To show that each of the following fixed points has an index equal to +1. a) stable spiral b) unstable spiral c) center d) star e) degenerate node Concept Introduction: Index theory provides the global information about the phase portrait. The index of the closed curve can be defined as the net number of counter-clockwise revolutions made by vector field. In mathematical form it can be written as, I c = 1 2π [ ϕ ] c . Where, I c is the index of the closed curve and [ ϕ ] c is the net change in ϕ around C.

Question

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Answer 1

Question

Chapter 6.8, Problem 1E

Interpretation Introduction

Interpretation:

To show that each of the following fixed points has an index equal to +1.

a) stable spiral b) unstable spiral c) center d) star e) degenerate node

Concept Introduction:

Index theory provides the global information about the phase portrait.

The index of the closed curve can be defined as the net number of counter-clockwise revolutions made by vector field. In mathematical form it can be written as,

$Ic=12π[ϕ]c$ .

Where, $Ic$ is the index of the closed curve and $[ϕ]c$ is the net change in $ϕ$ around C.

Expert Solution & Answer

Answer to Problem 1E

Solution:

For each of the fixed points, the index is equal to +1 is proved.

Explanation of Solution

Following fixed points have an index equal to $+1$ .

Sketch for the stable spiral,

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 1

The above sketch shows that every vector is always pointing inwards. It is seen from the direction of arrow that the vector field is rotating in counter-clockwise direction. Hence, the index for the stable spiral is +1.

Sketch for an unstable spiral

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 2

The above sketch shows that every vector is always pointing outwards. It is seen from the direction of arrow that the vector field is rotating in counter-clockwise direction. Hence, the index for the unstable spiral is +1.

Sketch for a Centre

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 3

The above sketch shows that every vector is making tangent to the closed orbit. It is seen from the direction of arrow that the vector field is rotating in counter-clockwise direction by $2π$ . Hence, the index for the centre is +1.

Sketch for a star

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 4

The above sketch shows that every vector is always pointing outwards. It is seen from the direction of arrow that the vector field is rotating in counter-clockwise direction by $2π$ . Hence, the index for the stable star is +1.

Sketch for a degenerate node:

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 5

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 6

The vector field at the point is pretty fastidious where the trajectory turns around and starts moving in opposite direction. Due to this, the graph of the only vector head is plotted instead of the individual vectors. These vectors bunch around the angle $π$ and the side directly opposite to it. It can be seen that the vector field is rotating in counterclockwise direction. Thus, the index for the degenerate node is +1.

Conclusion

Above sketches showthat each of the fixed point have an index equal to +1.

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can I see the steps for how you got the same answers already provided for μ1->μ4. this is a homework that provide you answers for question after attempting it three tries

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Answer 2

Question

Chapter 6.8, Problem 1E

Interpretation Introduction

Interpretation:

To show that each of the following fixed points has an index equal to +1.

a) stable spiral b) unstable spiral c) center d) star e) degenerate node

Concept Introduction:

Index theory provides the global information about the phase portrait.

The index of the closed curve can be defined as the net number of counter-clockwise revolutions made by vector field. In mathematical form it can be written as,

$Ic=12π[ϕ]c$ .

Where, $Ic$ is the index of the closed curve and $[ϕ]c$ is the net change in $ϕ$ around C.

Expert Solution & Answer

Answer to Problem 1E

Solution:

For each of the fixed points, the index is equal to +1 is proved.

Explanation of Solution

Following fixed points have an index equal to $+1$ .

Sketch for the stable spiral,

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 1

The above sketch shows that every vector is always pointing inwards. It is seen from the direction of arrow that the vector field is rotating in counter-clockwise direction. Hence, the index for the stable spiral is +1.

Sketch for an unstable spiral

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 2

The above sketch shows that every vector is always pointing outwards. It is seen from the direction of arrow that the vector field is rotating in counter-clockwise direction. Hence, the index for the unstable spiral is +1.

Sketch for a Centre

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 3

The above sketch shows that every vector is making tangent to the closed orbit. It is seen from the direction of arrow that the vector field is rotating in counter-clockwise direction by $2π$ . Hence, the index for the centre is +1.

Sketch for a star

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 4

The above sketch shows that every vector is always pointing outwards. It is seen from the direction of arrow that the vector field is rotating in counter-clockwise direction by $2π$ . Hence, the index for the stable star is +1.

Sketch for a degenerate node:

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 5

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 6

The vector field at the point is pretty fastidious where the trajectory turns around and starts moving in opposite direction. Due to this, the graph of the only vector head is plotted instead of the individual vectors. These vectors bunch around the angle $π$ and the side directly opposite to it. It can be seen that the vector field is rotating in counterclockwise direction. Thus, the index for the degenerate node is +1.

Conclusion

Above sketches showthat each of the fixed point have an index equal to +1.

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Answer 3

Question

Chapter 6.8, Problem 1E

Interpretation Introduction

Interpretation:

To show that each of the following fixed points has an index equal to +1.

a) stable spiral b) unstable spiral c) center d) star e) degenerate node

Concept Introduction:

Index theory provides the global information about the phase portrait.

The index of the closed curve can be defined as the net number of counter-clockwise revolutions made by vector field. In mathematical form it can be written as,

$Ic=12π[ϕ]c$ .

Where, $Ic$ is the index of the closed curve and $[ϕ]c$ is the net change in $ϕ$ around C.

Expert Solution & Answer

Answer to Problem 1E

Solution:

For each of the fixed points, the index is equal to +1 is proved.

Explanation of Solution

Following fixed points have an index equal to $+1$ .

Sketch for the stable spiral,

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 1

The above sketch shows that every vector is always pointing inwards. It is seen from the direction of arrow that the vector field is rotating in counter-clockwise direction. Hence, the index for the stable spiral is +1.

Sketch for an unstable spiral

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 2

The above sketch shows that every vector is always pointing outwards. It is seen from the direction of arrow that the vector field is rotating in counter-clockwise direction. Hence, the index for the unstable spiral is +1.

Sketch for a Centre

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 3

The above sketch shows that every vector is making tangent to the closed orbit. It is seen from the direction of arrow that the vector field is rotating in counter-clockwise direction by $2π$ . Hence, the index for the centre is +1.

Sketch for a star

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 4

The above sketch shows that every vector is always pointing outwards. It is seen from the direction of arrow that the vector field is rotating in counter-clockwise direction by $2π$ . Hence, the index for the stable star is +1.

Sketch for a degenerate node:

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 5

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 6

The vector field at the point is pretty fastidious where the trajectory turns around and starts moving in opposite direction. Due to this, the graph of the only vector head is plotted instead of the individual vectors. These vectors bunch around the angle $π$ and the side directly opposite to it. It can be seen that the vector field is rotating in counterclockwise direction. Thus, the index for the degenerate node is +1.

Conclusion

Above sketches showthat each of the fixed point have an index equal to +1.

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Answer 4

Question

Chapter 6.8, Problem 1E

Interpretation Introduction

Interpretation:

To show that each of the following fixed points has an index equal to +1.

a) stable spiral b) unstable spiral c) center d) star e) degenerate node

Concept Introduction:

Index theory provides the global information about the phase portrait.

The index of the closed curve can be defined as the net number of counter-clockwise revolutions made by vector field. In mathematical form it can be written as,

$Ic=12π[ϕ]c$ .

Where, $Ic$ is the index of the closed curve and $[ϕ]c$ is the net change in $ϕ$ around C.

Expert Solution & Answer

Answer to Problem 1E

Solution:

For each of the fixed points, the index is equal to +1 is proved.

Explanation of Solution

Following fixed points have an index equal to $+1$ .

Sketch for the stable spiral,

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 1

The above sketch shows that every vector is always pointing inwards. It is seen from the direction of arrow that the vector field is rotating in counter-clockwise direction. Hence, the index for the stable spiral is +1.

Sketch for an unstable spiral

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 2

The above sketch shows that every vector is always pointing outwards. It is seen from the direction of arrow that the vector field is rotating in counter-clockwise direction. Hence, the index for the unstable spiral is +1.

Sketch for a Centre

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 3

The above sketch shows that every vector is making tangent to the closed orbit. It is seen from the direction of arrow that the vector field is rotating in counter-clockwise direction by $2π$ . Hence, the index for the centre is +1.

Sketch for a star

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 4

The above sketch shows that every vector is always pointing outwards. It is seen from the direction of arrow that the vector field is rotating in counter-clockwise direction by $2π$ . Hence, the index for the stable star is +1.

Sketch for a degenerate node:

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 5

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 6

The vector field at the point is pretty fastidious where the trajectory turns around and starts moving in opposite direction. Due to this, the graph of the only vector head is plotted instead of the individual vectors. These vectors bunch around the angle $π$ and the side directly opposite to it. It can be seen that the vector field is rotating in counterclockwise direction. Thus, the index for the degenerate node is +1.

Conclusion

Above sketches showthat each of the fixed point have an index equal to +1.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 6.8, Problem 1E

Interpretation Introduction

Interpretation:

To show that each of the following fixed points has an index equal to +1.

a) stable spiral b) unstable spiral c) center d) star e) degenerate node

Concept Introduction:

Index theory provides the global information about the phase portrait.

The index of the closed curve can be defined as the net number of counter-clockwise revolutions made by vector field. In mathematical form it can be written as,

$Ic=12π[ϕ]c$ .

Where, $Ic$ is the index of the closed curve and $[ϕ]c$ is the net change in $ϕ$ around C.

Expert Solution & Answer

Answer to Problem 1E

Solution:

For each of the fixed points, the index is equal to +1 is proved.

Explanation of Solution

Following fixed points have an index equal to $+1$ .

Sketch for the stable spiral,

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 1

The above sketch shows that every vector is always pointing inwards. It is seen from the direction of arrow that the vector field is rotating in counter-clockwise direction. Hence, the index for the stable spiral is +1.

Sketch for an unstable spiral

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 2

The above sketch shows that every vector is always pointing outwards. It is seen from the direction of arrow that the vector field is rotating in counter-clockwise direction. Hence, the index for the unstable spiral is +1.

Sketch for a Centre

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 3

The above sketch shows that every vector is making tangent to the closed orbit. It is seen from the direction of arrow that the vector field is rotating in counter-clockwise direction by $2π$ . Hence, the index for the centre is +1.

Sketch for a star

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 4

The above sketch shows that every vector is always pointing outwards. It is seen from the direction of arrow that the vector field is rotating in counter-clockwise direction by $2π$ . Hence, the index for the stable star is +1.

Sketch for a degenerate node:

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 5

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 6

The vector field at the point is pretty fastidious where the trajectory turns around and starts moving in opposite direction. Due to this, the graph of the only vector head is plotted instead of the individual vectors. These vectors bunch around the angle $π$ and the side directly opposite to it. It can be seen that the vector field is rotating in counterclockwise direction. Thus, the index for the degenerate node is +1.

Conclusion

Above sketches showthat each of the fixed point have an index equal to +1.

Answer 6

Chapter 6.8, Problem 1E

Interpretation Introduction

Interpretation:

To show that each of the following fixed points has an index equal to +1.

a) stable spiral b) unstable spiral c) center d) star e) degenerate node

Concept Introduction:

Index theory provides the global information about the phase portrait.

The index of the closed curve can be defined as the net number of counter-clockwise revolutions made by vector field. In mathematical form it can be written as,

$Ic=12π[ϕ]c$ .

Where, $Ic$ is the index of the closed curve and $[ϕ]c$ is the net change in $ϕ$ around C.

Expert Solution & Answer

Answer to Problem 1E

Solution:

For each of the fixed points, the index is equal to +1 is proved.

Explanation of Solution

Following fixed points have an index equal to $+1$ .

Sketch for the stable spiral,

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 1

The above sketch shows that every vector is always pointing inwards. It is seen from the direction of arrow that the vector field is rotating in counter-clockwise direction. Hence, the index for the stable spiral is +1.

Sketch for an unstable spiral

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 2

The above sketch shows that every vector is always pointing outwards. It is seen from the direction of arrow that the vector field is rotating in counter-clockwise direction. Hence, the index for the unstable spiral is +1.

Sketch for a Centre

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 3

The above sketch shows that every vector is making tangent to the closed orbit. It is seen from the direction of arrow that the vector field is rotating in counter-clockwise direction by $2π$ . Hence, the index for the centre is +1.

Sketch for a star

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 4

The above sketch shows that every vector is always pointing outwards. It is seen from the direction of arrow that the vector field is rotating in counter-clockwise direction by $2π$ . Hence, the index for the stable star is +1.

Sketch for a degenerate node:

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 5

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 6

The vector field at the point is pretty fastidious where the trajectory turns around and starts moving in opposite direction. Due to this, the graph of the only vector head is plotted instead of the individual vectors. These vectors bunch around the angle $π$ and the side directly opposite to it. It can be seen that the vector field is rotating in counterclockwise direction. Thus, the index for the degenerate node is +1.

Conclusion

Above sketches showthat each of the fixed point have an index equal to +1.

Answer 7

Chapter 6.8, Problem 1E

Interpretation Introduction

Interpretation:

To show that each of the following fixed points has an index equal to +1.

a) stable spiral b) unstable spiral c) center d) star e) degenerate node

Concept Introduction:

Index theory provides the global information about the phase portrait.

The index of the closed curve can be defined as the net number of counter-clockwise revolutions made by vector field. In mathematical form it can be written as,

$Ic=12π[ϕ]c$ .

Where, $Ic$ is the index of the closed curve and $[ϕ]c$ is the net change in $ϕ$ around C.

Answer 8

Expert Solution & Answer

Answer to Problem 1E

Solution:

For each of the fixed points, the index is equal to +1 is proved.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Following fixed points have an index equal to $+1$ .

Sketch for the stable spiral,

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 1

The above sketch shows that every vector is always pointing inwards. It is seen from the direction of arrow that the vector field is rotating in counter-clockwise direction. Hence, the index for the stable spiral is +1.

Sketch for an unstable spiral

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 2

The above sketch shows that every vector is always pointing outwards. It is seen from the direction of arrow that the vector field is rotating in counter-clockwise direction. Hence, the index for the unstable spiral is +1.

Sketch for a Centre

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 3

The above sketch shows that every vector is making tangent to the closed orbit. It is seen from the direction of arrow that the vector field is rotating in counter-clockwise direction by $2π$ . Hence, the index for the centre is +1.

Sketch for a star

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 4

The above sketch shows that every vector is always pointing outwards. It is seen from the direction of arrow that the vector field is rotating in counter-clockwise direction by $2π$ . Hence, the index for the stable star is +1.

Sketch for a degenerate node:

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 5

Nonlinear Dynamics and Chaos, Chapter 6.8, Problem 1E , additional homework tip 6

The vector field at the point is pretty fastidious where the trajectory turns around and starts moving in opposite direction. Due to this, the graph of the only vector head is plotted instead of the individual vectors. These vectors bunch around the angle $π$ and the side directly opposite to it. It can be seen that the vector field is rotating in counterclockwise direction. Thus, the index for the degenerate node is +1.