# Linearisation and Stability of a system

Consider the system

$$\dot{x} = y$$ $$\dot{y} = x^{2} + x$$

Find the fixed points and the linearisation of the system at each. Identify the type and sketch a local phase portrait at each.

I am unsure how to get the fixed points. I get that it should be at $$\dot{x} = 0$$ and $$\dot{y} = 0$$, but that gives me $$y=0$$ and $$x = 0, -1$$, and I am unsure how to find the other co-ordinate point for each.

EDIT: Found the other coordinates for each, found the local phase portraits too. How to I now find the isoclines of the system?

The equilibrium points are located at $$(y = 0) \cap (x = \{-1,0\})$$

now calling $$X = (x,y)^{\dagger}$$ and $$F(X) = (y,x+x^2)^{\dagger}$$ we have

$$\dot X = F(X) = F(X_0) + \frac{\partial F_0}{\partial X}(X-X_0)+ O(|X-X_0|^2)$$

now choosing $$X_0 = \{(-1,0)^{\dagger},(0,0)^{\dagger}\}$$ at $$X_0^1$$ we have

$$\frac{d}{dt}(X-X_0^1) = \frac{\partial F_0^1}{\partial X}(X-X_0^1)+ O(|X-X_0^1|^2)$$

and near $$X_0^1$$

$$\frac{d}{dt}(X-X_0^1) = \frac{\partial F_0^1}{\partial X}(X-X_0^1)$$

with

$$\frac{\partial F_0^1}{\partial X} = \left(\begin{array}{cc}0 & 1\\ -1&0\end{array}\right)$$

and

$$\frac{\partial F_0^2}{\partial X} = \left(\begin{array}{cc}0 & 1\\ 1&0\end{array}\right)$$

then $$X_0^1$$ is a center and $$X_0^2$$ is a saddle point. Attached the stream plot

Near $$X_0^1$$ the orbits follow

$$(\delta x)' = \delta y\\ (\delta y)' = -\delta x$$

then

$$\delta x(\delta x)' = \delta x\delta y\\ \delta y(\delta y)' = -\delta y\delta x$$

$$\delta x(\delta x)' + \delta y(\delta y)' = \frac 12\frac{d}{dt}((\delta x)^2+(\delta y)^2)=0$$

or

$$(\delta x)^2+(\delta y)^2 = C_0$$

which means that in this case, the orbits remain circling, which characterizes this point as a center. NOTE

$$\frac{\partial F}{\partial X} = \left(\begin{array}{cc}0 & 1\\ 1+2x&0\end{array}\right)$$

plt = StreamPlot[{y, x^2 + x}, {x, -2, 2}, {y, -2, 2}];
pt1 = Graphics[{Red, Disk[{-1, 0}, 0.04]}];
pt2 = Graphics[{Red, Disk[{0, 0}, 0.04]}];
Show[plt, pt1, pt2]

• What software package did you use to make that vector field? – TSF Feb 21 at 12:02
• @TonyS.F. MATHEMATICA. I attached the script. – Cesareo Feb 21 at 12:07
• You cannot use linearization for the center. Lyapunov's linearization theorem does not allow any conclusion in this case. – MrYouMath Feb 22 at 18:47
• @MrYouMath Thanks for the hint. I had forgotten to include this detail. – Cesareo Feb 22 at 20:04
• @Cesareo: I like the idea with the circle. Do you have any resource that gives more examples of this method? But I find it a little bit strange that this method does neglect higher order terms. – MrYouMath Feb 22 at 21:00

A more elegant way for the center at $$\boldsymbol{x}_\text{eq}=[-1,0]^T$$ is to use the following Lyapunov function candidate

$$V(x,y)= 1/2y^2+1/6\left[1-x^2(2x+3)\right].$$

This function is positive definite in the neighborhood of the equilibrium point $$\boldsymbol{x}_\text{eq}$$. For a formal derivation, you can calculate the Hessian matrix and evaluate it at the equilibrium point. The eigenvalues will indicate a positive definite matrix. Hence, we know that $$V(\boldsymbol{x})>0$$ for all $$\boldsymbol{x}\neq\boldsymbol{x}_\text{eq}$$. For the equilibrium point we get $$V(\boldsymbol{x}_\text{eq})=0$$.

Now, differentiate $$V$$ to obtain

$$\dot{V}=y\dot{y}+1/6[-6x^2-6x]\dot{x}=y\left[x^2+x \right]+1/6[-6x^2-6x]y\equiv 0.$$

As $$V$$ is positive definite and $$\dot{V}\equiv 0$$, we can conclude that $$\boldsymbol{x}_\text{eq}$$ is indeed a center.

An alternative to the paid Mathematica is the free web-based Wolfram Alpha. Visit this link to see how you can display a stream plot.