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Exercise:
A system of linear differential s is given by the matrix bf A pmatrix - & & pmatrix abcliste abc Determine the eigenvalues. abc Verify that the eigenvectors are vec v_ pmatrix pmatrix quad textrmand quad vec v_ pmatrix pmatrix abc Discuss the stability of the system and sketch the phase portrait. abcliste

Solution:
abcliste abc The trace and determinant are tau -+ - Delta - - - and thus the eigenvalues are lambda_ fractaupmsqrttau^-Delta frac-pmsqrt+ frac-pm Longrightarrow lambda_ lambda_ - abc bf Avec v_ pmatrix - & & pmatrix pmatrix pmatrix pmatrix - + + pmatrix pmatrix pmatrix vec v_ bf Avec v_ pmatrix - & & pmatrix pmatrix pmatrix pmatrix - + + pmatrix pmatrix - pmatrix -vec v_ This proves that vec v_ is the eigenvector for eigenvalue lambda_ and vec v_ the eigenvector for eigenvalue lambda_-. abc Since both eigenvalues are real with lambda_ and lambda_ the fixed po at the origin is a saddle po. The eigenvector vec v_ defines the straight-line solution with exponential growth moving away from the origin whereas vec v_ defines the straight-line solution with exponential decrease moving towards the origin. vspacemm The orbites cannot cross the two straight-lines defined by the eigenvectors. Near the horizontal axis they are directed towards the origin but then turn away and approach the tilted line. center includegraphicswidthtextwidth#image_path:phasportrait-# center abcliste
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Exercise:
A system of linear differential s is given by the matrix bf A pmatrix - & & pmatrix abcliste abc Determine the eigenvalues. abc Verify that the eigenvectors are vec v_ pmatrix pmatrix quad textrmand quad vec v_ pmatrix pmatrix abc Discuss the stability of the system and sketch the phase portrait. abcliste

Solution:
abcliste abc The trace and determinant are tau -+ - Delta - - - and thus the eigenvalues are lambda_ fractaupmsqrttau^-Delta frac-pmsqrt+ frac-pm Longrightarrow lambda_ lambda_ - abc bf Avec v_ pmatrix - & & pmatrix pmatrix pmatrix pmatrix - + + pmatrix pmatrix pmatrix vec v_ bf Avec v_ pmatrix - & & pmatrix pmatrix pmatrix pmatrix - + + pmatrix pmatrix - pmatrix -vec v_ This proves that vec v_ is the eigenvector for eigenvalue lambda_ and vec v_ the eigenvector for eigenvalue lambda_-. abc Since both eigenvalues are real with lambda_ and lambda_ the fixed po at the origin is a saddle po. The eigenvector vec v_ defines the straight-line solution with exponential growth moving away from the origin whereas vec v_ defines the straight-line solution with exponential decrease moving towards the origin. vspacemm The orbites cannot cross the two straight-lines defined by the eigenvectors. Near the horizontal axis they are directed towards the origin but then turn away and approach the tilted line. center includegraphicswidthtextwidth#image_path:phasportrait-# center abcliste
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Attributes & Decorations
Branches
Differential equations
Tags
eigenvalue, eigenvector, phase portrait, stability
Difficulty
(3, default)
Points
7 (default)
Language
ENG (English)
Type
Calculative / Quantity
Decoration
Content image