For which values of the parameter a does the system…

For which values of the parameter $a$ does the system $$\dot{x}=y+ax,\qquad\dot{y}=-x$$ have a stable rest point?

Step 1 · Locating the rest point

Setting both right-hand sides to zero gives $x=0$, $y=0$: the unique rest point is $(0,0)$ for all values of $a$.

Step 2 · Coefficient matrix

$$A=\begin{pmatrix}a&1\\-1&0\end{pmatrix}.$$

Step 3 · Characteristic equation and eigenvalues

$$\det(A-\lambda I)=0\implies\lambda^2-a\lambda+1=0 \implies\lambda=\frac{a\pm\sqrt{a^2-4}}{2}.$$ We analyse the three cases determined by the sign of the discriminant $\Delta=a^2-4$.

Step 4 · Case $|a|>2$ (two distinct real eigenvalues)

Since $\lambda_1\lambda_2=1>0$ and $\lambda_1+\lambda_2=a$, both roots share the sign of $a$: for $a>2$ both are positive (unstable node); for $a<-2$ both are negative (stable node).

Step 5 · Case $|a|=2$ (repeated real eigenvalue)

$\lambda_{1,2}=a/2$. For $a=2$: $\lambda=1>0$, unstable (improper) node. For $a=-2$: $\lambda=-1<0$, stable (improper) node; checking $A+I=\begin{pmatrix}-1&1\\-1&1\end{pmatrix}$ has rank 1, confirming the node is degenerate, but stability is unaffected since $\lambda=-1<0$.

Step 6 · Case $|a|<2$ (complex conjugate eigenvalues)

$$\lambda_{1,2}=\frac{a}{2}\pm\frac{\sqrt{4-a^2}}{2}\,i,\qquad \operatorname{Re}(\lambda_{1,2})=\frac{a}{2}.$$ For $a>0$: unstable focus. For $a=0$: $\lambda_{1,2}=\pm i$, a centre (stable, not asymptotically stable) — the first integral $x^2+y^2=\text{const}$ shows all trajectories are circles. For $a<0$: stable focus.

Step 7 · Summary table

Step 8 · Stability conditions via Routh--Hurwitz

Applying the Routh--Hurwitz criterion to $\lambda^2+(-a)\lambda+1=0$, both roots have negative real part iff $-a>0$ and $1>0$, i.e.\ $a<0$ (asymptotic stability); Lyapunov stability holds for $a\leq0$.

Step 9 · Conclusion

$$\boxed{ \begin{aligned} &\text{The rest point $(0,0)$ is \textbf{asymptotically stable} if and only if }a<0.\\[4pt] &\text{It is \textbf{stable} (but not asymptotically) if and only if }a=0\text{ (centre)}.\\[4pt] &\text{It is \textbf{unstable} for all }a>0. \end{aligned} }$$ In particular, for asymptotic stability the necessary and sufficient condition is $a<0$, which ensures $\operatorname{Re}(\lambda_{1,2})=a/2<0$ regardless of whether the eigenvalues are real or complex.

Summary Table

Additional Notes

• The boundary values of the parameter (where the discriminant vanishes, or a real part passes through zero) mark the transitions between qualitatively different phase portraits, and deserve special attention alongside the open ranges.