Exercises — Linearization of nonlinear systems

An equilibrium needs both $\dot{x}=0$ and $\dot{y}=0$.

• $(1,1)$: $\dot{x}=1-1=0$ ✔, $\dot{y}=1^2-1=0$ ✔. Equilibrium.
• $(0,0)$: $\dot{x}=0$ ✔ but $\dot{y}=0-1=-1\ne 0$ ✘.
• $(-1,-1)$: $\dot{x}=-1-(-1)=0$ ✔, $\dot{y}=1-1=0$ ✔. Equilibrium.

Answer: $(1,1)$ and $(-1,-1)$.

$f=x-y\Rightarrow f_x=1,f_y=-1$. $g=x^2-1\Rightarrow g_x=2x,g_y=0$. $J=\left(\begin{array}{cc}1& -1\\ 2x& 0\end{array}\right).$ Note the $x$ still sitting there — this is not yet a linear system; it must be evaluated at a point.

Triangular matrix → eigenvalues are the diagonal: $\lambda =-2,-3$. Both real, both negative → stable node. Check via shortcut: $\tau =-5<0$, $\Delta =6>0$, ${\tau }^2-4\Delta =25-24=1>0$ → real same-sign, $\tau <0$ ⇒ stable node. ✔

Substitute $x=1$ into $J=\left(\begin{array}{cc}1& -1\\ 2x& 0\end{array}\right)$: $J{|}_{(1,1)}=\left(\begin{array}{cc}1& -1\\ 2& 0\end{array}\right).$ $\tau =1+0=1$, $\Delta =(1)(0)-(-1)(2)=2$. $\lambda =\frac{1\pm \sqrt{1-8}}{2}=\frac{1\pm \sqrt{-7}}{2}=\frac{1}{2}\pm \frac{\sqrt{7}}{2}i$. Complex with $\mathrm{Re}=\frac{1}{2}>0$ → unstable spiral.

Substitute $x=-1$: $J{|}_{(-1,-1)}=\left(\begin{array}{cc}1& -1\\ -2& 0\end{array}\right).$ $\tau =1$, $\Delta =(1)(0)-(-1)(-2)=-2<0$. $\Delta <0$ → real opposite-sign eigenvalues → saddle (unstable). Explicitly $\lambda =\frac{1\pm \sqrt{1+8}}{2}=\frac{1\pm 3}{2}=2,-1$. ✔

Equilibria: $y=0$ and $-x+x^3=x(x^2-1)=0\Rightarrow x=0,\pm 1$. Points: $(0,0),(1,0),(-1,0)$. Jacobian: $f=y\Rightarrow f_x=0,f_y=1$; $g=-x+x^3\Rightarrow g_x=-1+3x^2,g_y=0$. $J=\left(\begin{array}{cc}0& 1\\ -1+3x^2& 0\end{array}\right).$

• $(0,0)$: $J=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, $\tau =0,\Delta =1>0$ → $\lambda =\pm i$ → center (inconclusive).
• $(1,0)$: $g_x=-1+3=2$, $J=\left(\begin{array}{cc}0& 1\\ 2& 0\end{array}\right)$, $\Delta =-2<0$ → saddle. ($\lambda =\pm \sqrt{2}$.)
• $(-1,0)$: same $g_x=2$ (since $x^2=1$), identical → saddle.

Interior: divide out $x,y$ (nonzero): $3-x-2y=0$ and $2-x-y=0$. Subtract: $(3-x-2y)-(2-x-y)=1-y=0\Rightarrow y=1$. Then $x=2-y=1$. Point $(1,1)$. Jacobian. $f=3x-x^2-2xy\Rightarrow f_x=3-2x-2y,f_y=-2x$. $g=2y-xy-y^2\Rightarrow g_x=-y,g_y=2-x-2y$. At $(1,1)$: $f_x=3-2-2=-1$, $f_y=-2$, $g_x=-1$, $g_y=2-1-2=-1$. $J=\left(\begin{array}{cc}-1& -2\\ -1& -1\end{array}\right),\tau =-2,\Delta =(-1)(-1)-(-2)(-1)=1-2=-1<0.$ $\Delta <0$ → saddle (unstable). Explicitly $\lambda =\frac{-2\pm \sqrt{4+4}}{2}=-1\pm \sqrt{2}$, i.e. $\approx 0.414,-2.414$: opposite signs. ✔

$J=\left(\begin{array}{cc}0& 1\\ -1& -\mu \end{array}\right)$, $\tau =-\mu$, $\Delta =1>0$ always. Discriminant ${\tau }^2-4\Delta ={\mu }^2-4$.

Physical reading: $\mu$ is damping. Positive damping ($\mu >0$) drains energy → stable; $\mu =0$ is the undamped center; negative damping pumps energy → unstable. The stability flips exactly at $\mu =0$ where the eigenvalues cross the imaginary axis. See the crossing on the map figure.

Strategy: put the equilibrium at $(2,0)$ using shifted coordinate $u=x-2$, and pick a Jacobian with $\tau <0,\Delta >0,{\tau }^2<4\Delta$. Take target $J=\left(\begin{array}{cc}-1& -2\\ 2& -1\end{array}\right)$: $\tau =-2<0$, $\Delta =1+4=5>0$, ${\tau }^2-4\Delta =4-20<0$ → stable spiral. ✔ A linear-in-shift field with exactly this Jacobian: $\dot{x}=-(x-2)-2y,\dot{y}=2(x-2)-y.$ Check equilibrium: at $(2,0)$ both bracket terms vanish → $\dot{x}=\dot{y}=0$ ✔. Its Jacobian is the constant matrix above (partials: $f_x=-1,f_y=-2,g_x=2,g_y=-1$). Eigenvalues $\lambda =-1\pm 2i$ → decaying oscillation. Stable spiral at $(2,0)$. ✔

$J=\left(\begin{array}{cc}0& 1\\ -\cos x& -b\end{array}\right)$.

• Bottom $(0,0)$: $\cos 0=1$, $J=\left(\begin{array}{cc}0& 1\\ -1& -b\end{array}\right)$. $\tau =-b<0$, $\Delta =1>0$, disc $=b^2-4<0$ for small $b$ → complex, $\mathrm{Re}=-b/2<0$ → stable spiral. Damping turns the undamped center into a genuine sink — now Hartman–Grobman does apply because $\mathrm{Re}(\lambda )\ne 0$.
• Top $(\pi ,0)$: $\cos \pi =-1$, $J=\left(\begin{array}{cc}0& 1\\ 1& -b\end{array}\right)$. $\tau =-b$, $\Delta =(0)(-b)-(1)(1)=-1<0$ → saddle (unstable), unchanged by damping (still $\Delta <0$). $\lambda =\frac{-b\pm \sqrt{b^2+4}}{2}$: one positive, one negative. ✔

Moral: damping ($b>0$) breaks the marginal center at the bottom and makes it decisively stable, but cannot rescue the intrinsically unstable upright saddle.

Linearize. $f=-x^3\Rightarrow f_x=-3x^2,f_y=0$; $g=-y\Rightarrow g_x=0,g_y=-1$. At $(0,0)$: $J=\left(\begin{array}{cc}0& 0\\ 0& -1\end{array}\right)$, eigenvalues $\lambda =0,-1$. A zero eigenvalue → non-hyperbolic → linearization inconclusive in the $x$-direction. Lyapunov. Try $V=\frac{1}{2}{x}^2+\frac{1}{2}{y}^2\ge 0$, $V=0$ only at origin. $\dot{V}=x\dot{x}+y\dot{y}=x(-x^3)+y(-y)=-x^4-y^2\le 0,$ and $\dot{V}=0$ only at the origin. Since $V>0$ and $\dot{V}<0$ off the origin, the origin is asymptotically stable — even though linearization couldn't tell you. The nonlinear $-x^3$ term does pull $x\to 0$, just slowly.

Equilibria: $y=0$ and $x^2=\mu$.

• $\mu <0$: no real $x$ → no equilibria.
• $\mu =0$: single point $(0,0)$; $J=\left(\begin{array}{cc}-2x& 0\\ 0& -1\end{array}\right){|}_{(0,0)}=\left(\begin{array}{cc}0& 0\\ 0& -1\end{array}\right)$, $\lambda =0,-1$ → non-hyperbolic (birth of the pair).
• $\mu >0$: two points $(\pm \sqrt{\mu },0)$.
  • At $(+\sqrt{\mu },0)$: $-2x=-2\sqrt{\mu }<0$, eigenvalues $-2\sqrt{\mu },-1$ both $<0$ → stable node.
  • At $(-\sqrt{\mu },0)$: $-2x=+2\sqrt{\mu }>0$, eigenvalues $+2\sqrt{\mu },-1$ opposite signs → saddle.

This is a saddle–node bifurcation: as $\mu$ crosses $0$, a stable node and a saddle appear together out of nothing. The exact crossing point is non-hyperbolic — precisely where linearization goes silent.

Find every equilibrium → Jacobian → evaluate there → $(\tau ,\Delta )$ or eigenvalues → classify — and if any $\mathrm{Re}(\lambda )=0$, stop and reach for Lyapunov.