Partial solutions to 18.03 Problem Set 6 Part II

23. (a) The characteristic polynomial is λ^{2} + 2λ + a = 0. The eigenvalues are λ_{1} = −1 + \sqrt{1 − a}, λ_{2} = −1 − \sqrt{1 − a}. An eigenvector corresponding to λ_{1} is v_{1} = [ \begin{matrix}1 \\ \sqrt{1 − a} − 1 \end{matrix} ]. An eigenvector corresponding to λ_{2} is v_{2} = [ \begin{matrix}1 \\ −\sqrt{1 − a} − 1 \end{matrix} ].

• If a = 1, there is a repeated eigenvalue: λ_{1} = λ_{2} = 1. Then A = [ \begin{matrix}0 & 1 \\ −1 & −1 \end{matrix} ]. A doesn’t have independent eigenvectors, as A ≠ λ_{1}I.

• If a < 1, the eigenvalues are real and distinct.

• If a > 1, the eigenvalues are nonreal complex: −1 + \sqrt{a − 1}i and −1 − \sqrt{a − 1}i.

They are indeed complex conjugates. If a = 0, 0 is an eigenvalue. The other eigenvalue is -2.

(b) The general real solution is given as follows.

(i) If a < 1, [ \begin{matrix}x \\ y \end{matrix} ] = C_{1}e^{(−1+\sqrt{1−a})t} [ \begin{matrix}1 \\ \sqrt{1 − a} − 1 \end{matrix} ] + C_{2}e^{(−1−\sqrt{1−a})t} [ \begin{matrix}1 \\ −\sqrt{1 − a} − 1 \end{matrix} ].

(ii) If a > 1, [ \begin{matrix}x \\ y \end{matrix} ] is a linear combination of

e^{−t} (cos(\sqrt{a − 1}t) + i sin(\sqrt{a − 1}t)) [ \begin{matrix}1 \\ \sqrt{a − 1}i − 1 \end{matrix} ]

and its conjugate. Hence it is a linear combination of its real part and its imaginary part:

[ \begin{matrix}x \\ y \end{matrix} ] = e^{−t} (C_{1} (cos(\sqrt{a − 1}t) [ \begin{matrix}1 \\ −1 \end{matrix} ] + sin(\sqrt{a − 1}t) [ \begin{matrix}0 \\ −\sqrt{a − 1} \end{matrix} ])

+C_{2} (cos(\sqrt{a − 1}t) [ \begin{matrix}0 \\ √ & \overline{a − 1} \end{matrix} ] + sin(\sqrt{a − 1}t) [ \begin{matrix}1 \\ −1 \end{matrix} ])) .

(iii) If a = 1, one solution is e^{−t} [ \begin{matrix}1 \\ −1 \end{matrix} ]. The other solution is of the form e^{−t} (t [ \begin{matrix}1 \\ −1 \end{matrix} ] + β) for some vector β. β can be found (see Notes LS.15) by solving (A − λ_{1}I)β = [ \begin{matrix}1 \\ −1 \end{matrix} ]; one solution is β = [ \begin{matrix}1 \\ 0 \end{matrix} ]. Hence the general solution is

[ \begin{matrix}x \\ y \end{matrix} ] = C_{1}e^{−t} [ \begin{matrix}1 \\ −1 \end{matrix} ] + C_{2}e^{−t} (t [ \begin{matrix}1 \\ −1 \end{matrix} ] + [ \begin{matrix}1 \\ 0 \end{matrix} ]) .

24. (a) \begin{matrix}a & behavior \\ -8 & saddle point, unstable \\ -1 & saddle point, unstable \\ 0 & degenerate, comb, neutrally stable \\ .5 & asymptotically stable, proper node \\ 1 & asymptotically stable, improper node \\ 2 & spiral, asymptotically stable \\ 10 & spiral, asymptotically stable \end{matrix}

(b) If a < 1, we have real distinct roots. The two rays from the origin are the eigendirections: y = (\sqrt{1 − a} − 1)x and y = (−\sqrt{1 − a} − 1)x.

(c) In this case, [ \begin{matrix}x \\ y \end{matrix} ] = C_{1} [ \begin{matrix}1 \\ 0 \end{matrix} ] + C_{2}e^{−2t} [ \begin{matrix}1 \\ −2 \end{matrix} ] = [ \begin{matrix}C_{1} + C_{2}e^{−2t} \\ −2C_{2}e^{−2t} \end{matrix} ] . The long-term

behavior: the second normal mode dies out, so a particle will tend to the line y = 0 (this is visible on pplane5). The slope of the path is -2. The constant solutions occur when C_{2} = 0: x = C_{1}, y = 0. There are so many constant solutions because the matrix has 0 as an eigenvalue, i.e. is singular. Any first order homogeneous linear system has the constant solution (x, y) = (0, 0).

(d) The “straight line” solution corresponds to C_{2} in (iii) above. The line is y = −x. If C_{2} ≠ 0 (so the solution is neither the constant solution nor the rays), every solution crosses the x-axis exactly once (where y = 0, i.e. e^{−t}(−C_{1} − C_{2}t) = 0, i.e. t = −C_{1}/C_{2}), and the y-axis exactly once (where x = 0, i.e. e^{−t}(C_{1}+C_{2}t+C_{2}) = 0, i.e. t = −C_{1}/C_{2}−1). We also see that every solution (except the constant one and the rays) crosses the x-axis exaclty one time-unit after it crosses the y-axis.