# Differential Harnack Estimates for a Semilinear Parabolic System

Authors: Hui Wu
Journal: Journal of Function Spaces
 (2019)

---

## Abstract

In this paper, we prove differential Harnack inequalities for positive solutions of a semilinear parabolic system on hyperbolic space. We use the inequalities to construct classical Harnack estimates by integrating along space-time.


---

## Body

## 1. Introduction

In this paper, we study the following problem:(1)ft=Δf+eμtgp,Rn×0,+∞,gt=Δg+eνtfq,Rn×0,+∞,fx,0=f0,gx,0=g0,Rn,where p,q,μ,ν are positive constants.P. Li and S.-T.Yau in [1] were first pioneers to the study of differential Harnack inequalities which were brought to general parabolic geometric flows by R. Hamilton (see [2]). Using these inequalities can derive ancient solutions, bounds on gradient Ricci solitons, Holder continuity. Differential Harnack inequalities are important aspects of properties of partial differential equations. Paper [3] described differential Harnack inequalities to the initial value problem of a semilinear parabolic equation when the semilinear term is eμtfp,μ>0. There have been numerous interesting results on the properties of solutions of partial differential equations, such as existence of solutions [4–17], nonexistence and blow-up of solutions [18–22], and asymptotic behaviors of solutions [23–28].Let(f(x,t),g(x,t)) be positive smooth solutions to (1) and u,v≔(log⁡f,log⁡g). The main object of our study is the following Harnack quantities:(2)H1≡αΔu+β1∇u2+ceμt+pv-u+ψ1t+ϕ1x,H2≡αΔv+β2∇v2+ceνt+qu-v+ψ2t+ϕ2x,where α, βi, c∈R, α>max⁡β1,β2 and ψi, ϕi will be chosen suitably i=1,2.We will derive our differential Harnack estimate.Theorem 1.
Let(f(x,t),g(x,t)) be positive classical solutions to (1), and u,v≔(log⁡f,log⁡g). If α, βi, ki, and c satisfy(3)α>max⁡β1,β2≥0,k1-pk2≥0,k2-qk1≥0,ki≥nα22α-βi>0,-β2+αp+c1-p≥0,-β1+αq+c1-q≥0,4nα2β1c+1-p2α-β1-pβ2+αp2+cp1-p≥0,4nα2β2c+1-q2α-β2-qβ1+αq2+cq1-q≥0, then we have(4)H1≡αΔu+β1∇u2+ceμt+pv-u+k1t≥0,H2≡αΔv+β2∇v2+ceμt+qu-v+k2t≥0,for all t.The paper is organized as follows. In Section2 we prove Theorem 1 which describes differential Harnack estimate. There are applications of Theorem 1 in Section 3.


## 2. Harnack Estimate

In this section, we shall first obtain our differential Harnack inequalities, relying on the parabolic maximum principle.Lemma 2.
Suppose(f(x,t),g(x,t)) are positive solutions to (1) and u,v≔(log⁡f,log⁡g) and (H1,H2) are defined as in (2). Assume that α, β1, β2, and c satisfy(5)α>max⁡β1,β2≥0,-β2+αp+c1-p≥0,-β1+αq+c1-q≥0,4nα2β1c+1-p2α-β1-pβ2+αp2+cp1-p≥0,4nα2β2c+1-q2α-β2-qβ1+αq2+cq1-q≥0. Then we have (6)∂tH1≥ΔH1+2∇H1·∇u+h11H1+eμt+pv-upH2-H1+h12,∂tH2≥ΔH2+2∇H2·∇v+h21H2+eνt+qu-vqH1-H2+h22,where(7)h11=2α-β11nα2H1-2β1∇u2+ceμt+pv-u+ψ1+ϕ1,h12=∂tψ1-Δϕ1+2nα2α-β1ψ1+ϕ12-nα2∇ϕ124β1α-β1ψ1+ϕ1+eμt+pv-ucμ+ψ1+ϕ1-pψ2+ϕ2,h21=2α-β21nα2H2-2β2∇v2+ceνt+qu-v+ψ2+ϕ2,h22=∂tψ2-Δϕ2+2nα2α-β2ψ2+ϕ22-nα2∇ϕ224β2α-β2ψ2+ϕ2+eνt+qu-vcν+ψ2+ϕ2-qψ1+ϕ1.Proof.
Substituting(f,g)=(eu,ev) into (1), we have (8)ut=Δu+∇u2+eμt+pv-u,vt=Δv+∇v2+eνt+qu-v. Using the above equations, we have (9)∂t-ΔΔu=2∇∇u2+2∇u·∇Δu+eμt+pv-upΔv-Δu+p∇v-∇u2,∂t-Δ∇u2=2∇u·∇∇u2-2∇∇u2+2eμt+pv-u∇u·p∇v-∇u.Furthermore, applying (2) and Cauchy-Schwarz inequality ∇∇u2≥1/n(Δu)2 yields (10)∂tH1-ΔH1-2∇H1·∇u=2α-β1∇∇u2+eμt+pv-upH2-H1+eμt+pv-uαp2-pβ2+cp1-p∇v2-2pα-β1∇u·∇v+eμt+pv-uα-β1∇u2+eμt+pv-ucμ+ψ1+ϕ1-pψ2+ϕ2+∂tψ1-Δϕ1-2∇u·∇ϕ1≥2α-β11nα2H1H1-2β1∇u2+ceμt+pv-u+ψ1+ϕ1+2α-β11nα2β1β1∇u4+2∇u2ceμt+pv-u+ψ1+ϕ1+2α-β11nα2ceμt+pv-u+ψ1+ϕ12+eμt+pv-upH2-H1+eμt+pv-upαp-β2+c1-p∇v2-2pα-β1∇u·∇v+eμt+pv-uα-β1∇u2+eμt+pv-ucμ+ψ1+ϕ1-pψ2+ϕ2+∂tψ1-Δϕ1-2∇u·∇ϕ1 If αp-β2+c(1-p)≥0, the above inequality is (11)≥2α-β11nα2H1H1-2β1∇u2+ceμt+pv-u+ψ1+ϕ1+eμt+pv-upH2-H1+4β1nα2α-β1ψ1+ϕ1∇u2+2α-β11nα2β12∇u4+ceμt+pv-u+ψ1+ϕ12+4nα2β1c+1-pα-β1αp-β2+c1-pα-β1eμt+pv-u∇u2+eμt+pv-ucμ+ψ1+ϕ1-pψ2+ϕ2+∂tψ1-Δϕ1-2∇u·∇ϕ1First note the following inequality:(12)4β1nα2α-β1ψ1+ϕ1∇u2-2∇u·∇ϕ1≥-nα2∇ϕ124β1α-β1ψ1+ϕ1.If α>β1 and 4/nα2β1c+1-p2(α-β1)/-pβ2+αp2+cp(1-p)≥0, the above inequality is(13)≥2α-β11nα2H1H1-2β1∇u2+ceμt+pv-u+ψ1+ϕ1+eμt+pv-upH2-H1-nα2∇ϕ124β1α-β1ψ1+ϕ1+eμt+pv-ucμ+ψ1+ϕ1-pψ2+ϕ2+∂tψ1-Δϕ1+2nα2α-β1ψ1+ϕ12.For H2, we have similar results. (14)∂tH2-ΔH2-2∇H2·∇v≥2α-β21nα2H2H2-2β2∇v2+ceνt+qu-v+ψ2+ϕ2+2α-β21nα2β2β2∇v4+2∇v2ceνt+qu-v+ψ2+ϕ2+2α-β21nα2ceνt+qu-v+ψ2+ϕ22+eνt+qu-vqH1-H2+eνt+qu-vq-β1+αq+c1-q∇u2-2qα-β2∇u·∇v+eνt+qu-vα-β2∇v2+eνt+qu-vcν+ψ2+ϕ2-qψ1+ϕ1+∂tψ2-Δϕ2-2∇v·∇ϕ2 If αq-β1+c(1-q)≥0, the above inequality is (15)≥2α-β21nα2H2H2-2β2∇v2+ceνt+qu-v+ψ2+ϕ2+eνt+qu-vqH1-H2+4β2nα2α-β2ψ2+ϕ2∇v2+2α-β21nα2β22∇v4+ceνt+qu-v+ψ2+ϕ22+4nα2β2c+1-q2α-β2-qβ1+αq2+cq1-qα-β2eνt+qu-v∇v2+eνt+qu-vcν+ψ2+ϕ2-qψ1+ϕ1+∂tψ2-Δϕ2-2∇v·∇ϕ2. If 4/nα2β2c+1-q2(α-β2)/-qβ1+αq2+cq(1-q)≥0, the above inequality is (16)≥2α-β21nα2H2H2-2β2∇v2+ceνt+qu-v+ψ2+ϕ2+eνt+qu-vqH1-H2-nα2∇ϕ224β2α-β2ψ2+ϕ2+eνt+qu-vcν+ψ2+ϕ2-qψ1+ϕ1+∂tψ2-Δϕ2+2nα2α-β2ψ2+ϕ22.This completes the proof of Lemma 2.Next, we need to compute specificψi(t) and ϕi(x) that guarantee hi2>0 for the maximum principle to be applicable where i=1,2.Lemma 3.
Assumek1-pk2≥0, k2-qk1≥0, l1-pl2≥0, l2-ql1≥0 and (17)ki≥nα22α-βi>0,li≥nα22α-βi6+nα2βiα-βi,i=1,2.If ψi(t)=ki/t, ϕi(x)=∑m=1n(li/(xm-am)2+li/(bm-xm)2), i=1,2 for x∈Ω=[a1,b1]×[a2,b2]×···[an,bn], then for some x0∈Ω, h12x=x0>0 and h22x=x0>0.Proof.
Fromk1-pk2≥0, k2-qk1≥0 and l1-pl2≥0, l2-ql1≥0, we get cμ+ψ1+ϕ1-p(ψ2+ϕ2)≥0 and cν+ψ2+ϕ2-q(ψ1+ϕ1)≥0.
Now applying Lemma2 yields (18)h12≥∂tψ1-Δϕ1+2nα2α-β1ψ1+ϕ12-nα2∇ϕ124β1α-β1ψ1+ϕ1,h22≥∂tψ2-Δϕ2+2nα2α-β2ψ2+ϕ22-nα2∇ϕ224β2α-β2ψ2+ϕ2.
By the definitions ofψi and ϕi, we obtain (19)Δϕi=∑m=1n6lixm-am4+6libm-xm4,∇ϕi2=∑m=1n-2lixm-am3+2libm-xm32, and (20)∇ϕi2ψi+ϕi≤∑m=1n2lixm-am2+2libm-xm22.If (21)ki≥nα22α-βi>0,li≥nα22α-βi6+nα2βiα-βi,i=1,2,then h12>0 and h22>0.
This proves Lemma3.Proof of Theorem1.
Choose(22)ψit=kit,ϕix=∑m=1nlixm-am2+libm-xm2,i=1,2 for x∈Ω=[a1,b1]×[a2,b2]×···[an,bn]. Note that(23)limt→0Hi=∞=limxm→amHi=limxm→bmHi,m=1,2,···,n,i=1,2.Assume that there exists a first time t0 and point x0∈Ω where H1(x0,t0)=0 and H2(x0,t0)>0. At (x0,t0); we have(24)∇H1=0,ΔH1≥0,∂tH1≤0.Lemma 3 implies that(25)∂tH1-ΔH1-2∇H1·∇u-h11H1-eμt+pv-upH2-H1≥h12>0This is a contradiction. Assume that there exist a first time t0 and point x0∈Ω where H2(x0,t0)=0 and H1(x0,t0)>0. At (x0,t0), we have(26)∇H2=0,ΔH2≥0,∂tH2≤0.Lemma 3 implies that(27)∂tH2-ΔH2-2∇H2·∇v-h21H2-eνt+qu-vqH1-H2≥h22,This is a contradiction. Furthermore, H1(x0,t0)=0 and H2(x0,t0)=0 cause the same contradiction as H1(x0,t0)=0 and H2(x0,t0)>0. Thus H1(x,t)>0 and H2(x,t)>0 for all x,t>0.
TakingΩ→Rn which obtains ϕi→0 then gives the desired result.


## 3. Applications

In this section, we shall give an application of Theorem1. We integrate along space-time to derive a classical Harnack inequality.


### 3.1. Classical Harnack Inequality

In this subsection, we integrate our differential Harnack inequality of Theorem1 along space-time to derive a classical Harnack inequality.Proposition 4.
Let(f(x,t),g(x,t)) be positive classical solutions to (1) and ux,t,vx,t≔(log⁡f,log⁡g). Suppose that x1, x2∈Rn and t2>t1>0. Assume further that α≥2max⁡β1,β2, α≥c and ki=nα2/2(α-βi), i=1,2. Then we have(28)fx1,t1≤fx2,t2t2t1nexp⁡x2-x122t2-t1,gx1,t1≤gx2,t2t2t1nexp⁡x2-x122t2-t1.Proof.
Define the one-variable functionswi:[t1,t2]→R as(29)w1t≔uγt,t,w2t≔vγt,tfor any C1 path γ:[t1,t2]→Rn such that γ(t1)=x1, γ(t2)=x2.
Applying Theorem1, we have(30)Δu≥-α-1β1∇u2+ceμt+pv-u+k1t.It yields that (31)ddtw1t=ut+∇u·dγdt=Δu+∇u2+eμt+pv-u+∇u·dγdt≥-α-1β1∇u2+ceμt+pv-u+k1t+∇u2+eμt+pv-u+∇u·dγdt≥12-β1α∇u2+1-cαeμt+pv-u-k1αt-12dγdt2≥-12dγdt2-nt,where α≥0, α≥2β1, α≥c and k1=nα2/2(α-β1). Similarly,(32)ddtw2t≥-12dγdt2-nt,for α≥0, α≥2β2, α≥c and k2=nα2/2(α-β2).
By(33)∫t1t2dγdt2dt≥x2-x12t2-t1,∫t1t2-ddtwit≤infγt∫t1t212dγdt2+ntdt,i=1,2,applying (u,v)=(log⁡f,log⁡g) gives Proposition 4.


## 3.1. Classical Harnack Inequality

In this subsection, we integrate our differential Harnack inequality of Theorem1 along space-time to derive a classical Harnack inequality.Proposition 4.
Let(f(x,t),g(x,t)) be positive classical solutions to (1) and ux,t,vx,t≔(log⁡f,log⁡g). Suppose that x1, x2∈Rn and t2>t1>0. Assume further that α≥2max⁡β1,β2, α≥c and ki=nα2/2(α-βi), i=1,2. Then we have(28)fx1,t1≤fx2,t2t2t1nexp⁡x2-x122t2-t1,gx1,t1≤gx2,t2t2t1nexp⁡x2-x122t2-t1.Proof.
Define the one-variable functionswi:[t1,t2]→R as(29)w1t≔uγt,t,w2t≔vγt,tfor any C1 path γ:[t1,t2]→Rn such that γ(t1)=x1, γ(t2)=x2.
Applying Theorem1, we have(30)Δu≥-α-1β1∇u2+ceμt+pv-u+k1t.It yields that (31)ddtw1t=ut+∇u·dγdt=Δu+∇u2+eμt+pv-u+∇u·dγdt≥-α-1β1∇u2+ceμt+pv-u+k1t+∇u2+eμt+pv-u+∇u·dγdt≥12-β1α∇u2+1-cαeμt+pv-u-k1αt-12dγdt2≥-12dγdt2-nt,where α≥0, α≥2β1, α≥c and k1=nα2/2(α-β1). Similarly,(32)ddtw2t≥-12dγdt2-nt,for α≥0, α≥2β2, α≥c and k2=nα2/2(α-β2).
By(33)∫t1t2dγdt2dt≥x2-x12t2-t1,∫t1t2-ddtwit≤infγt∫t1t212dγdt2+ntdt,i=1,2,applying (u,v)=(log⁡f,log⁡g) gives Proposition 4.



---

*Source: 1314268-2019-03-10.xml*