Finally,

x

∗

is asymptotically stable if it is both attracting and Liapunov stable.

For each of the following systems, decide whether the origin is attracting, Liapunov stable, asymptotically stable, or none of the above.

a)

x

˙

=

y

,

y

˙

=

−

4

x

We are tasked with analyzing the stability of the origin for the given system of differential equations:

\[
\dot{x} = y, \quad \dot{y} = -4x
\]

This is a system of two first-order linear differential equations. To determine the nature of the origin (whether it is attracting, Lyapunov stable, asymptotically stable, or none of the above), we will proceed with the following steps:

### Step 1: Rewrite the system in matrix form
We can express the system in vector form as:

\[
\begin{pmatrix}
\dot{x} \\
\dot{y}
\end{pmatrix}
=
\begin{pmatrix}
0 & 1 \\
-4 & 0
\end{pmatrix}
\begin{pmatrix}
x \\
y
\end{pmatrix}
\]

### Step 2: Find the eigenvalues of the system
To analyze stability, we need to find the eigenvalues of the coefficient matrix. The matrix is:

\[
A = \begin{pmatrix}
0 & 1 \\
-4 & 0
\end{pmatrix}
\]

The characteristic equation for this matrix is:

\[
\det(A - \lambda I) = 0
\]

where \( I \) is the identity matrix and \( \lambda \) represents the eigenvalues. Solving for \( \lambda \):

\[
\det\begin{pmatrix}
0 - \lambda & 1 \\
-4 & 0 - \lambda
\end{pmatrix} = \lambda^2 + 4 = 0
\]

\[
\lambda^2 = -4 \quad \Rightarrow \quad \lambda = \pm 2i
\]

Thus, the system has purely imaginary eigenvalues \( \lambda = \pm 2i \).

### Step 3: Classify the stability based on eigenvalues

Since the eigenvalues are purely imaginary (\( \pm 2i \)), this suggests that the origin corresponds to a **center** in the phase plane, meaning the trajectories around the origin are **closed orbits** (i.e., periodic solutions).

### Step 4: Conclusion
Given that the origin is associated with closed orbits:
- The origin is **Lyapunov stable** because small perturbations from the origin will not cause trajectories to escape to infinity.
- However, the origin is **not attracting** because solutions do not approach the origin as time increases (they simply circle around it).
- Therefore, the origin is **not asymptotically stable**, as solutions do not decay to the origin over time.

### Final Answer:
For the system:

\[
\dot{x} = y, \quad \dot{y} = -4x
\]

- The origin is **Lyapunov stable**, but it is **not attracting**.
- Since it is not attracting, it is **not asymptotically stable**.

Would you like more details on any part of this process?

---

**Related Questions:**
1. What is the difference between Lyapunov stability and asymptotic stability?
2. How do eigenvalues affect the stability of linear systems?
3. What are the conditions for a fixed point to be classified as a saddle point?
4. Can a system with purely imaginary eigenvalues ever be attracting?
5. What does a phase portrait of this system look like?

**Tip:** Eigenvalues with positive real parts indicate instability, while negative real parts suggest stability. Purely imaginary eigenvalues often indicate neutral stability.

Finally, x* is asymptotically stable if it is both attracting and Liapunov stable. For each of the following systems, decide whether the origin is attracting, Liapunov stable, asymptotically stable, or none of the above. a) x˙ = y, y˙ = −4x

Explore the stability of the origin for the system x˙ = y, y˙ = −4x. This detailed analysis covers the use of eigenvalues to determine Lyapunov stability and asymptotic stability. Learn about key concepts like the characteristic equation and stability conditions. Suitable for college-level students.

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