ia-dyn-chapter11

Then it is a stable focus if α 0 ie if the eigen

Info icon This preview shows pages 3–5. Sign up to view the full content.

View Full Document Right Arrow Icon
then it is a stable focus . If α = 0 (i.e., if the eigen- values are purely imaginary) then the solution just goes round and round a closed loop (an ellipse with semi-axes a and b ): this is a centre . The sense of the rotation (i.e., either clockwise or anticlockwise) is best discovered on a case-by-case basis (usually by considering the sign of either f or g close to the equilibrium point). Note that e 1 and e 2 are not necessarily orthogonal. Similarly, for complex eigenvalues α ± i β , a and b are not necessarily orthogonal; so in this case the diagrams above might need to be skewed. In the case of a centre ( α = 0) this would result in a sheared ellipse: but a sheared ellipse is still an ellipse (albeit one with different axes of symmetry). The diagrams are qualitatively correct. Using the Trace and Determinant In practice, it is not necessary to find the exact values of the eigenvalues and eigenvectors; the signs of the eigenvalues (or of their real parts) are all that is required to perform the categorisation. The characteristic equation for J is ( f x - λ )( g y - λ ) - f y g x = 0 or, equivalently, λ 2 - + Δ = 0 where T = f x + g y is the trace and Δ = f x g y - f y g x the determinant of J . We note that the eigenvalues are real iff T 2 - 0; and that the product of the eigenvalues is Δ while their sum is T . This enables us to deduce the required signs. 77
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Summary of Results Δ T 2 - Eigenvalues of J Classification - ve [+ve] Real, opposite signs Saddle +ve +ve Real, same signs Node: T < 0 stable T > 0 unstable +ve - ve Complex conjugate pair Focus: T < 0 stable T = 0 centre T > 0 unstable In the above analysis we have ignored the following degenerate cases: One of the eigenvalues is zero, which occurs iff Δ = 0. In this situation we would have to consider second-order terms in the Taylor series, which we could neglect in the analysis above. These terms could either stabilise or destabilise the equilibrium point. The two eigenvalues are equal (and real), which occurs iff T 2 - 4Δ = 0. In this situation there may be two non-parallel eigenvectors, just as normal; or there may be only one eigenvector, in which case the general solution is not of the form given above. In either case, the classification still holds: the equilibrium point is an unstable node if λ 1 = λ 2 > 0 and a stable node if λ 1 = λ 2 < 0. The phase diagram looks somewhat different, however: a star node (in the case of two non-parallel eigenvectors) or an inflected node (in the case of only one). 11.3 The Phase Plane for a Conservative System A second-order differential equation for a variable x ( t ) can always be converted to two first-order differential equations by defining y = ˙ x . For example, a general force equation in one dimension, m ¨ x = F ( x, ˙ x ) , can be converted to ˙ x = y, ˙ y = 1 m F ( x, y ) which is of the form given in § 11.1 for a plane autonomous system.
Image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern