Blow-up rate estimate for degenerate parabolic equation with

Appl. Math. Mech. -Engl. Ed. 31(6), 787–796 (2010) DOI 10.1007/s10483-010-1313-6 c Shanghai University and Springer-Verlag Berlin Heidelberg 2010

Applied Mathematics and Mechanics (English Edition)

Blow-up rate estimate for degenerate parabolic equation with nonlinear gradient term∗ Zheng-ce ZHANG (),

Biao WANG ( )

(School of Science, Xi’an Jiaotong University, Xi’an 710049, P. R. China) (Communicated by Xing-ming GUO)

Abstract In this paper, the blow-up rate is obtained for a porous medium equation with a nonlinear gradient term and a nonlinear boundary flux. By using a scaling method and regularity estimates of parabolic equations, the blow-up rate determined by the interaction between the diffusion and the boundary flux is obtained. Compared with previous results, the gradient term, whose exponent does not exceed two, does not affect the blow-up rate of the solutions. Key words

degenerate parabolic equation, gradient, blow-up, nonlinear boundary flux

Chinese Library Classification O175.26, O175.29, O29 2000 Mathematics Subject Classification 35B33, 35K50, 35K55, 35K65

1

Introduction

In this paper, we study the following porous medium equation with a nonlinear gradient term: ⎧ m−1 m ⎪ ⎪ ut = (u )xx − (m − 1)um−p |ux |p ⎪ ⎪ m ⎪ ⎨ − (m − 1)um−2 |ux |2 , (x, t) ∈ (0, 1) × (0, T ), (1) ⎪ m ⎪ (u )x (0, t) = 0, (um )x (1, t) = muq (1, t), t ∈ (0, T ), ⎪ ⎪ ⎪ ⎩ u(x, 0) = u0 (x), x ∈ [0, 1], where q m > 1 and p < 2 are parameters, and u0 (x) > 1 is continuous and satisfies the compatibility conditions. Remark 1 The restriction of q m > 1 is reasonable since there exist global solutions if q < m (see [1]). The nonlinear parabolic equation ut =

m−1 m (u )xx − (m − 1)um−p |ux |p − (m − 1)um−2 |ux |2 m

(2)

with m = 0 is a mathematical model for many physical problems corresponding to nonlinear diffusion with convection. The source term on the right-hand side of (2) is of convective nature. In the theory of unsatured porous media, the convective part may represent the effect of gravity. ∗ Received Jan. 30, 2010 / Revised May 8, 2010 Project supported by the Youth Foundation of the National Natural Science Foundation of China (No. 10701061) Corresponding author Zheng-ce ZHANG, Associate Professor, E-mail: [email protected]

788

Zheng-ce ZHANG and Biao WANG

Moreover, (2) with m = 2 is also a Boussinesq equation of hydrology, which is involved in various fields of petroleum technology and ground water hydrology. The porous medium equation without convection has been considered extensively in the past few years (see [2–10]). For instance, in [2], Galaktionov and Levine studied the following equation: ⎧ m ⎪ ⎨ ut = (u )xx , (x, t) ∈ (0, +∞) × (0, T ), ⎪ ⎩

q − um t ∈ (0, T ), x (0, t) = u (0, t), u(x, 0) = u0 (x), x ∈ (0, +∞).

(3)

They proved that, if 0 < q m+1 2 , then all nonnegative solutions to (3) were global; while , the solutions to the equation would blow up in finite time. Moreover, if m+1 < for q > m+1 2 2 q m + 1, all nonnegative solutions blew up in finite time; if q > m + 1, global nontrivial nonnegative solutions existed. However, they did not consider the blow-up rate in their paper. In [3], Quir´ os and Rossi obtained the blow-up rates of the following equation: ⎧ m n ⎪ ⎨ ut = (u )xx , vt = (v )xx , (x, t) ∈ (0, +∞) × (0, T ), − (um )x (0, t) = v p (0, t), −(v n )x (0, t) = uq (0, t), t ∈ (0, T ), (4) ⎪ ⎩ u(x, 0) = u0 (x), v(x, 0) = v0 (x), x ∈ (0, +∞). Denote ⎧ 2p + n + 1 ⎪ ⎪ ⎨ α1 = (m + 1)(n + 1) − 4pq , ⎪ 2p + m + 1 ⎪ ⎩ α2 = , (m + 1)(n + 1) − 4pq

p(m − 1 − 2q) + m(n + 1) , (m + 1)(n + 1) − 4pq q(n − 1 − 2p) + n(m + 1) β2 = . (m + 1)(n + 1) − 4pq β1 =

They proved that the solutions to (4) were global if pq < (m+1)(n+1) and might blow up in finite 4 (m+1)(n+1) time if pq > (m+1)(n+1) . In the case of pq > , if α +β 0 or α2 +β2 0, then every 1 1 4 4 nonnegative nontrivial solution to (4) blew up in finite time; if α1 + β1 > 0 and α2 + β2 > 0, then there existed blow-up solutions for large initial data and global solutions for small initial data. Moreover, they obtained the blow-up rate O((T − t)−α1 ) for u and O((T − t)−α2 ) for v as t → T . Compared with the porous medium equation, the porous medium equation with convection is more difficult to be studied. There are many open problems left (see [11]). The elegant works for such problems have been done by Andreu et al.[12] , where the following equation with convection was studied: ⎧ m α q p ⎪ ⎨ ut = Δu − ∇u + u , (x, t) ∈ Ω × (0, +∞), u = 0, (x, t) ∈ ∂Ω × (0, +∞), (5) ⎪ ⎩ u(x, 0) = u0 (x), x ∈ Ω, in which Ω is a bounded smooth domain in RN (N 1), u0 (x) 0, m 1, α > 0, p 1, and q 1. However, they only obtained the conditions for the global existence of weak solutions. The blow-up conditions for the kind of porous medium systems with convection were investigated in [13], ⎧ ut = a1 Δum1 − b1 um1 −2 |∇u|2 + up1 v q1 , (x, t) ∈ Ω × (0, T ), ⎪ ⎪ ⎪ ⎨ v = a Δv m1 − b v m2 −2 |∇v|2 + v p2 v q2 , (x, t) ∈ Ω × (0, T ), t 2 2 ⎪ u(x, t) = v(x, t) = ε0 , (x, t) ∈ ∂Ω × (0, T ), ⎪ ⎪ ⎩ u(x, 0) = u0 (x), u(x, 0) = u0 (x), x ∈ Ω,

(6)

Blow-up rate estimate for degenerate parabolic equation with nonlinear gradient term

789

where Ω ⊂ RN was a bounded domain with the smooth boundary ∂Ω,

p1 , q1 , p2 , q2 , ε0 > 0, m i , ai , b i > 0

(i = 1, 2),

u0 (x), v0 (x) γ0 were smooth functions on Ω and were compatible on ∂Ω. They obtained the blow-up conditions for (6), while did not study the blow-up rate. Furthermore, in [14], Zhou and Mu considered the following porous medium equation with convection: ⎧ m−1 m m−2 ⎪ |ux |2 , (x, t) ∈ (0, 1) × (0, T ), ⎪ ⎨ ut = m (u )xx − (m − 1)u (7) (um )x (0, t) = 0, (um )x (1, t) = muq (1, t), t ∈ (0, T ), ⎪ ⎪ ⎩ u(x, 0) = u0 (x), x ∈ [0, 1], where the parameters q > m > 1, and u0 (x) 1 was continuous and satisfied the compatibility conditions. They mainly investigated the blow-up rate. They showed that the blow-up rate of the porous medium equation with convection was irrelevant to the convection. Recently, gradient blow-up solutions, i.e., the spatial derivative of solutions becomes unbounded in finite time while the solutions themselves remain bounded, have been studied by many mathematicians such as Souplet[11] , Hu and Yin[15] , Guo and Hu[16] , Fila and Lieberman[17] , Li and Souplet[18] , and Zhang and Hu[19–20] . Considerable attention has been drawn to the blow-up results and existence of global solutions for general nonlinear parabolic equations and systems. Numerous related works have been devoted to some of its variants such as more general quasilinear operators and domains. We refer the interested reader to [20–24] and some of the references therein. In this paper, we consider more general cases for q m > 1 and p 2, and study the blow-up rate of porous medium equations by the scaling method and the regularity estimates of parabolic equations. We get the same blow-up rate as in [14] when the exponent of the gradient term does not exceed two. For the case of p > 2, the solutions blow up. However, the same method cannot be used to get the rate because of the difficulty in Schauder estimates. To overcome this, we must apply a new technique, which will be presented in the next paper. This paper is organized as follows. In Section 2, we prove the blow-up condition and rate. In Section 3, we give a blow-up lemma for a more generalized degenerate parabolic equation with a nonlinear gradient term. Throughout the paper, C denotes different positive constants depending only on m, q, and p in varying places.

2

Blow-up condition and rate

In this section, we will establish four main theorems. Theorem 1 Let u be the solution to (1) on [0, 1]. Then, every solution to (1) will blow up in finite time for q > m > 1. Theorem 2 Let u be the solution to (1) with m−1 m (u0 )xx − (m − 1)um−p |(u0 )x |p − (m − 1)um−2 |(u0 )x |2 0 on 0 0 m and blow up at a finite time T . Then, 1

1

c(T − t)− 2q−m−1 max u(x, t) C(T − t)− 2q−m−1 x∈[0,1]

[0, 1]

790


for t ∈ (0, T ) and positive c, C. Theorem 3 Let u be the solution to (1) on [0, 1]. Then, every solution to (1) will blow up in finite time for q = m > 1 and p < 0. Theorem 4 Let u be the solution to (1) with m−1 m (u0 )xx − (m − 1)um−p |(u0 )x |p − (m − 1)um−2 |(u0 )x |2 0 0 0 m

on

[0, 1]

and blow up at a finite time T . Then, 1

1

c(T − t)− m−1 max u(x, t) C(T − t)− m−1 x∈[0,1]

for t ∈ (0, T ) and positive c, C. Remark 2 In fact, in Theorems 2 and 4, the initial conditions are easily satisfied, e.g., u0 (x) = AeBx + C, where the constants A, B, and C are determined by p, m, and q. In fluid mechanics, the problem deals with the filtration of an incompressible fluid (typically, water) through a porous medium such as sponges, soil, etc. For the initial conditions of exponential form, we assume that the flow has an almost vertical (or horizontal) speed so that the free boundary function u(x, t) has large (or small) gradients at the time t = 0. Remark 3 This blow-up rate estimate for the solution to (1) agrees with those without convection obtained in [25] and [26], namely, the gradient term in (1) makes no contribution to the blow-up rate. In other words, the convection is insufficient to affect the result of the intersection between the boundary flux and the diffusion. The proofs of Theorems 1 and 3 depend on the result of Lemma 1, while those of Theorems 2 and 4 are established by scaling arguments and regularity estimates of parabolic equations as in [15, 27–29]. Proof of Theorem 1 We make the transformation η = ln u and change (1) to the following form: ⎧ (m−1)η )xx − (m − 1)|ηx |p e(m−1)η , (x, t) ∈ (0, 1) × (0, T ), ⎪ ⎨ ηt = (e ⎪ ⎩ Then, we have

ηx (0, t) = 0, ηx (1, t) = e(q−m)η(1,t) , t ∈ (0, T ), η(x, 0) = η0 (0) = ln u0 (x), x ∈ [0, 1].

(8)

⎧ ⎪ ⎨ λ1 = 1 > 0, λ2 = (m − 1)|ηx |p 0, ⎪ ⎩ λ3 = 1 > 0,

which satisfy the first blow-up condition of Lemma 1 (we here deal with the symmetry case in one dimension). Thus, the solutions to (1) will blow up in finite time. Proof of Theorem 2 Let η be a solution to (8) with blow-up time T . Since (e(m−1)η0 )xx − (m − 1)|(η0 )x |p e(m−1)η0 0 on [0, 1], the maximum principle yields ηt 0. Then, (e(m−1)η )xx (m − 1)|ηx |p e(m−1)η 0. That it, (e(m−1)η )x is monotonic increasing with respect to x. Considering the boundary condition ηx (0, t) = 0, we have ηx (x, t) 0 for

(x, t) ∈ [0, 1] × [0, T ).


791

For any t > 0, we define M (t∗ ) = η(1, t∗ ) = max η(x, t∗ ), x∈[0,1]

t∗ ∈ (0, t),

⎧ ∗ ∗ ⎨ ψa,b (y, s) = e(m−1)(η(ay+1,bs+t )−M(t )) , ∗ ⎩ (y, s) ∈ [− 1 , 0] × [− t , 0], a b where a and b are parameters depending on m, q, and M . Clearly,

(9)

(10)

⎧ −(m−1)M(t∗ ) e ψa,b 1, ⎪ ⎪ ⎪ ⎨ ψ (0, 0) = 1, a,b ⎪ ⎪ ⎪ ⎩ ∂ψa,b 0. ∂s

Choose

a = e(m−q)M , b = e(1+m−2q)M .

A simple computation yields ⎧ 2 (ψa,b )s = (m − 1)ψa,b (ψa,b )yy + (m − 1)2 e(2−p)(m−q)M ψa,b , ⎪ ⎪ ⎪ ⎪ ∗ ⎪ t ⎪ ⎪ (y, s) ∈ (−1/a, 0) × (− , 0), ⎨ b (q−1)/(m−1) ⎪ (0, s), (ψa,b )y (−1/a, s) = 0, ⎪ ⎪ (ψa,b )y (0, s) = (m − 1)ψa,b ⎪ ⎪ ∗ ⎪ ⎪ t ⎩ s ∈ (− , 0). b

(11)

We remark that the positive parameters a and b go to zero as t∗ → T due to q > m > 1. Next, we claim that there exist positive constants C1 and C2 such that C1

∂ψa,b (0, 0) C2 holds for small a and b. ∂s

(12)

The proof of (12) relies on the uniform boundedness of {ψa,b } and {(ψa,b )y }. Actually, it is easy to see by (11) with 0 ψa,b 1 that (ψa,b )y is also uniformly bounded, while 2 e(2−p)(m−q)M ψa,b

(q > m, p < 2)

is still uniformly bounded. From the results for bounded solutions to porous medium type equations (see [30] and [31]), {ψa,b } is equicontinuous on compact subsets of their common domain. Let aj = a(t∗j ), bj = b(t∗j ) with t∗j → T as j → +∞. Passing to a subsequence if necessary, we have that ψa,b → ψ uniformly on the compact subsets of A = {y 0, s 0}. The limit function ψ is continuous with ψ(0, 0) = 1. Hence, for any ε0 ∈ (0, 1), there exists a neighborhood of (0, 0), denoted by U ⊂ A, such that ψ > ε0 in U and 1 2 ε0 ψaj bj 1 on U for sufficiently large j. By Schauder’s estimates in [32], we obtain ψaj bj C 2+α,1+α/2 (U) C for some constant C > 0 and 0 < α < 1. The second inequality in (12) follows immediately.

(13)

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Now, we come to the proof of the first inequality. On the contrary, if the first inequality in ∂ψ (12) is false, then there exists a subsequence (aj , bj ) → (0, 0) such that ∂sa,b (0, 0) → 0. We proceed as before and obtain that ψa,b → ψ and the estimate (13) holds on the compact subset 1 of {(y, s) : ψ > 0}. Thus, we have ψa,b → ψ in C 2+β,1+ 2 β for some β < α satisfying 0 ψ 1, ψ(0, 0) = 1, and ∂ψ ∂s 0, and ψ is a weak solution to ψs = (m − 1)ψψyy + (m − 1)2 e(2−p)(m−q)M ψ 2 , (14) ψy (0, s) = (m − 1)ψ (q−1)/(m−1) (0, s) in {y < 0} × (−∞, 0]. Define z = ψs . We get ⎧ 2 (2−p)(m−q)M ψz, ⎪ ⎨ zs = (m − 1)ψzyy + (m − 1)zψyy + 2(m − 1) e q−m zy (0, s) = (q − 1)(ψ m−1 z)(0, s), ⎪ ⎩ z(0, 0) = 0

(15)

in the positive set of ψ. It follows by the Hopf lemma[7, 33] that z ≡ 0, i.e., ψ is independent of s. Consequently, ψ = ψ(y) satisfies 0 = ψyy + (m − 1)e(2−p)(m−q)M ψ, (16) ψy (0) = m − 1, ψ(0) = 1. However, (16) does not have a bounded weak solution. This contradicts the boundedness of ψ. Therefore, the first inequality follows. In terms of ψ, it follows from (12) that C1 (m − 1)e(1+m−2q)M Mt (t∗ ) C2 .

(17)

Integrating (17) from t to T , we get 1

1

ln c(T − t)− 2q−m−1 max η(x, t) = M (t) ln C(T − t)− 2q−m−1 . x∈[0,1]

That is, ⎧ 1 1 − 2q−m−1 ⎪ max u(x, t) C(T − t)− 2q−m−1 , ⎪ ⎨ c(T − t) x∈[0,1]

C (2q − m − 1) − 2q−m−1 ⎪ 2 ⎪ ⎩c = , m−1 1

C=

1 C (2q − m − 1) − 2q−m−1 1 . m−1

The proof of Theorem 2 is completed. Proof of Theorem 3 Make the transformation η = ln u. Change (1) to the following form: ⎧ (m−1)η )xx − (m − 1)|ηx |p e(m−1)η , (x, t) ∈ (0, 1) × (0, T, ) ⎪ ⎨ ηt = (e (18) ηx (0, t) = 0, ηx (1, t) = e(q−m)η(1,t) , t ∈ (0, T ), ⎪ ⎩ η(x, 0) = η0 (0) = ln u0 (x), x ∈ [0, 1]. From the results of Lemma 1 (here, we deal with the symmetry case in one dimension), we obtain that ⎧ ⎪ ⎨ λ1 = 1 > 0, λ2 = (m − 1)|ηx |p 0, ⎪ ⎩ λ3 = 1 > 0.


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We can check that another condition is satisfied when q = m > 1 and p < 0. Thus, the solutions will blow up in finite time. Proof of Theorem 4 We can obtain the blow-up rate estimates if we repeat the same process as that in the proof of Theorem 2.

3

Blow-up lemma

In this section, for convenience, we present one more generalized blow-up lemma whose proof is taken from [1]. Lemma 1 Let u be the solution to the following quasilinear reaction-diffusion equations with nonlinear boundary conditions: ⎧ e ⎪ ut = λ1 Δemu − λ2 eqeu in Ω × (0, T ), ⎪ ⎪ ⎨ ∂u (19) = λ3 epeu on ∂Ω × (0, T ), ⎪ ∂n ⎪ ⎪ ⎩ u(x, 0) = u0 (x) on Ω, where

q > 0, λ1 , λ3 , m, λ2 , p 0,

Ω ⊂ RN is a bounded domain with smooth boundary ∂Ω, and u0 (x) is a positive function satisfying the compatibility conditions. (i) If q < 2 p + m, then the solutions to (19) will blow up in finite time. (ii) If q = 2 p+m with C m( m + p )λ1 λ23 > λ2 and C > 0 is a constant, then the solutions will blow up in finite time. Proof To prove these conclusions, we only need to find a suitable subsolution u to (19) which blows up in finite time. (i) We first introduce a function h(x) satisfying ⎧ |∂Ω| ⎪ ⎪ Δh = κ = in Ω, ⎪ ⎪ |Ω| ⎪ ⎪ ⎪ ⎪ ⎪ ∂h ⎪ ⎨ = 1 on ∂Ω, ∂n ⎪ ⎪ ⎪ L := max h(x), ⎪ ⎪ ⎪ x∈Ω ⎪ ⎪ ⎪ ⎪ ⎩ L1 := max |∇h(x)|. x∈Ω

By the maximum principle, we know that h(x) reaches its maximum at a point x0 ∈ ∂co(Ω)∩∂Ω, where ∂co(Ω) is the boundary of the closed convex hull of Ω. Let ⎧ ⎨ γ = n(x0 ), ⎩ μ(x) = (γ · x − min γ · x) x∈Ω

in Ω. Then, μ is positive in Ω and has a maximum at x0 . Construct aL1 μ(x) + h(x) u(x, t) = ϕ δ(t) + aL1 + 1 with ϕ (s) = λ3 epeϕ(s) , ϕ(0) > s0 > 0,

(20) (21)

794


δ (t) =

e p)ϕ(δ(t)) m( m + p )λ1 λ3 e(m+e , (aL1 + 1)2

δ(0) = 0,

(22)

where a > 2 is a constant. It is easy to see that ϕ and δ are positive, increasing, and convex functions. We have aL1 μ(x) + h(x) ϕ(0) > s0 u(x, t) = ϕ δ(t) + aL1 + 1

(23)

whenever it is defined. We assume that u is well defined in some interval of time. We will prove that it is a subsolution. Since e e + λ2 eqeu =λ3 epeu δ (t) − ∇(λ1 me mu ∇u) + λ2 eqeu ut − λ1 Δemu m( p+m)u e p)u e m + p )λ1 λ3 e(e mu λ2 e(eq−e κλ1 me λ3 epeu − − 2 (aL1 + 1) λ3 aL1 + 1 (e p+m)u e 2 2 qeu m( m + p )λ1 λ3 e (a − 1) L1 λ2 e − + (aL1 + 1)2 λ3 epeu 0

with constants

(24)

⎧ ⎪ ⎨ a > 2, L1 1, q > 0, λ1 , λ3 , m, ⎪ ⎩ λ2 , p 0.

Meanwhile, we have ∂h aL1 ∂μ ∂u ∂n + ∂n − λ3 epeu = λ3 epeu − λ3 epeu 0 ∂n aL1 + 1

(25)

on ∂Ω since u > s0 . It follows from (23)–(25) that u is a subsolution to (19). Having made the transformation s = ϕ(r), we obtain

∞

∞

=

dr e m)ϕ(r) m( e m+e e p)λ1 λ3 e(p+f (aL1 +1)2

−

e p)ϕ(r) e λ2 e(q− λ3

ds e m)s m( e m+e e p)λ1 λ3 2 e(2p+f (aL1 +1)2

− λ2 eqes

0 that ϕ1 is positive increasing with ϕ1 (0) > s0 . Because of the fact that s∗ is the maximum of ϕ1 with ϕ1 (s∗ ) min u0 (x), we have u(x, 0) ϕ1 (s∗ ). x∈Ω

Therefore, u is well defined. This completes the proof. (ii) In this case, the proof of (ii) is very similar to that of (i), which is omitted here.

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