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Blow-up problems for quasilinear reaction diffusion equations with weighted nonlocal source

Juntang Ding

B

and Xuhui Shen

School of Mathematical Sciences, Shanxi University, Taiyuan 030006, P.R. China Received 21 July 2017, appeared 8 January 2018

Communicated by Maria Alessandra Ragusa

Abstract.In this paper, we investigate the following quasilinear reaction diffusion equa- tions

(b(u))t=∇ · ρ |∇u|2u

+c(x)f(u) in×(0,t),

∂u

∂ν =0 on∂Ω×(0,t),

u(x, 0) =u0(x)≥0 inΩ.

Here is a bounded domain in Rn (n2) with smooth boundary ∂Ω. Weighted nonlocal source satisfies

c(x)f(u(x,t))≤a1+a2(u(x,t))p Z

(u(x,t))αdx m

,

where a2,p,α are some positive constants and a1,mare some nonnegative constants.

We make use of a differential inequality technique and Sobolev inequality to obtain a lower bound for the blow-up time of the solution. In addition, an upper bound for the blow-up time is also derived.

Keywords:blow-up problems, quasilinear reaction equation, weighted nonlocal source.

2010 Mathematics Subject Classification: 35B44, 35B40, 35K57.

1 Introduction

The blow-up problems to reaction diffusion equations has been extensively investigated by many researchers. Much of the work prior to the turn of the century is referenced in [1,9,10].

More recent work, we refer readers to [13–18,21]. In practical situations, one would like to know whether the solutions blows up and if so, at which time blow-up occurs. Hence, finding bounds for blow-up time has become the focus of the researchers, especially the search for lower bounds of blow-up time. Since Payne and Schaefer [20] introduced a first-order inequality technique and obtained a lower bound for blow-up time, many authors are devoted to the lower bounds of blow-up time for various reaction diffusion problems, (see, for instance,

BCorresponding author. Email: djuntang@sxu.edu.cn

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[3–7]). We note that above mentioned studies mainly aimed at seeking lower bounds for blow- up time of local reaction-diffusion equations. In this paper, we concern the reaction diffusion equations with weighted nonlocal source









(b(u))t =∇ · ρ |∇u|2∇u

+c(x)f(u) inΩ×(0,t),

∂u

∂ν =0 on∂Ω×(0,t),

u(x, 0) =u0(x)≥0 inΩ.

(1.1)

In (1.1), Ωis a bounded domain of Rn (n ≥ 2)with smooth boundary Ω, ν represents the unit normal vector to∂Ω,u0(x)∈C1()is a nonnegative function satisfying the compatibility condition,tis the blow-up time if blow-up occurs, or else t= ∞. Weighted nonlocal source satisfies

c(x)f(u(x,t))≤ a1+a2(u(x,t))p Z

(u(x,t))αdx m

,

where a2,p,α are some positive constants and a1,m are some nonnegative constants. Set R+ = (0,∞). Throughout this paper, we assume that bis a C2(R+) function withb0(s) > 0 fors >0,ρis a positiveC2(R+)function satisfying ρ(s) +2sρ0(s)>0 fors>0,cis a positive C()function, and f is a nonnegativeC(R+)function. By maximum principles [22], we know that the classical solutionuof (1.1) is nonnegative inΩ×[0,t).

For the information about the nonlocal reaction diffusion equations, we refer readers to [2,11,12,19,23]. Fang and Ma [11] dealt with the following problems

















 ut =

n i,j=1

(aij(x)uxi)xj−c(x)f(u) in Ω×(0,t),

n i,j=1

aij(x)uxiνj =g(u) on ∂Ω×(0,t), u(x, 0) =u0(x) in Ω,

whereΩ⊂Rn(n≥2)is a bounded star-shaped domain with smooth boundaryΩ, nonlocal source satisfies

f(u(x,t))≥a2(u(x,t))p Z

(u(x,t))αdx m

,

anda2,p,α, andm are positive constants. They derived conditions which imply the solution blows up in finite time or exists globally. Furthermore, upper and lower bounds for blow-up time are obtained.

As far as we know, there is little information on the bounds for blow-up time of problem (1.1). Motivated by the above work [11], we study the problem (1.1). Our results of this paper are based on some Sobolev type inequalities and differential inequality technique. In Section 2, when Ω ⊂ Rn (n ≥ 2), we obtain a criterion for blow-up of the solution of (1.1) and get an upper bound for blow-up time. In Section 3, whenΩ ⊂ Rn (n ≥ 3), we derive a lower bound for blow-up time. An example is presented in Section 4 to illustrate our abstract results derived in this paper.

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2 Blow-up solution

In this section, we establish conditions on data to ensure that the solution blows up att and obtain an upper bound fort. To accomplish these tasks, we introduce the following auxiliary functions

D(t) =

Z

G(u(x,t))dx, E(t) =−

Z

P(|∇u|2)dx+2 Z

c(x)F(u)dx, t ≥0, (2.1)

G(u) =2 Z u

0 sb0(s)ds, P(|∇u|2) =

Z |∇u|2

0

ρ(s)ds, F(u) =

Z u

0 f(s)ds, (2.2) where u is the classical solution of (1.1). Our main result of this section is the following Theorem2.1

Theorem 2.1. Let u be a classical solution of (1.1). We suppose that functions b,c,ρ, and f satisfy b00(s)<0, sρ(s)≤(1+β)P(s),

Z

c(x)s(x,t)f(s(x,t))dx ≥2(1+β)

Z

c(x)F(s(x,t))dx, s ≥0, (2.3) whereβis a nonnegative constant. In addition, initial data are assumed to satisfy

E(0) =−

Z

P(|∇u0|2)dx+2 Z

c(x)F(u0)dx>0. (2.4) Then u must blow up at t≤ T in measure D(t)with

T =





D(0)

2β(1+β)E(0), β>0,

∞, β=0.

Proof. It follows from Green’s formula and (2.3) that D0(t) =

Z

G0(u)utdx=2 Z

ub0(u)utdx

=2 Z

u

∇ · ρ(|∇u|2)∇u

+c(x)f(u)dx

=2

Z

∇ · uρ(|∇u|2)∇u dx−2

Z

ρ(|∇u|2)|∇u|2dx+2

Z

c(x)u f(u)dx

=2 Z

∂Ωuρ(|∇u|2)∂u

∂νdS−2 Z

ρ(|∇u|2)|∇u|2dx+2 Z

c(x)u f(u)dx

≥ −2(1+β)

Z

P(|∇u|2)dx+4(1+β)

Z

c(x)F(u)dx

=2(1+β)

Z

P(|∇u|2)dx+2 Z

c(x)F(u)dx

=2(1+β)E(t). (2.5)

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DifferentiatingE(t), we get E0(t) =−2

Z

ρ(|∇u|2) (∇u· ∇ut)dx+2 Z

c(x)f(u)utdx

=2 Z

∂Ωρ(|∇u|2)ut∂u

∂νdS−2 Z

ρ(|∇u|2) (∇u· ∇ut)dx+2 Z

c(x)f(u)utdx

=2 Z

∇ · ρ(|∇u|2)ut∇u dx−2

Z

ρ |∇u|2(∇u· ∇ut)dx+2 Z

c(x)f(u)utdx

=2 Z

ut

∇ · ρ(|∇u|2)∇u

+c(x)f(u)dx

=2 Z

b0(u)u2tdx≥0, (2.6)

which with (2.4) imply E(t) > 0 and D0(t) > 0 for all t ∈ (0,t). By the Hölder inequality, (2.5) andb0(s)>0 fors>0, we obtain

2(1+β)E(t)D0(t)≤ D0(t)2=

2 Z

b0(u)uutdx 2

4 Z

b0(u)u2dx Z

b0(u)u2tdx

. (2.7)

Using (2.3) and integrating by part, we lead to G(u) =2

Z u

0 sb0(s)ds =

Z u

0 b0(s)ds2 =b0(u)u2

Z u

0 s2b00(s)ds ≥b0(u)u2. (2.8) We combine (2.7) and (2.8) to derive

(1+β)E(t)D0(t)≤2 Z

G(u)dx Z

b0(u)u2tdx

=D(t)E0(t); that is

E(t)D−(1+β)(t)0 ≥0. (2.9) Integrating (2.9) over[0,t], we have

E(t)D−(1+β)(t)≥E(0)D−(1+β)(0). By (2.5), we can deduce

D0(t)D−(1+β)(t)≥2(1+β)E(0)D−(1+β)(0). (2.10) Ifβ>0, integrating (2.10) over[0,t], we derive

Dβ(t)≤Dβ(0)−2β(1+β)E(0)D−(1+β)(0)t. (2.11) This inequality can not hold for allt > 0. Hence,u(x,t)must blow up at some finite timet in the measureD(t). Furthermore, we conclude from (2.11)

t ≤T= D(0) 2β(1+β)E(0). Ifβ=0, we integrate (2.10) to get

D(t)≥D(0)e2E(0)D1(0)t, which impliesT=∞.

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3 Lower bound for blow-up time

In this section, we restrict Ω ⊂ Rn (n ≥ 3). Our goal is to determine a lower bound for blow-up time t. Here we impose the following constraints on data

ρ(s)≥b1+b2sq, b0(s)≥γ, c(x)f(s(x,t))≤a1+a2(s(x,t))p

Z

(s(x,t))αdx m

, s ≥0, (3.1)

where a2,b2,p,q,γ are positive constants, a1,b1,m are nonnegative constants, p > 2q+1, α=2r(q+1)−2q, and parameterr is restricted by the condition

r>max

1,n(p−2q−1) +4q 4(q+1)

. (3.2)

We introduce two auxiliary functions A(t) =

Z

B(u)dx, t ≥0, B(u) =α Z u

0

sα1b0(s)ds.

In this section, we also need to apply the following Sobolev inequality (see [8, Theorem 2, p. 265]) forn≥3,

Z

(vq+1)n2n2dx n2n2

≤C Z

v2(q+1)dx+

Z

|∇vq+1|2dx 12

, (3.3)

where C = C(n,Ω) is an embedding constant. The main result of this section is stated as follows.

Theorem 3.1. Let u be a classical solution of (1.1). Assume that(3.1)–(3.2) hold. If u blows up at finite time tin measure A(t), we then conclude that blow-up time tis bounded from blow by

t

Z

A(0)

dτ K1ταα1 +K2τ

[4r(q+1)+2q(n2)](1+m)−(n2)(p1) 4r(q+1)+2q(n2)−n(p1) +K3τ

2r(q+1) 2r(q+1)−2q

, where

K1= a1α||1αγ

1α

α , (3.4)

K2= a2α[4r(q+1) +2q(n−2)−n(p−1)]

4r(q+1) +2q(n−2) 2C

2 n

(p1) 4r(q+1)+2q(n2)−n(p1)

× 1+σ

4r(q+1)+2qn((pn12))−n(p1) 1

! γ

[4r(q+1)+2q(n2)](1+m)−(n2)(p1)

4r(q+1)+2q(n2)−n(p1) , (3.5)

K3=

b2qα(α−1)

r2(q+1) + a2(p−1)

4r(q+1) +2q(n−2) 2C

2 n

(p1) 4r(q+1)+2q(n2)−n(p1)

× 4r(q+1)−4q

2r(q+1) +q(n−2)(2C2)

nq 2r(q+1)−2qγ

2r(q+1) 2r(q+1)−2q

×

"

σ

2r(q+nq1)−2q

2 +

2r(q+1) +q(n−2) 2nq

2r(q+nq1)−2q#

, (3.6)

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σ1 = b2(α−1)[2r(q+1) +q(n−2)]

a2n(p−1)(q+1)r2(q+1) 2C24r(q+1)+n2q((pn12))−n(p1)

, (3.7)

σ2 = b2(α−1)[2r(q+1) +q(n−2)]

2nq(q+1)

×

"

2b2q(α−1) + a2n(p−1)r2(q+1) 2r(q+1) +q(n−2) 2C

2 n

(p1) 4r(q+1)+2q(n2)−n(p1)

#1

. (3.8)

Proof. By (3.2), we haveα>2. It follows from Green’s formula and (3.1) that A0(t) =

Z

B0(u)utdx= α Z

uα1b0(u)utdx

=α Z

uα1

∇ · ρ(|∇u|2)∇u

+c(x)f(u)dx

=α Z

∇ ·uα1ρ(|∇u|2)∇u

dx−α(α−1)

Z

ρ(|∇u|2)uα2|∇u|2dx +α

Z

uα1c(x)f(u)dx

α Z

∂Ωuα1ρ(|∇u|2)∂u

∂νdS−α(α−1)

Z

uα2 b1+b2|∇u|2q|∇u|2dx +a1α

Z

uα1dx+a2α Z

uα+p1dx Z

uαdx m

≤ −b2α(α−1)

Z

uα2|∇u|2(q+1)dx+a1α Z

uα1dx +a2α

Z

uα+p1dx Z

uαdx m

. (3.9)

We apply the Hölder inequality to get Z

uα1dx ≤ ||1α Z

uαdx α1

α

. (3.10)

For brevity, we denotev =ur and

|∇ur|2(q+1) =r2(q+1)u2(r1)(q+1)|∇u|2(q+1). (3.11) Hence, by (3.10)–(3.11), (3.9) can be rewritten as

A0(t)≤ −b2α(α−1) r2(q+1)

Z

|∇ur|2(q+1)dx+a1α||1α Z

uαdx α1

α

+a2α Z

uα+p1dx Z

uαdx m

= − b2α(α−1) r2(p+1)

Z

|∇v|2(q+1)dx+a1α||1α Z

vαrdx α1

α

+a2α Z

v2(q+1)+p2qr1dx Z

vαrdx m

. (3.12)

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Using the Hölder inequality and the Young inequality, we have Z

|∇vq+1|2dx= (q+1)2

Z

v2q|∇v|2dx

≤(q+1)2 Z

v2(q+1)dx

q+q1 Z

|∇v|2(q+1)dx q+11

≤q(q+1)

Z

v2(q+1)dx+ (q+1)

Z

|∇v|2(q+1)dx;

that is

Z

|∇v|2(q+1)dx≥ 1 q+1

Z

∇vq+1

2dx−q Z

v2(q+1)dx. (3.13) Substituting (3.13) into (3.12), we get

A0(t)≤ b2(α−1) r2(q+1)

Z

v2(q+1)dx− b2α(α−1) (q+1)r2(q+1)

Z

|∇vq+1|2dx+a1α||1α Z

vαrdx α1

α

+a2α Z

v2(q+1)+p2qr1dx Z

vαrdx m

. (3.14)

Now, we deal with the last term of (3.14). Applying the Hölder inequality and (3.3), we derive

Z

v2(q+1)+p2qr1dx Z

vαrdx m

Z

vαrdx

4r(q+1)+4r(2qq+(n1)+22q)−((nn22))(p1)+mZ

(vq+1)n2n2dx

4r((qn+12)+)(2qp(n1)2)

Z

vαrdx

4r(q+1)+4r(2qq+(n1)+22q)−((nn22))(p1)+m"

Cn2n2 Z

v2(q+1)dx+

Z

|∇vq+1|2dx

nn2#4r((qn+12)+)(2qp(1n)2)

= Z

vαrdx

4r(q+1)+4r(2qq+(n1)+22q)−((nn22))(p1)+m C2

Z

v2(q+1)dx+C2 Z

|∇vq+1|2dx

4r(q+n1()+p2q1)(n2)

, (3.15) where 0< 4r((qn+12)+)(2qp(n1)2) < 1 in view of (3.2). Using in (3.15) the basic inequality

(k1+k2)j ≤2j(k1j +kj2), k1, k2, j>0, (3.16) we have

Z

v2(q+1)+p2qr1dx Z

vαrdx m

≤ 2C2 n

(p1) 4r(q+1)+2q(n2)

Z

vαrdx

4r(q+1)+4r(2qq+(n1)+22q)−((nn22))(p1)+mZ

v2(q+1)dx

4r(q+n1()+p2q1)(n2)

+ 2C2

n(p1) 4r(q+1)+2q(n2)

Z

vαrdx

4r(q+1)+4r(2qq+(n1)+22q)−((nn22))(p1)+m

× Z

|∇vq+1|2dx

4r(q+n1()+p2q1)(n2)

. (3.17)

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An application of the Young inequality to the first term of (3.17) yields

Z

vαrdx

4r(q+1)+4r(2qq+(n1)+22q)−((nn22))(p1)+mZ

v2(q+1)dx

4r(q+n1()+p2q1)(n2)

=

Z

vαrdx

[4r(q+4r1)+(q2q+1(n)+2q2)]((n1+2m)−)−(n(np12))(p1)

4r(q+1)+2q(n2)−n(p1) 4r(q+1)+2q(n2)

Z

v2(q+1)dx

4r(q+n1()+p2q1)(n2)

4r(q+1) +2q(n−2)−n(p−1) 4r(q+1) +2q(n−2)

Z

vαrdx

[4r(q+14r)+(q2q+1()+n2q2)]((n1+2m)−)−(n(pn12))(p1)

+ n(p−1) 4r(q+1) +2q(n−2)

Z

v2(q+1)dx, (3.18)

where we use the fact that 0< 4r(q+n1()+p2q1)(n2) <1 due to (3.2). Similarly, for the second term of (3.17), we apply the Young inequality to obtain

Z

vαrdx

4r(q+1)+4r(2qq+(n1)+22q)−((nn22))(p1)+mZ

|∇vq+1|2dx

4r(q+n1()+p2q1)(n2)

=

Z

vαrdx

[4r(q+14r)+(q2q+1()+n2q2)]((n1+2m)−)−(n(pn12))(p1)

4r(q+1)+2q(n2)−n(p1) 4r(q+1)+2q(n2)

Z

|∇vq+1|2dx

4r(q+n1()+p2q1)(n2)

4r(q+1) +2q(n−2)−n(p−1) 4r(q+1) +2q(n−2) σ

4r(q+1)+2qn((pn12))−n(p1) 1

Z

vαrdx

[4r(q+4r1)+(q2q+1()+n2q2)]((n1+2m)−)−(n(pn12))(p1)

+ n(p−1)σ1 4r(q+1) +2q(n−2)

Z

|∇vq+1|2dx, (3.19)

whereσ1is given in (3.7). Substituting (3.18)–(3.19) into (3.17), we have Z

v2(q+1)+p2qr1dx Z

vαrdx m

4r(q+1) +2q(n−2)−n(p−1) 4r(q+1) +2q(n−2) 2C

2

n(p1) 4r(q+1)+2q(n2)−n(p1)

× 1+σ

4r(q+1)+2qn((pn12))−n(p1) 1

!Z

vαrdx

[4r(q+14r)+(q2q+1()+n2q2)]((n1+2m)−)−(n(pn12))(p1)

+ n(p−1)

4r(q+1) +2q(n−2) 2C

2 n

(p1) 4r(q+1)+2q(n2)−n(p1)

×

σ1 Z

|∇vq+1|2dx+

Z

v2(q+1)dx

. (3.20)

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