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Chapter 7: Problem 18

Solve the given initial value problem. Describe the behavior of the solution as \(t \rightarrow \infty\). $$ \mathbf{x}^{\prime}=\left(\begin{array}{rrr}{0} & {0} & {-1} \\ {2} & {0} & {0} \\ {-1} & {2} & {4}\end{array}\right) \mathbf{x}, \quad \mathbf{x}(0)=\left(\begin{array}{l}{7} \\ {5} \\ {5}\end{array}\right) $$

Short Answer

Expert verified
A: \(\left(\begin{array}{rrr}{0} & {0} & {-1}\\ {2} & {0} &{0} \\ {-1} & {2} & {4}\end{array}\right)\) and \(\mathbf{x}(0) = \left(\begin{array}{l}{7}\\ {5}\\ {5}\end{array}\right)\).

Step by step solution

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01

To find the eigenvalues of the matrix, we need to solve the following equation: $$ |A - \lambda I|=0, $$ where A is the given matrix, λ is the eigenvalue, and I is the identity matrix. After finding the eigenvalues, we will find the corresponding eigenvectors by solving the equation \((A - \lambda I) \mathbf{v} = \mathbf{0}\), where λ is an eigenvalue and \(\mathbf{v}\) is the corresponding eigenvector. The matrix for which we will find eigenvalues and eigenvectors is: \(\left(\begin{array}{rrr}{0} & {0} & {-1}\\ {2} & {0} &{0} \\ {-1} & {2} & {4}\end{array}\right)\). #Step 2: Calculate the matrix exponential#

After finding the eigenvalues and eigenvectors, we need to calculate the matrix exponential for A: $$ e^{tA} = P \cdot e^{tD} \cdot P^{-1}. $$ Here, P is the eigenvector matrix, D is the diagonal matrix formed by the eigenvalues, and \(P^{-1}\) is the inverse of the matrix P. #Step 3: Find the solution x(t)#
02

To find the solution at any time t, we will multiply the matrix exponential with the initial condition: $$ \mathbf{x}(t) = e^{tA} \cdot \mathbf{x}(0), $$ where \(\mathbf{x}(0)\) is the initial condition, given as: $$ \left(\begin{array}{l}{7}\\ {5}\\ {5}\end{array}\right). $$ #Step 4: Analyze the behavior of the solution as \(t \rightarrow \infty\)#

After finding the solution x(t), we will analyze the behavior of the solution as \(t \rightarrow \infty\). To do this, we will examine how each component of the solution x(t) behaves as time goes to infinity.

Key Concepts

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

Eigenvalues
Eigenvalues are pivotal in understanding the behavior of matrices. They are special numbers associated with a matrix that tell us a lot about the system the matrix represents.
When dealing with differential equations, such as the one given in the exercise, finding the eigenvalues is a crucial first step. To do this, we need to solve the characteristic equation \(|A - \lambda I| = 0\), where \(A\) is our given matrix, \(\lambda\) represents the eigenvalues, and \(I\) is the identity matrix of the same order as \(A\).
Solving this equation often involves calculating the determinant of \(A - \lambda I\) and setting it to zero, which gives us a polynomial. The roots of this polynomial are the eigenvalues.
  • Eigenvalues provide insights into the stability and dynamics of the system.
  • They can indicate whether solutions to the differential equations grow, shrink, or oscillate over time.
Once you have the eigenvalues, you're on your way to unlocking deep insights into the matrix's behavior and the system it represents.
Matrix Exponential
The matrix exponential, denoted as \(e^{tA}\), is a fundamental tool for solving systems of linear differential equations, like the one we have in the exercise.
In simple terms, it extends the concept of an exponential function to matrices.
To compute \(e^{tA}\), especially when \(A\) has real distinct eigenvalues, you use the formula \(e^{tA} = P \, e^{tD} \, P^{-1}\).
Here, \(P\) is the matrix composed of the eigenvectors of \(A\), \(D\) is a diagonal matrix constructed from the eigenvalues of \(A\), and \(P^{-1}\) is the inverse of \(P\).
  • The matrix exponential helps find the complete solution of the system over time.
  • It links the initial conditions to the solution \(x(t)\) at any time \(t\).
Understanding the matrix exponential will empower you to solve complex systems with ease.
Initial Value Problem
An initial value problem (IVP) in linear algebra involves finding a function that satisfies a given differential equation and meets an initial condition.
For the given exercise, you have a differential equation \(\mathbf{x}^{\prime} = A\mathbf{x}\) with an initial condition \(\mathbf{x}(0) = \begin{pmatrix} 7 \ 5 \ 5 \end{pmatrix}\).
Solving an IVP can be done by finding \(\mathbf{x}(t)\), which requires calculating the matrix exponential \(e^{tA}\) and using the initial condition.
The solution has the form \(\mathbf{x}(t) = e^{tA} \cdot \mathbf{x}(0)\).
  • The initial value specifies the starting point of the solution.
  • It ensures that the solution is unique, steering the system along a specific path determined by initial data.
Mastering the IVP concept allows you to determine the behavior of systems precisely from their given states.
Eigenvectors
Eigenvectors are vectors associated with a matrix that correspond to its eigenvalues.
They are non-zero vectors that satisfy the equation \((A - \lambda I) \mathbf{v} = \mathbf{0}\), where \(\lambda\) is an eigenvalue and \(\mathbf{v}\) is the eigenvector.
For a matrix \(A\), once you've found its eigenvalues, you naturally proceed to find its eigenvectors, which provide the direction to the "stretching" or "shrinking" initiated by the eigenvalues on the vector space.
Understanding the role of eigenvectors is crucial:
  • They define the directions in which the application of the matrix \(A\) scales the vector space.
  • They form the basis when we diagonalize a matrix (i.e., breaking the system into easier parts to handle).
  • They are essential in computing the matrix exponential \(e^{tA}\), as they form the matrix \(P\) of eigenvectors.
By grasping eigenvectors, you can gain insights into how transformations affect vector spaces.

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Most popular questions from this chapter

Consider a \(2 \times 2\) system \(\mathbf{x}^{\prime}=\mathbf{A} \mathbf{x}\). If we assume that \(r_{1} \neq r_{2}\), the general solution is \(\mathbf{x}=c_{1} \xi^{(1)} e^{t_{1}^{\prime}}+c_{2} \xi^{(2)} e^{\prime 2},\) provided that \(\xi^{(1)}\) and \(\xi^{(2)}\) are linearly independent In this problem we establish the linear independence of \(\xi^{(1)}\) and \(\xi^{(2)}\) by assuming that they are linearly dependent, and then showing that this leads to a contradiction. $$ \begin{array}{l}{\text { (a) Note that } \xi \text { (i) satisfies the matrix equation }\left(\mathbf{A}-r_{1} \mathbf{I}\right) \xi^{(1)}=\mathbf{0} ; \text { similarly, note that }} \\ {\left(\mathbf{A}-r_{2} \mathbf{I}\right) \xi^{(2)}=\mathbf{0}} \\ {\text { (b) Show that }\left(\mathbf{A}-r_{2} \mathbf{I}\right) \xi^{(1)}=\left(r_{1}-r_{2}\right) \mathbf{\xi}^{(1)}} \\\ {\text { (c) Suppose that } \xi^{(1)} \text { and } \xi^{(2)} \text { are linearly dependent. Then } c_{1} \xi^{(1)}+c_{2} \xi^{(2)}=\mathbf{0} \text { and at least }}\end{array} $$ $$ \begin{array}{l}{\text { one of } c_{1} \text { and } c_{2} \text { is not zero; suppose that } c_{1} \neq 0 . \text { Show that }\left(\mathbf{A}-r_{2} \mathbf{I}\right)\left(c_{1} \boldsymbol{\xi}^{(1)}+c_{2} \boldsymbol{\xi}^{(2)}\right)=\mathbf{0}} \\ {\text { and also show that }\left(\mathbf{A}-r_{2} \mathbf{I}\right)\left(c_{1} \boldsymbol{\xi}^{(1)}+c_{2} \boldsymbol{\xi}^{(2)}\right)=c_{1}\left(r_{1}-r_{2}\right) \boldsymbol{\xi}^{(1)} \text { . Hence } c_{1}=0, \text { which is }} \\\ {\text { a contradiction. Therefore } \xi^{(1)} \text { and } \boldsymbol{\xi}^{(2)} \text { are linearly independent. }}\end{array} $$ $$ \begin{array}{l}{\text { (d) Modify the argument of part (c) in case } c_{1} \text { is zero but } c_{2} \text { is not. }} \\ {\text { (e) Carry out a similar argument for the case in which the order } n \text { is equal to } 3 \text { ; note that }} \\ {\text { the procedure can be extended to cover an arbitrary value of } n .}\end{array} $$

Find the general solution of the given system of equations. $$ \mathbf{x}^{\prime}=\left(\begin{array}{rrr}{1} & {-1} & {4} \\ {3} & {2} & {-1} \\ {2} & {1} & {-1}\end{array}\right) \mathbf{x} $$

find a fundamental matrix for the given system of equations. In each case also find the fundamental matrix \(\mathbf{\Phi}(t)\) satisfying \(\Phi(0)=\mathbf{1}\) $$ \mathbf{x}^{\prime}=\left(\begin{array}{ll}{1} & {-1} \\ {5} & {-3}\end{array}\right) \mathbf{x} $$

In this problem we show that the eigenvalues of a Hermitian matrix \(\Lambda\) are real. Let \(x\) be an eigenvector corresponding to the eigenvalue \(\lambda\). (a) Show that \((A x, x)=(x, A x)\). Hint: See Problem 31 . (b) Show that \(\lambda(x, x)=\lambda(x, x)\), Hint: Recall that \(A x=\lambda x\). (c) Show that \(\lambda=\lambda\); that is, the cigenvalue \(\lambda\) is real.

Find the general solution of the given system of equations. $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{-\frac{5}{4}} & {\frac{3}{4}} \\\ {\frac{3}{4}} & {-\frac{5}{4}}\end{array}\right) \mathbf{x}+\left(\begin{array}{c}{2 t} \\ {e^{t}}\end{array}\right) $$
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