lecture2B

# lecture2B - Classification of Linear Systems I Given a...

This preview shows pages 1–5. Sign up to view the full content.

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

View Full Document

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

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

Unformatted text preview: Classification of Linear Systems I Given a linear system ˙ x = A x , the desired straight-line solutions ( x = e λt v ) exist if v is an eigenvector of A with eigenvalue λ , A v = λ v . I Recall: eigenvalues of A is given by characteristic equation det( A- λI ) which has solutions λ 1 = τ + p τ 2- 4 4 2 , λ 2 = τ- p τ 2- 4 4 2 where τ = trace( A ) = a + d and 4 = det( A ) = ad- bc . I If λ 1 6 = λ 2 (typical situation), eigenvectors its v 1 and v 2 are linear independent and span the entire plan. General Solution for Initial Value Problem I Can write any initial condition as a linear combination x = c 1 v 1 + c 2 v 2 I General solution is given by x ( t ) = c 1 e λ 1 t v 1 + c 2 e λ 2 v 2 I Solve IVP ˙ x = x + y, ˙ y = 4 x- 2 y with ( x , y ) = (2 ,- 3) and draw phase portrait. Both Eigenvalues are Real and Different I Have opposite signs: Fixed point is a saddle . I Same sign ( negative , positive ): nodes ( stable , unstable ). Eigenvalues are Complex Conjugates I Eigenvalues...
View Full Document

{[ snackBarMessage ]}

### Page1 / 9

lecture2B - Classification of Linear Systems I Given a...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online