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).

1. Introduction

Consider the following free boundary problem:
(1)ut=uxx+u(1-u),0<x<h(t),t>0,u(t,0)=u(t,h(t))=0,t>0,h′(t)=-μux(t,h(t))-μα,t>0,h(0)=h0,u(0,x)=u0(x),0≤x≤h0,
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 u0 belongs to Y(h0) for some h0>0, where
(2)Y(h0)∶={ϕ∈C2([0,h0])∶ϕ(0)=ϕ(h0)=0,ssϕ(x)≥(≢)0in(0,h0)}.

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 to infinity. 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 ux(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 given γ∈(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)×C1+γ/2([0,T]),
where DT:={(t,x)∈R2:x∈[0,h(t)],t∈(0,T]}, and the solution can be extended to some interval (0,T0) with T0>T as long as inf0<t<Th(t)>0.

Moreover, as in the proof of [9, Lemma 2.8], one can show that h∞:=limt→Th(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

spreading: h∞=+∞ and
(4)limt→∞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

vanishing: limt→Th(t)=0 and
(6)T<+∞,limt→Tmax0≤x≤h(t)u(t,x)=0
or

transition: 0<h∞<+∞ and
(7)limt→∞u(t,·)=v(·)locally uniformly in(0,h∞),
where v is the solution of
(8)v′′+v(1-v)=0,x∈(0,h∞),hhhhv(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 in a finite time and the free boundary converges to the point 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 when α<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 that T∈(0,∞), h¯∈C1([0,T]), and u¯∈C(D¯T)∩C1,2(DT) with DT={(t,x)∈R2:0<t≤T,0<x<h¯(t)} and
(9)u¯t≥u¯xx+u¯(1-u¯),0<t≤T,0<x<h¯(t),u¯(t,0)≥0,u¯(t,h¯(t))=0,0<t≤T,h¯′(t)≥-μu¯(t,h¯(t))-μα,0<t≤T.
If h0≤h¯(0) and u0(x)≤u¯(0,x) in [0,h0] and if (u,h) is a solution of (1), then
(10)h(t)≤h¯(t),u(x,t)≤u¯(x,t)fort∈(0,T],sssssssssssssssssssssssssssssshhggkkx∈(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 limt→T*h(t)=0, then T*<+∞ and
(11)limt→T*max0≤x≤h(t)u(t,x)=0.

Proof.

By [2, 10], one can prove that there exists a constant C1 such that u(t,x)≤C1. In order to prove that u converges to 0, we need to construct the function
(12)U(t,x):=C1[2M(h(t)-x)-M2(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+α2+22,4∥u0∥C1([-h0,h0])3C1}.

Clearly 0≤U≤C1 in Q. By the definitions of U and M, we have
(15)Ut-Uxx-U(1-U)≥C1(2M2-2Mα-1)≥0inQ.
Moreover,
(16)U(t,h(t))=u(t,h(t))=0fort∈(0,T*),U(t,0)>0=u(t,0)whenh(t)<M-1.
Therefore, u(t,x)≤U(t,x) in Q by the comparison principle Lemma 4. Note that limt→T*h(t)=0; then there exists T1<T* such that h(t)-M-1<0 for t>T1. Therefore, u(t,x)≤U(t,x) for t>T1 and x∈[0,h(t)]. For such t and x, we have
(17)U(t,x)≤2MC1h(t)⟶0ast⟶T*;
it follows that
(18)∥u(t,·)∥L∞([0,h(t)])⟶0ast⟶T*.

We now prove that T*<+∞. By limt→T*h(t)=0, there is some L*>0 such that
(19)h(t)≤L*fort∈[0,T*).
Set L:=2(1+L*) and
(20)ξ0(x):=2εL2(L2-x2),
where ε>0 is small such that
(21)8(α+α2+2)ε≤α,32ε≤α.
Consider the problem
(22)ξt=ξxx+2ξ(1-ξ2ε),0<x<h-(t),t>0,ξ(t,0)=ξ(t,h-(t))=0,t>0,h-′(t)=-μξx(t,h-(t))-μα,t>0,h-(0)=L,ξ(0,x)=ξ0(x),0≤x≤L.
It is obvious that ξ(t,x)≤2ε for all t≥0. Construct a function
(23)Uε(t,x):=2ε[2M(h-(t)-x)-M2(h-(t)-x)2]
over Q¯:={(t,x):t>0,max{0,h-(t)-M-1}≤x≤h-(t)}, where M:=max{α+α2+2,4}. Then Uε(t,x) is an upper solution of (22) over Q¯ and so
(24)-ξx(t,h-(t))≤-Uxε(t,h-(t))=4Mε≤α2.
Therefore, h-′(t)≤-αμ/2. Thus, h-(t)→0 as t→T¯*≤2L/αμ.

On the other hand, (18) implies that there exists some T0∈(0,T*) such that u(t,x)≤ε for all x∈[0,h(t)] and t>T0. Clearly ξ0(x)≥u(T0,x) for x∈[0,h(T0)]. By the comparison principle, we have h(t+T0)≤h-(t), and so T* cannot be ∞.

Theorem 7.

Assume that 0<α<3/3. Let (u,h) be a solution of (1). If 0<h∞<+∞, then
(25)h∞=Lα,limt→∞u(t,·)=vα(·)locallyuniformlyin(0,h∞),
where vα is a unique positive solution of
(26)v′′+v(1-v)=0,0<x<Lα,v(0)=v(Lα)=0,v′(0)=-v′(Lα)=α,
where
(27)Lα:=2∫0Bdrα2-r2+(2/3)r3
with B∈(0,1) given by α2=2∫0Bs(1-s)ds.

Remark 8.

This is a new phenomenon. It never happens when α=0. Moreover, by the phase plane method, one can prove that vα→0 and Lα→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∫01s(1-s)ds=3/3.

Proof of Theorem <xref ref-type="statement" rid="thm2.4">7</xref>.

For any ε>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)forx∈(0,h∞),ccu¯0(0)=u¯0(h∞+ε)=0.

By the comparison principle we have u(t,x)≤u¯(t,x) in (t*,∞)×(0,h(t)), where u¯(t,x) is the solution of
(29)u¯t=u¯xx+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

u¯→0 as t→∞ if h∞+ε≤π; or

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)limt→∞u(t,x)=0,orlimsupt→∞u(t,x)≤u¯ε*.
Similarly,
(32)liminf t→∞u(t,x)≥u_ε*(x)whenh∞-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)limt→∞u(t,x)=0ifh∞≤π,
or when h∞>π,
(35)limt→∞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 that limt→∞u(t,x)=0 is impossible when h∞>0. Suppose that this does not hold; there exists L0 such that h(t)≤L0. Then using the approach of proving T*<+∞ in Theorem 7, we can show that limt→Th(t)=0 for some 0<T<+∞; this contradicts the assumption h∞>0. Hence, limt→∞u(t,x)=u*(x), locally uniformly in (0,h∞); we next prove that u*(x)=vα(x).

Make a change of the variable x to reduce [0,h(t)] to the fixed interval [0,h0] and use Lp estimates as well as Sobolev embedding theorems on the reduced equation with Dirichlet boundary conditions to conclude that
(37)∥u(t,·)-u*(·)∥C1+(γ/2)([0,h(t)])⟶0(t⟶∞)
for some γ>0. It follows that h′(t)=-μux(t,h(t))-μα→-μ(u*)′(h∞)-μα 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)limt→∞u(t,x)=w(x)locallyuniformlyin[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 function W0(x)≥u0(x) for x∈[0,h0] and W0≥0 for x∈[0,+∞). Let W(t,x) be the unique solution of
(40)Wt=Wxx+W(1-W),t>0,x>0,W(t,0)=0,t>0,W(0,x)=W0(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)limsupt→∞u(t,x)≤limt→∞W(t,x)=w(x) forx∈[0,+∞).

On the other hand, since h∞=+∞, 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_xx+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)fort>τ,0≤x≤l.
By [11], one can obtain
(44)limt→∞u_l(t,x)=vl(x)uniformlyin[0,l],
where vl is the positive solution of
(45)v′′+v(1-v)=0,0<x<l,v(0)=v(l)=0,
It is well known that liml→∞vl(x)=w(x). Combining this with (43) and (44), we have
(46)w(x)≤liminft→∞u(t,x).
By (41) and (46), we have
(47)limt→∞u(t,x)=w(x).

Theorem 11.

Suppose that α≥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 limt→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 that α<3/3. Let h0>0 and u0(x)∈Y(h0); then the following properties hold:

vanishing happens when u0(x)<vα(x);

spreading happens if u0(x)>vα(x) for x∈[0,h0];

transition happens if u0(x)≡vα(x) for x∈[0,h0].

Proof.

(i) By [1], we see that vα1(x)<vα2(x) for α1<α2. Since u0(x)<vα(x), there is β<α such that u0(x)<vβ(x), by the comparison principle that u(t,x)<vβ(x), so h∞≠+∞ and h∞≠Lα. It then follows from Theorem 6 that vanishing happens.

(ii) Let (u,h) be a solution of (1) with initial data u0(x); by the phase plane analysis, there is γ>α such that u0(x)>vγ(x). It then follows from the comparison principle that u(t,x)>vγ(x), so Theorem 10 implies that h∞=+∞ and spreading happens.

(iii) It follows from the comparison principle Lemma 4 that u(t,x)≡vα(x) and h(t)≡Lα for all t>0.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This research was supported by Shanghai Natural Science Foundation (no. 13ZR1454900) and Shanghai University Young Teachers Training Scheme (no. ZZsdl13021).

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