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. 2014 Feb 16;2014:912356. doi: 10.1155/2014/912356

A Class of Nonlocal Coupled Semilinear Parabolic System with Nonlocal Boundaries

Hong Liu 1, Haihua Lu 2,*
PMCID: PMC3947733  PMID: 24696665

Abstract

We investigate the positive solutions of the semilinear parabolic system with coupled nonlinear nonlocal sources subject to weighted nonlocal Dirichlet boundary conditions. The blow-up and global existence criteria are obtained.

1. Introduction

In this paper, we consider the positive solutions of the semilinear parabolic system with coupled nonlinear nonlocal sources subject to weighted nonlocal Dirichlet boundary conditions:

uit=Δui+Ωuiqiui+1pi(x,t)dx,i=1,2,,k,uk+1=u1,xΩ,t>0,ui(x,t)=Ωφi(x,y)ui(y,t)dy,i=1,2,,k,xΩ,t>0,ui(x,0)=ui,0(x),i=1,2,,k,xΩ, (1)

where Ω is a bounded domain in ℝN, N ≥ 1, with smooth boundary ∂Ω. The exponents p i > 0, q i ≥ 0. The weighted functions φ i in the boundary conditions are continuous, nonnegative on Ω×Ω¯ and ∫Ω φ i(x, y)dy > 0 on ∂Ω. The initial data ui0(x)C2+ν(Ω¯) with 0 < ν < 1, u i0(x) ≥ 0, ≢0, and satisfy the compatibility conditions.

Many physical phenomena were formulated into nonlocal mathematical models and studied by many authors [113]. For example, in [1], Bebernes and Bressan studied an ignition model for a compressible reactive gas which is a nonlocal reaction-diffusion equation. Furthermore, Bebernes et al. [14] considered a more general model:

utΔu=f(u)+g(t),xΩ,t>0,u(x,0)=u0(x),xΩ,u(x,t)=0,xΩ,t>0, (2)

where u 0(x) ≥ 0, g(t) > 0 or g(t) = (k/|Ω|)∫Ω u t(x, t) d x with k > 0. Chadam et al. [15] studied another form of (2) with f(u) = 0 and g(t) = ∫Ω ψ(u(x, t))dx and proved that the blow-up set is the whole region (including the homogeneous Neumann boundary conditions). Souplet [16, 17] considered (2) with the general function g(t). Pao [18] discussed a nonlocal reaction-diffusion equation arising from the combustion theory.

The problems with both nonlocal sources and nonlocal boundary conditions have been studied as well. To motivate our study, we give a short review of examples of such parabolic equations or systems studied in the literature. For example, Lin and Liu [19] studied the following problem:

utΔu=Ωf(u(y,t))dy,xΩ,t>0,u(x,t)=Ωφ(x,y)u(y,t)dy,xΩ,t>0,u(x,0)=u0(x),xΩ; (3)

they established local existence, global existence, and nonexistence of solutions and discussed the blow-up properties of solutions.

Gladkov and Kim [20] considered the problem of the form

ut=Δu+c(x,t)up,xΩ,t>0,u(x,t)=Ωφ(x,y)ul(y,t)dy,xΩ,t>0,u(x,0)=u0(x),xΩ, (4)

with p, l > 0. And some criteria for the existence of global solution as well as for the solution to blow up in finite time were obtained.

In [21], Kong and Wang studied system (1) when k = 2:

ut=Δu+Ωum(x,t)vn(x,t)dx,xΩ,t>0,vt=Δv+Ωup(x,t)vq(x,t)dx,xΩ,t>0,u(x,t)=Ωφ(x,y)u(y,t)dy,v(x,t)=Ωψ(x,y)v(y,t)dy,xΩ,t>0,u(x,0)=u0(x),v(x,0)=v0(x),xΩ; (5)

they obtained the following results, and we extend them as follows.

  • (i)

    Assume that m, q < 1 and np ≤ (1 − m)(1 − q) hold; then the solution of (5) exists globally.

  • (ii)
    If one of the following conditions holds:
    (a)m>1,(b)q>1,(c)np>(1m)(1q), (6)
    then the solution of (5) blows up in a finite time for the sufficiently large initial data.
  • (iii)

    Assume that ∫Ω φ(x, y)dy ≥ 1 and ∫Ω ψ(x, y)dy ≥ 1 for all x ∈ ∂Ω and one of (6) holds; then the solution of problem (5) blows up in a finite time for any positive initial data.

Recently, Zheng and Kong [22] also studied the following problem:

utΔu=um(x,t)Ωvn(x,t)dx,xΩ,t>0,vtΔv=vq(x,t)Ωup(x,t)dx,xΩ,t>0, (7)

with the same initial and boundary conditions as (5), and they established similar conditions for global and nonglobal solutions and also blow-up solutions.

The main purpose of this paper is to get the blow-up criterion of problem (1) for any positive integer k.

In the following, we set Q T = Ω × (0, T), and  S T = ∂Ω × (0, T) with 0 < T < for convenience.

It is known by the standard theory [16, 23] that there exists a local positive solution to (1). Moreover, by the comparison principle (see Lemma 10 in the next section), the uniqueness of solutions holds if p i, q i ≥ 1, i = 1,2,…, k.

Theorem 1 —

Problem (1) has a positive classical solution (u1,u2,,uk)[C2+α^,1+α^/2(QT)C(Q¯T)]k for some α^:0<α^<1. Moreover, if T < , then

limtT(||u1(·,t)||++||uk(·,t)||)=. (8)

Theorem 2 —

If exponents p i, q i, i = 1,2,…, k satisfy

qi<1,i=1,2,,k,p1p2pk(1q1)(1q2)(1qk), (9)

the solution (u 1, u 2,…, u k) of (1) exists globally for any nontrivial nonnegative initial data.

Theorem 3 —

If exponents p i, q i, i = 1,2,…, k satisfy one of the following:

(a)qr>1,r{1,2,,k},(b)p1p2pk>(1q1)(1q2)(1qk) (10)

and if ∫Ω φ i(x, y) d y < 1, i = 1,2,…, k, for all x ∈ ∂Ω, then the solution of (1) exists globally for small nonnegative initial data.

Theorem 4 —

If exponents p i, q i, i = 1,2,…, k satisfy one of the following:

(a)qr>1,r{1,2,,k},(b)p1p2pk>(1q1)(1q2)(1qk), (11)

then the solution of (1) blows up in finite time for large initial data.

If the initial data u i,0(x) satisfies

(H)  Δui,0+Ωui,0qiui,0+1qi0,i=1,2,,k, (12)

we have another blow-up result.

Theorem 5 —

Assume that

    qr>1,Ωφr(x,y) d y1,r{1,2,,k} (13)

and the condition (H) holds. Then the solution of (1) blows up in finite time for any positive initial data.

This paper is organized as follows. Section 2 is devoted to some comparison principles. In Section 3, we prove two global existence results. The blow-up results are proved in the final section.

2. Comparison Principle

Before proving the main results, we give the maximum and comparison principles related to the problem. First, we give the following definition of the upper and lower solutions.

Definition 6 —

A pair of functions (u¯1(x,t),,u¯k(x,t)) is called an upper solution of (1), if, for every i = 1,2,…, k,  u¯i(x,t)C2,1(QT)C(Q¯T) and satisfies

u¯itΔu¯i+Ωu¯iqiu¯i+1pi(x,t)dx,u¯k+1=u¯1,xΩ,t>0,u¯i(x,t)Ωφi(x,y)u¯i(y,t)dy,xΩ,t>0,u¯i(x,0)ui,0(x),xΩ. (14)

Similarly, a lower solution of (1) is defined by the opposite inequalities.

Lemma 7 —

Suppose that aij,bi,fiC(Q¯T) and f i ≥ 0,  c i, d i ≥ 0 in Q T, g i(x, y) ≥ 0 on Ω×Ω¯, ∫Ω g i(x, y) d y > 0 on ∂Ω, i = 1,2,…, k, j = 1,2,…, N. If, for every i = 1,2,…, k, wiC2,1(QT)C(Q¯T) and satisfies

witΔwij=1Naijwixj+biwi+fi(x,t)Ω(ciwi+diwi+1) d x,(x,t)QT,wi(x,t)Ωgi(x,y)wi(y,t) d y,(x,t)ST,wi(x,0)>0,xΩ, (15)

where w k+1 = w 1, then w i(x, t) > 0, i = 1,2, on Q¯T.

Proof —

Set b¯i=supQ¯T|bi|, z i = e Kt w i with K>max{b¯i,i=1,2,,k}. Then

zitΔzi+(Kbi)zij=1Naijzixj+fi(x,t)Ω(cizi+dizi+1)dx,(x,t)QT,zi(x,t)Ωgi(x,y)zi(y,t)dy,(x,t)ST,zi(x,0)>0,xΩ. (16)

Since z i(x, 0) > 0, i = 1,2,…, there exists δ > 0 such that z i > 0 for (x,t)Ω¯×(0,δ). Suppose for a contradiction that t¯=sup{t(0,T):zi>0  on  Ω¯×[0,t),i=1,2,,k}<T. Then z i ≥ 0 on Q¯t¯, and at least one of z i vanishes at (x¯,t¯) for some x¯Ω¯. Without loss of generality, suppose that z1(x¯,t¯)=0=infQ¯t¯z1. If (x¯,t¯)Qt¯; by virtue of the first inequality of (16), we find that

z1tΔz1+(Kb1)z1j=1Naijzixj0,(x,t)Q¯t¯. (17)

This leads to the conclusion that z 1 ≡ 0 in Q¯t¯ by the strong maximum principle, a contradiction. If (x¯,t¯)St¯, this results in a contradiction too, that

0=z1(x¯,t¯)=Ωg1(x,y)z1(y,t)dy>0 (18)

due to ∫Ω g 1(x, y)dy > 0 on ∂Ω. This proves that z 1 > 0 and consequently w 1 > 0. We complete the proof.

Lemma 8 —

Suppose that, for every i = 1,2,…, k, wiC2,1(QT)C(Q¯T) and satisfies

witΔwiΩ(ai(x,t)wi+bi(x,t)wi+1) d x,(x,t)QT,wi(x,t)Ωgi(x,y)wi(y,t) d y,(x,t)ST,wi(x,0)0,xΩ, (19)

where w k+1 = w 1 and a i(x, t), b i(x, t) are continuous, nonnegative functions in Q¯T, g i(x, y) ≥ 0 on Ω×Ω¯ such that ∫Ω g i(x, y) d y < 1 on ∂Ω, and there exist positive constants C i such that ∫Ω(a i(x, t) + b i(x, t)) d xC i. Then w i(x, t) ≥ 0, i = 1,2, on Q¯T.

Proof —

Suppose that the strict inequalities of (19) hold; by Lemma 7, we have w i(x, t) > 0. Now we consider the general case. Set

vi=wi+ɛeKt, (20)

where ɛ is any fixed positive constant, and K = 1 + max⁡{∫Ω(a i(x, t) + b i(x, t))  d x, i = 1,2,…, k}. By (19), we get, for i = 1,2,…, k,

vitΔviΩ(ai(x,t)vi+bi(x,t)vi+1)dxɛeKt(KΩ(ai(x,t)+bi(x,t))dx)>0,(x,t)QT,vi(x,t)Ωgi(x,y)vi(y,t)dyɛeKt(1Ωgi(x,y)dy)>0,(x,t)ST,vi(x,0)ɛeKt>0,xΩ, (21)

Therefore, we have v i(x, t) ≥ 0 on Q T. Letting ɛ → 0+, we get the desired result.

If the boundary condition ∫Ω g i(x, y)dy < 1 is not necessarily valid, we have the following result. The argument of its proof can be referred to [22, Lemma 2.2].

Lemma 9 —

Suppose that aij,bi,fiC(Q¯T), f i ≥ 0, c i, d i, are nonnegative and bounded in Q T, g i(x, y) ≥ 0 on Ω×Ω¯, ∫Ω g i(x, y) d y > 0 on ∂Ω, i = 1,2,…, k, j = 1,2,…, N. If, for every i = 1,2,…, k,  wiC2,1(QT)C(Q¯T) and satisfies

witΔwij=1Naijwixj+biwi+fi(x,t)Ω(ciwi+diwi+1) d x,(x,t)QT,wi(x,t)Ωgi(x,y)wi(y,t) d y,(x,t)ST,wi(x,0)0,xΩ, (22)

where w k+1 = w 1, then w i(x, t) ≥ 0, i = 1,2, on Q¯T.

By Lemma 9, we can easily get the following result.

Lemma 10 —

Let (u¯1,u¯2,,u¯k) and (u_1,u_2,,u_k) be nonnegative upper and lower solution of system (1) on Q¯t, respectively. If one assumes that, for some r ∈ {1,2,…, k},

  1. u¯r+1>δ or u_r+1>δ when p r < 1,

  2. u¯r>δ or u_r>δ when q r < 1,

then (u¯1,u¯2,,u¯k)(u_1,u_2,,u_k) on Q T.

3. Global Existence Results

Before proving Theorem 2, we give a global existence result for a scalar equation.

Lemma 11 —

Let w 0(x) and φ(x, y) be continuous, nonnegative functions on Ω¯ and Ω×Ω¯, respectively, and let the nonnegative constants θ ij satisfy 0 < θ i1 + θ i2 ≤ 1. Then the solutions of the nonlocal problem

wtΔw=i=1kwθi1(x,t)Ωwθi2(x,t) d x,xΩ,t>0,w(x,t)=Ωφ(x,y)w(y,t) d y,xΩ,t>0,w(x,0)=w0(x),xΩ (23)

exist globally.

Proof —

The augment is similar to the proof of [22, Lemma 3.1] or [21, Lemma 6]. For the reader's convenience, we complete it. It is easy to prove that there exists a positive function ψC2(Ω¯) such that

minΩ¯ψ(x)>maxΩ¯w02(x),ψ(x)Ωφ2(x,y)dyΩψ(y)dy,xΩ. (24)

Let θ > 0 be large enough such that

2θminΩ¯ψ(x)(2k+1)max{maxΩ¯|Δψ(x)|,|Ω|[maxΩ¯ψ(x)](θi1+θi2+1)/2(i=1,2,,k)|Ω|}. (25)

Setting z(x, t) = e 2θt ψ(x) for (x, t) ∈ Ω × (0, ), one readily checks that

ztΔz2i=1kz(θi1+1)/2(x,t)Ωzθi2/2(x,t)dx,xΩ,t>0,z(x,t)Ωφ2(x,y)dyΩz(y,t)dy,xΩ,t>0,z(x,0)w02(x)+1,xΩ, (26)

Let w¯=z1/2(x,t); it follows that

w¯tΔw¯i=1kw¯θi1(x,t)Ωw¯θi2(x,t)dx,xΩ,t>0,w¯(x,t)Ωφ2(x,y)dyΩw¯(y,t)dy,xΩ,t>0,w¯(x,0)>w0(x),xΩ. (27)

This implies that w¯ is a global upper solution of (23). Clearly, 0 is a lower solution of it. So we complete the proof.

Proof of Theorem 2

By (11), we know that there exists a i ∈ (0,1), i = 1,2,…, k, such that

pi1qiaiai+1,i=1,2,,k,ak+1=a1. (28)

Define α = ∑i=1 k1/a i. Let Φ(x, y) ≥ max⁡{φ i(x, y), i = 1,2,…, k} be a continuous function defined for (x,y)Ω×Ω¯. Suppose that z solves

ztΔz=αi=1kz1ai(x,t)Ωzai(x,t)dx,xΩ,t>0,z(x,t)=i=1kgi(x)ΩΦ(x,y)z(y,t)dy,xΩ,t>0,z(x,0)=1+i=1kui,01/ai(x),xΩ, (29)

where

gi(x)=(ΩΦ(x,y)dy)(1ai)/ai. (30)

In view of Lemma 11, we know that z is global. Moreover, z > 1 in Ω¯×[0,) by the maximum principle. Set u¯i=zai, i = 1,2,…, k. By (28) and (29) and using Ho¨lder's inequality, we get

uitΔuiΩuiqiui+1pidx=aizai1ztaizai1Δzai(ai1)|z|2Ωzaiqi+ai+1pidxaizai1(ztΔz)Ωzaidx(αai1)Ωzaidx0,(x,t)QT,uiΩφi(x,y)ui(y,t)dy=zaiΩφi(x,y)zai(y,t)dy(Ωφi(x,y)dy)1ai(Ωφi(x,y)z(y,t)dy)aiΩφi(x,y)zai(y,t)dy0,(x,t)ST,ui(x,0)ui,0(x),xΩ. (31)

This means that (u¯1,u¯2,,u¯k) is a global upper solution of (1).

Proof of Theorem 3

Define

max{supΩΩφi(x,y)dy,i=1,2,,k}=δ0(0,1). (32)

Let w be the unique solution of the elliptic problem

Δw=1,xΩ;w=C0,  xΩ. (33)

Then there exists a constant M > 0 such that C 0w(x) ≤ C 0 + M in Ω¯. We choose C 0 to be large enough such that

1+C01+C0+Mδ0. (34)

Set u¯i(x,t)=bi(1+w(x)). When (x, t) ∈ S T, it follows that

u¯iΩφi(x,y)u¯i(y,t)dy=bi(1+C0)biΩφi(x,y)(1+w(y))dybi[1+C0(1+C0+M)δ0]0. (35)

Now we investigate (x, t) ∈ Q T. Set L i = (1 + C 0 + M)pi+qi | Ω| for convenience. A simple computation yields

u¯itΔu¯iΩu¯iqiu¯i+1pidx=bibiqibi+1piΩ(1+w(x))pi+qidxbiqi(bi1qibi+1piLi). (36)

(a) If q r > 1, no matter q r+1 > 1 or q r+1 ≤ 1, we can choose b r to be small enough such that b r 1−qrb r+1 pr L r. For fixed b r, there exist b i, i = 1,2,…, r − 1, r + 1,…, k, satisfying b i 1−qib i+1 pi L i, i = 1,2,…, k. It follows that

u¯itΔu¯iΩu¯iqiu¯i+1pidx0,i=1,2,,k. (37)

(b) If q i ≤ 1, i = 1,2,…, k and p 1 p 2p k > (1 − q 1)(1 − q 2)⋯(1 − q k), we can choose b 1 to be small enough such that

b1(1q1)(1q2)(1qk)>b1p1p2pkL1(1q2)(1qk)L2p1(1q3)(1qk)Lk1p1p2pk2(1qk1)Lkp1p2pk1. (38)

Consequently, there exist b i > 0, i = 2,3,…, k, b k+1 = b 1 satisfying b i 1−qib i+1 pi L i, i = 1,2,…, k. Hence (37) holds too.

By (35) and (37), in any case (a) or (b), we know that the solution of (1) must be global for small data u i,0(x) ≤ b i(1 + w(x)), i = 1,2,…, k for xΩ.

4. Blow-Up Results

In this section, we assume that (u(x, t), v(x, t)) is a positive solution of (1) on Ω¯×[0,T), where T is the maximal existence time.

Proof of Theorem 4

We denote by λ 1, ϕ 1(x) the first eigenvalue and the corresponding eigenfunction of the linear elliptic problem:

Δφ(x)=λφ(x),xΩ;φ(x)=0,xΩ, (39)

and ϕ 1(x) satisfies

φ1(x)>0,xΩ,maxΩ¯ϕ1(x)=1. (40)

Define γ = min⁡{α i(q i − 1) + α i+1 p i + 1, i = 1,2,…, k}.

(a) If q r ≥ 1, we claim that there exist positive constants α i > 1, i = 1,2,…, k, such that the inequality

αi(qi1)+αi+1pi>0 (41)

holds. First, when i = r, (41) holds for any α r, α r+1 > 1. When i = r + 1, if q r+1 ≥ 1, (41) holds for any α r+2 > 1; if q r+1 ≤ 1 we can choose α r+2 > max⁡{1, α r+1(1 − q r+1)/p r+1}. That is, (41) holds too. When i = r − 1, if q r−1 ≥ 1, (41) holds for any α r−1 > 1; if q r−1 < 1, we can choose 1 < α r−1 < (α i p r−1/1 − q r−1) such that (41) holds too.

(b) If q i < 1, i = 1,2,…, k, and p 1 p 2p k > (1 − q 1)(1 − q 2) ⋯ (1 − q k), we can choose α i > 1 such that

p11q1>α1α2,p21q2>α2α3,,pk1qk>αkα1. (42)

Hence (41) holds too.

Hence, for the case (a) or (b), we all have γ > 1. Now let s(t) be the unique solution of the ODE problem

s(t)=λs(t)+lsγ(t),t>0,s(0)=s0>1, (43)

where l = min⁡{(1/α i)∫Ω ϕ 1 αiqi+αi+1pi, i = 1,2,…, k}. Then s(t) blows up in finite time T(s 0) with s 0 being large enough.

Set

u_i=sαi(t)ϕ1αi(x),(x,t)Ω¯×[0,T(s0)),i=1,2,,k. (44)

We will show that (u_,v_) is a lower solution of problem (1). A direct computation yields

u_itΔu_iΩu_iqiu_i+1pidx=αilsαi1+γϕ1αiαi(αi1)sαiϕ1αi2|ϕ1|2Ωsαiqi+ai+1piϕ1αiqi+ai+1pidxαilsαi1+γsαiqi+ai+1piΩϕ1αiqi+ai+1pidx0,(x,t)Ω×[0,T(s0)),u_iΩφi(x,y)u_i(y,t)dy=0sαi(t)Ωφi(x,y)ϕ1αi(y)dy0,(x,t)Ω×(0,T(s0)). (45)

(u_1,,uk_) is a blowing up lower solution of (1) provided the initial data are so large that u i,0(x) ≥ s αi(0)ϕ 1 αi(x), i = 1,2,…, k for xΩ. We complete the proof.

Proof of Theorem 5

Since u i,0 > 0 in Ω, ∫Ω φ r(x, y) d y > 0 on ∂Ω, and

ui,0(x)=Ωφr(x,y)ui,0(y) d y,xΩ, (46)

by the compatibility conditions, we have u i,0 > 0 on ∂Ω. Denote by η the positive constant such that u i,0 > η on Ω¯. The assumption (H) implies that (u i)t > 0 by the comparison principle, and in turn u i > η, i = 1,2,…, k on Ω¯×[0,T). Furthermore, u r satisfies

(ur)tΔur+|Ω|ηprurqr,(x,t)QT,ur=Ωφr(x,y)ur(y,t) d y,(x,t)ST,ur(x,0)=ur,0(x),xΩ. (47)

Let z r(t) be the solution of the following Cauchy problem:

zr(t)=|Ω|ηprzrqr,zr(0)=12η>0. (48)

Clearly, z r(t) blows up under the condition

qr>1. (49)

On the other hand, since ∫Ω φ r(x, y) d y ≥ 1, by Lemma 9, we have u rz r as long as both u r and z r exist, and thus u r blows up for any positive initial data. The proof now is completed.

Acknowledgments

This work was supported by Natural Science Fund for Colleges and Universities in Jiangsu Province, 12KJB110018, and College Students' Innovative Projects, 2013.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

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