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,

${\text{I}}_{\text{c}}\text{=}\frac{\text{1}}{\text{2π}}{\left[\varphi \right]}_{\text{c}}$.

Where, ${\text{I}}_{\text{c}}$ is the index of the closed curve and ${\left[\varphi \right]}_{\text{c}}$ is the net change in $\varphi$ around C.

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,

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

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

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 $\text{2π}$. Hence, the index for the centre is +1.

Sketch for a star

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 $\text{2π}$. Hence, the index for the stable star is +1.

Sketch for a degenerate node:

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 $\text{π}$ 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.