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

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:

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

and

then we have

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

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

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

and

then we have

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

yielding

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

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

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

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

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

and

where we applied the formula

Hence we get

A direct calculation gives

and

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

By (2.8) and (2.7), we get

Direct computations show that

and

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

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

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

and

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

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

and

we arrive at

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

By (2.14), we get

and by rewriting (2.15), we obtain

and

Using (2.14), we have

and by rewriting (2.18), we obtain

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

The requirement of (1.3) in Theorem 1.1 also suggests

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

Rewriting (2.22), we obtain

Note the inequality

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

By (2.12), we can compute

and

For the sake of simplicity, we set

To arrive at a contradiction, we need

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

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

In summary, $a$ and $b$ satisfy

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

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

and then

Noting

by (2.31) and (2.32), we have

Furthermore, we observe that

which implies that

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

Therefore, for $t\geq1$, we obtain

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

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

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

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

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

Hence we have

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

and then

Using $u = \ln f$, we have

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:

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

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

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

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

and

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

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

For the sake of simplicity, we set

In order to obtain a contradiction, we need

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

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

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

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.