ODE, PDE & Applied Mathematics

Ordinary Differential Equations

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1. Existence–uniqueness, Picard, Lipschitz

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Lec-16 Gronwall's Lemma

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Grönwall is the tool behind uniqueness and continuous dependence.

Lec-18 Picard's Existence and Uniqueness Theorem

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Where the Lipschitz condition is used — and what breaks without it.

Lec-20 Cauchy Peano Existence Theorem

NPTEL · Ordinary Differential Equations and Applications

Existence from continuity alone, with no uniqueness.

Existence, uniqueness and how far the solution extends free

In plain words

Three questions get bundled together: does a solution exist, is it the only one, and how long does it last? Continuity gives the first, a Lipschitz condition the second, nothing here the third.

The three theorems

Theorem	Hypothesis on f(x, y)	Conclusion
Peano	continuous near $(x_{0}, y_{0})$	a local solution exists
Picard–Lindelöf	continuous + Lipschitz in y	local solution exists and is unique
Global	continuous + \|f(x,y)\| ≤ A + B\|y\|	solution extends to the whole interval

$f \in C^{1} \Rightarrow$ locally Lipschitz ⇒ uniqueness. Lipschitz is sufficient, not necessary.

The two spoilers

Non-uniqueness: y′ = $y^{1/3}$ , y(0) = 0. The right side is continuous but not Lipschitz at 0 (the derivative blows up). Solutions: y ≡ 0 and y = ± $(2x/3)^{3/2}$ , plus every "wait then leave" hybrid — infinitely many.

Finite-time blow-up: $y' = y^{2}, y(0) = 1$ has the unique solution y = 1/(1 − x), which escapes at x = 1. Uniqueness is local; superlinear growth destroys global existence.

Extension rules

Bounded f ⇒ |y′| bounded ⇒ no blow-up ⇒ global.
Linear growth |f| ≤ A + B|y| ⇒ global (Grönwall).
Equilibria trap solutions: if $f(y_{0}) = 0$ then $y \equiv y_{0}$ is a solution, and by uniqueness no other solution can cross that level. This is why $y' = (1 - y^{2})^{n} \cos y$ with y(0) = 0 stays in (−1, 1) for all time.

Grönwall's inequality

$u(x) \le A + B\int{}u \Rightarrow u(x) \le A$ $e^{Bx}$ . This is the engine behind continuous dependence on initial data and global existence.

Key takeaways

Continuity ⇒ existence; Lipschitz ⇒ uniqueness; linear growth ⇒ global.
$y^{1/3}$ kills uniqueness; $y^{2}$ kills global existence.
Equilibria act as barriers that solutions can never cross.

Picard iteration converginginteractive

This visual is part of Trap Analysis

Start from a constant and watch successive integrals close on the solution — agreement spreading outward, which is why existence is only local.

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Exam focus: 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.

Practice

Part B2025 Decembershow ▾

Consider the following statements: (P) The initial value problem y′ = f(y), where f(y) = y·cos(1/y) for y ≠ 0 and f(0) = 0, y(0) = 0, has at most one solution. (Q) If z(x) is the solution of the initial value problem

z' = (1-z^{4})^{100}/(1+z^{2}), z(0) = 0

, then for each

\alpha \in (0,2)

, there exists

x_\alpha \in \mathbb{R}

such that

z(x_\alpha) = 1-\alpha

. Then which of the following statements is true?

A.Both (P) and (Q) are true.✓ Correct
B.(P) is true, (Q) is FALSE.
C.(P) is FALSE, (Q) is true.
D.Both (P) and (Q) are FALSE.

The worked solution and the per-option analysis are in Trap Analysis.

Part C2025 Juneshow ▾

Let D = {

(x, y) \in \mathbb{R}^{2} : -1 \le x \le 1, -1 \le y \le 1

}, and

f : D \to \mathbb{R}

be the function defined by

f(x, y) = 1 + \surd(y

₊), where y₊ = max{y, 0}. Consider the initial value problem (IVP) dy/dx = f(x, y), y(0) = 0. Then which of the following statements are true?

A.f is a Lipschitz continuous function on D
B.f is NOT a Lipschitz continuous function on D✓ Correct
C.IVP has at least one solution✓ Correct
D.IVP has NO solution

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Decembershow ▾

The initial value problem dy/dx

= \surd

|x − 1| sin y, y(0) = 1 has

A.a unique solution on $\mathbb{R}$ ✓ Correct
B.infinitely many solutions on the interval (−2, 2)
C.a unique solution and its maximal interval of existence is $(-\infty, 1)$
D.no solution on the interval (−2, 2)

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Juneshow ▾

Consider the initial value problem (IVP

) y'(x) = \surd

y(x) + \varepsilon

x \in \mathbb{R}, y(0) = y_{0}

. Consider the following statements: S1: There is an

\varepsilon > 0

such that for all

y_{0} \in \mathbb{R}

, the IVP has more than one solution. S2: There is

a y_{0} \in \mathbb{R}

such that for all

\varepsilon > 0

, the IVP has more than one solution. Then

A.both $S_{1}$ and $S_{2}$ are true
B. $S_{1}$ is true but $S_{2}$ is false
C. $S_{1}$ is false but $S_{2}$ is true
D.both $S_{1}$ and $S_{2}$ are false✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2023 Juneshow ▾

Let

f \in C^{1}(\mathbb{R})

be bounded. Consider the initial-value problem (P): x′(t) = f(x(t)), t > 0, x(0) = 0. Which of the following statements are true?

A.(P) has solution(s) defined for all t > 0.✓ Correct
B.(P) has a unique solution.✓ Correct
C.(P) has infinitely many solutions.
D.The solution(s) of (P) is/are Lipschitz.✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2023 Decembershow ▾

Consider the initial value problem y′ = y + ½|

\sin(y^{2})

|, x > 0, y(0) = −1. Which of the following statements are true?

A.There exists an $\alpha \in (0, \infty)$ such that $lim_{x\to\alpha^{-}}$ |y(x)| $= \infty$ .
B.y(x) exists on $(0, \infty)$ and it is monotone.✓ Correct
C.y(x) exists on $(0, \infty)$ , but not bounded below.✓ Correct
D.y(x) exists on $(0, \infty)$ , but not bounded above.

The worked solution and the per-option analysis are in Trap Analysis.

Part C2023 Decembershow ▾

Consider the problem

y' = (1 - y^{2})^{10} \cos y, y(0) = 0

. Let J be the maximal interval of existence and K be the range of the solution. Which of the following statements are true?

A. $J = \mathbb{R}$ ✓ Correct
B.K = (−1, 1)✓ Correct
C.J = (−1, 1)
D.K = [−1, 1]

The worked solution and the per-option analysis are in Trap Analysis.

Part C2023 Decembershow ▾

Consider the initial value problem

x^{2}y'' - 2x^{2}y' + (4x - 2)y = 0, y(0) = 0

. Suppose

y = \varphi(x)

is a polynomial solution satisfying

\varphi(1) = 1

. Which of the following statements are true?

A. $\varphi(4) = 16$ ✓ Correct
B. $\varphi(2) = 2$
C. $\varphi(5) = 25$ ✓ Correct
D. $\varphi(3) = 3$

Solution

Try $y = x^{2}: 2x^{2} - 4x^{3} + 4x^{3} - 2x^{2} = 0$ .

So $\varphi(x) = x^{2}$ , giving $\varphi(4) = 16$ and $\varphi(5) = 25$ .

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Part Bshow ▾

The IVP y′ =

y^{1/3}

, y(0) = 0 has

A.exactly one solution
B.no solution
C.infinitely many solutions✓ Correct
D.exactly two solutions

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Consider the initial value problem y′ =

3y^{2/3}

, y(0) = 0. Which of the following are true?

A.y ≡ 0 is a solution.✓ Correct
B. $y(x) = x^{3}$ is a solution.✓ Correct
C.The IVP has infinitely many solutions.✓ Correct
D.The right-hand side is Lipschitz in y in a neighbourhood of y = 0.

The worked solution and the per-option analysis are in Trap Analysis.

2. Linear ODE, Wronskian, variation of parameters, systems

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Lec-12 Second Order Linear Equations

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Lec-24 General System and Diagonalizability

NPTEL · Ordinary Differential Equations and Applications

Systems ẋ = Ax solved through the eigenstructure of A.

Lec-27 General Systems

NPTEL · Ordinary Differential Equations and Applications

Linear ODE, Wronskian and systems free

In plain words

Solutions form a vector space whose dimension is the order, so enough independent ones finish the job. The Wronskian tests independence only for solutions of the same equation.

Wronskian

For solutions of the same linear ODE y⁽ $^{n}$ ⁾ + $a_{n-1}y$ ⁽ $^{n}$ ⁻ $^{1}$ ⁾ + ⋯ = 0 on an interval:

W ≡ 0 ⇔ the solutions are linearly dependent.
Abel's formula: $W(x) = W(x_{0}) \exp(-\int{}a_{n-1})$ . So W is either identically zero or never zero.

The trap: for arbitrary functions, W ≡ 0 does not imply dependence. $f = x^{2}$ and g = x|x| have W ≡ 0 on $\mathbb{R}$ but are independent. They are not solutions of a common second-order linear ODE with continuous coefficients.

Solving

Equation	Method
constant coefficients	characteristic roots; repeated root r of multiplicity m gives $e^{rx}$ , $xe^{rx}$ , …, $x^{m-1}e^{rx}$
Cauchy–Euler $x^{2}y'' +$ axy′ + by = 0	try y = x^r
non-homogeneous	undetermined coefficients or variation of parameters $y_p = -y_{1}\int{}y_{2}g/W + y_{2}\int{}y_{1}g/W$
reduction of order	one solution known: $y_{2} = y_{1}\int{}e^{-\int{}p}/y_{1}^{2}$

Systems ẋ = Ax

x(t) = $e^{At}x_{0}$ . Behaviour is read off the eigenvalues of A:

Eigenvalues	Origin is
all $Re \lambda < 0$	asymptotically stable (node/spiral)
some $Re \lambda > 0$	unstable
real, opposite signs	saddle (unstable)
pure imaginary	centre — stable but not asymptotically (for linear systems)
repeated with one eigenvector	degenerate node, $e^{At}$ has a t $e^{\lambda{}t}$ term

Growth: |x(t)| ~ $e^{\lambda_max t}$ . So $e^{-ct}$ |x(t)| → 0 for every $c > \max Re \lambda$ , and $\to \infty$ for $c < \max Re \lambda$ .

Key takeaways

Abel: the Wronskian of ODE solutions is never "sometimes zero".
W ≡ 0 ⇏ dependent for arbitrary functions $(x^{2}$ and x|x|).
Eigenvalue real parts decide stability and the exact exponential growth rate.

The Wronskian: all or nothing, by Abel's identityinteractivefree

Step 1 / 4Abel's identity

For $y'' + p(x)y' + q(x)y = 0$ with $p,q$ continuous on an interval $I$ , the Wronskian of two solutions satisfies $W' = -pW$ , so

$W(x) = W(x_0)\exp\left(-\int_{x_0}^x p\right).$

The exponential never vanishes.

Exam focus: 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}.

Practice

Part C2025 Decembershow ▾

Suppose

y_{1}(x)

and

y_{2}(x)

are two linearly independent solutions of the differential equation

x^{2}y'' + [(1+\sin x)x/2]y' - (3/2)(\cos x)y = 0, x>0

, satisfying

y_{2}(0)=0

. Then which of the following statements are true?

A. $\lim(x\to0+) y_{2}(x)/(x^{2}y_{1}(x))$ exists.✓ Correct
B. $\lim(x\to0+) y_{2}(x)/($ xy $_{1}(x))$ exists.✓ Correct
C.lim(x→0+) xy $_{1}(x)/y_{2}(x)$ does NOT exist.✓ Correct
D. $\lim(x\to0+) x^{2}y_{1}(x)/y_{2}(x)$ exists.

The worked solution and the per-option analysis are in Trap Analysis.

Part C2025 Juneshow ▾

For a continuous function q defined on

\mathbb{R}

, consider the ordinary differential equation (ODE

) d^{2}y/

^{2} + q(x)y = 0, x \in \mathbb{R}

. Then which of the following statements are FALSE?

A.There exists a q such that cos(x) and e^x cos(x) are solutions of ODE✓ Correct
B.There exists a q such that sin(x) and cos(x) are solutions of ODE
C.There exists a q such that e^x sin(x) and e^x cos(2x) are solutions of ODE✓ Correct
D.There exists a q such that xe^x and x(x − 1)e^x are solutions of ODE✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part B2025 Juneshow ▾

\varphi(x) = x

is a solution of the ordinary differential equation (ODE

) d^{2}y/

^{2} - (2/x^{2} + 1/x)(x

dy/dx

- y) = 0, 0 < x < \infty

, then the general solution of the ODE is given by

A.(a + $be^{-2x})x$ , $a, b \in \mathbb{R}$
B.(a + $be^{2x})x$ , $a, b \in \mathbb{R}$
C.ae^x + bx, $a, b \in \mathbb{R}$
D.(a + be $^x)x, a, b \in \mathbb{R}$ ✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2025 Juneshow ▾

Consider the ordinary differential equation (ODE

) d^{2}y/

^{2} + \cos(x)

dy/dx + sin(x) y = 0. Let

\varphi_{1}(x), \varphi_{2}(x)

be solutions of the ODE, satisfying

\varphi_{1}(0) = 1, d\varphi_{1}/

dx(0) = 0, and

\varphi_{2}(0) = 0, d\varphi_{2}/

dx(0) = 1. Then which of the following statements are true?

A. $\varphi_{1}(x + 2\pi)$ is also a solution of ODE✓ Correct
B. $\varphi_{2}(x + 4\pi)$ is also a solution of ODE✓ Correct
C.There are NO constants a, b such that $\varphi_{2}(x + 4\pi) = a\varphi_{1}(x) + b\varphi_{2}(x)$
D.There exist $a, b \in \mathbb{R}$ such that $\varphi_{1}(x + 2\pi) = a\varphi_{1}(x) + b\varphi_{2}(x)$ ✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part B2025 Decembershow ▾

For

a \in \mathbb{R}

, let

y_{1}(x)

and

y_{2}(x)

be solutions of the differential equation

y'' + (e^(x^{2}) + \cos x)y = 0

such that

y_{1}(0) = 3, y_{1}'(0) = -1, y_{2}(0) = -5, y_{2}'(0) = a

. Suppose

W(y_{1},y_{2})(x)

denotes the Wronskian of

y_{1}

and

y_{2}

. If

W(y_{1},y_{2})(1/2) = 4

, then the value of a is

A.2
B.3✓ Correct
C.4
D.−3

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Decembershow ▾

Given that

y_{1}(x) = e^(2x)

is a solution of the ordinary differential equation (ODE

) x d^{2}y/

^{2} - (3 + 4x)

dy/dx + (4x + 6)y = 0, x > 0. Let

y_{2} = y_{2}(x)

be the solution of the ODE satisfying the conditions

y_{2}(1) = e^{2}/4

and (dy

_{2}/

)(1) = 3e^{2}/2

. Then which of the following statements is true?

A. $y_{2}$ is a strictly increasing function on $(0, \infty)$ ✓ Correct
B. $e^(-2x)y_{2}(x) \to 1$ as $x \to \infty$
C. $y_{2}$ is a strictly decreasing function on $(0, \infty)$
D. $e^(-2x)y_{2}(x) \to 0$ as $x \to \infty$

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾

If x = x(t), y = y(t) is the solution of the initial value problem dx/dt = x − 4e^(−2t)y, dy/dt = e^(2t)x − y, x(0) = 1, y(0) = 1, then which of the following statements are true?

A. $\lim(t\to\infty) t$ ⁻ $^{2}x(t)y(t) = 0$
B.x(1) = 0, y(1/2) = 0
C.x(1/2) = 0, y(1) = 0✓ Correct
D. $\lim(t\to\infty) t$ ⁻ $^{2}x(t)y(t) = 2$ ✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾

If x = x(t), y = y(t) is the solution of the initial value problem dx/dt + dy/dt = 2x − 3y + eᵗ, dx/dt + 2(dy/dt) = 3x − 4y + 2eᵗ, x(0) = −1, y(0) = −1/2, then which of the following statements are true?

A. $x(\pi) = -e^\pi, y(\pi) = 1/2$ ✓ Correct
B. $x(-\pi) = e^(-\pi), y(-\pi) = 1/2$
C.x has infinitely many zeros in $\mathbb{R}$ ✓ Correct
D.y has infinitely many zeros in $\mathbb{R}$ ✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Juneshow ▾

Let

\varphi

denote the solution to the boundary value problem (BVP) (xy′)′ − 2y′ + y/x = 1 for

1 < x < e^{4}

, with y(1) = 0 and

y(e^{4}) = 4e^{4}

. Then the value of

\varphi(e)

A.−e/2✓ Correct
B.−e/3
C.e/3
D.e

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Juneshow ▾

x_{1} = x_{1}(t), x_{2} = x_{2}(t)

is the solution of the initial value problem e^(−t) dx

_{1}/

= -x_{1} + x_{2}, e^(-t)

_{2}/

= -x_{1} - x_{2}, x_{1}(0) = 1, x_{2}(0) = 0

, and

r(t) = \sqrt{x_{1}^{2}(t) + x_{2}^{2}(t)}

, then which of the following statements are true?

A.r(t) → 0 as $t \to +\infty$ ✓ Correct
B.r(ln 2) = e⁻ $^{1}$ ✓ Correct
C.r(ln 2) = 2e⁻ $^{1}$
D.r(t)eᵗ → 0 as $t \to +\infty$ ✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾

Consider the ordinary differential equation (ODE

) d^{2}y/

^{2} + 2(e^(2x) - 1)(

dy/dx

) + \alpha{}e^(4x)y = 0, x \in \mathbb{R}

, where

\alpha \in \mathbb{R}

. Note that the ODE transforms into an equation with constant coefficients under the change of independent variable given by t = e^(2x)/2. Then which of the following statements are true?

A.For $\alpha = 1$ , all the solutions of the ODE tend to zero as $x \to \infty$ ✓ Correct
B.For $\alpha = 0$ , there exists a solution of the ODE which tends to 1 as $x \to \infty$ ✓ Correct
C.For $\alpha = -1$ , there exists an unbounded solution of the ODE on $\mathbb{R}$ ✓ Correct
D.For $\alpha = 2$ , there exists an unbounded solution of the ODE on $\mathbb{R}$

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾

Consider the non-homogeneous ordinary differential equation (ODE

) d^{2}y/

^{2} + 5

dy/dx + 6y = sin(e^(−5x)), x > 0. Then which of the following statements are true?

A.Every solution of the ODE is bounded on $(0, \infty)$ ✓ Correct
B.There exists a solution of the ODE which is unbounded on $(0, \infty)$
C.Every solution of the ODE is unbounded on $(0, \infty)$
D.Every solution of the ODE tends to zero as $x \to \infty$ ✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Decembershow ▾

Suppose that the differential equation

d^{2}y/

^{2} + P(x)

dy/dx

+ e^(2x)y = 0, x \in \mathbb{R}

transforms into a second order differential equation with constant coefficients under the change of independent variable given by s = s(x) satisfying (ds/dx)(0) = 1. Then which of the following statements is true?

A.e⁻ˣ(P(x) + 1) is a constant function on $\mathbb{R}$ ✓ Correct
B.e^(−2x)P(x) is a constant function on $\mathbb{R}$
C. $s(x) = e^(2x)/2, x \in \mathbb{R}$
D.P(x) → 1 as $x \to \infty$

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Decembershow ▾

For any non-zero solution y = y(x) of the differential equation

(2x + 3)^{2}(d^{2}y/

^{2}) + 6(2x + 3)(

dy/dx) + 8y = 0, x > 0, denote S := {

x \in (0, \infty) : y(x) = 0

}. Then

A.S is an empty set.
B.S is a non-empty finite set.
C.S is a countably infinite set.✓ Correct
D.S is an uncountable set.

The worked solution and the per-option analysis are in Trap Analysis.

Part B2023 Juneshow ▾

Suppose x(t) is the solution of the initial value problem in

\mathbb{R}^{2}

: ẋ = Ax,

x(0) = x_{0}

, where A = [[5, 4], [1, 2]]. Which of the following statements is true?

A.x(t) is a bounded solution for some $x_{0} \ne 0$ .
B. $e^{-6t}$ |x(t)| → 0 as $t \to \infty$ , for all $x_{0} \ne 0$ .
C. $e^{-t}$ |x(t)| $\to \infty$ as $t \to \infty$ , for all $x_{0} \ne 0$ .
D. $e^{-10t}$ |x(t)| → 0 as $t \to \infty$ , for all $x_{0} \ne 0$ .✓ Correct

Solution

Eigenvalues of A are 6 and 1 (trace 7, det 6), both positive: every non-zero solution grows like $e^{6t}$ or $e^{t}$ .

So (1) fails, (2) fails along the 6-eigendirection, (3) fails along the 1-eigendirection, and (4) holds since 10 > 6.

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Part Cshow ▾

Let y″ + p(x)y′ + q(x)y = 0 with p, q continuous on an interval I, and let

y_{1}, y_{2}

be solutions with Wronskian W. Which of the following are true?

A.If $W(x_{0}) = 0$ for some $x_{0} \in I$ , then W ≡ 0 on I.✓ Correct
B. $y_{1}$ and $y_{2}$ are linearly independent on I if and only if W never vanishes on I.✓ Correct
C.W is constant on I.
D.Every solution on I is a linear combination of $y_{1}$ and $y_{2}$ .

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Consider the system x′ = Ax with A = [[0, 1], [−1, 0]]. Which of the following are true?

A.The eigenvalues of A are ±i.✓ Correct
B.Every solution is periodic with period $2\pi$ .✓ Correct
C.Trajectories are circles centred at the origin.✓ Correct
D.Some solutions spiral into the origin as $t \to \infty$ .

Solution

The characteristic polynomial is $\lambda^{2} + 1$ .

The flow is rotation: $(x, y)(t) = (x_{0}\cos t + y_{0}\sin t, -x_{0}\sin t + y_{0}\cos t)$ , so every orbit is a circle traversed with period $2\pi$ and ‖x(t)‖ is constant.

Purely imaginary eigenvalues give a centre; spiralling requires a nonzero real part, so (4) fails.

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Part Bshow ▾

The Wronskian of the functions eˣ and

e^{-x}

equals

A.0
B.−2✓ Correct
C.2
D. $e^{2x}$

The worked solution and the per-option analysis are in Trap Analysis.

Part Bshow ▾

Let y solve y′ + y =

e^{-x}

with y(0) = 0. Then y(1) equals

A.0
B.1/e✓ Correct
C.e
D.1 − 1/e

Solution

The integrating factor eˣ gives (eˣy)′ = 1, so eˣy = x and y = $xe^{-x}$ .

Hence y(1) = 1/e — the resonance case where the forcing solves the homogeneous equation, producing the factor x.

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3. Sturm–Liouville problems and Green's functions

Read this as a chapter →

Lec-38 Linear Second Order Equations

NPTEL · Ordinary Differential Equations and Applications

Lec-39 General Second Order Equations

NPTEL · Ordinary Differential Equations and Applications

Boundary value problems and eigenvalue structure.

Sturm–Liouville problems and Green's functions free

In plain words

The eigenvalues come out real, none repeats, and they run off to infinity. The eigenfunctions are orthogonal with respect to a weight, and together they form a complete basis.

The regular problem

$(p y')' + (q + \lambda{}w)y = 0$ on [a, b] with separated boundary conditions, p, w > 0.

Properties to quote:

Eigenvalues are real, simple, and form an increasing sequence $\lambda_{1} < \lambda_{2} < \cdots \to \infty$ .
Eigenfunctions for distinct eigenvalues are orthogonal with weight $w: \int y_{m} y_{n} w = 0$ .
The n-th eigenfunction has exactly n − 1 interior zeros (Sturm oscillation).
The eigenfunctions form a complete orthogonal basis of $L^{2}(w)$ .

The workhorse example

$y'' + \lambda{}y = 0$ on [0, L]:

Boundary conditions	Eigenvalues	Eigenfunctions
y(0) = y(L) = 0	$n^{2}\pi^{2}/L^{2}$	$\sin(n\pi{}x/L)$
y′(0) = y′(L) = 0	$n^{2}\pi^{2}/L^{2}, n \ge 0$	$\cos(n\pi{}x/L)$
y(0) = y′(L) = 0	$(2n-1)^{2}\pi^{2}/4L^{2}$	$\sin((2n-1)\pi{}x/2L)$
periodic	$4n^{2}\pi^{2}/L^{2}$	sin, cos (double eigenvalues)

For −y″ + qy $= \lambda{}y$ just shift: the smallest $\lambda$ for $-y'' + 3y = \lambda{}y$ with Dirichlet data on $[0, \pi]$ is 3 + 1 = 4.

Green's function

For Ly = f with homogeneous boundary conditions, $y(x) = \int{}G(x,t)f(t)$ dt where

$G(x, t) = y_{1}(t)y_{2}(x)/W$ for $t < x, y_{1}(x)y_{2}(t)/W$ for t > x,

with $y_{1}$ satisfying the left condition, $y_{2}$ the right, and W the (constant, by Abel) Wronskian times p.

Canonical case −y″ on [0,1] with y(0) = y(1) = 0: G(x,t) = t(1 − x) for t < x, x(1 − t) for t > x. Symmetric, continuous, with a unit jump in $\partial{}G/\partial{}x$ across x = t.

Green's function exists iff 0 is not an eigenvalue of the homogeneous problem.

Key takeaways

Real, simple, unbounded eigenvalues; orthogonal eigenfunctions with weight w.
Memorise the sin/cos eigenvalue table — most questions are a shift of it.
G is symmetric, continuous, with a jump of 1/p in its derivative at x = t.

Sturm–Liouville: what self-adjointness buysinteractive

This visual is part of Trap Analysis

Real eigenvalues, orthogonal eigenfunctions and simplicity — from one integration by parts.

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Exam focus: 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.

Practice

Part B2025 Juneshow ▾

Let

(\lambda_{n})_{n}\in\mathbb{N}

be the sequence of eigenvalues of the Sturm-Liouville problem (d/dx)(x dy/dx

) + (\lambda/x)y = 0, 1 < x <

e^{2\pi}

, y(1) = 0,

y(e^{2\pi})

= 0. Then

\sum_{n}

₌

_{1}^\infty 1/\lambda_{n}

is equal to

A. $\pi^{2}/12$
B. $2\pi^{2}/3$ ✓ Correct
C. $\pi^{2}/4$
D. $\pi^{2}/16$

The worked solution and the per-option analysis are in Trap Analysis.

Part C2025 Decembershow ▾

Consider the ordinary differential equation (ODE) y″ + r(x)y = 0, where

r(x) = x^{3}\sin(1/x)

for x≠0, and r(0)=0. Then which of the following statements are true?

A.There exists a non-trivial solution $\varphi$ of ODE, and $\alpha,\beta\in\mathbb{R}, \alpha<\beta$ such that $\varphi$ has infinitely many zeros in $[\alpha,\beta]$ .
B.There exists a non-trivial solution $\varphi$ of ODE, and a sequence { $x_{n}$ } such that $x_{n}\to\infty$ and $\varphi(x_{n})=0$ for all $n\in\mathbb{N}$ .✓ Correct
C.If $\varphi_{1},\varphi_{2}$ are two linearly independent solutions of ODE, then their Wronskian is a constant function on $\mathbb{R}$ .✓ Correct
D.There exists a non-trivial solution of ODE which vanishes at most at one point in $(0,\infty)$ .

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Decembershow ▾

Let

\lambda \in \mathbb{R}

, and

K : [0, 1] \times [0, 1] \to \mathbb{R}

be a function such that every solution of the boundary value problem

(d^{2}u/

^{2})(x) + \lambda{}u(x) = 0; (

du/dx)(0) = u(0), (du/dx)(1) = 0 satisfies the integral equation

u(x) + \lambda\int_{0}^{1}K(x, t)u(t)

dt = 0. Then

A.K(x, t) = (1 + x)(1 − t) for 0 ≤ x ≤ t ≤ 1, and (1 + t)(1 − x) for 0 ≤ t < x ≤ 1
B.K(x, t) = −1 − x for 0 ≤ x ≤ t ≤ 1, and −1 − t for 0 ≤ t < x ≤ 1✓ Correct
C. $K(x, t) = 1 - x^{2}$ for 0 ≤ x ≤ t ≤ 1, and $1 - t^{2}$ for 0 ≤ t < x ≤ 1
D.K(x, t) = (1 + x)(t − 1) for 0 ≤ x ≤ t ≤ 1, and (1 + t)(x − 1) for 0 ≤ t < x ≤ 1

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾

Consider the boundary value problem (BVP

) d^{2}y/

^{2} + 4(

dy/dx

) + (\lambda + 2)y = 0, y(0) = y(\pi) = 0

. Then which of the following statements are true?

A.The BVP has a non-zero solution if $\lambda = 2$
B.The BVP has a non-zero solution if $\lambda = 3$ ✓ Correct
C.The BVP has a non-zero solution y = y(x) satisfying $y(k\pi/2) = 0$ for every integer k if $\lambda = 6$ ✓ Correct
D.The BVP has a non-zero solution y = y(x) satisfying $y(k\pi/3) = 0$ for every integer k if $\lambda = 6$

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Juneshow ▾

Consider the boundary value problem (BVP) (e^(−5x)y′)′ + 6e^(−5x)y = −f(x), 0 < x < ln 2, with y(0) = 0, y(ln 2) = 0. If

G(x, \xi) = (e^(3x) +

Be^(2x))(Ce

^(2\xi) +

^(3\xi))

for

0 \le \xi \le x

, and

G(x, \xi) = (e^(3\xi) +

^(2\xi))(

Ce^(2x) + De^(3x)) for

x \le \xi \le \ln 2 (

Green's function) is such that

\int_{0}^(\ln 2) G(x, \xi)f(\xi)d\xi

is the solution of the BVP, then the values of B, C and D are

A.B = −2, C = −1, D = 1
B.B = −2, C = 1, D = −1✓ Correct
C.B = 2, C = 1, D = 1
D.B = 2, C = −1, D = −1

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Decembershow ▾

If u is a solution of the integral equation

u(x) = \lambda\int_{0}^{1}K(x, t)u(t)

dt, where K(x, t) := x(1 − t) for 0 ≤ x ≤ t ≤ 1 and t(1 − x) for 0 ≤ t ≤ x ≤ 1, then

A. $d^{2}u/$ dx $^{2} + \lambda{}u = 0, u(0) = 0 = u(1)$ ✓ Correct
B. $d^{2}u/$ dx $^{2} - \lambda{}u = 0, u(0) = 0 = u(1)$
C.du/dx $+ \lambda{}u = 0, u(0) = 0 = u(1)$
D.du/dx $- \lambda{}u = 0, u(0) = 0 = u(1)$

The worked solution and the per-option analysis are in Trap Analysis.

Part B2023 Decembershow ▾

The smallest real number

\lambda

for which the problem

-y'' + 3y = \lambda{}y, y(0) = 0, y(\pi) = 0

has a non-trivial solution is

A.3
B.2
C.1
D.4✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Consider

-y'' = \lambda{}y

[0, \pi]

with

y(0) = y(\pi) = 0

. Which of the following are true?

A.The eigenvalues are $\lambda_{n} = n^{2}, n = 1, 2, 3, \dots$ ✓ Correct
B.The eigenfunctions sin(nx) are pairwise orthogonal in $L^{2}(0, \pi)$ .✓ Correct
C. $\lambda = 0$ is an eigenvalue.
D.Every eigenvalue is simple.✓ Correct

Solution

The boundary conditions force y = sin(nx) with $\lambda = n^{2}, n \ge 1$ .

Orthogonality: $\int_{0}^\pi \sin($ mx)sin(nx) dx = 0 for m ≠ n.

For $\lambda = 0$ the general solution ax + b dies under both conditions, so 0 is not an eigenvalue.

Simplicity holds in every regular Sturm–Liouville problem with separated conditions: two independent eigenfunctions for one $\lambda$ would exhaust the solution space of the second-order ODE, contradicting the boundary conditions.

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Part Cshow ▾

For a regular Sturm–Liouville problem

-(p y')' + q y = \lambda w y

on [a, b] with p, w > 0 and separated boundary conditions, which of the following are true?

A.All eigenvalues are real.✓ Correct
B.Eigenfunctions belonging to distinct eigenvalues are orthogonal with respect to the weight w.✓ Correct
C.There are only finitely many eigenvalues.
D.The eigenvalues are bounded above.

The worked solution and the per-option analysis are in Trap Analysis.

4. Stability and phase portraits

Read this as a chapter →

Lec-25 2 by 2 systems and Phase Plane Analysis

NPTEL · Ordinary Differential Equations and Applications

The trace–determinant classification, drawn out in full.

Lec-30 Stability Equilibrium Points

NPTEL · Ordinary Differential Equations and Applications

Lec-34 Lyapunov Function

NPTEL · Ordinary Differential Equations and Applications

What to do when linearisation is inconclusive (the centre case).

Stability and phase portraits free

In plain words

Near an equilibrium, replace the system by its linear part. In two dimensions trace and determinant decide everything, except at a centre, where the terms you discarded decide instead.

Classification of ẋ = Ax in the plane

Let T = trace $A, D = \det A, \Delta = T^{2} - 4D$ .

Condition	Type	Stability
D < 0	saddle	unstable
$D > 0, \Delta > 0, T < 0$	stable node	asymptotically stable
$D > 0, \Delta > 0, T > 0$	unstable node	unstable
$D > 0, \Delta < 0, T < 0$	stable spiral	asymptotically stable
$D > 0, \Delta < 0, T > 0$	unstable spiral	unstable
D > 0, T = 0	centre	stable, not asymptotically
D = 0	degenerate (a line of equilibria)	—

Summary: asymptotically stable ⇔ T < 0 and D > 0 ⇔ both eigenvalues have negative real part.

Nonlinear systems

Linearise at an equilibrium: J = Df(x*).

Hartman–Grobman: if no eigenvalue of J has zero real part (a hyperbolic equilibrium), the nonlinear phase portrait is locally equivalent to the linear one.
The centre case is undecidable by linearisation: ẋ $= -y - x^{3}$ , ẏ $= x - y^{3}$ has a linear centre but is actually asymptotically stable, while flipping the cubic signs makes it unstable.

Lyapunov's direct method

Find V with V(x*) = 0, V > 0 nearby:

V̇ ≤ 0 ⇒ stable; V̇ < 0 ⇒ asymptotically stable; V̇ > 0 ⇒ unstable.

Works where linearisation fails. For a conservative system, the energy is the natural V.

Key takeaways

T < 0 and D > 0 ⇔ asymptotically stable; D < 0 ⇔ saddle.
Hyperbolic equilibria are settled by linearisation; centres are not.
Lyapunov functions handle the non-hyperbolic cases.

The trace–determinant planeinteractive

This visual is part of Trap Analysis

Drag a point across the plane and watch the portrait change: saddle, node, spiral, centre — a map instead of a list of rules.

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Exam focus: 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.

Practice

Part C2024 Decembershow ▾

For

b \in \mathbb{R}

, let y_b = y_b(x) be the unique solution of the initial value problem dy/dx

= y^{5} + y^{4} + y^{3} + y^{2} + y + 1, y(0) = b

defined on its maximal interval of existence I_b. Then which of the following statements are true?

A.There exists an $\alpha \in (0, \infty)$ such that for every $b \in \mathbb{R}$ with $b > \alpha$ , the solution y_b is bounded above on I_b
B.There exists an $\alpha \in (0, \infty)$ such that for every $b \in \mathbb{R}$ with $b > \alpha$ , the solution y_b is bounded below on I_b✓ Correct
C.There exists an $\alpha \in (-\infty, 0)$ such that for every $b \in \mathbb{R}$ with $b < \alpha$ , the solution y_b is bounded above on I_b✓ Correct
D.There exists an $\alpha \in (-\infty, 0)$ such that for every $b \in \mathbb{R}$ with $b < \alpha$ , the solution y_b is bounded below on I_b

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Juneshow ▾

Consider the initial value problem (IVP

) y'(x) = \sin(y(x))/(1 + y^{4}(x))

for

x \in \mathbb{R}

, with

y(0) = y_{0}

. Then which of the following statements are true?

A.There is a positive $y_{0}$ such that the solution of the IVP is unbounded
B.There is a negative $y_{0}$ such that the solution of the IVP is bounded✓ Correct
C.For every $y_{0} \in \mathbb{R}$ , every solution of the IVP is bounded✓ Correct
D.For every $y_{0} \in \mathbb{R}$ , there is a solution to the IVP for all $x \in \mathbb{R}$ ✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part Bshow ▾

For x′ = Ax with A a real 2 × 2 matrix having eigenvalues −1 ± 2i, the origin is

A.a stable spiral (asymptotically stable focus)✓ Correct
B.an unstable spiral
C.a centre
D.a saddle

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Consider x′ = Ax where A is a real 2 × 2 matrix with det A < 0. Which of the following are true?

A.A has one positive and one negative real eigenvalue.✓ Correct
B.The origin is a saddle point.✓ Correct
C.The origin is asymptotically stable.
D.Exactly two lines through the origin are invariant under the flow (generically, the eigendirections).✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part Bshow ▾

For the system ẋ = Ax with A = [[−2, 1], [1, −2]], the origin is

A.a saddle
B.an unstable node
C.an asymptotically stable node✓ Correct
D.a centre

Solution

Trace = −4 < 0, det = 3 > 0, discriminant 16 − 12 = 4 > 0: real negative eigenvalues (−1, −3).

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Part Cshow ▾

Consider ẋ = Ax in

\mathbb{R}^{2}

. Which of the following imply that the origin is asymptotically stable?

A.trace A < 0 and det A > 0✓ Correct
B.both eigenvalues have negative real part✓ Correct
C.det A > 0 alone
D.A is symmetric with negative diagonal entries

The worked solution and the per-option analysis are in Trap Analysis.