Linearizing it around

where

The reason I am getting this system is because I am considering

where

pacificak8m
2022-05-01
Answered

Linearization around the equilibrium point for a system of differential equations

$\frac{dN}{dt}=\mu N(1-\frac{N}{K})+N{\int}_{0}^{\mathrm{\infty}}p(a,t)da$

$\frac{\partial p}{\partial t}+\frac{\partial p}{\partial a}=kp(a,t).$

Linearizing it around$(N=K,p=0)$ , I am getting the system

$\frac{dN}{dt}=h-\mu N+N{\int}_{0}^{\mathrm{\infty}}p(a,t)da$

$\frac{\partial p}{\partial t}+\frac{\partial p}{\partial a}=kp(a,t),$

where$h=\mu K$ .

The reason I am getting this system is because I am considering$N{\int}_{0}^{\mathrm{\infty}}p(a,t)da$ as linear term. Is this linearisation right or the correct system is $\frac{dN}{dt}=h-\mu N$

$\frac{\partial p}{\partial t}+\frac{\partial p}{\partial a}=kp(a,t),$

where$h=\mu K$ .

Linearizing it around

where

The reason I am getting this system is because I am considering

where

You can still ask an expert for help

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 $N=K+\epsilon \stackrel{~}{N}$ and $p=\epsilon \stackrel{~}{p}$ into your equations and throw out everything that isn't order $\epsilon$. I got:

$\begin{array}{rl}\epsilon \frac{d\stackrel{~}{N}}{dt}=\frac{dN}{dt}& =\mu N(1-\frac{N}{K})+N{\int}_{0}^{\mathrm{\infty}}p(a,t)\phantom{\rule{0ex}{0ex}}da\\ & =-\mu (K+\epsilon \stackrel{~}{N})\epsilon \left(\frac{\stackrel{~}{N}}{K}\right)+\epsilon (K+\epsilon \stackrel{~}{N}){\int}_{0}^{\mathrm{\infty}}\stackrel{~}{p}(a,t)\phantom{\rule{0ex}{0ex}}da\\ & =(-\mu \stackrel{~}{N}+K{\int}_{0}^{\mathrm{\infty}}\stackrel{~}{p}(a,t\left)\phantom{\rule{0ex}{0ex}}da\right)\epsilon +o\left(\epsilon \right).\end{array}$

Step 2

Thus, the linearised equation is (dropping the tildes):

$\frac{dN}{dt}=-\mu N+K{\int}_{0}^{\mathrm{\infty}}p(a,t),da$.

Edit: Answer to question in the comments. Let $N\cdot ,p\cdot$ be steady state solutions to your system. Replacing p with ${p}^{\ast}+\epsilon p$ the equation for p becomes:

$\begin{array}{rl}k{p}^{\ast}+\epsilon kp=k({p}^{\ast}+\epsilon p)& =\frac{\mathrm{\partial}{p}^{\ast}}{\mathrm{\partial}t}+\epsilon \frac{\mathrm{\partial}p}{\mathrm{\partial}t}+\frac{\mathrm{\partial}{p}^{\ast}}{\mathrm{\partial}a}+\epsilon \frac{\mathrm{\partial}p}{\mathrm{\partial}a}\\ & =\frac{\mathrm{\partial}{p}^{\ast}}{\mathrm{\partial}a}+\epsilon (\frac{\mathrm{\partial}p}{\mathrm{\partial}t}+\frac{\mathrm{\partial}p}{\mathrm{\partial}a})\end{array}$

which implies the linearised equation is

$kp=\frac{\partial p}{\partial t}+\frac{\partial p}{\partial a}$

(not surprising since the original equation was linear). For the equation for $N$ replace $N$ and p with ${N}^{\ast}+\epsilon N$ and ${p}^{\ast}+\epsilon p$ respectively:

Step 3

$\begin{array}{rl}\epsilon \frac{dN}{dt}& =\frac{d({N}^{\ast}+\epsilon N)}{dt}\\ & =\mu ({N}^{\ast}+\epsilon N)(1-\frac{{N}^{\ast}+\epsilon N}{K})+({N}^{\ast}+\epsilon N){\int}_{0}^{\mathrm{\infty}}\left({p}^{\ast}\right(a)+\epsilon p(a,t\left)\right)\phantom{\rule{0ex}{0ex}}da\\ & =\mu {N}^{\ast}(1-\frac{{N}^{\ast}}{K})+{N}^{\ast}{\int}_{0}^{\mathrm{\infty}}{p}^{\ast}\phantom{\rule{0ex}{0ex}}da\\ & \phantom{\rule{2em}{0ex}}+\epsilon [{N}^{\ast}{\int}_{0}^{\mathrm{\infty}}p\phantom{\rule{0ex}{0ex}}da+N{\int}_{0}^{\mathrm{\infty}}{p}^{\ast}\phantom{\rule{0ex}{0ex}}da+\mu N(1-\frac{2{N}^{\ast}}{K}\left)\right]\\ & \phantom{\rule{2em}{0ex}}+{\epsilon}^{2}[N{\int}_{0}^{\mathrm{\infty}}p\phantom{\rule{0ex}{0ex}}da-\frac{\mu {N}^{2}}{K}]\\ & =\epsilon [{N}^{\ast}{\int}_{0}^{\mathrm{\infty}}p\phantom{\rule{0ex}{0ex}}da+N{\int}_{0}^{\mathrm{\infty}}{p}^{\ast}\phantom{\rule{0ex}{0ex}}da+\mu N(1-\frac{2{N}^{\ast}}{K}\left)\right]\\ & \phantom{\rule{2em}{0ex}}+{\epsilon}^{2}[N{\int}_{0}^{\mathrm{\infty}}p\phantom{\rule{0ex}{0ex}}da-\frac{\mu {N}^{2}}{K}]\end{array}$

using that $({N}^{\ast},{p}^{\ast})$ is the steady state solution. Thus, the linearised equation is:

$\frac{dN}{dt}={N}^{\ast}{\int}_{0}^{\mathrm{\infty}}p,da+N{\int}_{0}^{\mathrm{\infty}}{p}^{\ast},da+\mu N(1-\frac{2{N}^{\ast}}{K}).$

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