Q17E Consider the system dx鈫抎t=(-1K... [FREE SOLUTION]

91影视

Linear Algebra With Applications

Otto Bretscher

$Math Studyset 91影视 Explanations$ Math

5th Edition 路 442 pages

ISBN: 9780321796974

Chapter 9: Q17E (page 439)

Consider the system $\frac{d \overset{鈫�}{x}}{d t} = (\begin{matrix} - 1 & K \\ K & - 1 \end{matrix}) \overset{鈫�}{x}$ where k is an arbitrary constant. For which values of k is the zero state a stable equilibrium solution?

Short Answer

Expert verified

The value of k in the zero state a stable equilibrium solution is $|k| < 1$

Step by step solution

Explanation of the stability of a continuous dynamical system

For a system $\frac{d \overset{鈫�}{x}}{d t} = A \overset{鈫�}{x}$ , here A is the matrix form.

The zero state is an asymptotically stable equilibrium solution if and only if the real parts of all eigen values of A are negative

Step 2:Solution for the values of k is the zero state a stable equilibrium solution

Consider the system $\frac{d \overset{鈫�}{x}}{d t} = (\begin{matrix} - 1 & K \\ K & - 1 \end{matrix}) \overset{鈫�}{x}$ where k is an arbitrary constant

Here $A = (\begin{matrix} - 1 & K \\ K & - 1 \end{matrix}) \overset{鈫�}{x}$ which is the real $2 脳 2$ matrix.

$\begin{array}{rcl} A - 位 I & = & 0 \\ (\begin{matrix} - 1 & K \\ K & - 1 \end{matrix}) - 位 (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) & = & 0 \\ (\begin{matrix} - 1 & K \\ K & - 1 \end{matrix}) - (\begin{matrix} 位 & 0 \\ 0 & 位 \end{matrix}) & = & 0 \\ (\begin{matrix} - 1 - 位 & K \\ K & - 1 - 位 \end{matrix}) & = & 0 \\ (- 1 - 位) (- 1 - 位) - k^{2} & = & 0 \\ 1 + 位 + 位 + 位^{2} - k^{2} & = & 0 \\ 1 + 2 位 + 位^{2} - k^{2} & = & 0 \\ 位^{2} + 2 位 - k^{2} + 1 & = & 0 \end{array}$

Here $位$ is zero because it is zero state in the asymptotically stable equilibrium

$\begin{array}{rcl} 位^{2} + 2 位 - k^{2} + 1 & = & 0 \\ - k^{2} + 1 & = & 0 \\ - k^{2} & = & - 1 \\ k^{2} & = & 1 \\ |k| & < & 1 \end{array}$

Thus, the value for the zero state a stable equilibrium solution is $|k| < 1$

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91影视

Consider the system $\frac{d \overset{鈫�}{x}}{d t} = (\begin{matrix} - 1 & K \\ K & - 1 \end{matrix}) \overset{鈫�}{x}$ where k is an arbitrary constant. For which values of k is the zero state a stable equilibrium solution?

Short Answer

Step by step solution

Explanation of the stability of a continuous dynamical system

Step 2:Solution for the values of k is the zero state a stable equilibrium solution

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91影视

Consider the system dx鈫�dt=(-1KK-1)x鈫�where k is an arbitrary constant. For which values of k is the zero state a stable equilibrium solution?

Short Answer

Step by step solution

Explanation of the stability of a continuous dynamical system

Step 2:Solution for the values of k is the zero state a stable equilibrium solution

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Geometry

Mechanics Maths

Statistics

Pure Maths

Decision Maths

Study anywhere. Anytime. Across all devices.

Consider the system $\frac{d \overset{鈫�}{x}}{d t} = (\begin{matrix} - 1 & K \\ K & - 1 \end{matrix}) \overset{鈫�}{x}$ where k is an arbitrary constant. For which values of k is the zero state a stable equilibrium solution?