# Linearization of Differental system and stabilty

1. Jan 13, 2014

### prehisto

1. The problem statement, all variables and given/known data
So im looking at example where stabilty is determined.
And do not understand how linerazation is done.
x'=x+2y-ln(1+x)-siny+x^3
y'=4x+2y-sinx+y^2
2. Relevant equations

so the system after linearization looks like:
x'=y
y'=3x+2y

So i thought that higher order x and y are eliminated and other functions are just displaced with x and y.
I think it should not be true,could someone help me with this,please?

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 13, 2014

### BruceW

your answer looks correct to me. It is not right to say that other functions are just displaced with x and y. You write out the Taylor expansion, and then get rid of higher order terms in the expansion. For example, if you had a $\cos(x)$ term, then it would become just $1$ since there is no term linear in $x$ in the Taylor expansion of cosine.

So anyway, you have got the right answer, so next is to find out about the stability. In this case, it is simpler than usual, because you have found x'=y this means the set of equations are simpler than they generally would be for this kind of question.

edit: There is one other thing that you probably don't need to worry about. Which is that you can linearize around any point. In this case, you are Linearizing about the point x=0,y=0. But generally you can choose any point.

Last edited: Jan 13, 2014
3. Jan 13, 2014

### prehisto

4. Jan 13, 2014

### BruceW

the terms x^3 and y^2 disappear, as you said originally. the whole idea is that x and y are both very small, close to zero, which is why you can ignore the x^3 and y^2 terms (since they are much smaller than the x and y terms).

If I have trouble, I like to think like this: say y is a very small number 0.001, then y^2 will be 0.000001 Which is such a small number, we can just forget about it.

5. Jan 13, 2014

### epenguin

That is, for stability question any point for which x' = y' = 0 ?

6. Jan 13, 2014

### AlephZero

The question of stability about a point makes sense for any point, but the answer isn't very interesting unless x' = y' = 0 at that point.

Note there are other types of stability question - for example, stability of a periodic solution of the differential equations.

7. Jan 13, 2014

### epenguin

Yes I know :tongue: but I thought for where the questioner was at...

8. Jan 14, 2014

### ehild

X=0, Y=0 is solution of the original system of equations. That is a stationary point, and you need to investigate the stability of that point: moving out the system by a small Δx, Δy, how (Δx,Δy) changes with time. Will it go away from the stationary point or tending back to it... So Δx and Δy are small, and as the stationary point is (0,0) Δx=x and Δy=y. Do the Taylor expansion and keep only first-order terms.
The original equation can have other stationary points (X0,Y0) (where x'=0 and y'=0) then you need the Taylor expansion around (X0,Y0), assuming Δx=x-X0 and Δy=y-Y0 small.

ehild