- #1
Math100
- 802
- 222
- Homework Statement
- Show that the index of a limit cycle is ## 1 ##.
- Relevant Equations
- If ## \Gamma ## surrounds ## n ## equilibrium points ## P_{1}, P_{2}, ..., P_{n} ##, then ## I_{\Gamma}=\sum_{i=1}^{n}I_{i} ##, where ## I_{i} ## is the index of the point ## P_{i} ## for ## i=1, 2, ..., n ##.
Proof:
Consider the index of a limit cycle.
By definition, a limit cycle is an isolated periodic solution of an autonomous system represented in the phase plane by an isolated closed path.
The theorem states: If ## \Gamma ## surrounds ## n ## equilibrium points ## P_{1}, P_{2}, ..., P_{n} ##, then ## I_{\Gamma}=\sum_{i=1}^{n}I_{i} ##, where ## I_{i} ## is the index of the point ## P_{i} ## for ## i=1, 2, ..., n ##.
Let ## I_{\Gamma}=\frac{1}{2\pi}\triangle\theta_{\Gamma} ## where ## I_{\Gamma} ## is the index of the curve ## \Gamma ## and ## \triangle\theta_{\Gamma} ## is the total change in the angle of the vector field along ## \Gamma ##.
Note that the vector field along a limit cycle ## \Gamma ## behaves such that the direction of the vector field is always tangent to ## \Gamma ##.
Since the limit cycle traverses once, it follows that the vector field rotates once, and the total angular change of the vector field along the limit cycle is ## 2\pi ##.
Thus, ## I_{\Gamma}=\frac{1}{2\pi}\triangle\theta_{\Gamma}=\frac{1}{2\pi}(2\pi)=1 ##.
Therefore, the index of a limit cycle is ## 1 ##.
Above is the proof for this problem. May anyone please take a look and verify/confirm if it's accurate/correct?
Consider the index of a limit cycle.
By definition, a limit cycle is an isolated periodic solution of an autonomous system represented in the phase plane by an isolated closed path.
The theorem states: If ## \Gamma ## surrounds ## n ## equilibrium points ## P_{1}, P_{2}, ..., P_{n} ##, then ## I_{\Gamma}=\sum_{i=1}^{n}I_{i} ##, where ## I_{i} ## is the index of the point ## P_{i} ## for ## i=1, 2, ..., n ##.
Let ## I_{\Gamma}=\frac{1}{2\pi}\triangle\theta_{\Gamma} ## where ## I_{\Gamma} ## is the index of the curve ## \Gamma ## and ## \triangle\theta_{\Gamma} ## is the total change in the angle of the vector field along ## \Gamma ##.
Note that the vector field along a limit cycle ## \Gamma ## behaves such that the direction of the vector field is always tangent to ## \Gamma ##.
Since the limit cycle traverses once, it follows that the vector field rotates once, and the total angular change of the vector field along the limit cycle is ## 2\pi ##.
Thus, ## I_{\Gamma}=\frac{1}{2\pi}\triangle\theta_{\Gamma}=\frac{1}{2\pi}(2\pi)=1 ##.
Therefore, the index of a limit cycle is ## 1 ##.
Above is the proof for this problem. May anyone please take a look and verify/confirm if it's accurate/correct?
Last edited: