Nonlinear estimates for traveling wave solutions of reaction diffusion equations

TL;DR

This paper develops nonlinear bounds for traveling wave solutions of reaction-diffusion equations and applies these bounds to the Lotka-Volterra system, extending previous linear maximum principle methods.

Contribution

It introduces nonlinear a priori bounds for a broad class of reaction-diffusion equations, including the Lotka-Volterra system, using an extension of the linear N-barrier maximum principle.

Findings

Established nonlinear bounds for solutions of reaction-diffusion equations.

Applied bounds to the Lotka-Volterra system of two species.

Extended linear maximum principle techniques to nonlinear estimates.

Abstract

In this paper we will establish nonlinear a priori lower and upper bounds for the solutions to a large class of equations which arise from the study of traveling wave solutions of reaction-diffusion equations, and we will apply our nonlinear bounds to the Lotka-Volterra system of two competing species as examples. The idea used in a series of papers \cite{NBMP-Discrete,JDE-16,CPAA-16,DCDS-B-18,NBMP-n-species,DCDS-A-17} for the establishment of the linear N-barrier maximum principle will also be used in the proof.

Equations

$$\{\begin{array}{l}\text{vspace}3mm{d}_i(u_i{)}_{xx}+\theta (u_i{)}_x+u_i^{l_i}{f}_i(u_1,u_2,\cdots ,u_n)=0,x\in \mathbb{R},i=1,2,\cdots ,n,\\ (u_1,u_2,\cdots ,u_n)(-\infty )={\text{e}}_{-},(u_1,u_2,\cdots ,u_n)(\infty )={\text{e}}_{+}.\end{array}$$

$$\Big{\{}(u_{1},\cdots,u_{n})\;\Big{|}\;u_{i}^{l_{i}}\,f_{i}(u_{1},\cdots,u_{n})=0,\quad u_{i}\geq 0,\quad\forall i=1,\cdots,n\Big{\}}.$$

$$\sum _{i=1}^n{\alpha }_i{u}_i(x),\forall ({\alpha }_1,\cdots ,{\alpha }_n)\in ({\mathbb{R}}^{+}{)}^n$$

$$\prod _{i=1}^n(u_i(x)+k_i{)}^{{\alpha }_i},\forall ({\alpha }_1,\cdots ,{\alpha }_n)\in ({\mathbb{R}}^{+}{)}^n$$

$$\{\begin{array}{l}\text{vspace}3mm{d}_i(u_i{)}_{xx}+\theta (u_i{)}_x+u_i^{l_i}{f}_i(u_1,u_2,\cdots ,u_n)\le 0,x\in \mathbb{R},i=1,2,\cdots ,n,\\ (u_1,u_2,\cdots ,u_n)(-\infty )={\text{e}}_{-},(u_1,u_2,\cdots ,u_n)(\infty )={\text{e}}_{+},\end{array}$$

$$\begin{array}[]{l}f_{i}(u_{1},\cdots,u_{n})\geq 0,\hbox{ for all }\\ (u_{1},\cdots,u_{n})\in\underaccent{\bar}{\mathcal{R}}:=\big{\{}(u_{i})_{i=1}^{n}\in([0,\infty))^{n}\;|\;\sum_{i=1}^{n}\frac{\displaystyle u_{i}}{\displaystyle\underaccent{\bar}{u}_{i}}\leq 1\big{\}}.\end{array}$$

$$\prod_{i=1}^{n}(u_{i}(x)+k_{i})^{{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}d_{i}}\alpha_{i}}\geq e^{\lambda_{1}},\quad x\in\mathbb{R},$$

$${\lambda }_1$$

$$\eta$$

$${\lambda }_2$$

$$\lambda_{1}=\min_{1\leq j\leq n}\Bigl{(}\alpha_{j}\ln(\underaccent{\bar}{u}_{j}+k_{j})+\sum_{i=1,i\neq j}^{n}\alpha_{i}\ln k_{i}\Bigr{)}d=\lambda_{2}=d\eta,$$

$$\prod_{i=1}^{n}(u_{i}(x)+k_{i})^{\alpha_{i}}\geq\min_{1\leq j\leq n}\Bigl{(}(\underaccent{\bar}{u}_{j}+k_{j})^{\alpha_{j}}\prod_{i\neq j}k_{i}^{\alpha_{i}}\Bigr{)},\quad x\in\mathbb{R}.$$

$$\displaystyle\sum_{i=1}^{n}(u_{i}+k_{i})^{\alpha}\geq n\Bigl{(}\prod_{i=1}^{n}(u_{i}+k_{i})^{\alpha}\Bigr{)}^{\frac{1}{n}}\geq n\min_{1\leq j\leq n}\Bigl{(}(\underaccent{\bar}{u}_{j}+k_{j})^{\alpha}\prod_{i\neq j}k_{i}^{\alpha}\Bigr{)}^{\frac{1}{n}}.$$

$$\{\begin{array}{l}\text{vspace}3mm{d}_i(u_i{)}_{xx}+\theta (u_i{)}_x+u_i^{l_i}{f}_i(u_1,u_2,\cdots ,u_n)\ge 0,x\in \mathbb{R},i=1,2,\cdots ,n,\\ (u_1,u_2,\cdots ,u_n)(-\infty )={\text{e}}_{-},(u_1,u_2,\cdots ,u_n)(\infty )={\text{e}}_{+},\end{array}$$

$$\begin{array}[]{l}f_{i}(u_{1},\cdots,u_{n})\leq 0,\hbox{ for all }\\ (u_{1},\cdots,u_{n})\in\bar{\mathcal{R}}:=\big{\{}(u_{i})_{i=1}^{n}\in([0,\infty))^{n}\;|\;\sum_{i=1}^{n}\frac{\displaystyle u_{i}}{\displaystyle\bar{u}_{i}}\geq 1\big{\}}.\end{array}$$

$$\sum _{i=1}^n{\alpha }_i(u_i(x){)}^{m_i}\le \left(\underset{1\le i\le n}{\max }{\alpha }_i({\bar{u}}_i{)}^{m_i}\right)\frac{\underset{1\le i\le n}{\max }{d}_i}{\underset{1\le i\le n}{\min }{d}_i},x\in \mathbb{R},$$

$$\prod _{i=1}^n(u_i(x){)}^{m_i/n}\le \frac{\underset{1\le i\le n}{\max }{\alpha }_i{\bar{u}}_i^{m_i}}{n{\left(\prod _{i=1}^n{\alpha }_i\right)}^{1/n}}\frac{\underset{1\le i\le n}{\max }{d}_i}{\underset{1\le i\le n}{\min }{d}_i},x\in \mathbb{R}.$$

$$\prod _{i=1}^n(u_i(x){)}^{m_i/n}\le \frac{\underset{1\le i\le n}{\max }{\bar{u}}_i^{m_i}}{n}\frac{\underset{1\le i\le n}{\max }{d}_i}{\underset{1\le i\le n}{\min }{d}_i},x\in \mathbb{R}.$$

$$\{\begin{array}{ll}\text{vspace}3mm{d}_1{u}_{xx}+\theta u_x+u(1-u-a_{1}v)=0,& x\in \mathbb{R},\\ \text{vspace}3mm{d}_2{v}_{xx}+\theta v_x+\kappa v(1-a_{2}u-v)=0,& x\in \mathbb{R},\\ (u,v)(-\infty )={\text{e}}_i,(u,v)(+\infty )={\text{e}}_j,& \end{array}$$

$$\underline{u}=\min \left(1,\frac{1}{a_2}\right),\bar{u}=\max \left(1,\frac{1}{a_2}\right),$$

$$\underline{v}=\min \left(1,\frac{1}{a_1}\right),\bar{v}=\max \left(1,\frac{1}{a_1}\right),$$

$$(u+k_1)(v+k_2)\ge \min \left((\underline{u}+k_1)k_2,k_1(\underline{v}+k_2)\right),\forall k_1,k_2>0.$$

$$\min \left(\frac{k_2}{a_2},\frac{k_1}{a_1}\right)\le k_{2}u+k_{1}v\le \max \left(k_1,k_2\right),$$

$$uv+\max \left(k_1,k_2\right)\ge uv+k_{2}u+k_{1}v\ge \min \left(k_2\underline{u},k_1\underline{v}\right).$$

$$uv+k_1\ge k_1\min (\frac{1}{a_1},\frac{1}{a_2})\Rightarrow uv\ge 0.$$

$$\sqrt{u(x)v(x)}\le \frac{1}{2}\max \left(\bar{u},\bar{v}\right)\frac{\max \left(d_1,d_2\right)}{\min \left(d_1,d_2\right)},x\in \mathbb{R}$$

$$u(x)v(x)\le \frac{1}{4}{\left(\max \left(\bar{u},\bar{v}\right)\right)}^2{\left(\frac{\max \left(d_1,d_2\right)}{\min \left(d_1,d_2\right)}\right)}^2,x\in \mathbb{R}.$$

$$u(x)v(x)\le \frac{1}{4},x\in \mathbb{R}.$$

$$(u^{\ast },v^{\ast })=\left(\frac{1-a_1}{1-a_1{a}_2},\frac{1-a_2}{1-a_1{a}_2}\right)=\left(\frac{1}{1+a},\frac{1}{1+a}\right)\to \left(1/2,1/2\right).$$

$$(\ln (u(x)+k){)}^{\prime }$$

$$(\ln (u(x)+k){)}^{\prime \prime }$$

$$d_i(\ln (u_i+k_i){)}^{\prime \prime }+d_i\frac{((u_i{)}^{\prime }{)}^2}{(u_i+k_i{)}^2}+\theta (\ln (u_i+k_i){)}^{\prime }+\frac{u_i^{l_i}}{u_i+k_i}{f}_i\le 0.$$

Full text

*∗*Department of Mathematics, National Taiwan University, Taipei, Taiwan

*♮*Institute of Analysis, Karlsruhe Institute of Technology, Karlsruhe, Germany

1. Introduction

The present paper is devoted to nonlinear a priori upper and lower bounds for the solutions $u_{i}=u_{i}(x):\mathbb{R}\mapsto[0,\infty)$, $i=1,\cdots,n$ to the following boundary value problem of $n$ equations

[TABLE]

In the above, $d_{i}$, $l_{i}>0$, $\theta\in\mathbb{R}$ are parameters, $f_{i}\in C^{0}([0,\infty)^{n})$ are given functions and the boundary values $\textbf{e}_{-},\textbf{e}_{+}$ take value in the following constant equilibria set

[TABLE]

Equations (1) arise from the study of traveling waves solutions of reaction-diffusion equations (see [16, 18]). A series of papers [2, 3, 4, 5, 6, 7] by Hung et al. have been contributed to the linear (N-barrier) maximum principle for the $n$ equations (1), and in particular the lower and upper bounds for any linear combination of the solutions

[TABLE]

have been established in terms of the parameters $d_{i},l_{i},\theta$ in (1).

Here we aim to derive nonlinear estimates for the polynomials of the solutions:

[TABLE]

for some $k_{i}\geq 0$, which is related to the diversity indices of the species in ecology: $D^{q}=(\sum_{i=1}^{n}(u_{i})^{q})^{1/(1-q)}$, $q\in[1,\infty)$. Observe that when either $\text{\bf e}_{+}=(0,\cdots,0)$ or $\text{\bf e}_{-}=(0,\cdots,0)$, the trivial lower bound of $\displaystyle\prod_{i=1}^{n}(u_{i}(x)+k_{i})^{\alpha_{i}}$ is $\prod_{i=1}^{n}k_{i}^{\alpha_{i}}$. For $k_{i}>0$ the following lower bound for the upper solutions of (1) holds.

Theorem 1.1** (Lower bound).**

\thlabel

prop: lower bed Suppose that $(u_{i}(x))_{i=1}^{n}\in(C^{2}(\mathbb{R}))^{n}$ with $u_{i}(x)\geq 0$, $\forall i=1,\cdots,n$ is an upper solution of (1):

[TABLE]

and that there exist (\underaccent{\bar}{u}_{i})_{i=1}^{n}\in(\mathbb{R}^{+})^{n} such that

[TABLE]

Then we have for any $(k_{i})_{i=1}^{n}\in(\mathbb{R}^{+})^{n}$ and $(\alpha_{i})_{i=1}^{n}\in(\mathbb{R}^{+})^{n}$,

[TABLE]

where

[TABLE]

Remark 1.2** (Equal diffusion).**

\thlabel

rem: lower bound equal diffusion When $d_{i}=d$ for all $i=1,2,\cdots,n$, then

[TABLE]

and the lower bound (5) becomes

[TABLE]

If furthermore $\alpha_{i}=\alpha$, $\forall i=1,\cdots,n$, then the inequality of arithmetic and geometric averages yields

[TABLE]

On the other hand, we can find an upper bound of $\displaystyle\prod_{i=1}^{n}(u_{i}(x))^{\alpha_{i}}$ for the lower solutions of (1).

Theorem 1.3** (Upper bound).**

\thlabel

prop: upper bed Suppose that $(u_{i}(x))_{i=1}^{n}\in(C^{2}(\mathbb{R}))^{n}$ with $u_{i}(x)\geq 0$ $\forall i=1,\cdots,n$ is a lower solution of (1):

[TABLE]

and there exist $\bar{u}_{i}>0,i=1,\cdots,n,$ such that

[TABLE]

Then we have for any $m_{i}\geq 1$ and $\alpha_{i}>0$ $(i=1,2,\cdots,n)$

[TABLE]

and hence

[TABLE]

In particular, when $\alpha_{i}=\alpha$ for all $i=1,\cdots,n$, (10) becomes

[TABLE]

In order to prove \threfprop: lower bed, we will first rewrite the system (3) into the system for the new unknowns $(U_{i})_{i=1}^{n}:=(\ln(u_{i}+k_{i}))_{i=1}^{n}$. Then we will follow the ideas in [2, 3, 4, 5, 6, 7] to establish the lower bound for the linear combination of $(U_{i})_{i=1}^{n}$, which implies the nonlinear lower bound (5) correspondingly. Similarly, we will consider the new unknowns $(U_{i})_{i=1}^{n}:=(u_{i}^{m_{i}})_{i=1}^{n}$ to establish the upper bound (9). The proofs will be found in Section 2.

As an example to illustrate our main result, we use the Lotka-Volterra system of two competing species to conclude with Section 1. This example provides an intuitive idea of the construction of the N-barrier in multi-species cases.

To illustrate \threfprop: upper bed for the case $n=2$, we use the Lotka-Volterra system of two competing species coupled with Dirichlet boundary conditions:

[TABLE]

where $a_{1}$, $a_{2}$, $\kappa>0$ are constants. In (12), the constant equilibria are $\textbf{e}_{1}=(0,0)$, $\textbf{e}_{2}=(1,0)$, $\textbf{e}_{3}=(0,1)$ and $\textbf{e}_{4}=(u^{\ast},v^{\ast})$, where $(u^{\ast},v^{\ast})=\left(\displaystyle\frac{1-a_{1}}{1-a_{1}\,a_{2}},\displaystyle\frac{1-a_{2}}{1-a_{1}\,a_{2}}\right)$ is the intersection of the two straight lines $1-u-a_{1}\,v=0$ and $1-a_{2}\,u-v=0$ whenever it exists. We call the solution $(u(x),v(x))$ of (12) an $(\textbf{e}_{i},\textbf{e}_{j})$-wave.

Tang and Fife ([17]), and Ahmad and Lazer ([1]) established the existence of the $(\textbf{e}_{1},\textbf{e}_{4})$-waves. Kan-on ([10, 11]), Fei and Carr ([8]), Leung, Hou and Li ([15]), and Leung and Feng ([14]) proved the existence of $(\textbf{e}_{2},\textbf{e}_{3})$-waves using different approaches. $(\textbf{e}_{2},\textbf{e}_{4})$-waves were studied for instance, by Kanel and Zhou ([13]), Kanel ([12]), and Hou and Leung ([9]).

For the above-mentioned $(\textbf{e}_{1},\textbf{e}_{4})$-waves, $(\textbf{e}_{2},\textbf{e}_{3})$-waves, and $(\textbf{e}_{2},\textbf{e}_{4})$-waves, we show a lower and an upper bounds of $u(x)v(x)$ by \threfprop: lower bed and \threfprop: upper bed respectively. To this end, let

[TABLE]

then the hypothesis (4) and (8) are satisfied. If $d_{1}=d_{2}=1$ and $\alpha_{1}=\alpha_{2}=1$, then by \threfprop: lower bed (or by \threfrem: lower bound equal diffusion),

[TABLE]

Recall the maximum principle in Theorem 1.1 in [3]:

[TABLE]

then we have

[TABLE]

Under the bistable condition $a_{1},a_{2}>1$, we derive the following “trivial” lower bound by taking $k_{1}=k_{2}$,

[TABLE]

According to (10), letting $\alpha_{1}=\alpha_{2}=m_{1}=m_{2}=1$ leads to

[TABLE]

or

[TABLE]

For the equal diffusion case $d_{1}=d_{2}=1$ with the bistable condition $a_{1},a_{2}>1$, (14) is simplified to

[TABLE]

If we further consider the boundary conditions in the $(\textbf{e}_{2},\textbf{e}_{4})$-waves (also $(\textbf{e}_{3},\textbf{e}_{4})$-waves) or the $(\textbf{e}_{4},\textbf{e}_{4})$-waves, the upper bound given by (15) is optimal for the case $a:=a_{1}=a_{2}>1$ since as $a\to 1^{+}$, we have

[TABLE]

2. Proofs of \threfprop: lower bed and \threfprop: upper bed

Proof of \threfprop: lower bed.

We first rewrite the inequality $d_{i}(u_{i})^{\prime\prime}+\theta(u_{i})^{\prime}+u_{i}^{l_{i}}f_{i}\leq 0$ in (3). If $u(x)\geq 0$, then for any $k>0$, a straightforward calculation gives

[TABLE]

Hence we divide the inequality by $u_{i}+k_{i}>0$ with $k_{i}>0$ to arrive at

[TABLE]

Thus $(U_{i})_{i=1}^{n}:=(\ln(u_{i}+k_{i}))_{i=1}^{n}$ satisfies the following inequalities:

[TABLE]

For any $(\alpha_{i})_{i=1}^{n}\in(\mathbb{R}^{+})^{n}$, let

[TABLE]

then the above inequality (17) reads as

[TABLE]

We are going to derive a lower bound for

[TABLE]

and hence a lower bound for $\prod_{i=1}^{n}(u_{i}+k_{i})^{d_{i}\alpha_{i}}$. The idea is similar as in the papers [2, 3, 4, 5, 6, 7], namely we are going to determine three parameters

[TABLE]

to construct an N-barrier consisting of three hypersurfaces

[TABLE]

such that the following inclusion relations hold:

[TABLE]

It will turn out that if $\lambda_{1}$, $\eta$, and $\lambda_{2}$ are given respectively by (6a), (6b), and (6c), then $\lambda_{1}$ determines a lower bound of $q(x)$: $q(x)\geq\lambda_{1}$, which is exactly (5).

More precisely, we follow the steps as in [2, 3, 4, 5, 6, 7] to determine $\lambda_{2}$, $\eta$, $\lambda_{1}$ such that the above inclusion relations \mathcal{Q}_{1}\subset\mathcal{P}\subset\mathcal{Q}_{2}\subset\underaccent{\bar}{\mathcal{R}} hold:

1. $(i)$

Determine $\lambda_{2}$ The hypersurface $Q_{2}$ intersects the $u_{j}$-axis: $\{(u_{i})_{i=1}^{n}\,|\,u_{i}=0,\,\forall i\neq j\}$ at the point

[TABLE]

If u_{2,j}\leq\underaccent{\bar}{u}_{j}, $\forall j=1,\cdots,n$, then by the monotonicity of the function $\ln(\cdot+k)$, \mathcal{Q}_{2}\subset\underaccent{\bar}{\mathcal{R}}. That is, \mathcal{Q}_{2}\subset\underaccent{\bar}{\mathcal{R}} if $\lambda_{2}$ is chosen as in (6c):

[TABLE] 2. $(ii)$

Determine $\eta$ As above, the hypersurface $P$ intersects the $u_{j}$-axis at

[TABLE]

If $u_{0,j}\leq u_{2,j}$, $\forall j=1,\cdots,n$, then $\mathcal{P}\subset\mathcal{Q}_{2}$ and the hypersurface $Q_{2}$ is above the hypersurface $P$. That is, $\mathcal{P}\subset\mathcal{Q}_{2}$ if $\eta$ is chosen as in (6b):

[TABLE] 3. $(iii)$

Determine $\lambda_{1}$ Replacing $\lambda_{2}$ by $\lambda_{1}$ in step $(i)$, the $u_{j}$-intercept of the hypersurface $Q_{1}$ is given by $u_{1,j}=e^{\frac{\lambda_{1}-\sum_{i=1,i\neq j}^{n}\alpha_{i}\,d_{i}\ln k_{i}}{\alpha_{j}\,d_{j}}}-k_{j}$. Hence if we take $\lambda_{1}$ as in (6a):

[TABLE]

then $u_{1,j}\leq u_{0,j}$, $\forall j=1,\cdots,n$ and hence $\mathcal{Q}_{1}\subset\mathcal{P}$.

We now show $q(x)\geq\lambda_{1}$, $x\in\mathbb{R}$ by a contradiction argument. Suppose by contradiction that there exists $z\in\mathbb{R}$ such that $q(z)<\lambda_{1}$. Since $u_{i}(x)\in C^{2}(\mathbb{R})$ $(i=1,\cdots,n)$ and $(u_{1},u_{2},\cdots,u_{n})(\pm\infty)=\textbf{e}_{\pm}$, we may assume $\displaystyle\min_{x\in\mathbb{R}}q(x)=q(z)$. We denote respectively by $z_{2}$ and $z_{1}$ the first points at which the solution trajectory $\{(u_{i}(x))_{i=1}^{n}\,|\,x\in\mathbb{R}\}$ intersects the hypersurface $Q_{2}$ when $x$ moves from $z$ towards $\infty$ and $-\infty$. For the case where $\theta\leq 0$, we integrate (18) with respect to $x$ from $z_{1}$ to $z$ and obtain

[TABLE]

We also have the following facts from the construction of the hypersurfaces $Q_{1},Q_{2},P$:

• •

$q^{\prime}(z)=0$ because of $\displaystyle\min_{x\in\mathbb{R}}q(x)=q(z)$;

• •

$q(z_{1})=\lambda_{2}$ because of $(u_{i}(z_{1}))_{i=1}^{n}\in Q_{2}$.

• •

$q^{\prime}(z_{1})<0$ because $z_{1}$ is the first point for $q(x)$ taking the value $\lambda_{2}$ when $x$ moves from $z$ to $-\infty$, such that $q(z_{1}+\delta)<\lambda_{2}$ for $z-z_{1}>\delta>0$;

• •

$p(z)<\eta$ since $(u_{i}(z))_{i=1}^{n}$ is below the hypersurface $P$;

• •

$p(z_{1})>\eta$ since $(u_{i}(z_{1}))_{i=1}^{n}$ is above the hypersurface $P$;

• •

$F(u_{1}(x),\cdots,u_{n}(x))=\displaystyle\sum_{i=1}^{n}\frac{\alpha_{i}\,u_{i}^{l_{i}}}{u_{i}+k_{i}}\,f_{i}(u_{1},\cdots,u_{n})\geq 0$, $\forall x\in[z_{1},z]$. Indeed, since (u_{i}(z_{1}))_{i=1}^{n}\in Q_{2}\subset\mathcal{Q}_{2}\subset\underaccent{\bar}{\mathcal{R}} and (u_{i}(z))_{i=1}^{n}\in\mathcal{Q}_{1}\subset\underaccent{\bar}{\mathcal{R}}, we derive that $F(u_{1}(x),\cdots,u_{n}(x))|_{x\in[z_{1},z]}\geq 0$ by the hypothesis (4).

We hence have the following inequality from the above facts when $\theta\leq 0$

[TABLE]

which contradicts (19). Therefore when $\theta\leq 0$, $q(x)\geq\lambda_{1}$ for $x\in\mathbb{R}$. For the case where $\theta\geq 0$, we simply integrate (18) with respect to $x$ from $z$ to $z_{2}$ to arrive at

[TABLE]

Then we apply the facts that $q^{\prime}(z_{2})>0$, $q^{\prime}(z)=0$, $p(z_{2})>\eta$, $p(z)<\eta$ and $F(u_{1}(x),\cdots,u_{n}(x))|_{x\in[z,z_{2}]}\geq 0$, as well as a similar contradiction argument as above, to derive $q\geq\lambda_{1}$.

∎

Proof of \threfprop: upper bed.

We prove \threfprop: upper bed in a similar manner to the proof of \threfprop: lower bed. We first rewrite the inequality $d_{i}(u_{i})^{\prime\prime}+\theta(u_{i})^{\prime}+u_{i}^{l_{i}}f_{i}\geq 0$ in (7). A straightforward calculation shows

[TABLE]

Hence we multiply the inequality by $m_{i}\,u^{m_{i}-1}(x)$ to arrive at

[TABLE]

For notational simplicity, we will adopt the same notations as in the proof of \threfprop: lower bed. Since $u_{i}\geq 0$, $\forall i=1,\cdots,n$, for any $(m_{i})_{i=1}^{n}\in([1,\infty))^{n}$, the vector field $(U_{i})_{i=1}^{n}:=(u_{i}^{m_{i}})_{i=1}^{n}$ satisfies the following inequalities

[TABLE]

For any $(\alpha_{i})_{i=1}^{n}\in(\mathbb{R}^{+})^{n}$, $p(x)=\sum_{i=1}^{n}\alpha_{i}\,U_{i}$ and $q(x)=\sum_{i=1}^{n}\alpha_{i}\,d_{i}U_{i}$ satisfy

[TABLE]

We are going to show the upper bound $q\leq\lambda_{1}$ by employing the N-barrier method as in the proof of Proposition LABEL:prop:_lower_bed. That is, we are going to construct the three hyperellipsoids

[TABLE]

such that the following inclusion relations hold:

[TABLE]

and the upper bound $q\leq\lambda_{1}$ follows by a contradiction argument. More precisely, we take

[TABLE]

such that the $u_{j}$-intercept of the hyperellipsoid $Q_{2}$

[TABLE]

Then we take

[TABLE]

such that the $u_{j}$-intercept of the hyperellipsoid $P$

[TABLE]

Finally we take

[TABLE]

such that the $u_{j}$-intercept of the hyperellipsoid $Q_{1}$

[TABLE]

Combining (22), (23), and (24), we have

[TABLE]

We follow exactly the same contradiction argument to prove $q(x)\leq\lambda_{1}$ for $x\in\mathbb{R}$ as in the proof of \threfprop: lower bed, which is omitted here. Since $(\alpha_{i})_{i=1}^{n}\in(\mathbb{R}^{+})^{n}$ is arbitrary, $q(x)=\displaystyle\sum_{i=1}^{n}\alpha_{i}\,d_{i}(u_{i}(x))^{m_{i}}\leq\lambda_{1}$ implies the upper bound (9). Now we use the inequality of arithmetic and geometric means to obtain

[TABLE]

which together with (9) yields (10).

∎

Acknowledgements

The authors are grateful to the anonymous referees for many helpful comments and valuable suggestions on this paper. L.-C. Hung thanks for the hospitality he received from KIT while visiting KIT. The research of L.-C. Hung is partly supported by the grant 106-2115-M-011-001-MY2 of Ministry of Science and Technology, Taiwan.