$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

4.4E: Autonomous Second Order Equations (Exercises)

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

In Exercises [exer:4.4.1} –[exer:4.4.4} find the equations of the trajectories of the given undamped equation. Identify the equilibrium solutions, determine whether they are stable or unstable, and plot some trajectories.

[exer:4.4.1] $$y''+y^3=0$$

[exer:4.4.2] $$y''+y^2=0$$

[exer:4.4.3] $$y''+y|y|=0$$

[exer:4.4.4] $$y''+ye^{-y}=0$$

[exer:4.4.5] $$y''-y^3+4y=0$$

[exer:4.4.6] $$y''+y^3-4y=0$$

[exer:4.4.7] $$y''+y(y^2-1)(y^2-4)=0$$

[exer:4.4.8] $$y''+y(y-2)(y-1)(y+2)=0$$

[exer:4.4.9] $$y''+y^2-a=0$$

[exer:4.4.10] $$y''+y^3-ay=0$$

[exer:4.4.11] $$y''-y^3+ay=0$$

[exer:4.4.12] $$y''+y-ay^3=0$$

[exer:4.4.13] $$y''+cy'+y^3=0$$

[exer:4.4.14] $$y''+cy'-y=0$$

[exer:4.4.15] $$y''+cy'+y^3=0$$

[exer:4.4.16] $$y''+cy'+y^2=0$$

[exer:4.4.17] $$y''+cy'+y|y|=0$$

[exer:4.4.18] $$y''+y(y-1)+cy=0$$

[exer:4.4.19] The van der Pol equation

$y''-\mu(1-y^2)y'+y=0, \tag{\rm(A)}$

where $$\mu$$ is a positive constant and $$y$$ is electrical current (Section 6.3), arises in the study of an electrical circuit whose resistive properties depend upon the current. The damping term
$$-\mu(1-y^2)y'$$ works to reduce $$|y|$$ if $$|y|<1$$ or to increase $$|y|$$ if $$|y|>1$$. It can be shown that van der Pol’s equation has exactly one closed trajectory, which is called a limit cycle. Trajectories inside the limit cycle spiral outward to it, while trajectories outside the limit cycle spiral inward to it (Figure [figure:4.4.16}). Use your favorite differential equations software to verify this for $$\mu=.5,1.1.5,2$$. Use a grid with $$-4<y<4$$ and $$-4<v<4$$.

[exer:4.4.20] Rayleigh’s equation,

$y''-\mu(1-(y')^2/3)y'+y=0$

also has a limit cycle. Follow the directions of Exercise [exer:4.4.19} for this equation.

[exer:4.4.21] In connection with Eqn Equation \ref{eq:4.4.15}, suppose $$y(0)=0$$ and $$y'(0)=v_0$$, where $$0<v_0<v_c$$.

Let $$T_1$$ be the time required for $$y$$ to increase from zero to $$y_{\max}=2\sin^{-1}(v_0/v_c)$$. Show that

${dy\over dt}=\sqrt{v_0^2-v_c^2\sin^2y/2},\quad 0\le t<T_1. \tag{\rm(A)}$

Separate variables in (A) and show that

$T_1=\int_0^{y_{\max}}{du\over\sqrt{v_0^2-v_c^2\sin^2u/2}} \tag{\rm(B)}$

Substitute $$\sin u/2=(v_0/v_c)\sin\theta$$ in (B) to obtain

$T_1=2\int_0^{\pi/2}{d\theta\over\sqrt{v_c^2-v_0^2\sin^2\theta}}. \tag{\rm(C)}$

Conclude from symmetry that the time required for $$(y(t),v(t))$$ to traverse the trajectory

$v^2=v_0^2-v_c^2\sin^2y/2$

is $$T=4T_1$$, and that consequently $$y(t+T)=y(t)$$ and $$v(t+T)=v(t)$$; that is, the oscillation is periodic with period $$T$$.

Show that if $$v_0=v_c$$, the integral in (C) is improper and diverges to $$\infty$$. Conclude from this that $$y(t)<\pi$$ for all $$t$$ and $$\lim_{t\to\infty}y(t)=\pi$$.

[exer:4.4.22] Give a direct definition of an unstable equilibrium of $$y''+p(y)=0$$.

[exer:4.4.23] Let $$p$$ be continuous for all $$y$$ and $$p(0)=0$$. Suppose there’s a positive number $$\rho$$ such that $$p(y)>0$$ if $$0<y\le \rho$$ and $$p(y)<0$$ if $$-\rho\le y<0$$. For $$0<r\le\rho$$ let

$\alpha(r)=\min\left\{\int_0^r p(x)\,dx,\ \int_{-r}^0 |p(x)|\,dx\right\} \mbox{\quad and \quad} \beta(r)=\max\left\{\int_0^r p(x)\,dx,\ \int_{-r}^0 |p(x)|\,dx\right\}.$

Let $$y$$ be the solution of the initial value problem

$y''+p(y)=0,\quad y(0)=v_0,\quad y'(0)=v_0,$

and define $$c(y_0,v_0)=v_0^2+2\int_0^{y_0}p(x)\,dx$$.

Show that

$0<c(y_0,v_0) <v_0^2+2\beta(|y_0|)\mbox{\quad if \quad} 0<|y_0|\le\rho.$

Show that

$v^2+2\int_0^y p(x)\,dx=c(y_0,v_0),\quad t>0.$

Conclude from (b) that if $$c(y_0,v_0)<2\alpha(r)$$ then $$|y|<r,\ t>0$$.

Given $$\epsilon>0$$, let $$\delta>0$$ be chosen so that

$\delta^2+2\beta(\delta)<\max\left\{\epsilon^2/2,2\alpha(\epsilon/\sqrt2) \right\}.$

Show that if $$\sqrt{y_0^2+v_0^2}<\delta$$ then $$\sqrt{y^2+v^2}<\epsilon$$ for $$t>0$$, which implies that $$\overline y=0$$ is a stable equilibrium of $$y''+p(y)=0$$.

Now let $$p$$ be continuous for all $$y$$ and $$p(\overline y)=0$$, where $$\overline y$$ is not necessarily zero. Suppose there’s a positive number $$\rho$$ such that $$p(y)>0$$ if $$\overline y<y\le \overline y+\rho$$ and $$p(y)<0$$ if $$\overline y-\rho\le y<\overline y$$. Show that $$\overline y$$ is a stable equilibrium of $$y''+p(y)=0$$.

[exer:4.4.24] Let $$p$$ be continuous for all $$y$$.

Suppose $$p(0)=0$$ and there’s a positive number $$\rho$$ such that $$p(y)<0$$ if $$0<y\le \rho$$. Let $$\epsilon$$ be any number such that $$0<\epsilon<\rho$$. Show that if $$y$$ is the solution of the initial value problem

$y''+p(y)=0,\quad y(0)=y_0,\quad y'(0)=0$

with $$0<y_0<\epsilon$$, then $$y(t)\ge\epsilon$$ for some $$t>0$$. Conclude that $$\overline y=0$$ is an unstable equilibrium of $$y''+p(y)=0$$.

Now let $$p(\overline y)=0$$, where $$\overline y$$ isn’t necessarily zero. Suppose there’s a positive number $$\rho$$ such that $$p(y)<0$$ if $$\overline y<y\le \overline y+\rho$$. Show that $$\overline y$$ is an unstable equilibrium of $$y''+p(y)=0$$.

Modify your proofs of (a) and (b) to show that if there’s a positive number $$\rho$$ such that $$p(y)>0$$ if $$\overline y-\rho\le y<\overline y$$, then $$\overline y$$ is an unstable equilibrium of $$y''+p(y)=0$$.