In addition, Ding obtained a lower bound for the blow-up time when Ω⊆R3is a bounded convex domain

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

LOWER BOUNDS FOR THE BLOW-UP TIME OF NONLINEAR PARABOLIC PROBLEMS WITH ROBIN BOUNDARY

CONDITIONS

KHADIJEH BAGHAEI, MAHMOUD HESAARAKI

Abstract. In this article, we find a lower bound for the blow-up time of solu- tions to some nonlinear parabolic equations under Robin boundary conditions in bounded domains ofRⁿ.

1. Introduction

In this article, we consider the nonlinear initial-boundary value problem (b(u))t=∇ ·(g(u)∇u) +f(u), x∈Ω, t >0

∂u

∂ν +γu= 0, x∈∂Ω, t >0, u(x,0) =u0(x)≥0, x∈Ω

(1.1)

where Ω⊆Rⁿ, n≥3, is a bounded domain with smooth boundary,νis the outward normal vector to∂Ω,γis a positive constant andu₀(x)∈C¹(Ω) is the initial value.

We assume thatf is a nonnegative C(R⁺) function and the nonnegative functions g andbsatisfy

g∈C¹(R⁺), g(s)≥g_m>0, g⁰(s)≤0, ∀s >0,

b∈C²(R⁺), 0< b⁰(s)≤b⁰_M, b⁰⁰(s)≤0, ∀s >0, (1.2) wheregm andb⁰_M are positive constants.

The reader is referred to [1, 3, 4, 5, 6, 8] for results on bounds for blow-up time in nonlinear parabolic problems. Ding [2] studied problem (1.1) under assumptions (1.2) and derived conditions on the data which imply blow-up or the global existence of solutions. In addition, Ding obtained a lower bound for the blow-up time when Ω⊆R³is a bounded convex domain. Here we obtain a lower bound for the blow-up time for (1.1) in general bounded domains Ω⊆Rⁿ, n≥3.

2. A lower bound for the blow-up time

In this section we find a lower bound for the blow-up time T in an appropriate measure. The idea of the proof of the following theorem comes from [1].

2000Mathematics Subject Classification. 35K55, 35B44.

Key words and phrases. Parabolic equation; Robin boundary condition; blow-up; lower bound.

c

2014 Texas State University - San Marcos.

Submitted June 6, 2013. Published April 16, 2014.

1

Theorem 2.1. Let Ω be a bounded domain in Rⁿ, n ≥ 3, and let the functions f, g, b satisfy

0< f(s)≤cg(s)Z s 0

b⁰(y) g(y)dy^p+1

, s >0, (2.1)

for some constantsc >0 andp≥1. Ifu(x, t)is a nonnegative classical solution to problem (1.1), which becomes unbounded in the measure

Φ(t) = Z

Ω

Z u(x,t) 0

b⁰(y) g(y)dy2k

dx,

wherek is a parameter restricted by the condition k >max

p(n−2),1 , (2.2)

thenT is bounded from below by Z +∞

Φ(0)

dξ

k1+k2ξ^{3(n−2)3n−8} +k3ξ^{2(n−2)2n−3}

, (2.3)

where k1, k2 and k3 are positive constants which will be determined later in the proof.

Proof. To simplify our computations we define v(s) =

Z s 0

b⁰(y)

g(y)dy, s >0. (2.4)

Hence, dΦ

dt = d dt

Z

Ω

v^2kdx= 2k Z

Ω

v^2k−1b⁰(u) g(u)utdx

= 2k Z

Ω

v^2k−1(b(u))t

g(u) dx

= 2k Z

Ω

v^2k−1 1 g(u)

h∇ ·(g(u)∇u) +f(u)i dx

=−2k(2k−1) Z

Ω

v^2k−2v⁰(u)|∇u|²dx+ 2k Z

Ω

v^2k−1g⁰(u)

g(u)|∇u|²dx

−2kγ Z

∂Ω

v^2k−1u ds+ 2k Z

Ω

v^2k−1f(u) g(u)dx

≤ −2k(2k−1) Z

Ω

v^2k−2b⁰(u)

g(u)|∇u|²dx+ 2k Z

Ω

v^2k−1f(u) g(u)dx,

where in the above inequality we usedu≥0 andg⁰(u)≤0 from (1.2). From (2.4), we have

|∇u|²=g(u) b⁰(u)

²

|∇v|². (2.5)

By (1.2), (2.5), and (2.1) we have dΦ

dt ≤ −2k(2k−1) Z

Ω

v^2k−2g(u)

b⁰(u)|∇v|²dx+ 2k Z

Ω

v^2k−1f(u) g(u)dx

≤ −2(2k−1)gm

kb⁰_M Z

Ω

|∇v^k|²dx+ 2kc Z

Ω

v^2k+pdx.

(2.6)

From (2.2), H¨older, and Young inequalities, we infer Z

Ω

v^2k+pdx≤ |Ω|^m1Z

Ω

v^{k(2n−3)n−2} dxm2

≤m₁|Ω|+m₂ Z

Ω

v^{k(2n−3)n−2} dx,

(2.7)

where

m₁= k(2n−3)−(n−2)(2k+p)

k(2n−3) , m₂=(n−2)(2k+p) k(2n−3) . From (2.7) and the Cauchy-Schwartz inequality we have:

Z

Ω

v^{k(2n−3)n−2} dx≤Z

Ω

v^2kdx^1/2Z

Ω

v^{2k(n−1)n−2} dx^1/2

≤Z

Ω

v^2kdx³₄Z

Ω

(v^k)ⁿ⁻²²ⁿ dx^1/4 .

(2.8)

Applying the Sobolev inequality (see [7]) to the last term in (2.8), for n >3, we obtain

kv^kk

n 2(n−2)

L

2n n−2(Ω)

≤(c_s)^2(n−2)n kv^kk

n 2(n−2)

W^1,2(Ω)

≤(cs)^2(n−2)n k∇v^kk

n 2(n−2)

L²(Ω) +kv^kk

n 2(n−2)

L²(Ω)

(2.9)

In the case,n= 3, we have kv^kk

n 2(n−2)

Lⁿ⁻²²ⁿ (Ω)

≤(cs)^2(n−2)n kv^kk

n 2(n−2)

W^1,2(Ω)

≤2^{2(n−2)4−n} (cs)^2(n−2)n k∇v^kk

n 2(n−2)

L²(Ω) +kv^kk

n 2(n−2)

L²(Ω)

.

(2.10)

Here,csis the best constant in the Sobolev inequality.

By inserting (2.9) in (2.8) forn >3 and (2.10) in (2.8) forn= 3, we have Z

Ω

v^{k(2n−3)n−2} dx

≤c0

Z

Ω

v^2kdx³₄ k∇v^kk

n 2(n−2)

L²(Ω) +kv^kk

n 2(n−2)

L²(Ω)

=c0

Z

Ω

v^2kdx³₄Z

Ω

|∇v^k|²dx_4(n−2)ⁿ +c0

Z

Ω

v^2kdx_2(n−2)²ⁿ⁻³ ,

(2.11)

where

c₀=







2^{2(n−2)4−n} (cs)^2(n−2)n , forn= 3, (c_s)^2(n−2)n , forn >3.

Now, using Young’s inequality we obtain Z

Ω

v^{k(2n−3)n−2} dx

≤ c

4(n−2) 3n−8

0 (3n−8)

4(n−2)³ⁿ⁻⁸ⁿ Φ^{3(n−2)3n−8} + n 4(n−2)

Z

Ω

|∇v^k|²dx+c0Φ^{2(n−2)2n−3} ,

(2.12)

whereis a positive constant to be determined later. Substituting (2.12) into (2.7) yields

2kc Z

Ω

v^2k+pdx≤2kcm2

n (3n−8) 4(n−2)³ⁿ⁻⁸ⁿ c

4(n−2) 3n−8

0 Φ^{3(n−2)3n−8} + n

4(n−2) Z

Ω

|∇v^k|²dx

+c0Φ^{2(n−2)2n−3} o

+ 2kcm1|Ω|.

By inserting the last inequality in (2.6), we have dΦ

dt ≤

−2(2k−1)g_m

kb⁰_M + nkcm₂ 2(n−2)

Z

Ω

|∇v^k|²dx+k₁+k₂Φ^{3(n−2)3n−8} +k₃Φ^{2(n−2)2n−3} , where

k1= 2kcm1|Ω|, k2=2kcm2(c0)^{4(n−2)3n−8} (3n−8)

4(n−2)³ⁿ⁻⁸ⁿ , k3= 2kcc0m2. For

=4(n−2)(2k−1)gm

nk²cm2b⁰_M , the above inequality becomes

dΦ

dt ≤k1+k2Φ^{3(n−2)3n−8} +k3Φ^{2(n−2)2n−3} . Thus,

dΦ

k1+k2Φ^{3(n−2)3n−8} +k3Φ^{2(n−2)2n−3}

≤dt. (2.13)

We integrate from 0 totto obtain Z Φ(t)

Φ(0)

dξ

k₁+k₂ξ^{3(n−2)3n−8} +k₃ξ^{2(n−2)2n−3}

≤t,

where

Φ(0) = Z

Ω

Z u₀(x) 0

b⁰(y) g(y)dy^2k

dx.

Passing to the limit ast→T⁻, we conclude that Z +∞

Φ(0)

dξ

k1+k2ξ^{3(n−2)3n−8} +k3ξ^{2(n−2)2n−3}

≤T.

The proof is complete.

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Khadijeh Baghaei

Department of mathematics, Iran University of Science and Technology, Tehran, Iran E-mail address:[email protected]

Mahmoud Hesaaraki

Department of mathematics, Sharif University of Technology, Tehran, Iran E-mail address:[email protected]