Partial Solutions to 18.03 Practice Hour Exam III

Spring, 1999

1. (a), (b) Let A = [ \begin{matrix}3 & −4 \\ 6 & −7 \end{matrix} ]. The eigenvalues of A are the solutions to det [ \begin{matrix}3 − λ & −4 \\ 6 & −7 − λ \end{matrix} ] =

0, so λ_{1} = −1, and λ_{2} = −3. The eigenvectors are v_{1} = [ \begin{matrix}1 \\ 1 \end{matrix} ] and v_{2} = [ \begin{matrix}2 \\ 3 \end{matrix} ].

(c) The adjectives which correctly describe the phase portrait of the autonomous system ˙x⃗ = Ax⃗ are: proper node, asymptotically stable, structurally stable.

2. (a) We can use the eigenvalues and eigenvectors to construct a fundamental matrix:

Φ(t) = [ \begin{matrix}e^{t} & e^{2t} \\ e^{t} & 2e^{2t} \end{matrix} ] .

Then Φ(0)^{−1} = [ \begin{matrix}2 & −1 \\ −1 & 1 \end{matrix} ], so

e^{At} = Φ(t)Φ(0)^{−1} = [ \begin{matrix}e^{t} & e^{2t} \\ e^{t} & 2e^{2t} \end{matrix} ] [ \begin{matrix}2 & −1 \\ −1 & 1 \end{matrix} ] = [ \begin{matrix}2e^{t} − e^{2t} & −e^{t} + e^{2t} \\ 2e^{t} − 2e^{2t} & −e^{t} + 2e^{2t} \end{matrix} ] .

One can also do this directly: the matrix of eigenvectors is P = [ \begin{matrix}1 & 1 \\ 1 & 2 \end{matrix} ], so P^{−1} =

[ \begin{matrix}2 & −1 \\ −1 & 1 \end{matrix} ] (look familiar?), and the matrix of eigenvectors is D = [ \begin{matrix}1 & 0 \\ 0 & 2 \end{matrix} ]. Then

e^{At} = P e^{Dt}P^{−1} = [ \begin{matrix}1 & 1 \\ 1 & 2 \end{matrix} ] [ \begin{matrix}e^{t} & 0 \\ 0 & e^{2t} \end{matrix} ] [ \begin{matrix}2 & −1 \\ −1 & 1 \end{matrix} ] = [ \begin{matrix}2e^{t} − e^{2t} & −e^{t} + e^{2t} \\ 2e^{t} − 2e^{2t} & −e^{t} + 2e^{2t} \end{matrix} ] .

(b)

x⃗(t) = e^{At}x⃗(0) = [ \begin{matrix}2e^{t} − e^{2t} & −e^{t} + e^{2t} \\ 2e^{t} − 2e^{2t} & −e^{t} + 2e^{2t} \end{matrix} ] [ \begin{matrix}1 \\ −1 \end{matrix} ] = [ \begin{matrix}3e^{t} − 2e^{2t} \\ 3e^{t} − 4e^{2t} \end{matrix} ] .

(c)

A = P DP^{−1} = [ \begin{matrix}1 & 1 \\ 1 & 2 \end{matrix} ] [ \begin{matrix}1 & 0 \\ 0 & 2 \end{matrix} ] [ \begin{matrix}2 & −1 \\ −1 & 1 \end{matrix} ] = [ \begin{matrix}0 & 1 \\ −2 & 3 \end{matrix} ] .

3. (a) The critical points are where ẋ = ẏ = 0, i.e. where y − x^{2} = x − y^{2} = 0. Then y = x^{2} = y^{4}, so y^{4} − y = 0, so y = 0 or 1. Then x = y^{2}, so (x, y) = (0, 0) or (1, 1).

(b) Near (0, 0), we can linearize (i.e. approximate) to get ẋ = y, ẏ = x. The corresponding matrix is [ \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ] which has eigenvalues 1 and −1. As the eigenvalues are real and of opposite sign, this is a saddle, which is unstable.

(c) Let f(x, y) = y−x^{2} and g(x, y) = x−y^{2}, so f_{x}(x, y) = −2x, f_{y}(x, y) = 1, g_{x}(x, y) = 1, g_{y}(x, y) = −2y. If x = 1 + u and y = 1 + v, then when u and v are very small (i.e. x and y are very close to 1), the following is approximately true:

u̇ = f_{x}(1, 1)u + f_{y}(1, 1)v,

v̇ = g_{x}(1, 1)u + g_{y}(1, 1)v.

Hence u̇ = −2u + v, v̇ = u − 2v. The corresponding matrix of coefficients is [ \begin{matrix}−2 & 1 \\ 1 & −2 \end{matrix} ],

which has eigenvalues −1 and −3 and corresponding eigenvectors [ \begin{matrix}1 \\ 1 \end{matrix} ] and [ \begin{matrix}1 \\ −1 \end{matrix} ].

(d) This is a proper node.

4. Let y = x′, so we have a system x′ = y, y′ = −2y − 2x. The corresponding matrix of coefficients is [ \begin{matrix}0 & 1 \\ −2 & −2 \end{matrix} ], which has eigenvalues −1 + i and −1 − i. The phase portrait is a spiral. The spiral is inwards, as the real part of the eigenvalues is negative. To see if it is clockwise or counterclockwise, we plug in a random point (x, y) = (1, 0) where (x′, y′) = (0, −2), so we see that the spiral is clockwise.