Exercise:
The eigenvalues for a system of linear differential s are lambda_ and lambda_-. The figure shows the straight-line solutions textcircled and textcircled for this system. center includegraphicswidth.mm#image_path:phasportrait-# center abcliste abc Assign the straight-line solutions to the eigenvalues lambda_ and lambda_. Give a short reason for your solution. hfill abc Determine the eigenvectors from the diagram. hfill abc Sketch the orbits including direction for solutions starting at pos A to E. hfill abc Bonus question: Derive the coefficient matrix from the eigenvalues and eigenvectors. hfill abcliste

Solution:
abcliste abc The origin is a saddle po. The straight-line solution textcircled corresponds to the unstable part of the solution i.e. it is the eigenvector for the positive eigenvalue lambda_. The straight-line solution textcircled corresponds to the stable part of the solution i.e. it is the eigenvector for the negative eigenvalue lambda_. abc vec v_ pmatrix pmatrix vec v_ pmatrix pmatrix abc The phase portrait below displays the orbits starting at pos A to E: center includegraphicswidth.mm#image_path:phasportrait-solution-# center abc The coefficient matrix has to fulfil the conditions bf M vec v_ lambda_ vec v_ The matrix elements a b c and d are therefore given by pmatrix a & b c & d pmatrix pmatrix pmatrix pmatrix pmatrix pmatrix pmatrix &quad textrmand pmatrix a & b c & d pmatrix pmatrix pmatrix - pmatrix pmatrix pmatrix - - pmatrix This leads to the following system of simultaneous s: a+b labeleqn: c+d a+b - c+d - Subtracting the third from three times the first yields b-b + Longrightarrow b Longrightarrow bfrac Longrightarrow a -bfrac-frac-frac In the same way we find the matrix elements c and d. The coefficient matrix is bf M pmatrix -frac & frac -frac & frac pmatrix abcliste