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旧 2009-04-30, 06:54   #1
kathy_hanyu
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默认 能否帮小妹看看这题啊。。。?在线等~谢谢!!!

Consider the following predator-prey model for sharks (S) and fish (F)
F’ = a F (1-F/c) - v F S - d(1- e-hF),
S’ = -b S + w F S,
where ‘ stands for the derivative d/dt with respect to time t. The coefficients a,b,c,d,h,v,w are all positive. The last term,
d(1- e-hF), in the first equation models people fishing. We have a fishing quota (d) which we shall vary (we would not want
the fish to become extinct) and the exponential term describes the fact that it is hard to catch fish if the population is very
small.
(i) Discuss the meaning of the individual terms that appear in the differential-equations system and argue why this
system models the predator-prey situation between two species (sharks and fish).
(ii) For a=1200, c=40, v=30 and b=400, d=8000, w=30, h= 1/8,
a) show that F=40/3 and S = 20/3 + 20*e-5/3 is a time-independent solution (equilibria);
b) use pplane7 in Matlab to investigate the system: Describe how solutions behave for large values of t and interpret
these results in terms of shark and fish populations. Support your study by appropriate plots of solutions.
(iii) Repeat the steps from part (ii) where we change d to d=10900 while keeping the other parameters as in part (ii).
(iv) Discuss the differences between the cases (ii) and (iii).
Note: Experiment with the minimum and maximum values of F,S in pplane7 to make sure you capture the region of
interest.
麻烦各位大侠了~~~
kathy_hanyu 当前离线   回复时引用此帖
旧 2009-04-30, 09:22   #2
anbcjys
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默认 回复: 能否帮小妹看看这题啊。。。?在线等~谢谢!!!

引用:
作者: kathy_hanyu 查看帖子
Consider the following predator-prey model for sharks (S) and fish (F)
F’ = a F (1-F/c) - v F S - d(1- e-hF),
S’ = -b S + w F S,
where ‘ stands for the derivative d/dt with respect to time t. The coefficients a,b,c,d,h,v,w are all positive. The last term,
d(1- e-hF), in the first equation models people fishing. We have a fishing quota (d) which we shall vary (we would not want
the fish to become extinct) and the exponential term describes the fact that it is hard to catch fish if the population is very
small.
(i) Discuss the meaning of the individual terms that appear in the differential-equations system and argue why this
system models the predator-prey situation between two species (sharks and fish).
(ii) For a=1200, c=40, v=30 and b=400, d=8000, w=30, h= 1/8,
a) show that F=40/3 and S = 20/3 + 20*e-5/3 is a time-independent solution (equilibria);
b) use pplane7 in Matlab to investigate the system: Describe how solutions behave for large values of t and interpret
these results in terms of shark and fish populations. Support your study by appropriate plots of solutions.
(iii) Repeat the steps from part (ii) where we change d to d=10900 while keeping the other parameters as in part (ii).
(iv) Discuss the differences between the cases (ii) and (iii).
Note: Experiment with the minimum and maximum values of F,S in pplane7 to make sure you capture the region of
interest.
麻烦各位大侠了~~~
对微分方程计算 .
__________________
qq604443022
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旧 2009-04-30, 10:18   #3
laosam280
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默认 回复: 能否帮小妹看看这题啊。。。?在线等~谢谢!!!

我求解你给出的模型,很漂亮的一个蝴蝶型。从相图上可以明显的看到吸引子,特别是在步长为0.001求解的时候,可以看到相图的中央有个小小的空白区域(一个小白点)
上传的图像
文件类型: jpg phase map of this equation with step t=0.0001.jpg (28.9 KB, 8 次查看)
文件类型: jpg step t=0.001.jpg (29.3 KB, 6 次查看)
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坚持就是胜利,努力就有奇迹。
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旧 2009-04-30, 23:54   #4
麒麟子
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默认 回复: 能否帮小妹看看这题啊。。。?在线等~谢谢!!!

我怎么看你这题好像很像混沌啊,(ii)中证明a中各量与时间无关是不是用分离变量法可以解决?我觉的应该可以用分离变量法解决
__________________
:水榭焚香听琴事,浪荡江湖不系舟:
感谢请点thanks
麒麟子 当前离线   回复时引用此帖
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