Consider the modification to the Malthusian equation d N d t = r S ( N ) N ,where r&gt;0 is the per capita growth rate, and S(N) is a survival fraction. For some organisms, finding a mate at low population densities may be difficult. In such cases, the survival fraction can take the form S ( N ) = N A + N , where A&gt;0 is a constant.(a)i. By examining what happens to S(N) for N≫A and N≪A explain why this fraction models the situation outlined above.Attempt:When N≫A (my assumption is that the sign '≫'means significantly greater than ) so A is much smaller than N then the constant A.This would effect the survival fraction in a way which it becomes roughly equal to 1 since A is really small.When N≪A (my assumption is that the sign '≪' means significantly smaller than) so N is much smaller than A will impact the survival fraction in a way which will make it become small.The survival fraction models matches with the situation above since:For a low population finding a mate is difficult (a.k.a N≪A )For a large population finding a mate is more likely (a.k.a N≫A )ii. By examining the form of the equation, determine the long-term behaviour of a population for an initial condition N(0)&gt;0.Attempt: For the initial conditions stated above the Malthusian equation has the form: d N d t = r N A + N N ,which has the solution. N ( t ) = N 0 e λ t If λ &gt; 0 then exponential growthIf λ &lt; 0 then exponential decayPlease could you check if this is correct. If it is please can you suggest any improvement I could add to my attempts.

Question

Consider the modification to the Malthusian equation            d      N              d      t        =  r  S  (  N  )  N  ,where r&amp;gt;0 is the per capita growth rate, and S(N) is a survival fraction. For some organisms, finding a mate at low population densities may be difficult. In such cases, the survival fraction can take the form   S  (  N  )  =      N          A      +      N      , where A&amp;gt;0 is a constant.(a)i. By examining what happens to S(N) for N≫A and N≪A explain why this fraction models the situation outlined above.Attempt:When N≫A (my assumption is that the sign &#039;≫&#039;means significantly greater than ) so A is much smaller than N then the constant A.This would effect the survival fraction in a way which it becomes roughly equal to 1 since A is really small.When N≪A (my assumption is that the sign &#039;≪&#039; means significantly smaller than) so N is much smaller than A will impact the survival fraction in a way which will make it become small.The survival fraction models matches with the situation above since:For a low population finding a mate is difficult (a.k.a N≪A )For a large population finding a mate is more likely (a.k.a N≫A )ii. By examining the form of the equation, determine the long-term behaviour of a population for an initial condition N(0)&amp;gt;0.Attempt: For the initial conditions stated above the Malthusian equation has the form:            d      N              d      t        =  r        N          A      +      N          N  ,which has the solution.  N  (  t  )  =      N          0            e          λ      t      If   λ  &amp;gt;  0 then exponential growthIf   λ  &amp;lt;  0 then exponential decayPlease could you check if this is correct. If it is please can you suggest any improvement I could add to my attempts.

Kash Osborn · Accepted Answer

You don&#039;t need to find the exact solution in the second part. The right-hand side is positive for all N&amp;gt;0, so there are no equilibria except N=0. Hence the solution will increase indefinitely, and for N≫A you have             N      ˙        ≈  k  N(as you showed in the first part), so you will end up with approximately exponential growth. It&#039;s just that the population grows slower to begin with (if N(0)≪A), since you have             N      ˙        ≈      r    A        N    2   for small N (as you also, more or less, showed in the first part).

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