GATE Mathematics Ordinary Differential Equations Flashcards

24 sample cards from the Ordinary Differential Equations deck

Sampled from the end of the deck, so these are different cards from the ones shown on the syllabus page.

1. State the eigenvalue criterion for (Lyapunov) stability of the origin of $\dot{\mathbf{x}}=A\mathbf{x}$.

   The origin is stable (in the sense of Lyapunov) iff all eigenvalues satisfy $\operatorname{Re}(\lambda)\leq 0$ AND every eigenvalue with $\operatorname{Re}(\lambda)=0$ is semisimple (its algebraic and geometric multiplicities are equal, i.e. no defective Jordan blocks).

2. State the eigenvalue criterion for instability of the origin of $\dot{\mathbf{x}}=A\mathbf{x}$.

   The origin is unstable if at least one eigenvalue has positive real part $\operatorname{Re}(\lambda)>0$, or if a purely imaginary/zero eigenvalue is defective (algebraic multiplicity exceeds geometric multiplicity).

3. For the 2D system $\dot{\mathbf{x}}=A\mathbf{x}$, classify the equilibrium when $A$ has two real eigenvalues of the same sign.

   A node. Both negative $\Rightarrow$ stable node (sink); both positive $\Rightarrow$ unstable node (source). Trajectories approach/leave the origin tangent to the slow (smaller magnitude) eigendirection.

4. For the 2D system $\dot{\mathbf{x}}=A\mathbf{x}$, classify the equilibrium when $A$ has two real eigenvalues of opposite signs.

   A saddle point, which is always unstable. Trajectories approach along the stable eigendirection (negative eigenvalue) and depart along the unstable eigendirection (positive eigenvalue).

5. For the 2D system $\dot{\mathbf{x}}=A\mathbf{x}$, classify the equilibrium when $A$ has complex eigenvalues $\alpha\pm i\beta$ with $\alpha\neq0$.

   A spiral (focus). $\alpha<0\Rightarrow$ stable spiral (trajectories spiral inward); $\alpha>0\Rightarrow$ unstable spiral (spiral outward). $\beta$ governs the rotation rate.

6. For the 2D system $\dot{\mathbf{x}}=A\mathbf{x}$, classify the equilibrium when $A$ has purely imaginary eigenvalues $\pm i\beta$.

   A center: trajectories are closed ellipses around the origin. It is stable (Lyapunov) but not asymptotically stable.

7. Using trace $T=\operatorname{tr}A$ and determinant $D=\det A$ of a $2\times2$ matrix, give the conditions for the origin to be asymptotically stable.

   Asymptotically stable iff $T=\lambda_1+\lambda_2<0$ and $D=\lambda_1\lambda_2>0$ (both eigenvalues have negative real part).

8. Using $T=\operatorname{tr}A$, $D=\det A$, and discriminant $\Delta=T^2-4D$, how do you distinguish nodes from spirals in 2D?

   If $D>0$: $\Delta=T^2-4D>0$ gives a node (real eigenvalues), $\Delta<0$ gives a spiral (complex eigenvalues), and $\Delta=0$ gives a degenerate/improper node. If $D<0$, it is a saddle regardless of $T$.

9. What does $\det A<0$ tell you immediately about the equilibrium of a planar linear system?

   It is a saddle point (eigenvalues are real with opposite signs), and therefore unstable.

10. Define a stable node versus an unstable node in terms of eigenvalues for a planar system.

    Stable (improper/proper) node: both eigenvalues real and negative, $\lambda_2\leq\lambda_1<0$. Unstable node: both real and positive, $0<\lambda_1\leq\lambda_2$. Trajectories converge to / diverge from origin without spiraling.

11. State the general nonlinear autonomous system whose equilibrium stability is studied by linearization.

    $$\dot{\mathbf{x}}=\mathbf{f}(\mathbf{x}),\qquad \mathbf{f}(\mathbf{x}^*)=\mathbf{0},$$ where $\mathbf{x}^*$ is an equilibrium (stationary) point.

12. What is the Jacobian (linearization) matrix used to analyze stability of an equilibrium $\mathbf{x}^*$ of $\dot{\mathbf{x}}=\mathbf{f}(\mathbf{x})$?

    $$J=D\mathbf{f}(\mathbf{x}^*)=\left[\frac{\partial f_i}{\partial x_j}\right]_{\mathbf{x}=\mathbf{x}^*}.$$ The linearized system is $\dot{\mathbf{u}}=J\mathbf{u}$ with $\mathbf{u}=\mathbf{x}-\mathbf{x}^*$.

13. State the Linearized Stability (Hartman-Grobman based) theorem for a hyperbolic equilibrium.

    If all eigenvalues of $J=D\mathbf{f}(\mathbf{x}^*)$ have nonzero real part (the equilibrium is hyperbolic), then the nonlinear system's stability matches the linear one: all $\operatorname{Re}\lambda<0\Rightarrow$ asymptotically stable; some $\operatorname{Re}\lambda>0\Rightarrow$ unstable. Near $\mathbf{x}^*$ the nonlinear and linear phase portraits are topologically conjugate.

14. When does linearization FAIL to determine the stability of an equilibrium?

    When the equilibrium is non-hyperbolic, i.e. the Jacobian has at least one eigenvalue with zero real part ($\operatorname{Re}\lambda=0$, e.g. a center case). Then higher-order/nonlinear terms decide stability, and a Lyapunov function is needed.

15. Define a Lyapunov-stable equilibrium $\mathbf{x}^*$ of $\dot{\mathbf{x}}=\mathbf{f}(\mathbf{x})$.

    $\mathbf{x}^*$ is stable if for every $\varepsilon>0$ there exists $\delta>0$ such that $\|\mathbf{x}(0)-\mathbf{x}^*\|<\delta$ implies $\|\mathbf{x}(t)-\mathbf{x}^*\|<\varepsilon$ for all $t\geq0$. It is asymptotically stable if additionally $\mathbf{x}(t)\to\mathbf{x}^*$ as $t\to\infty$.

16. Define a Lyapunov function $V(\mathbf{x})$ for an equilibrium at the origin of $\dot{\mathbf{x}}=\mathbf{f}(\mathbf{x})$.

    A continuously differentiable function $V$ on a neighborhood of $\mathbf{0}$ with $V(\mathbf{0})=0$ and $V(\mathbf{x})>0$ for $\mathbf{x}\neq\mathbf{0}$ (positive definite), whose orbital derivative $\dot{V}$ satisfies a sign condition along trajectories.

17. How is the orbital derivative $\dot{V}$ of a Lyapunov function computed along solutions of $\dot{\mathbf{x}}=\mathbf{f}(\mathbf{x})$?

    $$\dot{V}(\mathbf{x})=\nabla V(\mathbf{x})\cdot \mathbf{f}(\mathbf{x})=\sum_{i=1}^{n}\frac{\partial V}{\partial x_i}\,f_i(\mathbf{x}).$$ No need to solve the ODE; it is evaluated using the vector field directly.

18. State Lyapunov's theorem on (simple) stability via a Lyapunov function.

    If there is a positive definite $V$ with $\dot{V}(\mathbf{x})\leq 0$ (negative semidefinite) in a neighborhood of the origin, then the origin is stable (in the sense of Lyapunov).

19. State Lyapunov's theorem on asymptotic stability.

    If there is a positive definite $V$ with $\dot{V}(\mathbf{x})<0$ for all $\mathbf{x}\neq\mathbf{0}$ in a neighborhood (negative definite), then the origin is asymptotically stable.

20. State Chetaev's / Lyapunov's instability theorem.

    If there is a $C^1$ function $V$ with $V(\mathbf{0})=0$ that takes positive values arbitrarily close to the origin and has $\dot{V}>0$ (positive definite) there, then the origin is unstable.

21. What additional condition (LaSalle's invariance principle) lets $\dot{V}\leq0$ still imply asymptotic stability?

    If $\dot{V}\leq0$ and the only complete trajectory contained in the set $\{\dot{V}=0\}$ is the equilibrium itself, then the origin is asymptotically stable.

22. For a linear system $\dot{\mathbf{x}}=A\mathbf{x}$, what quadratic form is commonly used as a Lyapunov function, and what equation must $P$ satisfy?

    Use $V(\mathbf{x})=\mathbf{x}^T P\mathbf{x}$ with $P$ symmetric positive definite. Then $\dot{V}=\mathbf{x}^T(A^TP+PA)\mathbf{x}$. Stability follows if $P$ solves the Lyapunov equation $$A^TP+PA=-Q$$ for some positive definite $Q$, which has a positive definite solution $P$ iff $A$ is a stability (Hurwitz) matrix.

23. Compare the strengths of the eigenvalue (linearization) method versus the Lyapunov function method for stability.

    Eigenvalue/linearization: easy and definitive for hyperbolic equilibria, but fails when $\operatorname{Re}\lambda=0$ and is only local. Lyapunov functions: can handle non-hyperbolic cases and give global/large basin estimates, but there is no general recipe for constructing $V$, so it requires ingenuity.

24. For the undamped pendulum / center-type system, why is $V=$ total energy a useful Lyapunov function, and what does it show?

    Energy $V$ is positive definite near the stable equilibrium and conserved, giving $\dot{V}=0$. By Lyapunov's stability theorem this proves the equilibrium is stable (a center), even though linearization gives purely imaginary eigenvalues and is inconclusive on its own.

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