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Chapter 7: Problem 10

find a fundamental matrix for the given system of equations. In each case also find the fundamental matrix \(\mathbf{\Phi}(t)\) satisfying \(\Phi(0)=\mathbf{1}\) $$ \mathbf{x}^{\prime}=\left(\begin{array}{rrr}{1} & {-1} & {4} \\ {3} & {2} & {-1} \\ {2} & {1} & {-1}\end{array}\right) \mathbf{x} $$

Short Answer

Expert verified
Answer: The fundamental matrix \(\mathbf{\Phi}(t)\) is given by: \(\mathbf{\Phi}(t) = \begin{pmatrix}0 & e^{2t} & 2e^{-t} \\ 0 & (2-t)e^{2t} & 2(-t-1)e^{-t} \\ 0 & (-1-t)e^{2t} & (e^{-t}-t-1)\end{pmatrix}\).

Step by step solution

01

Finding Eigenvalues and Eigenvectors

Given the matrix: $$ A = \left(\begin{array}{rrr}{1} & {-1} & {4} \\\ {3} & {2} & {-1} \\\ {2} & {1} & {-1}\end{array}\right) $$ We can find the eigenvalues by solving the following equation: $$ \det(A - \lambda I) = 0 $$ This results in the characteristic equation: $$ -(\lambda - 1)\left[(2-\lambda)(-1-\lambda) + 1\right] + (-1)\left(3(-1-\lambda)+2\right) = 0 $$ Solving this equation, we get the eigenvalues: $$ \lambda_1 = 1, \lambda_2 = 2, \lambda_3 = -1. $$ Next, we find the eigenvectors associated with each eigenvalue by solving the equation: $$ (A - \lambda_k I)\mathbf{v}_k = 0 $$ where \(k=1,2,3\). The eigenvectors are: $$ \mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \mathbf{v}_2 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}, \mathbf{v}_3 = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}. $$
02

Constructing the General Solution

To construct the general solution to the system, we will first form the matrix \(\mathbf{V}(t)\) whose columns are the solutions of the form \(\mathbf{v}_k e^{\lambda_k t}\): $$ \mathbf{V}(t) = \begin{pmatrix} \ e^{t} & \ \ e^{2t} & \ e^{-t} \\ \ e^{t} & \ \ 0 & \ -2e^{-t} \\ \ 0 & \ -e^{2t} & \ \ e^{-t} \\ \end{pmatrix} $$ Next, we will compute the matrix exponential of \(A\), denoted as \(e^{At}\): $$ e^{At} = \mathbf{V}(t) \begin{pmatrix} e^{\lambda_1t} & 0 & 0 \\ 0 & e^{\lambda_2t} & 0 \\ 0 & 0 & e^{\lambda_3t} \end{pmatrix} \mathbf{V}(t)^{-1} $$ The matrix \(\mathbf{V}(t)^{-1}\) can be computed using Gaussian elimination or Cramer's rule: $$ \mathbf{V}(t)^{-1} = \dfrac{1}{t-1}\begin{pmatrix}2 & e^{t} & e^{-t} \\ -1 & 0 & -e^{t}+e^{-t} \\ 0 & -e^{t} & -1-e^{t}+e^{-t}\end{pmatrix} $$ The general solution is then given by: $$ \mathbf{x}(t) = \mathbf{V}(t) \begin{pmatrix} e^{\lambda_1t} \\ e^{\lambda_2t} \\ e^{\lambda_3t} \end{pmatrix} = \begin{pmatrix}xe^{t}+ye^{2t}+2ze^{-t}\\xe^{t}+y(2-t)e^{2t}+2z(-t-1)e^{-t}\\y(-1-t)e^{2t}+z(e^{-t}-t-1)\end{pmatrix} $$ where \(x,y,z\) are constants.
03

Finding the Fundamental Matrix Satisfying the Initial Conditions

To find the fundamental matrix \(\mathbf{\Phi}(t)\) satisfying \(\Phi(0)=\mathbf{1}\), we need to determine the constants \(x,y,z\) such that: $$ \mathbf{1} = \mathbf{x}(0) $$ This gives us the system of equations: $$ \begin{cases} x+y+\phantom{t}2z = \phantom{-}1 \\ x-2y+(1-2t)z = -1 \\ (-1-t)y+(1-t-z)z = \phantom{-}1 \\ \end{cases} $$ At \(t = 0\), we have: $$ \begin{cases} x+y+2z = 1 \\ x-2y+z = -1 \\ -y+z = 1 \\ \end{cases} $$ Solving this system of equations, we obtain \(x=0, y=-1, z=0\). Thus, the fundamental matrix satisfying the initial condition is given by: $$ \mathbf{\Phi}(t) = \begin{pmatrix}0 & e^{2t} & 2e^{-t} \\ 0 & (2-t)e^{2t} & 2(-t-1)e^{-t} \\ 0 & (-1-t)e^{2t} & (e^{-t}-t-1)\end{pmatrix}. $$

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Key Concepts

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

Eigenvalues and Eigenvectors
Understanding eigenvalues and eigenvectors is crucial in the study of linear algebra and its application to systems of differential equations. When working with a system of linear equations, eigenvalues represent how the matrix of the system can scale a vector, while eigenvectors are the specific vectors that are scaled by the matrix.

Eigenvalues are found by solving the characteristic equation, which leads to values that express certain fundamental properties of the linear transformation described by the matrix. For a given eigenvalue, the corresponding eigenvector is computed by solving the equation \(A - \(lambda_k\) I)\mathbf{v}_k = 0\), where \((\lambda_k\)) is an eigenvalue, \(I\) is the identity matrix, and \((\mathbf{v}_k\)) is the eigenvector linked to \((\lambda_k\)). In the context of the given exercise, computing these values allows us to determine the behavior of the system over time.

Eigenvectors can also be particularly helpful when forming the general solution to a system of differential equations, as they can be combined with the eigenvalues in an exponential function to express how the system evolves.
Characteristic Equation
The characteristic equation is a fundamental concept in determining the eigenvalues of a matrix. It is obtained by setting the determinant of \( (A - \(lambda\) I) \) equal to zero, where \(A\) is the matrix in question, \(lambda\) represents a scalar (eigenvalue), and \(I\) is the identity matrix of corresponding size. This equation results in a polynomial whose roots are the eigenvalues of the matrix.

In our exercise, the characteristic equation has the form \(\det(A - \lambda I) = 0\), which leads to a cubic polynomial. Solving this polynomial entails finding the values of \((\lambda\)) that satisfy the equation. These values are the keys to unlocking the system's dynamics as they provide the necessary constants to construct the fundamental matrix.

Once the eigenvalues are computed, eigenvectors are subsequently found, which in turn assist in determining the solution to the system of linear differential equations by directly influencing the structure of the solution.
Matrix Exponential
The concept of the matrix exponential is a powerful tool that arises in the analysis of linear systems of differential equations. The matrix exponential, denoted as \( e^{At} \) for a matrix \(A\), is used to find solutions of such systems over time. It represents the continuous-time version of raising a matrix to a power when dealing with dynamical systems.

In the exercise provided, \( e^{At} \) is found by multiplying the matrix of eigenvectors, \( \mathbf{V}(t) \), by a diagonal matrix with the exponential eigenvalues along the diagonal, and then multiplying this by the inverse of \( \mathbf{V}(t) \).

The calculation of the matrix exponential provides us with the fundamental matrix of the system, which encapsulates the full behavior of the system over time when coupled with initial conditions. By analyzing the matrix exponential, we gain insight into the long-term behavior of the system and how perturbations in the initial state will evolve as time progresses.

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Most popular questions from this chapter

Consider the system $$ x^{\prime}=A x=\left(\begin{array}{rrr}{1} & {1} & {1} \\ {2} & {1} & {-1} \\\ {-3} & {2} & {4}\end{array}\right) x $$ (a) Show that \(r=2\) is an eigenvalue of multiplicity 3 of the coefficient matrix \(\mathbf{A}\) and that there is only one corresponding cigenvector, namely, $$ \xi^{(1)}=\left(\begin{array}{r}{0} \\ {1} \\ {-1}\end{array}\right) $$ (b) Using the information in part (a), write down one solution \(\mathbf{x}^{(1)}(t)\) of the system (i). There is no other solution of the purely exponential form \(\mathbf{x}=\xi e^{y t}\). (c) To find a second solution assume that \(\mathbf{x}=\xi t e^{2 t}+\mathbf{\eta} e^{2 t} .\) Show that \(\xi\) and \(\mathbf{\eta}\) satisfy the equations $$ (\mathbf{A}-2 \mathbf{I}) \xi=\mathbf{0}, \quad(\mathbf{A}-2 \mathbf{I}) \mathbf{n}=\mathbf{\xi} $$ since \(\xi\) has already been found in part (a), solve the second equation for \(\eta\). Neglect the multiple of \(\xi^{(1)}\) that appears in \(\eta\), since it leads only to a multiple of the first solution \(\mathbf{x}^{(1)}\). Then write down a second solution \(\mathbf{x}^{(2)}(t)\) of the system (i). (d) To find a third solution assume that \(\mathbf{x}=\xi\left(t^{2} / 2\right) e^{2 t}+\mathbf{\eta} t e^{2 t}+\zeta e^{2 t} .\) Show that \(\xi, \eta,\) and \(\zeta\) satisfy the equations $$ (\mathbf{A}-2 \mathbf{l}) \xi=\mathbf{0}, \quad(\mathbf{\Lambda}-2 \mathbf{I}) \mathbf{\eta}=\mathbf{\xi}, \quad(\mathbf{A}-2 \mathbf{l}) \zeta=\mathbf{\eta} $$ The first two equations are the same as in part (c), so solve the third equation for \(\zeta,\) again neglecting the multiple of \(\xi^{(1)}\) that appears. Then write down a third solution \(\mathbf{x}^{(3)}(t)\) of the system (i). (e) Write down a fundamental matrix \(\boldsymbol{\Psi}(t)\) for the system (i). (f) Form a matrix \(\mathbf{T}\) with the cigenvector \(\xi^{(1)}\) in the first column, and the generalized eigenvectors \(\eta\) and \(\zeta\) in the second and third columns. Then find \(T^{-1}\) and form the product \(\mathbf{J}=\mathbf{T}^{-1} \mathbf{A} \mathbf{T}\). The matrix \(\mathbf{J}\) is the Jordan form of \(\mathbf{A}\).

Deal with the problem of solving \(\mathbf{A x}=\mathbf{b}\) when \(\operatorname{det} \mathbf{A}=0\) Suppose that det \(\mathbf{A}=0\) and that \(y\) is a solution of \(\mathbf{A}^{*} \mathbf{y}=\mathbf{0} .\) Show that if \((\mathbf{b}, \mathbf{y})=0\) for every such \(\mathbf{y},\) then \(\mathbf{A} \mathbf{x}=\mathbf{b}\) has solutions. Note that the converse of Problem \(27 ;\) the form of the solution is given by Problem \(28 .\)

Find all eigenvalues and eigenvectors of the given matrix. $$ \left(\begin{array}{ccc}{11 / 9} & {-2 / 9} & {8 / 9} \\ {-2 / 9} & {2 / 9} & {10 / 9} \\ {8 / 9} & {10 / 9} & {5 / 9}\end{array}\right) $$

Express the general solution of the given system of equations in terms of real-valued functions. In each of Problems 1 through 6 also draw a direction field, sketch a few of the trajectories, and describe the behavior of the solutions as \(t \rightarrow \infty\). $$ \mathbf{x}^{\prime}=\left(\begin{array}{ll}{1} & {-1} \\ {5} & {-3}\end{array}\right) \mathbf{x} $$

The coefficient matrix contains a parameter \(\alpha\). In each of these problems: (a) Determine the eigervalues in terms of \(\alpha\). (b) Find the critical value or values of \(\alpha\) where the qualitative nature of the phase portrait for the system changes. (c) Draw a phase portrait for a value of \(\alpha\) slightly below, and for another value slightly above, each crititical value. $$ x^{\prime}=\left(\begin{array}{cc}{4} & {\alpha} \\ {8} & {-6}\end{array}\right) \mathbf{x} $$
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