Electronic Journal of Differential Equations, Vol. 2020 (2020), No. 49, pp. 1–14.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

SUPERCOOLED STEFAN PROBLEM WITH A NEUMANN TYPE

BOUNDARY CONDITION

ADRIANA C. BRIOZZO

Abstract. We consider a supercooled one-dimensional Stefan problem with a
Neumann boundary condition and a variable thermal diffusivity. We establish

a necessary and sufficient condition for the heat flux at the fixed face x = 0, in

order to obtain existence and uniqueness of a similarity type solution. More-
over we over-specified the fixed face x = 0 by a Dirichlet boundary condition

aiming at the simultaneous determination of one or two thermal coefficients.

1. Introduction

We consider the following Stefan problem that describes the freezing of a super-
cooled liquid,

wt = (D(w)wx)x , 0 < x < s(t), t > 0 (1.1)

D(w(0, t))wx(0, t) = q∗(t) > 0 , t > 0 (1.2)

D(w(s(t), t))wx(s(t), t) = − l
c
ṡ(t) , t > 0 (1.3)

w(s(t), t) = w0 , t > 0 (1.4)

w(x, 0) = g(x) < w0 , 0 < x < 1 (1.5)

s(0) = 1 , (1.6)

where w = w(x, t) represents the temperature of the material and s = s(t) is the
moving interface to be determined, c is the specific heat, l is the latent heat of
fusion of the medium, w0 is the phase change temperature and g = g(x) is the
initial temperature of the material. We impose a Neumann type condition at the
fixed face x = 0 characterized by a heat flux q∗ = q∗(t) > 0 [1], which corresponds
to a supercooled liquid. We assume that the thermal diffusivity D is given by

D(w) =
a

(b+ w)2
(1.7)

and we assume that the positive parameters a, b and the initial temperature g satisfy
the following condition

b+ w0 6=
l

c
, −b < g(x) < w0, for 0 < x < 1. (1.8)

2010 Mathematics Subject Classification. 35R35, 80A22, 35K55.
Key words and phrases. Stefan problem; supercooling; non-linear thermal diffusivity;

similarity solution; determination of thermal coefficient.
c©2020 Texas State University.

Submitted January 9, 2020. Published May 22, 2020.

1



2 A. C. BRIOZZO EJDE-2020/49

In [5] a supercooled one-phase Stefan problem with temperature-dependent thermal
conductivity and a Dirichlet condition at fixed face was studied. Existence of a
similarity type solution was obtained through the unique solution of an integral
equation and it was showed that finite time blow-up occurs.

Free boundary problems with diffusion coefficient given by (1.7) or with temper-
ature dependent conductivity were considered in [2, 4, 6, 7, 16, 19, 23, 30].

Free boundary problems which involves the freezing of a supercooled liquid can
be seen in [12, 13, 14, 17, 18, 22, 24]. The problems with phase change occur
in several and important biological, industrial and technological processes among
others [8, 9, 10, 11, 15, 20, 21, 31]. A large bibliography on the subject is given in
[29].

The organization of the paper is as follow: In Section 2 the Stefan problem is
reduced through reciprocal transformations [25, 26, 27] to another one free bound-
ary problem which admits a similarity type solution. We prove that there exists
a similarity type solution if and only if a Neumann-type boundary condition given
by (1.2) with q∗(t) = Q0/s(t) is considered, where Q0 > 0 and s(t) is the free
boundary. In Section 3 we prove that the similarity solution to (1.1)-(1.6) can
be obtained through the unique solution of an integral equation. In Section 4 we
establish an over-condition of Dirichlet type at the fixed face x = 0 in order to
determine one or two unknown thermal coefficients according to whether the free
boundary (unknown) or the movil boundary (known) is considered.

2. Necessary and sufficient conditions for the existence and
uniqueness of similarity type solution

To obtain an equivalent Stefan problem to (1.1)-(1.7) which admits a similarity
type solution we give preliminaries results [5, 25, 26, 27, 28].

Lemma 2.1. Under the reciprocal transformation

u(x, t) =

√
a

b+ w(x, t)
, (2.1)

problem (1.1)-(1.7) is equivalent to the Stefan problem

ut = u2uxx , 0 < x < s(t) , t > 0 , (2.2)

ux(0, t) = −r(t) , t > 0 , (2.3)

ux(s(t), t) =
l

c
√
a
ṡ(t) , t > 0 , (2.4)

u(s(t), t) = γ , t > 0 , (2.5)

u(x, 0) =

√
a

b+ g(x)
, 0 < x < 1 , (2.6)

s(0) = 1 , (2.7)

where

r(t) =
q∗(t)√
a
, γ =

√
a

b+ w0
. (2.8)

The proof of the above lemma is standard; we omit it here. Now we introduce
the transformation

v(y, t) = u(x, t), y(x, t) =

∫ x

0

dη

u(η, t)
, 0 < x < s(t) , t > 0. (2.9)



EJDE-2020/49 SUPERCOOLED STEFAN PROBLEM 3

Lemma 2.2. From (2.9) the free boundary problem (2.2)–(2.7) is equivalent to

vt = vyy + r(t)vy , 0 < y < Y (t) , t > 0 , (2.10)

vy(0, t) = −r(t)v(0, t), t > 0 , (2.11)

vy(Y (t), t) = β[Ẏ (t) + r(t)] , t > 0 , (2.12)

v(Y (t), t) = γ , t > 0 (2.13)

v(y, 0) = G(y) , (2.14)

Y (0) = y0 , (2.15)

where

Y (t) = y(s(t), t) =

∫ s(t)

0

dη

u(η, t)
, t > 0 (2.16)

is the free boundary and

β =
lγ

c(b+ w0)− l
, y0 =

1√
a

(b+

∫ 1

0

g(η)dη), (2.17)

G(y) =

√
a

b+ g(φ(y))
, φ(y) =

∫ y

0

v(η, 0)dη . (2.18)

Proof. From (2.9), using (2.2)-(2.3) we have

ux =
vy
v
, uxx =

vyy
v2
−
v2y
v3
, ut = −

v2y
v
− r(t)vy + vt .

Then, from (2.2) we obtain (2.10).
To prove (2.12), we differentiate Y (t) in (2.16) and take into account the condi-

tions (2.3)-(2.5), to obtain

Ẏ (t) =
1

u(s(t), t)
ṡ(t) +

∫ s(t)

0

−ut(η, t)
u2(η, t)

dη

=
1

γ
ṡ(t)−

∫ s(t)

0

uηη(η, t)dη

=
1

γ
ṡ(t)− ux(s(t), t) + ux(0, t)

=
( 1

γ
− l

c
√
a

)
ṡ(t)− r(t)

or equivalently

Ẏ (t) =
(c(b+ w0)− l

c
√
a

)
ṡ(t)− r(t) . (2.19)

Therefore, from (2.19) and (2.4) we obtain (2.12). From (2.3), (2.6), (2.7) and
(2.17)-(2.18) follow immediately the conditions (2.11), (2.14) and (2.15). �

Now, for the free boundary problem (2.10)-(2.15) we propose a similarity type
solution given by

T (z) = v(y, t), z =
y

Y (t)
. (2.20)

We have the following results.



4 A. C. BRIOZZO EJDE-2020/49

Theorem 2.3. The free boundary problem (2.10)-(2.15) admits a solution of type
(2.20) if and only if there exist constants µ and R such that

Y (t)Ẏ (t) = µ , ∀t > 0 , (2.21)

Y (t)r(t) = R, ∀t > 0. (2.22)

Then the free boundary problem (2.10)-(2.15) is equivalent to the problem

T ′′(z) + T ′(z)(µz +R) = 0 , 0 < z < 1 , (2.23)

T ′(0) = −RT (0) , (2.24)

T (1) = γ , (2.25)

T ′(1) = β(µ+R) , (2.26)

T (
y

y0
) = G(y) , 0 < y < y0 , (2.27)

whose solution is given by

T (z) = γ
−R

∫ z
0

exp
(
− µ

2 (η + R
µ )2
)
dη + exp(−R

2

2µ )

exp(−R2

2µ )−R
∫ 1

0
exp

(
− µ

2 (η + R
µ )2
)
dη

, 0 < z < 1 , (2.28)

and the coefficient µ must be determined from the equation

−Rγ exp
(
− µ

2 (1 + R
µ )2
)

exp(−R2

2µ )−R
∫ 1

0
exp

(
− µ

2 (η + R
µ )2
)
dη

= β(µ+R). (2.29)

Moreover, the function ϕ(z) := G(zy0) is a solution of (2.23).

Proof. Taking into account (2.20) we have

vy =
T ′(z)

Y (t)
, vyy =

T ′′(z)

Y 2(t)
, vt = −T ′(z)z Ẏ (t)

Y (t)
.

Then (2.10) is equivalent to

T ′′(z) + [r(t)Y (t) + Y (t)Ẏ (t)z]T ′(z) = 0.

Moreover from (2.11) we have T ′(0) = −r(t)Y (t)T (0) and, by (2.12), we obtain

T ′(1) = β
(
Y (t)Ẏ (t) + Y (t)r(t)

)
.

Therefore, we can obtain a similarity type solution T = T (z) if and only if there
exist constants µ and R such that (2.21) and (2.22) hold.

The conditions (2.26) and (2.27) follow from (2.13) and (2.14), respectively. It is
easy to see that the solution to (2.23)-(2.25) is given by (2.28), and equation (2.29)
for µ follows inmediately from (2.26).

Moreover, taking into account (2.27) we obtain that the function ϕ(z) = G(zA),
0 < z < 1 solves (2.23). �

Corollary 2.4. The heat flux at fixed face x = 0 of problem (1.1)-(1.6) satisfies

q∗(t) =
R
√
a

Y (t)
, R > 0. (2.30)



EJDE-2020/49 SUPERCOOLED STEFAN PROBLEM 5

Remark 2.5. The solution to the supercooled Stefan problem with tempera-
ture condition at fixed face given by w(0, t) = −B, a phase-change temperature

w(s(t), t) = 0 and a variable diffusivity given by D(w) = k(w)
ρc = 1

(a+bw)2 obtained

in [5] satisfies

D(w(0, t))wx(0, t) =
Q0

s(t)
, (2.31)

where Q0 is a positive constant which depends on the data.

In what follows for each R > 0 we will analyze the existence of the solution µ to
(2.29). We have the following results.

Lemma 2.6. There is no solution µ > 0 for (2.29).

Proof. If we suppose that there exists µ > 0 which satisfies (2.21). Then

Y (t) =
√

2µt+ y20 > 0, t > 0, Ẏ (t) > 0. (2.32)

Taking into account (2.19) and (2.21) we have(c(b+ w0)− l
c
√
a

)
ṡ(t) =

µ+R

Y (t)
(2.33)

and from the boundary conditions of problem (2.2)-(2.7) we deduce ṡ(t) < 0 which
implies c(b+ w0)− l < 0. Moreover we obtain

s(t) = 1 +
c
√
a(µ+R)

µ(c(b+ w0)− l)

(√
y20 + 2µt− y0

)
, t > 0, (2.34)

and there exists t0 > 0 such that s(t0) = 0 where

t0 =
m2 − y20

2µ
> 0, m = y0 −

µ(c(b+ w0)− l)
(µ+R)c

√
a

.

Then we have a contradiction: s(t0) = 0 implies Y (t0) = 0, but Y (t) > 0 for all
t > 0. Therefore there is no solution µ > 0 for (2.29). �

Next, we study the existence of solution to equation (2.29) for µ < 0. For this,
it is suitable to define

σ = −µ > 0, f(σ) =
R√
2σ
, j(σ) = f(σ)−

√
σ

2
, (2.35)

U(x) =

∫ x

0

exp(z2)dz. (2.36)

Then equation (2.29) can be rewritten as

exp(j2(σ)− f2(σ))

1 + 2f(σ) exp(−f2(σ))[U(j(σ))− U(f(σ))]
= p(R− σ) (2.37)

in the unknown σ > 0, where

p =
−β
Rγ

=
l

lR− cR(b+ w0)
. (2.38)

Lemma 2.7. The real function

F1(σ) :=
exp(j2(σ)− f2(σ))

1 + 2f(σ) exp(−f2(σ))[U(j(σ))− U(f(σ))]
(2.39)



6 A. C. BRIOZZO EJDE-2020/49

satisfies

F1(0) = 1, F1(+∞)−∞ . (2.40)

If R 6= 2x21, then

F1(R) =
exp(−R/2)

F ′(
√

R
2 )

. (2.41)

If R = 2x21, then

F1(R−) = +∞, F1(R+) = −∞ , (2.42)

where

F (x) = exp(−x2)

∫ x

0

exp(z2)dz (2.43)

is the Dawson’s function and x1 is such that F ′(x1) = 0.
Moreover:

(a) If R < 2x21 there exists σ0 > R such that

F1(σ)

{
> 0, σ < σ0

< 0, σ > σ0

and F1(σ−0 ) = +∞, F1(σ+
0 ) = −∞.

(b) If R > 2x21 then there exists σ0 < R such that

F1(σ)

{
> 0, 0 < σ < σ0

< 0, σ0 < σ

and in both cases F1(σ−0 ) = +∞, F1(σ+
0 ) = −∞

(c) If R = 2x21, then

F1(σ)

{
> 0, σ < R

< 0, σ > R

Proof. We have

F1(σ) =
exp(j2(σ)− f2(σ))

1− 2f(σ)F (f(σ)) + 2f(σ) exp(−f2(σ))U(j(σ))

=
exp(j2(σ)− f2(σ))

F ′(f(σ)) + 2f(σ) exp(−f2(σ))U(j(σ))

Taking into account that Dawson’s function satisfies [3]

lim
x→+∞

xF (x) =
1

2
,

lim
σ→0

f(σ) exp(−f2(σ))U(j(σ)) =
exp(−R)

2
.

Then we have F1(0) = 1. By applying L’Hopital’s rule we show that F1(+∞) =
−∞.

We know that

F ′(x) = 1− 2xF (x)


> 0 if 0 < x < x1

= 0 if x = x1

< 0 if x > x1 ,



EJDE-2020/49 SUPERCOOLED STEFAN PROBLEM 7

where x1 ' 0.924, F (x1) ' 0.541. Then if R < 2x21, this is
√
R/2 < x1, we have

F ′(
√
R/2) > 0 and

F1(R) =
exp(−R/2)

F ′(
√
R/2)

> 0.

Since F1 does not cancel and F1(+∞) = −∞, there exists σ0 > R such that

F1(σ)

{
> 0, σ < σ0

< 0, σ > σ0,

F1(σ−0 ) = +∞, F1(σ+
0 ) = −∞.

If R > 2x21 this is
√
R/2 > x1, we have F1(R) < 0. Then, there exists σ0 < R

such that

F1(σ)

{
> 0, 0 < σ < σ0

< 0, σ0 < σ,

F1(σ−0 ) = +∞, F1(σ+
0 ) = −∞

If R = 2x21, then F ′(
√
R/2) = 0 and

F1(R−) = +∞, F1(R+) = −∞ (2.44)

In this case,

F1(σ)

{
> 0, σ < R

< 0, σ > R ,

and the claim holds. �

Lemma 2.8. The function

F2(σ) = p(R− σ), σ > 0 (2.45)

satisfies

(a) If b+ w0 > l/c then F2(0) = pR < 0 and F ′2(σ) = −p > 0
(b) If b+ w0 < l/c then F2(0) = pR > 1 and F ′2(σ) = −p < 0,

and F2(R) = 0.

Lemma 2.9. (a) If b+ w0 > l/c there is no solution µ < 0 to (2.29).
(b) If b+ w0 < l/c there exists unique −R < µ < 0 solution to (2.29).

Proof. Taking into account properties of functions F1 and F2 it is easy to see that
there exist at least one 0 < σ < R solution of (2.37) only in the case p < 0 this is
b+ w0 < l/c.

To prove uniqueness of the solution to (2.37) we rewrite it as

M1(σ) = M2(σ) (2.46)

for 0 < σ < R, where

M1(σ) = exp(j2(σ)) + p(R− σ)2f(σ)U(f(σ)),

M2(σ) = p(R− σ)
[

exp(f2(σ)) + 2f(σ)U(j(σ))
]
.

This functions satisfy

M1(0) = +∞, M1(R) = 1, M ′1(σ) < 0,

M2(0) = +∞, M2(R) = 0, M ′2(σ) < 0, M2(σ) > M1(σ)



8 A. C. BRIOZZO EJDE-2020/49

if and only if F2(σ) > F1(σ) when σ approaches 0, which is satisfied by the previous
lemma.

Then there exists unique 0 < σ < R solution of (2.37) and therefore there exists
unique −R < µ < 0 solution to (2.29). �

Theorem 2.10. Under assumption (1.8), if b + w0 < l/c, for each R > 0 then
problem (2.23)-(2.27) has a unique solution given by

T (z) = C1

[
U
(µz +R√
−2µ

)
− U

( R√
−2µ

)]
+D1, 0 < z < 1 (2.47)

with

C1 =
Rγ
√
−2/µ

exp(−R2

2µ ) +R
√

2
−µ
[
U( µ+R√

−2µ )− U( R√
−2µ )

] , (2.48)

D1 =
exp(−R

2

2µ )γ

exp(−R2

2µ ) +R
√

2
−µ
[
U( µ+R√

−2µ )− U( R√
−2µ )

] (2.49)

where −R < µ < 0 is the unique solution of (2.29).

Theorem 2.11. Assuming (1.8), b + w0 < l/c and r(t) = R/Y (t), with R > 0,
there exists a unique similarity type solution to (2.10)-(2.15) given by

v(y, t) = T
( y

Y (t)

)
= C1

[
U
(µ y

Y (t) +R
√
−2µ

)
− U

( R√
−2µ

)]
+D1,

0 < y < Y (t), t > 0 ,

(2.50)

where

Y (t) =
√

2µt+ y20 , 0 ≤ t ≤ −y
2
0

2µ
(2.51)

is the free boundary and −R < µ < 0 is the solution of equation (2.29).

The proof of the above theorem follows from Theorems 2.3 and 2.10, and (2.20).
With these results we have proved that the free boundary problem (2.10)-(2.15),
with a particular heat flow in the fixed face x = 0, has a unique similarity type
solution. Then, taking into account the transformation (2.9) we can enunciate the
following result.

Theorem 2.12. If we assume (1.8), b+ w0 < l/c and q∗(t) = R
y0s(t))

with R > 0,

then there exists a unique solution to problem (2.2)-(2.7) which is given by

u(x, t) = C1

[
U
( µ

∫ x
0

dη
u(η,t)√

−2µ
√

2µt+ y20
+

R√
−2µ

)
− U(

R√
−2µ

)
]

+D1, (2.52)

for 0 ≤ x ≤ s(t) where the coefficient −R < µ < 0 is the unique solution of (2.29),
and the free boundary is

s(t) =
1

y0

√
y20 + 2µt , 0 ≤ t < −y

2
0

2µ
. (2.53)



EJDE-2020/49 SUPERCOOLED STEFAN PROBLEM 9

Moreover the constant y0 defined by (2.15) satisfies

y0 =
µ((b+ w0)c− l)
c
√
a(µ+R)

. (2.54)

Proof. Taking into account (2.9) and Theorem 2.11 we obtain (2.52)-(2.53). From
the above theorem and (2.19) we have

s(t)− 1 =
(µ+R)c

√
a

µ
(
(b+ w0)c− l

)(√2µt+ y20 − y0
)

From (2.16), it follows that s(
y
2

0

−2µ ) = 0, then (2.54) holds. �

Corollary 2.13. Under the claim of Theorem 2.11, the free boundary given by
(2.53) satisfies

lim
t→(

−y20
2µ )−

s(t) = 0, lim
t→(

−y20
2µ )−

ṡ(t) = −∞ (2.55)

so finite time blow-up occurs.

Remark 2.14. The higher value of R decreases the value of µ, this implies that
the free boundary s = sR(t) given by (2.53) decreases, for each t > 0. This is
sR1

(t) > sR2
(t) when R1 < R2.

3. Solving the equivalent integral equation

In the previous section we have shown that if u = u(x, t) is a solution to problem
(2.2)-(2.7), then it must be solution in variable x, of the integral equation (2.52)
and the free boundary s = s(t) is given by (2.53). Reciprocally we will demonstrate
that if u is solution of integral equation (2.52) then (u, s) is the solution of the free
boundary problem (2.2)-(2.7).

Theorem 3.1. Let b + w0 < l/c and q∗(t) = R
y0s(t)

, for R > 0 and y0 defined by

(2.15). If u = u(x, t) is a solution of the integral equation (2.52) with s given by
(2.53) and function

V (x, t) =

µ
∫ x
0

dη
u(η,t)√

2µτ+y20
+R

√
−2µ

, (3.1)

where µ is the unique solution of (2.29), satisfies the following conditions

∂V

∂x
(x, t) =

−
√
−µ/2√

2µt+ y20

(
C1

[
U(V (x, t))− U( R√

−2µ )
]

+D1

) , (3.2)

V (0, t) =
R√
−2µ

, (3.3)

∂V

∂t
(x, t) =

√
−µ/2

(2µt+ y20)
(−ux(x, t)

√
2µt+ y20 −

√
−2µV (x, t)) , (3.4)

V (s(t), t) =
µ+R√
−2µ

, (3.5)

V (x, 0) =

√
−µ
∫ x
0

(b+ g(z))dz
√

2ay0
+

R√
−2µ

. (3.6)

Then (u, s) is a solution of the free boundary problem (2.2)-(2.7).



10 A. C. BRIOZZO EJDE-2020/49

The proof of the above theorem follows by taking into account the above devel-
opments and elementary computations. Following the ideas developed in [5], we
will prove that the integral equation (2.52) has a unique solution, showing that it
is equivalent to a Cauchy differential problem

Theorem 3.2. The integral equation (2.52) has a unique solution for 0 ≤ t ≤ t1 <
y20/(−2µ) where t1 is an arbitrary positive time.

Proof. We define V (x, t) by (3.1). Then (2.52) is equivalent to the Cauchy differ-
ential problem

∂V

∂x
(x, t) =

−
√
−µ/2√

2µt+ y20
(
C1

[
U(V (x, t))− U( R√

−2µ )
]

+D1

) , (3.7)

V (0, t) =
R√
−2µ

, (3.8)

with a parameter 0 ≤ t ≤ t1 < −y20/(2µ), the coefficients C1, D1 are given by
(2.48), (2.49). The claim follows by the same method as in [5]. �

Theorem 3.3. The free boundary problem (1.1)-(1.6) with b+w0 < l/c and q∗(t) =
R/(y0s(t)), for R > 0 and y0 defined by (2.17), has a unique similarity type solution
(w, s) which is given by

w(x, t) =

√
a

u(x, t)
− b, 0 < x < s(t), 0 ≤ t ≤ t1 < −

y20
2µ

(3.9)

where s(t) is the free boundary given by (2.53), u is solution of the integral equation
(2.52) and −R < µ < 0 is the unique solution of equation (2.29). Moreover we have
that finite blow-up occurs at t = −y20/(2µ).

4. Determination of thermal coefficients

Taking into account the results obtained in previous sections we have that the
solution given by Theorem 3.3 satisfies

w(0, t) =

√
a

γ

{
1 + exp(R2/2µ)R

√
2
−µ
[
U
(
µ+R√
−2µbig)− U

(
R√
−2µ

)]}
− b

= (b+ w0)N − b ,
(4.1)

where

N = 1 + exp(R2/2µ)R
√

2
−µ
[
U
(
µ+R√
−2µ

)
− U

(
R√
−2µ

)]
< 1

it is to see that the temperature at fixed face is not time dependent and it satisfies
w(0, t) < w0.

Therefore, if we add to problem (1.1)-(1.7) a Dirichlet condition of the type

w(0, t) = wi < w0 (4.2)

we have an overspecified condition at fixed face x = 0 that allows us to determine
one or two unknown thermal coefficients chosen among l, c, b, w0, wi as a function
of data depending if s = s(t) is a free boundary (unknown function) or a moving
boundary (known function) , which satisfy the problem (1.1)-(1.7).



EJDE-2020/49 SUPERCOOLED STEFAN PROBLEM 11

From (2.37) and (4.1) we have that the unknown thermal coefficients and σ = −µ
(the coefficient which characterize the free boundary s = s(t)) must satisfy the
system of the equations

wi + b

w0 + b
= 1 + R√

2σ
exp(−R

2

2σ )
[
U(R−σ√

2σ
)− U( R√

2σ
)
]
, (4.3)

exp(σ2 −R)

1 + R√
2σ

exp(−R
2

2σ )
[
U(R−σ√

2σ
)− U( R√

2σ
)
] = p(R− σ) (4.4)

or the equivalent system of equations

wi + b

w0 + b
= N(R, σ) , (4.5)

F1(σ) =
l

R[l − c(b+ w0)]
(R− σ) (4.6)

where N(R, σ) is the right hand of (4.3) which satisfies

N(R, σ) =
exp(σ2 −R)

F1(σ)
(4.7)

Firstly, we assume that s = s(t) is the free boundary, then from previous sections
we have that

s(t) =
1

y0

√
y20 − 2σt

with unknown σ. Therefore, through (4.5)-(4.6) we can determine σ and one un-
known thermal coefficient chosen among l, c, b, wi.

From the results given in Section 3 is easy to see that there exists a unique σ so-
lution to (4.5). Therefore, from (4.6) we determine an unknown thermal coefficient
chosen among l, c, b. We get the following formula for them:

Case 1: Unknown l.

l =
c(b+ w0)M(R, σ)

M(R, σ)− 1
(4.8)

where

M(R, σ) =
RF1(σ)

R− σ
. (4.9)

Case 2: Unknown c.

c =
l [M(R, σ)− 1]

(b+ w0)M(R, σ)
(4.10)

Case 3: Unknown b.

b =
l [M(R, σ)− 1]

c M(R, σ)N(R, σ)
− wi (4.11)

Case 4: Unknown wi. For the case where the unknowns coefficients are σ and wi
we obtain a unique solution σ from (4.6) and wi is determined from (4.5) and is

wi = (b+ w0)N(R, σ)− b . (4.12)

Now, if we assume that s(t) is known and it is defined by

s(t) =
1

y0

√
y20 + 2µt =

1

y0

√
y20 − 2σt



12 A. C. BRIOZZO EJDE-2020/49

for a given −R < µ < 0 (0 < σ < R), then two unknown coefficients can be
chosen among l, c, b, w0, wi which must satisfy equations (4.3)-(4.4). We get the
following cases and formulas for them:

Case 5: Unknowns wi and l.

wi = N(R, σ)(b+ w0)− b, l =
c(b+ w0)M(R, σ)

M(R, σ)− 1
(4.13)

Case 6: Unknowns w0 and l.

w0 =
wi + b

N(R, σ)
− b, l =

c(b+ w0)M(R, σ)

M(R, σ)− 1
(4.14)

Case 7: Unknowns b and l.

b =
wi −N(R, σ)w0

N(R, σ)− 1
, l =

c(b+ w0)M(R, σ)

M(R, σ)− 1
(4.15)

Case 8: Unknowns wi and c.

wi = N(R, σ)(b+ w0)− b, c =
l[M(R, σ)− 1]

(b+ w0)M(R, σ)
(4.16)

Case 9: Unknowns w0 and c.

w0 =
wi + b

N(R, σ)
− b, c =

l[M(R, σ)− 1]

(b+ w0)M(R, σ)
(4.17)

Case 10: Unknowns b and c.

b =
wi −N(R, σ)w0

N(R, σ)− 1
, c =

l[M(R, σ)− 1]

(b+ w0)M(R, σ)
(4.18)

Conclusions. We studied a supercooled one-phase Stefan problem for a semi-
infinite material with temperature-dependent diffusivity at the fixed face x = 0 and
a Neumann type condition at fixed face. We established a necessary and sufficient
condition for the heat flux q∗(t) = Q0/s(t) where s(t) is the free boundary, in
order to obtain existence and uniqueness of the solution of similarity type, local in
time. This explicit solution was obtained through the unique solution of an integral
equation with the time as a parameter. Moreover it is showed that finite time
blow-up occurs. At last, an over-specified boundary condition of Dirichlet type
is considered at the fixed face and one or two thermal coefficients are obtained,
depending if s = s(t) is a free boundary (unknown function) or a moving boundary
(known function. Formulae for these thermal coefficients are given.

Acknowledgements. This work was partially sponsored by the Project PIP No
0275 from CONICET-UA, Rosario, Argentina, by the ANPCyT PICTO Austral
2016 No 0090, and by the European Union’s Horizon 2020 Research and Innova-
tion Programme under the Marie Sklodowska-Curie grant agreement 823731 CON-
MECH.



EJDE-2020/49 SUPERCOOLED STEFAN PROBLEM 13

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Adriana C. Briozzo

CONICET and Depto. Matemática, FCE, Univ. Austral, Paraguay 1950, S2000FZF

Rosario, Argentina
Email address: abriozzo@austral.edu.ar


	1. Introduction
	2. Necessary and sufficient conditions for the existence and uniqueness of similarity type solution
	3. Solving the equivalent integral equation
	4. Determination of thermal coefficients
	Conclusions
	Acknowledgements

	References