Chapter 9: Nonlinear systemsContent check — Differential Equations (Lebl), Chapter 9: Nonlinear systems

a) Critical points $(\mathrm{1,0})$ and $(\mathrm{1,1})$⁠. At $(\mathrm{1,0})$ using $u=x-1$⁠, $v=y$ the linearization is $u^{\prime }=\pi v$⁠, $v^{\prime }=-v$⁠. At $(\mathrm{1,1})$ using $u=x-1$⁠, $v=y-1$ the linearization is $u^{\prime }=-\pi v$⁠, $v^{\prime }=v$⁠.
b) Critical points $(\mathrm{0,0})$ and $(0,-1)$⁠. At $(\mathrm{0,0})$ using $u=x$⁠, $v=y$ the linearization is $u^{\prime }=u+v$⁠, $v^{\prime }=u$⁠. At $(0,-1)$ using $u=x$⁠, $v=y+1$ the linearization is $u^{\prime }=u-v$⁠, $v^{\prime }=u$⁠.
c) Critical point $({}^1/{}_2,{}^{-1}/{}_4)$⁠. Using $u=x-{}^1/{}_2$⁠, $v=y+{}^1/{}_4$ the linearization is $u^{\prime }=-u+v$⁠, $v^{\prime }=u+v$⁠.

1) is c), 2) is a), 3) is b)

Critical points are $(\mathrm{0,0,0})$⁠, and $(-1,1,-1)$⁠. The linearization at the origin using variables $u=x$⁠, $v=y$⁠, $w=z$ is $u^{\prime }=u$⁠, $v^{\prime }=-v$⁠, $w^{\prime }=w$⁠. The linearization at the point $(-\mathrm{1,1},-1)$ using variables $u=x+1$⁠, $v=y-1$⁠, $w=z+1$ is $u^{\prime }=u-2w$⁠, $v^{\prime }=-v-2w$⁠, $w^{\prime }=w-2u$⁠.

$u^{\prime }=f(u,v,w)$⁠, $v^{\prime }=g(u,v,w)$⁠, $w^{\prime }=1$⁠.

a) $(\mathrm{0,0})$⁠: saddle (unstable), $(\mathrm{1,0})$⁠: source (unstable), b) $(\mathrm{0,0})$⁠: spiral sink (asymptotically stable), $(\mathrm{0,1})$⁠: saddle (unstable), c) $(\mathrm{1,0})$⁠: saddle (unstable), $(\mathrm{0,1})$⁠: source (unstable)

a) $\frac{1}{2}{y}^2+\frac{1}{3}{x}^3-4x=C$⁠, critical points: $(-\mathrm{2,0})$⁠, an unstable saddle, and $(\mathrm{2,0})$⁠, a stable center. b) $\frac{1}{2}{y}^2+e^x=C$⁠, no critical points. c) $\frac{1}{2}{y}^2+x{e}^x=C$⁠, critical point at $(-\mathrm{1,0})$ is a stable center.

Critical point at $(\mathrm{0,0})$⁠. Trajectories are $y=\pm \sqrt{2C-({}^1/{}_2)x^4}$⁠, for $C>0$⁠, these give closed curves around the origin, so the critical point is a stable center.

A critical point $x_0$ is stable if $f^{\prime }(x_0)<0$ and unstable when $f^{\prime }(x_0)>0$⁠.

a) Critical points are $\omega =0$⁠, $\theta =k\pi$ for any integer $k$⁠. When $k$ is odd, we have a saddle point. When $k$ is even we get a sink. b) The findings mean the pendulum will simply go to one of the sinks, for example $(\mathrm{0,0})$ and it will not swing back and forth. The friction is too high for it to oscillate, just like an overdamped mass-spring system.

a) Solving for the critical points we get $(0,-{}^h/{}_d)$ and $(\frac{bh+ad}{ac},\frac{a}{b})$⁠. The Jacobian matrix at $(0,-{}^h/{}_d)$ is $\left[\begin{array}{cc}a+bh/d& 0\\ -ch/d& -d\end{array}\right]$ whose eigenvalues are $a+bh/d$ and $-d$⁠. The eigenvalues are real of opposite signs and we get a saddle. (In the application, however, we are only looking at the positive quadrant so this critical point is irrelevant.) At $(\frac{bh+ad}{ac},\frac{a}{b})$ we get Jacobian matrix $\left[\begin{array}{cc}0& -\frac{b(bh+ad)}{ac}\\ \frac{ac}{b}& \frac{bh+ad}{a}-d\end{array}\right]$⁠. b) For the specific numbers given, the second critical point is $(\frac{550}{3},40)$ the matrix is $\left[\begin{array}{cc}0& -11/6\\ 3/25& 1/4\end{array}\right]$⁠, which has eigenvalues $\frac{5\pm i\sqrt{327}}{40}$⁠. Therefore there is a spiral source; the solution spirals outwards. The solution eventually hits one of the axes, $x=0$ or $y=0$⁠, so something will die out in the forest.

The critical points are on the line $x=0$⁠. In the positive quadrant the $y^{\prime }$ is always positive and so the fox population always grows. The constant of motion is $C=y^a{e}^{-cx-by}$⁠. For any $C$⁠, this curve must hit the $y$⁠-axis (why?), so the trajectory will simply approach a point on the $y$ axis somewhere and the number of hares will go to zero.

Use Bendixson–Dulac Theorem. a) $f_x+g_y=1+1>0$⁠, so no closed trajectories. b) $f_x+g_y=-{\sin }^2(y)+0<0$ for all $x,y$ except the lines given by $y=k\pi$ (where we get zero), so no closed trajectories. c) $f_x+g_y=y^2+0>0$ for all $x,y$ except the line given by $y=0$ (where we get zero), so no closed trajectories.

Using Poincaré–Bendixson Theorem, the system has a limit cycle, which is the unit circle centered at the origin, as $x=\cos (t)(1+e^{-t})$⁠, $y=\sin (t)(1+e^{-t})$ gets closer and closer to the unit circle. Thus $x=\cos (t)$⁠, $y=\sin (t)$ is the periodic solution.

$f(x,y)=y$⁠, $g(x,y)=\mu (1-x^2)y-x$⁠. So $f_x+g_y=\mu (1-x^2)$⁠. The Bendixson–Dulac Theorem says there is no closed trajectory lying entirely in the set $x^2<1$⁠.

The closed trajectories are those where $\sin (r)=0$⁠, therefore, all the circles centered at the origin with radius that is a positive multiple of $\pi$ are closed trajectories.

Critical points: $(\mathrm{0,0,0})$⁠, $(6\sqrt{2},6\sqrt{2},27)$⁠, $(-6\sqrt{2},-6\sqrt{2},27)$⁠. Linearization at $(\mathrm{0,0,0})$ using $u=x$⁠, $v=y$⁠, $w=z$ is $u^{\prime }=-10u+10v$⁠, $v^{\prime }=28u-v$⁠, $w^{\prime }=-({}^8/{}_3)w$⁠. Linearization at $(6\sqrt{2},6\sqrt{2},27)$ using $u=x-6\sqrt{2}$⁠, $v=y-6\sqrt{2}$⁠, $w=z-27$ is $u^{\prime }=-10u+10v$⁠, $v^{\prime }=u-v-6\sqrt{2}w$⁠, $w^{\prime }=6\sqrt{2}u+6\sqrt{2}v-({}^8/{}_3)w$⁠. Linearization at $(-6\sqrt{2},-6\sqrt{2},27)$ using $u=x+6\sqrt{2}$⁠, $v=y+6\sqrt{2}$⁠, $w=z-27$ is $u^{\prime }=-10u+10v$⁠, $v^{\prime }=u-v+6\sqrt{2}w$⁠, $w^{\prime }=-6\sqrt{2}u-6\sqrt{2}v-({}^8/{}_3)w$⁠.