Extinction and Decay Estimates of Solutions for the p-Laplacian Equations with Nonlinear Absorptions and Nonlocal Sources

Lemma 1 (Gagliardo-Nirenberg inequality). Suppose that β ≥ 0, N > p ≥ 1, β + 1 ≤ q ≤ (β + 1)Np/(N − p); then for u such that |u|^βu ∈ W^1,p(Ω), one has

with θ = ((β + 1)r⁻¹ − q⁻¹)/(N⁻¹ − p⁻¹ + (β + 1)r⁻¹), where C is a constant depending only on N, p, and r.

Theorem 2. Assume that p − 1 = q with r < 1; then the non-negative nontrivial weak solution of Problem (1)-(2) vanishes in finite time for any non-negative initial data provided that |Ω| or λ is sufficiently small.

• (1)

  For the case  2N/(N + 2) ≤ p < 2, one has

  $$ ()

•  

  where  k₁,   M₁, and  T₁  are given by (11), (16), and (17), respectively.

• (2)

  For the case  1 < p < 2N/(N + 2), one has

  $$ ()

•  

  where  s,  k₂,  M₂, and  T₂  are given by (18), (22), (26), and (28), respectively.

Proof. (1) For the case 2N/(N + 2) ≤ p < 2, multiplying (1) by u and integrating over Ω, we deduce from the Hölder inequality that

inequality where B denotes the optimal embedding constant, combining (6) and (7) we have By Lemma 1, we have where θ₁ = (1/(1 + r) − 1/2)(1/N − 1/p + 1/(1 + r)) ⁻¹.

It is easy to check that θ₁ ∈ (0,1]; using Young′s inequality with ε, it follows from (9) that

where ε₁ > 0 and k₁ > 0 will be determined later. We choose Then we can conclude that k₁ ∈ (1,2) and pk₁(1 − θ₁)/(p − k₁θ₁) = 1 + r. Therefore, it follows from (10) that

By combining (8) and (12), we have

Choosing ε₁ small enough such that 1 − kε₁/C(ε₁) > 0 and |Ω | ≤ (1 − kε₁/C(ε₁))/λB^p, then we have 1 − kε₁/C(ε₁) − B^pλ | Ω | > 0. Therefore, we deduce from k₁ ∈ (1,2) that which implies that where (2) For the case 1 < p < 2N/(N + 2), multiplying (1) by u^s, where integrating over Ω, we deduce from the Hölder inequality that By Lemma 1 and s > 1, we have where θ₂ = N(1 − r)(p + s − 1)/(s + 1)[p(s + r) + N(p − 1 − r)]. By (18) and r < 1, it is easy to check that θ₂ ∈ (0,1). By Young′s inequality with ε, it follows from (19) that where ε₂ > 0 and k₂ > 0 will be determined later. We choose then it follows that k₂ ∈ (s, s + 1) and (p + s − 1)k₂(1 − θ₂)/(p + s − 1 − k₂θ₂) = s + r. Therefore, it follows from (21) that

By combining (19) and (23), we have by poincare inequality

Choosing ε₂ > 0 small enough such that sp^p/(p + s − 1) ^p − kε₂/C(ε₂) > 0 and |Ω | ≤ (sp^p/(p + s − 1) ^p − kε₂/C(ε₂))/λ | Ω | B^p, then we have sp^p/(p + s − 1) ^p − kε₂/C(ε₂) − λ | Ω | B^p > 0. Therefore, we deduce from k₂ ∈ (s, s + 1) that where which implies that where The proof of Theorem 2 is complete.

Theorem 3. Assume that r < 1.

• (1)

  If  2N/(N + 2) ≤ p < 2  with  q > k₁ − 1 = (2rp + N(p − 1 − r))/(2p + N(p − 1 − r)), then the non-negative nontrivial weak solution of Problem (1)-(2) vanishes in finite time provided that  u₀  (or  |Ω|  or  λ) is sufficiently small and

  $$ ()

•  

  where  k₁,  M₃, and  T₃  are given by (11), (35), and (33), respectively.

• (2)

  If  1 < p < 2N/(N + 2)  with  q > k₂ − s = ((s + 1)rp + N(p − 1 − r))/((s + 1)p + N(p − 1 − r)), then the non-negative nontrivial weak solution of Problem (1)-(2) vanishes if finite time provided that  u₀  (or  |Ω|  or  λ) is sufficiently small and

  $$ ()

•  

  where  s,  k₂,  M₄, and  T₄  are given by (18), (22), (39), and (41), respectively.

Proof. (1)  If 2N/(N + 2) ≤ p < 2, multiplying (1) by u and integrating over Ω, we deduce from (12) and the Hölder inequality that

By choosing ε₁ > 0 small enough such that 1 − kε₁/C(ε₁) ≥ 0, we obtain that provided that $$ and q > k₁ − 1 = (2rp + N(p − 1 − r))/(2p + N(p − 1 − r)), where

From (32) and k₁ ∈ (1,2), we can derive that

where (2)  If 1 < p < 2N/(N + 2), multiplying (1) by u^s, where s is given by (18) and integrating over Ω, we deduce from the Hölder inequality and (23) that Choosing ε₂ > 0 small enough such that sp^p/(p + s − 1) ^p − kε₂/C(ε₂) > 0, we have

Therefore, we have

provided that $$ and q > k₂ − s = ((s + 1)rp + N(p − 1 − r))/((s + 1)p + N(p − 1 − r)), where It follows from (38) and k₂ ∈ (s, s + 1) that where The proof of Theorem 3 is complete.