本文研究了由Boulanouar在文献[1-2]中提出来的以细菌的成熟度和成熟速度为特征的另一类具结构化的细菌种群模型

(1.1) $ \begin{equation} \frac{\partial\psi(u, v, t)}{\partial t} = -h(v)\frac{\partial\psi(u, v, t)} {\partial u}-\sigma(u, v)\psi(u, v, t) +\int_{a}^{b}r(u, v, v')\psi(u, v', t){\rm d}v'. \end{equation} $

其中$ \psi(u, v, t) $表示由细菌成熟度$ u\in(0, 1) $和细菌成熟速度$ v\in(a, b) $在时间$ t $构成的细菌密度函数, $ h(v) $表示细菌成熟速度的权重因子, $ r(u, v, v') $表示细菌成熟速度从$ v' $到$ v $改变时的转变速率, $ \sigma(u, v) $为总转变截面,且

(1.2) $ \begin{equation} \sigma(u, v) = \int_{a}^{b}r(u, v', v){\rm d}v'. \end{equation} $

在生物学上,每一有丝分裂时,子细菌被看成种群细菌的一部分,它们之间存在相互关系$ k(u, v, v') $,在数学上表示为下列具转变规则的边界条件

(1.3) $ \begin{equation} \psi(0, v) = \frac{\alpha}{h(v)}\int^{b}_{a}\int^{1}_{0}k(u', v, v') \psi(u', v', t){\rm d}u'{\rm d}v'. \end{equation} $

这里常数$ \alpha\geq0 $表示每一有丝分裂子细菌的平均数.

文献[1-2]分别讨论了该模型(1.1)–(1.3)相应的迁移算子生成正不可约$ C_0 $半群和它的渐近行为等.之后关于这类模型的研究工作较少,文献[3-4]对该模型在更一般的边界条件下进行了讨论,文献[3]得到了该模型生成半群的Dyson-phillips展式的9阶余项$ R_9(t) $在$ L^1 $空间上是弱紧的和在$ L^{p}(1< p<+\infty) $空间上是紧的,从而获得了该迁移算子的谱在某右半平面上仅由有限个具有限代数重数的离散本征值组成等结果;文献[4]讨论了该模型生成正不可约$ C_0 $半群和它的解在一致拓扑意义下的渐近行为等.但是对这类具总转变规则的结构化细菌种群模型的迁移算子生成$ C_0 $半群的紧性和它的谱分析如何?目前还未见研究成果.本文在$ L^1 $空间上对这类模型进行研究,讨论了该模型相应的迁移算子生成正不可约$ C_0 $半群的弱紧性,得到了该迁移算子的谱仅由至多可数个具有限代数重数的离散本征值组成, $ -\infty $是唯一可能的聚点以及该模型的解在一致算子拓扑意义下的渐近行为,从而给出了该细菌种群的异步生长特性等结果.

设$ X = L^1(\Omega)(\Omega = (0, 1)\times(a, b) = I\times J, \; 0\leq a<b\leq \infty) $和索伯列夫空间

$ W = \bigg\{\psi\in X|h(v)\frac{\partial\psi}{\partial u}\in X\bigg\} $

及迹空间$ Y = L^1(J, h(v){\rm d}v) $,它们分别按范数

$ \|\psi\| = \int_{\Omega}|\psi(u, v)|{\rm d}u{\rm d}v = \int_a^b\int_0^{1}|\psi(u, v)|{\rm d}u{\rm d}v $

和

$ \|\varphi\|_{W} = \|\varphi\|+\|h(v)\frac{\partial\varphi}{\partial u}\| $

及

$ \|\varphi\|_{Y} = \int_a^{b}|\varphi(v)|h(v){\rm d}v $

构成Banach空间,边界空间为

$ X^{0} = L^{1} (\{0\}\times(a, b);h(v){\rm d}v);\; \; X^{1} = L^{1} (\{1\}\times(a, b);h(v){\rm d}v), $

其中$ h(v) $为有界可测函数.假设$ (A_{1}) $

$ \overline{k} = {\rm ess}\sup\bigg\{\int_a^b|k(u', v, v')|{\rm d}v:(u', v')\in\Omega\bigg\}<\infty; \; \; k(u, v, v')>0, \; (u, v, v')\in \Omega\times J, \ {\rm a.e.;} $

$ \overline{k}(v) = {\rm ess}\sup\{|k(u', v, v')|:(u', v')\in\Omega\}\in L^1(J). $

设$ (T(t))_{t\geq 0} $是Banach空间$ X $上的$ C_0 $半群, $ B $是它的母元,半群$ (T(t))_{t\geq 0} $的增长界和本质增长界分别定义为^[2]

(1.4) $ \begin{equation} \omega(T(t)) = \lim\limits_{t \rightarrow \infty}\frac{\ln\|T(t)\|}{t};\; \; \; \; \omega_{{\rm ess}}(T(t)) = \lim\limits_{t \rightarrow \infty}\frac{\ln\|T(t)\|_{{\rm ess}}}{t}, \end{equation} $

且有

(1.5) $ \begin{equation} \omega_{{\rm ess}}(T(t))\leq \omega(T(t));\; \; \; \; r_{{\rm ess}}(T(t)) = e^{\omega_{{\rm ess}}(T(t))t}, \; \; \forall t\geq 0. \end{equation} $

注意$ \|C\|_{{\rm ess}} = 0 $当且仅当$ C $是紧算子.母元$ B $的谱界$ s(B) $定义为

(1.6) $ \begin{equation} s(B) = \left\{ \begin{array}{ll} {\rm sup}\{{\rm Re}\lambda: \lambda\in \sigma(B)\}, &{\rm if}\; \sigma(B)\neq \emptyset, \\ -\infty, &{\rm if}\; \sigma(B) = \emptyset. \end{array} \right. \end{equation} $

当$ X = L^p(1\leq p<\infty) $,则^{[2, 5]}

(1.7) $ \begin{equation} \omega(T(t)) = s(B). \end{equation} $

引理1.1^{[2, $ \rm Lemma $1.1]}  设$ (T(t))_{t\geq 0} $是$ \rm Banach $格空间$ X $上的正不可约强连续半群,且有$ \omega_{{\rm ess}}(T(t))<\omega(T(t)) $,则存在$ X $上秩1投影算子$ P $和$ \varepsilon>0 $,使得$ \forall \eta\in(0, \varepsilon) $,存在$ M(\eta)\geq 1 $,有

$ \parallel e^{-\omega(T(t))t}T(t)-P\parallel_{L(X)}\leq M(\eta)e^{-\eta t}, \; \; t\geq 0. $

引理1.2^{[2, $ \rm Lemma $2.1]}  设$ S $和$ T $是$ X $上的有界算子,

1)若$ S $和$ T $是弱紧算子,则$ ST $是紧算子;

2)若$ T $是弱紧算子和$ 0\leq S\leq T $,则$ S $也是弱紧算子;

3)若$ T $是弱紧算子和$ 0\leq |S|\leq T $,则$ S $也是弱紧算子.

引理1.3^{[2, $ \rm Lemma $2.2]}  设$ (T(t))_{t\geq 0} $和$ (U(t))_{t\geq 0} $是$ X $上的强连续半群,若对某$ t_0>0 $, $ T(t_0)-U(t_0) $是弱紧算子,则$ \omega_{{\rm ess}}U(t) = \omega_{{\rm ess}}T(t) $.

引理1.4^{[2, $ \rm Lemma $2.3]}  设$ A $和$ A+B $是$ X $上的强连续半群$ (T(t))_{t\geq 0} $和$ (U(t))_{t\geq 0} $的生成元, $ B $是$ X $上的有界算子.若$ BT(t)B $是弱紧算子,则$ \omega_{{\rm ess}}U(t) = \omega_{{\rm ess}}T(t) $.

定义下列边界算子

(1.8) $ \begin{equation} H_{\alpha}\varphi = \frac{\alpha}{h(v)}\int_0^1\int_{a}^{b}k(u', \cdot, v')\varphi(u', v'){\rm d}u'{\rm d}v'. \end{equation} $

设

(1.9) $ \begin{equation} \left\{ \begin{array}{ll} { } T_{\alpha} \varphi(u, v) = -h(v)\frac{\partial\varphi(u, v)}{\partial u}, \\ { } D(T_{\alpha}) = \bigg\{\varphi\in X : h(v)\varphi\in X, h(v)\frac{\partial\varphi}{\partial u}\in X;\varphi(0, \cdot) = H_{\alpha}\varphi\bigg\}. \end{array} \right. \end{equation} $

引理1.5^{[2, $ \rm Lemma 3.1;\;Lemma 3.2 $]}  假设$ (A_1) $成立,则

1) $ H_{\alpha} $是$ X $到$ Y $上的有界正算子;

2)算子$ T_{\alpha} $在$ X $上产生一个正$ C_0 $半群$ U_{\alpha}(t) $,且

(1.10) $ \begin{equation} \|U_{\alpha}(t)\|_{L(X)}\leq e^{(\alpha \overline{k})t}, \; \; 0\leq U_{0}(t)\leq U_{\alpha}(t), \; \; t\geq 0 \end{equation} $

和

(1.11) $ \begin{equation} U_{\alpha}(t) = U_{0}(t)+B_{\alpha}(t)+D_{\alpha}(t), \; \; t\geq 0, \end{equation} $

其中

(1.12) $ \begin{equation} U_{0}(t)\varphi(u, v) = \chi(u, h(v), t)\varphi(u-th(v), v), \end{equation} $

(1.13) $ \begin{equation} B_{\alpha}(t)\varphi(u, v) = \alpha\frac{\xi (u, h(v), t)}{h(v)}\int_{0}^{1}\int_{a}^{b}k(u', v, v')U_{0}(t-\frac{u}{h(v)})\varphi(u', v'){\rm d}u'{\rm d}v', \end{equation} $

(1.14) $ \begin{eqnarray} D_{\alpha}(t)\varphi(u, v) & = &\alpha^2\frac{\xi (u, h(v), t)}{h(v)}\int_{0}^{1}\int_{a}^{b}\int_{0}^{1}\int_{a}^{b} \xi (u', h(v'), (t-\frac{u}{h(v)})){}\\ && \times \frac{k(u', v, v')k(u'', v', v'')}{h(v')} U_{\alpha}(t-\frac{u}{h(v)}-\frac{u'}{h(v')})\varphi(u'', v''){\rm d}u''{\rm d}v''{\rm d}u'{\rm d}v', {\qquad} \end{eqnarray} $

对几乎所有$ (u, v)\in \Omega $,有

(1.15) $ \begin{equation} \chi(u, y, t) = \left\{ \begin{array}{ll} 1, \; \; \; \; {\rm if}\; ty\leq u, \\ 0, \; \; \; \; {\rm if}\; ty>u. \end{array} \right. \; \; \; \; \xi(u, y, t) = \left\{ \begin{array}{ll} 1, \; \; \; \; {\rm if}\; ty> u, \\ 0, \; \; \; \; {\rm if}\; ty\leq u. \end{array} \right. \end{equation} $

定义算子

(1.16) $ \begin{equation} S\varphi(u, v) = -\sigma(u, v)\varphi(u, v), \end{equation} $

(1.17) $ \begin{equation} B_{\alpha} = T_{\alpha}+S, \; \; \; D(B_{\alpha}) = D(T_{\alpha}). \end{equation} $

假设$ (A_{2}) $

(1.18) $ \begin{equation} \underline{r} = {\rm ess}\inf\limits_{(u, v)\in \Omega}\int_{a}^{b}|r(u, v, v')|{\rm d}v'; \; \overline{r} = {\rm ess}\sup\limits_{(u, v)\in \Omega}\int_{a}^{b}|r(u, v, v')|{\rm d}v'<\infty. \end{equation} $

则$ S $是$ X $上的有界算子,且根据扰动定理^[12]可知如下引理.

引理1.6^{[2, $ \rm Lemma $3.3]}  假设$ (A_1)(A_{2}) $成立,则算子$ B_{\alpha} $在$ X $上产生一个正$ C_0 $半群$ V_{\alpha}(t) $,且

(1.19) $ \begin{equation} 0\leq V_{\alpha}(t)\leq e^{-t\underline{r}}U_{\alpha}(t), \; \; t\geq 0 \end{equation} $

和

(1.20) $ \begin{equation} 0\leq V_{\alpha}(t)-V_{0}(t)\leq U_{\alpha}(t)-U_{0}(t), \; \; t\geq 0. \end{equation} $

引理1.7^{[2, $ \rm Corollary $3.5]}  假设$ (A_{1})(A_{2}) $成立,若$ \underline{h} = {\rm ess}\inf\{h(v):v\in J\}>0 $,则

(1.21) $ \begin{equation} \omega_{{\rm ess}}(V_0(t)) = -\infty. \end{equation} $

定义

$ P_{r}\varphi = \int_a^b r(u, v, v')\varphi(u, v', t){\rm d}v', $

则$ P_{r} $是$ X $上的有界算子,且迁移算子$ A_{\alpha} $定义为

(1.22) $ \begin{equation} A_{\alpha} = B_{\alpha}+P_{r}, \; \; \; \; D(A_{\alpha}) = D(B_{\alpha}). \end{equation} $

根据引理1.6和扰动定理^[6]知如下定理.

定理1.8^{[2, $ \rm Lemma $3.6]}  假设$ (A_{1}) $–$ (A_{2}) $成立,则算子$ A_{\alpha} $在$ X $上产生一个正$ C_0 $半群$ W_{\alpha}(t) $,且

(1.23) $ \begin{equation} -\overline{r}+\omega(U_{\alpha}(t))\leq \omega(W_{\alpha}(t)). \end{equation} $

若$ \alpha>0 $,则正半群$ (W_{\alpha}(t))_{t\geq 0} $是不可约的.

文献[1-2]对该模型证明其产生正不可约$ C_0 $半群和它的渐近行为,而本文在此基础上进一步证明了该迁移半群$ (W_{\alpha}(t))_{t\geq 0} $是弱紧的,从而获得了其相应的迁移算子的谱仅由至多可数个具有限代数重数的离散本征值组成, $ -\infty $是唯一可能的聚点等结果.为此先引入辅助算子

(2.1) $ \begin{equation} L_{\alpha, \lambda}\psi(v) = \frac{\alpha}{h(v)}\int_0^1\int_a^b e^{-\frac{\lambda u'}{h(v')}}k(u', v, v')\psi(v'){\rm d}u'{\rm d}v'. \end{equation} $

引理2.1^{[2, $ \rm Lemma $4.1]}  假设$ (A_{1}) $成立,则当$ \lambda>0 $时, $ L_{\alpha, \lambda} $是$ Y $上的有界算子,且

(2.2) $ \begin{equation} \| L_{\alpha, \lambda}\|_{L(Y)}\leq \frac{\alpha \overline{k}}{\lambda}. \end{equation} $

又若$ \alpha>0 $,则$ L_{\alpha, \lambda} $是$ Y $上的严格正算子.

定理2.2  假设$ (A_1)(A_{2}) $成立,若$ \alpha>0 $,则$ L_{\alpha, \lambda} $是$ Y $上的正弱紧算子.

证  设$ \lambda> 0 $,令$ \psi\in (Y)_{+} $,则对几乎所有的$ v\in (a, b) $,有

(2.3) $ \begin{eqnarray} |L_{\alpha, \lambda}\psi(v)|&\leq&\frac{\alpha}{h(v)} \int_a^b\int_0^1 e^{-\frac{\lambda u'}{h(v')}}|k(u', v, v')|\psi(v'){\rm d}u'{\rm d}v'{}\\ &\leq&\frac{\alpha \overline{k}(v)}{h(v)} \int_a^b \bigg[\int_0^1 e^{-\frac{\lambda u'}{h(v')}}{\rm d}u'\bigg]\psi(v'){\rm d}v'{}\\ & = &\frac{\alpha \overline{k}(v)}{h(v)} \int_a^b\bigg[\frac{h(v')}{\lambda} (1-e^{-\frac{\lambda}{h(v')}})\bigg]\psi(v'){\rm d}v'{}\\ &\leq&\frac{\alpha\overline{k}(v)}{\lambda h(v)}\int_0^1\int_a^b\psi(v')h(v'){\rm d}v'{\rm d}u. \end{eqnarray} $

因为

$ \bigg\|\frac{\alpha\overline{k}(v)}{\lambda h(v)}\bigg\|_{Y} = \frac{\alpha}{\lambda}\int_a^b\frac{\overline{k}(v)}{h(v)}h(v){\rm d}v = \frac{\alpha}{\lambda}\|\overline{k}\|_{Y}<\infty. $

所以(2.3)式的右边是秩1算子,从而知$ L_{\alpha, \lambda} $是$ Y $上的正弱紧算子.

定理2.3  假设$ (A_1)(A_{2}) $成立,若$ \alpha >0 $, $ {\rm Re}\lambda>0 $,则

(2.4) $ \begin{equation} \lambda\in\sigma(T_{\alpha})\Rightarrow 1\in\sigma_p(L^2_{\alpha, \lambda}). \end{equation} $

证  设$ 1\in \rho(L^2_{\alpha, \lambda}) $,则$ \forall g\in Y $,由引理1.5和定理2.1,有

$ (I_{Y}+L_{\alpha, \lambda})H_{\alpha}(\lambda-T_{0})^{-1}g\in Y, $

则存在唯一$ \psi_{\lambda}\in Y $,使得

$ \psi_{\lambda} = L^{2}_{\alpha, \lambda}\psi_{\lambda}+(I_{Y}+L_{\alpha, \lambda})H_{\alpha}(\lambda-T_{0})^{-1}g. $

从而

$ (I_{Y}+L_{\alpha, \lambda})(\psi_{\lambda}-L_{\alpha, \lambda}\psi_{\lambda}-H_{\alpha}(\lambda-T_{0})^{-1}g) = 0. $

所以

(2.5) $ \begin{equation} \psi_{\lambda} = L_{\alpha, \lambda}\psi_{\lambda}+H_\alpha(\lambda-T_{0})^{-1}g. \end{equation} $

令

(2.6) $ \begin{equation} \varphi_{\lambda}(u, v) = \theta(u, v)\psi_{\lambda}(v)+(\lambda-T_{0})^{-1}g, \end{equation} $

其中$ \theta(u, v) = e^{-\frac{\lambda u}{h(v)}} $,则$ \varphi_{\lambda}\in D(T_{\alpha}) $.事实上,因为

$ \begin{eqnarray*} \parallel\varphi_{\lambda}\parallel_{X}&\leq& \int_a^b\bigg[\int_0^1e^{-\frac{\lambda u}{h(v)}}{\rm d}u\bigg] |\psi_{\lambda}(v)|{\rm d}v+\|(\lambda-T_0)^{-1}g\|_{X}\\ & = &\int_a^b\frac{h(v)}{\lambda}(1-e^{-\frac{\lambda}{h(v)}}) |\psi_{\lambda}(v)|{\rm d}v+\|(\lambda-T_0)^{-1}g\|_{X}\\ &\leq&\frac{1}{\lambda}\int_a^b{h(v)} |\psi_{\lambda}(v)|{\rm d}v+\|(\lambda-T_0)^{-1}g\|_{X}\\ &\leq&\frac{1}{\lambda}\|\psi_{\lambda}\|_{Y}+\frac{1}{\lambda}\|g\|_{X}<\infty. \nonumber \end{eqnarray*} $

类似地,有

$ \begin{eqnarray*} \parallel h\varphi_{\lambda}\parallel_{X}&\leq &\int_a^b\int_0^1h(v)e^{-\frac{\lambda u}{h(v)}} |\psi_{\lambda}(v)|{\rm d}u{\rm d}v+\|h(\lambda-T_0)^{-1}g\|_{X}\\ &\leq &\int_a^b{h(v)} |\psi_{\lambda}(v)|{\rm d}v+\|h(\lambda-T_0)^{-1}g\|_{X}\\ &\leq&\|\psi_{\lambda}\|_{Y}+\|g\|_{X} <\infty. \nonumber \end{eqnarray*} $

所以$ \varphi_{\lambda}\in X $.又因为对几乎所有的$ (u, v)\in\Omega $,有

$ \begin{eqnarray*} h(v)\frac{\partial\varphi_{\lambda}(u, v)}{\partial u}& = & h(v)\frac{\partial (\theta_{\lambda}\psi_{\lambda}+(\lambda-T_0)^{-1}g)(u, v)}{\partial u}\\ & = &-\lambda e^{-\frac{\lambda u}{h(v)}}\psi_{\lambda}(u, v)-T_0(\lambda-T_0)^{-1}g(u, v)\\ & = &-\lambda \theta_{\lambda}\psi_{\lambda}(u, v)+(g-\lambda(\lambda-T_0)^{-1}g)(u, v)\\ & = &-\lambda (\theta_{\lambda}\psi_{\lambda}+(\lambda-T_0)^{-1}g)(u, v)+g(u, v)\\ & = &-\lambda \varphi_{\lambda}(u, v)+g(u, v), \nonumber \end{eqnarray*} $

所以

$ \bigg\| h\frac{\partial\varphi_{\lambda}}{\partial u}\bigg\|_{X} \leq \lambda\parallel \varphi_{\lambda}\parallel_{X}+\parallel g\parallel_{X}. $

最后,由(2.6)式知

$ \varphi_{\lambda}(0, \cdot) = \theta_{\lambda}\psi_{\lambda}(0, \cdot)+(\lambda-T_0)^{-1}g(0, \cdot) = \psi_{\lambda}+0 $

和

$ \begin{eqnarray*} H_{\alpha}\varphi_{\lambda}& = & H_{\alpha}(\theta_{\lambda}\psi_{\lambda}+(\lambda-T_0)^{-1}g)\\ & = &H_{\alpha}\theta_{\lambda}\psi_{\lambda}+H_{\alpha}((\lambda-T_0)^{-1}g)\\ & = &L_{\alpha, \lambda}\psi_{\lambda}+H_{\alpha}((\lambda-T_0)^{-1}g)\\ & = &\psi_{\lambda}. \nonumber \end{eqnarray*} $

因此$ \varphi_{\lambda}(0, \cdot) = H_{\alpha}\varphi_{\lambda} $,故有$ \varphi_{\lambda}\in D(T_{\alpha}) $.所以方程

$ (\lambda-T_{\alpha})\varphi = g $

有唯一解$ \psi\in Y $,故$ \lambda\in \rho(T_{\alpha, p}) $.因为$ L_{\alpha, \lambda}^2 $是紧算子,则由谱映象定理^[7]知

$ 1\in [\sigma(L_{\alpha, \lambda})]^2 = \sigma(L_{\alpha, \lambda}^2) = \sigma_p(L_{\alpha, \lambda}^2). $

所以定理2.2获证.

定理2.4  假设$ (A_1) $–$ (A_{2}) $成立, $ \alpha>0 $,令$ \lambda_{r} = \overline{r}-\underline{r}>0 $,则算子$ L_{\alpha, \lambda} $的谱半径$ r(L_{\alpha, \lambda}) $是一个严格递减的连续函数,且

(2.7) $ \begin{equation} \omega(U_{\alpha}(t))>0, \; \; \; \; r(L_{\alpha, \omega(U_{\alpha})}) = 1. \end{equation} $

证  设$ \lambda_2>\lambda_1\geq \lambda_{r} $, $ \varphi\in Y_{+}, \varphi\neq 0 $,则对几乎所有的$ v\in (a, b) $,有

(2.8) $ \begin{equation} L_{\alpha, \lambda_2}\varphi(v)< L_{\alpha, \lambda_1}\varphi(v). \end{equation} $

因为$ L_{\alpha, \lambda}(\lambda\geq\lambda_r) $是$ Y $上的正弱紧算子,根据文献[8], $ r(L_{\alpha, \lambda})>0 $且存在严格正函数$ \varphi_{\lambda}\in Y_+ $, $ \varphi_{\lambda}^*\in Y^*_+ $,使得

(2.9) $ \begin{equation} L_{\alpha, \lambda}\varphi_{\lambda} = r(L_{\alpha, \lambda})\varphi_{\lambda}, \; \; L^*_{\alpha, \lambda}\varphi^*_{\lambda} = r(L_{\alpha, \lambda})\varphi^*_{\lambda}, \end{equation} $

其中$ \|\varphi_{\lambda}\| = \|\varphi^*_{\lambda}\| = 1 $, $ L^*_{\alpha, \lambda} $是$ L_{\alpha, \lambda} $的共轭算子.所以

$ \begin{eqnarray*} r(L_{\alpha, \lambda_1})& = &\frac{(L^*_{\alpha, \lambda_1} \varphi^*_{\lambda_1}, \varphi_{\lambda_2})}{(\varphi^*_{\lambda_1}, \varphi_{\lambda_2})} = \frac{(\varphi^*_{\lambda_1}, L_{\alpha, \lambda_1}\varphi_{\lambda_2})}{(\varphi^*_{\lambda_1}, \varphi_{\lambda_2})}\\ & = &\frac{(\varphi^*_{\lambda_1}, L_{\alpha, \lambda_2}\varphi_{\lambda_2})}{(\varphi^*_{\lambda_1}, \varphi_{\lambda_2})}+ \frac{(\varphi^*_{\lambda_1}, (L_{\alpha, \lambda_1}-L_{\alpha, \lambda_2})\varphi_{\lambda_2})}{(\varphi^*_{\lambda_1}, \varphi_{\lambda_2})}\\ & = &\frac{(\varphi^*_{\lambda_1}, r(L_{\alpha, \lambda_2})\varphi_{\lambda_2})}{(\varphi^*_{\lambda_1}, \varphi_{\lambda_2})}+ \frac{(\varphi^*_{\lambda_1}, (L_{\alpha, \lambda_1}-L_{\alpha, \lambda_2})\varphi_{\lambda_2})}{(\varphi^*_{\lambda_1}, \varphi_{\lambda_2})}\\ & = &r(L_{\alpha, \lambda_2})+ \frac{(\varphi^*_{\lambda_1}, (L_{\alpha, \lambda_1}-L_{\alpha, \lambda_2})\varphi_{\lambda_2})}{(\varphi^*_{\lambda_1}, \varphi_{\lambda_2})}. \nonumber \end{eqnarray*} $

从而由(2.8)式和$ \varphi_{\lambda_1} $的严格正性知

(2.10) $ \begin{equation} r(L_{\alpha, \lambda_2})- r(L_{\alpha, \lambda_1}) = \frac{(\varphi^*_{\lambda_1}, (L_{\alpha, \lambda_1}-L_{\alpha, \lambda_2})\varphi_{\lambda_2})}{(\varphi^*_{\lambda_1}, \varphi_{\lambda_2})}>0. \end{equation} $

故$ r(L_{\alpha, \lambda}) $是$ (\lambda_r, \infty) $上的严格递减函数.又由(2.10)式知

(2.11) $ \begin{equation} |r(L_{\alpha, \lambda_1})- r(L_{\alpha, \lambda_2})|\leq \frac{1}{(\varphi^*_{\lambda_1}, \varphi_{\lambda_2})} \parallel L_{\alpha, \lambda_1}\varphi_{\lambda_2}-L_{\alpha, \lambda_2}\varphi_{\lambda_2}\parallel_{Y}. \end{equation} $

又因为

$ \begin{eqnarray*} \parallel L_{\alpha, \lambda_1}\varphi_{\lambda_2}-L_{\alpha, \lambda_2}\varphi_{\lambda_2}\parallel_{Y} &\leq& \alpha\overline{k}\int_a^b\int_0^1(e^{-\frac{\lambda_1 u'}{h(v')}}-e^{-\frac{\lambda_2 u'}{h(v')}})|\varphi_{\lambda_2(v')}|{\rm d}v'\\ & = &\alpha\overline{k}\int_a^b\bigg[\frac{1-e^{-\frac{\lambda_2}{h(v')}}}{\lambda_2}-\frac{1-e^{-\frac{\lambda_1}{h(v')}}} {\lambda_1}\bigg]|\varphi_{\lambda_2(v')}|h(v'){\rm d}v', \nonumber \end{eqnarray*} $

所以由Lebesgue控制收敛定理知

$ \lim\limits_{\lambda_1 \rightarrow \lambda_2}\parallel L_{\alpha, \lambda_1}\varphi_{\lambda_2}-L_{\alpha, \lambda_2}\varphi_{\lambda_2}\parallel_{Y} = 0. $

从而由(2.11)式知

$ \lim\limits_{\lambda_1 \rightarrow \lambda_2}| r(L_{\alpha, \lambda_1})-r(L_{\alpha, \lambda_2})| = 0, $

故$ r(L_{\alpha, \lambda}) $是$ (\lambda_r, \infty) $上的连续函数.根据(2.2)式可得

(2.12) $ \begin{equation} \lim\limits_{\lambda \rightarrow \infty} r(L_{\alpha, \lambda}) \leq \lim\limits_{\lambda \rightarrow \infty} \parallel L_{\alpha, \lambda}\parallel_{Y} = 0. \end{equation} $

所以由$ r(L_{\alpha, \lambda_r})>1 $和(2.12)式及$ r(L_{\alpha, \lambda}) $是$ (\lambda_r, \infty) $上的严格递减连续函数知:存在唯一$ \lambda_0>\lambda_r>0 $,使得

(2.13) $ \begin{equation} r(L_{\alpha, \lambda_0}) = 1. \end{equation} $

设$ \lambda\in\sigma(T_{\alpha}), {\rm Re}\lambda>\lambda_r $,由定理2.3,存在$ \psi_{\lambda}\neq 0 $,使得$ L^2_{\alpha, \lambda}\psi_{\lambda} = \psi_{\lambda} $.则由(2.1)式知

$ |L_{\alpha, \lambda}\psi_{\lambda}|\leq \frac{\alpha}{h} \int_0^1\int_a^b e^{-{\rm Re}\lambda\frac{u'}{h(v')}}k(u', v, v')|\psi_{\lambda}(v')|{\rm d}u'{\rm d}v' = L_{\alpha, {\rm Re}\lambda}|\psi_{\lambda}|. $

从而

(2.14) $ \begin{equation} |\psi_{\lambda}| = \mid L_{\alpha, \lambda}^2\psi_{\lambda}\mid \leq L_{\alpha, {\rm Re}\lambda}^2\mid\psi_{\lambda}\mid. \end{equation} $

所以$ \forall n\geq 1 $,有$ L_{\alpha, {\rm Re}\lambda}^{2n}|\psi_{\lambda}|\geq |\psi_{\lambda}| $,因此, $ r(L_{\alpha, {\rm Re}\lambda})\geq 1 $.

因为$ r(L_{\alpha, \lambda}) $是严格退减的,则由(2.13)式知: $ {\rm Re}\lambda\leq \lambda_0 $,且由(1.6)式知

(2.15) $ \begin{equation} s(T_{\alpha})\leq \lambda_0. \end{equation} $

反过来,由(2.9)式, (2.13)式及文献[9, Prop 2.1]知:存在$ \psi_{\lambda_0}\in( Y)_{+} $和$ \psi_{\lambda_0}\neq 0 $,使得

(2.16) $ \begin{equation} L_{\alpha, \lambda_0}\psi_{\lambda_0} = \psi_{\lambda_0}. \end{equation} $

令

(2.17) $ \begin{equation} \varphi_{\lambda_0}(u, v) = \theta_{\lambda_0}(u, v)\psi_{\lambda_0}, \end{equation} $

其中$ \theta_{\lambda_0}(u, v) = e^{-\frac{\lambda_0 u}{h(v)}} $.则由(2.16)式和(2.17)式,即为(2.5)式和(2.6)式中$ g = 0 $的情况,所以$ \varphi_{\lambda_0}\in D(T_{\alpha}) $.由此对几乎所有$ (u, v)\in \Omega $,有

$ T_{\alpha}\varphi_{\lambda_0}(u, v) = -h(v)\frac{\partial\varphi_{\lambda_0}(u, v)}{\partial u} = \lambda_0\theta_{\lambda_0}(u, v)\psi_{\lambda_0}(v) = \lambda_0\varphi_{\lambda_0}(u, v). $

则$ \lambda_0\in \sigma_p(T_{\alpha})\subset \sigma(T_{\alpha}), $从而

(2.18) $ \begin{equation} \lambda_0\leq s(T_{\alpha}). \end{equation} $

所以,由(2.15)式和(2.18)式知$ s(T_{\alpha}) = \lambda_0>\lambda_{r}>0 $.根据(1.7)式知$ \omega(U_{\alpha}) = s(T_{\alpha}) $,故$ \omega(U_{\alpha}) = \lambda_0>0 $.从而定理2.3获证.

定理2.5  假设$ (A_1)-(A_{2}) $成立,则$ \forall t\geq 0 $, $ U_{\alpha}(t)-U_{0}(t) $是$ X $上的弱紧算子.

证  设$ t\geq 0 $,则由(1.11)式知

(2.19) $ \begin{equation} U_{\alpha}(t)-U_{0}(t) = B_{\alpha}(t)+D_{\alpha}(t), \end{equation} $

其中$ U_{0}(t), B_{\alpha}(t) $和$ D_{\alpha}(t) $分别由(1.12)式, (1.13)式和(1.14)式给出.

1)先证$ B_{\alpha}(t) $是$ X $上的弱紧算子.设$ \varphi\in X $,对几乎所有的$ (u, v)\in \Omega $,有

(2.20) $ \begin{eqnarray} |B_{\alpha}(t)\varphi(u, v)| &\leq& \frac{\alpha\overline{\chi}(u, h(v), t)}{h(v)}\int_0^1\int_a^{b}|k(u', v, v')||U_{0}(t-\frac{u}{h(v)})\varphi(u', v')|{\rm d}u'{\rm d}v'{}\\ &\leq&\frac{\alpha\overline{\chi}(u, h(v), t)\overline{k}(v)}{h(v)}\int_0^1\int_a^{b}|U_{0}(t-\frac{u}{h(v)})\varphi(u', v')|{\rm d}u'{\rm d}v' {}\\ &\leq&\frac{\alpha\overline{\chi}(u, h(v), t)\overline{k}(v)}{h(v)}\int_0^1\int_a^{b} \chi(u', h(v'), t-\frac{u}{h(v)}){}\\ &&\times |\varphi(u'-(t-\frac{u}{h(v)})h(v'), v')|{\rm d}u'{\rm d}v'. \end{eqnarray} $

令$ x = u'-(t-\frac{u}{h(v)})h(v') $,则$ {\rm d}x = {\rm d}u' $,所以

(2.21) $ \begin{equation} |B_{\alpha}(t)\varphi(u, v)| \leq\frac{\alpha\overline{\chi}(u, h(v), t)\overline{k}(v)}{h(v)}\int_0^1\int_a^{b}|\varphi(x, v')|{\rm d}x{\rm d}v'. \end{equation} $

又令

(2.22) $ \begin{equation} f_{t}(u, v) = \frac{\overline{k}(v)\overline{\chi}(u, h(v), t)}{h(v)}, \end{equation} $

则

(2.23) $ \begin{equation} \parallel f_{t}\parallel_{X} = \int_a^b\bigg[\int_0^{1}\overline{\chi}(u, h(v), t){\rm d}u\bigg]\frac{\overline{k}(v)}{h(v)}{\rm d}v = t\|\overline{k}\|_{L(J)}<\infty. \end{equation} $

即$ f_{t} $是$ X $上的一个正秩1算子,从而由(2.21)式和[2, Lemma 2.1]知:$ B_{\alpha}(t) $是$ X $上的弱紧算子.

2)再证$ D_{\alpha}(t) $是$ X $上的弱紧算子.设$ \varphi\in X $,对几乎所有的$ (u, v)\in \Omega $,有

(2.24) $ \begin{eqnarray} |D_{\alpha}(t)\varphi(u, v)| &\leq& \frac{\alpha^2\overline{\chi}(u, h(v), t)}{h(v)}\int_0^1\int_a^{b}\int_0^1\int_a^{b}\overline{\chi}(u', h(v'), t-\frac{u}{h(v)}) \frac{|k(u', v, v')||k(u'', v', v'')|}{h(v')}{}\\ &&\times|U_{\alpha}(t-\frac{u}{h(v)}-\frac{u'}{h(v')})\varphi(u'', v'')|{\rm d}u''{\rm d}v''{\rm d}u'{\rm d}v' {}\\ &\leq&\frac{\alpha^2\overline{\chi}(u, h(v), t)\overline{k}(v)}{h(v)}\int_0^1\int_a^{b}\int_0^1\int_a^{b}\overline{\chi}(u', h(v'), t-\frac{u}{h(v)}){}\\ &&\times\frac{|k(u'', v', v'')|}{h(v')}|U_{\alpha}(t-\frac{u}{h(v)}-\frac{u'}{h(v')})\varphi(u', v')|{\rm d}u''{\rm d}v''{\rm d}u'{\rm d}v' {}\\ & = &\alpha^2 f_{t}\int_0^1\int_a^{b}\int_0^1\int_a^{b}\overline{\chi}(u', h(v'), t-\frac{u}{h(v)})\frac{|k(u'', v', v'')|}{h(v')}{}\\ &&\times|U_{\alpha}(t-\frac{u}{h(v)}-\frac{u'}{h(v')})\varphi(u', v')|{\rm d}u''{\rm d}v''{\rm d}u'{\rm d}v', \end{eqnarray} $

其中$ f_{t} $为(2.22)式给定.令$ x = t-\frac{u}{h(v)}-\frac{u'}{h(v')} $,则$ {\rm d}x = -\frac{1}{h(v')}{\rm d}u' $,所以

(2.25) $ \begin{eqnarray} |D_{\alpha}(t)\varphi(u, v)| &\leq& \alpha^2 f_{t}\int_0^t\int_a^{b}\int_0^1\bigg[\int_a^{b}|k(u'', v', v'')|{\rm d}v' |U_{\alpha}(x)\varphi(u'', v'')|{\rm d}u''{\rm d}v''\bigg]{\rm d}x{}\\ &\leq& \alpha^2 f_{t}\overline{k}\int_0^t\bigg[\int_a^{b}\int_0^1 |U_{\alpha}(x)\varphi(u'', v'')|{\rm d}u''{\rm d}v''\bigg]{\rm d}x{}\\ &\leq &\alpha^2 f_{t}\overline{k}\int_0^t \|U_{\alpha}(x)\varphi\|_{X}{\rm d}x \leq \alpha^2 f_{t}\overline{k}\int_0^t e^{(\alpha\overline{k})x}\|\varphi\|_{X}{\rm d}x. \end{eqnarray} $

从而

(2.26) $ \begin{equation} |D_{\alpha}(t)\varphi(u, v)| \leq \alpha f_{t}(e^{(\alpha\overline{k})t}-1)\int_0^1\int_a^{b}\varphi(u, v) {\rm d}u{\rm d}v, \; \; \forall \varphi\geq 0. \end{equation} $

由于$ f_{t} $是$ X $上的一个正秩1算子,所以由(2.26)式和[2, Lemma2.1]知:$ D_{\alpha}(t) $是$ X $上的弱紧算子.

3)最后由(2.19)式知:$ U_{\alpha}(t)-U_{0}(t) $是$ X $上的弱紧算子,从而定理2.4获证.

定理2.6  假设$ (A_1) $–$ (A_{2}) $成立,则$\forall t>0, P_{r}U_{\alpha}(t) P_{r}$是$ X$上的弱紧算子.

证  设$ t> 0, \varphi\in X $,则对几乎所有的$ (u, v)\in \Omega $,有

$ \begin{eqnarray*} P_{r}U_{0}(t) P_{r}\varphi(u, v) = \int_a^b\int_a^{b}r(u, v, v')r(u-th(v'), v', v'')\chi(u, h(v'), t)\varphi(u-th(v'), v''){\rm d}v''{\rm d}v' \nonumber \end{eqnarray*} $

所以

$ \begin{eqnarray*} | P_{r}U_{0}(t) P_{r}\varphi(u, v)| &\leq& \widetilde{r}\int_a^b\int_a^{b}|r(u-th(v'), v', v'')|\chi(u, h(v'), t)|\varphi(u-th(v'), v'')|{\rm d}v''{\rm d}v'\\ &\leq &\widetilde{r}(v)\|r\|_{\infty}\int_a^b\int_a^{b}\chi(u, h(v'), t)|\varphi(u-th(v'), v'')|{\rm d}v''{\rm d}v'. \nonumber \end{eqnarray*} $

令$ x = u-th(v') $,则$ {\rm d}x = -th'(h^{-1}(\frac{u-x}{t})){\rm d}v' $,所以

$ \begin{eqnarray*} | P_{r}U_{0}(t) P_{r}\varphi(u, v)| &\leq &\frac{\widetilde{r}(v)\|r\|_{\infty}}{t}\int_{u-th(b)}^{u-th(a)}\int_a^{b}|\varphi(x, v'')| \frac{\chi(u, \frac{u-x}{t}, t)}{h'(v)(h^{-1}(\frac{u-x}{t}))}{\rm d}v''{\rm d}x\\ &\leq& \frac{\widetilde{r}(v)\|r\|_{\infty}}{th'(v)}\int_{u-th(b)}^{u-th(a)}\int_a^{b}|\varphi(x, v'')| \chi(u, \frac{u-x}{t}, t){\rm d}v''{\rm d}x. \nonumber \end{eqnarray*} $

根据(1.15)式知:$ \chi(u, \frac{u-x}{t}, t) = 1 $的充要条件是$ x>0 $,所以

(2.27) $ \begin{equation} | P_{r}U_{0}(t) P_{r}\varphi| \leq \frac{\widetilde{r}\|r\|_{\infty}}{th'}\int_{0}^{1}\int_a^{b}\varphi(x, v''){\rm d}v''{\rm d}x, \; \; \forall \varphi\geq 0. \end{equation} $

当$ \widetilde{r}\geq 0 $时,又因为

$ \|\widetilde{r}\|_{X} = \int_{0}^{1}\int_a^{b}\widetilde{r}(v){\rm d}v{\rm d}u = \int_a^{b}\widetilde{r}(v){\rm d}v = \|\widetilde{r}\|_{L(a, b)}<\infty. $

所以(2.27)式的右边是正秩一算子,因此由定理2.3即知本定理成立.

定理2.7  假设$ (A_{1}) $–$ (A_{2}) $成立,则

(2.28) $ \begin{equation} \omega_{{\rm ess}}(W_{\alpha}(t)) = -\infty, \end{equation} $

且迁移算子$ A_{\alpha} $的谱$ \sigma(A_{\alpha}) $仅由至多可数个具有限代数重数的离散本征值组成, $ -\infty $是唯一可能的聚点.

证  因为由(1.20)式知

$ 0\leq V_{\alpha}(t)-V_{0}(t)\leq U_{\alpha}(t)-U_{0}(t), \; \; t\geq 0. $

所以由定理2.4和引理1.2知:$ V_{\alpha}(t)-V_{0}(t) $是$ X $上的弱紧算子,从而引理1.3知

(2.29) $ \begin{equation} \omega_{{\rm ess}}(V_{\alpha}(t)) = \omega_{{\rm ess}}(V_{0}(t)). \end{equation} $

又因为

(2.30) $ \begin{equation} 0\leq P_{r}V_{\alpha}(t)P_{r}\leq e^{-t\underline{r}}P_{r} U_{\alpha}(t)P_{r}, \; \; t\geq 0. \end{equation} $

所以由引理1.2知:$ P_{r}V_{\alpha}(t)P_{r} $也是$ X $上的弱紧算子,且由引理1.4知

(2.31) $ \begin{equation} \omega_{{\rm ess}}(W_{\alpha}(t)) = \omega_{{\rm ess}}(V_{\alpha}(t)). \end{equation} $

因此由(2.29)式和(2.31)式及引理1.7即知(2.28)式成立.从而由(1.4)式即知$ \|W(t)\|_{{\rm ess}} = 0 $,即半群$ W(t) $是(弱)紧算子,故由谱映像定理知:其母元即迁移算子$ A_{\alpha} $的谱$ \sigma(A_{\alpha}) $仅由至多可数个具有限代数重数的离散本征值组成, $ -\infty $是唯一可能的聚点.从而定理2.6获证.

定理2.8  假设$ (A_1) $–$ (A_{2}) $成立,设$ \alpha>0, \lambda_{r}>0 $,有

(2.32) $ \begin{equation} r(L_{\alpha, \lambda_r})>1, \end{equation} $

则

(2.33) $ \begin{equation} \omega_{{\rm ess}}(W_{\alpha}(t))<\omega(W_{\alpha}(t)), \end{equation} $

且存在$ X $上秩1投影$ P $和$ \varepsilon>0 $,使得$ \forall \eta\in(0, \varepsilon) $,有$ M(\eta)\geq 1 $,满足

$ \parallel e^{-t\omega(W_{\alpha}(t))}W_{\alpha}(t)-P\parallel_{L(X)}\leq M(\eta)e^{-\eta t}, \; \; t\geq 0. $

证  由(2.7)式和(2.32)式,有

$ r(L_{\alpha, \omega(U_{\alpha}(t))}) = 1 <r(L_{\alpha, \overline{r}-\underline{r}}). $

从而由定理2.3知

(2.34) $ \begin{equation} \overline{r}-\underline{r}<\omega(U_{\alpha}(t)). \end{equation} $

所以由(2.29)式, (2.34)式和(1.23)式知

(2.35) $ \begin{equation} \omega_{{\rm ess}}(W_{\alpha}(t)) = -\infty\leq -\underline{r} <-\overline{r}+\omega(U_{\alpha}(t)) \leq \omega(W_{\alpha}(t)). \end{equation} $

因此(2.33)式成立.由于$ (W_{\alpha}(t))_{t\geq 0} $是正不可约$ C_0 $半群,故由引理1.1即知定理2.7成立.

根据文献[10, Theorem 8.7]知:投影算子$ P $可写成$ P\varphi = (\varphi_0, \varphi_0^{*})\varphi_0 $,其中$ \varphi_0\in X_+ $是一个拟内元和$ \varphi_0^{*}\in (X^*){_+} $是一个严格正元使得$ (\varphi_0, \varphi_0^{*}) = 1 $,从而由定理2.7结论的第二部分说明迁移方程的解关于内在生长常数$ \omega(W_{\alpha}(t)) $具有异步指数增长特性,而且还描述了具有独特方向$ P $的细菌种群的分布图等.