Question

Populationkinetics: progress of measles. I have to check whether there are oscillations regarding the small changes of the stationary equilibrium. The given equations follow in next post.

Answer 1

\[\frac{ dN _{A} }{ dt }= v - k _{A}N _{A}N _{I}\]
\[\frac{ dN _{I} }{ dt }= k _{A}N _{A}N _{I}-k _{I}N _{I}\]

In stationary equilibrium (no change over time) these equations become

\[N _{A}=\frac{ k _{I} }{ k _{A} }\]
\[N _{I}=\frac{ v }{ k _{I} }\]

I'm asked to set up the equations for
\[\frac{ d(\Delta N _{A}) }{ dt }=\]
\[\frac{ d(\Delta N _{I}) }{ dt }=\]

But not sure how to go about that

Answer 2

v= growth through births

the term
\[k _{A}N _{A}N _{I}\]
represents decrease through infection via contact with infected people

the term
\[k _{I}N _{I}\]
represents the decrease of infected people through healing or death.

Answer 3

I'm assuming: NA represents the number of healthy people, NI represents the number of infected people, v represents the initial population, and the k's represent the various growth/decay constants.

This is an interesting question, but I'm a bit confused about what exactly your question is asking. If the question is simply asking what the differential equations become at stationary equilibrium, they would both obviously equal 0. However, I'm not sure if your question is asking that...

Answer 4

If your question is "whether there are oscillations regarding the small changes of the stationary equilibrium," you can answer this by substituting some random numbers for v, kA, and KI and checking out the phase portrait for the system of DEs. I've done so using PPlane (free software online). Using the equations for stationary equilibrium, the equilibrium point for the numbers I chose came out to be (1.5, 4). Near the equilibrium point, solutions were not circular/periodic, and so solutions near the equilibrium point do not oscillate. Instead, solutions near the stationary equilibrium immediately converge toward the equilibrium.

Answer 5

@Rogue I explained the terms as they were given to me in the second extra comment of mine. Yes, the equilibrium means that Na and NI are constants so their derivations over time would be 0. I found the second set of equilibrium equations that way in order to calculate Na and Ni with a given v and ks (which are coefficients), and I mentioned them in my first comment.

The problem I had was to find the Laplacians for the small changes from equilibrium and use these to prove it oscillates at little changes of the equilbrium or not. Intuitively I thought not but I need the Laplacians to prove it mathematically.

The course has an example of an oscillation with little changes for hares and lynxes, but I'm confused on what she uses to set up those Laplacians: she skips a whole set of steps (I find her doing this often with derivations, but other sources tend to help. Unfortunately I didn't come across online sources to help me on this) and it's a pain to figure out what she used or altered, etc. Perhaps it's better to ask help in figuring it out in the physics section?

Answer 6

@Rogue I have the solution mathematically: first I took the partial derivations for both Delta N_A and Delta N_I.

\[\frac{ d(\Delta N_{A}) }{ dt } = -k_{A}N_{I}\Delta N_{A} - k_{A}N_{A} \Delta N_{I}\]

Substituting the equilibrium values for N_I and N_A makes it
\[=-k _{A}\frac{ v }{ k_{I} } \Delta N_{A} - k _{A}\frac{ k_{I} }{ k_{A} } \Delta N_{I}\]
\[-\frac{k_{A}v}{k_{I}} \Delta N_{A}-k_{I}\Delta N_{I}\]

similar method for the second function

\[\frac{ d(\Delta N_{I}) }{dt }= \frac{ k_{A}v }{ k_{I} } \Delta N_{A}+(k_{A}\frac{ k_{I} }{ k_{A}} - k_{I})\Delta N_{I}\]
the second term cancels out
\[=\frac{ k_{A}v }{ k_{I} }\Delta N_{A}\]

we can see the derived equations as
\[\frac{ dx }{ dt }=ax+by\]
\[\frac{ dy }{ dt }=cx+dy\]

with

\[a=-\frac{ k_{A}v }{ k_{I} }\]
\[b=-k_{I}\]
\[c=\frac{ k_{A}v }{ k_{I} }\]
\[d=0\]

There is an oscillation if (a-d)²+4bc <0 then there is an oscillation

\[(a-d)^{2}+4bc=(\frac{ k_{A}v }{ k_{I} })^{2}-4k_{A}v= -39.75<0\]

So there is an oscillation with a frequency

\[\omega = \left[ \left( \left( a-d \right)^{2}+4bc \right)^{1/2} \right]/2=\pm 3.15i\]

And the period

\[T=2\pi/\omega \approx 2 \]

So, every 2 years there is an outbreak of the measles to be expected.