For which values of the parameter a is the trivial solution…
For which values of the parameter $a$ is the trivial solution $x=0,\;y=0,\;z=0$ of the system $$\frac{dx}{dt}=z,\qquad\frac{dy}{dt}=-3x,\qquad\frac{dz}{dt}=ax+2y-3z$$ asymptotically 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 · Coefficient matrix
$$A=\begin{pmatrix}0&0&1\\-3&0&0\\a&2&-3\end{pmatrix}.$$
Step 2 · Characteristic polynomial
$$\det(A-\lambda I)=0\implies P(\lambda)=\lambda^3+3\lambda^2-a\lambda+6, \qquad p_0=1,p_1=3,p_2=-a,p_3=6.$$
Step 3 · Routh--Hurwitz criterion
Necessary: $p_2=-a>0\implies a<0$. Hurwitz condition: $$\Delta_2=p_1p_2-p_0p_3=-3a-6=-3(a+2)>0\implies a<-2.$$ $\Delta_3=p_3\Delta_2=-18(a+2)>0$ gives the same condition.
Step 4 · Boundary case $a=-2$
$$P(\lambda)=\lambda^3+3\lambda^2+2\lambda+6=(\lambda+3)(\lambda^2+2),$$ with roots $\lambda_1=-3<0$ and $\lambda_{2,3}=\pm i\sqrt2$ (simple). The trivial solution is stable but not asymptotically stable at $a=-2$.
Step 5 · Summary table
| Range of $a$ | $\Delta_2=-3(a+2)$ | Stability |
|---|---|---|
| $a<-2$ | $>0$ | Asymptotically stable |
| $a=-2$ | $=0$ | Stable (not asympt.) |
| $a>-2$ | $<0$ | Unstable |
Step 6 · Conclusion
$$\boxed{ \text{The trivial solution $(0,0,0)$ is \textbf{asymptotically stable} if and only if }\;a<-2. }$$ The Routh--Hurwitz condition $\Delta_2=-3(a+2)>0$ gives $a<-2$; at the boundary $a=-2$ the system has simple purely imaginary roots $\pm i\sqrt2$ and is stable but not asymptotically stable.
Summary Table
| Range of $a$ | $\Delta_2=-3(a+2)$ | Stability |
|---|---|---|
| $a<-2$ | $>0$ | Asymptotically stable |
| $a=-2$ | $=0$ | Stable (not asympt.) |
| $a>-2$ | $<0$ | 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: