## Positive Solutions of a Predator-Prey Model with Cross Diffusion

Yuan Hailong,1,2, Wang Yuping1, Li Yanling3

 基金资助: 国家自然科学基金.  11271236国家自然科学基金.  61672021国家自然科学基金.  61872227陕西科技大学博士科研启动基金.  2017BJ-44

 Fund supported: the NSFC.  11271236the NSFC.  61672021the NSFC.  61872227the Natural Science Foundation of Shaanxi University of Science and Technology.  2017BJ-44

Abstract

A predator-prey model with cross diffusion under homogeneous Dirichlet boundary conditions is investigated. Firstly, the existence of positive solutions can be established by the Leray-Schauder degree theory. Secondly, we consider that the existence of positive solutions of the regular perturbation system and the singular perturbation system when m=β is sufficiently large, respectively, and moreover, we show that the positive solutions of the singular perturbation system will blow up along the continuum at a* by the bifurcation theory. Finally, the multiplicity results of positive solutions of system is also considered.

Keywords： Cross diffusion ; Bifurcation ; Positive solutions

Yuan Hailong, Wang Yuping, Li Yanling. Positive Solutions of a Predator-Prey Model with Cross Diffusion. Acta Mathematica Scientia[J], 2019, 39(3): 545-559 doi:

## 1 引言

$$$\left\{\begin{array}{ll} -\Delta[(1+\alpha v) u] = u \Big(a-u-\frac{cv}{1+mu}\Big), & x \in \Omega, \\ -\Delta \Big [\Big(\mu + \frac{1}{1 + \beta u}\Big)v\Big ] = v\Big (b - v + \frac{du}{1+mu}\Big), \quad & x \in \Omega, \\ u = v = 0, & x\in\partial\Omega, \end{array}\right.$$$

$\alpha = \beta = 0$时,则系统(1.1)变成了经典的捕食-食饵模型,其已经被许多生物数学家所研究.特别地, Du和Lou[4]讨论了当参数$m$不太小时,系统(1.1)正解的存在性、唯一性与多解性.接着,文献[5]证明:在合适的条件下,该系统存在Hopf分歧解的发生,且至少存在3个非退化的正解.对于其他的模型的研究,可参见文献[6-11, 25]及其参考文献.

$$$- \mu \Delta u = u (a - u), \; u|_{\partial \Omega} = 0$$$

## 2 预备工作

$$$U = (1+\alpha v) u, \; V = \Big(\mu + \frac{1}{1 + \beta u}\Big)v,$$$

$$$\left\{\begin{array}{ll} -\Delta U = u\Big(a-u-\frac{cv}{1+mu}\Big), & x \in \Omega, \\ -\Delta V = v\Big(b-v+\frac{du}{1+mu}\Big), \quad & x \in \Omega, \\ U = V = 0, & x\in\partial\Omega. \end{array}\right.$$$

(ⅰ) $a^{*}(b)$关于参数$b \in [\mu\lambda_{1}, +\infty)$是严格单调递增的;

(ⅱ) $a^{*}(\mu\lambda_{1}) = \lambda_{1}; \lim\limits_{b \rightarrow +\infty} a^{*}(b) = \infty$.

(ⅰ) $a^{**}(b)$关于参数$b \in [(\mu+1)\lambda_{1}, +\infty)$是严格单调递增的;

(ⅱ) $a^{**}((\mu+1)\lambda_{1}) = \lambda_{1}; \lim\limits_{b \rightarrow +\infty} a^{**}(b) = \infty$.

(ⅰ) $L $$W_{y_{0}} 上具有 \alpha 性质,则 {\rm index}_{W} (A, y_{0}) = 0 ; (ⅱ) L$$ W_{y_{0}}$不上具有$\alpha$性质,则${\rm index}_{W} (A, y_{0}) = {\rm index}_{E}(L, 0) = (-1)^{\sigma}$,其中, $\sigma $$L 大于1的特征值的代数重数之和. ## 3 正解的存在性 在本节中,我们将通过Leray-Schauder度理论来建立系统(2.2)的正解的存在性.首先,我们给出下列记号: E = C_{0} (\bar{\Omega}) \times C_{0} (\bar{\Omega}) . P = K \times K, \; \mbox{其中}, \; K = \{U \in C_{0} (\bar{\Omega}): U (x) \geq 0\; \mbox{当}\; x \in \bar{\Omega} \} . D = \{(U, V) \in P : U < M (a, \alpha) + 1, \; V < (\mu + 1)(b + \frac{d}{m}) + 1\} . 我们定义正的紧算子 A_{t} : \bar{D} \rightarrow E 其中, p 是充分大的数满足 A_{t} 是正的紧算子 为了计算算子 A_{t} 的不动点指标,我们在此给出算子 A_{t} 的Fréchet导数.令 Q = (1+\alpha v)(\mu + \frac{1}{1+\beta u}) + \frac{\alpha \beta u v}{(1 + \beta u)^{2}} ,则 根据(2.1)式,我们有 进一步,我们有 因此 现在,我们建立算子 A 的不动点指标. 引理3.1 假设 a > \lambda_{1} ,则 (ⅰ) deg_{P} (I - A, D) = 1 ; (ⅱ) ind_{P} (A, (0, 0)) = 0 如果 b \not = (\mu+1) \lambda_{1} ; (ⅲ) ind_{P} (A, (\theta_{a}, 0)) = 0 如果 \lambda_{1} (- \frac{(b(1 + m \theta_{a}) + d \theta_{a}) (1 + \beta \theta_{a})}{(1 + m \theta_{a}) (\mu (1 + \beta \theta_{a}) + 1)}) < 0 ; (ⅳ) ind_{P} (A, (\theta_{a}, 0)) = 1 如果 \lambda_{1} (- \frac{(b(1 + m \theta_{a}) + d \theta_{a}) (1 + \beta \theta_{a})}{(1 + m \theta_{a}) (\mu (1 + \beta \theta_{a}) + 1)}) > 0 .进一步,如果 b > (\mu + 1) \lambda_{1} ,则 (ⅴ) ind_{P} (A, (0, (\mu + 1)^{2} \theta_{b/(\mu + 1)})) = 0 如果 \lambda_{1} (\frac{c (\mu + 1) \theta_{b/(\mu + 1)} -a}{1 + \alpha (\mu + 1) \theta_{b/(\mu + 1)}}) < 0 ; (ⅵ) ind_{P} (A, (0, (\mu + 1)^{2}\theta_{b/(\mu + 1)})) = 1 如果 \lambda_{1} (\frac{c (\mu + 1)\theta_{b/(\mu + 1)} -a}{1 + \alpha (\mu + 1) \theta_{b/(\mu + 1)}}) > 0 . (ⅰ)显然,通过引理2.1我们知道算子 A_{t} 在边界 \partial D 无不动点.因此,我们知道Leray-Schauder度 deg_{P}(I - A_{t}, D) 有定义,且我们有 特别地 经过简单计算我们有 r(A_{0}'(0, 0)) < 1 .因此,算子 I - A_{0}'(0, 0)$$ \bar{P}_{(0, 0)}$是可逆的且算子$A_{0}'(0, 0) $$\bar{P}_{(0, 0)}$$ \alpha$性质.因此, $ind_{P} (A_{0}, (0, 0)) = 1$.也就是说, $deg_{P} (I - A, D) = 1$.

(ⅱ)显然,我们有$A'(0, 0) (U, V) = (-\Delta + p I)^{-1} ((p + a) U, (p + \frac{b}{\mu + 1}) V)$,假设存在一些$(U, V) \in \bar{P}_{(0, 0)}$使得$A'(0, 0) (U, V) = (U, V)$成立,则$- \Delta U = a U, \; U|_{\partial \Omega} = 0$.如果$U > 0$,则$a = \lambda_{1}$,矛盾.因此, $U \equiv 0$.类似地, $V \equiv 0$.从而, $I - A'(0, 0) $$\bar{P}_{(0, 0)} 是可逆的. 由于 a > \lambda_{1} ,从而 r_{a} = r [(- \Delta + p)^{-1}(a + p)] > 1 ,且 r_{a} 是算子 (-\Delta + p)^{-1}(a + p) 的主特征值,其对应的主特征函数 U > 0 . t_{0} = r_{a}^{-1} ,则 0 < t_{0} < 1$$ (I - t_{0} A'(0, 0)) (U, 0) = (0, 0) \in S_{(0, 0)}$.从而, $A'(0, 0) $$\alpha 性质.因此, ind_{P} (A, (0, 0)) = 0 . (ⅲ)经过简单的计算我们有 假设存在一些 (U, V) \in \bar{P}_{(\theta_{a}, 0)} 使得 A'(\theta_{a}, 0) (U, V) = (U, V) 成立,则 如果 V \geq\not\equiv 0 ,则 \lambda_{1} (- \frac{(b(1 + m \theta_{a}) + d \theta_{a}) (1 + \beta \theta_{a})}{(1 + m \theta_{a}) (\mu (1 + \beta \theta_{a}) + 1)}) = 0 .由于 \lambda_{1} (- \frac{(b(1 + m \theta_{a}) + d \theta_{a}) (1 + \beta \theta_{a})}{(1 + m \theta_{a}) (\mu (1 + \beta \theta_{a}) + 1)}) < 0 ,从而 V \equiv 0 ,进而 U \equiv 0 .因此, (U, V) = (0, 0) .也就是说, I - A'(\theta_{a}, 0)$$ \bar{P}_{(\theta_{a}, 0)}$是可逆的.

$r_{b} = r(L) > 1$是算子$L$的特征值,且其对应的特征函数$V > 0$.$t_{0} = r_{b}^{-1}$,则$t_{0} \in (0, 1)$.由于$(0, V) \in \bar{P}_{(\theta_{a}, 0)} \setminus S_{(\theta_{a}, 0)}$,我们有

我们采用反证法.假设当$a > \lambda_{1} (c \frac{(\mu + 1)^{2}}{\mu} \theta_{b/\mu})$时,系统(4.2)有正解$(w, v)$,则

$\phi = (1 + \alpha v) w$,则$w = \frac{\phi}{1 + \alpha v}$.因此

$$$-\Delta [(1 + \alpha v_{i}) \tilde{w}_{i}] = \tilde{w}_{i}\Big (a_{i} - \frac{cv_{i}}{1 + w_{i}}\Big), \; \tilde{w}_{i}|_{\partial \Omega} = 0.$$$

$\phi_{i} = (1 + \alpha v_{i}) \tilde{w}_{i}$,则$\tilde{w}_{i} = \frac{\phi_{i}}{1 + \alpha v_{i}}$.因此,系统(4.3)可以改写为

(ⅲ)存在$a = \hat{a}, Z \in \overline{Y} \setminus \{0\}$满足$(\hat{a}; Z) \in C$,其中, $\overline{Y} $$\{(- \Delta \phi^{**}, 0)\} 的补空间,其中, \{(- \Delta \phi^{**}, 0)\} 由定理4.1给出. 如果 C - \{ (a^{**}; 0, (\mu + 1)^{2} \theta_{b/(\mu + 1)}) \} \not\subset P_{0} ,则存在 (\tilde{a}; \tilde{W}, \tilde{V}) \in \{C - (a^{**}; 0, (\mu + 1)^{2} \theta_{b/(\mu + 1)}) \}$$ \cap \partial P_{0}$是下列序列的极限$\{ (\tilde{a}_{i}; \tilde{W}_{i}, \tilde{V}_{i}) \} \subset C \cap P_{0}, \; \tilde{W}_{i} > 0, \; \tilde{V}_{i} > 0 $$\Omega .由于 (\tilde{a}; \tilde{W}, \tilde{V}) \in \partial P_{0} ,我们知道存在点 x_{0} \in \Omega 使得 \tilde{W}(x_{0}) = 0 或存在点 x_{0} \in \partial \Omega 使得 \frac{\partial \tilde{W}}{\partial \upsilon} (x_{0}) = 0 .根据强最大值原理,我们知道 \tilde{W} \equiv 0$$ \Omega$上成立.类似地,对于其他情形我们有$\tilde{V} \equiv 0 $$\Omega 上成立. 假设 \tilde{W} \equiv 0, \tilde{V} = (\mu + 1)^{2} \theta_{b/(\mu + 1)} ,令 \tilde{W}_{i}^{*} = \frac{\tilde{W}_{i}}{\|\tilde{W}_{i}\|_{\infty}} ,则 根据二阶椭圆型正则化理论我们可以假设 \tilde{W}_{i}^{*} \rightarrow \tilde{W}^{*} , \tilde{V}_{i} \rightarrow \tilde{V}$$ C^{1}$空间上成立且$\|\tilde{W}^{*}\|_{\infty} = 1$, $\tilde{v}_{i} \rightharpoonup \tilde{v} $$L^{p} 空间弱收敛成立.在上述方程两边同时取极限,我们有 因此, a = a^{**} ,矛盾. 假设 \tilde{W} > 0, \tilde{V} \equiv 0 ,令 \tilde{V}_{i}^{*} = \frac{\tilde{V}_{i}}{\|\tilde{V}_{i}\|_{\infty}} ,则 根据二阶椭圆型正则化理论我们可以假设 \tilde{W}_{i} \rightarrow \tilde{W} , \tilde{V}_{i}^{*} \rightarrow \tilde{V}^{*}$$ C^{1}$空间成立且满足$\| \tilde{V}^{*} \|_{\infty} = 1$, $\tilde{w}_{i} \rightharpoonup \tilde{w} $$L^{p} 空间弱收敛成立.在上述方程两边同时取极限,我们有 \tilde{W} = s \phi_{1} ,其中, s \geq 0 .因此, \lambda_{1} (- b \frac{1 + \beta s \phi_{1}}{\mu (1 + \beta s \phi_{1}) + 1}) = 0 ,矛盾. 假设 (\tilde{W}, \tilde{V}) \equiv (0, 0) ,类似地,我们可以得到矛盾. 因此,我们必有 C - \{ (a^{**}; 0, (\mu + 1)^{2} \theta_{b/(\mu + 1)}) \} \subset P_{0} .从而,情形(ⅰ)不会发生.由于 \phi^{**} > 0$$ \Omega$上成立,补空间$\overline{Y}$不可能包含不变号的元素,从而,情形(ⅲ)也不会发生.因此,情形(ⅱ)必然发生.也就是说,解曲线$C $$R \times X 空间延伸至无穷.根据引理4.3,我们知道参数 a 是有界的,从而我们有 \lim\limits_{i \rightarrow \infty} \|W_{i}\|_{\infty} = \infty .现在,我们证明: \lim\limits_{i \rightarrow \infty} a_{i} = a^{*} .显然, \lim\limits_{i \rightarrow \infty} v_{i} = \mu \theta_{b/\mu} ,因此, \lim\limits_{i \rightarrow \infty} V_{i} = \mu^{2} \theta_{b/\mu} . W_{i}^{*} = \frac{W_{i}}{\|W_{i}\|_{\infty}} ,则 根据二阶椭圆型正则化理论我们可以假设 W_{i}^{*} \rightarrow W^{*}$$ C^{1}$空间成立且$\frac{1}{1 + \alpha v_{i} + W_{i}} \rightharpoonup h $$L^{2} 空间弱收敛成立.在上述方程两边同时取极限,我们有 根据强最大值原理,我们有 W^{*} > 0$$ \Omega$上成立.因此, $h = 0, -\Delta W^{*} = \frac{a}{1 + \alpha \mu \theta_{b/\mu}} $$W^{*} .从而, \lambda_{1} (- \frac{a}{1 + \alpha \mu \theta_{b/\mu}}) = 0 .也就是说, a = a^{*} .证毕. 下面的定理表明:当 a \in [a^{*} + \epsilon, a^{**}), b > (\mu + 1) \lambda_{1}$$ m = \beta$充分大时,则系统(1.1)有两种类型的正解.

我们采用反证法.假设存在$m_{i} \rightarrow \infty$, $a_{i} \rightarrow a \in [a^{*} + \epsilon, a^{**}]$且当$(a, m) = (a_{i}, m_{i})$时,设$(u_{i}, v_{i})$是系统(1.1)的解使得$(u_{i}, v_{i})$满足$\|u_{i} - \tilde{\theta}_{a, \alpha} \|_{C^{1}} + \|v_{i} - \mu \theta_{b/\mu} \|_{C^{1}} \geq \delta$,且$\|m_{i} u_{i} - \tilde{w}\|_{C^{1}} + \|v_{i} - \tilde{v}\|_{C^{1}} \geq \delta$.这里有两种情形我们需要考虑:

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