For Bessel’s equation x²y″ + xy′ + (x² − ν²)y = 0, the point x = 0 is

an ordinary point only when ν = 0

For the equation x²(x − 1)²y″ + xy′ + y = 0, which statements are true?

x = 0 is a regular singular point

x = 1 is an irregular singular point

x = 0 is a regular singular point

x = 1 is a regular singular point

x = 1 is an irregular singular point

Which statements about Bessel functions of the first kind are true?

J_{−n}(x) = (−1)ⁿJₙ(x) for every positive integer n

For X′ = AX with A a real 2 × 2 matrix, which statements are true?

if trace A 0, the origin is asymptotically stable

if det A < 0, the origin is a saddle point

if the eigenvalues are purely imaginary, the origin is stable but not asymptotically stable

if trace A 0, the origin is asymptotically stable

if det A < 0, the origin is a saddle point

if the eigenvalues are purely imaginary, the origin is stable but not asymptotically stable

if trace A = 0, the origin is a centre

For the system x′ = −x + y, y′ = −x − y³ and V(x, y) = x² + y², which statements are true?

along solutions, dV/dt = −2x² − 2y⁴

Which statements are true? (q denotes a continuous function.)

between two consecutive zeros of a non-trivial solution of y″ + y = 0 there is exactly one zero of any linearly independent solution

every non-trivial solution of y″ + (1 + x²)y = 0 has a zero in every open interval of length π

the zeros of a non-trivial solution of y″ + q(x)y = 0 are isolated

between two consecutive zeros of a non-trivial solution of y″ + y = 0 there is exactly one zero of any linearly independent solution

every non-trivial solution of y″ + (1 + x²)y = 0 has a zero in every open interval of length π

every non-trivial solution of y″ − y = 0 has infinitely many zeros

the zeros of a non-trivial solution of y″ + q(x)y = 0 are isolated

Series Solutions, Legendre and Bessel Functions, Systems, Sturm Theory and Stability

The second ODE chapter covers the rest of Section 5. Series solutions handle equations with variable coefficients: a power series at an ordinary point, and the Frobenius method with its indicial equation at a regular singular point. Two of the equations so solved define the Legendre polynomials and the Bessel functions, whose orthogonality relations are standard NAT material. Systems of linear first-order equations are solved by the matrix exponential. Sturm’s oscillation and separation theorems say where zeros of solutions can lie without solving anything, and Sturm–Liouville problems produce real eigenvalues and orthogonal eigenfunctions. Finally, for planar autonomous systems, the eigenvalues of the linear part classify the stationary point, linearisation transfers the verdict to non-linear systems when no eigenvalue has zero real part, and a Lyapunov function settles the cases linearisation cannot.

1. Power series and the Frobenius method

For y″ + P(x)y′ + Q(x)y = 0, x₀ is an ordinary point if P and Q are analytic at x₀; then two independent power-series solutions exist, converging at least up to the nearest singular point. x₀ is a regular singular point if it is not ordinary but (x − x₀)P and (x − x₀)²Q are analytic there; otherwise it is irregular. At a regular singular point (say 0) with xP → p₀ and x²Q → q₀, try y = Σ aₙx^n+r; the lowest power gives the indicial equation r(r − 1) + p₀r + q₀ = 0.

If the roots r₁ ≥ r₂ do not differ by an integer, there are two Frobenius solutions x^r₁Σaₙxⁿ and x^r₂Σbₙxⁿ.
If r₁ = r₂, the second solution is y₁ ln x + x^r₁Σbₙxⁿ; if r₁ − r₂ is a positive integer, the second may or may not contain a logarithm.
Example: 2x²y″ + xy′ − (x + 1)y = 0 has p₀ = 1/2, q₀ = −1/2, indicial equation 2r² − r − 1 = 0, roots 1 and −1/2, which differ by 3/2: two Frobenius series.

2. Legendre and Bessel functions

Legendre’s equation (1 − x²)y″ − 2xy′ + n(n + 1)y = 0 has ordinary point 0 and regular singular points ±1. For integer n ≥ 0 one solution is the polynomial Pₙ with Pₙ(1) = 1, given by Rodrigues’ formula Pₙ = (1/(2ⁿn!)) dⁿ/dxⁿ (x² − 1)ⁿ: P₀ = 1, P₁ = x, P₂ = (3x² − 1)/2, P₃ = (5x³ − 3x)/2. Orthogonality: ∫₋₁¹ PₘPₙ dx = 0 for m ≠ n and 2/(2n + 1) for m = n. Every polynomial of degree n is a combination of P₀, …, Pₙ; for instance x² = (2P₂ + P₀)/3, so ∫₋₁¹ x²P₂ dx = (2/3)(2/5) = 4/15.

Bessel’s equation x²y″ + xy′ + (x² − ν²)y = 0 has a regular singular point at 0 with indicial roots ±ν. The Frobenius solution for r = ν is J_ν(x) = Σ (−1)ᵏ(x/2)^2k+ν/(k! Γ(k + ν + 1)). Facts: J₀(0) = 1 and Jₙ(0) = 0 for n ≥ 1; J_−n = (−1)ⁿJₙ for integer n (so a second solution Yₙ is needed); J₀′ = −J₁; (xᵛJ_ν)′ = xᵛJ_ν−1; J_1/2(x) = √(2/(πx)) sin x. Each J_ν has infinitely many positive zeros, and if α ≠ β are two of them, ∫₀¹ x J_ν(αx)J_ν(βx) dx = 0: orthogonality with weight x.

3. Systems of linear first-order equations

X′ = A(t)X with A continuous on I has an n-dimensional solution space; a fundamental matrix Φ has independent solutions as columns and det Φ satisfies Liouville’s formula det Φ(t) = det Φ(t₀)exp ∫ tr A. For constant A the solution is X(t) = e^tAX(0). If A = PDP⁻¹ is diagonalisable, e^tA = Pe^tDP⁻¹; if A = λI + N with N nilpotent, e^tA = e^λt(I + tN + t²N²/2 + …), a finite sum.

Example. A = [[1, 1], [0, 1]] = I + N with N² = 0, so e^tA = eᵗ[[1, t], [0, 1]]. From X(0) = (1, 1), X(t) = eᵗ(1 + t, 1); at t = 1 the first component is 2e ≈ 5.44.

4. Sturm’s theorems and Sturm–Liouville problems

Separation: if y₁, y₂ are independent solutions of y″ + q(x)y = 0, their zeros strictly interlace: between two consecutive zeros of one there is exactly one zero of the other (sin x and cos x).
Comparison (oscillation): if q₂ > q₁ on an interval, then between two zeros of a non-trivial solution of y″ + q₁y = 0 lies a zero of every solution of y″ + q₂y = 0. Since sin(x − a) solves y″ + y = 0, every solution of y″ + (1 + x²)y = 0 has a zero in every open interval of length π.
If q ≤ 0, a non-trivial solution has at most one zero (y″ − y = 0: cosh, sinh, eˣ); zeros of non-trivial solutions are always isolated.

A regular Sturm–Liouville problem is (p(x)y′)′ + (q(x) + λr(x))y = 0 on [a, b], p > 0, r > 0, with separated boundary conditions α₁y(a) + α₂y′(a) = 0, β₁y(b) + β₂y′(b) = 0. Its eigenvalues are real and simple, form an increasing sequence λ₁ < λ₂ < … → ∞, the n-th eigenfunction has exactly n − 1 zeros in (a, b), and eigenfunctions of distinct eigenvalues are orthogonal with weight r. For y″ + λy = 0: with y(0) = y(π) = 0, λₙ = n² and yₙ = sin nx; with y(0) = 0 and y′(π) = 0, cos(√λ π) = 0 so λₙ = (n − 1/2)², the smallest being 1/4.

5. Planar autonomous systems: stability, linearisation and Lyapunov functions

X′ = AX, A real 2 × 2, by the eigenvalues of A (τ = trace, Δ = det)
Eigenvalues	Stationary point	Stability of the origin
real, both negative (Δ > 0, τ < 0, τ² ≥ 4Δ)	stable node	asymptotically stable
real, opposite signs (Δ < 0)	saddle	unstable
complex, negative real part (τ < 0, τ² < 4Δ)	stable spiral (focus)	asymptotically stable
purely imaginary (τ = 0, Δ > 0)	centre	stable, not asymptotically stable
positive real parts (τ > 0, Δ > 0)	unstable node or spiral	unstable

Linearisation. For X′ = F(X) with F(X∗) = 0 and F of class C¹, let J = DF(X∗). If every eigenvalue of J has negative real part, X∗ is asymptotically stable; if some eigenvalue has positive real part, X∗ is unstable. If the eigenvalues are purely imaginary the linearisation decides nothing. For the damped pendulum x′ = y, y′ = −sin x − y: at (0, 0), J = [[0, 1], [−1, −1]], τ = −1, Δ = 1, a stable spiral; at (π, 0), J = [[0, 1], [1, −1]], Δ = −1, a saddle.

🎯 Lyapunov’s direct method

If V is C¹, positive definite near X∗ (V(X∗) = 0, V > 0 elsewhere) and V̇ = ∇V·F ≤ 0, then X∗ is stable; if V̇ < 0 away from X∗, it is asymptotically stable. For x′ = −x + y, y′ = −x − y³ take V = x² + y²: V̇ = 2x(−x + y) + 2y(−x − y³) = −2x² − 2y⁴ < 0 for (x, y) ≠ 0, so the origin is asymptotically stable. The method needs no solution and works where linearisation is silent, e.g. x′ = −x³ (J = 0, V = x²).

Key takeaways

Regular singular point: (x − x₀)P and (x − x₀)²Q analytic; the indicial equation r(r − 1) + p₀r + q₀ = 0 gives the Frobenius exponents.
∫₋₁¹ Pₙ² = 2/(2n + 1); Bessel functions are orthogonal on [0, 1] with weight x at their zeros; J₀′ = −J₁.
X′ = AX is solved by e^tA; for A = λI + N use the finite nilpotent series.
Sturm: zeros of independent solutions interlace, larger q oscillates faster, q ≤ 0 allows at most one zero; Sturm–Liouville eigenvalues are real and simple with weighted-orthogonal eigenfunctions.
Trace and determinant classify planar linear systems; linearisation decides unless an eigenvalue has zero real part, and Lyapunov functions decide without solving.

Practice questions (14)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

For Bessel’s equation x²y″ + xy′ + (x² − ν²)y = 0, the point x = 0 is
1. a regular singular point
2. an ordinary point
3. an irregular singular point
4. an ordinary point only when ν = 0
Show answer
Answer: A — a regular singular point
In standard form P = 1/x and Q = 1 − ν²/x², both singular at 0, so 0 is not ordinary for any ν. But xP = 1 and x²Q = x² − ν² are analytic, so it is a regular singular point, with indicial equation r² − ν² = 0.
For 2x²y″ + xy′ − (x + 1)y = 0, the difference between the two roots of the indicial equation at x = 0 (larger minus smaller) is ____.
Numerical answer — type the value.
Show answer
Answer: 1.5
Dividing by 2x²: P = 1/(2x), Q = −(x + 1)/(2x²), so p₀ = lim xP = 1/2 and q₀ = lim x²Q = −1/2. The indicial equation r(r − 1) + r/2 − 1/2 = 0 is 2r² − r − 1 = (2r + 1)(r − 1) = 0: roots 1 and −1/2, difference 1.5, not an integer, so two Frobenius series exist.
For the equation x²(x − 1)²y″ + xy′ + y = 0, which statements are true?
1. x = 0 is a regular singular point
2. x = 1 is a regular singular point
3. x = 1 is an irregular singular point
4. x = −1 is an ordinary point
Show answer
Answer: A — x = 0 is a regular singular point; C — x = 1 is an irregular singular point; D — x = −1 is an ordinary point
P = 1/(x(x − 1)²), Q = 1/(x²(x − 1)²). At 0: xP = 1/(x − 1)² and x²Q = 1/(x − 1)² are analytic, so regular singular. At 1: (x − 1)P = 1/(x(x − 1)) is not analytic, so irregular. At −1, P and Q are analytic, so ordinary.
The value of ∫₋₁¹ [P₃(x)]² dx, where P₃ is the Legendre polynomial of degree 3, correct to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.29
∫₋₁¹ Pₙ² dx = 2/(2n + 1) = 2/7 = 0.2857 for n = 3, i.e. 0.29. Directly, P₃ = (5x³ − 3x)/2 and ∫(25x⁶ − 30x⁴ + 9x²)/4 dx over [−1, 1] = (50/7 − 12 + 6)/4 = 2/7.
The value of the Legendre polynomial P₂ at x = 0.5 is ____.
Numerical answer — type the value.
Show answer
Answer: -0.125
P₂(x) = (3x² − 1)/2, so P₂(0.5) = (0.75 − 1)/2 = −0.125. Check the normalisation P₂(1) = 1, which the unnormalised 3x² − 1 would fail.
The value of ∫₋₁¹ x² P₂(x) dx is
1. 4/15
2. 2/5
3. 0
4. 2/3
Show answer
Answer: A — 4/15
x² = (2P₂ + P₀)/3. By orthogonality ∫P₀P₂ = 0, so the integral is (2/3)∫P₂² = (2/3)(2/5) = 4/15. The value 0 would hold for ∫x P₂, since x = P₁; 2/5 is ∫P₂² itself.
Which statements about Bessel functions of the first kind are true?
1. J_1/2(x) = √(2/(πx)) sin x
2. J₀(0) = 1
3. J_−n(x) = (−1)ⁿJₙ(x) for every positive integer n
4. J₀′(x) = J₁(x)
Show answer
Answer: A — J_{1/2}(x) = √(2/(πx)) sin x; B — J₀(0) = 1; C — J_{−n}(x) = (−1)ⁿJₙ(x) for every positive integer n
(1) The series for ν = 1/2 sums to √(2/(πx)) sin x. (2) Only the k = 0 term of J₀ survives at 0. (3) For integer n the first n terms of J_−n vanish (1/Γ of a non-positive integer is 0) and the rest reproduce (−1)ⁿJₙ. (4) is false: J₀′ = −J₁, from (x^−νJ_ν)′ = −x^−νJ_ν+1 with ν = 0.
For X′ = AX with A = [[1, 1], [0, 1]] and X(0) = (1, 1)ᵀ, the first component of X(1), correct to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 5.44
A = I + N with N = [[0, 1], [0, 0]], N² = 0, and I, N commute, so e^tA = eᵗ(I + tN) = eᵗ[[1, t], [0, 1]]. X(t) = eᵗ(1 + t, 1), and at t = 1 the first component is 2e = 5.4366, i.e. 5.44. Directly: x₂ = eᵗ, then x₁′ − x₁ = eᵗ gives x₁ = (1 + t)eᵗ.
For the system x′ = y, y′ = −2x − 3y, the origin is
1. a stable node
2. a saddle point
3. a centre
4. an unstable spiral
Show answer
Answer: A — a stable node
A = [[0, 1], [−2, −3]] has trace −3 and determinant 2, so the eigenvalues solve λ² + 3λ + 2 = 0: −1 and −2, real, distinct and negative. That is a stable node, asymptotically stable. It is the system form of y″ + 3y′ + 2y = 0.
For X′ = AX with A a real 2 × 2 matrix, which statements are true?
1. if trace A < 0 and det A > 0, the origin is asymptotically stable
2. if det A < 0, the origin is a saddle point
3. if the eigenvalues are purely imaginary, the origin is stable but not asymptotically stable
4. if trace A = 0, the origin is a centre
Show answer
Answer: A — if trace A < 0 and det A > 0, the origin is asymptotically stable; B — if det A < 0, the origin is a saddle point; C — if the eigenvalues are purely imaginary, the origin is stable but not asymptotically stable
(1) Both eigenvalues then have negative real part (sum < 0, product > 0). (2) det < 0 means real eigenvalues of opposite sign. (3) Orbits are closed ellipses: bounded, not tending to 0. (4) is false: A = [[1, 0], [0, −1]] has trace 0 and det −1, a saddle; a centre needs trace 0 and det > 0.
For the damped pendulum x′ = y, y′ = −sin x − y, the stationary point (π, 0) is
1. a saddle point, hence unstable
2. a stable spiral
3. a centre
4. a stable node
Show answer
Answer: A — a saddle point, hence unstable
The Jacobian is [[0, 1], [−cos x, −1]]; at x = π it is [[0, 1], [1, −1]] with determinant −1 < 0, so the eigenvalues are real of opposite sign, (−1 ± √5)/2. Linearisation is decisive (no zero real part): a saddle, the inverted pendulum. The stable spiral is at (0, 0).
For the system x′ = −x + y, y′ = −x − y³ and V(x, y) = x² + y², which statements are true?
1. V is positive definite
2. along solutions, dV/dt = −2x² − 2y⁴
3. the origin is asymptotically stable
4. the origin is unstable
Show answer
Answer: A — V is positive definite; B — along solutions, dV/dt = −2x² − 2y⁴; C — the origin is asymptotically stable
V > 0 except at 0. dV/dt = 2x(−x + y) + 2y(−x − y³) = −2x² + 2xy − 2xy − 2y⁴ = −2x² − 2y⁴, which is negative except at the origin, so Lyapunov’s theorem gives asymptotic stability, and (4) is false.
Which statements are true? (q denotes a continuous function.)
1. between two consecutive zeros of a non-trivial solution of y″ + y = 0 there is exactly one zero of any linearly independent solution
2. every non-trivial solution of y″ + (1 + x²)y = 0 has a zero in every open interval of length π
3. every non-trivial solution of y″ − y = 0 has infinitely many zeros
4. the zeros of a non-trivial solution of y″ + q(x)y = 0 are isolated
Show answer
Answer: A — between two consecutive zeros of a non-trivial solution of y″ + y = 0 there is exactly one zero of any linearly independent solution; B — every non-trivial solution of y″ + (1 + x²)y = 0 has a zero in every open interval of length π; D — the zeros of a non-trivial solution of y″ + q(x)y = 0 are isolated
(1) is Sturm separation (sin x and cos x). (2) Compare with y″ + y = 0: its solution sin(x − a) vanishes at a and a + π, and 1 + x² ≥ 1 with equality only at x = 0, so every solution of the equation with the larger coefficient vanishes strictly between a and a + π. (3) is false: solutions are c₁eˣ + c₂e^−x, with at most one zero. (4) A zero with y′ = 0 too would force y ≡ 0 by uniqueness, so y′ ≠ 0 at each zero.
The smallest eigenvalue λ of y″ + λy = 0 on [0, π] with y(0) = 0 and y′(π) = 0 is ____.
Numerical answer — type the value.
Show answer
Answer: 0.25
λ ≤ 0 gives only y ≡ 0. For λ = μ² > 0, y(0) = 0 gives y = sin μx, and y′(π) = μ cos μπ = 0 needs μ = n − 1/2. So λₙ = (n − 1/2)² and the smallest is 1/4 = 0.25, with eigenfunction sin(x/2). With y(π) = 0 instead, the answer would be 1.