Science:Math Exam Resources/Courses/MATH307/December 2005/Question 06 (a)

MATH307 December 2005

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Question 06 (a)
Consider the differential equation $x'(t)={\begin{bmatrix}-1&-\alpha \\1&-1\end{bmatrix}}x(t)$ (a) For what values of $\alpha$ is the system stable, neutrally stable or unstable?

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Hint

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. To determine if a system is stable, neutrally stable or unstable, look at the eigenvalue. A system is: - stable if $\displaystyle \lambda _{1}=\lambda _{2}<0$ - neutrally stable (or asymptotically stable) if $\displaystyle \lambda _{2}<\lambda _{1}<0$ - unstable if any eigenvalue $\displaystyle \lambda >0$ ${\begin{aligned}{\text{det}}(\lambda I-A)&={\text{det}}{\begin{bmatrix}\lambda +1&\alpha \\-1&\lambda +1\end{bmatrix}}\\&=(\lambda +1)^{2}+\alpha \\&=\lambda ^{2}+2\lambda +(\alpha +1)\end{aligned}}$ and thus ${\begin{aligned}\lambda &={\frac {-2\pm {\sqrt {2^{2}-4(\alpha +1)}}}{2}}\\&={\frac {-2\pm {\sqrt {-4\alpha }}}{2}}\\&{\frac {-2\pm 2i\alpha }{2}}\\&=-1\pm \alpha i\end{aligned}}$ stable if: $\displaystyle -1+\alpha i=-1-\alpha i$; this implies that $\displaystyle \alpha =0$ neutrally stable if $\displaystyle \lambda _{2}<\lambda _{1}<0$. Since -1<0, this occurs for any $\displaystyle \alpha \neq 0$ the system is never unstable because -1 <0