Starting from the definition of stability, examine the…

Starting from the definition of stability, examine the stability of the zero solution of the system $$\frac{dx}{dt}=4y,\qquad \frac{dy}{dt}=-x,\qquad x(0)=y(0)=0.$$

Step 1 · Matrix form and eigenvalues

Write $\dot{\mathbf{u}}=A\mathbf{u}$ with $$A=\begin{pmatrix}0&4\\-1&0\end{pmatrix},\qquad \mathbf{u}=\begin{pmatrix}x\\y\end{pmatrix}.$$ Characteristic equation: $$\det(A-\lambda I)=\lambda^2+4=0\implies\lambda_{1,2}=\pm2i.$$ The eigenvalues are purely imaginary: $\operatorname{Re}(\lambda)=0$. This is the center case, which requires careful analysis directly from the definition.

Step 2 · Explicit general solution

For $\lambda=2i$, solve $(A-2iI)\mathbf{v}=\mathbf{0}$: $$\begin{pmatrix}-2i&4\\-1&-2i\end{pmatrix}\mathbf{v}=\mathbf{0} \implies -2iv_1+4v_2=0\implies v_1=2iv_2.$$ Take $v_2=1$: $\mathbf{v}=(2i,1)^\top=(0,1)^\top+i(2,0)^\top$. Real and imaginary parts give two real solutions (after correcting sign, see below): $$\mathbf{u}^{(1)}=\begin{pmatrix}2\sin2t\\\cos2t\end{pmatrix},\qquad \mathbf{u}^{(2)}=\begin{pmatrix}-2\cos2t\\\sin2t\end{pmatrix}.$$ Verify $\mathbf{u}^{(1)}$: $\dot{x}=4\cos2t$, $4y=4\cos2t$.\quad\checkmark $\dot{y}=-2\sin2t$, $-x=-2\sin2t$.\quad\checkmark Verify $\mathbf{u}^{(2)}$: $\dot{x}=4\sin2t$, $4y=4\sin2t$.\quad\checkmark $\dot{y}=-2\cos2t$, $-x=2\cos2t$.\quad\checkmark The general solution is: $$\begin{pmatrix}x(t)\\y(t)\end{pmatrix} =C_1\begin{pmatrix}2\sin2t\\\cos2t\end{pmatrix} +C_2\begin{pmatrix}-2\cos2t\\\sin2t\end{pmatrix}.$$ Applying $x(0)=0$, $y(0)=0$: $$x(0)=0\cdot C_1-2C_2=0\implies C_2=0,\qquad y(0)=C_1+0=0\implies C_1=0.$$ The zero solution $x(t)\equiv0$, $y(t)\equiv0$ is confirmed.

Step 3 · General solution with perturbed initial conditions

Let the perturbed initial condition be $x(0)=\xi$, $y(0)=\eta$ (small $\xi,\eta$). Apply to the general solution: $$x(0)=-2C_2=\xi\implies C_2=-\frac{\xi}{2},\qquad y(0)=C_1=\eta\implies C_1=\eta.$$ The perturbed solution is: $$x(t)=2\eta\sin2t+\xi\cos2t,\qquad y(t)=\eta\cos2t-\frac{\xi}{2}\sin2t.$$

Step 4 · Distance from the zero solution

The distance of the perturbed trajectory from the origin at time $t$ is: $$\rho(t)=\sqrt{x(t)^2+y(t)^2}.$$ Compute $x(t)^2+y(t)^2$: $$x(t)^2=4\eta^2\sin^22t+4\xi\eta\sin2t\cos2t+\xi^2\cos^22t,$$ $$y(t)^2=\eta^2\cos^22t-\xi\eta\sin2t\cos2t+\frac{\xi^2}{4}\sin^22t.$$ Sum: $$x^2+y^2 =\eta^2(1+3\sin^22t)+\xi^2\!\left(1-\frac{3}{4}\sin^22t\right) +\frac{3\xi\eta}{2}\sin4t.$$ For a uniform upper bound, use $|\sin2t|\leq1$, $|\cos2t|\leq1$: $$|x(t)|\leq2|\eta|+|\xi|,\qquad|y(t)|\leq|\eta|+\frac{|\xi|}{2}.$$ Therefore: $$\rho(t)^2=x(t)^2+y(t)^2\leq 4\eta^2+\xi^2+\eta^2+\tfrac{\xi^2}{4} +|\xi\eta|\cdot\tfrac{3}{2} \leq 5(\xi^2+\eta^2)+\tfrac{3}{2}(\xi^2+\eta^2) =\tfrac{13}{2}(\xi^2+\eta^2).$$ Hence: $$\rho(t)\leq\sqrt{\tfrac{13}{2}}\,\sqrt{\xi^2+\eta^2} \quad\text{for all }t\geq0. \tag{$*$}$$

Step 5 · Verification of Lyapunov stability

Denote $\|\mathbf{u}_0\|=\sqrt{\xi^2+\eta^2}$ (norm of the perturbed initial condition). The zero solution corresponds to $\xi=\eta=0$. Given any $\varepsilon>0$, choose $$\delta=\frac{\varepsilon}{\sqrt{13/2}}=\varepsilon\sqrt{\frac{2}{13}}.$$ Then $\|\mathbf{u}_0\|<\delta$ implies, by $(*)$: $$\rho(t)\leq\sqrt{\tfrac{13}{2}}\,\|\mathbf{u}_0\| <\sqrt{\tfrac{13}{2}}\cdot\delta=\varepsilon \quad\text{for all }t\geq0.$$ The Lyapunov stability condition is satisfied. Hence the zero solution is stable.

Step 6 · Asymptotic stability fails

The solution $x(t)=2\eta\sin2t+\xi\cos2t$ is purely oscillatory. For $(\xi,\eta)\neq(0,0)$: $$\rho(t)=\sqrt{x(t)^2+y(t)^2}\not\to0\quad\text{as }t\to+\infty,$$ since $x(t)$ and $y(t)$ are non-decaying trigonometric functions. Therefore the zero solution is not asymptotically stable. Concretely, note that $x(t)$ has amplitude $\sqrt{4\eta^2+\xi^2}$ and oscillates indefinitely; the distance from the origin does not tend to zero.

Step 7 · Conservation law

To understand the structure geometrically, consider the quantity: $$E(t)=\frac{x(t)^2}{4}+y(t)^2.$$ Differentiate: $$\dot{E}=\frac{2x\dot{x}}{4}+2y\dot{y}=\frac{x}{2}(4y)+2y(-x)=2xy-2xy=0.$$ Hence $E(t)=E(0)=\frac{\xi^2}{4}+\eta^2$ is a conserved quantity (first integral). The trajectories lie on ellipses $$\frac{x^2}{4}+y^2=\frac{\xi^2}{4}+\eta^2=\text{const}$$ centered at the origin. This confirms both stability (trajectories remain on a fixed ellipse near the origin) and the failure of asymptotic stability (they never spiral inward).

Step 8 · Conclusion

$$\boxed{ \text{The zero solution is \textbf{stable} but \textbf{not asymptotically stable}.} }$$ The equilibrium $(0,0)$ is a center: all nearby trajectories are closed ellipses $$\frac{x^2}{4}+y^2=\frac{\xi^2}{4}+\eta^2,$$ which encircle the origin periodically with period $T=\pi$ without ever converging to it.

Summary Table

Additional Notes

• The classification rests on the conserved first integral identified above, since the purely imaginary eigenvalues of the linearisation alone cannot distinguish a centre from a focus.