Problem 13 - Differential Equations and Linear Algebra

Question

Characterize the equilibrium point for the system \(\mathbf{x}^{\prime}=A \mathbf{x}\) and sketch the phase portrait. $$A=\left[\begin{array}{ll} 3 & -2 \\ 2 & -1 \end{array}\right]$$

Step-by-Step Solution

Answer

The equilibrium point for the given system is at the origin, \(\mathbf{x^*} = \begin{pmatrix}0\\0\end{pmatrix}\), with eigenvalues \(\lambda_1 = 1\) (positive) and \(\lambda_2 = -1\) (negative), indicating a saddle point behavior. The phase portrait will show trajectories starting from initial conditions along the eigenvectors, converging or diverging towards the equilibrium point, reflecting the saddle point behavior. The corresponding eigenvectors are \(\mathbf{v}_1 = \begin{pmatrix}1\\1\end{pmatrix}\) and \(\mathbf{v}_2 = \begin{pmatrix}1\\2\end{pmatrix}\).

Step 1: Find the eigenvalues and eigenvectors of matrix A

We can start by finding the characteristic equation of matrix \(A\): $$\text{det}(A-\lambda I) = \text{det}\begin{pmatrix} 3-\lambda & -2 \\ 2 & -1-\lambda \end{pmatrix}$$ Calculate the determinant: \((3-\lambda)((-1-\lambda))-(-2)(2) = (\lambda^2-2\lambda -1). \) Now, we need to find the eigenvalues by solving the quadratic equation: \(\lambda^2-2\lambda -1 = 0\) Solving this equation, we obtain the eigenvalues: \(\lambda_1 = 1\) and \(\lambda_2 = -1\). Next, find the eigenvectors for each eigenvalue: For \(\lambda_1 = 1\), $$(A-\lambda_1 I)\mathbf{v}_1 = \begin{pmatrix} 2 & -2 \\ 2 & -2 \end{pmatrix}\mathbf{v}_1 = \mathbf{0}$$ We can choose \(\mathbf{v}_1 = \begin{pmatrix}1\\1\end{pmatrix}\) as an eigenvector. For \(\lambda_2 = -1\), $$(A-\lambda_2 I)\mathbf{v}_2 = \begin{pmatrix} 4 & -2 \\ 2 & 0 \end{pmatrix}\mathbf{v}_2 = \mathbf{0}$$ We can choose \(\mathbf{v}_2 = \begin{pmatrix}1\\2\end{pmatrix}\) as an eigenvector.

Step 2: Characterize the equilibrium point

To find the equilibrium point, we need to determine when \(\mathbf{x}^{\prime}=A\mathbf{x}\) is equal to the zero vector: $$A\mathbf{x} = \mathbf{0}$$ $$\left[\begin{array}{ll} 3 & -2 \\ 2 & -1 \end{array}\right]\begin{pmatrix}x_1\\x_2\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}$$ Solving this system of linear equations, we find that the equilibrium point is at the origin: $$\mathbf{x^*} = \begin{pmatrix}0\\0\end{pmatrix}.$$

Step 3: Classify system behavior and sketch phase portrait

Considering the eigenvalues previously found: \(\lambda_1 = 1\) and \(\lambda_2 = -1\), we have a saddle point as both eigenvalues have different signs (one positive, one negative). To sketch the phase portrait, we will plot solution trajectories for a few initial conditions for \(\mathbf{x}\). Consider that the general solution for this system is given by: \[\mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2,\] With different initial conditions, we can create a sketch that will include enough trajectories to characterize the phase portrait. The phase portrait will have trajectories starting from initial conditions along the eigenvectors converging or diverging towards the equilibrium point at \(\mathbf{x^*} = \begin{pmatrix}0\\0\end{pmatrix}\), showing that points on this eigenvector direction will tend to move closer or farther from the equilibrium point, highlighting the saddle point behavior.

Key Concepts

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental concepts in the study of linear algebra and differential equations. They provide insight into the behavior of a system. By analyzing the matrix \(A\), we find values called eigenvalues and vectors called eigenvectors. These help us understand how transformations affect the system's state.

To find eigenvalues, we solve the characteristic equation derived from \(\text{det}(A - \lambda I)\). For the given matrix \(A\), the characteristic equation is \(\lambda^2 - 2\lambda - 1 = 0\). Solving this, we find the eigenvalues: \(\lambda_1 = 1\) and \(\lambda_2 = -1\).

With each eigenvalue, we compute its respective eigenvector by solving \((A - \lambda I)\mathbf{v} = \mathbf{0}\). For example:

• For \(\lambda_1 = 1\), the eigenvector is \(\mathbf{v}_1 = \begin{pmatrix}1\1\end{pmatrix}\).
• For \(\lambda_2 = -1\), the eigenvector is \(\mathbf{v}_2 = \begin{pmatrix}1\2\end{pmatrix}\).

Eigenvectors give us directions along which the system exhibits stretched behavior, and eigenvalues tell us about the stretching or compression factor.

Equilibrium Points

Equilibrium points in the context of differential equations are the points where the system remains constant over time. For a system described by \(\mathbf{x}^{\prime}=A\mathbf{x}\), the equilibrium point occurs when \(A\mathbf{x} = \mathbf{0}\).

In our example, solving the equation \(A\mathbf{x} = \mathbf{0}\) for the given matrix \(A\) provides:\
\[ \left[\begin{array}{ll} 3 & -2\ 2 & -1 \end{array}\right]\begin{pmatrix}x_1\x_2\end{pmatrix} = \begin{pmatrix}0\0\end{pmatrix} \]

Solving this system, we find that the equilibrium point is at the origin, \(\mathbf{x^*} = \begin{pmatrix}0\0\end{pmatrix}\). This is the point where the system will settle if undisturbed, regardless of its initial position. Detecting equilibrium points helps in predicting long-term behavior and understanding stability.

Saddle Points

Saddle points in differential equations are special types of equilibrium points characterized by a certain instability. When we observe the eigenvalues, if there are both positive and negative values, the equilibrium point behaves like a saddle.

In this exercise, the eigenvalues \(\lambda_1 = 1\) and \(\lambda_2 = -1\) indicate a saddle point because they have different signs:

• The positive eigenvalue indicates that the system grows without bound along one direction.
• The negative eigenvalue implies contraction along another direction.

This leads to trajectories moving away from some regions and towards others when visualized. Saddle points show how an equilibrium can be stable in some directions yet unstable in others, which is crucial in understanding dynamic systems.

Phase Portraits

Phase portraits are graphical representations that illustrate how dynamical systems behave over time. They provide a visual insight into the paths taken by the system states.

For the matrix \(A\), with the given eigenvalues and eigenvectors, the phase portrait can be drawn by plotting trajectories that represent solutions to the system \(\mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2\).

• Trajectories along the direction of \(\mathbf{v}_2\) (eigenvector corresponding to \(\lambda_2 = -1\)) tend to converge towards the equilibrium point.
• Trajectories along the \(\mathbf{v}_1\) direction, due to \(\lambda_1 = 1\), diverge away from the equilibrium.

Phase portraits of systems displaying saddle point behavior show hyperbolic trajectories, clearly demonstrating stable and unstable directions. They are important tools for predicting system behavior in scenarios with various initial conditions.