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In Exercises \(9-18\) , construct the general solution of \(\mathbf{x}^{\prime}=A \mathbf{x}\) involving complex eigenfunctions and then obtain the general real solution. Describe the shapes of typical trajectories. $$ A=\left[\begin{array}{rl}{3} & {1} \\ {-2} & {1}\end{array}\right] $$

Short Answer

Expert verified
The general real solution is a spiral outward from the origin, due to exponential growth and cyclical movement.

Step by step solution

01

Find the Eigenvalues

The system is defined by \( \mathbf{x'} = A \mathbf{x} \), where \( A = \begin{bmatrix} 3 & 1 \ -2 & 1 \end{bmatrix} \). First, we need to find the eigenvalues by solving the characteristic equation \( |A - \lambda I| = 0 \). This gives us \( |\begin{bmatrix} 3-\lambda & 1 \ -2 & 1-\lambda \end{bmatrix}| = 0 \), which simplifies to \((3-\lambda)(1-\lambda) + 2 = 0 \). Expand this to obtain \( \lambda^2 - 4\lambda + 5 = 0 \). Use the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), with \(a = 1\), \(b = -4\), \(c = 5\), resulting in eigenvalues \( \lambda = 2 \pm i \).
02

Find the Complex Eigenvectors

For \( \lambda = 2 + i \), substitute back into \((A-\lambda I)\mathbf{v} = \mathbf{0}\): \( \begin{bmatrix} 1-i & 1 \ -2 & -1-i \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \). This system reduces to \( x(1-i) + y = 0 \). Let \( y = 1 \), hence \( x = i-1 \). So, an eigenvector can be \( \mathbf{v} = \begin{bmatrix} i-1 \ 1 \end{bmatrix} \).
03

Formulate the General Complex Solution

The general solution is \( \mathbf{x}(t) = c_1 e^{(2+i)t} \begin{bmatrix} i-1 \ 1 \end{bmatrix} \). This simplifies to \( c_1 e^{2t} e^{it} \begin{bmatrix} i-1 \ 1 \end{bmatrix} \). Writing \( e^{it} = \cos(t) + i \sin(t) \), split into real and imaginary parts: \( \mathbf{x}_1(t) = e^{2t} \left( \cos(t) \begin{bmatrix} 1 \ 0 \end{bmatrix} - \sin(t) \begin{bmatrix} 0 \ 1 \end{bmatrix} \right) \) and \( \mathbf{x}_2(t) = e^{2t} \left( \sin(t) \begin{bmatrix} 1 \ 0 \end{bmatrix} + \cos(t) \begin{bmatrix} 0 \ 1 \end{bmatrix} \right) \).
04

Write the General Real Solution

The general real solution is obtained by combining the real and imaginary parts that stem from the complex solution. Thus, \( \mathbf{x}(t) = c_1 \mathbf{x}_1(t) + c_2 \mathbf{x}_2(t) \), where \( \mathbf{x}(t) = c_1 e^{2t} \begin{bmatrix} \cos(t) \ \sin(t) \end{bmatrix} + c_2 e^{2t} \begin{bmatrix} \sin(t) \ \cos(t) \end{bmatrix} \).
05

Describe the Trajectories

The trajectories are spirals moving outward from the origin, as indicated by the \( e^{2t} \) term (meaning they grow exponentially over time). The periodic sine and cosine components cause the solution to rotate around the origin. The eigenvalues' real part, positive at \(2\), confirms outward spiraling, showing growth without oscillation damping.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

General Solution
When we are tasked to find the general solution to a system of differential equations like \( \mathbf{x'} = A \mathbf{x} \), it involves finding the eigenvalues and eigenvectors of matrix \( A \). The general solution provides a comprehensive form of how solutions behave over time. Since matrix \( A \) here gives complex eigenvalues \( \lambda = 2 \pm i \), we can expect the solutions to exhibit oscillatory behavior while their amplitude increases.
To find the general complex solution, we first solve the characteristic equation, resulting in complex conjugate eigenvalues indicating oscillation. Next, we find the corresponding complex eigenvectors, here shown as \( \mathbf{v} = \begin{bmatrix} i-1 \ 1 \end{bmatrix} \). The general complex solution takes form \( \mathbf{x}(t) = c_1 e^{(2+i)t} \mathbf{v} \), combining both real and imaginary exponential and trigonometric components.
We express the complex exponential part using Euler's formula \( e^{it} = \cos(t) + i\sin(t) \). Thus, by separating into real and imaginary parts, we extract a solution structure that incorporates both exponential growth and rotational aspects, fundamental to systems characterized by complex eigenvalues.
Real Solution
Once the general complex solution is found, it is crucial to convert it into a real solution, especially when dealing extensively with real-world phenomena. To do this, we separate the complex expression of the solution into two distinct real parts by utilizing the terms from the identities involving sine and cosine.
The real solution of the system is formed by taking the real and imaginary parts of the generalized complex solution:
  • The first component \( \mathbf{x}_1(t) = e^{2t} \left( \cos(t) \begin{bmatrix} 1 \ 0 \end{bmatrix} - \sin(t) \begin{bmatrix} 0 \ 1 \end{bmatrix} \right) \) captures cosine-related behavior.
  • The second component \( \mathbf{x}_2(t) = e^{2t} \left( \sin(t) \begin{bmatrix} 1 \ 0 \end{bmatrix} + \cos(t) \begin{bmatrix} 0 \ 1 \end{bmatrix} \right) \) reflects sine-oriented effects.
The combination of these components leads to the general real solution \( \mathbf{x}(t) = c_1 \mathbf{x}_1(t) + c_2 \mathbf{x}_2(t) \). It reveals both oscillatory characteristics from \( \sin \) and \( \cos \) terms, alongside the exponential factor \( e^{2t} \) dictating radial expansion over time.
Trajectories
The trajectories of a system like \( \mathbf{x'} = A\mathbf{x} \) depict the paths followed by the system's state over time. In this context, with eigenvalues \( \lambda = 2 \pm i \), we observe that trajectories are spiral outward, showcasing both growth and rotational dynamics.
The exponential term \( e^{2t} \) from the solution increases the amplitude of the state vector, causing the trajectories to expand outward from the origin as time progresses. This means that, without any restricting forces, trajectories move away from the equilibrium at an increasing rate.
Simultaneously, the sine and cosine components impart a rotational movement, ensuring that the state rotates around the origin, creating a spiraling effect. The real part of the eigenvalues, \( 2 \), confirms that these spirals are not merely stationary but expand progressively, without oscillations dying out. Thus, these trajectories signify a dynamic shift involving both directional rotation and radial growth out from the starting point.

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Most popular questions from this chapter

Exercises \(9-14\) require techniques from Section \(3.1 .\) Find the characteristic polynomial of each matrix, using either a cofactor expansion or the special formula for \(3 \times 3\) determinants described prior to Exercises \(15-18\) in Section \(3.1 .\) INote: Finding the characteristic polynomial of a \(3 \times 3\) matrix is not easy to do with just row operations, because the variable \(\lambda\) is involved. $$ \left[\begin{array}{rrr}{1} & {0} & {-1} \\ {2} & {3} & {-1} \\ {0} & {6} & {0}\end{array}\right] $$

Show that if \(A\) has \(n\) linearly independent eigenvectors, then so does \(A^{T} .[\text { Hint: Use the Diagonalization Theorem. }]\)

In Exercises 30 and \(31,\) find the \(\mathcal{B}\) -matrix for the transformation \(\mathbf{x} \mapsto A \mathbf{x}\) when \(\mathcal{B}=\left\\{\mathbf{b}_{1}, \mathbf{b}_{2}, \mathbf{b}_{3}\right\\}\) $$ \begin{array}{l}{A=\left[\begin{array}{rrr}{-7} & {-48} & {-16} \\ {1} & {14} & {6} \\ {-3} & {-45} & {-19}\end{array}\right]} \\\ {\mathbf{b}_{1}=\left[\begin{array}{r}{-3} \\ {1} \\ {-3}\end{array}\right], \mathbf{b}_{2}=\left[\begin{array}{r}{-2} \\ {-2} \\ {-3}\end{array}\right], \mathbf{b}_{3}=\left[\begin{array}{r}{3} \\ {-1} \\\ {0}\end{array}\right]}\end{array} $$

Let \(A\) be an \(n \times n\) matrix, and suppose \(A\) has \(n\) real eigenvalues, \(\lambda_{1}, \ldots, \lambda_{n},\) repeated according to multiplicities, so that \(\operatorname{det}(A-\lambda I)=\left(\lambda_{1}-\lambda\right)\left(\lambda_{2}-\lambda\right) \cdots\left(\lambda_{n}-\lambda\right)\) Explain why det \(A\) is the product of the \(n\) eigenvalues of \(A .\) (This result is true for any square matrix when complex eigenvalues are considered.)

Let \(A\) be a complex (or real) \(n \times n\) matrix, and let \(\mathbf{x}\) in \(\mathbb{C}^{n}\) be an eigenvector corresponding to an eigenvalue \(\lambda\) in \(\mathbb{C} .\) Show that for each nonzero complex scalar \(\mu\) , the vector \(\mu \mathbf{x}\) is an eigenvector of \(A .\)

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