Linearization around the equilibrium point for a system of differential equations \frac{dN}{d

pacificak8m 2022-05-01 Answered
Linearization around the equilibrium point for a system of differential equations
dNdt=μN(1NK)+N0p(a,t)da
pt+pa=kp(a,t).
Linearizing it around (N=K,p=0), I am getting the system
dNdt=hμN+N0p(a,t)da
pt+pa=kp(a,t),
where h=μK.
The reason I am getting this system is because I am considering N0p(a,t)da as linear term. Is this linearisation right or the correct system is dNdt=hμN
pt+pa=kp(a,t),
where h=μK.
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Answers (1)

eslasadanv3
Answered 2022-05-02 Author has 20 answers

Step 1
I agree with the linearisation of the second equation. For the first, you are close but not quite correct. To find the linearisation around N=K,p=0, substitute N=K+εN~ and p=εp~ into your equations and throw out everything that isn't order ε. I got:
εdN~dt=dNdt=μN(1NK)+N0p(a,t) da =μ(K+εN~)ε(N~K)+ε(K+εN~)0p~(a,t) da =(μN~+K0p~(a,t) da)ε+o(ε).
Step 2
Thus, the linearised equation is (dropping the tildes):
dN dt =μN+K0p(a,t),da.
Edit: Answer to question in the comments. Let N,p be steady state solutions to your system. Replacing p with p+εp the equation for p becomes:
kp+εkp=k(p+εp)=pt+εpt+pa+εpa =pa+ε(pt+pa)
which implies the linearised equation is
kp=pt+pa
(not surprising since the original equation was linear). For the equation for N replace N and p with N+εN and p+εp respectively:
Step 3
εdNdt=d(N+εN)dt =μ(N+εN)(1N+εNK)+(N+εN)0(p(a)+εp(a,t)) da =μN(1NK)+N0p da  +ε[N0p da+N0p da+μN(12NK)]  +ε2[N0p daμN2K] =ε[N0p da+N0p da+μN(12NK)]  +ε2[N0p daμN2K]
using that (N,p) is the steady state solution. Thus, the linearised equation is:
dN dt =N0p,da+N0p,da+μN(12NK).

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