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CSIR NET Mathematical Sciences · revision sheet

ODE, PDE & Applied Mathematics

16 subtopics · roughly 45 marks · 12 counterexamples

What the exam asks of each subtopic

Ordinary Differential Equations

Existence–uniqueness, Picard, Lipschitz.: Peano gives existence from continuity alone; uniqueness needs a Lipschitz condition in y. Global existence needs the right side to grow at most linearly. y′ = y^{1/3} and y′ = y² are the two spoilers.
Linear ODE, Wronskian, variation of parameters, systems.: Wronskian non-zero ⇒ independent, but the converse needs the functions to solve a common linear ODE. For systems, the eigenvalues of A decide the growth rate of e^{At}.
Sturm–Liouville problems and Green's functions.: Eigenvalues of a regular Sturm–Liouville problem are real, simple and unbounded above; eigenfunctions for distinct eigenvalues are orthogonal with respect to the weight. Green's function is built from the two boundary solutions.
Stability and phase portraits.: Classify a 2×2 linear system from trace and determinant alone; for nonlinear systems linearise and use Hartman–Grobman — but remember the centre case is the one linearisation cannot decide.

Partial Differential Equations

First-order PDE: Lagrange, Charpit, characteristics.: Write the characteristic ODEs, carry the initial data along them, and check whether characteristics cross — that crossing is what makes the solution fail to exist globally.
Classification and canonical forms.: B² − 4AC decides the type pointwise; the sign can change with the point (Tricomi). Reduce to canonical form by solving the characteristic ODE dy/dx = (B ± √(B²−4AC))/2A.
Laplace, heat and wave equations: separation of variables.: Know d'Alembert cold, the separation-of-variables series for the three classical equations, and the qualitative differences: smoothing (heat), finite speed (wave), mean value (Laplace).

Numerical Analysis

Root finding: bisection, Newton–Raphson, fixed point, order of convergence.: Order of convergence is the whole topic: bisection 1, secant φ ≈ 1.618, Newton 2 (but only 1 at a multiple root), fixed point 1 unless g′(r) = 0.
Interpolation and numerical integration with error terms.: Degree of precision is the key number: trapezoidal 1, Simpson 3, n-point Gauss 2n−1. Interpolation error carries f⁽ⁿ⁺¹⁾/(n+1)! times the node product.
Numerical ODE: Euler, Runge–Kutta.: Know the local vs global order (Euler: local O(h²), global O(h)), the RK2 order conditions, and the stability interval for Euler applied to y′ = λy.

Calculus of Variations

Euler–Lagrange equation and standard functionals.: Write the Euler–Lagrange equation; use the Beltrami identity when F has no explicit x. Terms that are exact derivatives (null Lagrangians) do not change the extremals.
Isoperimetric problems.: Introduce a Lagrange multiplier, solve the modified Euler–Lagrange equation, then use the constraint plus boundary conditions to pin every constant.

Linear Integral Equations

Fredholm and Volterra equations.: Volterra always has a unique solution; Fredholm depends on whether λ is an eigenvalue, and then the alternative decides solvability by an orthogonality condition.
Separable kernels and resolvent kernels.: Separable kernels turn integral equations into matrix problems. The resolvent kernel sums the Neumann series and gives the solution in closed form.

Classical Mechanics

Lagrangian formalism and generalised coordinates.: Write T and V in the generalised coordinate, form L = T − V, and apply d/dt(∂L/∂q̇) = ∂L/∂q. Cyclic coordinates give conserved momenta immediately.
Hamiltonian formalism and conservation laws.: H = Σpq̇ − L expressed in (q, p). Poisson brackets test canonicity: {Qᵢ, Pⱼ} = δᵢⱼ, {Qᵢ, Qⱼ} = {Pᵢ, Pⱼ} = 0.

Tempting, and false

Each claim below feels true and is not. The object beside it is the one that settles it.

Every IVP has a unique solution
↳ y′ = y^{1/3}, y(0) = 0 — Not Lipschitz at 0; y ≡ 0 and y = (2x/3)^{3/2} both solve it.
A continuous right-hand side gives a unique solution
↳ y′ = y^{1/3}, y(0) = 0 — Not Lipschitz at 0: y ≡ 0 and y = (2x/3)^{3/2} both solve it, as do infinitely many hybrids.
W(f, g) ≡ 0 implies f and g are linearly dependent
↳ f(x) = x², g(x) = x|x| on ℝ — The Wronskian vanishes identically but no constant multiple relates them. The implication holds only for solutions of a common linear ODE.
Every boundary value problem has a Green's function
↳ y″ + π²y = f on [0,1] with y(0) = y(1) = 0 — λ = π² is an eigenvalue of the homogeneous problem, so the operator is not invertible and no Green's function exists.
Linearisation determines stability at every equilibrium
↳ ẋ = −y − x³, ẏ = x − y³ at the origin — The linearisation is a centre (eigenvalues ±i), but V = x² + y² gives V̇ = −2(x⁴ + y⁴) < 0: asymptotically stable. Non-hyperbolic equilibria escape Hartman–Grobman.
A first-order quasilinear Cauchy problem with smooth data has a global smooth solution
↳ u uₓ + u_y = 0 with u(x, 0) = −x — Characteristics x = x₀(1 − y) all meet at y = 1: the solution blows up in gradient (a shock).
A PDE has a single type throughout its domain
↳ The Tricomi equation y uₓₓ + u_yy = 0 — Elliptic in the upper half-plane, hyperbolic in the lower, parabolic on the axis.
Newton's method converges quadratically to any root
↳ f(x) = x² at the root 0 — The root is double: xₙ₊₁ = xₙ/2, only linear convergence.
Increasing the number of equally spaced interpolation nodes improves the approximation
↳ Runge's function 1/(1 + 25x²) on [−1, 1] — The interpolants diverge near the endpoints as n → ∞. Chebyshev nodes restore convergence.
A weak minimum of a functional is a strong minimum
↳ J[y] = ∫₀^π (1 − (y′)²)y² dx at y ≡ 0 — In the C¹-ball the integrand is non-negative so J ≥ 0; in the C⁰-ball a steeply oscillating small y makes J negative.
Every integral equation of the second kind has a unique solution
↳ y(x) − 3∫₀¹ tx y(t)dt = eˣ — λ = 3 is an eigenvalue of the kernel tx, and eˣ is not orthogonal to t — no solution exists.
The Hamiltonian always equals the total energy
↳ A bead on a wire rotating at forced angular velocity ω — The constraint is time-dependent, so H is conserved but differs from T + V.

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