/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 The text solved \(d u / d t=J u\... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 37

The text solved \(d u / d t=J u\) for a 3 by 3 Jordan block \(J\). Add a fourth equation \(d w / d t=5 w+x\). Follow the pattern of solutions for \(z, y, x\) to find \(w\).

Short Answer

Expert verified
The solution for \( w(t) \) is \( e^{5t} \left( D + C_1 t + C_2 \frac{t^2}{2} + C_3 \frac{t^3}{6} \right) \).

Step by step solution

01

Understand the System

We have a system of equations defined by a differential equation of the form \( \frac{du}{dt} = Ju \) for a 3x3 Jordan block \( J \). The additional equation \( \frac{dw}{dt} = 5w + x \) is added to this system. We need to solve this system for \( w \) following the pattern used for the variables \( z, y, \) and \( x \).
02

Define the Jordan Block

A Jordan block \( J \) is a matrix where each diagonal element is equal and the superdiagonal elements (right above the main diagonal) are 1, with 0s elsewhere. For a 3x3 block associated with eigenvalue \( \lambda \), \( J \) can be represented as: \[ J = \begin{bmatrix} \lambda & 1 & 0 \ 0 & \lambda & 1 \ 0 & 0 & \lambda \end{bmatrix} \].
03

Solving for Each Component

The solution for a Jordan block system \( \frac{du}{dt} = Ju \) involves an exponential term \( e^{\lambda t} \) and polynomial terms due to the Jordan structure. The solutions typically take the form: \[ x(t) = e^{\lambda t}(C_1 + C_2 t + C_3 \frac{t^2}{2}) \] \[ y(t) = e^{\lambda t}(C_2 + C_3 t) \] \[ z(t) = e^{\lambda t} C_3 \] where \( C_1, C_2, C_3 \) are constants.
04

Solve the Fourth Equation

For the additional equation \( \frac{dw}{dt} = 5w + x \), we can use an integrating factor. The solution pattern involves using the exponential from \( x(t) = e^{\lambda t}(C_1 + C_2 t + C_3 \frac{t^2}{2}) \). Assuming \( \lambda = 5 \), multiply both sides by the integrating factor \( e^{-5t} \): \[ e^{-5t} \frac{dw}{dt} - 5w e^{-5t} = x e^{-5t} \]. This simplifies to \( \frac{d}{dt}(we^{-5t}) = xe^{-5t} \). Integrating gives \( we^{-5t} = \int x(t)e^{-5t} \, dt \), then solve for \( w \).
05

Evaluate the Integral

The integral requires substitution of \( x(t) = e^{5t}(C_1 + C_2 t + C_3 \frac{t^2}{2}) \), making the integral \[ \int (C_1 + C_2 t + C_3 \frac{t^2}{2}) \, dt \]. Evaluate this integral to find \( w = e^{5t} \left( D + C_1 t + C_2 \frac{t^2}{2} + C_3 \frac{t^3}{6} \right) \), where \( D \) is a constant of integration.

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Key Concepts

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

Differential Equations
Differential equations are equations that relate a function to its derivatives, which describe the rate of change of quantities. In the context of the problem, we have a system governed by a differential equation
  • \( \frac{du}{dt} = Ju \): Showcases the relationship between the derivative of vector \( u \) and the effects produced by a 3x3 Jordan block matrix \( J \).
  • \( \frac{dw}{dt} = 5w + x \): Represents an additional relationship where the rate of change of \( w \) depends on itself and another variable \( x \).
Differential equations can often be challenging, as they require understanding how functions behave as they change. Solving these equations often involves finding functions that satisfy the given relation, which is what we aim to do here with the solutions for \( w, x, y, \) and \( z \).
Eigenvalue
An eigenvalue is a number that characteristically restricts or scales a function when taken through a specific linear transformation, commonly represented in matrix forms. In the provided exercise, our Jordan block \( J \) is centered around a particular eigenvalue \( \lambda \).
  • A Jordan block is structured with \( \lambda \) repeated along the diagonal, such that for a 3x3 matrix: \[ J = \begin{bmatrix} \lambda & 1 & 0 \ 0 & \lambda & 1 \ 0 & 0 & \lambda \end{bmatrix} \]
  • Eigenvalues in a Jordan block affect the solutions to the differential equations. In our solution, determining the behavior of each component \( x, y, \) and \( z \) heavily relies on \( e^{\lambda t} \), where \( e \) is the base of the natural logarithm.
Understanding the eigenvalue helps us see how each part of the system will exponentially grow or shrink over time.
Integrating Factor
An integrating factor is a mathematical tool used to solve linear first-order differential equations by making them easier to integrate. This concept was used to solve for \( w \) in the additional equation.
  • For the equation \( \frac{dw}{dt} = 5w + x \), choosing an integrating factor helps simplify integration. We chose \( e^{-5t} \) to simplify the differential equation.
  • Multiplying both sides by this integrating factor yields: \[ e^{-5t} \frac{dw}{dt} - 5w e^{-5t} = x e^{-5t} \]
  • This transforms the equation to a derivative of a product, simply denoted as \( \frac{d}{dt}(we^{-5t}) = xe^{-5t} \).
Using an integrating factor turns a complicated equation into one that is straightforward to integrate, which is critical for finding solutions in complex systems like this one.

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Most popular questions from this chapter

Diagonalize \(A\) and compute \(S \Lambda^{k} S^{-1}\) to prove this formula for \(A^{k}\) : $$ A=\left[\begin{array}{ll} 3 & 2 \\ 2 & 3 \end{array}\right] \quad \text { has } \quad A^{k}=\frac{1}{2}\left[\begin{array}{ll} 5^{k}+1 & 5^{k}-1 \\ 5^{k}-1 & 5^{k}+1 \end{array}\right] $$

Every matrix \(Z\) can be split into a Hermitian and a skew-Hermitian part, \(Z=A+K\), just as a complex number \(z\) is split into \(a+i b\). The real part of \(z\) is half of \(z+\bar{z}\), and the "real part" of \(Z\) is half of \(Z+Z^{H}\). Find a similar formula for the "imaginary part" \(K\), and split these matrices into \(A+K\) : $$ Z=\left[\begin{array}{cc} 3+i & 4+2 i \\ 0 & 5 \end{array}\right] \quad \text { and } \quad Z=\left[\begin{array}{rr} i & i \\ -i & i \end{array}\right] $$

The eigenvalues of \(A\) are 1 and 9 , the eigenvalues of \(B\) are \(-1\) and 9 : $$ A=\left[\begin{array}{ll} 5 & 4 \\ 4 & 5 \end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{ll} 4 & 5 \\ 5 & 4 \end{array}\right] $$ Find a matrix square root of \(A\) from \(R=S \sqrt{\Lambda} S^{-1}\). Why is there no real matrix square root of \(B\) ?

Show that if \(U\) and \(V\) are unitary, so is \(U V\). Use the criterion \(U^{\mathrm{H}} U=I\).

Substitute \(y=e^{\lambda t}\) into \(y^{\prime \prime}=6 y^{\prime}-9 y\) to show that \(\lambda=3\) is a repeated root. This is trouble; we need a second solution after \(e^{3 t}\). The matrix equation is $$ \frac{d}{d t}\left[\begin{array}{l} y \\ y^{\prime} \end{array}\right]=\left[\begin{array}{rr} 0 & 1 \\ -9 & 6 \end{array}\right]\left[\begin{array}{l} y \\ y^{\prime} \end{array}\right] $$ Show that this matrix has \(\lambda=3,3\) and only one line of eigenvectors. Trouble here too. Show that the second solution is \(y=t e^{3 t}\).
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