## Section8.2Stability and classification of isolated critical points

Note: 1.5–2 lectures, §6.1–§6.2 in [EP], §9.2–§9.3 in [BD]

### Subsection8.2.1Isolated critical points and almost linear systems

A critical point is isolated if it is the only critical point in some small “neighborhood” of the point. That is, if we zoom in far enough it is the only critical point we see. In the example above, the critical point was isolated. If on the other hand there would be a whole curve of critical points, then it would not be isolated.

A system is called almost linear at a critical point $$(x_0,y_0)\text{,}$$ if the critical point is isolated and the Jacobian matrix at the point is invertible, or equivalently if the linearized system has an isolated critical point. In such a case, the nonlinear terms are very small and the system behaves like its linearization, at least if we are close to the critical point.

For example, the system in Examples Example 8.1.1 and Example 8.1.2 has two isolated critical points $$(0,0)$$ and $$(0,1)\text{,}$$ and is almost linear at both critical points as the Jacobian matrices at both points, $$\left[ \begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix} \right]$$ and $$\left[ \begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix} \right]\text{,}$$ are invertible.

On the other hand, the system $$x' = x^2\text{,}$$ $$y' = y^2$$ has an isolated critical point at $$(0,0)\text{,}$$ however the Jacobian matrix

\begin{equation*} \begin{bmatrix} 2x & 0 \\ 0 & 2y \end{bmatrix} \end{equation*}

is zero when $$(x,y) = (0,0)\text{.}$$ So the system is not almost linear. Even a worse example is the system $$x' = x\text{,}$$ $$y' = x^2\text{,}$$ which does not have isolated critical points; $$x'$$ and $$y'$$ are both zero whenever $$x=0\text{,}$$ that is, the entire $$y$$-axis.

Fortunately, most often critical points are isolated, and the system is almost linear at the critical points. So if we learn what happens there, we will have figured out the majority of situations that arise in applications.

### Subsection8.2.2Stability and classification of isolated critical points

Once we have an isolated critical point, the system is almost linear at that critical point, and we computed the associated linearized system, we can classify what happens to the solutions. We more or less use the classification for linear two-variable systems from Section 3.5, with one minor caveat. Let us list the behaviors depending on the eigenvalues of the Jacobian matrix at the critical point in Table 9.2.1. This table is very similar to Table 4.5.9, with the exception of missing “center” points. We will discuss centers later, as they are more complicated.

In the third column, we mark points as asymptotically stable or unstable. Formally, a stable critical point $$(x_0,y_0)$$ is one where given any small distance $$\epsilon$$ to $$(x_0,y_0)\text{,}$$ and any initial condition within a perhaps smaller radius around $$(x_0,y_0)\text{,}$$ the trajectory of the system never goes further away from $$(x_0,y_0)$$ than $$\epsilon\text{.}$$ An unstable critical point is one that is not stable. Informally, a point is stable if we start close to a critical point and follow a trajectory we either go towards, or at least not away from, this critical point.

A stable critical point $$(x_0,y_0)$$ is called asymptotically stable if given any initial condition sufficiently close to $$(x_0,y_0)$$ and any solution $$\bigl( x(t), y(t) \bigr)$$ satisfying that condition, then

\begin{equation*} \lim_{t \to \infty} \bigl( x(t), y(t) \bigr) = (x_0,y_0) . \end{equation*}

That is, the critical point is asymptotically stable if any trajectory for a sufficiently close initial condition goes towards the critical point $$(x_0,y_0)\text{.}$$

#### Example8.2.1.

Consider $$x'=-y-x^2\text{,}$$ $$y'=-x+y^2\text{.}$$ See Figure 8.3 for the phase diagram. Let us find the critical points. These are the points where $$-y-x^2 = 0$$ and $$-x+y^2=0\text{.}$$ The first equation means $$y = -x^2\text{,}$$ and so $$y^2 = x^4\text{.}$$ Plugging into the second equation we obtain $$-x+x^4 = 0\text{.}$$ Factoring we obtain $$x(1-x^3)=0\text{.}$$ Since we are looking only for real solutions we get either $$x=0$$ or $$x=1\text{.}$$ Solving for the corresponding $$y$$ using $$y = -x^2\text{,}$$ we get two critical points, one being $$(0,0)$$ and the other being $$(1,-1)\text{.}$$ Clearly the critical points are isolated.

Let us compute the Jacobian matrix:

\begin{equation*} \begin{bmatrix} -2x & -1 \\ -1 & 2y \end{bmatrix} . \end{equation*}

At the point $$(0,0)$$ we get the matrix $$\left[ \begin{smallmatrix} 0 & -1 \\ -1 & 0 \end{smallmatrix} \right]$$ and so the two eigenvalues are $$1$$ and $$-1\text{.}$$ As the matrix is invertible, the system is almost linear at $$(0,0)\text{.}$$ As the eigenvalues are real and of opposite signs, we get a saddle point, which is an unstable equilibrium point.

At the point $$(1,-1)$$ we get the matrix $$\left[ \begin{smallmatrix} -2 & -1 \\ -1 & -2 \end{smallmatrix} \right]$$ and computing the eigenvalues we get $$-1\text{,}$$ $$-3\text{.}$$ The matrix is invertible, and so the system is almost linear at $$(1,-1)\text{.}$$ As we have real eigenvalues and both negative, the critical point is a sink, and therefore an asymptotically stable equilibrium point. That is, if we start with any point $$(x_i,y_i)$$ close to $$(1,-1)$$ as an initial condition and plot a trajectory, it approaches $$(1,-1)\text{.}$$ In other words,

\begin{equation*} \lim_{t \to \infty} \bigl( x(t), y(t) \bigr) = (1,-1) . \end{equation*}

As you can see from the diagram, this behavior is true even for some initial points quite far from $$(1,-1)\text{,}$$ but it is definitely not true for all initial points.

#### Example8.2.2.

Let us look at $$x'=y+y^2e^x\text{,}$$ $$y'=x\text{.}$$ First let us find the critical points. These are the points where $$y+y^2e^x = 0$$ and $$x=0\text{.}$$ Simplifying we get $$0=y+y^2 = y(y+1)\text{.}$$ So the critical points are $$(0,0)$$ and $$(0,-1)\text{,}$$ and hence are isolated. Let us compute the Jacobian matrix:

\begin{equation*} \begin{bmatrix} y^2e^x & 1+2ye^x \\ 1 & 0 \end{bmatrix}. \end{equation*}

At the point $$(0,0)$$ we get the matrix $$\left[ \begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix} \right]$$ and so the two eigenvalues are $$1$$ and $$-1\text{.}$$ As the matrix is invertible, the system is almost linear at $$(0,0)\text{.}$$ And, as the eigenvalues are real and of opposite signs, we get a saddle point, which is an unstable equilibrium point.

At the point $$(0,-1)$$ we get the matrix $$\left[ \begin{smallmatrix} 1 & -1 \\ 1 & 0 \end{smallmatrix} \right]$$ whose eigenvalues are $$\frac{1}{2} \pm i \frac{\sqrt{3}}{2}\text{.}$$ The matrix is invertible, and so the system is almost linear at $$(0,-1)\text{.}$$ As we have complex eigenvalues with positive real part, the critical point is a spiral source, and therefore an unstable equilibrium point.

See Figure 8.4 for the phase diagram. Notice the two critical points, and the behavior of the arrows in the vector field around these points.

### Subsection8.2.3The trouble with centers

Recall, a linear system with a center means that trajectories travel in closed elliptical orbits in some direction around the critical point. Such a critical point we call a center or a stable center. It is not an asymptotically stable critical point, as the trajectories never approach the critical point, but at least if you start sufficiently close to the critical point, you stay close to the critical point. The simplest example of such behavior is the linear system with a center. Another example is the critical point $$(0,0)$$ in Example 8.1.1.

The trouble with a center in a nonlinear system is that whether the trajectory goes towards or away from the critical point is governed by the sign of the real part of the eigenvalues of the Jacobian matrix, and the Jacobian matrix in a nonlinear system changes from point to point. Since this real part is zero at the critical point itself, it can have either sign nearby, meaning the trajectory could be pulled towards or away from the critical point.

#### Example8.2.3.

An example of such a problematic behavior is the system $$x'=y, y' = -x+y^3\text{.}$$ The only critical point is the origin $$(0,0)\text{.}$$ The Jacobian matrix is

\begin{equation*} \begin{bmatrix} 0 & 1 \\ -1 & 3 y^2 \\ \end{bmatrix} . \end{equation*}

At $$(0,0)$$ the Jacobian matrix is $$\left[ \begin{smallmatrix} 0 & 1 \\ -1 & 0 \\ \end{smallmatrix} \right]\text{,}$$ which has eigenvalues $$\pm i\text{.}$$ So the linearization has a center.

Using the quadratic equation, the eigenvalues of the Jacobian matrix at any point $$(x,y)$$ are

\begin{equation*} \lambda = \frac{3}{2}y^2 \pm i \frac{\sqrt{4-9y^4}}{2} . \end{equation*}

At any point where $$y \not= 0$$ (so at most points near the origin), the eigenvalues have a positive real part ($$y^2$$ can never be negative). This positive real part pulls the trajectory away from the origin. A sample trajectory for an initial condition near the origin is given in Figure 8.5.

The moral of the example is that further analysis is needed when the linearization has a center. The analysis will in general be more complicated than in the example above, and is more likely to involve case-by-case consideration. Such a complication should not be surprising to you. By now in your mathematical career, you have seen many places where a simple test is inconclusive, recall for example the second derivative test for maxima or minima, and requires more careful, and perhaps ad hoc analysis of the situation.

### Subsection8.2.4Conservative equations

An equation of the form

\begin{equation*} x'' + f(x) = 0 \end{equation*}

for an arbitrary function $$f(x)$$ is called a conservative equation. For example the pendulum equation is a conservative equation. The equations are conservative as there is no friction in the system so the energy in the system is “conserved.” Let us write this equation as a system of nonlinear ODE.

\begin{equation*} x' = y, \qquad y' = -f(x) . \end{equation*}

These types of equations have the advantage that we can solve for their trajectories easily.

The trick is to first think of $$y$$ as a function of $$x$$ for a moment. Then use the chain rule

\begin{equation*} x'' = y' = \frac{dy}{dx} x' = y \frac{dy}{dx} , \end{equation*}

where the prime indicates a derivative with respect to $$t\text{.}$$ We obtain $$y \frac{dy}{dx} + f(x) = 0\text{.}$$ We integrate with respect to $$x$$ to get $$\int y \frac{dy}{dx} \,dx + \int f(x)\, dx = C\text{.}$$ In other words

\begin{equation*} \frac{1}{2} y^2 + \int f(x)\, dx = C . \end{equation*}

We obtained an implicit equation for the trajectories, with different $$C$$ giving different trajectories. The value of $$C$$ is conserved on any trajectory. This expression is sometimes called the Hamiltonian or the energy of the system. If you look back to Section 1.8, you will notice that $$y\frac{dy}{dx} + f(x) = 0$$ is an exact equation, and we just found a potential function.

#### Example8.2.4.

Let us find the trajectories for the equation $$x'' + x-x^2 = 0\text{,}$$ which is the equation from Example 8.1.1. The corresponding first order system is

\begin{equation*} x' = y , \qquad y' = -x+x^2 . \end{equation*}

Trajectories satisfy

\begin{equation*} \frac{1}{2} y^2 + \frac{1}{2} x^2 - \frac{1}{3} x^3 = C . \end{equation*}

We solve for $$y$$

\begin{equation*} y = \pm \sqrt{-x^2 + \frac{2}{3} x^3 + 2C} . \end{equation*}

Plotting these graphs we get exactly the trajectories in Figure 8.1. In particular we notice that near the origin the trajectories are closed curves: they keep going around the origin, never spiraling in or out. Therefore we discovered a way to verify that the critical point at $$(0,0)$$ is a stable center. The critical point at $$(0,1)$$ is a saddle as we already noticed. This example is typical for conservative equations.

Consider an arbitrary conservative equation $$x'' + f(x) = 0\text{.}$$ All critical points occur when $$y=0$$ (the $$x$$-axis), that is when $$x' = 0\text{.}$$ The critical points are those points on the $$x$$-axis where $$f(x) = 0\text{.}$$ The trajectories are given by

\begin{equation*} y = \pm \sqrt{ - 2 \int f(x)\, dx + 2C} . \end{equation*}

So all trajectories are mirrored across the $$x$$-axis. In particular, there can be no spiral sources nor sinks. The Jacobian matrix is

\begin{equation*} \begin{bmatrix} 0 & 1 \\ -f'(x) & 0 \end{bmatrix} . \end{equation*}

The critical point is almost linear if $$f'(x) \not= 0$$ at the critical point. Let $$J$$ denote the Jacobian matrix. The eigenvalues of $$J$$ are solutions to

\begin{equation*} 0 = \det(J - \lambda I) = \lambda^2 + f'(x) . \end{equation*}

Therefore $$\lambda = \pm \sqrt{-f'(x)}\text{.}$$ In other words, either we get real eigenvalues of opposite signs (if $$f'(x) < 0$$), or we get purely imaginary eigenvalues (if $$f'(x) > 0$$). There are only two possibilities for critical points, either an unstable saddle point, or a stable center. There are never any sinks or sources.

### Subsection8.2.5Exercises

#### Exercise8.2.1.

For the systems below, find and classify the critical points, also indicate if the equilibria are stable, asymptotically stable, or unstable.

1. $$\displaystyle x'=-x+3x^2, y'=-y$$

2. $$x'=x^2+y^2-1\text{,}$$ $$y'=x$$

3. $$x'=ye^x\text{,}$$ $$y'=y-x+y^2$$

#### Exercise8.2.2.

Find the implicit equations of the trajectories of the following conservative systems. Next find their critical points (if any) and classify them.

1. $$\displaystyle x''+ x+x^3 = 0$$

2. $$\displaystyle \theta''+\sin \theta = 0$$

3. $$\displaystyle z''+ (z-1)(z+1) = 0$$

4. $$\displaystyle x''+ x^2+1 = 0$$

#### Exercise8.2.3.

Find and classify the critical point(s) of $$x' = -x^2\text{,}$$ $$y' = -y^2\text{.}$$

#### Exercise8.2.4.

Suppose $$x'=-xy\text{,}$$ $$y'=x^2-1-y\text{.}$$

1. Show there are two spiral sinks at $$(-1,0)$$ and $$(1,0)\text{.}$$

2. For any initial point of the form $$(0,y_0)\text{,}$$ find what is the trajectory.

3. Can a trajectory starting at $$(x_0,y_0)$$ where $$x_0 > 0$$ spiral into the critical point at $$(-1,0)\text{?}$$ Why or why not?

#### Exercise8.2.5.

In the example $$x'=y\text{,}$$ $$y'=y^3-x$$ show that for any trajectory, the distance from the origin is an increasing function. Conclude that the origin behaves like is a spiral source. Hint: Consider $$f(t) = {\bigl(x(t)\bigr)}^2 + {\bigl(y(t)\bigr)}^2$$ and show it has positive derivative.

#### Exercise8.2.6.

Suppose $$f$$ is always positive. Find the trajectories of $$x''+f(x') = 0\text{.}$$ Are there any critical points?

#### Exercise8.2.7.

Suppose that $$x' = f(x,y)\text{,}$$ $$y' = g(x,y)\text{.}$$ Suppose that $$g(x,y) > 1$$ for all $$x$$ and $$y\text{.}$$ Are there any critical points? What can we say about the trajectories at $$t$$ goes to infinity?

#### Exercise8.2.101.

For the systems below, find and classify the critical points.

1. $$x'=-x+x^2\text{,}$$ $$y'=y$$

2. $$x'=y-y^2-x\text{,}$$ $$y'=-x$$

3. $$x'=xy\text{,}$$ $$y'=x+y-1$$

a) $$(0,0)\text{:}$$ saddle (unstable), $$(1,0)\text{:}$$ source (unstable),        b) $$(0,0)\text{:}$$ spiral sink (asymptotically stable), $$(0,1)\text{:}$$ saddle (unstable),        c) $$(1,0)\text{:}$$ saddle (unstable), $$(0,1)\text{:}$$ source (unstable)

#### Exercise8.2.102.

Find the implicit equations of the trajectories of the following conservative systems. Next find their critical points (if any) and classify them.

1. $$\displaystyle x''+ x^2 = 4$$

2. $$\displaystyle x''+ e^x = 0$$

3. $$\displaystyle x''+ (x+1)e^x = 0$$

a) $$\frac{1}{2}y^2 + \frac{1}{3}x^3 -4x = C\text{,}$$ critical points: $$(-2,0)\text{,}$$ an unstable saddle, and $$(2,0)\text{,}$$ a stable center.     b) $$\frac{1}{2}y^2 + e^x = C\text{,}$$ no critical points.     c) $$\frac{1}{2}y^2 + xe^x = C\text{,}$$ critical point at $$(-1,0)$$ is a stable center.

#### Exercise8.2.103.

The conservative system $$x''+x^3 = 0$$ is not almost linear. Classify its critical point(s) nonetheless.

Critical point at $$(0,0)\text{.}$$ Trajectories are $$y = \pm \sqrt{2C-(\nicefrac{1}{2})x^4}\text{,}$$ for $$C > 0\text{,}$$ these give closed curves around the origin, so the critical point is a stable center.
Derive an analogous classification of critical points for equations in one dimension, such as $$x'= f(x)$$ based on the derivative. A point $$x_0$$ is critical when $$f(x_0) = 0$$ and almost linear if in addition $$f'(x_0) \not= 0\text{.}$$ Figure out if the critical point is stable or unstable depending on the sign of $$f'(x_0)\text{.}$$ Explain. Hint: see Section 1.6.
A critical point $$x_0$$ is stable if $$f'(x_0) < 0$$ and unstable when $$f'(x_0) > 0\text{.}$$