Electronic Journal of Differential Equations, Vol. 2022 (2022), No. 08, pp. 1–17. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu BLOW-UP FOR PARABOLIC EQUATIONS IN NONLINEAR DIVERGENCE FORM WITH TIME-DEPENDENT COEFFICIENTS XUHUI SHEN, JUNTANG DING Abstract. In this article, we study the blow-up of solutions to the nonlinear parabolic equation in divergence form,( h(u) ) t = n∑ i,j=1 ( aij(x)uxi ) xj − k(t)f(u) in Ω × (0, t∗), n∑ i,j=1 aij(x)uxiνj = g(u) on ∂Ω × (0, t∗), u(x, 0) = u0(x) ≥ 0 in Ω, where Ω is a bounded convex domain in Rn (n ≥ 2) with smooth boundary ∂Ω. By constructing suitable auxiliary functions and using a differential inequality technique, when Ω ⊂ Rn (n ≥ 2), we establish conditions for the solution blow up at a finite time, and conditions for the solution to exist for all time. Also, we find an upper bound for the blow-up time. In addition, when Ω ⊂ Rn with (n ≥ 3), we use a Sobolev inequality to obtain a lower bound for the blow-up time. 1. Introduction There are many results about the blow-up of solutions to nonlinear parabolic problems; see for example [4, 6, 16, 18, 19] and the references therein. A variety of methods have been used to study the blow-up phenomena of the solutions to parabolic equations in a bounded domain Ω ⊂ Rn (n ≥ 2). Authors often derive lower bounds for the blow-up time by restricting Ω ⊂ R3 (see [12, 13, 14]). Recently, some studies determined lower bounds for the blow-up time when Ω ⊂ Rn (n ≥ 3), see [1, 3, 8, 9, 10]. In this article, we investigate the blow-up of solutions to the nonlinear parabolic equation in divergence form, 2010 Mathematics Subject Classification. 35K55, 35B44. Key words and phrases. Nonlinear parabolic equation; blow-up; upper bound; lower bound. ©2022. This work is licensed under a CC BY 4.0 license. Submitted January 27, 2021. Published January 25, 2022. 1 2 X. SHEN, J. DING EJDE-2022/08 ( h(u) ) t = n∑ i,j=1 ( aij(x)uxi ) xj − k(t)f(u) in Ω× (0, t∗), n∑ i,j=1 aij(x)uxi νj = g(u) on ∂Ω× (0, t∗), u(x, 0) = u0(x) ≥ 0 in Ω, (1.1) where Ω is a bounded convex domain in Rn (n ≥ 2) with smooth boundary ∂Ω, (aij(x))n×n is a differentiable positive definite matrix, ν is the outward normal vector to ∂Ω, u0(x) is the initial value, t∗ is the maximal existence time of u, and Ω is the closure of Ω. Set R+ = (0,+∞). We assume, in this paper, that h is a C2(R+) function with h′(s) > 0 for all s ≥ 0, k is a positive C1(R+) function, g and f are nonnegative C(R+) functions, and u0 is a nonnegative C1(Ω) function. The blow-up phenomena in parabolic equations with nonlinear boundary con- ditions have been studied in [2, 7, 11, 16]. Payne, Philippin and Vernier Piro [15] studied a special case of (1.1), ut = ∆u− f(u) in Ω× (0, t∗), ∂u ∂ν = g(u) on ∂Ω× (0, t∗), u(x, 0) = u0(x) ≥ 0 in Ω, (1.2) where Ω is a bounded convex domain in Rn (n ≥ 2) with smooth boundary ∂Ω. When Ω ⊂ Rn (n ≥ 2), some conditions on data were established to ensure that u(x, t) exists for all time or blows up at some finite time. Moreover, they also derived an upper bound for blow-up time. In particular, when Ω ⊂ R3, they obtained a lower bound for blow-up time under more appropriate hypotheses. In [2, 11], the following special case of (1.1) has been discussed, ut = n∑ i,j=1 ( aij(x)uxi ) xj − f(u) in Ω× (0, t∗), n∑ i,j=1 aij(x)uxi νj = g(u) on ∂Ω× (0, t∗), u(x, 0) = u0(x) ≥ 0 in Ω, (1.3) where Ω is a bounded convex domain in Rn (n ≥ 2) with smooth boundary ∂Ω. Under certain conditions on data, Li and Li [11] showed that the solution blows up or remains global when Ω ⊂ Rn (n ≥ 2). For Ω ⊂ R3, a lower bound for blow-up time was also derived. By restricting Ω ⊂ Rn (n ≥ 3), Baghaei and Hesaaraki [2] derived a lower bound for blow-up time when blow-up occurs. Motivated by above works, we study the more general problem (1.1). It seems that the auxiliary functions defined in [2, 11, 15] are no longer applicable for problem (1.1). By constructing suitable auxiliary functions and using a differential inequality technique, we establish conditions on the data for the solution u(x, t) to blow up at a finite time, and for the solution to exist for all time when Ω ⊂ Rn (n ≥ 2). Also, we obtain an upper bound for the blow-up time. When Ω ⊂ Rn (n ≥ 3), we use a Sobolev inequality to derive a lower bound for the blow-up time. Note that if h(u) ≡ u, (aij(x))n×n is a unit matrix, and k(t) ≡ 1, problem (1.1) is the same as EJDE-2022/08 BLOW-UP FOR PARABOLIC EQUATIONS 3 problem (1.2); if h(u) ≡ u and k(t) ≡ 1, problem (1.1) becomes problem (1.3). In the above two cases, our results derived in this paper still hold. Hence, our results can be regarded as an extension of the results in [2, 11, 15]. This article is organized as follows. In Section 2, we establish the conditions on the data sufficient to guarantee that the solution u(x, t) exists for all time. In Section 3, we obtain an upper bound for the blow-up time under some appropriate assumptions. In Section 4, we obtain a lower bound for blow-up time. Section 5, we give two examples that illustrate the results obtained. 2. Global solution In this section, we establish a sufficient condition for the existence of a global solution. We define the auxiliary functions Φ(t) = ∫ Ω H(u(x, t)) dx, t ≥ 0; H(s) = 2 ∫ s 0 yh′(y) dy, s ≥ 0. (2.1) Since (aij(x))n×n is a positive definite matrix, there exists a constant θ > 0 such that n∑ i,j=1 aij(x)ξiξj ≥ θ|ξ|2 (2.2) for all x ∈ Ω and all ξ ∈ Rn. Theorem 2.1. Let u(x, t) be the nonnegative classical solution of problem (1.1). Suppose that functions f , g, h, and k satisfy f(s) ≥ γ1s p, g(s) ≤ γ2s q, h′(s) ≤ ζ0, s ≥ 0, (2.3) k(t) ≥ m, t ≥ 0, (2.4) where p, q, γ1, γ2, ζ0, and m are some positive constants, and p > 1, q > 1, p+ 1 > 2q. (2.5) Then u(x, t) exists for all t > 0 in the measure Φ(t). Proof. Using the divergence theorem and assumptions (2.2)–(2.4), we have Φ′(t) = ∫ Ω H ′ (u(x, t))ut dx = 2 ∫ Ω uh′(u)ut dx = 2 ∫ Ω u ( n∑ i,j=1 ( aij(x)uxi ) xj − k(t)f(u) ) dx = 2 ∫ ∂Ω u n∑ i,j=1 aij(x)uxi νj ds− 2 ∫ Ω n∑ i,j=1 aij(x)uxi uxj dx − 2k(t) ∫ Ω uf(u) dx ≤ 2 ∫ ∂Ω ug(u) ds− 2θ ∫ Ω |∇u|2 dx− 2k(t) ∫ Ω uf(u) dx ≤ 2γ2 ∫ ∂Ω uq+1 ds− 2θ ∫ Ω |∇u|2 dx− 2mγ1 ∫ Ω up+1 dx. (2.6) 4 X. SHEN, J. DING EJDE-2022/08 Owing to [11, Lemma 2.1], we deduce that∫ ∂Ω uq+1 ds ≤ n ρ0 ∫ Ω uq+1 dx+ (q + 1)d ρ0 ∫ Ω uq|∇u|dx, (2.7) where ρ0 = min ∂Ω (x · ν), d = max ∂Ω |x|. (2.8) Substituting (2.7) into (2.6), we obtain Φ′(t) ≤ 2nγ2 ρ0 ∫ Ω uq+1 dx+ 2γ2(q + 1)d ρ0 ∫ Ω uq|∇u|dx− 2θ ∫ Ω |∇u|2 dx − 2mγ1 ∫ Ω up+1 dx. (2.9) Using Hölder’s and Young’s inequalities in the second term of (2.9), we have∫ Ω uq|∇u|dx ≤ ( ε ∫ Ω u2q dx )1/2(1 ε ∫ Ω |∇u|2 dx )1/2 ≤ ε 2 ∫ Ω u2q dx+ 1 2ε ∫ Ω |∇u|2 dx, (2.10) where ε = γ2(q + 1)d 2ρ0θ > 0. (2.11) Inserting (2.10) into (2.9) and using (2.11), we can rewrite (2.9) as Φ′(t) ≤ 2nγ2 ρ0 ∫ Ω uq+1 dx+ εγ2(q + 1)d ρ0 ∫ Ω u2q dx + (γ2(q + 1)d ρ0ε − 2θ )∫ Ω |∇u|2 dx− 2mγ1 ∫ Ω up+1 dx = 2nγ2 ρ0 ∫ Ω uq+1 dx+ 2θε2 ∫ Ω u2q dx− 2mγ1 ∫ Ω up+1 dx. (2.12) It follows from (2.5) that 0 < p+1−2q p−q < 1. We apply Hölder’s and Young’s inequal- ities to obtain∫ Ω u2q dx ≤ (∫ Ω uq+1 dx ) p+1−2q p−q (∫ Ω up+1 dx ) q−1 p−q = ( σ 1−q p+1−2q ∫ Ω uq+1 dx ) p+1−2q p−q ( σ ∫ Ω up+1 dx ) q−1 p−q ≤ p+ 1− 2q p− q σ 1−q p+1−2q ∫ Ω uq+1 dx+ q − 1 p− q σ ∫ Ω up+1 dx, (2.13) where 0 < σ < mγ1(p− q) ε2θ(q − 1) . (2.14) Next, we substitute (2.13) into (2.12) to obtain Φ′(t) ≤ I1 ∫ Ω uq+1 dx− I2 ∫ Ω up+1 dx (2.15) with I1 = 2nγ2 ρ0 + 2θε2(p+ 1− 2q) p− q σ 1−q p+1−2q , I2 = 2mγ1 − 2θε2(q − 1) p− q σ. EJDE-2022/08 BLOW-UP FOR PARABOLIC EQUATIONS 5 In view of (2.14) and (2.5), we have I1, I2 > 0. Using Hölder’s inequality, we have∫ Ω uq+1 dx ≤ (∫ Ω up+1 dx ) q+1 p+1 |Ω| p−q p+1 , (2.16)∫ Ω u2 dx ≤ (∫ Ω up+1 dx ) 2 p+1 |Ω| p−1 p+1 , (2.17) where |Ω| is the volume of Ω. Thanks to (2.3), H(u) = 2 ∫ u 0 yh′(y) dy ≤ 2ζ0 ∫ u 0 y dy = ζ0u 2; that is u2 ≥ 1 ζ0 H(u). (2.18) Combining (2.15)-(2.18), we obtain Φ′(t) ≤ I1 (∫ Ω up+1 dx ) q+1 p+1 ( |Ω| p−q p+1 − I2 I1 (∫ Ω up+1 dx ) p−q p+1 ) ≤ I1 (∫ Ω up+1 dx ) q+1 p+1 ( |Ω| p−q p+1 − I2 I1 (( |Ω| 1−p p+1 ∫ Ω u2 dx ) p+1 2 ) p−q p+1 ) ≤ I1 (∫ Ω up+1 dx ) q+1 p+1 ( |Ω| p−q p+1 − I2 I1 |Ω| (p−q)(1−p) 2(p+1) ( 1 ζ0 ) p−q 2 (∫ Ω H(u) dx ) p−q 2 ) = I1 (∫ Ω up+1 dx ) q+1 p+1 ( |Ω| p−q p+1 − I2 I1 |Ω| (p−q)(1−p) 2(p+1) ( 1 ζ0 ) p−q 2 Φ p−q 2 (t) ) . (2.19) Thus, u(x, t) cannot blow up in measure Φ(t) for all time t > 0. In fact, if u(x, t) blows up at finite time t∗ in measure Φ(t), by passing to the limit as t → t∗−, we have limt→t∗− Φ(t) = +∞ and lim t→t∗− ( |Ω| p−q p+1 − I2 I1 |Ω| (p−q)(1−p) 2(p+1) ( 1 ζ0 ) p−q 2 Φ p−q 2 (t) ) = −∞. (2.20) In view of (2.19) and (2.20), we deduce Φ′(t) < 0 in some interval [t0, t ∗). Hence, for any t ∈ [t0, t ∗), we have Φ(t) ≤ Φ(t0). Taking the limits as t→ t∗−, we obtain +∞ = lim t→t∗− Φ(t) ≤ Φ(t0) which is a contradiction. The proof is complete. � 3. Blow-up solution In this section, we establish conditions for the solution of (1.1) to blow up in finite time, and give an upper bound for the blow-up time. We set the following auxiliary functions: F (s) = ∫ s 0 f(y) dy, G(s) = ∫ s 0 g(y) dy, s ≥ 0, (3.1) Ψ(t) = 2 ∫ ∂Ω G(u) ds− ∫ Ω n∑ i,j=1 aij(x)uxi uxj dx− 2k(t) ∫ Ω F (u) dx, t ≥ 0, (3.2) 6 X. SHEN, J. DING EJDE-2022/08 where u is the nonnegative classical solution of (1.1). We also use the auxiliary function Φ(t) defined by (2.1). Our main result reads as follows. Theorem 3.1. Let u(x, t) be the nonnegative classical solution of problem (1.1). Suppose that functions f , g, h, and k satisfy sf(s) ≤ 2(1 + α)F (s), sg(s) ≥ 2(1 + β)G(s), h′′(s) ≤ 0, s ≥ 0, (3.3) k′(t) ≤ 0, t ≥ 0, (3.4) where α and β are nonnegative constants with 0 ≤ α ≤ β. In addition, assume that the initial value u0 satisfies Ψ(0) = 2 ∫ ∂Ω G(u0) ds− ∫ Ω n∑ i,j=1 aij(x)u0xi u0xj dx − 2k(0) ∫ Ω F (u0) dx > 0. (3.5) Then u(x, t) blows up at a finite time t∗ in the measure Φ(t), and t∗ ≤ Φ(0) 2β(1 + β)Ψ(0) , β > 0. When β = 0, we have t∗ =∞. Proof. Using the divergence theorem and (3.3), we have Φ′(t) = ∫ Ω H ′(u(x, t))utdx = 2 ∫ Ω uh′(u)ut dx = 2 ∫ Ω u ( n∑ i,j=1 ( aij(x)uxi ) xj − k(t)f(u) ) dx = 2 ∫ ∂Ω u n∑ i,j=1 aij(x)uxiνj ds− 2 ∫ Ω n∑ i,j=1 aij(x)uxiuxj dx − 2k(t) ∫ Ω uf(u) dx = 2 ∫ ∂Ω ug(u) ds− 2 ∫ Ω n∑ i,j=1 aij(x)uxiuxj dx− 2k(t) ∫ Ω uf(u) dx ≥ 4(1 + β) ∫ ∂Ω G(u) ds− 2 ∫ Ω n∑ i,j=1 aij(x)uxiuxj dx − 4(1 + α)k(t) ∫ Ω F (u) dx = 2(1 + β) ( 2 ∫ ∂Ω G(u) ds− 1 1 + β ∫ Ω n∑ i,j=1 aij(x)uxi uxj dx − 2(1 + α) 1 + β k(t) ∫ Ω F (u) dx ) ≥ 2(1 + β)Ψ(t). (3.6) EJDE-2022/08 BLOW-UP FOR PARABOLIC EQUATIONS 7 Furthermore, from (3.4) and the divergence theorem, Ψ′(t) = 2 ∫ ∂Ω G′(u)ut ds− ∫ Ω ( n∑ i,j=1 aij(x)uxiuxj ) t dx− 2k′(t) ∫ Ω F (u) dx − 2k(t) ∫ Ω F ′(u)ut dx ≥ 2 ∫ ∂Ω g(u)ut ds− 2 ∫ Ω n∑ i,j=1 aij(x)uxi (uxj )t dx− 2k(t) ∫ Ω f(u)ut dx = 2 ∫ ∂Ω g(u)ut ds− 2 ∫ Ω n∑ j=1 ( n∑ i=1 aij(x)uxiut ) xj dx + 2 ∫ Ω ut n∑ i,j=1 ( aij(x)uxi ) xj dx− 2k(t) ∫ Ω f(u)ut dx = 2 ∫ ∂Ω g(u)ut ds− 2 ∫ ∂Ω ut n∑ i,j=1 aij(x)uxi νj ds + 2 ∫ Ω ut n∑ i,j=1 ( aij(x)uxi ) xj dx− 2k(t) ∫ Ω f(u)ut dx = 2 ∫ Ω ut ( n∑ i,j=1 ( aij(x)uxi ) xj − k(t)f(u) ) dx = 2 ∫ Ω h′(u)u2 t dx ≥ 0. (3.7) Hence, Ψ(t) is a nondecreasing function in t. By (3.5), we know that Ψ(t) ≥ Ψ(0) > 0 for all t ∈ (0, t∗). It follows from (3.6) that Φ′(t) > 0. (3.8) Employing Hölder’s inequality, (3.6)–(3.8), and the fact that h′(s) > 0 for all s ≥ 0, we have Ψ(t)Φ′(t) ≤ 1 2(1 + β) (Φ′(t))2 = 2 1 + β (∫ Ω uh′(u)ut dx )2 ≤ 2 1 + β ∫ Ω h′(u)u2 t dx ∫ Ω h′(u)u2 dx ≤ 1 1 + β Ψ′(t) ∫ Ω h′(u)u2 dx. (3.9) Using assumption (3.3) and integration by parts, we obtain H(u) = 2 ∫ u 0 yh′(y) dy = ∫ u 0 h′(y) dy2 = h′(u)u2 − ∫ u 0 y2h′′(y) dy ≥ h′(u)u2. (3.10) Combining this and (3.9), we have Ψ(t)Φ′(t) ≤ 1 1 + β Ψ′(t) ∫ Ω H(u) dx = 1 1 + β Ψ′(t)Φ(t). 8 X. SHEN, J. DING EJDE-2022/08 Multiplying the above inequality by Φ−2−β(t), we obtain( Ψ(t)Φ−1−β(t) )′ ≥ 0. (3.11) We integrate (3.11) from 0 to t to obtain Ψ(t)Φ−1−β(t) ≥ Ψ(0)Φ−1−β(0) = M > 0. (3.12) Now (3.6) and (3.12) imply Φ′(t) ≥ 2(1 + β)Ψ(t) ≥ 2M(1 + β)Φ1+β(t). (3.13) If β > 0, it follows from (3.13) that (Φ−β(t))′ = −βΦ−1−β(t)Φ′(t) ≤ −2Mβ(1 + β). (3.14) Integrating (3.14) over [0, t], we obtain Φ−β(t) ≤ Φ−β(0)− 2Mβ(1 + β)t. (3.15) It is obvious that (3.15) cannot hold for all time t. Consequently, u(x, t) blows up at some finite time t∗ in the measure Φ(t) and t∗ ≤ Φ(0) 2β(1 + β)Ψ(0) . For β = 0, we have α = 0. It follows from (3.13) that Φ(t) ≥ Φ(0)e2Mt, which implies that t∗ =∞. The proof is complete. � 4. Lower bound for the blow-up time In this section, we consider Ω ⊂ Rn (n ≥ 3), and assume that f, g, h, and k to satisfy the following conditions: f(s) ≤ γ1s p, g(s) ≤ γ2s q, h′(s) ≥ ζ, s ≥ 0, (4.1) k(t) ≥ m, k′(t) k(t) ≤ η, t ≥ 0, (4.2) where p, q, γ1, γ2, ζ, and m are positive constants, and η is a nonnegative constant. Moreover, we assume that p > 1, q > 1, 2q > p+ 1. (4.3) We define the auxiliary functions A(t) = k 2r p−1 (t) ∫ Ω B(u) dx, t ≥ 0, B(s) = 2r ∫ s 0 h′(y)y2r−1 dy, s ≥ 0, (4.4) where r is a constant such that r > max{1, 1 2 n(q − 1)}. (4.5) In this section, we need the Sobolev inequality(∫ Ω (ur) 2n n−2 dx )n−2 2n ≤ c (∫ Ω u2r dx+ ∫ Ω |∇ur|2 dx )1/2 , (4.6) where c = c(n,Ω) is the best Sobolev constant depending on n (n ≥ 3) and Ω. For the more details we refer reader to [5, Corollary 9.14]. We state our result as follows. EJDE-2022/08 BLOW-UP FOR PARABOLIC EQUATIONS 9 Theorem 4.1. Let u(x, t) be the nonnegative classical solution of (1.1). Assume that (4.1)–(4.3) and (4.5) hold, and u(x, t) becomes unbounded in the measure A(t) at a finite time t∗. Then t∗ ≥ ∫ ∞ A(0) dτ J1τ + J2τ 2r−(q−1)(n−2) 2r−n(q−1) , where J1 = 2rη p− 1 + rγ2 ζ , (4.7) J2 = c1 ( (2q − p− 1)[2r − n(q − 1)] 2r(2q − p− 1) + n(q − 1)(p− 1) c − 2n(q−1)2 (2q−p−1)[2r−n(q−1)] 2 + 2r − n(q − 1) 2r c − n(q−1) 2r−n(q−1) 3 )(1 ζ ) 2r−(q−1)(n−2) 2r−n(q−1) , (4.8) c1 = rγ2m − 2(q−1) p−1 c n(q−1) r 2 n(q−1) 2r (( n ρ0 )2 + γ2(2r + q − 1)2d2 (2r − 1)θρ2 0 ) , (4.9) c2 = rγ1|Ω|− p−1 2r [2r(2q − p− 1) + n(q − 1)(p− 1)] n(q − 1)2c1 , c3 = 2(2r − 1)θ n(q − 1)c1 , (4.10) and ρ0 and d are defined by (2.8). Proof. By the divergence theorem and assumptions (2.2), (4.1),(4.2), and (4.5), we obtain A′(t) = 2r p− 1 k 2r p−1−1(t)k′(t) ∫ Ω B(u) dx+ k 2r p−1 (t) ∫ Ω B′(u)ut dx = 2r p− 1 k′(t) k(t) k 2r p−1 (t) ∫ Ω B(u) dx+ 2rk 2r p−1 (t) ∫ Ω u2r−1h′(u)ut dx ≤ 2rη p− 1 k 2r p−1 (t) ∫ Ω B(u) dx + 2rk 2r p−1 (t) ∫ Ω u2r−1 ( n∑ i,j=1 (aij(x)uxi )xj − k(t)f(u) ) dx ≤ 2rη p− 1 A(t) + 2rk 2r p−1 (t) ∫ ∂Ω u2r−1 n∑ i,j=1 aij(x)uxi νj ds − 2r(2r − 1)θk 2r p−1 (t) ∫ Ω u2r−2|∇u|2 dx− 2rk 2r+p−1 p−1 (t) ∫ Ω u2r−1f(u) dx = 2rη p− 1 A(t) + 2rk 2r p−1 (t) ∫ ∂Ω u2r−1g(u) ds− 2r(2r − 1)θk 2r p−1 (t) ∫ Ω u2r−2|∇u|2 dx − 2rk 2r+p−1 p−1 (t) ∫ Ω u2r−1f(u) dx ≤ 2rη p− 1 A(t) + 2rγ2k 2r p−1 (t) ∫ ∂Ω u2r+q−1 ds− 2(2r − 1)θ r k 2r p−1 (t) ∫ Ω |∇ur|2 dx − 2rγ1k 2r+p−1 p−1 (t) ∫ Ω u2r+p−1 dx. (4.11) 10 X. SHEN, J. DING EJDE-2022/08 It follows from (4.3) and (4.5) that 0 < 2r 2r+p−1 < 1. Applying Hölder’s inequality, we deduce that k 2r p−1 (t) ∫ Ω u2r dx ≤ ( k 2r+p−1 p−1 (t) ∫ Ω u2r+p−1 dx ) 2r 2r+p−1 |Ω| p−1 2r+p−1 , or equivalently, k 2r+p−1 p−1 (t) ∫ Ω u2r+p−1 dx ≥ |Ω|− p−1 2r ( k 2r p−1 (t) ∫ Ω u2r dx ) 2r+p−1 2r . (4.12) Then we substitute (4.12) into (4.11) to obtain A′(t) ≤ 2rη p− 1 A(t) + 2rγ2k 2r p−1 (t) ∫ ∂Ω u2r+q−1 ds − 2(2r − 1)θ r k 2r p−1 (t) ∫ Ω |∇ur|2 dx − 2rγ1|Ω|− p−1 2r ( k 2r p−1 (t) ∫ Ω u2r dx ) 2r+p−1 2r . (4.13) Now, we deal with the second term on the right-hand side of (4.13). Using (4.5) and [11, Lemma 2.1 ], we obtain k 2r p−1 (t) ∫ ∂Ω u2r+q−1 ds ≤ n ρ0 k 2r p−1 (t) ∫ Ω u2r+q−1 dx+ (2r + q − 1)d ρ0 k 2r p−1 (t) ∫ Ω u2r+q−2|∇u|dx. (4.14) Utilizing Hölder’s inequality and Young’s inequality, we have n ρ0 ∫ Ω u2r+q−1 dx ≤ (n2 ρ2 0 ∫ Ω u2r+2q−2 dx )1/2(∫ Ω u2r dx )1/2 ≤ 1 2 ( n ρ0 )2 ∫ Ω u2r+2q−2 dx+ 1 2 ∫ Ω u2r dx (4.15) and (2r + q − 1)d ρ0 ∫ Ω u2r+q−2|∇u|dx ≤ (2r + q − 1)d ρ0 ( 1 r2 ∫ Ω |∇ur|2 dx )1/2(∫ Ω u2r+2q−2 dx )1/2 = (ε1 r2 ∫ Ω |∇ur|2 dx )1/2( (2r + q − 1)2d2 ρ2 0ε1 ∫ Ω u2r+2q−2 dx )1/2 ≤ ε1 2r2 ∫ Ω |∇ur|2 dx+ (2r + q − 1)2d2 2ρ2 0ε1 ∫ Ω u2r+2q−2 dx, (4.16) where ε1 = (2r − 1)θ γ2 > 0. EJDE-2022/08 BLOW-UP FOR PARABOLIC EQUATIONS 11 Inserting (4.15) and (4.16) into (4.14), we have k 2r p−1 (t) ∫ ∂Ω u2r+q−1 ds ≤ 1 2 k 2r p−1 (t) ∫ Ω u2r dx + 1 2 (( n ρ0 )2 + (2r + q − 1)2d2 ρ2 0ε1 ) k 2r p−1 (t) ∫ Ω u2r+2q−2 dx + ε1 2r2 k 2r p−1 (t) ∫ Ω |∇ur|2 dx. (4.17) From this and (4.13), we deduce A′(t) ≤ 2rη p− 1 A(t) + rγ2k 2r p−1 (t) ∫ Ω u2r dx + rγ2 (( n ρ0 )2 + (2r + q − 1)2d2 ρ2 0ε1 ) k 2r p−1 (t) ∫ Ω u2r+2q−2 dx + (−2(2r − 1)θ r + ε1γ2 r ) k 2r p−1 (t) ∫ Ω |∇ur|2 dx − 2rγ1|Ω|− p−1 2r ( k 2r p−1 (t) ∫ Ω u2r dx ) 2r+p−1 2r . (4.18) Then, we pay our attention to the integral k 2r p−1 (t) ∫ Ω u2r+2q−2 dx in (4.18). Noticing that 0 < (q−1)(n−2) 2r < 1 in view of (4.5), and using Hölder’s inequal- ity, we obtain k 2r p−1 (t) ∫ Ω u2r+2q−2 dx ≤ ( k 2r p−1 (t) ∫ Ω (ur) 2n n−2 dx ) (q−1)(n−2) 2r ( k 2r p−1 (t) ∫ Ω u2r dx ) 2r−(q−1)(n−2) 2r . (4.19) Owing to the Sobolev inequality given in (4.6), we obtain∫ Ω (ur) 2n n−2 dx ≤ c 2n n−2 (∫ Ω u2r dx+ ∫ Ω |∇ur|2 dx ) n n−2 . (4.20) Substituting (4.20) into (4.19) and using (4.2) and the inequality (a+ b)µ ≤ 2µ(aµ + bµ), a, b, µ > 0, we derive k 2r p−1 (t) ∫ Ω u2r+2q−2 dx ≤ ( k 2r p−1 (t)c 2n n−2 (∫ Ω u2r dx+ ∫ Ω |∇ur|2 dx ) n n−2 ) (q−1)(n−2) 2r × ( k 2r p−1 (t) ∫ Ω u2r dx ) 2r−(q−1)(n−2) 2r ≤ k (q−1)(n−2) p−1 (t)c n(q−1) r 2 n(q−1) 2r ((∫ Ω u2r dx )n(q−1) 2r + (∫ Ω |∇ur|2 dx )n(q−1) 2r ) × ( k 2r p−1 (t) ∫ Ω u2r dx ) 2r−(q−1)(n−2) 2r 12 X. SHEN, J. DING EJDE-2022/08 = k− 2(q−1) p−1 (t)c n(q−1) r 2 n(q−1) 2r (( k 2r p−1 (t) ∫ Ω u2r dx )n(q−1) 2r + ( k 2r p−1 (t) ∫ Ω |∇ur|2 dx )n(q−1) 2r )( k 2r p−1 (t) ∫ Ω u2r dx ) 2r−(q−1)(n−2) 2r ≤ m− 2(q−1) p−1 c n(q−1) r 2 n(q−1) 2r ( k 2r p−1 (t) ∫ Ω u2r dx ) r+q−1 r +m− 2(q−1) p−1 c n(q−1) r 2 n(q−1) 2r × ( k 2r p−1 (t) ∫ Ω |∇ur|2 dx )n(q−1) 2r ( k 2r p−1 (t) ∫ Ω u2r dx ) 2r−(q−1)(n−2) 2r . (4.21) From (4.5), we can easily see that 0 < n(q−1) 2r < 1. It follows from Young’s inequality that( k 2r p−1 (t) ∫ Ω |∇ur|2 dx )n(q−1) 2r ( k 2r p−1 (t) ∫ Ω u2r dx ) 2r−(q−1)(n−2) 2r = ( k 2r p−1 (t) ∫ Ω |∇ur|2 dx )n(q−1) 2r (( k 2r p−1 (t) ∫ Ω u2r dx ) 2r−(q−1)(n−2) 2r−n(q−1) ) 2r−n(q−1) 2r = ( c3k 2r p−1 (t) ∫ Ω |∇ur|2 dx )n(q−1) 2r × ( c − n(q−1) 2r−n(q−1) 3 ( k 2r p−1 (t) ∫ Ω u2r dx ) 2r−(q−1)(n−2) 2r−n(q−1) ) 2r−n(q−1) 2r ≤ n(q − 1)c3 2r k 2r p−1 (t) ∫ Ω |∇ur|2 dx + 2r − n(q − 1) 2r c − n(q−1) 2r−n(q−1) 3 ( k 2r p−1 (t) ∫ Ω u2r dx ) 2r−(q−1)(n−2) 2r−n(q−1) , (4.22) where c3 is defined in (4.10). Combining (4.21) and (4.22) with (4.18), we have A′(t) ≤ 2rη p− 1 A(t) + rγ2k 2r p−1 (t) ∫ Ω u2r dx+ rγ2m − 2(q−1) p−1 c n(q−1) r 2 n(q−1) 2r × (( n ρ0 )2 + (2r + q − 1)2d2 ρ2 0ε1 )( k 2r p−1 (t) ∫ Ω u2r dx ) r+q−1 r + rγ2m − 2(q−1) p−1 c n(q−1) r 2 n(q−1) 2r (( n ρ0 )2 + (2r + q − 1)2d2 ρ2 0ε1 ) × c − n(q−1) 2r−n(q−1) 3 2r − n(q − 1) 2r ( k 2r p−1 (t) ∫ Ω u2r dx ) 2r−(q−1)(n−2) 2r−n(q−1) + [ − 2(2r − 1)θ r + ε1γ2 r + rγ2m − 2(q−1) p−1 c n(q−1) r 2 n(q−1) 2r × (( n ρ0 )2 + (2r + q − 1)2d2 ρ2 0ε1 )n(q − 1) 2r c3 ] k 2r p−1 (t) ∫ Ω |∇ur|2 dx − 2rγ1|Ω|− p−1 2r ( k 2r p−1 (t) ∫ Ω u2r dx ) 2r+p−1 2r = 2rη p− 1 A(t) + rγ2k 2r p−1 (t) ∫ Ω u2r dx+ c1 ( k 2r p−1 (t) ∫ Ω u2r dx ) r+q−1 r EJDE-2022/08 BLOW-UP FOR PARABOLIC EQUATIONS 13 + 2r − n(q − 1) 2r c1c − n(q−1) 2r−n(q−1) 3 ( k 2r p−1 (t) ∫ Ω u2r dx ) 2r−(q−1)(n−2) 2r−n(q−1) − 2rγ1|Ω|− p−1 2r ( k 2r p−1 (t) ∫ Ω u2r dx ) 2r+p−1 2r , (4.23) where c1 is given in (4.9). In view of (4.3) and (4.5), we have 0 < (2q − p− 1)[2r − n(q − 1)] 2r(2q − p− 1) + n(q − 1)(p− 1) < 1. Then Young’s inequality implies that ( k 2r p−1 (t) ∫ Ω u2r dx ) r+q−1 r = ( c − 2n(q−1)2 (2q−p−1)[2r−n(q−1)] 2 ( k 2r p−1 (t) ∫ Ω u2r dx ) 2r−(q−1)(n−2) 2r−n(q−1) ) (2q−p−1)[2r−n(q−1)] 2r(2q−p−1)+n(q−1)(p−1) × ( c2 ( k 2r p−1 (t) ∫ Ω u2r dx ) 2r+p−1 2r ) 2n(q−1)2 2r(2q−p−1)+n(q−1)(p−1) (4.24) ≤ (2q − p− 1)[2r − n(q − 1)] 2r(2q − p− 1) + n(q − 1)(p− 1) c − 2n(q−1)2 (2q−p−1)[2r−n(q−1)] 2 × ( k 2r p−1 (t) ∫ Ω u2r dx ) 2r−(q−1)(n−2) 2r−n(q−1) + 2n(q − 1)2c2 2r(2q − p− 1) + n(q − 1)(p− 1) ( k 2r p−1 (t) ∫ Ω u2r dx ) 2r+p−1 2r , where c2 is defined by (4.10). Finally, we insert (4.24) into (4.23) to obtain A′(t) ≤ 2rη p− 1 A(t) + rγ2k 2r p−1 (t) ∫ Ω u2r dx + c1 ( (2q − p− 1)[2r − n(q − 1)] 2r(2q − p− 1) + n(q − 1)(p− 1) c − 2n(q−1)2 (2q−p−1)[2r−n(q−1)] 2 + 2r − n(q − 1) 2r c − n(q−1) 2r−n(q−1) 3 )( k 2r p−1 (t) ∫ Ω u2r dx ) 2r−(q−1)(n−2) 2r−n(q−1) . (4.25) From (4.1), we obtain B(u) = 2r ∫ u 0 h′(y)y2r−1 dy ≥ 2rζ ∫ u 0 y2r−1 dy = ζu2r; that is u2r ≤ 1 ζ B(u). (4.26) 14 X. SHEN, J. DING EJDE-2022/08 By (4.26), we rewrite (4.25) as A′(t) ≤ 2rη p− 1 A(t) + rγ2 ζ k 2r p−1 (t) ∫ Ω B(u) dx + c1 ( (2q − p− 1)[2r − n(q − 1)] 2r(2q − p− 1) + n(q − 1)(p− 1) c − 2n(q−1)2 (2q−p−1)[2r−n(q−1)] 2 + 2r − n(q − 1) 2r c − n(q−1) 2r−n(q−1) 3 )(1 ζ ) 2r−(q−1)(n−2) 2r−n(q−1) × ( k 2r p−1 (t) ∫ Ω B(u) dx ) 2r−(q−1)(n−2) 2r−n(q−1) = J1A(t) + J2A(t) 2r−(q−1)(n−2) 2r−n(q−1) , (4.27) where J1 and J2 are defined in (4.7) and (4.8), respectively. The integration of (4.27) from 0 to t results in∫ A(t) A(0) dτ J1τ + J2τ 2r−(q−1)(n−2) 2r−n(q−1) ≤ t. Since u(x, t) blows up in measure A(t) at finite time t∗, we pass to the limits as t→ t∗− to obtain a lower bound t∗ ≥ ∫ ∞ A(0) dτ J1τ + J2τ 2r−(q−1)(n−2) 2r−n(q−1) . The proof is complete. � 5. Applications We provide two applications of Theorems 2.1, 3.1, and 4.1. Example 5.1. Let u(x, t) be a nonnegative classical solution of( u+ ln(1 + u) ) t = 3∑ i=1 ((31 16 + |x|2 ) uxi ) xi − (1 + e−t)u2 in Ω× (0, t∗), 3∑ i=1 (31 16 + |x|2 ) uxi νi = u2 on ∂Ω× (0, t∗), u(x, 0) = 15 16 + |x|2 in Ω, where Ω = {x = (x1, x2, x3) : |x|2 = ∑3 i=1 x 2 i < 1/16} a ball of R3. Now we have ( aij(x) ) 3×3 =  31 16 + |x|2 0 0 0 31 16 + |x|2 0 0 0 31 16 + |x|2  , h(u) = u+ ln(1 + u), k(t) = 1 + e−t, f(u) = u2, g(u) = u2, u0(x) = 15 16 + |x|2, n = 3. From (2.1), (3.1) and (3.2), it follows that F (u) = ∫ u 0 f(y) dy = ∫ u 0 y2 dy = 1 3 u3, G(u) = ∫ u 0 g(y) dy = ∫ u 0 y2 dy = 1 3 u3, EJDE-2022/08 BLOW-UP FOR PARABOLIC EQUATIONS 15 H(u) = 2 ∫ u 0 yh′(y) dy = 2 ∫ u 0 y ( 1 + 1 1 + y ) dy = u2 + 2u− 2 ln(1 + u), Φ(t) = ∫ Ω H(u) dx = ∫ Ω ( u2 + 2u− 2 ln(1 + u) ) dx, and Ψ(t) = 2 ∫ ∂Ω G(u) ds− ∫ Ω 3∑ i=1 aii(x)(uxi )2 dx− 2k(t) ∫ Ω F (u) dx = 2 3 ∫ ∂Ω u3 ds− ∫ Ω 3∑ i=1 (31 16 + |x|2 ) (uxi )2 dx− 2 3 (1 + e−t) ∫ Ω u3 dx. By choosing α = β = 1/2, it is easy to check that (3.3) and (3.4) hold. We then calculate Φ(0) = ∫ Ω ( u2 0 + 2u0 − 2 ln(1 + u0) ) dx = ∫ Ω ((15 16 + |x|2 )2 + 2 (15 16 + |x|2 ) − 2 ln (31 16 + |x|2 )) dx = 0.1008 and Ψ(0) = 2 3 ∫ ∂Ω u3 0 ds− ∫ Ω 3∑ i=1 ( 31 16 + |x|2 ) (u0xi )2 dx− 4 3 ∫ Ω u3 0 dx = 2 3 ∫ ∂Ω ( 15 16 + |x|2 )3 ds− 4 ∫ Ω (31 16 + |x|2 ) |x|2 dx− 4 3 ∫ Ω (15 16 + |x|2 )3 dx = 0.4232. It follows from Theorem 3.1 that u(x, t) blows up at t∗ in measure Φ(t), and t∗ ≤ Φ(0) 2β(1 + β)Ψ(0) = 0.1588. (5.1) To use Theorem 4.1 in obtaining a lower bound for the blow-up time t∗, we select γ1 = γ2 = 1, p = q = 2, ζ = 1, m = 1, η = 0, θ = 31/16, and r = 3. Here |Ω| = π/48, ρ0 = 1/4, and |d| = 1/4. It is easy to verify that (4.1)–(4.3) and (4.5) hold. The best Sobolev’s constant c = 3−1/241/3π−2/3 is given in [17]. Inserting the above paraments into (4.7)–(4.10), we obtain c1 = 270.2244, c2 = 0.0525, c3 = 0.0239, J1 = 3, and J2 = 3.8333× 104. By (4.4), we have B(u) = 2r ∫ u 0 h′(y)y2r−1 dy = 6 ∫ u 0 y5 ( 1 + 1 1 + y ) dy = u6 + 6 5 u5 − 3 2 u4 + 2u3 − 3u2 + 6u− 6 ln(1 + u), A(t) = k 2r p−1 (t) ∫ Ω B(u) dx = (1 + e−t)6 ∫ Ω ( u6 + 6 5 u5 − 3 2 u4 + 2u3 − 3u2 + 6u− 6 ln(1 + u) ) dx, 16 X. SHEN, J. DING EJDE-2022/08 A(0) = 64 ∫ Ω ( u6 0 + 6 5 u5 0 − 3 2 u4 0 + 2u3 0 − 3u2 0 + 6u0 − 6 ln(1 + u0) ) dx = 64 ∫ Ω ((15 16 + |x|2 )6 + 6 5 (15 16 + |x|2 )5 − 3 2 (15 16 + |x|2 )4 + 2 (15 16 + |x|2 )3 − 3 (15 16 + |x|2 )2 + 6 (15 16 + |x|2 ) − 6 ln (31 16 + |x|2 )) dx = 5.5901. Since u(x, t) blows up in measure Φ(t) at finite time t∗, u(x, t) must blow up in measure A(t) at t∗. From Theorem 4.1, we obtain a lower bound t∗ ≥ ∫ ∞ A(0) dτ J1τ + J2τ 2r−(q−1)(n−2) 2r−n(q−1) = ∫ ∞ 5.5901 dτ 3τ + 3.8333× 104τ5/3 = 1.2423× 10−5. Combining this with (5.1), we obtain 1.2423× 10−5 ≤ t∗ ≤ 0.1588. Example 5.2. Let u(x, t) be a nonnegative classical solution of (u+ ln(1 + u))t = 3∑ i=1 (( 1 + |x|2 ) uxi ) xi − etu4 in Ω× (0, t∗), 3∑ i=1 ( 1 + |x|2 ) uxi νi = u2 on ∂Ω× (0, t∗), u(x, 0) = 1 + |x|2 in Ω, where Ω = {x = (x1, x2, x3) : |x|2 = ∑3 i=1 x 2 i < 1}, a ball of R3. Now ( aij(x) ) 3×3 = 1 + |x|2 0 0 0 1 + |x|2 0 0 0 1 + |x|2  , h(u) = u+ ln(1 + u), k(t) = et, f(u) = u4, g(u) = u2, u0(x) = 1 + |x|2, n = 3. Here we choose γ1 = γ2 = 1, p = 4, q = 2, ζ0 = 2, θ = 1, and m = 1. It is easy to see that (2.2)–(2.5) are valid. Consequently, by Theorem 2.1, u(x, t) exists for all time t > 0 in measure Φ(t) with Φ(t) = ∫ Ω H(u) dx = ∫ Ω ( u2 + 2u− 2 ln(1 + u) ) dx. Acknowledgements. This work was supported by the National Natural Science Foundation of China (No. 61473180). References [1] K. Baghaei, M. B. Ghaemi, M. Hesaaraki; Lower bounds for the blow-up time in a semilinear parabolic problem involving a variable source, Appl. Math. Lett., 27 (2014), no. 1, 49–52. [2] K. Baghaei, M. 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Xuhui Shen School of Mathematical Sciences, Shanxi University, Taiyuan 030006, China Email address: xhuishen@163.com Juntang Ding (corresponding author) School of Mathematical Sciences, Shanxi University, Taiyuan 030006, China Email address: djuntang@sxu.edu.cn 1. Introduction 2. Global solution 3. Blow-up solution 4. Lower bound for the blow-up time 5. Applications Acknowledgements References