Question

Biophysics, population kinetics... I have trouble understanding what my prof used to derive the Laplacian regarding population equations. Can someone here help me figure it out? (The equations follow in next post)

Answer 1

The course uses the example of lynxes-hare connection to explain the equations. First the initial part I get:
\[(1) \frac{ dN _{H} }{ dt }=k_{H}N_{H}−k_{HL}N_{L}N_{H}\]
\[(2)\frac{dN_{L}} {dt}=k_{LH}N_{H}N_{L}−k_{L,dth}N_{L}\]
stationary equilibrium (these I figured out myself
\[\frac{ dN_{H} }{ dt } =0 \rightarrow N_{L}=\frac{k_{H}}{k_{HL}}\]
\[\frac{ dN_{L} }{ dt } =0−>N_{H}=\frac{k_{L,dth}}{k_{LH}}\]

N_h: number of hares
N_lL: number of lynxes
k_h. N_h: number of births for hares
k_hl . N_l. N_h: number of hares killed by lynxes

The course continues to including the hunt on hares and gives the following equations
\[(3)N_{L,dth}=\frac{(k_{H}−k_{H,dth})}{k_{HL}}\]
and
\[N_{H,dth}=\frac{k_{L,dth}}{k_{LH}}\]

So it seems to me that the stationary equations I found are the death equations, except that the hare's also includes human hunting, and not just lynxes hunting them. Now the part that confuses me: we investigate what happens in case of small distortions to the equilibrium
\[ΔN_{H}=N_{H}−N_{H,dth}\]
\[ΔN_{L}=N_{L}−N_{L,dth}\]

Then my course says "from (1) (2) and (3) it follows that"
\[(4)\frac{d(ΔN_{H})}{dt}=−\frac{k_{HL}k_{L,dth}}{k_{LH}}ΔNL\]
\[(5)\frac{d(ΔN_{L})}{dt}=\frac{k_{LH}k_{H}}{k_{HL}}ΔNH\]

The course does mention that for the last two she uses...
\[\frac{dN_{H}}{dt}=f1(N_{H},N_{L})\]
\[\frac{dN_{L}}{dt}=f2(N_{H},N_{L})\]
and so
\[\frac{d(ΔN_{H})}{dt}=(\frac{δf_{1}}{δN_{H}})_{dth}ΔN_{H}+(\frac{δf1}{δN_{L}})_{dth}ΔN_{L}\]
\[\frac{d(ΔN_{L})}{dt}=(\frac{δf_{2}}{δN_{H}})_{dth}ΔN_{H}+(\frac{δf_{2}}{δN_{L}})_{dth}ΔN_{L}\]

I'm confused which equations she uses exactly, because when I use (1) and (2) I don't end up with (4) and (5).

Answer 2

I'm not really familiar with this, I remember doing something similar to this in the past but I might be able to help with a smaller, more specific question.

Answer 3

Ok, I simply used the initial equations and plugged in the equilibrium equations, and that at least gives an answer that looks similar to the result I'm supposed to get, except I have an extra term.

\[\frac{ d(\Delta N_{H}) }{ dt }=(k_{H}-k_{H}N_ {L})\Delta N_{H}-(k_{HL}N_{H})\Delta N_{L}\]
\[=(k_{H}-\frac{k_{HL}k_ {H}}{k_{HL}})\Delta N_{H}-(\frac{k_{HL}k_{L,dth}}{k_{LH}})\Delta N_{L}\]
\[=(k_{H}-k_ {H})\Delta N_{H}-(k_{HL}N_{H})\Delta N_{L}\]

Since the first term =0 I do find

\[=-(k_{HL}N_{H})\Delta N_{L}\]

So figured out why the first partial derivative is what is is. But for the second one I'm still puzzled. I have
\[\frac{ d(\Delta N_{L}) }{ dt }=(k_{LH}N_ {L})\Delta N_{H}-k_{L,dth}\Delta N_{L}\]
\[=(\frac{k_{LH}k_ {H}}{k_{HL}})\Delta N_{H}-k_{L,dth}\Delta N_{L}\]

It seems I'm supposed to drop the second term, but I don't know why

Answer 4

I emailed my professor about it, and it turns out that my second partial derivation was wrong. I had forgotten to partially derive the first term for N_L. So it becomes

\[\frac{ d(\Delta N_{L} )}{ dt }=(k _{LH}N_{L})\Delta N_{H}+(k_{LH}N_{H}-k_{L,dth})\Delta N_{L}\]

Substituting the equilibrium values for N_L and N_H it becomes

\[=\frac{ k_{LH}k_{H} }{ k_{H} }\Delta N_{H}+(k_{LH}\frac{k_{L,dth}}{k_{LH}}-k_{L,dth})\Delta N_{L}\]

And so you can cancel out the second term

\[=\frac{ k_{LH} k_{H}}{ k_{HL} }\Delta N_{H}\]