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1、Leslie种群年龄结构的差分方程模型已知一种昆虫每两周产卵一次，六周以后死亡（给出了变化过程的基本规律）。孵化后的幼虫2周后成熟，平均产卵100个，四周龄的成虫平均产卵150个。假设每个卵发育成2周龄成虫的概率为0.09，（称为成活率），2周龄成虫发育成4周龄成虫的概率为0.2。假设开始时，0~2，2~4，4~6周龄的昆虫数目相同，计算2周、4周、6周后各种周龄的昆虫数目；讨论这种昆虫各种周龄的昆虫数目的演变趋势：各周龄的昆虫比例是否有一个稳定值？昆虫是无限地增长还是趋于灭亡？假设使用了除虫剂，已知使用了除虫剂后
2、各周龄的成活率减半，问这种除虫剂是否有效？解：将两周分成一个时段，设k时段2周后幼虫数量为：x1(k),2到4周虫的数量为：x2(K),4到6周虫数量为：x3(K)。据题意可列出下列差分方程：x1(k+1)=x2(k)*100+x3(k)*150x2(k+1)=x1(k)*0.09x3(k+1)=x2(k)*0.2运用matlab编写的程序如下：functionz=diliuti(a,r1,r2,n)x(1)=a;y(1)=a;w(1)=a;fork=1:nx(k+1)=y(k)*100+w(k)*150;y(k+
3、1)=x(k)*r1;w(k+1)=y(k)*r2;endz=[x',y',w'];fork=1:n+1m=x(k)+y(k)+w(k)endplot(1:n+1,x);holdonplot(1:n+1,y,'r');holdonplot(1:n+1,w,'k'),grid计算前三年的结果为：z=diliuti(100,0.009,0.2,2)m=300m=2.5021e+004m=3.3152e+003z=1.0e+004*0.01000.01000.01002.50000.00010.00200.30900.0
4、2250.0000（蓝线为0~2周的虫，红线为2~4周的虫，黑线为4~6周的虫）其中，m表示三个不同生长周期的虫的总数，可见虫并未灭绝。当年份足够长时，可观察到各年龄段虫的数量变化：>>z=diliuti(100,0.009,0.2,20)m=300m=2.5021e+004m=3.3152e+003m=2.2600e+004m=9.7393e+003m=2.1235e+004m=1.4867e+004m=2.1741e+004m=1.9114e+004m=2.3581e+004m=2.3073e+004m=2.6
5、384e+004m=2.7132e+004m=2.9975e+004m=3.1543e+004m=3.4303e+004m=3.6482e+004m=3.9389e+004m=4.2095e+004m=4.5301e+004m=4.8521e+004z=1.0e+004*0.01000.01000.01002.50000.00010.00200.30900.02250.00002.25270.00280.00450.95310.02030.00062.11090.00860.00411.46600.01900.00
6、172.15710.01320.00381.88930.01940.00262.33720.01700.00392.28280.02100.00342.61360.02050.00422.68560.02350.00412.96860.02420.00473.12270.02670.00483.39690.02810.00533.61200.03060.00563.90030.03250.00614.16790.03510.00654.48550.03750.00704.80420.04040.0075由此可见，0
7、~2周的虫的数量急剧增多，2~4周的虫的数量也增多，而4~6周的虫的数量相对很少。三者并无太多比例关系。最终整个种群数量增多。当使用杀虫剂时：z=diliuti(100,0.0045,0.1,20)m=300m=2.5010e+004m=1.6575e+003m=1.1275e+004m=2.4341e+003m=5.1856e+003m=1.8564e+003m=2.4978e+003m=1.1854e+003m=1.2493e+003m=702.0383m=642.2136m=400.2470m=336.383
8、7m=223.4606m=178.3893m=123.2632m=95.3588m=67.5097m=51.2317m=36.8161z=1.0e+004*0.01000.01000.01002.50000.00000.00100.15450.01120.00001.12570.00070.00110.23830.00510.00010.51700.0011
