# Singularity and Decay Estimates for a Degenerate Parabolic Equation

Authors: Dongyan Li
Journal: Complexity
 (2021)

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## Abstract

In this paper, a degenerate parabolic equationut−divxθ∇u=xaup with p>1 and θ<2,a∈ℝ, is considered. Based on rescaling arguments combined with a doubling property, the space-time singularity and decay estimates are established. Moreover, a universal and a priori bound of global nonnegative solutions for the corresponding initial boundary value problem is derived.


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## Body

## 1. Introduction

In this paper, we consider the degenerate parabolic equation of the form(1)ut−divxθ∇u=xaup,x,t∈Ω×I,where θ<2,a∈ℝ, Ω is a domain of ℝN with N≥2, and I is an interval of ℝ. Throughout this paper, solutions are considered in the class (unless otherwise specified) C2,1Ω\0×I∩C0,0Ω×I. We denote(2)N′=N+θ>2,τ≔a−θ>−2,pcτ≔N′+2+2τN′−2>1,pB≔NN+2N−12.In order to prove the existence and multiplicity of positive solutions, it is important to obtain a priori bounds for their positive solutions. It is well known that Liouville theorem would imply optimal universal estimates for solutions of related initial or initial boundary value problems, including estimates of their singularities and decay via the rescaling method (see [1, 2]). The following Liouville result [3] will play an important role in our proof of singularity estimates.Theorem 1.
Letθ=a=0 and 1<p<pB. Equation (1) has no nontrivial solution in ℝN×ℝ.In the special caseθ=a=0, equation (1) has been widely studied by many authors. Here, the space-time singularity and decay estimates of positive solutions of equation (1) have been established by Peter et al. in [2] for the case θ=a=0 and by Phan [4] for the case θ=0. Motivated by the aforementioned work, we establish the space-time singularity and decay estimates of solutions of equation (1).Theorem 2.
(i)
Assume thatN′>2,τ>−2, and 1<p<pB. Let u be a nonnegative solution of (1) on Ω×0,T, where Ω=x∈ℝN:0<x<ρρ>0. Then, for all 0<x<ρ/2 and t∈0,T, there holds(3)xa/p−1ux,t+xa/p−1∇ux2/p+1≤Ct−1/p−1+T−t−1/p−1+x−2−θ/p−1.(ii)
Assume thatN′>2,τ>−2, and 1<p<pB. Let u be a nonnegative solution of (1) on Ω×0,T, where Ω=x∈ℝN:x>ρρ≥0. Then, for all x>2ρ and t∈0,T, there holds(4)xa/p−1ux,t+xa/p−1∇ux2/p+1≤Ct−1/p−1+T−t−1/p−1+x−2−θ/p−1.The proof of Theorem1 is based on rescaling arguments combined with a key doubling property. The classical rescaling technique argument was first introduced by Gidas and Spruck [5] for the elliptic problem; it was then significantly improved in [2, 6–11] for elliptic and parabolic that enables one to obtain a variety of important results. Moreover, it also needs the nonexistence of nontrivial solutions on the whole space for equation (1) with θ=a=0.Remark 1.
In Theorem2, if we replace the interval 0,T by ℝ, then we have the spatial decay estimate(5)ux,t≤Cx−2+τ/p−1,∇ux,t≤Cx−p+1+τ/p−1,x>0,t∈ℝ.
As the next topic, we consider the associated initial boundary value problem:(6)ut−divxθ∇u=xaup,x∈Ω,0<t<T,u=0,x∈∂Ω,0<t<T,ux,0=u0x,x∈Ω.
Here, we assume thatΩ⊂ℝN is a smoothly bounded domain containing the origin. We have the following result.Theorem 3.
LetN′≥2,τ>−2, and 1<p<minN+2/N−2,pcτ. Assume that u is any global solution of (6) with initial data u0≥0 and u satisfies the estimate(7)∫0T∫Ωut2dxdt<K<∞.Then, there exists a constantC independent of u,u0, and T such that(8)supt≥0ut∞≤C.Remark 2.
Phan [2] proved a priori bound of nonnegative solutions for the parabolic problem ut−Δu=xaup under the condition p<minN+2/N−2,pcτ. Theorem 3 says that a priori bound for a degenerate parabolic equation in the form of (6) also holds under condition (7). It should be mentioned that the universal and a priori bound of global solutions was shown by Quittner [12], Xing [13], and Földers [14] for sign changing to (6) in the special case θ=a=0.
Our proof of Theorem3 appeals to rescaling techniques and argues by contradiction following the ideas introduced by Phan [2] for the parabolic equation. We also use this powerful idea and introduce some rescalings to deal with some new difficulties arising due to the degeneracy and singularity.
The rest of the paper is organized as follows. Section2 is devoted to the proof of singularity and decay estimates (Theorem 2). Section 3 contains the proof of Theorem 3 which is an a priori bound of global solutions and blow-up estimates.


## 2. Singularity and Decay Estimates

In this section, we give the proof of Theorem2.Proof of Theorem 2.
Assume eitherΩ=x∈ℝN:0<x<ρ and 0<x0<ρ/2 or Ω=x∈ℝN:x>ρ and x0>2ρ. Denote R=2/3x0 and observe that, for all y∈C=x∈ℝN:1<x<2, 2x0/3<Ry<4x0/3 so that Ry∈Ω. Let us define Uy,s=R2+τ/p−1uRy,R−2+θs. Then, Uy,s is a solution of(9)∂sU−divyθ∇U=yaUp,y,s∈C×0,R−2+θT.
Notice thaty∈1,2 for all y∈C¯. We first claim that, for all y,s∈C×0,R−2+θT, there exists a constant C=CN,p,τ>0 (independent of Ω and U) such that(10)Uy,s+∇Uy,s2/p+1≤C1+s−1/p−1+R−2+θT−s−1/p−1+dist−2/p−1y,∂C.
Estimate (10) can be rewritten in a more concise equivalent form by using the parabolic distance dPx,t,y,s≔x−y+t−s1/2, namely, let D=C×0,R−2+θT∈ℝN+1, ∂D=∂Ω×0,R−2+θT∪Ω¯×0,R−2+θT, and dx,t,∂D=infy,s∈∂DdPx,t,y,s. Then, estimate (10) can be restated as(11)Uy,s+∇Uy,s2/p+1≤C1+d−2/p−1y,s,∂D,y,s∈D.
Assume that estimate (10) fails. Then, there exist sequences Tk, points zk,γk∈Dk, and Uk that solves(12)∂sUk−divyθ∇Uk=yaUkp,y,s∈Dk=C×0,Rk−2+θTk,where Rk=2/3xk such that the functions Mk≔Ukp−1/2+∇Ukp−1/p+1 satisfy(13)Mkzk,γk>2k1+d−1zk,γk,∂Dk>2kd−1zk,γk,∂Dk.
By the doubling lemma in [11], it follows that there exists yk,sk∈Dk such that(14)Mkyk,sk≥Mkzk,γk,Mkyk,sk>2kd−1zk,γk,∂Dk,(15)Mky,s≤2Mkyk,sk,for ally,ssuch thatdy,s,yk,sk≤kMk−1yk,sk.
It follows from (14) that dzk,γk,∂Dk>2kMk−1yk,sk. Therefore, any y,s satisfying the condition dy,s,yk,sk≤kMk−1yk,sk is automatically contained in Dk. Since Mkyk,sk≥Mkzk,γk>2k, we have(16)λk≔Mk−1yk,sk⟶0,k⟶∞.
Now, we define the scaled function(17)Vkz,γ=λk2/p−1Ukyk+λkz,sk+λk2γ,z,γ∈D˜k,where D˜k=z<k/2×−k2/4,k2/4. Clearly,(18)Vk0,0p−1/2+∇Vk0,0p−1/p+1=λkMkyk,sk=1,(19)Vkp−1/2+∇Vkp−1/p+1z,γ≤2,z,γ∈D˜k,due to (15). We see that Vk satisfies(20)∂γVk−divyk+λkzθ∇Vk=yk+λkzaVkp,z,γ∈D˜k.
Now, for eachA>0 and 1<q<∞, by (19), (20), and interior parabolic Lq estimates, the sequence Vk is uniformly bounded in Wq2,1BA×−A,A. Note that yk+λkz∈1,2 for all k. By the interior parabolic Lq estimates and standard embeddings, we can find subsequences (still denoted by yk,Vk), and we may assume that yk⟶y0∈C¯ and Vk⟶VV>0 in Cloc2,1ℝN×ℝ. Therefore, V is a classical solution of ∂γV−y0θΔV=y0aVp,z,γ∈ℝN×ℝ, and V0,0p−1/2+∇V0,0p−1/p+1=1. Since p<pB, after a space shift, we know that this contradicts Theorem 1 and concludes our claim (11). This also implies (10) holds. Therefore,(21)UR−1x0,R−2+θt+∇UR−1x0,R−2+θt2/p+1≤C1+R−2+θt−1/p−1+R−2+θT−R−2+θt−1/p−1,that is,(22)Ra/p−1ux0,t+Ra/p−1∇ux0,t2/p+1≤Ct−1/p−1+T−t−1/p−1+R−2−θ/p−1,which completes the proof of Theorem 2.


## 3. A Priori Bound

Proof of Theorem 3.
Our proof is by contradiction. Suppose that Theorem3 is false. Then, there exist ukx,t satisfying (6) and sequences of points xk,tktk>0 such that(23)Nk≔uxk,tk=supukx,t:x∈Ω,t∈0,tk⟶∞,ask⟶∞.
SinceΩ¯ is compact, we may assume that xk⟶x0∈Ω¯ as k⟶∞. Denote dk=distxk,∂Ω. We divide the situation into three cases:Case 1:x0∈Ω\0. Let λk=Nk−p−1/2. Since Nk⟶∞,p>1, evidently, λk⟶0. Define the scaled function(24)vky,s=λk2/p−1ukxk+λky,tk+λk2s,y,s∈Qk,whereQk=y,s:xk+λky,tk+λk2s∈Ω×0,tk. Clearly, vk0,0=1and0≤vky,s≤1. It follows that vk is a solution to(25)∂svk−divxk+λkyθ∇vk=xk+λkyavkp,y,s∈Q˜k,whereQ˜k=y,s:y<dk/λk,−tk/2λk2<s<0. Using parabolic Lp estimates and standard embedding theorems, we can find a subsequence (still denoted by vk) which converges uniformly to some function v≥0 in Cloc0,0ℝN×−∞,0, and(26)∂sv−x0θΔv=x0avp,y,s∈ℝN×−∞,0.Moreover,v0,0=1and0≤vy,s≤1. By our assumption (7), we have that(27)∫Q˜k∂svk2dyds=λk4/p−1−N+2∫tk/2tk∫x−xk<dk∂tuk2dxdt,≤λk4/p−1−N+2∫0T∫Ω∂tuk2dxdt<λk4/p−1−N+2K⟶0,ask⟶∞.Since1<p<minN+2/N−2,pcτ, we get 4/p−1−N+2>0. This yields ∂sv=0 because ∂svk⟶∂sv in ℝN×−∞,0. After a space shift, this gives a contradiction with Theorem 1. Hence, v≡0. This contradicts the fact v0,0=1. Hence, case 1 cannot occur.Case 2:x0∈∂Ω. Without loss of generality, we may assume that, near x0∈∂Ω, the boundary ∂Ω is contained in the hyperplane xN=0. We rescale the solution according to(28)vky,s=λk2/p−1ukxk+λky,tk+λk2s,λk=Nk−p−1/2.Note that, for largek, vky,s is well defined in Dk=y:y<δ/λk∩yN>−dk/λk×−tk/2λk2<s<0 for some δ>0 and satisfies(29)∂svk−divxk+λkyθ∇vk=xk+λkyavkp,y,s∈Dk.Clearly,(30)vk0,0=1and0≤vky,s≤1.By (30) and the fact vk=0 on yN=0, we have dk/λk≥A>0. If dk/λk⟶∞, then after a choice of the subsequence, we have the same contradiction as in the first case. If dk/λk⟶c>0, using parabolic Lp estimates and standard embedding theorems, we can find a subsequence (still denoted by vk) which converges uniformly to some function v satisfying 0≤v≤v0,0=1, and(31)∂sv−x0θΔv=x0avp,inyN>c×−∞,0,v=0,onyN=c×−∞,0.As in case 1, with the necessary changes of the domain of integration, we yield∂sv=0. This contradicts Theorem 2.1 in [2].Case 3:x0=0. We discuss the situation in two possibilities:(a)
IfNkxk2+τ/p−1≤C, let λk=Nk−2+τ/p−1; then, we have λk−1xk is bounded. We may assume that λk−1xk⟶x¯ as k⟶∞ by choosing subsequences. Denote by vky=λk2+τ/p−1ukxk+λky,tk+λk2−θs; then, vk is a solution to(32)∂vk−divy+xkλkθ∇vk=y+xkλkavkp,y,s∈Hk,whereHk=y,s:y<dk/λk,−tk/2λk2−θ<s<0. A similar limiting procedure as in case 1 then produces a solution v of(33)∂v−divy+x¯θ∇v=y+x¯avp,y,s∈ℝN×−∞,0.Moreover,(34)v0,0=1and0≤vy,s≤1.Using (7) and p<pcτ, we have(35)∫Hk∂svk2dyds=λk4+2τ/p−1−N+2−θ⋅∫tk/2tk∫x−xk<dk∂tuk2dxdt,≤λk4+2τ/p−1−N+2−θ∫0T∫Ω∂tuk|2dxdt<λk4+2τ/p−1−N+2−θK⟶0,ask⟶∞.Since1<p<minN+2/N−2,pcτ≤pcτ, we get 4+2τ/p−1−N+2−θ>0. This yields ∂sv=0 because ∂svk⟶∂sv in ℝN×−∞,0. Hence,(36)−divy+x¯θ∇v=y+x¯avp,y,s∈ℝN×−∞,0.After a space shift, this gives a contradiction with Theorem2 in [15]. Hence, v≡0.(b)
If there exists a subsequence ofk, still denoted by k, such that Nkxk2+τ/p−1⟶∞, then we can choose bk>1 such that(37)Nkxk2bk+τ/p−1=1.
Let(38)wky=λk2bk+τ/bkp−1ukxk+λky,tk+λk2−θ/bks,where λk=Nk−bkp−1/2bk+τ; then, wk solves the problem(39)∂wk−divλk−1/bkxk+λk1−1/bkyθ∇wk=λk−1/bkxk+λk1−1/bkyawkp,y,s∈Gk=y,s:y<dkλk,−tk2λk2−θ/bk<s<0.
Sinceλk−1/bkxk=1 and 0<λk1−1/bk<1, we may assume that λk−1/bkxk⟶P with P=1 and λk1−1/bk⟶l∈0,1. A similar limiting procedure as in case 1 then produces a solution w of(40)∂w−divP+lyθ∇w=P+lyawp,y,s∈ℝN×−∞,0.
We will show that∂w=0; indeed,(41)∫Gk∂swk2dyds=λk4bk+2τ/bkp−1−N+2−θ/bk∫tk/2tk∫x−xk<dk∂tuk2dxdt,≤λk4bk+2τ/bkp−1−N+2−θ/bk∫0T∫Ω∂tuk2dxdt<λk4bk+2τ/bkp−1−N+2−θ/bkK⟶0,ask⟶∞.
Since(42)4bk+2τbkp−1−N+2−θbk≥min4p−1−N+2−θbk,4+2τp−1−N+2−θbk>0,thus, ∂ws=0. This implies that w is a solution of(43)−divP+lyθ∇w=P+lyawp,y,s∈ℝN×−∞,0.
This contradicts with Theorem2 in [15].



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*Source: 5517008-2021-03-11.xml*