For which values of the parameters α and β is the trivial…
For which values of the parameters $\alpha$ and $\beta$ is the trivial solution of the equation $$y^{(4)}+\alpha y'''+2y''+\beta y'+y=0$$ 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
$$P(\lambda)=\lambda^4+\alpha\lambda^3+2\lambda^2+\beta\lambda+1,\qquad a_0=1,a_1=\alpha,a_2=2,a_3=\beta,a_4=1.$$
Step 2 · Hurwitz determinants
$$\Delta_1=\alpha,\qquad\Delta_2=2\alpha-\beta,\qquad \Delta_3=2\alpha\beta-\alpha^2-\beta^2=-(\alpha-\beta)^2.$$
Step 3 · Key observation
Since $\Delta_3=-(\alpha-\beta)^2\leq0$ for all real $\alpha,\beta$, with equality only when $\alpha=\beta$, the condition $\Delta_3>0$ required for asymptotic stability is never satisfied: the trivial solution is never asymptotically stable, for any choice of $\alpha,\beta$.
Step 4 · Case $\alpha=\beta>0$
Then $\Delta_2=\alpha>0$ and $\Delta_3=0$ (critical). The polynomial factors as $$P(\lambda)=(\lambda^2+1)(\lambda^2+\alpha\lambda+1),$$ with roots $\lambda_{1,2}=\pm i$ (simple, purely imaginary) and, for $\alpha>0$, $\lambda_{3,4}$ with strictly negative real part (Routh-- Hurwitz for the quadratic). The trivial solution is stable but not asymptotically stable.
Step 5 · Case $\alpha=\beta=0$
$P(\lambda)=(\lambda^2+1)^2$: purely imaginary roots $\pm i$ each with multiplicity 2, producing unbounded terms $t\cos t,t\sin t$: unstable.
Step 6 · Case $\alpha=\beta<0$ or $\alpha\neq\beta$
For $\alpha=\beta<0$ the quadratic factor has roots with positive real part: unstable. For $\alpha\neq\beta$, $\Delta_3<0$ strictly, giving a sign change in the Routh array and hence roots with positive real part: unstable.
Step 7 · Summary table
| Condition | $\Delta_3$ | Stability |
|---|---|---|
| $\alpha=\beta>0$ | $=0$ | Stable (not asympt.) |
| $\alpha=\beta=0$ | $=0$ | Unstable (repeated $\pm i$) |
| $\alpha=\beta<0$ | $=0$ | Unstable |
| $\alpha\neq\beta$ | $<0$ | Unstable |
Step 8 · Conclusion
$$\boxed{ \text{The trivial solution is \textbf{never asymptotically stable} for any }(\alpha,\beta)\in\mathbb{R}^2. }$$ $$\boxed{ \text{The trivial solution is \textbf{stable} (not asymptotically) if and only if }\alpha=\beta>0. }$$ The obstruction is $\Delta_3=-(\alpha-\beta)^2\leq0$ always; equality (and hence marginal stability) holds exactly on the line $\alpha=\beta>0$, where $P(\lambda)=(\lambda^2+1)(\lambda^2+\alpha\lambda+1)$ has simple purely imaginary roots $\pm i$.
Summary Table
| Condition | $\Delta_3$ | Stability |
|---|---|---|
| $\alpha=\beta>0$ | $=0$ | Stable (not asympt.) |
| $\alpha=\beta=0$ | $=0$ | Unstable (repeated $\pm i$) |
| $\alpha=\beta<0$ | $=0$ | Unstable |
| $\alpha\neq\beta$ | $<0$ | Unstable |
Additional Notes
- The boundary curves or points where a Hurwitz determinant vanishes mark the transition between asymptotic stability and instability; on such a boundary the characteristic polynomial typically factors to reveal simple purely imaginary roots, giving stability without asymptotic stability exactly on that boundary.
Result
The final classification is: