ODE, PDE & Applied Mathematics

Linear Integral Equations

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1. Fredholm and Volterra equations

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Lec-21 Calculus of Variations and Integral Equations

NPTEL · Calculus of Variations and Integral Equations

Integral equations begin here: Fredholm and Volterra, first and second kind.

Lec-22 Calculus of Variations and Integral Equations

NPTEL · Calculus of Variations and Integral Equations

Conversion between differential and integral equations.

Integral equations: Fredholm vs Volterra free

In plain words

Both have the unknown sitting inside an integral, and only the upper limit differs. A fixed limit brings eigenvalues and can fail. A moving limit always has exactly one solution.

The two types

Fredholm: $y(x) = f(x) + \lambda\int_{a}$ ᵇ K(x,t)y(t)dt — fixed limits.
Volterra: $y(x) = f(x) + \lambda\int_{a}$ ˣ K(x,t)y(t)dt — variable upper limit.

Second kind has y outside the integral; first kind does not (and is ill-posed).

Volterra

The Neumann series always converges: a Volterra equation of the second kind has a unique solution for every $\lambda$ , with no eigenvalues. Converting to an ODE is usually fastest: differentiate and use y(a) = f(a).

Example. y = eˣ $+ \int_{0}$ ˣ $((1+x^{2})/(1+t^{2}))y$ dt. Put $I(x) = \int_{0}$ ˣ $y/(1+t^{2})$ ; then I′ − I = eˣ $/(1+x^{2})$ , giving I = eˣ arctan x and y = eˣ $(1 + (1+x^{2})$ arctan x).

Fredholm with a separable (degenerate) kernel

$K(x,t) = \sum a_{i}(x)b_{i}(t)$ reduces the equation to a finite linear system. Eigenvalues are where that system is singular.

Example. K = xt on $[0,1]: y = f + \lambda{}x\int_{0}^{1} t y$ dt. Setting $c = \int$ ty gives $c(1 - \lambda/3) = \int$ tf, so $\lambda = 3$ is the eigenvalue.

Fredholm alternative

For $y - \lambda\int$ Ky = f with $\lambda$ an eigenvalue:

a solution exists iff f is orthogonal to every solution of the homogeneous adjoint equation;
then the solution is not unique (add any eigenfunction).

If $\lambda$ is not an eigenvalue, there is a unique solution for every f.

Example. $y - 3\int_{0}^{1}$ tx y dt = f is solvable iff $\int_{0}^{1} t f(t)$ dt = 0 — true for $f = x^{2} -$ ½ and f = 2 − 3x, false for eˣ and x − 1.

Relation to ODE

A boundary value problem converts to a Fredholm equation with the Green's function as kernel; an initial value problem converts to a Volterra equation.

Key takeaways

Volterra: always uniquely solvable. Fredholm: check for eigenvalues first.
Separable kernels reduce everything to linear algebra.
The alternative's solvability test is an orthogonality condition on f.

Volterra always solvable, Fredholm not — and whyinteractive

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The factorial in the iterated kernel that makes one type unconditionally well behaved.

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Exam focus: Volterra always has a unique solution; Fredholm depends on whether λ is an eigenvalue, and then the alternative decides solvability by an orthogonality condition.

Practice

Part C2025 Juneshow ▾

Let u(x) be the solution to the Volterra integral equation

u(x) = x^{2} + 4\int_{0}

(t - x)^{2}u(t)

dt. Then which of the following statements are true?

A.u(0) = 0✓ Correct
B. $u(2\pi/\sqrt{3}) =$ $(1/6)(e^{4\pi/\sqrt{3}}$ − $e^{-2\pi/\sqrt{3}})$ ✓ Correct
C. $u(\pi/(2\sqrt{3})) =$ $(1/6)(e^{\pi/\sqrt{3}}$ $- \sqrt{3}$ $e^{-\pi/(2\sqrt{3})})$ ✓ Correct
D. $u(\pi/(2\sqrt{3})) =$ $(1/6)(e^{\pi/\sqrt{3}}$ $+ \sqrt{3}$ $e^{-\pi/\sqrt{3}})$

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

Part C2025 Decembershow ▾

Let y(x) be the solution of the integral equation (IE) y(x) = eˣ

+ e + 1 + (1/3)\int_{0}^{1}y(t)

dt, and R(x,t,1/3) be the resolvent kernel associated to IE. Then which of the following statements are true?

A.R(0,1,1/3) = 3/2✓ Correct
B.R(1/2,1/2,1/3) = 2/3
C.y(1) = 3e+1✓ Correct
D.y(1) = e+3

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

Part B2025 Decembershow ▾

Suppose u(x) is the solution of the integral equation

u(x) = 3 + \int_{0}

ˣ (x−t)u(t) dt. Then which of the following statements is true?

A. $u(\pi) = 2e^\pi$ .
B. $u'(\pi) = e^\pi$ .
C. $u(\pi) + u'(\pi) = 3e^\pi$ .✓ Correct
D. $u(\pi) - u'(\pi) = e^\pi$ .

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

Part C2025 Juneshow ▾

Let f and K be such that the solution of the initial value problem y″ − 3y′ + 2y = 4sin(x), y(0) = 1, y′(0) = −2 satisfies the Volterra integral equation

y(x) = f(x) + \int_{0}

ˣ K(x, t)y(t) dt. Then which of the following statements are true?

A. $f'(\pi) = 3$ ✓ Correct
B. $f(\pi) + f'(\pi) = 4 - \pi$ ✓ Correct
C. $f(\pi) + f'(\pi) = 2 - \pi$
D.f(0) + f′(0) = −4✓ Correct

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

Part C2025 Decembershow ▾

Let

\lambda\in\mathbb{R}

be such that the integral equation

y(x) = \lambda\int

₋

_{1}^{1}(5

^{3}+4x^{2}t+3

xt)y(t)dt admits a non-trivial solution y(x) such that y(1)=5/2. Then which of the following statements are true?

A.y(0)+y′(0) = 3/2✓ Correct
B.y(1/2)+y′(1/2) = 7/2✓ Correct
C.y(−1)+y′(−1) = −1✓ Correct
D.y(1/3)+y′(1/3) = 14/9

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

Part C2024 Decembershow ▾

The integral equation

u(x) = f(x) + (2/\pi)\int_{0}^\pi \sin(x - t)u(t)

dt has a unique solution if

A.f(x) = cos x✓ Correct
B.f(x) = cos 5x✓ Correct
C.f(x) = sin x✓ Correct
D.f(x) = sin 5x✓ Correct

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

Part C2024 Decembershow ▾

If u is the solution of the Volterra integral equation

u(x) = 3 + \sin x + \int_{0}

ˣ [(3 + sin x)/(3 + sin t)]u(t)dt, then

A. $u(\pi/2) = 4e^(\pi/2)$ ✓ Correct
B. $u(\pi) = 3e^\pi$ ✓ Correct
C. $u(-\pi) = 4e^(-\pi)$
D. $u(-\pi/2) = 4e^\pi$

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

Part C2024 Decembershow ▾

The integral equation

u(x) = f(x) + (2/\pi)\int_{0}^\pi \cos(x + t)u(t)

dt has infinitely many solutions if

A.f(x) = cos x
B.f(x) = cos 5x✓ Correct
C.f(x) = sin x✓ Correct
D.f(x) = sin 5x✓ Correct

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

Part B2024 Juneshow ▾

Let u be the solution of the Volterra integral equation

\int_{0}

ᵗ [½

+ \sin(t - \tau)] u(\tau) d\tau = \sin t

. Then the value of u(1) is

A.0✓ Correct
B.1
C.2
D.2e⁻ $^{1}$

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

Part C2024 Juneshow ▾

For

c \in \mathbb{R}

, consider the following Fredholm integral equation y(x) = 1 + x + cx

^{2} + 2\int_{0}^{1} (1 - 3

xt)y(t)dt. Then the values of c for which the integral equation admits a solution are

A.−8✓ Correct
B.−6
C.2
D.6

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

Part B2023 Juneshow ▾

For the unknown

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

, consider the boundary value problem y″(x) + 2y(x) = 0 for x ∈ (0,1), y(0) = y(1) = 0. It is given that it corresponds to the integral equation

y(x) = 2\int_{0}^{1} K(x, t) y(t)

dt. Which of the following is the kernel K(x, t)?

A.K(x,t) = t(1 − x) for t < x; x(1 − t) for t > x✓ Correct
B. $K(x,t) = t^{2}(1 - x)$ for $t < x; x^{2}(1 - t)$ for t > x
C. $K(x,t) = \sqrt{t}(1 - x)$ for $t < x; \sqrt{x}(1 - t)$ for t > x
D. $K(x,t) = \sqrt{t^{3}}(1 - x)$ for $t < x; \sqrt{x^{3}}(1 - t)$ for t > x

Solution

The Green's function of −y″ with Dirichlet conditions on [0,1] is G(x,t) = t(1 − x) for t < x and x(1 − t) for t > x; then $y = 2\int{}G y$ .

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Part B2023 Decembershow ▾

The value of

\lambda

for which the integral equation

y(x) = \lambda\int_{0}^{1} x^{2}

e^{x+t}

y(t) dt has a non-zero solution is

A. $4/(1 + e^{2})$
B. $2/(1 + e^{2})$
C. $4/(e^{2} - 1)$ ✓ Correct
D. $2/(e^{2} - 1)$

Solution

Separable kernel: $y = c x^{2}e$ ˣ with $c = \lambda{}c \int_{0}^{1} t^{2}$ $e^{2t}$ dt $= \lambda{}c (e^{2} - 1)/4$ , so $\lambda = 4/(e^{2} - 1)$ .

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Part C2023 Decembershow ▾

Let y be the solution to the Volterra integral equation y(x) = eˣ

+ \int_{0}

((1 + x^{2})/(1 + t^{2})) y(t)

dt. Which of the following statements are true?

A. $y(1) = (1 + \pi/4)e$
B. $y(1) = (1 + \pi/2)e$ ✓ Correct
C. $y(\sqrt{3}) = (1 +$ $3\pi/4)e^{\sqrt{3}}$
D. $y(\sqrt{3}) = (1 +$ $4\pi/3)e^{\sqrt{3}}$ ✓ Correct

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

Part C2023 Decembershow ▾

Consider the Fredholm integral equation

y(x) - 3\int_{0}^{1} t x y(t)

dt = f(x), with f continuous on [0, 1]. Which of the following choices for f(x) have the property that the equation admits at least one solution?

A. $f(x) = x^{2} - 1/2$ ✓ Correct
B.f(x) = eˣ
C.f(x) = 2 − 3x✓ Correct
D.f(x) = x − 1

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

Part Cshow ▾

Consider the Volterra equation

u(x) = f(x) + \lambda\int_{0}

ˣ K(x, t)u(t) dt with K and f continuous on a bounded interval. Which of the following are true?

A.A unique continuous solution exists for every value of $\lambda$ .✓ Correct
B.The Neumann (successive-approximation) series converges for every $\lambda$ .✓ Correct
C.The analogous Fredholm equation (fixed integration limits) is also uniquely solvable for every $\lambda$ .
D.If K ≡ 0, the solution is u = f.✓ Correct

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

Part Bshow ▾

The solution of the Volterra equation

u(x) = x + \int_{0}

ˣ u(t) dt is

A.u(x) = eˣ − 1✓ Correct
B.u(x) = eˣ
C. $u(x) = x + x^{2}/2$
D.u(x) = sinh x

Solution

Differentiate: u′ = 1 + u with u(0) = 0.

Solving gives u = Ceˣ − 1 and the initial condition makes C = 1: u = eˣ − 1.

Check in the original equation: $x + \int_{0}$ ˣ(eᵗ − 1)dt = x + eˣ − 1 − x = eˣ − 1 ✓.

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2. Separable kernels and resolvent kernels

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Lec-25 Calculus of Variations and Integral Equations

NPTEL · Calculus of Variations and Integral Equations

Separable (degenerate) kernels reduce the problem to linear algebra.

Separable kernels and resolvent kernels free

In plain words

If the kernel splits into a short sum of products, the whole problem collapses into a small matrix equation, and the eigenvalues are simply where a determinant vanishes.

Separable (degenerate) kernels

$K(x, t) = \sum_{i=1}^{n}$ $a_{i}(x)b_{i}(t)$ . Substituting into $y = f + \lambda\int$ Ky and setting $c_{i} = \int{}b_{i} y$ turns the problem into

$(I - \lambda{}A)c = d$ , where $A_{ij} = \int b_{i} a_{j}$ and $d_{i} = \int b_{i} f$ .

$- \lambda$ is an eigenvalue $\Leftrightarrow \det(I - \lambda{}A) = 0, i.e. 1/\lambda$ is an eigenvalue of A.

At most n eigenvalues for a rank-n kernel.
Then $y = f + \lambda \sum c_{i} a_{i}(x)$ .

Example. $K = x^{2}e^{x+t}$ on $[0,1]: a(x) = x^{2}e$ ˣ, b(t) = eᵗ. $A = \int_{0}^{1} t^{2}e^{2t}$ dt $= (e^{2} - 1)/4$ , so the eigenvalue is $\lambda = 4/(e^{2} - 1)$ .

Neumann series and the resolvent

Iterating $y = f + \lambda\int$ Ky gives $y = f + \lambda\int{}R(x,t;\lambda)f(t)$ dt with

$R = K_{1} + \lambda{}K_{2} + \lambda^{2}K_{3} + \cdots$ , where $K_{n}$ ₊ $_{1}(x,t) = \int{}K(x,s)K_{n}(s,t)$ ds are the iterated kernels.

Converges for | $\lambda$ | < 1/‖K‖ (Fredholm), and for all $\lambda$ in the Volterra case.
For Volterra with $K = 1: K_{2} = (x - t), K_{3} = (x - t)^{2}/2$ !, so R = $e^{\lambda(x-t)}$ .
For separable K, the resolvent is a rational function of $\lambda$ whose poles are the eigenvalues.

Symmetric kernels

K(x,t) = K(t,x) real ⇒ eigenvalues are real, eigenfunctions for distinct eigenvalues are orthogonal, and (Hilbert–Schmidt $) K(x,t) = \sum \varphi_{n}(x)\varphi_{n}(t)/\lambda_{n}$ . This is the integral-operator analogue of the spectral theorem.

Key takeaways

Rank-n separable kernel ⇒ an n × n matrix problem with at most n eigenvalues.
Resolvent = sum of iterated kernels; always convergent for Volterra.
Symmetric kernels behave like symmetric matrices.

Separable kernels turn an integral equation into linear algebrainteractive

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A degenerate kernel reduces the whole problem to a finite system — worked.

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Exam focus: Separable kernels turn integral equations into matrix problems. The resolvent kernel sums the Neumann series and gives the solution in closed form.

Practice

Part B2025 Juneshow ▾

If y(x) is the solution of the integral equation

y(x) = x^{2} + 2\int_{0}^{1}

xt y(t) dt, then which of the following statements is true?

A.y(0) + y(1) = 1/2
B.y(−1) + y(1) = 1
C.y′(0) + y′(1) = 3/2
D.y′(−1) + y′(1) = 3✓ Correct

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

Part C2024 Juneshow ▾

For

\lambda \in \mathbb{R}

such that |

\lambda

| < 5/32, let

R(x, t, \lambda)

and u denote the resolvent kernel and the solution, respectively, of the Fredholm integral equation

u(x) = x + (\lambda/2)\int

₋

_{2}^{2} (

+ x^{2}t^{2})u(t)

dt. Then which of the following statements are true?

A. $R(x, t, \lambda) = 3$ xt $/(3 - 8\lambda) - 5x^{2}t^{2}/(5 - 32\lambda)$
B. $R(x, t, \lambda) = 3$ xt $/(3 - 8\lambda) + 5x^{2}t^{2}/(5 - 32\lambda)$ ✓ Correct
C. $u(1) = -5/(5 - 32\lambda)$
D. $u(1) = 3/(3 - 8\lambda)$ ✓ Correct

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

Part C2024 Decembershow ▾

Let R(x, t) and u(x) denote the resolvent kernel and the solution, respectively, of the Volterra integral equation u(x) = eˣ

+ 2\int_(\ln 6)^x e^(2(t-x))u(t)

dt. Then which of the following statements are true?

A.R(x, t) = 1✓ Correct
B.R(x, t) = e^(t−x)
C.u(ln 8) = 10
D.u(ln 7) = 9✓ Correct

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

Part Cshow ▾

Consider

y(x) = f(x) + \lambda\int_{0}^{1}

xt·y(t) dt, a Fredholm equation of the second kind with separable kernel K(x, t) = xt. Which of the following are true?

A.The separable kernel reduces the equation to an algebraic equation for a single unknown constant.✓ Correct
B.The equation has a unique solution for every $\lambda \ne 3$ .✓ Correct
C. $\lambda = 3$ is an eigenvalue of the kernel: the homogeneous equation $y(x) = \lambda\int_{0}^{1}$ xt·y(t) dt has a nontrivial solution exactly when $\lambda = 3$ .✓ Correct
D.The equation has a unique solution for every value of $\lambda$ , including $\lambda = 3$ .

Solution

Writing $y(x) = f(x) + \lambda$ xC with $C = \int_{0}^{1}t\cdot{}y(t)$ dt reduces the equation to $C(1 - \lambda/3) = \int_{0}^{1}$ tf(t)dt, giving $C = 3\int$ tf(t)dt $/(3 - \lambda)$ — solvable and unique whenever $\lambda \ne 3$ — (1)–(2). At $\lambda = 3$ the coefficient $(3 - \lambda)$ vanishes, exactly the eigenvalue case in (3), so (4) is false: existence and uniqueness break down at $\lambda = 3$ , per the Fredholm alternative.

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Part Cshow ▾

Which of the following are true for the Volterra equation of the second kind

y(x) = f(x) + \lambda\int_{0}

ˣ K(x,t)y(t)dt with K continuous?

A.It has a unique solution for every $\lambda$ ✓ Correct
B.It may have eigenvalues
C.The Neumann series converges for all $\lambda$ ✓ Correct
D.It can be converted to an initial value problem when K is a polynomial in x − t✓ Correct

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

Part Bshow ▾

The eigenvalue

\lambda

for which

y(x) = \lambda\int_{0}^{1} x t y(t)

dt has a non-trivial solution is

A.1
B.2
C.3✓ Correct
D.4

Solution

Put $c = \int_{0}^{1} t y$ .

Then $y = \lambda$ cx and $c = \lambda{}c\int_{0}^{1}t^{2}$ dt $= \lambda{}c/3$ , so $\lambda = 3$ .

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Part Cshow ▾

Let

K(x, t) = \sum_{i} g_{i}(x)h_{i}(t)

be a separable (degenerate) kernel with finitely many terms. Which of the following are true?

A.The Fredholm equation with a separable kernel always reduces to a finite linear system for constants, rather than the original integral equation.✓ Correct
B.The resolvent kernel of a separable kernel is itself separable.✓ Correct
C.K(x, t) = sin(x + t) is not separable (degenerate).
D.K(x, t) = eˣᵗ is not separable — it cannot be written as a finite sum of products g(x)h(t).✓ Correct

Solution

(1)–(2) are the standard method for degenerate kernels and a fact about the resolvent it produces. (4) eˣᵗ $= \sum($ xt $)^{n}/n$ ! is a genuine infinite series with no finite reduction, so it is not degenerate. (3) is false: sin(x + t) = sin x cos t + cos x sin t IS separable, with rank 2 — a kernel built from a sum of arguments is not automatically non-separable.

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