GATE Mathematics Ordinary Differential Equations Syllabus

In the Frobenius method, what is the form of the solutions when the indicial roots are equal, $r_1=r_2=r$?

$$y_1=x^{r}\sum_{n\geq0} a_n x^n,\qquad y_2=y_1\ln x + x^{r}\sum_{n\geq1} b_n x^n.$$ A logarithmic term always appears in the second solution.

In the Frobenius method, what is the form of the solutions when the indicial roots differ by a positive integer, $r_1-r_2=N\in\mathbb{Z}^+$?

$$y_1=x^{r_1}\sum_{n\geq0} a_n x^n,\qquad y_2=C\,y_1\ln x + x^{r_2}\sum_{n\geq0} b_n x^n,$$ where the constant $C$ may be zero (no log term) or nonzero, depending on the equation.

Which root of the indicial equation is used first to guarantee a Frobenius series solution always exists?

The larger root $r_1$ (with $r_1\geq r_2$ for real roots) always yields a Frobenius series solution $y_1=x^{r_1}\sum a_n x^n$.

What is Bessel's equation, and what determines whether its solutions involve a logarithm via Frobenius?

$$x^2y''+xy'+(x^2-\nu^2)y=0.$$ Indicial roots are $r=\pm\nu$. They differ by $2\nu$; when $2\nu$ is a non-negative integer (e.g. integer order $\nu$), the second solution $Y_\nu$ involves a logarithmic term.

Write the general homogeneous linear system with constant coefficients whose stability is studied.

$$\dot{\mathbf{x}}=A\mathbf{x},\qquad \mathbf{x}\in\mathbb{R}^n,$$ where $A$ is a constant $n\times n$ matrix; the origin $\mathbf{x}=\mathbf{0}$ is the stationary (equilibrium) point.

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

The origin is asymptotically stable if and only if every eigenvalue $\lambda$ of $A$ has negative real part, $\operatorname{Re}(\lambda)<0$.

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).

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).

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.

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).

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.

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.

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).

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 $T$.

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.

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.

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.

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 $\mathbf{u}=\mathbf{x}-\mathbf{x}^*$.

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.

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.

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 $t\to\infty$.

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.

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.

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).

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.

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.

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.

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.

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.

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.