9.1

Suppose an object moves in the 2D plane (the \(x_1, x_2\) plane) so that it is at the point \(\left(x_1(t), x_2(t)\right)\) at time \(t\). Suppose the object's velocity is given by

\[ \notag
\begin{gathered}
x_1^{\prime}(t)=a x_1+b x_2 \\
x_2^{\prime}(t)=c x_1+d x_2
\end{gathered}
\]

Or in matrix form \(\binom{x_1}{x_2}=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)\binom{x_1}{x_2}\)
To solve, find eigenvalues and corresponding eigenvectors:
\[ \notag
\left|\begin{array}{cc}
a-r & b \\
c & d-r
\end{array}\right|=(a-r)(d-r)-b c=r^2-(a+d) r+a d-b c=0 .
\]

Thus \( r=\frac{(a+d) \pm \sqrt{(a+d)^2-4(a d-b c)}}{2} =\frac{p \pm \sqrt{p^2-4 q}}{2}\)

Let \(p=\operatorname{trace}(A)=a+d\) and le \(q=\operatorname{det} A=a d-b c\)

Thus the type of solution depends on \((p, q)\)

NOTE: NEED TO ADD FIGURES see slides9_1A in COURSES/100/FALL20/3600.html

Case 2: \((a+d)^2-4(a d-b c)=0\)
Case 2i: Two independent eigenvectors:
The general solution is \(\binom{x_1}{x_2}=c_1\binom{v_1}{v_2} e^{r t}+c_2\binom{w_1}{w_2} e^{r t}\)
Case 2ii: One independent eigenvectors:
The general solution is \(\binom{x_1}{x_2}=c_1\binom{v_1}{v_2} e^{r t}+c_2\left[\binom{v_1}{v_2} t+\binom{w_1}{w_2}\right] e^{r t}\)
Case 2a: \(r>0\)
Case 2b: \(r<0\)

 

Case 3: \((a+d)^2-4(a d-b c)<0\). I.e., \(r=\lambda \pm i \mu\)
Suppose eigenvector corresponding to eigenvalue is
\[
\binom{v_1 \pm i w_1}{v_2 \pm i w_2}=\binom{v_1}{v_2} \pm i\binom{w_1}{w_2}
\]

Then general solution is
\[ \notag
\begin{aligned}
{\left[\begin{array}{l}
x_1 \\
x_2
\end{array}\right]=c_1 e^{\lambda t}\left(\left[\begin{array}{l}
v_1 \\
v_2
\end{array}\right] \cos (\mu t)-\right.} & {\left.\left[\begin{array}{l}
w_1 \\
w_2
\end{array}\right] \sin (\mu t)\right) } \\
& +c_2 e^{\lambda t}\left(\left[\begin{array}{l}
v_1 \\
v_2
\end{array}\right] \sin (\mu t)+\left[\begin{array}{l}
w_1 \\
w_2
\end{array}\right] \cos (\mu t)\right)
\end{aligned}
\]