If has linearly independent eigenvectors then any

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: rs, then any combination of these can be written as a linear , where can be regarded as the coordinates of the point in eigenvector coordinates. The matrix transforms from eigenvector coordinates to standard coordinates. Correspondingly, the matrix transforms back from standard coordinates to eigenvector coordinates. With this in mind, consider [1]. Multiply on the left by to get: 4 [2] where are the coordinates of the point in the basis of eigenvectors. Since diagonal matrix, the equations [2] are not coupled. The i^th component is just solution for some constant . In matrix notation this solution is written is a , with , where . Summary Our results in this subsection allow us to solve the following initial value problem in the case that the real matrix has a complete set of eigenvectors : . The matrix initial condition by is nonsingular. Multiply both the differential equation and the on the left to get . Solve this uncoupled system to get [3] , and then multiply on the left by to get , where . 2.2 Two-Dimensional Linear Systems Classify the two-dimensional linear systems according to the qualitative nature of their solutions. These are systems of the form , where is a matrix. Eigenvalues are the roots of the characteristic polynomial . Here, The roots are is the trace of and is the determinant of . 5 [6] where the , is the discriminant of . Qualitatively different cases of eigenvalues divide plane into 5 regions: Sketch of plane and the parabola complex plane. This is Meiss Fig. 2.1 . Include in each of the 5 regions a sketch of the Note that the eigenvalues are distinct off of the parabola . According to a theorem of linear algebra, when the eigenvalues are distinct the corresponding eigenvectors will be too. Then is diagonalizable and the general solution of the initial value problem has the form of [3]. For the 2D case this may be written [4] The 5 regions of the plane correspond to 5 geometrically distinct phase portraits: [A] Unstable node: . There are „straight line‟ solutions moving away from the origin that correspond to initial conditions on the eigenspaces and . Other solutions are asymptotically parallel to as and asymptotically parallel to as . Sketch -plane, and , several trajectories. Like Meiss Fig. 2.2 [B] Stable node: . This is like [A] with arrows reversed and with and interchanged. Straight line solutions move toward the origin on and . Other solutions are asymptotically parallel to as and asymptotica...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online