CDS - Linearization

How I think the Linearization/Linearize System should be defined:

(My Notes / My Take / My Thoughts, TAKE THEM WITH A GRAIN OF SALT!) (Under this section you’ll find the professor’s notes.)

Let $x^{\ast }$ a steady state of the non-linear, monodimensional system $\dot{x}=f(x)$, then $f(x^{\ast })=0$.

1. If we perform the Taylor Series around $x^{\ast }$ we will obtain:$\dot{x}=f(x^{\ast })+f^{\prime }(x^{\ast })(x-x^{\ast })+O(x^2)$Where:

• $f(x^{\ast })=0$
• $f^{\prime }(x^{\ast })\in \mathbb{C}$, we can also call it $a$. (it can be an imaginary number)
• $O(x^2)=0$, (we ignore it)

2. We define a new system (transposed) near the steady state, so: $y(t)=x(t)-x^{\ast }$, such that $\dot{y}=\dot{x}$, and we can define the linearized system by substituting to the Taylor series as:$\dot{y}=f^{\prime }(x^{\ast })y\text{or}y=ay$
3. So we can find the solution of the linearized system $(\dot{y}=ay)$ preatty easly:$y(t)=y(0)e^{at}$
4. Notice how:

• If $a>0$, so if we have $f^{\prime }(x)>0$, then this linearized system will diverge.
• While if $a<0$ (meaning $f^{\prime }(x)<0$) it will converge.

5. For higher-dimensions systems, we will see that $f^{\prime }(x^{\ast })$ becomes a matrix, more specifically the Jacobian Matrix, and we will analyze like we did for linear systems its eigenvalues.

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Remember:

(Professor’s Notes)

The linearization is a method for characterizing locally the geometry of the phase space (flow), locally meaning in the neighborhood of $x^{\ast }$.

Let $x^{\ast }$ a steady state of the non-linear system $\dot{x}=f(x)$, then $f(x^{\ast })=0$. Let $\eta$ be a perturbation of the steady state $x^{\ast }$. We want to study the “perturbated solution” $x(t)=x^{\ast }+\eta$.

1. From: $x(t)=x^{\ast }+\eta$ we can say that $\eta =x(t)-x^{\ast }$.
2. And if we derive both part in time, we have that: $\dot{\eta }=\dot{x}=f(x)$
3. If we expand the $f(x)$ with Taylor, near $x^{\ast }$, we have:$\dot{x}=f(x^{\ast })+f^{\prime }(x^{\ast })(x-x^{\ast })+O(x^2)$Where:

• $f(x^{\ast })=0$.
• $\eta =x(t)-x^{\ast }$

4. Finally we have that $\dot{\eta }=f^{\prime }(x^{\ast })\eta$, then if we study this “linearized system”, and we see that $\eta$ grows, meaning for $f^{\prime }(x^{\ast })>0$, then we have that the original system will diverge from the steady state $x^{\ast }$, meaning that $x^{\ast }$ is an unstable steady state.
   While if we have that $f^{\prime }(x^{\ast })<0$ then this $\eta$ will go toward zero, meaning that the perturbated system will converge to the steady state, $x^{\ast }$ is a stable steady state.
   However if $f^{\prime }(x^{\ast })=0$ we cannot say anything about the original system $f(x)$.

Given a system:$\dot{x}=f(x)$And a steady steate $x^{\ast }$ of this system, if this steady state is hyperbolic then we can study the stability of the linearized system: $f^{\prime }(x)=\frac{d}{dx}f(x)$And its stability or instabilty will be the same as the original $f(x)$ system. ==However if $f^{\prime }(x)$ is marginally stable, then we cannot say anything on the stability of $f(x)$==.

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• Mathematical proof and explanation on how we can linarize numerically.