# Asymptotic Behavior of Solutions of Free Boundary Problem with Logistic Reaction Term

Authors: Jingjing Cai
Journal: Journal of Applied Mathematics
 (2014)

---

## Abstract

We study a free boundary problem for a reaction diffusion equation modeling the spreading of a biological or chemical species. In this model, the free boundary represents the spreading front of the species. We discuss the asymptotic behavior of bounded solutions and obtain a trichotomy result: spreading (the free boundary tends to+
∞ and the solution converges to a stationary solution defined on [
0
+
∞
)), transition (the free boundary stays in a bounded interval and the solution converges to a stationary solution with positive compact support), and vanishing (the free boundary converges to 0 and the solution tends to 0 within a finite time).


---

## Body

## 1. Introduction

Consider the following free boundary problem:(1)
u
t
=
u
x
x
+
u
(
1
-
u
)
,
0
<
x
<
h
(
t
)
,
t
>
0
,
u
(
t
,
0
)
=
u
(
t
,
h
(
t
)
)
=
0
,
t
>
0
,
h
′
(
t
)
=
-
μ
u
x
(
t
,
h
(
t
)
)
-
μ
α
,
t
>
0
,
h
(
0
)
=
h
0
,
u
(
0
,
x
)
=
u
0
(
x
)
,
0
≤
x
≤
h
0
,
where x
=
h
(
t
) is a moving boundary to be determined together with u
(
t
,
x
) and α
>
0 is a given constant. The initial function u
0 belongs to Y
(
h
0
) for some h
0
>
0, where
(2)
Y
(
h
0
)
∶
=
{
ϕ
∈
C
2
(
[
0
,
h
0
]
)
∶
ϕ
(
0
)
=
ϕ
(
h
0
)
=
0
,
s
s
ϕ
(
x
)
≥
(
≢
)
0
in
(
0
,
h
0
)
}
.Recently, problem (1) with α
=
0 was studied by [1–3] and so forth. They used this model to describe the spreading of a new or invasive species; they used the free boundary h
(
t
) which represents the expanding front of the species whose density is represented by u
(
t
,
x
). They obtained a spreading-vanishing dichotomy result; namely, the species either spreads to the whole environment and stabilizes at the positive state 1 (i.e., u
→
1) or vanishes (i.e., u
→
0) as time goes toinfinity. Such a result shows that problem (1) with α
=
0 has advantages comparing with the Cauchy problems (the Cauchy problems have hair-trigger effect: any positive solution which converges to a positive constant; cf. [4, 5]). In the last two years, [6] also studied the corresponding problem of (1) with α
=
0 in high dimension spaces.In this paper, we mainly study problem (1) with α
>
0; such a boundary condition represents that there is a spreading resistant force at the front for some species. Intuitively, the presence of α
>
0 makes the solution more difficult to spread than the case where α
=
0. Indeed, h
′
(
t
)
>
0 only if u
x
(
t
,
h
(
t
)
)
<
-
α. This boundary condition is widely used in many biological models. For example, it is often used in protocell models (cf. [7, 8]).We give the following theorem whose proof is similar to that of [1, 2]. It suffices to repeat their arguments with obvious modification.Theorem 1.
For any givenγ
∈
(
0,1
), there is a T
∈
(
0
,
+
∞
) such that free boundary problem (1) has a solution
(3)
(
u
,
h
)
∈
C
(
(
1
+
γ
)
/
2
)
,
1
+
γ
(
D
¯
T
)
×
C
1
+
γ
/
2
(
[
0
,
T
]
)
,
where D
T
:
=
{
(
t
,
x
)
∈
R
2
:
x
∈
[
0
,
h
(
t
)
]
,
t
∈
(
0
,
T
]
}, and the solution can be extended to some interval (
0
,
T
0
) with T
0
>
T as long as inf
⁡
0
<
t
<
T
h
(
t
)
>
0.Moreover, as in the proof of [9,  Lemma 2.8], one can show that h
∞
:
=
lim
⁡
t
→
T
h
(
t
)
∈
[
0
,
+
∞
] exist.The main purpose of this paper is to study the asymptotic behavior of bounded solutions of (1) and obtain trichotomy result. We will prove that, for a solution (
u
,
h
) of (1), one has either(i)
spreading:h
∞
=
+
∞ and
(4)
lim
⁡
t
→
∞
u
(
t
,
x
)
=
w
(
x
)
locally uniformly in
(
0
,
+
∞
)
,
where w is the unique positive solution of
(5)
q
′′
+
q
(
1
-
q
)
=
0
,
x
>
0
,
q
(
0
)
=
0
,
or
(ii)
vanishing:lim
⁡
t
→
T
h
(
t
)
=
0 and
(6)
T
<
+
∞
,
lim
⁡
t
→
T
max
⁡
0
≤
x
≤
h
(
t
)
u
(
t
,
x
)
=
0
or
(iii)
transition:0
<
h
∞
<
+
∞ and
(7)
lim
⁡
t
→
∞
u
(
t
,
·
)
=
v
(
·
)
locally uniformly in
(
0
,
h
∞
)
,
where v is the solution of
(8)
v
′′
+
v
(
1
-
v
)
=
0
,
x
∈
(
0
,
h
∞
)
,
h
h
h
h
v
(
0
)
=
v
(
h
∞
)
=
0
,
-
v
′
(
h
∞
)
=
α
.Remark 2.
Comparing with the results in [1–3], the phenomenon (iii) is a new one, since it does not happen in case α
=
0.Remark 3.
(ii) shows that vanishing happens ina finite time and the free boundary converges to thepoint
0; those phenomena are also new and do not happen in case α
=
0.


## 2. Asymptotic Behavior of Solutions

In this section, we study the asymptotic behavior of solutions and obtain trichotomy result whenα
<
3
/
3; namely, the solution of (1) is either vanishing (Theorem 6) or transition (Theorem 7) or spreading (Theorem 10). Then, we prove that only vanishing happens if α
≥
3
/
3 (Theorem 11) for the completeness of the paper.We first prepare the following comparison theorems which can be proved similarly as in [2, Lemma 3.5].Lemma 4.
Suppose thatT
∈
(
0
,
∞
), h
¯
∈
C
1
(
[
0
,
T
]
), and u
¯
∈
C
(
D
¯
T
)
∩
C
1,2
(
D
T
) with D
T
=
{
(
t
,
x
)
∈
R
2
:
0
<
t
≤
T
,
0
<
x
<
h
¯
(
t
)
} and
(9)
u
¯
t
≥
u
¯
x
x
+
u
¯
(
1
-
u
¯
)
,
0
<
t
≤
T
,
0
<
x
<
h
¯
(
t
)
,
u
¯
(
t
,
0
)
≥
0
,
u
¯
(
t
,
h
¯
(
t
)
)
=
0
,
0
<
t
≤
T
,
h
¯
′
(
t
)
≥
-
μ
u
¯
(
t
,
h
¯
(
t
)
)
-
μ
α
,
0
<
t
≤
T
.
If h
0
≤
h
¯
(
0
) and u
0
(
x
)
≤
u
¯
(
0
,
x
) in [
0
,
h
0
] and if (
u
,
h
) is a solution of (1), then
(10)
h
(
t
)
≤
h
¯
(
t
)
,
u
(
x
,
t
)
≤
u
¯
(
x
,
t
)
f
o
r
t
∈
(
0
,
T
]
,
ssssssssssssssssssssssssssssss
h
h
g
g
k
k
x
∈
(
0
,
h
(
t
)
)
.Remark 5.
The pair(
u
¯
,
h
¯
) is usually called an upper solution of problem (1) and one can define a lower solution by revising all the inequalities.Theorem 6.
Let(
u
,
h
) be a solution of (1) on [
0
,
T
*
). If lim
⁡
t
→
T
*
h
(
t
)
=
0, then T
*
<
+
∞ and
(11)
lim
⁡
t
→
T
*
max
⁡
0
≤
x
≤
h
(
t
)
u
(
t
,
x
)
=
0
.Proof.
By [2, 10], one can prove that there exists a constant C
1 such that u
(
t
,
x
)
≤
C
1. In order to prove that u converges to 0, we need to construct the function
(12)
U
(
t
,
x
)
:
=
C
1
[
2
M
(
h
(
t
)
-
x
)
-
M
2
(
h
(
t
)
-
x
)
2
]
over the region
(13)
Q
:
=
{
(
t
,
x
)
:
0
<
t
<
T
*
,
max
⁡
{
h
(
t
)
-
M
-
1
,
0
}
<
x
<
h
(
t
)
}
,
where
(14)
M
:
=
max
⁡
{
α
+
α
2
+
2
2
,
4
∥
u
0
∥
C
1
(
[
-
h
0
,
h
0
]
)
3
C
1
}
.
Clearly0
≤
U
≤
C
1 in Q. By the definitions of U and M, we have
(15)
U
t
-
U
x
x
-
U
(
1
-
U
)
≥
C
1
(
2
M
2
-
2
M
α
-
1
)
≥
0
in
Q
.
Moreover,
(16)
U
(
t
,
h
(
t
)
)
=
u
(
t
,
h
(
t
)
)
=
0
for
t
∈
(
0
,
T
*
)
,
U
(
t
,
0
)
>
0
=
u
(
t
,
0
)
when
h
(
t
)
<
M
-
1
.
Therefore, u
(
t
,
x
)
≤
U
(
t
,
x
) in Q by the comparison principle Lemma 4. Note that lim
⁡
t
→
T
*
h
(
t
)
=
0; then there exists T
1
<
T
* such that h
(
t
)
-
M
-
1
<
0 for t
>
T
1. Therefore, u
(
t
,
x
)
≤
U
(
t
,
x
) for t
>
T
1 and x
∈
[
0
,
h
(
t
)
]. For such t and x, we have
(17)
U
(
t
,
x
)
≤
2
M
C
1
h
(
t
)
⟶
0
as
t
⟶
T
*
;
it follows that
(18)
∥
u
(
t
,
·
)
∥
L
∞
(
[
0
,
h
(
t
)
]
)
⟶
0
as
t
⟶
T
*
.
We now prove thatT
*
<
+
∞. By lim
⁡
t
→
T
*
h
(
t
)
=
0, there is some L
*
>
0 such that
(19)
h
(
t
)
≤
L
*
for
t
∈
[
0
,
T
*
)
.
Set L
:
=
2
(
1
+
L
*
) and
(20)
ξ
0
(
x
)
:
=
2
ε
L
2
(
L
2
-
x
2
)
,
where ε
>
0 is small such that
(21)
8
(
α
+
α
2
+
2
)
ε
≤
α
,
32
ε
≤
α
.
Consider the problem
(22)
ξ
t
=
ξ
x
x
+
2
ξ
(
1
-
ξ
2
ε
)
,
0
<
x
<
h
-
(
t
)
,
t
>
0
,
ξ
(
t
,
0
)
=
ξ
(
t
,
h
-
(
t
)
)
=
0
,
t
>
0
,
h
-
′
(
t
)
=
-
μ
ξ
x
(
t
,
h
-
(
t
)
)
-
μ
α
,
t
>
0
,
h
-
(
0
)
=
L
,
ξ
(
0
,
x
)
=
ξ
0
(
x
)
,
0
≤
x
≤
L
.
It is obvious that ξ
(
t
,
x
)
≤
2
ε for all t
≥
0. Construct a function
(23)
U
ε
(
t
,
x
)
:
=
2
ε
[
2
M
(
h
-
(
t
)
-
x
)
-
M
2
(
h
-
(
t
)
-
x
)
2
]
over Q
¯
:
=
{
(
t
,
x
)
:
t
>
0
,
max
⁡
{
0
,
h
-
(
t
)
-
M
-
1
}
≤
x
≤
h
-
(
t
)
}, where M
:
=
max
⁡
{
α
+
α
2
+
2
,
4
}. Then U
ε
(
t
,
x
) is an upper solution of (22) over Q
¯ and so
(24)
-
ξ
x
(
t
,
h
-
(
t
)
)
≤
-
U
x
ε
(
t
,
h
-
(
t
)
)
=
4
M
ε
≤
α
2
.
Therefore, h
-
′
(
t
)
≤
-
α
μ
/
2. Thus, h
-
(
t
)
→
0 as t
→
T
¯
*
≤
2
L
/
α
μ.
On the other hand, (18) implies that there exists some T
0
∈
(
0
,
T
*
) such that u
(
t
,
x
)
≤
ε for all x
∈
[
0
,
h
(
t
)
] and t
>
T
0. Clearly ξ
0
(
x
)
≥
u
(
T
0
,
x
) for x
∈
[
0
,
h
(
T
0
)
]. By the comparison principle, we have h
(
t
+
T
0
)
≤
h
-
(
t
), and so T
* cannot be ∞.Theorem 7.
Assume that0
<
α
<
3
/
3. Let (
u
,
h
) be a solution of (1). If 0
<
h
∞
<
+
∞, then
(25)
h
∞
=
L
α
,
lim
⁡
t
→
∞
u
(
t
,
·
)
=
v
α
(
·
)
l
o
c
a
l
l
y
u
n
i
f
o
r
m
l
y
i
n
(
0
,
h
∞
)
,
where v
α is a unique positive solution of
(26)
v
′′
+
v
(
1
-
v
)
=
0
,
0
<
x
<
L
α
,
v
(
0
)
=
v
(
L
α
)
=
0
,
v
′
(
0
)
=
-
v
′
(
L
α
)
=
α
,
where
(27)
L
α
:
=
2
∫
0
B
‍
d
r
α
2
-
r
2
+
(
2
/
3
)
r
3
with B
∈
(
0,1
) given by α
2
=
2
∫
0
B
‍
s
(
1
-
s
)
d
s.Remark 8.
This is a new phenomenon. It never happens whenα
=
0. Moreover, by the phase plane method, one can prove that v
α
→
0 and L
α
→
π as α
→
0. This conclusion gives an explanation of Lemma 3.1 in [2]; that is, vanishing happens if h
∞
≤
π.Remark 9.
It is easily seen that (26) has no positive solution when α
≥
2
∫
0
1
‍
s
(
1
-
s
)
d
s
=
3
/
3.Proof of Theorem7.
For anyε
>
0, there exists t
*
>
0 such that h
∞
-
ε
<
h
(
t
)
<
h
∞
+
ε for t
>
t
*. Let u
¯
0
(
x
) be a function defined on (
0
,
h
∞
+
ε
) and satisfies
(28)
u
¯
0
(
x
)
≥
u
(
t
*
,
x
)
for
x
∈
(
0
,
h
∞
)
,
c
c
u
¯
0
(
0
)
=
u
¯
0
(
h
∞
+
ε
)
=
0
.
By the comparison principle we haveu
(
t
,
x
)
≤
u
¯
(
t
,
x
) in (
t
*
,
∞
)
×
(
0
,
h
(
t
)
), where u
¯
(
t
,
x
) is the solution of
(29)
u
¯
t
=
u
¯
x
x
+
u
¯
(
1
-
u
¯
)
,
t
>
t
*
,
0
<
x
<
h
∞
+
ε
,
u
¯
(
t
,
0
)
=
u
¯
(
t
,
h
∞
+
ε
)
=
0
,
t
>
t
*
,
u
¯
(
t
*
,
x
)
=
u
¯
0
(
x
)
,
0
<
x
<
h
∞
+
ε
.
It is well known that (i)
u
¯
→
0 as t
→
∞ if h
∞
+
ε
≤
π; or
(ii)
u
¯
→
u
¯
ε
* as t
→
∞ if h
∞
+
ε
>
π, where u
¯
ε
* is a positive function. More precisely, when h
∞
+
ε
>
π, it follows from [11, Corollary 3.4] that u
¯
ε
* is the unique positive solution of
(30)
(
u
¯
ε
*
)
′′
+
u
¯
ε
*
(
1
-
u
¯
ε
*
)
=
0
,
0
<
x
<
h
∞
+
ε
,
u
¯
ε
*
(
h
∞
+
ε
)
=
u
¯
ε
*
(
0
)
=
0
.
Hence,
(31)
lim
⁡
t
→
∞
u
(
t
,
x
)
=
0
,
or
lim
⁡
sup
⁡
t
→
∞
⁡
u
(
t
,
x
)
≤
u
¯
ε
*
.
Similarly,
(32)
lim
⁡
inf
t
→
∞
u
(
t
,
x
)
≥
u
_
ε
*
(
x
)
when
h
∞
-
ε
>
π
,
where u
_
ε
*
(
x
) is a positive solution of
(33)
(
u
_
ε
*
)
′′
+
u
_
ε
*
(
1
-
u
_
ε
*
)
=
0
,
0
<
x
<
h
∞
-
ε
,
u
_
ε
*
(
h
∞
-
ε
)
=
u
_
ε
*
(
0
)
=
0
.
We conclude from (31) and (32) that
(34)
lim
⁡
t
→
∞
u
(
t
,
x
)
=
0
if
h
∞
≤
π
,
or when h
∞
>
π,
(35)
lim
⁡
t
→
∞
u
(
t
,
x
)
=
u
*
(
x
)
locally uniformly in
(
0
,
h
∞
)
,
where u
*
(
x
) is the unique positive solution of
(36)
(
u
*
)
′′
+
u
*
(
1
-
u
*
)
=
0
,
0
<
x
<
h
∞
,
u
*
(
h
∞
)
=
u
*
(
0
)
=
0
.
We now show thatlim
⁡
t
→
∞
u
(
t
,
x
)
=
0 is impossible when h
∞
>
0. Suppose that this does not hold; there exists L
0 such that h
(
t
)
≤
L
0. Then using the approach of proving T
*
<
+
∞ in Theorem 7, we can show that lim
⁡
t
→
T
h
(
t
)
=
0 for some 0
<
T
<
+
∞; this contradicts the assumption h
∞
>
0. Hence, lim
⁡
t
→
∞
u
(
t
,
x
)
=
u
*
(
x
), locally uniformly in (
0
,
h
∞
); we next prove that u
*
(
x
)
=
v
α
(
x
).
Make a change of the variablex to reduce [
0
,
h
(
t
)
] to the fixed interval [
0
,
h
0
] and use L
p estimates as well as Sobolev embedding theorems on the reduced equation with Dirichlet boundary conditions to conclude that
(37)
∥
u
(
t
,
·
)
-
u
*
(
·
)
∥
C
1
+
(
γ
/
2
)
(
[
0
,
h
(
t
)
]
)
⟶
0
(
t
⟶
∞
)
for some γ
>
0. It follows that h
′
(
t
)
=
-
μ
u
x
(
t
,
h
(
t
)
)
-
μ
α
→
-
μ
(
u
*
)
′
(
h
∞
)
-
μ
α as t
→
∞. Hence, we conclude that (
0
,
h
∞
) is not a finite interval unless -
(
u
*
)
′
(
h
∞
)
=
α.Theorem 10.
Let(
u
,
h
) be a solution of (1). If h
∞
=
+
∞, then
(38)
lim
⁡
t
→
∞
u
(
t
,
x
)
=
w
(
x
)
l
o
c
a
l
l
y
u
n
i
f
o
r
m
l
y
i
n
[
0
,
+
∞
)
,
where w is the unique positive solution of
(39)
q
′′
+
q
(
1
-
q
)
=
0
,
x
>
0
,
q
(
0
)
=
0
.Proof.
Choose a bounded continuous functionW
0
(
x
)
≥
u
0
(
x
) for x
∈
[
0
,
h
0
] and W
0
≥
0 for x
∈
[
0
,
+
∞
). Let W
(
t
,
x
) be the unique solution of
(40)
W
t
=
W
x
x
+
W
(
1
-
W
)
,
t
>
0
,
x
>
0
,
W
(
t
,
0
)
=
0
,
t
>
0
,
W
(
0
,
x
)
=
W
0
(
x
)
,
x
>
0
.
Then the comparison principle theorem shows that u
(
t
,
x
)
≤
W
(
t
,
x
) for t
>
0, x
>
0. Using [11, Lemma 3.4], we see that
(41)
lim
⁡
sup
⁡
t
→
∞
⁡
u
(
t
,
x
)
≤
lim
⁡
t
→
∞
W
(
t
,
x
)
=
w
(
x
)
for
x
∈
[
0
,
+
∞
)
.
On the other hand, sinceh
∞
=
+
∞, for any large l
>
π, there is τ
>
0 such that h
(
τ
)
=
l and h
(
t
)
≥
l for all t
>
τ. Let u
_
l
(
t
,
x
) be the solution of the following problem:
(42)
u
_
t
=
u
_
x
x
+
u
_
(
1
-
u
_
)
,
t
>
τ
,
0
<
x
<
l
,
u
_
(
t
,
0
)
=
u
_
(
t
,
l
)
=
0
,
t
>
τ
,
u
_
(
0
,
x
)
=
ψ
(
x
)
,
0
<
x
<
l
,
where ψ is a nonnegative continuous function satisfying ψ
(
x
)
≤
u
(
τ
,
x
) for 0
<
x
<
l. The comparison principle implies
(43)
u
_
l
(
t
,
x
)
≤
u
(
t
,
x
)
for
t
>
τ
,
0
≤
x
≤
l
.
By [11], one can obtain
(44)
lim
⁡
t
→
∞
u
_
l
(
t
,
x
)
=
v
l
(
x
)
uniformly
in
[
0
,
l
]
,
where v
l is the positive solution of
(45)
v
′′
+
v
(
1
-
v
)
=
0
,
0
<
x
<
l
,
v
(
0
)
=
v
(
l
)
=
0
,
It is well known that lim
⁡
l
→
∞
v
l
(
x
)
=
w
(
x
). Combining this with (43) and (44), we have
(46)
w
(
x
)
≤
lim
⁡
inf
⁡
t
→
∞
⁡
u
(
t
,
x
)
.
By (41) and (46), we have
(47)
lim
⁡
t
→
∞
u
(
t
,
x
)
=
w
(
x
)
.Theorem 11.
Suppose thatα
≥
3
/
3 and (
u
,
h
) is a solution of (1) defined on some maximal existence interval [
0
,
T
*
); then T
*
<
+
∞, u converges to 0 as t
→
T
*, and lim
⁡
t
→
T
*
h
(
t
)
=
0.Proof.
The proof of this theorem is similar to [10]; it suffices to repeat their arguments with obvious modification.


## 3. Example

In this section, we give some sufficient conditions for vanishing, spreading, and transition.Example 1.
Suppose thatα
<
3
/
3. Let h
0
>
0 and u
0
(
x
)
∈
Y
(
h
0
); then the following properties hold: (i)
vanishing happens whenu
0
(
x
)
<
v
α
(
x
);
(ii)
spreading happens ifu
0
(
x
)
>
v
α
(
x
) for x
∈
[
0
,
h
0
];
(iii)
transition happens ifu
0
(
x
)
≡
v
α
(
x
) for x
∈
[
0
,
h
0
].Proof.
(i) By [1], we see that v
α
1
(
x
)
<
v
α
2
(
x
) for α
1
<
α
2. Since u
0
(
x
)
<
v
α
(
x
), there is β
<
α such that u
0
(
x
)
<
v
β
(
x
), by the comparison principle that u
(
t
,
x
)
<
v
β
(
x
), so h
∞
≠
+
∞ and h
∞
≠
L
α. It then follows from Theorem 6 that vanishing happens.
(ii) Let(
u
,
h
) be a solution of (1) with initial data u
0
(
x
); by the phase plane analysis, there is γ
>
α such that u
0
(
x
)
>
v
γ
(
x
). It then follows from the comparison principle that u
(
t
,
x
)
>
v
γ
(
x
), so Theorem 10 implies that h
∞
=
+
∞ and spreading happens.
(iii) It follows from the comparison principle Lemma4 that u
(
t
,
x
)
≡
v
α
(
x
) and h
(
t
)
≡
L
α for all t
>
0.



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*Source: 724582-2014-06-02.xml*