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Mathematics LibreTexts

4.4E: Autonomous Second Order Equations (Exercises)

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    18323
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    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 \(-4<v<4\).

    Trajectories of van der Pol’s equation
    Trajectories of van der Pol’s equation

    [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 \(y''+p(y)=0\).