A p-Laplacian type problem with a source reaction term involving the product of the function and its gradient is considered in this paper. A Harnack inequality is proved, and the main idea is based on de Giorgi-Nash-Moser iteration and Moser's iteration technique. As a consequence, H¨older continuity and boundness for the solution of this problem also are obtained.
Citation: Bo Chen, Junhui Xie. Harnack inequality for a p-Laplacian equation with a source reaction term involving the product of the function and its gradient[J]. Electronic Research Archive, 2023, 31(2): 1157-1169. doi: 10.3934/era.2023059
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A p-Laplacian type problem with a source reaction term involving the product of the function and its gradient is considered in this paper. A Harnack inequality is proved, and the main idea is based on de Giorgi-Nash-Moser iteration and Moser's iteration technique. As a consequence, H¨older continuity and boundness for the solution of this problem also are obtained.
In this paper, we consider a p-Laplacian Dirichlet boundary problem with a source reaction term involving the product of the function and its gradient as follows:
{−Δpu=uα1|∇u|α2,u>0,u∈Hp0(Ω). | (1.1) |
We mainly focus on the Harnack inequality for nonnegative solutions of problem (1.1), when Ω is a bounded domain in RN, N>2, and 0<α1,α2<p−1 are positive exponents.
Particularly, if p=2,α1=1, and α2=2, the equation in problem (1.1) becomes
−Δu=u|∇u|2, |
which is associated with the Euler-Lagrange equation of the Dirichlet energy of mappings between two Riemannian manifolds; see [1]. From this point of view, we call it a p-Harmonic Mappings type equation. If α1=0, we recall the equation in problem (1.1) is the Hamilton-Jacobi equation:
−Δpu=|∇u|α2. |
In [2], the author proved that any C1 solution of the Hamilton-Jacobi equation in an arbitrary domain Ω⊂RN with N≥p>1 and α2>p−1 satisfies
|∇u|≤CN,p,α2(dist(x,∂Ω))−1α2−p+1. |
Similar estimates and related results can be found in [3,4,5,6,7], and see also the references therein. If α2=0, then problem (1.1) reduces to the subcase Lane-Emden equation,
{−Δpu=uα1,u>0,u∈Hp0(Ω), | (1.2) |
which has been widely studied in the literature [8,9,10,11]. For the generalized case of problem (1.1), when p=2,α1+α2>1,α1≥0, and 0≤α2≤2, the authors in [12] proved a local Harnack inequality and nonexistence of positive solutions in RN under the condition α1(N−2)+α2(N−1)<1. In addition, the local and global estimates of the solution are also obtained in [12]. For p=2,α1>0, and α2>2, Liouville type theorems of problem (1.1) were studied in [13], and the proof technique is based on monotonicity properties for the spherical averages of sub- and super-harmonic functions, combined with a gradient bound obtained by a local Bernstein argument.
As is known to all, the Harnack inequality plays an important role in the theory of regularity for elliptic and parabolic partial differential equations, and it was originally defined for harmonic functions in the plane and much later became an important tool in the general theory of harmonic functions. In [14], the German mathematician C-G. Axel von Harnack formulated and proved the following theorem in the case N=2.
Theorem. A (Harnack inequality [14]) Let u be a nonnegative harmonic function in an open set E⊂RN. Then, for all x∈Br(x0)⊂BR(x0)⊂E,
(RR+r)N−2R−rR+ru(x0)≤u(x)≤(RR−r)N−2R+rR−ru(x0), |
where BR(x0)={x∈RN:|x−x0|<R}.
The estimate above is scale invariant in the sense that it does not change for various choices of R, with r=mR,m∈(0,1) fixed. Furthermore, it depends neither on the position of the ball BR(x0) nor on u itself. In its modern version, the currently used Harnack inequality for harmonic functions is given in the following form:
Theorem. B (Harnack inequality [15]) Let N≥2 and E⊂RN be an open set. Then, there exists a constant C>1, dependent only on the dimension N, such that
C−1supBr(x0)u≤u(x0)≤CinfBr(x0)u, |
for every nonnegative harmonic function u:E→R and for every ball Br(x0)={x∈RN:|x−x0|<r} such that B2r(x0) is contained in E.
Generally speaking, the Harnack inequality asserts that the upper and lower bounds of the solution for a specific initial-boundary value problem can be locally estimated by each other. Although such an estimate seems trivial according to its two forms in Theorems A and B, it plays a fundamental role in many other aspects, such as in the proof of the Liouville Theorem (see [16,17]), in the construction of solutions to the Dirichlet problem for the Laplace equation using Perron's method (see [18] for more details) and so on. It is worth pointing out that one can also refer to [19,20] for the applications of Harnack inequalities in geometry analysis and the estimates of heat kernel on manifolds.
We will use an iteration technique to obtain a Harnack inequality for solutions of problem (1.1) in this paper; it was originally developed by de Giorgi[21]-Nash[22]-Moser[23] to deal with the regularity of solutions for general linear elliptic and parabolic equations. More specifically, this method originally derived from the study of regularity of calculus of variations problems, which mainly focus on the H¨older continuity for solutions of linear uniformly elliptic equations. Here, we follow the method developed by Moser in [23], in which the following type weak Harnack inequality was proved:
Theorem. C (Moser's Harnack inequality [23]) Let u∈H1(Ω) be a non-negative weak solution to the linear equation
div(ai,j(x)u(x))=0 in Ω |
with measurable coefficients ai,j(x):Ω→RN×N being elliptic and bounded. Then, for every ball B√NR(x0)⊂Ω⊂RN, there holds
infBR/2(x0)u≥CsupBR/2(x0)u |
where C is a constant depending on N and the coefficients ai,j(x). In addition, the H¨older continuity is obtained as a direct correspondence.
Nevertheless, it should be pointed out that this iteration method can also deal with the regularity of quasilinear operators, especially for the p-Laplacian operator. Furthermore, the iteration technique developed by Moser can overcome the difficulties caused by the degeneracy of the p-Laplacian operator, and more details about this issue can be found in [24,25]. The standard techniques of this iteration method associated with p-Laplacian operators in the homogeneous case
div(|∇u|p−2)∇u=0 |
can be found in [26].
Now, we state the main result in this paper.
Theorem 1.1. (Harnack inequality) If u∈Hp0(Ω) is a solution of problem (1.1), then for any ball Br(x)⊂BR(x)⊂⊂Ω,0<r<R,x∈Ω, there exists C=C(N,ε,p) such that
supBr(x)u≤CinfBR(x)u. |
Remark 1.1. Throughout this paper, we assume that ε>0 is small enough.
This paper is organized as follows. We first give some lemmas and preliminaries in Section 2 which will be used in the proof of Theorem 1.1. Then, the Harnack inequality for the solution of problem (1.1) is addressed in Section 3 using de Giorgi-Nash-Moser iteration and Moser's iteration technique. In addition, H¨older continuity and boundness for the solution of problem (1.1) also are obtained in Section 3.
In this section, we give some preliminaries and lemmas needed in the proof of the main result.
Definition 2.1. We say u∈Hp0(Ω) is a sub-solution (sup-solution) of problem (1.1) if u is positive a.e. in Ω and
∫Ω|∇u|p−2∇u⋅∇φdx≤(≥)∫Ωuα1|∇u|α2φdx |
for any φ∈Hp0(Ω), φ≥0 in Ω.
Lemma 2.1. (John-Nirenberg Lemma [27,28]) Suppose w∈L1loc(Ω) satisfies
∫Br(x)|w(x)−wBr|dx≤KrN for any Br(x)∈Ω, |
where wBr=1|Br(x)|∫Br(x)u(y)dy. Then, there holds, for any Br(x)⊂Ω,
∫Br(x)ep0|w(x)−wBr|/Kdx≤CrN |
for some positive p0 and C depending only on N.
Lemma 2.2. ([28]) Let ω and σ be nondecreasing functions in an interval (0,R]. Suppose there holds
ω(τr)≤γω(r)+σ(r) |
for all r≤R and some 0<γ,τ<1. Then, for any μ∈(0,1) and r≤R, we have
ω(r)≤C{(rR)αω(R)+σ(rμR1−μ)} |
where C=C(γ,τ) and α=α(γ,τ,μ) are positive constants. In fact, α= (1−μ)logγ/logτ.
Lemma 2.3. (Leray-Schauder's principle [29]) Assume that X is a real Banach space, Ω is an open bounded subset of X, and Φ:[a,b]×ˉΩ⟶X is given by Φ(λ,u)=u−T(λ,u) with T being a compact map. Suppose also that
Φ(λ,u)=u−T(λ,u)≠0, ∀(λ,u)∈[a,b]×∂Ω. |
We define Φa=I−T(a,u), where I denotes the identity map in X. If deg(Φa,Ω,0)≠0, then,
(i) Φ(λ,u)=u−T(λ,u)=0 has a solution in Ω for every a≤λ≤b.
(ii) Furthermore, define Σ={(λ,u)∈[a,b]×ˉΩ:Φ(λ,u)=0} and Σλ={u∈ˉΩ:(λ,u)∈Σ}. Thus, there exists a compact connected set Q⊂Σ such that
Q∩({a}×Σa)≠∅ and Q∩({b}×Σb)≠∅. |
In this section, we give a Harnack inequality for the solution of problem (1.1) by de Giorgi-Nash-Moser iteration and Moser's iteration technique. First, we deduce the following lemmas.
Lemma 3.1. Suppose u∈Hp0(Ω) is a super-solution of problem (1.1). Then, for any ball Br(x)⊂BR(x)⊂⊂Ω,0<r<R<1,p−1<β≤p, there holds,
infBr(x)u≥C(N,ε){1(R−r)p∫BR(x)u−βdx}−1/β, |
where C(N,ε) is a constant depending on ε and N, and ε>0 is small enough.
Proof. Choose a test function as follows: φ=ηpus, where η∈C∞0(BR(x)),η≡1 in Br(x),0<η≤1,|∇η|≤1R−r. It is not difficult to calculate that
∇φ=pηp−1us∇η+sηpus−1∇u. |
Multiplying on both sides of the first equation for problem (1.1) by φ and integrating by parts, since u is a super-solution of problem (1.1), we get that
p∫BR(x)ηp−1us|∇u|p−2∇u⋅∇ηdx+s∫BR(x)ηpus−1|∇u|pdx≥∫BR(x)uα1+s|∇u|α2ηpdx. |
Hence, by H¨older inequality and Young's inequality with ε, we obtain that
−s∫BR(x)ηpus−1|∇u|pdx≤p∫BR(x)ηp−1us|∇u|p−2∇u⋅∇ηdx−∫BR(x)uα1+s|∇u|α2ηpdx≤p∫BR(x)ηp−1us|∇u|p−2∇u⋅∇ηdx≤p∫BR(x)ηp−1|∇u|p−1u(s−1)p−1pus−(s−1)p−1p|∇η|dx≤p(∫BR(x)ηpus−1|∇u|pdx)p−1p(∫BR(x)u−β|∇η|pdx)1p≤εppp−1∫BR(x)ηpus−1|∇u|pdx+C(ε)∫BR(x)u−β|∇η|pdx, | (3.1) |
where s=−β−(p−1). Choose ε small enough and consider the definition of φ, and we can get that
∫BR(x)ηpus−1|∇u|pdx≤C(ε)∫BR(x)u−β|∇η|pdx≤C(ε)1(R−r)p∫BR(x)u−βdx. | (3.2) |
Also, we define κ=NN−p and deduce the following conclusion by Sobolev inequality:
(∫Br(x)u−κβdx)1κ=(∫Br(x)u−NβN−pdx)N−pN≤C(N)∫Br(x)|∇(u−βp)|pdx=C(N)∫Br(x)|u−βp−1∇u|pdx=C(N)∫Br(x)us−1|∇u|pdx≤C(N,ε)1(R−r)p∫BR(x)u−βdx, | (3.3) |
where C(N) is a constant depending on N, C(N,ε) is a constant depending on ε and N, and the last inequality is based on the conclusion of (3.2). Define ri=r+2−i(R−r),βi=−κiβ,Aβi,ri=∫Bri(x)uβidx, i=1,2,....n...; and owing to (3.3), we have
A1κiβi,ri≤{C(N,ε)1(ri−1−ri)p}1ki−1A1ki−1βi−1,ri−1≤⋯⋯≤{C(N,ε)2p(R−r)p}1κi−1+1κi−2+⋯+1κ0Aβ0,r0. | (3.4) |
Let i⟶∞, and it is obvious that
supBr(x)u−β≤C(N,ε)1(R−r)p∫BR(x)u−βdx. | (3.5) |
Thus,
infBr(x)u≥C(N,ε){1(R−r)p∫BR(x)u−βdx}−1β. |
This completes the proof.
Lemma 3.3. Suppose u ∈Hp0(Ω) is a sub-solution of problem (1.1). Then, for any ball Br(x)⊂BR(x)⊂⊂Ω,0<r<R<1,p−1<β≤p, there exists a constant C=C(N,ε) such that,
supBr(x)u≤{C{1(R−r)p∫BR(x)uβdx}1β,if ∫BR(x)uβdx≥1,C(1R−r)pβ,if∫BR(x)uβdx<1. |
Proof. Choose a test function ϕ=ηput, where η∈ C∞0(BR(x)), η≡1 in Br(x), 0≤η≤1, |∇η|≤1R−r and t=β−(p−1). Multiplying by ϕ on both sides of the first equation in problem (1.1) and integrating by parts, noting that u is a sub-solution, one obtains that
p∫BR(x)ηp−1ut|∇u|p−2∇u⋅∇ηdx+t∫BR(x)ηput−1|∇u|pdx≤∫BR(x)uα1+t|∇u|α2ηpdx. | (3.6) |
By H¨older inequality and Young's inequality with ε, we infer that
t∫BR(x)ηput−1|∇u|pdx≤p∫BR(x)ηp−1ut|∇η||∇u|p−1dx+∫BR(x)uα1+t|∇u|α2ηpdx≤εppp−1∫BR(x)ηput−1|∇u|pdx+C(ε)∫BR(x)uβ|∇η|pdx+∫BR(x)uα1+t|∇u|α2ηpdx. | (3.7) |
Choose ε small enough, and we obtain that
∫BR(x)ηput−1|∇u|pdx≤C(ε)∫BR(x)uβ|∇η|pdx+∫BR(x)uα1+t|∇u|α2ηpdx. | (3.8) |
Again, by H¨older inequality and Young's inequality with ε, we have
∫BR(x)uα1+t|∇u|α2ηpdx=∫BR(x)ηp|∇u|α2u(t−1)α2puα1+t−(t−1)α2pdx≤(∫BR(x)ηp2α2ut−1|∇u|pdx)α2p(∫BR(x)up(α1+t)−(t−1)α2p−α2dx)1−α2p≤ε∫BR(x)ηp2α2ut−1|∇u|pdx+C(ε)∫BR(x)up(α1+t)−(t−1)α2p−α2dx≤ε∫BR(x)ηp2α2ut−1|∇u|pdx+C(ε)(∫BR(x)uβdx)β−δβ, | (3.9) |
and the last inequality holds owing to 0<p(α1+t)−(t−1)α2p−α2<β. Thus, we define p(α1+t)−(t−1)α2p−α2≜β−δ, where δ>0. Obviously, due to (3.7) and (3.9), we claim the following result
∫Br(x)ut−1|∇u|pdx≤{C(ε)1(R−r)p∫BR(x)uβdx,if∫BR(x)uβdx≥1,C(ε)(∫BR(x)uβdx)β−δβ,if∫BR(x)uβdx<1. | (3.10) |
We assume κ=NN−p as in Lemma 3.1. Therefore, Sobolev inequality and (3.10) imply that
(∫Br(x)uκβdx)1κβ={∫Br(x)(uβp)NpN−pdx}N−pNβ≤{C(N)∫Br(x)|∇(uβp)|pdx}1β={C(N)∫Br(x)ut−1|∇u|pdx}1β≤{C(N,ε){1(R−r)p∫BR(x)uβdx}1β,if∫BR(x)uβdx≥1,C(N,ε)(∫BR(x)uβdx)β−δβ2,if∫BR(x)uβdx<1, | (3.11) |
where C(N) is a constant depending on N, and C(N,ε) is a constant depending on N and ε. Similarly, consider ri,βi as ri=r+2−i(R−r),βi=κiβ. We deduce the following claims from (3.11)
‖u‖βi,ri≤{C1βi(N,ε)(1ri−ri−1)pβi−1‖u‖βi−1,ri−1,if ‖u‖βi−1,ri−1≥1,C1βi(N,ε)(1ri−ri−1)pβi−1‖u‖βi−2,ri−2,if ‖u‖βi−1,ri−1<1, | (3.12) |
where ‖u‖βi,ri=(∫Bri(x)uβidx)1βi. Actually, we assume that ‖u‖βi−2,ri−2≥1. Nevertheless, the conclusion still holds because we can omit the intermediate terms if they are less than 1. Therefore, we have the following conclusions by iterating (3.12)
‖u‖βi,ri≤Ci−1∑m=01βm(N,ε)2pi−1∑m=0mκm(1R−r)pi−1∑m=01κm‖u‖β0,r0 |
Let i⟶+∞, and we have
supBr(x)u≤{C(N,ε){1(R−r)p∫BR(x)uβdx}1β,if ∫BR(x)uβdx≥1,C(N,ε)(1R−r)pβ,if∫BR(x)uβdx<1. | (3.13) |
The proof is complete.
Lemma 3.3. Suppose u∈Hp0(Ω) is a super-solution of problem (1.1). Then for any ball Br(x)⊂BR(x)⊂⊂Ω,0<r<R<1,x∈Ω,0<β≤p−1, there exist p−1≤ˆβ<p∗ and constant C(N,ε) depending on ε and N such that
(∫Br(x)uˆβdx)1/ˆβ≤C(N,ε){1(R−r)p∫BR(x)uβdx}1/β. |
Proof. Again, pick the same test function ϕ=ηput, where η is a cutoff function in BR(x) as in Lemma 3.2, and let t=β−(p−1). Multiplying the first equation of problem (1.1) by ϕ and integrating by parts, noticing that u is a super-solution of problem (1.1), the following result can be obtained:
−t∫BR(x)ηput−1|∇u|pdx≤p∫BR(x)ηp−1ut|∇η||∇u|p−1dx≤ε∫BR(x)ηput−1|∇u|pdx+C(ε)p∫BR(x)uβ|∇η|pdx. | (3.14) |
Thus,
(∫Br(x)uκβ)1κβ≤C(N){∫Br(x)|∇uβp|p}1β=C(N){∫BR(x)ηput−1|∇u|p}1β≤C(N,ε){1(R−r)p∫BR(x)uβ}1β. | (3.15) |
The proof can be completed by iterating finite steps over (3.15). For κ>1, the desired p−1≤ˆβ<p∗ exists such that
(∫Br(x)uˆβdx)1/ˆβ≤C(N,ε){1(R−r)p∫BR(x)uβdx}1/β. |
Lemma 3.4. Suppose u∈Hp0(Ω) is a super-solution of problem (1.1). Then for any ball BR(x)⊂⊂Ω,x∈Ω,0<R<1, there exists ε0>0 small enough such that
(∫BR(x)uε0)1/ε0≤C(N,ε,p)(∫BR(x)u−ε0)−1/ε0, |
where C(N,ε,p) is a constant depending on N, p and ε.
Proof. 1) Suppose w=logu satisfies
∫BR(x)|w(x)−wBr|dx≤CRN for any BR(x)∈Ω, |
where wBr=−∫Br(x)wdx=1|Br(x)|∫Br(x)wdx. According to Lemma 2.1, the following result holds:
(−∫BR(x)ewε0dx)⋅(−∫BR(x)e−wε0dx)≤C(N), |
which can be rewritten as
(∫BR(x)uε0dx)⋅(∫BR(x)u−ε0dx)≤C(N). |
Consequently, the desired inequality follows immediately.
2) Now, we prove that w does belong to BMO(Ω). We choose the similar test function ψ=ηpu1−p and multiply both sides of the first equation in problem (1.1) by ψ, and we integrate by parts, with η∈ C∞0(Ω), η≡1 in BR(x), 0≤η≤1. Observing that u is a super-solution of problem (1.1), we have that
(p−1)∫BR(x)ηpu−p|∇u|pdx≤p∫BR(x)ηp−1u1−p|∇η||∇u|p−1≤p(∫BR(x)u−p|∇u|pηp)1−1p(∫BR(x)|∇η|p)1p≤εp∫BR(x)ηpu−p|∇u|p+C(ε)p∫BR(x)|∇η|p. | (3.16) |
This means that
∫BR(x)|∇logu|p≤C(ε,p)∫BR(x)|∇η|p≤C(ε,p)R−p|BR(x)|≤C(N,ε,p)RN−p. | (3.17) |
Thanks to the Poincarˊe inequality, the following result can be claimed:
∫BR(x)|logu−(logu)x,R|≤C(N)Rp∫BR(x)|∇logu|p≤C(N,ε,p)RN. | (3.18) |
This completes the proof of Lemma 3.4.
Proof of Theorem 1.1. Without loss of generality, we may assume ε0<p. From Lemmas 3.1–3.4, there holds
supBr(x)u≤C(N,ε){1(R−r)p∫BR(x)updx}1p(Lemma 3.2)≤C(N,ε){1(R−r)p∫BR(x)uε0dx}1ε0(Lemma 3.3)≤C(N,ε,p){1(R−r)p∫BR(x)u−ε0dx}−1/ε0(Lemma 3.4)≤C(N,ε,p)infBR(x)u(Lemma 3.1). | (3.19) |
As a consequence of the Harnack inequality, we can obtain the following existence and regularity results.
Corollary 3.1. If u is a solution of problem (1.1) in the space Hp0(Ω), then u locally belongs to the H¨older space C0,α(Ω) for some α∈(0,1).
Proof. For Br(x)⊂⊂Ω, define the oscillation of u as w=M(r)−m(r), where M(r)=supBr(x)u,m(r)=infBr(x)u. According to Theorem 1.1, we can make the following claim:
M(r)−m(r2)=supBr2(x){M(r)−u}≤C(N,ε,p)infBr2(x){M(r)−u}=C(N,ε,p){M(r)−M(r2)}. | (3.20) |
Similarly, we have the following result:
M(r2)−m(r)≤C(N,ε,p){m(r2)−m(r)}. | (3.21) |
Obviously, by adding up (3.20) and (3.21), the following claims can be guaranteed:
w(r2)≤γw(r),for γ=C−1C+1∈(0,1). | (3.22) |
Using Lemma 2.1 and (3.22), we can obtain the following result:
w(r2)≤C(N,ε,p)rαw(R),α∈(0,1). |
The proof is completed.
Corollary 3.2. There exists a solution u of problem (1.1) in the space Hp0(Ω) with |u|C0(Ω)≤C, for some constant C>0.
Proof. 1) Owing to Corollary 3.1, we can obtain that if u is a solution of problem (1.1) in Hp0(Ω), then there exists a constant C(N,ε,Ω,p), such that ‖u‖Hp0(Ω)≤C(N,ε,Ω,p).
2) Define an operator T:Hp0(Ω)⟶Hp0(Ω),T(t,u)=u−Φ(t,u), with Φ(t,u)=−Δ−1p∘A(t,u), with A(t,u)=tuα1|∇u|α2+(1−t). Consider the problem as follows:
{−Δpu=tuα1|∇u|α2+(1−t),u>0,u∈Hp0(Ω). | (3.23) |
Obviously, we have the following result for the ball B={u∈Hp0(Ω):‖u‖Hp0(Ω)≤C(N,ε,Ω,p)+1} similar to the estimate above, this means that Φ(t,u) is compact on B, for A(t,u) is bounded on B. In addition, T(t,u)≠0 on ∂B; thus, Lemma 2.3 holds that
deg(T(0,u),B,0)=deg(T(1,u),B,0). |
This completes the proof of Corollary 3.2.
J. Xie is supported by the National Natural Science Foundation of China (No. 11761030) and Cultivation Project for High-Level Scientific Research Achievements of Hubei Minzu University (No. PY20002).
The authors declare there is no conflicts of interest.
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