Problem 52
Easy Lyapunov Stability

For which values of the parameter α is the trivial solution…

For which values of the parameter $\alpha$ is the trivial solution of the equation $$y'''+\alpha y''+2y'+y=0$$ stable?

Explanation of Method

The Routh–Hurwitz criterion is applied with the coefficients of the characteristic polynomial left as functions of the free parameter(s). Requiring every coefficient and every Hurwitz determinant to be strictly positive translates into a system of inequalities on the parameter(s), carving out the region of asymptotic stability; the boundary of that region, where a determinant vanishes, is checked separately and typically corresponds to simple purely imaginary roots (marginal stability).

Step 1 · Characteristic polynomial

$$P(\lambda)=\lambda^3+\alpha\lambda^2+2\lambda+1,\qquad a_0=1,\;a_1=\alpha,\;a_2=2,\;a_3=1.$$

Step 2 · Routh--Hurwitz criterion

For a cubic, asymptotic stability requires $a_k>0$ for all $k$ and $\Delta_2=a_1a_2-a_0a_3>0$: $$\alpha>0,\qquad\Delta_2=2\alpha-1>0\implies\alpha>\tfrac12.$$ The binding condition is $\alpha>\tfrac12$.

Step 3 · Boundary case $\alpha=\tfrac12$

At $\alpha=\tfrac12$, $\Delta_2=0$ and $$P(\lambda)=\left(\lambda+\tfrac12\right)(\lambda^2+2),$$ with roots $\lambda_1=-\tfrac12<0$ and $\lambda_{2,3}=\pm i\sqrt2$ (simple, purely imaginary). The trivial solution is stable but not asymptotically stable at $\alpha=\tfrac12$.

Step 4 · Other ranges

For $0<\alpha<\tfrac12$: one sign change in the Routh array, unstable. For $\alpha\leq0$: necessary condition fails, unstable.

Step 5 · Summary table

Range of $\alpha$Sign changesStability
$\alpha\leq0$$\geq1$Unstable
$0<\alpha<\tfrac12$$1$Unstable
$\alpha=\tfrac12$$0$ (zero row)Stable (not asympt.)
$\alpha>\tfrac12$$0$Asymptotically stable

Step 6 · Conclusion

$$\boxed{ \text{The trivial solution is \textbf{stable} if and only if } \alpha\geq\frac{1}{2}. }$$ More precisely: $\alpha>\tfrac12$ gives asymptotic stability, $\alpha=\tfrac12$ gives stability without asymptotic stability (roots $-\tfrac12,\pm i\sqrt2$), and $\alpha<\tfrac12$ is unstable.


Summary Table

Range of $\alpha$Sign changesStability
$\alpha\leq0$$\geq1$Unstable
$0<\alpha<\tfrac12$$1$Unstable
$\alpha=\tfrac12$$0$ (zero row)Stable (not asympt.)
$\alpha>\tfrac12$$0$Asymptotically stable

Additional Notes

Result

The final classification is:

$$\boxed{ \text{The trivial solution is \textbf{stable} if and only if } \alpha\geq\frac{1}{2}. }$$