Problem 28
Easy Linearization

Examine the character of the rest point (0,0) of the system…

Examine the character of the rest point $(0,0)$ of the system $$\dot{x}=x+y,\qquad\dot{y}=-x+y.$$

Explanation of Method

Since the system is already linear and homogeneous, its coefficient matrix is written down directly. The eigenvalues are computed from the characteristic equation, and the rest point is classified according to the standard theory of planar linear systems: a node, saddle, focus, or centre, depending on the sign and nature of the eigenvalues.

Step 1 · Coefficient matrix

$$A=\begin{pmatrix}1&1\\-1&1\end{pmatrix}.$$

Step 2 · Characteristic equation and eigenvalues

$$\det(A-\lambda I)=0\implies(1-\lambda)^2+1=0\implies\lambda_{1,2}=1\pm i.$$

Step 3 · Classification of the rest point

The eigenvalues are complex with positive real part: $\operatorname{Re}(\lambda_{1,2})=1>0$. The rest point $(0,0)$ is an unstable focus: all trajectories spiral away from the origin as $t\to+\infty$.

Step 4 · Direction of rotation

Evaluating $\dot y$ at $(1,0)$ gives $\dot y=-1<0$, so the trajectories rotate clockwise.

Step 5 · Polar form and exact spiral

Introducing polar coordinates $x=r\cos\theta$, $y=r\sin\theta$ and differentiating $r^2=x^2+y^2$ and $\theta=\arctan(y/x)$ along the flow gives the decoupled system $$\dot r=r,\qquad\dot\theta=-1.$$ Solving exactly: $r(t)=r_0e^{t}$, $\theta(t)=\theta_0-t$, so eliminating $t$ gives the family of logarithmic spirals $$r=C\,e^{-\theta},\quad C>0.$$

Step 6 · Behaviour of trajectories

From $\dot r=r>0$: as $t\to+\infty$, $r\to+\infty$ (outward spiral); as $t\to-\infty$, $r\to0$ (trajectories approach the origin while completing infinitely many clockwise revolutions). The angular speed is constant, $|\dot\theta|=1$, giving period $T=2\pi$ and exponential radial growth rate $\alpha=1$.

Step 7 · Stability analysis

Since $\operatorname{Re}(\lambda_{1,2})=1>0$, the rest point is unstable: it is not Lyapunov stable.

Step 8 · Conclusion

$$\boxed{ (0,0)\text{ is an \textbf{unstable focus}.}\quad \lambda_{1,2}=1\pm i,\quad\operatorname{Re}(\lambda)=1>0. }$$ In polar coordinates the system reduces to $\dot{r}=r$, $\dot{\theta}=-1$, giving the exact family of logarithmic spirals $r=Ce^{-\theta}$ ($C>0$). All trajectories spiral clockwise outward with angular frequency $\omega=1$ and exponential growth rate $\alpha=1$, confirming instability.


Summary Table

QuantityValue
Eigenvalues$\det(A-\lambda I)=0\implies(1-\lambda)^2+1=0\implies\lambda_{1,2}=1\pm i.$
Classificationunstable focus

Additional Notes

Result

The final classification is:

$$\boxed{ (0,0)\text{ is an \textbf{unstable focus}.}\quad \lambda_{1,2}=1\pm i,\quad\operatorname{Re}(\lambda)=1>0. }$$