For which values of the parameter a is the trivial solution…
For which values of the parameter $a$ is the trivial solution of the equation $$y^{(4)}+3y'''+ay''+2y'+y=0$$ 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
$$P(\lambda)=\lambda^4+3\lambda^3+a\lambda^2+2\lambda+1,\qquad a_0=1,a_1=3,a_2=a,a_3=2,a_4=1.$$
Step 2 · Hurwitz determinants
$$\Delta_1=3>0,\qquad \Delta_2=\begin{vmatrix}3&2\\1&a\end{vmatrix}=3a-2,\qquad \Delta_3=\begin{vmatrix}3&2&0\\1&a&1\\0&3&2\end{vmatrix}=6a-13,\qquad \Delta_4=a_4\Delta_3=6a-13.$$
Step 3 · Conditions
$\Delta_2>0\iff a>\tfrac23$; $\Delta_3>0\iff a>\tfrac{13}{6}$. Since $\tfrac{13}{6}>\tfrac23$, the binding condition is $a>\tfrac{13}{6}$.
Step 4 · Boundary case $a=\tfrac{13</h2>
<p>{6}$} $$P(\lambda)=\left(\lambda^2+\tfrac23\right)\left(\lambda^2+3\lambda+\tfrac32\right),$$ with roots $\lambda_{1,2}=\pm i\sqrt{2/3}$ (simple, purely imaginary) and $\lambda_{3,4}=\tfrac{-3\pm\sqrt3}{2}<0$. The trivial solution is stable but not asymptotically stable at $a=\tfrac{13}{6}$.
Step 5 · Summary table
| Range of $a$ | $\Delta_3=6a-13$ | Stability |
|---|---|---|
| $a<\tfrac{13}{6}$ | $<0$ | Unstable |
| $a=\tfrac{13}{6}$ | $=0$ | Stable (not asympt.) |
| $a>\tfrac{13}{6}$ | $>0$ | Asymptotically stable |
Step 6 · Conclusion
$$\boxed{ \text{The trivial solution is \textbf{stable} if and only if } a\geq\frac{13}{6}. }$$ At the boundary $a=\tfrac{13}{6}$ the polynomial factors as $(\lambda^2+\tfrac23)(\lambda^2+3\lambda+\tfrac32)$, giving simple purely imaginary roots $\pm i\sqrt{2/3}$ and two negative real roots.
Summary Table
| Range of $a$ | $\Delta_3=6a-13$ | Stability |
|---|---|---|
| $a<\tfrac{13}{6}$ | $<0$ | Unstable |
| $a=\tfrac{13}{6}$ | $=0$ | Stable (not asympt.) |
| $a>\tfrac{13}{6}$ | $>0$ | Asymptotically stable |
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: