ ) between s2 (t) − s2 (τ ) and
s1 (t) − s1 (τ ) such that
ft,τ (s2 (t) − s2 (τ )) − ft,τ (s1 (t) − s1 (τ ))
0
= ft,τ
(c)(s2 (t) − s2 (τ ) − s1 (t) + s1 (τ ))
=
−c
c2
exp(−
)(s2 (t) − s2 (τ ) − s1 (t) + s1 (τ ))
2(t − τ )
4(t − τ )
Taking into account that
|c| ≤ max{|si (t) − si (τ )|, i = 1, 2} ≤ M (t − τ )
it results
M
[|s2 (t)) − s1 (t)| + |s2 (τ ) − s1 (τ )|]
2
≤ M 2 kv2 − v1 kt .
|ft,τ (s2 (t) − s2 (τ )) − ft,τ (s1 (t) − s1 (τ ))| ≤
Then we have
M2
kv2 − v1 kt .
|K(s2 (t), t, s2 (τ ), τ ) − K(s1 (t), t, s1 (τ ), τ )| ≤ p
2 π(t − τ )
In the same way we have
ft,τ (s2 (t) + s2 (τ )) − ft,τ (s1 (t) + s1 (τ ))
0
= ft,τ
(c∗ )(s2 (t) + s2 (τ ) − s1 (t) − s1 (τ ))
=
−c∗
c∗2
exp(−
)(s2 (t) + s2 (τ ) − s1 (t) − s1 (τ ))
2(t − τ )
4(t − τ )
where c∗ = c∗ (t, τ ) is between s2 (t) + s2 (τ ) and s1 (t) + s1 (τ ). Since s1 (t) + s1 (τ ) ≤
c∗ ≤ s2 (t)+s2 (τ ), (or viceversa), we deduce that b ≤ c∗ ≤ 3b, that is exp(−c∗2 /4(t−
τ )) ≤ exp(−b2 /4(t − τ )). Then we obtain
|K(−s2 (t), t, s2 (τ ), τ ) − K(−s1 (t), t, s1 (τ ), τ )|
(t − τ )−1/2
√
|ft,τ (s2 (t) + s2 (τ )) − ft,τ (s1 (t) + s1 (τ ))|
2 π
3b
b2 ≤ √
exp
−
2M kv2 − v1 kt
4(t − τ )
4 π(t − τ )3/2
3bM
6
≤ ( 2 )3/2 √ kv2 − v1 kt
eb
2 π
=
and
|N (s2 (t), t, s2 (τ ), τ ) − N (s1 (t), t, s1 (τ ), τ )|
M2
6
3bM
≤( p
+ ( 2 )3/2 √ )kv2 − v1 kt .
eb
2 π
2 π(t − τ )
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Therefore,
Z
t
|N (s2 (t), t, s2 (τ ), τ ) − N (s1 (t), t, s1 (τ ), τ )||F (V1 (τ ))| dτ
Z t
M2
6
3bM p
≤
+ ( 2 )3/2 √ kv2 − v1 kt |F (V1 (τ ))| dτ
eb
2 π
2 π(t − τ )
0
M 2 √t
3bM 6
+ ( 2 )3/2 √ t kv2 − v1 kt
≤ LM √
eb
π
2 π
√
= (C6 (L, M ) t + C7 (L, M, b)t)kv2 − v1 kt .
0
To prove (2.25), we take into account (2.10):
s2 (τ )
t−τ
s22 (τ ) (t − τ )−3/2
√
s2 (τ )
= exp −
4(t − τ )
2 π
1
24 3/2
3b 24 3
≤ √
s2 (τ ) ≤ √ ( 2 ) 2 = C8 (b).
2 π eb2
4 π eb
Gx (0, t, s2 (τ ), τ ) = K(0, t, s2 (τ ), τ )
To prove (2.26), we have
|Gx (0, t, s2 (τ ), τ )v2 (τ ) − Gx (0, t, s1 (τ ), τ )v1 (τ )|
≤ |Gx (0, t, s2 (τ ), τ )||v2 (τ ) − v1 (τ )|
+ |Gx (0, t, s2 (τ ), τ ) − Gx (0, t, s1 (τ ), τ )||v1 (τ )|.
Using the mean value theorem there exists c = c(τ ) between s2 (τ ) and s1 (τ ) such
that Gx (0, t, s2 (τ ), τ )−Gx (0, t, s1 (τ ), τ ) = Gxξ (0, t, c, τ )(s2 (τ )−s1 (τ )). Taking into
account the following properties
c2
K(0, t, c, τ )
+1 ,
t−τ
2(t − τ )
2
K(0, t, c, τ )
c
24
1
1
= √ exp −
(t − τ )−3/2 ≤ √ ( 2 )3/2 ,
t−τ
4(t − τ )
2 π
2 π eb
5
c2
1
c2 9b2 40
K(0, t, c, τ )
= √ exp −
(t − τ )− 2 c2 ≤ √ ( 2 )5/2
2
2(t − τ )
4(t − τ )
4 π
16 π eb
Gxξ (0, t, c, τ ) =
we have
|Gx (0, t, s2 (τ ), τ ) − Gx (0, t, s1 (τ ), τ )||v1 (τ )|
1
24 3
9b2 40 5
≤ ( √ ( 2 ) 2 + √ ( 2 ) 2 )|s2 (τ ) − s1 (τ )||v1 (τ )|
2 π eb
16 π eb
1
24 3
9b2 40 5
≤ M ( √ ( 2 ) 2 + √ ( 2 ) 2 )τ |v2 (τ ) − v1 (τ )| .
2 π eb
16 π eb
Then
Z
t
|Gx (0, t, s2 (τ ), τ ) − Gx (0, t, s1 (τ ), τ )||v1 (τ )| dτ
0
1
24 3
9b2 40 5 M t2
kv2 − v1 kt .
≤ ( √ ( 2 )2 + √ ( 2 )2 )
2
2 π eb
16 π eb
12
A. C. BRIOZZO, D. A. TARZIA
EJDE-2006/21
Then (2.26) holds by using (2.25). To prove (2.27), we have
[N (0, t, s2 (τ ), τ ) − N (0, t, 0, τ )]F (V2 (τ ))
− [N (0, t, s1 (τ ), τ ) − N (0, t, 0, τ )]F (V1 (τ ))
= [N (0, t, s2 (τ ), τ ) − N (0, t, 0, τ )][F (V2 (τ )) − F (V1 (τ ))]
(2.29)
+ [N (0, t, s2 (τ ), τ ) − N (0, t, s1 (τ ), τ )]F (V1 (τ ))
Using |N (0, t, s2 (τ ), τ ) − N (0, t, 0, τ )| ≤ √
2
π(t−τ )
we get
2
|N (0, t, s2 (τ ), τ ) − N (0, t, 0, τ )||F (V2 (τ )) − F (V1 (τ ))| ≤ p
π(t − τ )
L|V2 (τ ) − V1 (τ )|,
and
t
Z
|N (0, t, s2 (τ ), τ ) − N (0, t, 0, τ )||F (V2 (τ )) − F (V1 (τ ))| dτ
√
√
4 t
≤ √ LkV2 − V1 kt = C4 (L) tkV2 − V1 kt .
π
0
(2.30)
Furthermore,
|N (0, t, s2 (τ ), τ ) − N (0, t, s1 (τ ), τ )| = |Nξ (0, t, c, τ )||s2 (τ ) − s1 (τ )|
where c = c(τ ) is between s2 (τ ) and s1 (τ ) and
|Nξ (0, t, c, τ )||s2 (τ ) − s1 (τ )| = | − Gx (0, t, c, τ )||s2 (τ ) − s1 (τ )|
|c| 24
≤ √ ( 2 )3/2 τ |v2 (τ ) − v1 (τ )|
2 π eb
3b 24
≤ √ ( 2 )3/2 τ |v2 (τ ) − v1 (τ )|.
4 π eb
Then
Z
t
|N (0, t, s2 (τ ), τ ) − N (0, t, s1 (τ ), τ )||F (V1 (τ ))| dτ
0
3b 24
t2
≤ LM √ ( 2 )3/2 kv2 − v1 kt = C5 (L, M, b)t2 kv2 − v1 kt
2
4 π eb
Therefore, by (2.29), (2.30), and (2.31)), the inequality (2.27) holds.
(2.31)
Theorem 2.6. The map A : CM,σ → CM,σ is well defined and is a contraction
map if σ satisfies the following inequalities:
σ ≤ 1 , 2M σ ≤ b
√
2kf˙kσ
(2kf˙kσ C1 (b) + M C3 (b))σ + (2M 2 C2 (b) + √
+ 3M C4 (L)) σ ≤ 1
π
√
D(b, f, h, L, M ) σ < 1,
(2.32)
(2.33)
(2.34)
where
M = 1 + 3kh0 k
(2.35)
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ONE-PHASE STEFAN PROBLEMS
13
and
D1 (b, f, h, L, M ) = E(b, f, h) + 2C6 (L, M ) + 3C4 (L)
D2 (b, L, M ) = 2[C5 (b, L, M ) + 2C7 (b, L, M ) + C8 (b)]
D3 (b, L, M ) = C9 (b, M ) + C5 (b, L, M )
D(b, f, h, L, M ) = D1 (b, f, h, L, M ) + D2 (b, L, M ) + D3 (b, L, M ).
Then there exists a unique solution on CM,σ to the system of integral equations
(2.4), (2.5).
Proof. Firstly we demonstrate that A maps Cσ,M into itself, that is
kA(w)k
~ σ = max |A1 (v(t), V (t))| + max |A2 (v(t), V (t))| ≤ M
t∈[0,σ]
(2.36)
t∈[0,σ]
Using the Lemmas 2.3, 2.4 and the definitions (2.8)-(2.9), we have
√
√
|A1 (v(t), V (t))| ≤ 2kf˙kσ C1 (b)t + 2M 2 C2 (b) t + 2kh0 k + 2C4 (L)M t,
√
2kf˙kσ
+ C4 (L)M ) t + C3 (b)M t.
|A2 (v(t), V (t))| ≤ kh0 k + ( √
π
Then
kA(w)k
~ σ = max |A1 (v(t), V (t))| + max |A2 (v(t), V (t))|
t∈[0,σ]
t∈[0,σ]
≤ 3kh k + (2kf˙kσ C1 (b) + C3 (b)M )σ
√
2kf˙kσ
+ 2M 2 C2 (b) + √
+ 3M C4 (L) σ.
π
0
Selecting M by (2.35) and σ such that (2.32) and (2.33) hold, we obtain (2.36).
Now, we prove that
√
−→
kA(w~2 ) − A( w1 )kσ ≤ D(b, h, f, L, M ) σkw~2 − w~1 kσ
where w~1 = Vv11 , w~2 = Vv22 . By selecting σ such that (2.34) holds, A becomes a
contraction mapping on Cσ,M and therefore it has a unique fixed point. To prove
this assertion we consider
A1 (v2 (t), V2 (t)) − A1 (v1 (t), V1 (t))
A(w~1 )(t) − A(w~2 )(t) =
A2 (v2 (t), V2 (t)) − A2 (v1 (t), V1 (t))
where
A1 (v2 (t), V2 (t)) − A1 (v1 (t), V1 (t))
Z t
= F0 (v2 (t)) − F0 (v1 (t)) + 2
[N (s2 (t), t, s2 (τ ), τ ) − N (s2 (t), t, 0, τ )]F (V2 (τ )) dτ
0
Z t
−2
[N (s1 (t), t, s1 (τ ), τ ) − N (s1 (t), t, 0, τ )]F (V1 (τ )) dτ
0
14
A. C. BRIOZZO, D. A. TARZIA
EJDE-2006/21
and
A2 (v2 (t), V2 (t)) − A2 (v1 (t), V1 (t))
Z t
[Gx (0, t, v2 (τ ), τ )v2 (τ ) − Gx (0, t, v1 (τ ), τ )v1 (τ )] dτ
=
0
Z t
[N (0, t, s2 (τ ), τ ) − N (0, t, 0, τ )]F (V2 (τ ))
+
0
− [N (0, t, s1 (τ ), τ ) − N (0, t, 0, τ )]F (V1 (τ )) dτ .
Taking into account the Lemmas 2.4 and 2.5 it results
|A1 (v2 (t), V2 (t)) − A1 (v1 (t), V1 (t))|
√
√
≤ E(b, h, f ) tkv2 − v1 kt + 2C4 (L) tkV2 − V1 kt
√
+ 2C5 (b, L, M )tkv2 − v1 kt + 2[C6 (L, M ) t + C7 (b, L, M )t]kv2 − v1 kt ,
and
|A2 (v2 (t), V2 (t)) − A2 (v1 (t), V1 (t))|
≤ (C8 (b)t + C9 (b, M )t2 )kv2 − v1 kt
√
+ C4 (L) tkV2 − V1 kt + C5 (b, L, M )t2 kv2 − v1 kt .
Therefore,
kA(w~2 ) − A(w~1 )kσ
≤ max |A1 (v2 (t), V2 (t)) − A1 (v1 (t), V1 (t))|
t∈[0,σ]
+ max |A2 (v2 (t), V2 (t)) − A2 (v1 (t), V1 (t))|
t∈[0,σ]
√
≤ {D1 (b, f, h, L, M ) σ + D2 (b, L, M )σ + D3 (b, L, M )σ 2 }kw~2 − w~1 kσ
√
≤ D(b, f, h, L, M ) σkw~2 − w~1 kσ .
By hypothesis (2.34) we have that A is a contraction.
Remark. If F satisfies the conditions
(H2) F (V ) > 0, for all V 6= 0 and F (0) = 0,
then by the maximum principle [5], u is a sub-solution for the same problem with
F ≡ 0, that is
u(x, t) ≤ u0 (x, t) , s(t) ≤ s0 (t)
where u0 (x, t) and s0 (t) solve the classical Stefan problem
u0t − u0xx = 0,
0 < x < s0 (t), 0 < t < T,
u0 (0, t) = f (t) > 0,
u0 (s0 (t), t) = 0,
0 < t < T,
u0 x (s0 (t), t) = −s˙0 (t).0 < t < T,
u0 (x, 0) = h(x)
0 6 x 6 b = s0 (0) .
Acknowledgments. The authors have been partially sponsored by Project PIP
No. 5379 from CONICET - UA (Rosario, Argentina), by Project #53900/4 from
Fundación Antorchas (Argentina), and and by ANPCYT PICT # 03-11165 from
Agencia (Argentina).
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Adriana C. Briozzo
Departamento de Matemática, FCE, Universidad Austral, Paraguay 1950, S2000FZF
Rosario, Argentina
E-mail address: Adriana.Briozzo@fce.austral.edu.ar
Domingo Alberto Tarzia
Departamento de Matemática - CONICET, FCE, Universidad Austral, Paraguay 1950,
S2000FZF Rosario, Argentina
E-mail address: Domingo.Tarzia@fce.austral.edu.ar