Differential Harnack estimates for the semilinear parabolic equation with three exponents on $ \mathbb{R}^{n} $

Fanqi Zeng; Wenli Geng; Ke An Liu; Boya Wang; Fanqi Zeng; Wenli Geng; Ke An Liu; Boya Wang

doi:10.3934/era.2025008

Electronic Research Archive

2025, Volume 33, Issue 1: 142-157. doi: 10.3934/era.2025008

Previous Article Next Article

Research article

Differential Harnack estimates for the semilinear parabolic equation with three exponents on $\mathbb{R}^{n}$

School of Mathematics and Statistics, Xinyang Normal University, Xinyang 464000, China

Received: 20 September 2024 Revised: 26 November 2024 Accepted: 09 January 2025 Published: 16 January 2025

In this paper, we thought about the positive solutions to the semilinear parabolic equation with three exponents, and obtained several differential Harnack estimates of the positive solutions to the equation. As applications of the main theorems, we found blow-up solutions for the equation and classical Harnack inequalities. Our results generalize some recent works in this direction.

Keywords:

Citation: Fanqi Zeng, Wenli Geng, Ke An Liu, Boya Wang. Differential Harnack estimates for the semilinear parabolic equation with three exponents on $\mathbb{R}^{n}$ [J]. Electronic Research Archive, 2025, 33(1): 142-157. doi: 10.3934/era.2025008

Related Papers:

[1]	Yaning Li, Yuting Yang . The critical exponents for a semilinear fractional pseudo-parabolic equation with nonlinear memory in a bounded domain. Electronic Research Archive, 2023, 31(5): 2555-2567. doi: 10.3934/era.2023129
[2]	Guifen Liu, Wenqiang Zhao . Regularity of Wong-Zakai approximation for non-autonomous stochastic quasi-linear parabolic equation on $ {\mathbb{R}}^N $. Electronic Research Archive, 2021, 29(6): 3655-3686. doi: 10.3934/era.2021056
[3]	Yong Zhou, Jia Wei He, Ahmed Alsaedi, Bashir Ahmad . The well-posedness for semilinear time fractional wave equations on $ \mathbb R^N $. Electronic Research Archive, 2022, 30(8): 2981-3003. doi: 10.3934/era.2022151
[4]	Junsheng Gong, Jiancheng Liu . A Liouville-type theorem of a weighted semilinear parabolic equation on weighted manifolds with boundary. Electronic Research Archive, 2025, 33(4): 2312-2324. doi: 10.3934/era.2025102
[5]	Yitian Wang, Xiaoping Liu, Yuxuan Chen . Semilinear pseudo-parabolic equations on manifolds with conical singularities. Electronic Research Archive, 2021, 29(6): 3687-3720. doi: 10.3934/era.2021057
[6]	Changling Xu, Tianliang Hou . Superclose analysis of a two-grid finite element scheme for semilinear parabolic integro-differential equations. Electronic Research Archive, 2020, 28(2): 897-910. doi: 10.3934/era.2020047
[7]	Weiwei Qi, Yongkun Li . Weyl almost anti-periodic solution to a neutral functional semilinear differential equation. Electronic Research Archive, 2023, 31(3): 1662-1672. doi: 10.3934/era.2023086
[8]	Mingyou Zhang, Qingsong Zhao, Yu Liu, Wenke Li . Finite time blow-up and global existence of solutions for semilinear parabolic equations with nonlinear dynamical boundary condition. Electronic Research Archive, 2020, 28(1): 369-381. doi: 10.3934/era.2020021
[9]	Anh Tuan Nguyen, Chao Yang . On a time-space fractional diffusion equation with a semilinear source of exponential type. Electronic Research Archive, 2022, 30(4): 1354-1373. doi: 10.3934/era.2022071
[10]	Changmu Chu, Shan Li, Hongmin Suo . Existence of a positive radial solution for semilinear elliptic problem involving variable exponent. Electronic Research Archive, 2023, 31(5): 2472-2482. doi: 10.3934/era.2023125

Abstract

1. Introduction

The differential Harnack estimate is a fundamental and powerful technique in the study of partial differential equations on $\mathbb{R}^{n}$ (see ^[1,2]). Gaussian bounds for the heat kernel follow immediately from the differential Harnack estimate. The Hölder continuity is also a direct consequence of the differential Harnack estimate. Numerous other conclusions about the fundamental geometry of space can also be deduced by differential Harnack estimates. Many mathematicians have paid attention to the study on this topic (see, for example, ^[3,4,5] and the references therein).

In this paper, we consider differential Harnack estimates for the following Cauchy problem:

$\begin{equation} \begin{cases} \frac{\partial}{\partial t}f(x,t) = \Delta f+h_{1}(x,t)f^{p}+h_{2}(x,t)f^{q}+h_{3}(x,t)f^{s}\quad in \quad \mathbb{R}^{n}\times[0,\infty),\\ f(x,0) = f_{0}(x)\quad in \quad \mathbb{R}^{n}, \end{cases} \end{equation}$

(1.1)

where the functions $h_{1}$ , $h_{2}$ , and $h_{3}$ are $C^{2}$ in $x$ and $C^{0}$ in $t$ with $h_{1} > 0$ , $h_{2} > 0$ , and $h_{3} > 0$ , and $p$ , $q$ , and $s$ are positive constants with $p \; \geq \; q \; \geq \; s \; > \; 1$ . Equation (1.1) arises from many classical equations (see ^[6,7,8]) and there are many questions related to Eq (1.1) (see ^[9,10]).

Now, let us recall some relevant work with the above Eq (1.1). In the case where $h_{1}(x, t) = 1$ and $h_{2}(x, t) = h_{3}(x, t) = 0$ , Eq (1.1) reduces to the endangered species equation. Cao et al. ^[8] proved a differential Harnack estimate for positive solutions of the Cauchy problem for the endangered species equation. In the case where $h_{1}(x, t) = c$ , $h_{2}(x, t) = -c$ , $h_{3}(x, t) = 0$ , $p = 1$ , and $q = 2$ , Eq (1.1) reduces to the Fisher-Kolmogorov-Petrovsky-Piskunov (Fisher-KPP) equation. Cao et al. ^[9] proved a differential Harnack estimate for positive solutions of the Fisher-KPP equation on an $n$ -dimensional Riemannian manifold $M$ with non-negative Ricci curvature, where $c$ is a positive constant. If $h_{1}(x, t) = -1$ , $h_{2}(x, t) = 1$ , $h_{3}(x, t) = 0$ , $p = 3$ , and $q = 1$ , then Eq (1.1) reduces to the parabolic Allen-Cahn equation. Bǎileşteanu ^[6] proved a differential Harnack estimate for the solution of the parabolic Allen-Cahn equation on a closed $n$ -dimensional manifold. When $h_{1}(x, t) = a$ , $h_{2}(x, t) = -b$ , $h_{3}(x, t) = 0$ , $p = 1$ , and $q = 3$ , where $a$ and $b$ are two constants, Eq (1.1) reduces to the Newell-Whitehead-Segel equation. The differential Harnack estimate for the Newell-Whitehead-Segel equation was obtained by the authors in ^[7]. Hou ^[10] proved a differential Harnack estimate for positive solutions of equation (1.1) when $h_{3}(x, t) = 0$ . For more results on differential Harnack estimates of Eq (1.1), see ^{[11,12,13,14,15]}.

The motivation of this article is to develop some differential Harnack estimates for positive solutions to Eq (1.1) on $\mathbb{R}^{n}$ . The method we employ is the parabolic maxinum principle. We are now ready to state our main results.

Theorem 1.1. Assume that $f(x, t)$ is a positive solution of Eq (1.1) and $u = \ln f$ . If $\alpha$ , $\beta$ , $c$ , $d$ , $k$ , $a$ , and $h_{i} \; (i = 1, 2, 3)$ satisfy

$\begin{equation} \alpha\geq2\beta\geq0,\quad\alpha > 0, \end{equation}$

(1.2)

$\begin{equation} \begin{cases} \frac{\alpha(p-1)+2\beta}{p}\geq c\geq \max\left\{\frac{(p-1)n\alpha^{2}}{4(\alpha-\beta)},\frac{\alpha(p-1)+\beta}{p}\right\},\\ \frac{\alpha(q-1)+2\beta}{q}\geq d\geq \max\left\{\frac{(q-1)n\alpha^{2}}{4(\alpha-\beta)},\frac{\alpha(q-1)+\beta}{q}\right\},\\ \frac{\alpha(s-1)+2\beta}{s}\geq k\geq \max\left\{\frac{(s-1)n\alpha^{2}}{4(\alpha-\beta)},\frac{\alpha(s-1)+\beta}{s}\right\}, \end{cases} \end{equation}$

(1.3)

$\begin{equation} c\geq d\geq k\geq\beta,\quad a\geq\frac{n\alpha^{2}}{2(\alpha-\beta)} > 0, \end{equation}$

(1.4)

and

$\begin{equation} \left(\frac{\partial}{\partial t}-\Delta\right)h_{i}\geq0,\quad \Delta h_{i}\geq0,\quad i = 1,2,3, \end{equation}$

(1.5)

then we have

$\begin{equation} H_{0}\equiv\alpha\Delta u+\beta |\nabla u|^{2}+ch_{1}e^{u(p-1)}+dh_{2}e^{u(q-1)}+kh_{3}e^{u(s-1)}+\frac{a}{t}\geq0 \end{equation}$

(1.6)

for all t.

Remark 1.1. (1) Compared with the previous work established in ^[7,8,11], here we do not assume the coefficients of equations are constant, and therefore our results can be regarded as an extension of several classical estimates.

(2) When $h_{3}(x, t) = 0$ , the estimate (1.6) above can be reduced to the formulas (1.6) in Theorem 1.1 of ^[10]. Hence the above Theorem 1.1 generalizes the result in ^[10].

As applications of this estimate (1.6), we derive the blow-up of the solutions for Eq (1.1) and a classical Harnack inequality by integrating along space-time paths.

Corollary 1.2. Let $f$ be a positive solution of equation (1.1) with $h_{i} \; (i = 1, 2, 3)$ satisfying (1.5), and $c$ is a constant satisfying $0 \; < \; n(p-1) \; \leq \; c \; < \; 2$ and $c \; \geq \; d \; \geq \; k \; \geq \; 1$ . Then $f$ blows up in finite time provided that

$\begin{equation} f(x_{0},t_{0})\geq\left(\frac{4n}{(2-c)h_{1}(x_{0},t_{0})}\right)^{\frac{1}{p-1}} \end{equation}$

(1.7)

at some point $(x_{0}, t_{0})$ .

Corollary 1.3. Let $f$ be a positive solution of Eq (1.1) with $h_{i} \; (i = 1, 2, 3)$ satisfying (1.5) and $u = \ln f$ . Let $\gamma(t) = (x(t), t)$ , $t\in[t_{1}, t_{2}]$ , be a space-time curve joining two given points $(x_{1}, t_{1})$ , $(x_{2}, t_{2}) \; \in \; \mathbb{R}^{n} \; \times \; [0, \infty)$ with $0 < t_{1} < t_{2}$ . Assume further that $a = \frac{n\alpha^{2}}{2(\alpha-\beta)}\leq n\alpha$ . Then we get

$\begin{equation} f(x_{1},t_{1})\leq f(x_{2},t_{2})\left(\frac{t_{2}}{t_{1}}\right)^{n}\exp\left[\frac{|x_{2}-x_{1}|^{2}}{2(t_{2}-t_{1})}\right]. \end{equation}$

(1.8)

We also get the following differential Harnack estimate, which is different from (1.6).

Theorem 1.4. Assume $f(x, t)$ is a positive solution of Eq (1.1) and $u = \ln f$ . If $\tilde{\alpha}$ , $\tilde{\beta}$ , $\tilde{c}$ , $\tilde{d}$ , $\tilde{k}$ , $\tilde{a}$ , $m$ , and $h_{i} \; (i = 1, 2, 3)$ satisfy

$\begin{equation} \tilde{\alpha}\geq2\tilde{\beta}\geq0,\quad m > 0, \end{equation}$

(1.9)

$\begin{equation} \begin{cases} \frac{\tilde{\alpha}(p-1)+2\tilde{\beta}}{p}\geq \tilde{c}\geq \max\left\{\frac{(p-1)n\tilde{\alpha}^{2}}{4(\tilde{\alpha}-\tilde{\beta})},\frac{\tilde{\alpha}(p-1)+\tilde{\beta}}{p}\right\},\\ \frac{\tilde{\alpha}(q-1)+2\tilde{\beta}}{q}\geq \tilde{d}\geq \max\left\{\frac{(q-1)n\tilde{\alpha}^{2}}{4(\tilde{\alpha}-\tilde{\beta})},\frac{\tilde{\alpha}(q-1)+\tilde{\beta}}{q}\right\},\\ \frac{\tilde{\alpha}(s-1)+2\tilde{\beta}}{s}\geq \tilde{k}\geq \max\left\{\frac{(s-1)n\tilde{\alpha}^{2}}{4(\tilde{\alpha}-\tilde{\beta})},\frac{\tilde{\alpha}(s-1)+\tilde{\beta}}{s}\right\}, \end{cases} \end{equation}$

(1.10)

$\begin{equation} \tilde{c}\geq \tilde{d}\geq \tilde{k}\geq\tilde{\beta},\quad \tilde{a}\geq\frac{nm\tilde{\alpha}^{2}}{2(\tilde{\alpha}-\tilde{\beta})} > 0, \end{equation}$

(1.11)

and

$\begin{equation} \left(\frac{\partial}{\partial t}-\Delta\right)h_{i}\geq0,\quad \Delta h_{i}\geq0,\quad i = 1,2,3, \end{equation}$

(1.12)

then we have

$\begin{equation} \widetilde{H}_{0}\equiv\tilde{\alpha}\Delta u+\tilde{\beta} |\nabla u|^{2}+\tilde{c}h_{1}e^{u(p-1)}+\tilde{d}h_{2}e^{u(q-1)}+\tilde{k}h_{3}e^{u(s-1)}+\frac{\tilde{a}}{1-e^{-mt}}\geq0 \end{equation}$

(1.13)

for all t.

Remark 1.2. (1) When $h_{1}(x, t) = e^{\gamma t}$ with a constant $\gamma$ and $h_{2}(x, t) = h_{3}(x, t) = 0$ , Theorem 1.4 reduces to Theorem 1 in ^[14]. Hence the above Theorem 1.4 generalizes the result in ^[14].

(2) The case of $n = 1$ , $p = 2$ , and $h_{2} = h_{3} = 0$ was studied by Hamilton in ^[16]. Particularly, we apply Theorem 1.4 with $n = 1$ and $p = 2$ , and by picking $\tilde{\alpha} = 1$ , $\tilde{\beta} = 0$ , $h_{1} = 1$ , $h_{2} = h_{3} = 0$ , $\tilde{a} = \frac{m}{2}$ , and $\tilde{c} = \frac{1}{4}$ , we conclude that

$\begin{equation*} u_{xx}+\frac{1}{4}e^{u}+\frac{m}{2(1-e^{-mt})}\geq0, \end{equation*}$

yielding

$\begin{equation*} f_{t}+\frac{m}{2(1-e^{-mt})}f\geq\frac{f^{2}_{x}}{f}+\frac{3}{4}f^{2}. \end{equation*}$

If $m$ is small enough, the estimate in ^[16] will be improved.

Corollary 1.5. Let $f$ be a positive solution of Eq (1.1) with $h_{i} \; (i = 1, 2, 3)$ satisfying (1.12) and $u = \ln f$ . Let $\tau(t) = (x(t), t)$ , $t\in[t_{1}, t_{2}]$ , be a space-time curve joining two given points $(x_{1}, t_{1})$ , $(x_{2}, t_{2}) \; \in \; \mathbb{R}^{n} \; \times \; [0, \infty)$ with $0 < t_{1} < t_{2}$ . Assume further that $\tilde{a} = \frac{nm\tilde{\alpha}^{2}}{2(\tilde{\alpha}-\tilde{\beta})}\leq nm\tilde{\alpha}$ . Then we get

$\begin{equation} f(x_{1},t_{1})\leq f(x_{2},t_{2})\left(\frac{e^{mt_{2}}-1}{e^{mt_{1}}-1}\right)^{n}\exp\left[\frac{|x_{2}-x_{1}|^{2}}{2(t_{2}-t_{1})}\right]. \end{equation}$

(1.14)

The paper is structured as follows. In Section 2, we prove Theorem 1.1, Corollary 1.2 and Corollary 1.3. In Section 3, we prove Theorem 1.4 and Corollary 1.5.

2. The proofs of Theorem 1.1, Corollary 1.2 and Corollary 1.3

Using the parabolic maximum principle, we will first derive our differential Harnack estimate in this section. We always write $u_{t}$ for the partial derivative of $u$ with respect to $t$ and omit the time variable $t$ for simplicity.

Let $f(x, t)\in C^{\infty} \; (\mathbb{R}^{n}\times[0, \infty))$ be a positive solution of (1.1) and $u = \ln f$ . Substituting $f = e^{u}$ into Eq (1.1), we have

$\begin{equation} u_{t} = \Delta u+|\nabla u|^{2}+h_{1}e^{u(p-1)}+h_{2}e^{u(q-1)}+h_{3}e^{u(s-1)}. \end{equation}$

(2.1)

Based on this observation, a Harnack quantity $H$ is defined as

$\begin{equation} H: = \alpha\Delta u+\beta|\nabla u|^{2}+ch_{1}e^{u(p-1)}+dh_{2}e^{u(q-1)}+kh_{3}e^{u(s-1)}+\phi, \end{equation}$

(2.2)

where $\alpha$ , $\beta$ , $c$ , $d$ , $k\in \mathbb{R}$ and $\phi :\mathbb{R}^{n}\times[0, \infty)\rightarrow [0, \infty)$ will be determined later. To support our primary findings, we first assert and prove a technical lemma.

Lemma 2.1. $u = \ln f$ and $H$ are defined as in (2.2). Assume that $f(x, t)$ is a positive solution of Eq (1.1). Then we have

$\begin{equation} \begin{aligned} H_{t} & = \Delta H+2\nabla H\cdot\nabla u+(p-1)h_{1}e^{u(p-1)}H+(q-1)h_{2}e^{u(q-1)}H+(s-1)h_{3}e^{u(s-1)}H\\ &\quad+2(\alpha-\beta)|\nabla\nabla u|^{2}+[\alpha(p-1)+\beta-cp](p-1)h_{1}e^{u(p-1)}|\nabla u|^{2}\\ &\quad+[\alpha(q-1)+\beta-dq](q-1)h_{2}e^{u(q-1)}|\nabla u|^{2}\\ &\quad+[\alpha(s-1)+\beta-ks](s-1)h_{3}e^{u(s-1)}|\nabla u|^{2}\\ &\quad+\big[(\alpha-c)\Delta h_{1}+2(\alpha(p-1)+\beta-cp)\nabla h_{1}\cdot\nabla u+c(h_{1})_{t}-h_{1}(p-1)\phi\big]e^{u(p-1)}\\ &\quad+\big[(\alpha-d)\Delta h_{2}+2(\alpha(q-1)+\beta-dq)\nabla h_{2}\cdot\nabla u+d(h_{2})_{t}-h_{2}(q-1)\phi\big]e^{u(q-1)}\\ &\quad+\big[(\alpha-k)\Delta h_{3}+2(\alpha(s-1)+\beta-ks)\nabla h_{3}\cdot\nabla u+k(h_{3})_{t}-h_{3}(s-1)\phi\big]e^{u(s-1)}\\ &\quad+[(c-d)(p-q)]h_{1}h_{2}e^{u(p-1)}e^{u(q-1)}\\ &\quad+[(c-k)(p-s)]h_{1}h_{3}e^{u(p-1)}e^{u(s-1)}\\ &\quad+[(d-k)(q-s)]h_{2}h_{3}e^{u(q-1)}e^{u(s-1)}-2\nabla\phi\cdot\nabla u-\Delta\phi+\phi_{t}. \end{aligned} \end{equation}$

(2.3)

Proof. Using (2.1), we can compute the following evolution equations:

$H_{t} = \alpha(\Delta u)_{t}+\beta(|\nabla u|^{2})_{t}+c(h_{1}e^{u(p-1)})_{t}+d(h_{2}e^{u(q-1)})_{t}+k(h_{3}e^{u(s-1)})_{t}+\phi_{t},$

$\begin{equation*} \begin{aligned} (\Delta u)_{t} & = \Delta (u_{t})\\ & = \Delta\left(\Delta u+|\nabla u|^{2}+h_{1}e^{u(p-1)}+h_{2}e^{u(q-1)}+h_{3}e^{u(s-1)}\right)\\ & = \Delta(\Delta u)+\Delta|\nabla u|^{2}+\Delta(h_{1}e^{u(p-1)})+\Delta(h_{2}e^{u(q-1)})+\Delta(h_{3}e^{u(s-1)}) \end{aligned} \end{equation*}$

and

$\begin{equation*} \begin{aligned}(|\nabla u|^{2})_{t} & = 2\nabla(u_{t})\cdot\nabla u\\ & = 2\nabla\left(\Delta u+|\nabla u|^{2}+h_{1}e^{u(p-1)}+h_{2}e^{u(q-1)}+h_{3}e^{u(s-1)}\right)\cdot \nabla u\\ & = \Delta|\nabla u|^{2}-2|\nabla\nabla u|^{2}+2\nabla|\nabla u|^{2}\cdot\nabla u\\ &\quad+2\nabla(h_{1}e^{u(p-1)})\cdot\nabla u+2\nabla(h_{2}e^{u(q-1)})\cdot\nabla u+2\nabla(h_{3}e^{u(s-1)})\cdot\nabla u, \end{aligned} \end{equation*}$

where we applied the formula

$\begin{equation} \Delta|\nabla u|^{2} = 2\nabla u\cdot\nabla\Delta u+2|\nabla\nabla u|^{2}. \end{equation}$

(2.4)

Hence we get

$\begin{equation} \begin{aligned} H_{t} & = \alpha[\Delta(\Delta u)+\Delta|\nabla u|^{2}+\Delta(h_{1}e^{u(p-1)})+\Delta(h_{2}e^{u(q-1)})+\Delta(h_{3}e^{u(s-1)})]\\ &\quad+\beta[\Delta|\nabla u|^{2}-2|\nabla\nabla u|^{2}+2\nabla|\nabla u|^{2}\cdot\nabla u\\ &\quad+2\nabla(h_{1}e^{u(p-1)})\cdot\nabla u+2\nabla(h_{2}e^{u(q-1)})\cdot\nabla u+2\nabla(h_{3}e^{u(s-1)})\cdot\nabla u]\\ &\quad+ce^{u(p-1)}(h_{1})_{t}+ch_{1}(p-1)e^{u(p-1)}[\Delta u+|\nabla u|^{2}+h_{1}e^{u(p-1)}+h_{2}e^{u(q-1)}+h_{3}e^{u(s-1)}]\\ &\quad+de^{u(q-1)}(h_{2})_{t}+dh_{2}(q-1)e^{u(q-1)}[\Delta u+|\nabla u|^{2}+h_{1}e^{u(p-1)}+h_{2}e^{u(q-1)}+h_{3}e^{u(s-1)}]\\ &\quad+ke^{u(s-1)}(h_{3})_{t}+kh_{3}(s-1)e^{u(s-1)}[\Delta u+|\nabla u|^{2}+h_{1}e^{u(p-1)}+h_{2}e^{u(q-1)}+h_{3}e^{u(s-1)}]\\ &\quad+\phi_{t}. \end{aligned} \end{equation}$

(2.5)

A direct calculation gives

$\begin{equation} \begin{aligned} \Delta H & = \Delta(\alpha\Delta u+\beta|\nabla u|^{2}+ch_{1}e^{u(p-1)}+dh_{2}e^{u(q-1)}+kh_{3}e^{u(s-1)}+\phi)\\ & = \alpha\Delta(\Delta u)+\beta\Delta|\nabla u|^{2}+c\Delta(h_{1}e^{u(p-1)})+d\Delta(h_{2}e^{u(q-1)})+k\Delta(h_{3}e^{u(s-1)})+\Delta\phi \end{aligned} \end{equation}$

(2.6)

and

$\begin{equation} \begin{aligned} \nabla H & = \nabla(\alpha\Delta u+\beta|\nabla u|^{2}+ch_{1}e^{u(p-1)}+dh_{2}e^{u(q-1)}+kh_{3}e^{u(s-1)}+\phi)\\ & = \alpha\nabla\Delta u+\beta\nabla|\nabla u|^{2}+c\nabla(h_{1}e^{u(p-1)})+d\nabla(h_{2}e^{u(q-1)})+k\nabla(h_{3}e^{u(s-1)})+\nabla\phi. \end{aligned} \end{equation}$

(2.7)

Using (2.4), (2.6), and (2.5), we obtain

$\begin{equation} \begin{aligned} H_{t} & = \Delta H+2(\alpha-\beta)|\nabla\nabla u|^{2}+2\alpha\nabla u\cdot\nabla\Delta u\\ &\quad+(\alpha-c)\Delta(h_{1}e^{u(p-1)})+(\alpha-d)\Delta(h_{2}e^{u(q-1)})+(\alpha-k)\Delta(h_{3}e^{u(s-1)})\\ &\quad+\beta[2\nabla|\nabla u|^{2}\cdot\nabla u+2\nabla(h_{1}e^{u(p-1)})\cdot\nabla u+2\nabla(h_{2}e^{u(q-1)})\cdot\nabla u\\ &\quad+2\nabla(h_{3}e^{u(s-1)})\cdot\nabla u]-\Delta\phi\\ &\quad+ce^{u(p-1)}(h_{1})_{t}+ch_{1}(p-1)e^{u(p-1)}[\Delta u+|\nabla u|^{2}+h_{1}e^{u(p-1)}+h_{2}e^{u(q-1)}+h_{3}e^{u(s-1)}]\\ &\quad+de^{u(q-1)}(h_{2})_{t}+dh_{2}(q-1)e^{u(q-1)}[\Delta u+|\nabla u|^{2}+h_{1}e^{u(p-1)}+h_{2}e^{u(q-1)}+h_{3}e^{u(s-1)}]\\ &\quad+ke^{u(s-1)}(h_{3})_{t}+kh_{3}(s-1)e^{u(s-1)}[\Delta u+|\nabla u|^{2}+h_{1}e^{u(p-1)}+h_{2}e^{u(q-1)}+h_{3}e^{u(s-1)}]\\ &\quad+\phi_{t}. \end{aligned} \end{equation}$

(2.8)

By (2.8) and (2.7), we get

$\begin{equation} \begin{aligned} H_{t} & = \Delta H+2(\alpha-\beta)|\nabla\nabla u|^{2}+2\nabla H\cdot\nabla u\\ &\quad+(\alpha-c)\Delta(h_{1}e^{u(p-1)})+(\alpha-d)\Delta(h_{2}e^{u(q-1)})+(\alpha-k)\Delta(h_{3}e^{u(s-1)})\\ &\quad+2(\beta-c)\nabla(h_{1}e^{u(p-1)})\cdot\nabla u+2(\beta-d)\nabla(h_{2}e^{u(q-1)})\cdot\nabla u\\ &\quad+2(\beta-k)\nabla(h_{3}e^{u(s-1)})\cdot\nabla u-\Delta\phi-2\nabla\phi\cdot\nabla u\\ &\quad+ce^{u(p-1)}(h_{1})_{t}+ch_{1}(p-1)e^{u(p-1)}[\Delta u+|\nabla u|^{2}+h_{1}e^{u(p-1)}+h_{2}e^{u(q-1)}+h_{3}e^{u(s-1)}]\\ &\quad+de^{u(q-1)}(h_{2})_{t}+dh_{2}(q-1)e^{u(q-1)}[\Delta u+|\nabla u|^{2}+h_{1}e^{u(p-1)}+h_{2}e^{u(q-1)}+h_{3}e^{u(s-1)}]\\ &\quad+ke^{u(s-1)}(h_{3})_{t}+kh_{3}(s-1)e^{u(s-1)}[\Delta u+|\nabla u|^{2}+h_{1}e^{u(p-1)}+h_{2}e^{u(q-1)}+h_{3}e^{u(s-1)}]\\ &\quad+\phi_{t}. \end{aligned} \end{equation}$

(2.9)

Direct computations show that

$\begin{equation} \begin{cases} \begin{aligned} \Delta(h_{1}e^{u(p-1)}) & = e^{u(p-1)}\Delta h_{1}+2(p-1)e^{u(p-1)}\nabla h_{1}\cdot\nabla u\\ &\quad+h_{1}(p-1)^{2}e^{u(p-1)}|\nabla u|^{2}+h_{1}(p-1)e^{u(p-1)}\Delta u,\\ \end{aligned}\\ \begin{aligned} \Delta(h_{2}e^{u(q-1)}) & = e^{u(q-1)}\Delta h_{2}+2(q-1)e^{u(q-1)}\nabla h_{2}\cdot\nabla u\\ &\quad+h_{2}(q-1)^{2}e^{u(q-1)}|\nabla u|^{2}+h_{2}(q-1)e^{u(q-1)}\Delta u,\\ \end{aligned}\\ \begin{aligned} \Delta(h_{3}e^{u(s-1)}) & = e^{u(s-1)}\Delta h_{3}+2(s-1)e^{u(s-1)}\nabla h_{3}\cdot\nabla u\\ &\quad+h_{3}(s-1)^{2}e^{u(s-1)}|\nabla u|^{2}+h_{3}(s-1)e^{u(s-1)}\Delta u \end{aligned} \end{cases} \end{equation}$

(2.10)

and

$\begin{equation} \begin{cases} \nabla(h_{1}e^{u(p-1)})\cdot\nabla u = e^{u(p-1)}\nabla h_{1}\cdot\nabla u+(p-1)h_{1}e^{u(p-1)}|\nabla u|^{2},\\ \nabla(h_{2}e^{u(q-1)})\cdot\nabla u = e^{u(q-1)}\nabla h_{2}\cdot\nabla u+(q-1)h_{2}e^{u(q-1)}|\nabla u|^{2},\\ \nabla(h_{3}e^{u(s-1)})\cdot\nabla u = e^{u(s-1)}\nabla h_{3}\cdot\nabla u+(s-1)h_{3}e^{u(s-1)}|\nabla u|^{2}. \end{cases} \end{equation}$

(2.11)

Substituting (2.10) and (2.11) into (2.9), we get (2.3). This completes the proof of Lemma 2.1. □

We can now validate Theorem 1.1.

Proof of Theorem 1.1. Define the $n$ -rectangle $R: = \Pi^{n}_{i = 1}[p_{i}, q_{i}] \; \subset \; \mathbb{R}^{n}$ , and set

$\begin{equation} \phi_{R}(x,t) = \frac{a}{t}+\sum\limits^{n}_{k = 1}\left(\frac{b}{(x_{k}-p_{k})^{2}}+\frac{b}{(q_{k}-x_{k})^{2}}\right) \end{equation}$

(2.12)

for $t > 0$ , $a > 0$ , $b > 0$ , and $x = (x_{1}, ..., x_{n})\in R$ , while $\phi_{R}\rightarrow +\infty$ as $x_{i}\rightarrow p_{i}$ , $q_{i}$ or $t\rightarrow 0$ .

The corresponding Harnack quantity is

$\begin{equation*} H_{R} = \alpha\Delta u+\beta|\nabla u|^{2}+ch_{1}e^{u(p-1)}+dh_{2}e^{u(q-1)}+kh_{3}e^{u(s-1)}+\phi_{R}(x,t). \end{equation*}$

Note that $H_{R}\rightarrow H_{0}$ as $R\rightarrow \mathbb{R}^{n}$ , and $H_{R} > 0$ for small $t$ .

So as to obtain a contradiction, assume that there is a first time $t_{0}$ and point $x_{0}\in R$ such that $H_{R}(x_{0}, t_{0}) = 0$ . Then at $(x_{0}, t_{0})$ , we have

$\begin{equation*} (H_{R})_{t}\leq0,\quad \nabla H_{R} = 0,\quad \Delta H_{R}\geq0 \end{equation*}$

and

$\begin{equation*} \Delta u = -\frac{1}{\alpha}(\beta|\nabla u|^{2}+ch_{1}e^{u(p-1)}+dh_{2}e^{u(q-1)}+kh_{3}e^{u(s-1)}+\phi_{R}). \end{equation*}$

Then using Lemma 2.1 and the Cauchy-Schwarz inequality $|\nabla\nabla u|^{2}\geq\frac{1}{n}(\Delta u)^{2}$ , we can get

$\begin{equation*} \begin{aligned} 0 &\geq \frac{2(\alpha-\beta)}{n\alpha^{2}}(\beta|\nabla u|^{2}+ch_{1}e^{u(p-1)}+dh_{2}e^{u(q-1)}+kh_{3}e^{u(s-1)}+\phi_{R})^{2}\\ &\quad+[\alpha(p-1)+\beta-cp](p-1)h_{1}e^{u(p-1)}|\nabla u|^{2}+[\alpha(q-1)+\beta-dq](q-1)h_{2}e^{u(q-1)}|\nabla u|^{2}\\ &\quad+[\alpha(s-1)+\beta-ks](s-1)h_{3}e^{u(s-1)}|\nabla u|^{2}\\ &\quad+[(\alpha-c)\Delta h_{1}+2(\alpha(p-1)+\beta-cp)\nabla h_{1}\cdot\nabla u+c(h_{1})_{t}-h_{1}(p-1)\phi_{R}]e^{u(p-1)}\\ &\quad+[(\alpha-d)\Delta h_{2}+2(\alpha(q-1)+\beta-dq)\nabla h_{2}\cdot\nabla u+d(h_{2})_{t}-h_{2}(q-1)\phi_{R}]e^{u(q-1)}\\ &\quad+[(\alpha-k)\Delta h_{3}+2(\alpha(s-1)+\beta-ks)\nabla h_{3}\cdot\nabla u+k(h_{3})_{t}-h_{3}(s-1)\phi_{R}]e^{u(s-1)}\\ &\quad+[(c-d)(p-q)]h_{1}h_{2}e^{u(p-1)}e^{u(q-1)}+[(c-k)(p-s)]h_{1}h_{3}e^{u(p-1)}e^{u(s-1)}\\ &\quad+[(d-k)(q-s)]h_{2}h_{3}e^{u(q-1)}e^{u(s-1)}-2\nabla\phi_{R}\cdot\nabla u-\Delta\phi_{R}+(\phi_{R})_{t}. \end{aligned} \end{equation*}$

Setting $X = e^{u(p-1)}$ , $Y = e^{u(q-1)}$ , $Z = e^{u(s-1)}$ , and $W = |\nabla u|^{2}$ , and using

$\begin{equation*} 2\nabla h_{1}\cdot\nabla u\geq-2|\nabla h_{1}|^{2}-\frac{1}{2}|\nabla u|^{2}, \end{equation*}$

$\begin{equation*} 2\nabla h_{2}\cdot\nabla u\geq-\frac{1}{2}|\nabla h_{2}|^{2}-2|\nabla u|^{2}, \end{equation*}$

and

$\begin{equation*} 2\nabla h_{3}\cdot\nabla u\geq-|\nabla h_{3}|^{2}-|\nabla u|^{2}, \end{equation*}$

we arrive at

$\begin{equation} \begin{aligned} 0 &\geq \frac{2(\alpha-\beta)}{n\alpha^{2}}(c^{2}h_{1}^{2}X^{2}+d^{2}h_{2}^{2}Y^{2}+k^{2}h_{3}^{2}Z^{2}+\beta^{2}W^{2})\\ &\quad+\left[(\alpha(p-1)+\beta-cp)\left(1-\frac{1}{h_{1}(p-1)}\right)+\frac{4(\alpha-\beta)\beta c}{n(p-1)\alpha^{2}}\right](p-1)h_{1}XW\\ &\quad+\left[(\alpha(q-1)+\beta-dq)\left(1-\frac{1}{h_{2}(q-1)}\right)+\frac{4(\alpha-\beta)\beta d}{n(q-1)\alpha^{2}}\right](q-1)h_{2}YW\\ &\quad+\left[(\alpha(s-1)+\beta-ks)\left(1-\frac{1}{h_{3}(s-1)}\right)+\frac{4(\alpha-\beta)\beta k}{n(s-1)\alpha^{2}}\right](s-1)h_{3}ZW\\ &\quad+\bigg[(\alpha-c)\Delta h_{1}-4(\alpha(p-1)+\beta-cp)|\nabla h_{1}|^{2}+c(h_{1})_{t}+\left(\frac{4(\alpha-\beta)c}{n\alpha^{2}}-(p-1)\right)h_{1}\phi_{R}\bigg]X\\ &\quad+\bigg[(\alpha-d)\Delta h_{2}-4(\alpha(q-1)+\beta-dq)|\nabla h_{2}|^{2}+d(h_{2})_{t}+\left(\frac{4(\alpha-\beta)d}{n\alpha^{2}}-(q-1)\right)h_{2}\phi_{R}\bigg]Y\\ &\quad+\bigg[(\alpha-k)\Delta h_{3}-4(\alpha(s-1)+\beta-ks)|\nabla h_{3}|^{2}+k(h_{3})_{t}+\left(\frac{4(\alpha-\beta)k}{n\alpha^{2}}-(s-1)\right)h_{3}\phi_{R}\bigg]Z\\ &\quad+\left[(c-d)(p-q)+\frac{4(\alpha-\beta)cd}{n\alpha^{2}}\right]h_{1}h_{2}XY+\left[(c-k)(p-s)+\frac{4(\alpha-\beta)ck}{n\alpha^{2}}\right]h_{1}h_{3}XZ\\ &\quad+\left[(d-k)(q-s)+\frac{4(\alpha-\beta)dk}{n\alpha^{2}}\right]h_{2}h_{3}YZ-2\nabla\phi_{R}\cdot\nabla u-\Delta\phi_{R}\\ &\quad+(\phi_{R})_{t}+\frac{2(\alpha-\beta)}{n\alpha^{2}}\phi^{2}_{R}+\frac{4(\alpha-\beta)\beta\phi_{R}}{n\alpha^{2}}W. \end{aligned} \end{equation}$

(2.13)

By demonstrating that the right-hand side of (2.13) is positive, we can then obtain a contradiction. The assumption of (1.3) in Theorem 1.1 implies

$\begin{equation} \begin{cases} c\geq\max\left\{\frac{(p-1)n\alpha^{2}}{4(\alpha-\beta)},\frac{\alpha(p-1)+\beta}{p}\right\},\\ d\geq\max\left\{\frac{(q-1)n\alpha^{2}}{4(\alpha-\beta)},\frac{\alpha(q-1)+\beta}{q}\right\},\\ k\geq\max\left\{\frac{(s-1)n\alpha^{2}}{4(\alpha-\beta)},\frac{\alpha(s-1)+\beta}{s}\right\}. \end{cases} \end{equation}$

(2.14)

By (2.14), we get

$\begin{equation} c\geq\frac{(p-1)n\alpha^{2}}{4(\alpha-\beta)},\quad d\geq\frac{(q-1)n\alpha^{2}}{4(\alpha-\beta)},\quad k\geq\frac{(s-1)n\alpha^{2}}{4(\alpha-\beta)}, \end{equation}$

(2.15)

and by rewriting (2.15), we obtain

$\begin{equation} \frac{4(\alpha-\beta)c}{n\alpha^{2}}-(p-1)\geq0,\quad \frac{4(\alpha-\beta)d}{n\alpha^{2}}-(q-1)\geq0,\quad \frac{4(\alpha-\beta)k}{n\alpha^{2}}-(s-1)\geq0, \end{equation}$

(2.16)

and

$\begin{equation} \frac{4(\alpha-\beta)c}{(p-1)n\alpha^{2}}\geq1,\quad \frac{4(\alpha-\beta)d}{(q-1)n\alpha^{2}}\geq1,\quad \frac{4(\alpha-\beta)k}{(s-1)n\alpha^{2}}\geq1. \end{equation}$

(2.17)

Using (2.14), we have

$\begin{equation} c\geq\frac{\alpha(p-1)+\beta}{p},\quad d\geq\frac{\alpha(q-1)+\beta}{q},\quad k\geq\frac{\alpha(s-1)+\beta}{s}, \end{equation}$

(2.18)

and by rewriting (2.18), we obtain

$\begin{equation} \alpha(p-1)+\beta-cp\leq0,\quad \alpha(q-1)+\beta-dq\leq0,\quad \alpha(s-1)+\beta-ks\leq0. \end{equation}$

(2.19)

By combining (2.19) and (2.16), we get

$\begin{equation*} \label{add7:Ha:9} \alpha(p-1)+\beta-cp\leq0,\quad \frac{4(\alpha-\beta)c}{n\alpha^{2}}-(p-1)\geq0,\\ \end{equation*}$

$\begin{equation} \alpha(q-1)+\beta-dq\leq0,\quad \frac{4(\alpha-\beta)d}{n\alpha^{2}}-(q-1)\geq0,\\ \end{equation}$

(2.20)

$\begin{equation*} \label{add9:Ha:9} \alpha(s-1)+\beta-ks\leq0,\quad \frac{4(\alpha-\beta)k}{n\alpha^{2}}-(s-1)\geq0.\\ \end{equation*}$

The requirement of (1.3) in Theorem 1.1 also suggests

$\begin{equation} \begin{cases} \frac{\alpha(p-1)+2\beta}{p}\geq c,\\ \frac{\alpha(q-1)+2\beta}{q}\geq d,\\ \frac{\alpha(s-1)+2\beta}{s}\geq k. \end{cases} \end{equation}$

(2.21)

Then, combining (2.21) and (2.17), we have

$\begin{equation*} \label{add11:Ha:9} \alpha(p-1)+\beta-cp+\frac{4(\alpha-\beta)\beta c}{n(p-1)\alpha^{2}}\geq\alpha(p-1)+2\beta-cp\geq0,\\ \end{equation*}$

$\begin{equation} \alpha(q-1)+\beta-dq+\frac{4(\alpha-\beta)\beta d}{n(q-1)\alpha^{2}}\geq\alpha(q-1)+2\beta-dq\geq0,\\ \end{equation}$

(2.22)

$\begin{equation*} \label{add13:Ha:9} \alpha(s-1)+\beta-ks+\frac{4(\alpha-\beta)\beta k}{n(s-1)\alpha^{2}}\geq\alpha(s-1)+2\beta-ks\geq0. \end{equation*}$

Rewriting (2.22), we obtain

$\begin{equation*} \label{add14:Ha:9} (\alpha(p-1)+\beta-cp)\left(1-\frac{1}{h_{1}(p-1)}\right)+\frac{4(\alpha-\beta)\beta c}{n(p-1)\alpha^{2}}\geq0,\\ \end{equation*}$

$\begin{equation} (\alpha(q-1)+\beta-dq)\left(1-\frac{1}{h_{2}(q-1)}\right)+\frac{4(\alpha-\beta)\beta d}{n(q-1)\alpha^{2}}\geq0,\\ \end{equation}$

(2.23)

$\begin{equation*} \label{add16:Ha:9} (\alpha(s-1)+\beta-ks)\left(1-\frac{1}{h_{3}(s-1)}\right)+\frac{4(\alpha-\beta)\beta k}{n(s-1)\alpha^{2}}\geq0. \end{equation*}$

Note the inequality

$\begin{equation*} \frac{4(\alpha-\beta)\beta\phi_{R}}{n\alpha^{2}}W-2\nabla\phi_{R}\cdot\nabla u\geq\frac{-n\alpha^{2}|\nabla\phi_{R}|^{2}}{4(\alpha-\beta)\beta\phi_{R}}. \end{equation*}$

Combining (1.2), (1.4), (1.5), (2.20), and (2.23) and removing a number of non-negative terms from the right side of (2.13), we have

$\begin{equation} 0\geq(\phi_{R})_{t}-\Delta\phi_{R}-\frac{n\alpha^{2}|\nabla\phi_{R}|^{2}}{4(\alpha-\beta)\beta\phi_{R}}+\frac{2(\alpha-\beta)}{n\alpha^{2}}\phi^{2}_{R}. \end{equation}$

(2.24)

By (2.12), we can compute

$\begin{equation} \Delta\phi_{R} = \sum\limits^{n}_{k = 1}\left(\frac{6b}{(x_{k}-p_{k})^{4}}+\frac{6b}{(q_{k}-x_{k})^{4}}\right), \end{equation}$

(2.25)

$\begin{equation*} |\nabla\phi_{R}|^{2} = \sum\limits^{n}_{k = 1}\left(-\frac{2b}{(x_{k}-p_{k})^{3}}-\frac{2b}{(q_{k}-x_{k})^{3}}\right)^{2}, \end{equation*}$

and

$\begin{equation} \begin{aligned} \frac{|\nabla\phi_{R}|^{2}}{\phi_{R}} & = \sum\limits^{n}_{k = 1}\left(-\frac{2b}{(x_{k}-p_{k})^{3}\sqrt{\phi_{R}}}-\frac{2b}{(q_{k}-x_{k})^{3}\sqrt{\phi_{R}}}\right)^{2}\\ &\leq \sum\limits^{n}_{k = 1}\left(\frac{4b}{(x_{k}-p_{k})^{4}}+\frac{4b}{(q_{k}-x_{k})^{4}}\right). \end{aligned} \end{equation}$

(2.26)

For the sake of simplicity, we set

$\begin{equation*} A: = \frac{2(\alpha-\beta)}{n\alpha^{2}} > 0,\quad B: = \frac{n\alpha^{2}}{4(\alpha-\beta)\beta} > 0. \end{equation*}$

To arrive at a contradiction, we need

$\begin{equation} A\phi_{R}^{2}-\Delta\phi_{R}-B\frac{|\nabla\phi_{R}|^{2}}{\phi_{R}}+(\phi_{R})_{t} > 0. \end{equation}$

(2.27)

Next, plugging (2.12), (2.25), and (2.26) into the left-hand side of (2.27), we get

$\begin{equation} \begin{aligned} &A\left[\frac{a}{t}+\sum\limits^{n}_{k = 1}\left(\frac{b}{(x_{k}-p_{k})^{2}}+\frac{b}{(q_{k}-x_{k})^{2}}\right)\right]^{2}-\left[\sum\limits^{n}_{k = 1}\left(\frac{6b}{(x_{k}-p_{k})^{4}}+\frac{6b}{(q_{k}-x_{k})^{4}}\right)\right]\\ &-B\left[\sum\limits^{n}_{k = 1}\left(-\frac{2b}{(x_{k}-p_{k})^{3}\sqrt{\phi_{R}}}-\frac{2b}{(q_{k}-x_{k})^{3}\sqrt{\phi_{R}}}\right)^{2}\right]-\frac{a}{t^{2}}\\ \geq&\; \frac{Aa^{2}-a}{t^{2}}+(Ab^{2}-6b-4bB)\left[\sum\limits^{n}_{k = 1}(\frac{1}{(x_{k}-p_{k})^{4}}+\frac{1}{(q_{k}-x_{k})^{4}})\right]. \end{aligned} \end{equation}$

(2.28)

By (1.4), we have $Aa^{2}-a\geq0$ . To prove (2.27), we need

$\begin{equation*} Ab^{2}-b(6+4B) > 0. \end{equation*}$

In summary, $a$ and $b$ satisfy

$\begin{equation*} a\geq\frac{n\alpha^{2}}{2(\alpha-\beta)},\quad b\geq \frac{n\alpha^{2}}{2(\alpha-\beta)}\left[6+\frac{n\alpha^{2}}{(\alpha-\beta)\beta}\right]. \end{equation*}$

Then, we can demonstrate that the inequality on the right side is positive. Thus, there is a contradiction.

We obtain $\phi_{R}\rightarrow \frac{a}{t}$ , $H_{R}\rightarrow H_{0}$ if $R\rightarrow \mathbb{R}^{n}$ , assuming that the solution is present in the complete space $\mathbb{R}^{n}$ . This suggests $H_{0}\geq0$ and completes the proof.

Proof of Corollary 1.2. We pick $\alpha = 2$ , $\beta = 1$ , $a = 2n$ , and $c$ such that $0 < n(p-1)\leq c < 2$ and $c\geq d\geq k\geq\beta$ in Theorem 1.1. Since $u = \ln f$ , we get

$\begin{equation} \Delta u = \frac{f\Delta f -|\nabla f|^{2}}{f^{2}}, \end{equation}$

(2.29)

$\begin{equation} |\nabla u|^{2} = \frac{|\nabla f|^{2}}{f^{2}}. \end{equation}$

(2.30)

By substituting (2.29) and (2.30) into (1.6), we can calculate

$\begin{equation*} 2\frac{\Delta f}{f}-\frac{|\nabla f|^{2}}{f^{2}}+ch_{1}f^{p-1}+dh_{2}f^{q-1}+kh_{3}f^{s-1}+\frac{2n}{t}\geq0, \end{equation*}$

and then

$\begin{equation} 2\Delta f-\frac{|\nabla f|^{2}}{f}+ch_{1}f^{p}+dh_{2}f^{q}+kh_{3}f^{s}+\frac{2n}{t}f\geq0. \end{equation}$

(2.31)

Noting

$\begin{equation} f_{t} = \Delta f+h_{1}f^{p}+h_{2}f^{q}+h_{3}f^{s}, \end{equation}$

(2.32)

by (2.31) and (2.32), we have

$\begin{equation*} 2f_{t}-\frac{|\nabla f|^{2}}{f}+\frac{2n}{t}f\geq (2-c)h_{1}f^{p}+(2-d)h_{2}f^{q}+(2-k)h_{3}f^{s}. \end{equation*}$

Furthermore, we observe that

$2f_{t}+\frac{2n}{t}f\geq(2-c)h_{1}f^{p}+(2-d)h_{2}f^{q}+(2-k)h_{3}f^{s},$

which implies that

$\begin{equation} \begin{aligned} 2\left(\frac{1}{f}\right)_{t} & = -2\frac{1}{f^{2}}\cdot f_{t}\\ &\leq-\frac{1}{f^{2}}\left(-\frac{2n}{t}f+(2-c)h_{1}f^{p}+(2-d)h_{2}f^{q}+(2-k)h_{3}f^{s}\right)\\ & = \frac{1}{f}\left(\frac{2n-(2-d)th_{2}f^{q-1}-(2-k)th_{3}f^{s-1}}{t}-(2-c)h_{1}f^{p-1}\right)\\ & = \frac{1}{f^{2-p}}\left(\frac{2n-(2-d)th_{2}f^{q-1}-(2-k)th_{3}f^{s-1}}{tf^{p-1}}-(2-c)h_{1}\right)\\ &\leq\frac{1}{f^{2-p}}\left(\frac{2n}{tf^{p-1}}-(2-c)h_{1}\right). \end{aligned} \end{equation}$

(2.33)

We might presume that $f\geq(\frac{4n}{(2-c)h_{1})})^{\frac{1}{p-1}}$ at the origin $x_{0} = 0$ for $t_{0} = 1$ , and hence we have

$\begin{equation*} \frac{1}{f^{p-1}}\leq\frac{(2-c)h_{1}}{4n}, \end{equation*}$

$\begin{equation} \frac{2n}{tf^{p-1}}\leq\frac{(2-c)h_{1}}{2t}. \end{equation}$

(2.34)

Therefore, for $t\geq1$ , we obtain

$\begin{equation*} \begin{aligned} 2\left(\frac{1}{f}\right)_{t}(0,t) &\leq\frac{1}{f^{2-p}}\left(\frac{2n}{tf^{p-1}}-(2-c)h_{1}\right)\\ &\leq\frac{(2-c)h_{1}}{f^{2-p}(0,t)}(\frac{1}{2t}-1)\\ & < 0, \end{aligned} \end{equation*}$

such that $f(0, t)$ is strictly increasing when $f(0, t)$ is finite.

(i) If $p > 2$ , then $f^{p-2}(0, t)\geq f^{p-2}(0, 1)$ for $t\geq1$ and (2.33) simplifies to

$\begin{equation} 2\left(\frac{1}{f}\right)_{t}(0,t)\leq\frac{2n}{tf(0,1)}-(2-c)h_{1}f^{p-2}(0,1). \end{equation}$

(2.35)

(ii) If $1 < p\leq2$ , it is easy to obtain that

$\begin{equation} \frac{2}{p-1}\left[\left(\frac{1}{f}\right)^{p-1}\right]_{t}(0,t) = 2f^{2-p}\left(\frac{1}{f}\right)_{t}(0,t)\leq\frac{2n}{tf^{p-1}(0,1)}-(2-c)h_{1}. \end{equation}$

(2.36)

Therefore, there is $\delta > 0$ such that when $t$ is sufficiently large, the right-hand side of (2.35) and (2.36) are smaller than $-\delta < 0$ , and therefore $\frac{1}{f}\rightarrow0$ in finite time. This completes the proof.

Proof of Corollary 1.3. We obtain $H_{0}\geq0$ by the differential Harnack estimate (1.6), which indicates that

$\begin{equation*} \Delta u\geq\frac{1}{\alpha}\left(-\beta|\nabla u|^{2}-ch_{1}e^{u(p-1)}-dh_{2}e^{u(q-1)}-kh_{3}e^{u(s-1)}-\frac{a}{t}\right). \end{equation*}$

Then, combined with (2.1), we calculate the evolution of $u$ along $\gamma$ , i.e.,

$\begin{equation*} \begin{aligned} (u(x(t),t))_{t} & = \nabla u\cdot\dot{x}+u_{t}\\ & = \nabla u\cdot\dot{x}+\Delta u+|\nabla u|^{2}+h_{1}e^{u(p-1)}+h_{2}e^{u(q-1)}+h_{3}e^{u(s-1)}\\ &\geq\nabla u\cdot\dot{x}+|\nabla u|^{2}(1-\frac{\beta}{\alpha})-\frac{a}{\alpha t}+(1-\frac{c}{\alpha})h_{1}e^{u(p-1)}\\ &\quad+(1-\frac{d}{\alpha})h_{2}e^{u(q-1)}+(1-\frac{k}{\alpha})h_{3}e^{u(s-1)}\\ &\geq|\nabla u|^{2}(\frac{1}{2}-\frac{\beta}{\alpha})-\frac{1}{2}|\dot{x}|^{2}-\frac{a}{\alpha t}+(1-\frac{c}{\alpha})h_{1}e^{u(p-1)}\\ &\quad+(1-\frac{d}{\alpha})h_{2}e^{u(q-1)}+(1-\frac{k}{\alpha})h_{3}e^{u(s-1)}\\ &\geq-\frac{1}{2}|\dot{x}|^{2}-\frac{a}{\alpha t}, \end{aligned} \end{equation*}$

where we have used the assumption $\alpha\geq2\beta$ and $k\leq d\leq c\leq\alpha$ .

Hence we have

$\begin{equation} \begin{aligned}(-u(x(t),t))_{t} &\leq\frac{1}{2}|\dot{x}|^{2}+\frac{a}{\alpha t}\\ &\leq\frac{1}{2}|\dot{x}|^{2}+\frac{n}{t}. \end{aligned} \end{equation}$

(2.37)

Integrating the previously mentioned quality (2.37) along $\gamma$ , and taking the infimum of all such space-time pathways, we get

$\begin{equation*} \int^{t_{2}}_{t_{1}}d(-u(x(t),t))\leq\inf\limits_{\gamma(t) = (x(t),t)}\int^{t_{2}}_{t_{1}}(\frac{1}{2}|\dot{x}|^{2}+\frac{n}{t})dt, \end{equation*}$

and then

$\begin{equation*} u(x_{1},t_{1})-u(x_{2},t_{2})\leq\inf\limits_{\gamma(t) = (x(t),t)}\int^{t_{2}}_{t_{1}}(\frac{1}{2}|\dot{x}|^{2}+\frac{n}{t})dt. \end{equation*}$

Using $u = \ln f$ , we have

$\begin{equation*} \frac{f(x_{1},t_{1})}{f(x_{2},t_{2})}\leq\exp\left[\inf\limits_{\gamma(t) = (x(t),t)}\int^{t_{2}}_{t_{1}}(\frac{1}{2}|\dot{x}|^{2}+\frac{n}{t})dt\right]. \end{equation*}$

Hence we can arrive at (1.8). This finishes the proof.

3. The proofs of Theorem 1.4 and Corollary 1.5

Estimating the following Harnack quantity is our main method of research:

$\begin{equation} \widetilde{H}: = \tilde{\alpha}\Delta u+\tilde{\beta}|\nabla u|^{2}+\tilde{c}h_{1}e^{u(p-1)}+\tilde{d}h_{2}e^{u(q-1)}+\tilde{k}h_{3}e^{u(s-1)}+\theta, \end{equation}$

(3.1)

where $\tilde{\alpha}$ , $\tilde{\beta}$ , $\tilde{c}$ , $\tilde{d}$ , $\tilde{k}\in \mathbb{R}$ and $\theta :\mathbb{R}^{n}\times[0, \infty)\rightarrow [0, \infty)$ will be determined later. We now derive the derivation of $\widetilde{H}$ in $t$ .

Next, similar to the proof of Lemma 2.1, we can get Lemma 3.1.

Lemma 3.1. Suppose that $f(x, t)$ is a positive solution of (1.1), $u = \ln f$ , and the definition of $\widetilde{H}$ is stated in (3.1). Then we obtain

$\begin{equation} \begin{aligned} \widetilde{H}_{t} & = \Delta \widetilde{H}+2\nabla \widetilde{H}\cdot\nabla u+(p-1)h_{1}e^{u(p-1)}\widetilde{H}+(q-1)h_{2}e^{u(q-1)}\widetilde{H}+(s-1)h_{3}e^{u(s-1)}\widetilde{H}\\ &\quad+2(\tilde{\alpha}-\tilde{\beta})|\nabla\nabla u|^{2}+[\tilde{\alpha}(p-1)+\tilde{\beta}-\tilde{c}p](p-1)h_{1}e^{u(p-1)}|\nabla u|^{2}\\ &\quad+[\tilde{\alpha}(q-1)+\tilde{\beta}-\tilde{d}q](q-1)h_{2}e^{u(q-1)}|\nabla u|^{2}\\ &\quad+[\tilde{\alpha}(s-1)+\tilde{\beta}-\tilde{k}s](s-1)h_{3}e^{u(s-1)}|\nabla u|^{2}\\ &\quad+[(\tilde{\alpha}-\tilde{c})\Delta h_{1}+2(\tilde{\alpha}(p-1)+\tilde{\beta}-\tilde{c}p)\nabla h_{1}\cdot\nabla u+\tilde{c}(h_{1})_{t}-h_{1}(p-1)\phi]e^{u(p-1)}\\ &\quad+[(\tilde{\alpha}-\tilde{d})\Delta h_{2}+2(\tilde{\alpha}(q-1)+\tilde{\beta}-\tilde{d}q)\nabla h_{2}\cdot\nabla u+\tilde{d}(h_{2})_{t}-h_{2}(q-1)\phi]e^{u(q-1)}\\ &\quad+[(\tilde{\alpha}-\tilde{k})\Delta h_{3}+2(\tilde{\alpha}(s-1)+\tilde{\beta}-\tilde{k}s)\nabla h_{3}\cdot\nabla u+\tilde{k}(h_{3})_{t}-h_{3}(s-1)\phi]e^{u(s-1)}\\ &\quad+[(\tilde{c}-\tilde{d})(p-q)]h_{1}h_{2}e^{u(p-1)}e^{u(q-1)}\\ &\quad+[(\tilde{c}-\tilde{k})(p-s)]h_{1}h_{3}e^{u(p-1)}e^{u(s-1)}\\ &\quad+[(\tilde{d}-\tilde{k})(q-s)]h_{2}h_{3}e^{u(q-1)}e^{u(s-1)}-2\nabla\theta\cdot\nabla u-\Delta\theta+\theta_{t}. \end{aligned} \end{equation}$

(3.2)

Proof of Theorem 1.4. Define the $n$ -rectangle $R: = \Pi^{n}_{i = 1}[p_{i}, q_{i}] \; \subset \; \mathbb{R}^{n}$ , and set

$\begin{equation} \theta_{R}(x,t) = \frac{\tilde{a}}{1-e^{-mt}}+\sum\limits^{n}_{k = 1}\left(\frac{\tilde{b}}{(x_{k}-p_{k})^{2}}+\frac{\tilde{b}}{(q_{k}-x_{k})^{2}}\right) \end{equation}$

(3.3)

for $t > 0$ , $\tilde{a} > 0$ , $\tilde{b} > 0$ , $m\geq\frac{n\tilde{\alpha}^{2}}{2(\tilde{\alpha}-\tilde{\beta})}[6+\frac{n\tilde{\alpha}^{2}}{(\tilde{\alpha}-\tilde{\beta})\tilde{\beta}}]$ , and $x = (x_{1}, ..., x_{n})\in R$ , while $\theta_{R}\rightarrow +\infty$ as $x_{i}\rightarrow p_{i}$ , $q_{i}$ or $t\rightarrow 0$ .

The corresponding Harnack quantity is defined as

$\begin{equation*} \widetilde{H}_{R} = \tilde{\alpha}\Delta u+\tilde{\beta}|\nabla u|^{2}+\tilde{c}h_{1}e^{u(p-1)}+\tilde{d}h_{2}e^{u(q-1)}+\tilde{k}h_{3}e^{u(s-1)}+\theta_{R}(x,t). \end{equation*}$

Note that $\widetilde{H}_{R}\rightarrow \widetilde{H}_{0}$ as $R\rightarrow \mathbb{R}^{n}$ , and $\widetilde{H}_{R} > 0$ for small $t$ .

In order to obtain a contradiction, assume that there is a first time $t_{0}$ and point $x_{0}\in R$ such that $\widetilde{H}_{R}(x_{0}, t_{0}) = 0$ . Then at $(x_{0}, t_{0})$ , we have

$\begin{equation*} (\widetilde{H}_{R})_{t}\leq0,\quad \nabla \widetilde{H}_{R} = 0,\quad \Delta \widetilde{H}_{R}\geq0, \end{equation*}$

and

$\begin{equation*} \Delta u = -\frac{1}{\tilde{\alpha}}(\tilde{\beta}|\nabla u|^{2}+\tilde{c}h_{1}e^{u(p-1)}+\tilde{d}h_{2}e^{u(q-1)}+\tilde{k}h_{3}e^{u(s-1)}+\theta_{R}). \end{equation*}$

Similar to the proof of (2.13), we can obtain

$\begin{equation} \begin{aligned} 0 &\geq \frac{2(\tilde{\alpha}-\tilde{\beta})}{n\tilde{\alpha}^{2}}(\tilde{c}^{2}h_{1}^{2}X^{2}+\tilde{d}^{2}h_{2}^{2}Y^{2}+\tilde{k}^{2}h_{3}^{2}Z^{2}+\tilde{\beta}^{2}W^{2})\\ &\quad+\left[(\tilde{\alpha}(p-1)+\tilde{\beta}-\tilde{c}p)\left(1-\frac{1}{h_{1}(p-1)}\right)+\frac{4(\tilde{\alpha}-\tilde{\beta})\tilde{\beta} \tilde{c}}{n(p-1)\tilde{\alpha}^{2}}\right](p-1)h_{1}XW\\ &\quad+\left[(\tilde{\alpha}(q-1)+\tilde{\beta}-\tilde{d}q)\left(1-\frac{1}{h_{2}(q-1)}\right)+\frac{4(\tilde{\alpha}-\tilde{\beta})\tilde{\beta} \tilde{d}}{n(q-1)\tilde{\alpha}^{2}}\right](q-1)h_{2}YW\\ &\quad+\left[(\tilde{\alpha}(s-1)+\tilde{\beta}-\tilde{k}s)\left(1-\frac{1}{h_{3}(s-1)}\right)+\frac{4(\tilde{\alpha}-\tilde{\beta})\tilde{\beta} \tilde{k}}{n(s-1)\tilde{\alpha}^{2}}\right](s-1)h_{3}ZW\\ &\quad+\bigg[(\tilde{\alpha}-\tilde{c})\Delta h_{1}-4(\tilde{\alpha}(p-1)+\tilde{\beta}-\tilde{c}p)|\nabla h_{1}|^{2}+\tilde{c}(h_{1})_{t} +\left(\frac{4(\tilde{\alpha}-\tilde{\beta})\tilde{c}}{n\tilde{\alpha}^{2}}-(p-1)\right)h_{1}\theta_{R}\bigg]X\\ &\quad+\bigg[(\tilde{\alpha}-\tilde{d})\Delta h_{2}-4(\tilde{\alpha}(q-1)+\tilde{\beta}-\tilde{d}q)|\nabla h_{2}|^{2}+\tilde{d}(h_{2})_{t}+\left(\frac{4(\tilde{\alpha}-\tilde{\beta})\tilde{d}}{n\tilde{\alpha}^{2}}-(q-1)\right)h_{2}\theta_{R}\bigg]Y\\ &\quad+\bigg[(\tilde{\alpha}-\tilde{k})\Delta h_{3}-4(\tilde{\alpha}(s-1)+\tilde{\beta}-\tilde{k}s)|\nabla h_{3}|^{2}+\tilde{k}(h_{3})_{t}\left(\frac{4(\tilde{\alpha}-\tilde{\beta})\tilde{k}}{n\tilde{\alpha}^{2}}-(s-1)\right)h_{3}\theta_{R}\bigg]Z\\ &\quad+\left[(\tilde{c}-\tilde{d})(p-q)+\frac{4(\tilde{\alpha}-\tilde{\beta})\tilde{c}\tilde{d}}{n\tilde{\alpha}^{2}}\right]h_{1}h_{2}XY\\ &\quad+\left[(\tilde{c}-\tilde{k})(p-s)+\frac{4(\tilde{\alpha}-\tilde{\beta})\tilde{c}\tilde{k}}{n\tilde{\alpha}^{2}}\right]h_{1}h_{3}XZ\\ &\quad+\left[(\tilde{d}-\tilde{k})(q-s)+\frac{4(\tilde{\alpha}-\tilde{\beta})\tilde{d}\tilde{k}}{n\tilde{\alpha}^{2}}\right]h_{2}h_{3}YZ-2\nabla\theta_{R}\cdot\nabla u-\Delta\theta_{R}\\ &\quad+(\theta_{R})_{t}+\frac{2(\tilde{\alpha}-\tilde{\beta})}{n\tilde{\alpha}^{2}}\theta^{2}_{R}+\frac{4(\tilde{\alpha}-\tilde{\beta})\tilde{\beta}\theta_{R}}{n\tilde{\alpha}^{2}}W, \end{aligned} \end{equation}$

(3.4)

where $X = e^{u(p-1)}$ , $Y = e^{u(q-1)}$ , $Z = e^{u(s-1)}$ , and $W = |\nabla u|^{2}$ .

By demonstrating that the right-hand side of (3.4) is positive, we can then obtain a contradiction. Similar to the proof of (2.24), we can get

$\begin{equation} 0\geq(\theta_{R})_{t}-\Delta\theta_{R}-\frac{n\tilde{\alpha}^{2}|\nabla\theta_{R}|^{2}}{4(\tilde{\alpha}-\tilde{\beta})\tilde{\beta}\theta_{R}}+\frac{2(\tilde{\alpha}-\tilde{\beta})}{n\tilde{\alpha}^{2}}\theta^{2}_{R}. \end{equation}$

(3.5)

For the sake of simplicity, we set

$\begin{equation*} \tilde{A}: = \frac{2(\tilde{\alpha}-\tilde{\beta})}{n\tilde{\alpha}^{2}} > 0,\quad \tilde{B}: = \frac{n\tilde{\alpha}^{2}}{4(\tilde{\alpha}-\tilde{\beta})\tilde{\beta}} > 0. \end{equation*}$

In order to obtain a contradiction, we need

$\begin{equation} \tilde{A}\theta_{R}^{2}-\Delta\theta_{R}-\tilde{B}\frac{|\nabla\theta_{R}|^{2}}{\theta_{R}}+(\theta_{R})t > 0. \end{equation}$

(3.6)

Next, similar to the calculation of (2.29), we can get

$\begin{equation*} \begin{aligned} &\tilde{A}\left[\frac{\tilde{a}}{1-e^{-mt}}+\sum\limits^{n}_{k = 1}\left(\frac{\tilde{b}}{(x_{k}-p_{k})^{2}}+\frac{\tilde{b}}{(q_{k}-x_{k})^{2}}\right)\right]^{2}-\left[\sum\limits^{n}_{k = 1}\left(\frac{6\tilde{b}}{(x_{k}-p_{k})^{4}}+\frac{6\tilde{b}}{(q_{k}-x_{k})^{4}}\right)\right]\\ &-\tilde{B}\left[\sum\limits^{n}_{k = 1}\left(-\frac{2\tilde{b}}{(x_{k}-p_{k})^{3}\sqrt{\theta_{R}}}-\frac{2\tilde{b}}{(q_{k}-x_{k})^{3}\sqrt{\theta_{R}}}\right)^{2}\right]-\frac{m\tilde{a}}{(1-e^{-mt})^{2}e^{mt}}\\ \geq&\; \frac{\tilde{A}\tilde{a}^{2}e^{mt}-m\tilde{a}}{(1-e^{-mt})^{2}e^{mt}}+(\tilde{A}\tilde{b}^{2}-6\tilde{b}-4\tilde{b}\tilde{B})\left[\sum\limits^{n}_{k = 1}(\frac{1}{(x_{k}-p_{k})^{4}}+\frac{1}{(q_{k}-x_{k})^{4}})\right]. \end{aligned} \end{equation*}$

By (1.9), we have $\tilde{A}\tilde{a}^{2}e^{mt}-m\tilde{a}\geq0$ . To prove (3.6), we need

$\begin{equation*} \tilde{A}\tilde{b}^{2}-\tilde{b}(6+4\tilde{B}) > 0. \end{equation*}$

In summary, $\tilde{a}$ and $\tilde{b}$ satisfy

$\begin{equation*} \tilde{a}\geq\frac{nm\tilde{\alpha}^{2}}{2(\tilde{\alpha}-\tilde{\beta})},\quad \tilde{b}\geq \frac{n\tilde{\alpha}^{2}}{2(\tilde{\alpha}-\tilde{\beta})}\left[6+\frac{n\tilde{\alpha}^{2}}{(\tilde{\alpha}-\tilde{\beta})\tilde{\beta}}\right]. \end{equation*}$

Then, we can demonstrate that the inequality on the right side is positive. Thus, we obtain a contradiction.

Assuming that the solution exists in the whole space $\mathbb{R}^{n}$ , we get $\theta_{R}\rightarrow \frac{\tilde{a}}{1-e^{-mt}}$ , $\widetilde{H}_{R}\rightarrow \widetilde{H}_{0}$ if $R\rightarrow \mathbb{R}^{n}$ . This implies $\widetilde{H}_{0}\geq0$ and completes the proof.

Proof of Corollary 1.5. Corollary 1.5 follows immediately from Theorem 1.4 by using a similar method to that in the proof of Corollary 1.3. We omit the proof of Corollary 1.5.

4. Conclusions

In this paper, some new types of differential Harnack estimates were established for positive solutions of the semilinear parabolic equation with three exponents on $\mathbb{R}^{n}$ . Additionally, as applications, we found the blow-up of the solutions and classical Harnack inequalities for this equation. Our results generalize some known results.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

This research is supported by NSFC (No. 12101530) and the Natural Science Foundation of Henan Province (No. 232300420363).

Conflict of interest

The authors declare there is no conflict of interest.

References

[1]	R. Hamilton, The Harnack estimate for the Ricci flow, J. Differential Geom., 37 (1993), 225–243. https://doi.org/10.4310/jdg/1214453430 doi: 10.4310/jdg/1214453430
[2]	J. Li, Gradient estimates and Harnack inequalities for nonlinear parabolic and nonlinear elliptic equations on Riemannian manifolds, J. Funct. Anal., 100 (1991), 233-256. https://doi.org/10.1016/0022-1236(91)90110-Q doi: 10.1016/0022-1236(91)90110-Q
[3]	P. Li, S. T. Yau, On the parabolic kernel of the Schrödinger operator, Acta Math., 156 (1986), 153–201. https://doi.org/10.1007/BF02399203 doi: 10.1007/BF02399203
[4]	J. Moser, A Harnack inequality for parabolic differential equations, Comm. Pure Appl. Math., 17 (1964), 101–134. https://doi.org/10.1002/cpa.3160170106 doi: 10.1002/cpa.3160170106
[5]	J. Moser, Correction to: "A Harnack inequality for parabolic differential equations", Comm. Pure Appl. Math., 20 (1967), 231–236.
[6]	M. Bǎileşteanu, A Harnack inequality for the parabolic Allen-Cahn equation, Ann. Global Anal. Geom., 51 (2017), 367–378. https://doi.org/10.1007/s10455-016-9540-2 doi: 10.1007/s10455-016-9540-2
[7]	D. Booth, J. Burkart, X. Cao, M. Hallgren, Z. Munro, J. Snyder, et al., A differential Harnack inequality for the Newell-Whitehead-Segel equation, Anal. Theory Appl., 35 (2019), 192–204. https://doi.org/10.4208/ata.OA-0005 doi: 10.4208/ata.OA-0005
[8]	X. Cao, M. Cerenzia, D. Kazaras, Harnack estimate for the endangered species equation, Proc. Amer. Math. Soc., 143 (2015), 4537–4545. https://doi.org/10.1090/S0002-9939-2015-12576-2 doi: 10.1090/S0002-9939-2015-12576-2
[9]	X. Cao, B. Liu, I. Pendleton, A. Ward, Differential Harnack estimates for Fisher's equation, Pacific J. Math., 290 (2017), 273–300. https://doi.org/10.2140/pjm.2017.290.273 doi: 10.2140/pjm.2017.290.273
[10]	S. Hou, L. Zou, Harnack estimate for a semilinear parabolic equation, Sci. China Math., 60 (2017), 833–840. https://doi.org/10.1007/s11425-016-0270-6 doi: 10.1007/s11425-016-0270-6
[11]	A. Abolarinwa, A. Osilagun, S. Azami, A Harnack inequality for a class of 1D nonlinear reaction-diffusion equations and applications to wave solutions, Int. J. Geom. Methods Mod. Phys., 21 (2024), 2450111. https://doi.org/10.1142/S0219887824501111 doi: 10.1142/S0219887824501111
[12]	H. Fujita, On the blowing up of solutions of the Cauchy problem for $u_{t} = \Delta u+u^{1+\alpha}$ , J. Fac. Sci. Univ. Tokyo Sect. I, 13 (1966), 109–124. https://doi.org/10.1090/proc/14297 doi: 10.1090/proc/14297
[13]	G. Huang, Z. Huang, H. Li, Gradient estimates and differential Harnack inequalities for a nonlinear parabolic equation on Riemannian manifolds, Ann. Global Anal. Geom., 43 (2013), 209–232. https://doi.org/10.1007/s10455-012-9342-0 doi: 10.1007/s10455-012-9342-0
[14]	H. Wu, C. Kong, Differential Harnack estimate of solutions to a class of semilinear parabolic equation, Math. Inequal. Appl., 25 (2022), 397–405. https://doi.org/10.7153/mia-2022-25-24 doi: 10.7153/mia-2022-25-24
[15]	H. Wu, L. Min, Differential Harnack estimate for a semilinear parabolic equation on hyperbolic space, Appl. Math. Lett., 50 (2015), 69–77. https://doi.org/10.1016/j.aml.2015.06.002 doi: 10.1016/j.aml.2015.06.002
[16]	R. Hamilton, Li-Yau estimates and their Harnack inequalities, Adv. Lect. Math., 17 (2011), 329–362.

Reader Comments

Your name:*

Email:*
© 2025 the Author(s), licensee AIMS Press. This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0)

通讯作者: 陈斌, bchen63@163.com

1.
沈阳化工大学材料科学与工程学院沈阳 110142

Electronic Research Archive

1.1 1.7

Metrics

Article views(624) PDF downloads(26) Cited by(0)

Preview PDF

Download XML

Export Citation

Article outline

Show full outline

Electronic Research Archive

Differential Harnack estimates for the semilinear parabolic equation with three exponents on $\mathbb{R}^{n}$

Related Papers:

Abstract

1. Introduction

2. The proofs of Theorem 1.1, Corollary 1.2 and Corollary 1.3

3. The proofs of Theorem 1.4 and Corollary 1.5

4. Conclusions

Use of AI tools declaration

Acknowledgments

Conflict of interest

References

Reader Comments

通讯作者: 陈斌, bchen63@163.com

Metrics

Other Articles By Authors

Catalog

Electronic Research Archive

Differential Harnack estimates for the semilinear parabolic equation with three exponents on Rn \mathbb{R}^{n}

Related Papers:

Abstract

1. Introduction

2. The proofs of Theorem 1.1, Corollary 1.2 and Corollary 1.3

3. The proofs of Theorem 1.4 and Corollary 1.5

4. Conclusions

Use of AI tools declaration

Acknowledgments

Conflict of interest

References

Reader Comments

通讯作者: 陈斌, bchen63@163.com

Metrics

Other Articles By Authors

Related pages

Tools

Export File

Citation

Format

Content

Catalog

Differential Harnack estimates for the semilinear parabolic equation with three exponents on $\mathbb{R}^{n}$