Linearisation and Stability of a system

Question

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?

Answer 1

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

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

after adding we have

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

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]

Answer 2

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.