 # 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
$\frac{dN}{dt}=\mu N\left(1-\frac{N}{K}\right)+N{\int }_{0}^{\mathrm{\infty }}p\left(a,t\right)da$
$\frac{\partial p}{\partial t}+\frac{\partial p}{\partial a}=kp\left(a,t\right).$
Linearizing it around $\left(N=K,p=0\right)$, I am getting the system
$\frac{dN}{dt}=h-\mu N+N{\int }_{0}^{\mathrm{\infty }}p\left(a,t\right)da$
$\frac{\partial p}{\partial t}+\frac{\partial p}{\partial a}=kp\left(a,t\right),$
where $h=\mu K$.
The reason I am getting this system is because I am considering $N{\int }_{0}^{\mathrm{\infty }}p\left(a,t\right)da$ as linear term. Is this linearisation right or the correct system is $\frac{dN}{dt}=h-\mu N$
$\frac{\partial p}{\partial t}+\frac{\partial p}{\partial a}=kp\left(a,t\right),$
where $h=\mu K$.
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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:

Step 2
Thus, the linearised equation is (dropping the tildes):
.
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:

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

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