Linear Integral Equations Questions and Answers

Q: Iterated kernel \(K_{m}(t, s)\) of the integral equation\[ u(t) = 1 + \lambda\int_{0}^{t} e^{t - s}u(s)\,ds \]is :

The iterated kernel is \(K_{m}(t, s) = \dfrac{(t - s)^{m - 1}}{(m - 1)!}e^{t - s}\).Here \(K_{1}(t, s) = e^{t - s}\), and \(K_{m}(t, s) = \int_{s}^{t} K(t, r)K_{m-1}(r, s)\,dr\). The exponentials combine to \(e^{t - s}\) inside every integral:\[ \begin{aligned} K_{2} &= \int_{s}^{t} e^{t - r}e^{r - s}dr = (t - s)e^{t - s} \\ K_{3} &= e^{t - s}\int_{s}^{t}(r - s)dr = \tfrac{(t - s)^{2}}{2}e^{t - s} \end{aligned} \]By induction \(K_{m} = \tfrac{(t - s)^{m-1}}{(m - 1)!}e^{t - s}\). Option A drops the exponential, option B drops the power, and option D is only \(K_{2}\).

Q: Solution of the integral equation\[ u(t) = a\sin t - 2\int_{0}^{t} \cos(t - s)u(s)\,ds \]is :

The solution is \(u(t) = ate^{-t}\).The integral is a convolution, so take Laplace transforms, with \(U = L\{u\}\):\[ U = \frac{a}{s^{2} + 1} - \frac{2s}{s^{2} + 1}U \]\[ \begin{aligned} U\frac{s^{2} + 2s + 1}{s^{2} + 1} &= \frac{a}{s^{2} + 1} \\ U &= \frac{a}{(s + 1)^{2}} \end{aligned} \]Inverting, \(u(t) = ate^{-t}\). Check at \(t = 0\): both sides vanish. The other options do not satisfy the equation: \(a\sin t\) ignores the integral term, and \(at^{2}e^{-t}\) would need \(2a/(s + 1)^{3}\).

Q: For the Fredholm integral equation\(u(t) = \lambda\displaystyle\int_{a}^{b} k(t, s)\, u(s)\, ds\)with separable kernel, Fredholm determinant \(D(\lambda) \ne 0\). Then the integral equation has :

The equation has only the zero solution.For a homogeneous Fredholm equation with separable kernel \(k(t, s) = \sum a_{i}(t)b_{i}(s)\), any solution has the form \(u = \lambda\sum c_{i}a_{i}(t)\), and the constants satisfy a homogeneous linear system \((I - \lambda A)c = 0\). Its determinant is the Fredholm determinant \(D(\lambda)\).If \(D(\lambda) \ne 0\), the system has only \(c = 0\), so \(u \equiv 0\) is the unique solution. Non-zero solutions, infinitely many of them, exist only when \(D(\lambda) = 0\), that is, when \(\lambda\) is an eigenvalue. A homogeneous equation always has the zero solution, so 'no solution' is impossible.

Q: Eigen value and eigen function of the homogeneous integral equation\(u(t) = \lambda\displaystyle\int_{0}^{1} e^{t+s}\, u(s)ds\)is :

The eigenvalue and eigenfunction are \(\lambda = \dfrac{2}{e^{2} - 1}\), \(u(t) = e^{t}\).The kernel \(e^{t+s} = e^{t}e^{s}\) is separable, so \(u(t) = \lambda e^{t}C\) with \(C = \int_{0}^{1} e^{s}u(s)\,ds\). Substituting \(u = \lambda Ce^{t}\):\[ \begin{aligned} C &= \lambda C\int_{0}^{1} e^{2s}\,ds = \lambda C\,\frac{e^{2} - 1}{2} \\ \lambda &= \frac{2}{e^{2} - 1} \end{aligned} \]for a non-zero C. The eigenfunction is a multiple of \(e^{t}\), not \(e^{t^{2}}\), and the reciprocal value \((e^{2} - 1)/2\) is the integral itself, not the eigenvalue.

Q: The solution of the Volterra integral equation of first kind\(f(t) = \displaystyle\int_{0}^{t} e^{t-s}\, u(s)\, ds,\)where \(f(t)\) is known function, is :

The solution is \(u(t) = f'(t) - f(t)\).Write \(f(t) = e^{t}\int_{0}^{t} e^{-s}u(s)\,ds\) and differentiate:\[ \begin{aligned} f'(t) &= e^{t}\int_{0}^{t} e^{-s}u(s)\,ds + e^{t}e^{-t}u(t) \\ &= f(t) + u(t) \end{aligned} \]so \(u = f' - f\). Check with Laplace transforms: \(F(s) = U(s)/(s - 1)\), so \(U = (s - 1)F = sF - F\), which is the transform of \(f' - f\) when \(f(0) = 0\) (as the equation forces). The options with a factor \(t\) do not come from this differentiation.

Q: The initial value problem\(y'' + y = 0\), \(y(0) = y'(0) = 0\)is equivalent to the integral equation :

The equivalent integral equation is \(y(t) = \int_{0}^{t} (s - t)\, y(s)\, ds\).Write \(y'' = -y\). Integrating twice from 0 and using \(y(0) = y'(0) = 0\):\[ \begin{aligned} y(t) &= \int_{0}^{t} (t - s)\, y''(s)\, ds \\ &= -\int_{0}^{t} (t - s)\, y(s)\, ds = \int_{0}^{t} (s - t)\, y(s)\, ds \end{aligned} \]The first step is the standard formula for a repeated integral. Option A has the sign wrong, which would correspond to \(y'' = y\), and options B and D carry an extra factor \(t\) that the double integration does not produce. (Of course the only solution here is \(y \equiv 0\), but the question asks for the equivalent equation.)

Q: If K(\(t\), \(s\)) is symmetric kernel of the Fredholm integral equation. Define the operator K by :\[ K\phi = \int_{a}^{b} K(t, s)\, \phi(s)\, ds \]and consider the following statements :(I) K is self-adjoint operator.(II) Inner product \(\langle K\phi, \phi \rangle\) is always real.Then :

Both (I) and (II) are true.For a real symmetric kernel, \(K(t, s) = K(s, t)\), so\[ \begin{aligned} \langle K\phi, \psi \rangle &= \int\!\!\int K(t, s)\phi(s)\psi(t)\, ds\, dt \\ &= \langle \phi, K\psi \rangle \end{aligned} \]which makes K self-adjoint (I). For a self-adjoint operator, \(\langle K\phi, \phi \rangle = \langle \phi, K\phi \rangle\), which is \(\overline{\langle K\phi, \phi \rangle}\), so the inner product is real (II). This is why symmetric kernels have real characteristic values and orthogonal eigenfunctions.Since both statements hold, the other options are wrong.

Q: Solution of the Fredholm integral equation\[ \int_{0}^{\infty} \cos \lambda t\, u(t)\, dt = e^{-\lambda} \]is :

The solution is \(u(t) = \dfrac{2}{\pi}\dfrac{1}{1 + t^{2}}\).The equation says the Fourier cosine transform of \(u\) is \(e^{-\lambda}\). Inverting:\[ \begin{aligned} u(t) &= \frac{2}{\pi}\int_{0}^{\infty} e^{-\lambda}\cos\lambda t\, d\lambda \\ &= \frac{2}{\pi}\cdot\frac{1}{1 + t^{2}} \end{aligned} \]using \(\int_{0}^{\infty} e^{-\lambda}\cos\lambda t\, d\lambda = 1/(1 + t^{2})\). Check: \(\int_{0}^{\infty} \frac{\cos\lambda t}{1 + t^{2}}\, dt\) equals \(\frac{\pi}{2}e^{-\lambda}\), so the factor \(2/\pi\) is right. The options with \(1 - t^{2}\) or \((1 \pm t)^{2}\) do not come from this integral.

Q: The kernel of the linear integral equation\[ x(t) = \lambda \int_{0}^{1} (t^{2} - s^{2})\, x(s)\, ds \]is :

The kernel \(t^{2} - s^{2}\) is separable.A separable (degenerate) kernel is a finite sum of products \(a_{i}(t) b_{i}(s)\). Here \(t^{2} - s^{2} = t^{2} \cdot 1 + 1 \cdot (-s^{2})\), a sum of two such products, so the integral equation reduces to a 2 × 2 linear system.It is not symmetric: swapping \(t\) and \(s\) gives \(s^{2} - t^{2}\), the negative, so the kernel is skew-symmetric. It is not a convolution kernel, which would be a function of \(t - s\) alone, and it is bounded and continuous on the square, so it is not singular.

11.

Iterated kernel \(K_{m}(t, s)\) of the integral equation

\[ u(t) = 1 + \lambda\int_{0}^{t} e^{t - s}u(s)\,ds \]

is :

A.\(\dfrac{(t - s)^{m - 1}}{(m - 1)!}\)
B.\(\dfrac{e^{t - s}}{(m - 1)!}\)
C.\(\dfrac{(t - s)^{m - 1}}{(m - 1)!}\,e^{t - s}\)
D.\((t - s)e^{t - s}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The iterated kernel is \(K_{m}(t, s) = \dfrac{(t - s)^{m - 1}}{(m - 1)!}e^{t - s}\).

Here \(K_{1}(t, s) = e^{t - s}\), and \(K_{m}(t, s) = \int_{s}^{t} K(t, r)K_{m-1}(r, s)\,dr\). The exponentials combine to \(e^{t - s}\) inside every integral:

\[ \begin{aligned} K_{2} &= \int_{s}^{t} e^{t - r}e^{r - s}dr = (t - s)e^{t - s} \\ K_{3} &= e^{t - s}\int_{s}^{t}(r - s)dr = \tfrac{(t - s)^{2}}{2}e^{t - s} \end{aligned} \]

By induction \(K_{m} = \tfrac{(t - s)^{m-1}}{(m - 1)!}e^{t - s}\). Option A drops the exponential, option B drops the power, and option D is only \(K_{2}\).

Discuss in Forum

12.

Solution of the integral equation

\[ u(t) = a\sin t - 2\int_{0}^{t} \cos(t - s)u(s)\,ds \]

is :

A.\(u(t) = a\sin t\)
B.\(u(t) = a\cos t\)
C.\(u(t) = ate^{-t}\)
D.\(u(t) = at^{2}e^{-t}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The solution is \(u(t) = ate^{-t}\).

The integral is a convolution, so take Laplace transforms, with \(U = L\{u\}\):

\[ U = \frac{a}{s^{2} + 1} - \frac{2s}{s^{2} + 1}U \]

\[ \begin{aligned} U\frac{s^{2} + 2s + 1}{s^{2} + 1} &= \frac{a}{s^{2} + 1} \\ U &= \frac{a}{(s + 1)^{2}} \end{aligned} \]

Inverting, \(u(t) = ate^{-t}\). Check at \(t = 0\): both sides vanish. The other options do not satisfy the equation: \(a\sin t\) ignores the integral term, and \(at^{2}e^{-t}\) would need \(2a/(s + 1)^{3}\).

Discuss in Forum

13.

For the Fredholm integral equation

\(u(t) = \lambda\displaystyle\int_{a}^{b} k(t, s)\, u(s)\, ds\)

with separable kernel, Fredholm determinant \(D(\lambda) \ne 0\). Then the integral equation has :

A.unique zero solution
B.unique non-zero solution
C.infinitely many solutions
D.no solution

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The equation has only the zero solution.

For a homogeneous Fredholm equation with separable kernel \(k(t, s) = \sum a_{i}(t)b_{i}(s)\), any solution has the form \(u = \lambda\sum c_{i}a_{i}(t)\), and the constants satisfy a homogeneous linear system \((I - \lambda A)c = 0\). Its determinant is the Fredholm determinant \(D(\lambda)\).

If \(D(\lambda) \ne 0\), the system has only \(c = 0\), so \(u \equiv 0\) is the unique solution. Non-zero solutions, infinitely many of them, exist only when \(D(\lambda) = 0\), that is, when \(\lambda\) is an eigenvalue. A homogeneous equation always has the zero solution, so 'no solution' is impossible.

Discuss in Forum

14.

Eigen value and eigen function of the homogeneous integral equation

\(u(t) = \lambda\displaystyle\int_{0}^{1} e^{t+s}\, u(s)ds\)

is :

A.\(\lambda = \dfrac{2}{e^{2} - 1}, u(t) = e^{t^{2}}\)
B.\(\lambda = \dfrac{2}{e^{2} - 1}, u(t) = e^{t}\)
C.\(\lambda = \dfrac{e^{2} - 1}{2}, u(t) = e^{t}\)
D.\(\lambda = \dfrac{e^{2} - 1}{2}, u(t) = e^{t^{2}}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The eigenvalue and eigenfunction are \(\lambda = \dfrac{2}{e^{2} - 1}\), \(u(t) = e^{t}\).

The kernel \(e^{t+s} = e^{t}e^{s}\) is separable, so \(u(t) = \lambda e^{t}C\) with \(C = \int_{0}^{1} e^{s}u(s)\,ds\). Substituting \(u = \lambda Ce^{t}\):

\[ \begin{aligned} C &= \lambda C\int_{0}^{1} e^{2s}\,ds = \lambda C\,\frac{e^{2} - 1}{2} \\ \lambda &= \frac{2}{e^{2} - 1} \end{aligned} \]

for a non-zero C. The eigenfunction is a multiple of \(e^{t}\), not \(e^{t^{2}}\), and the reciprocal value \((e^{2} - 1)/2\) is the integral itself, not the eigenvalue.

Discuss in Forum

15.

The solution of the Volterra integral equation of first kind

\(f(t) = \displaystyle\int_{0}^{t} e^{t-s}\, u(s)\, ds,\)

where \(f(t)\) is known function, is :

A.\(tf(t) - f'(t)\)
B.\(f'(t) - f(t)\)
C.\(f'(t) - tf(t)\)
D.\(t\left(f'(t) - f(t)\right)\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The solution is \(u(t) = f'(t) - f(t)\).

Write \(f(t) = e^{t}\int_{0}^{t} e^{-s}u(s)\,ds\) and differentiate:

\[ \begin{aligned} f'(t) &= e^{t}\int_{0}^{t} e^{-s}u(s)\,ds + e^{t}e^{-t}u(t) \\ &= f(t) + u(t) \end{aligned} \]

so \(u = f' - f\). Check with Laplace transforms: \(F(s) = U(s)/(s - 1)\), so \(U = (s - 1)F = sF - F\), which is the transform of \(f' - f\) when \(f(0) = 0\) (as the equation forces). The options with a factor \(t\) do not come from this differentiation.

Discuss in Forum

16.

The initial value problem

\(y'' + y = 0\), \(y(0) = y'(0) = 0\)

is equivalent to the integral equation :

A.\(y(t) = \int_{0}^{t} (t - s)\, y(s)\, ds\)
B.\(y(t) = \int_{0}^{t} t(t - s)\, y(s)\, ds\)
C.\(y(t) = \int_{0}^{t} (s - t)\, y(s)\, ds\)
D.\(y(t) = \int_{0}^{t} t(s - t)\, y(s)\, ds\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The equivalent integral equation is \(y(t) = \int_{0}^{t} (s - t)\, y(s)\, ds\).

Write \(y'' = -y\). Integrating twice from 0 and using \(y(0) = y'(0) = 0\):

\[ \begin{aligned} y(t) &= \int_{0}^{t} (t - s)\, y''(s)\, ds \\ &= -\int_{0}^{t} (t - s)\, y(s)\, ds = \int_{0}^{t} (s - t)\, y(s)\, ds \end{aligned} \]

The first step is the standard formula for a repeated integral. Option A has the sign wrong, which would correspond to \(y'' = y\), and options B and D carry an extra factor \(t\) that the double integration does not produce. (Of course the only solution here is \(y \equiv 0\), but the question asks for the equivalent equation.)

Discuss in Forum

17.

If K(\(t\), \(s\)) is symmetric kernel of the Fredholm integral equation. Define the operator K by :

\[ K\phi = \int_{a}^{b} K(t, s)\, \phi(s)\, ds \]

and consider the following statements :

(I) K is self-adjoint operator.
(II) Inner product \(\langle K\phi, \phi \rangle\) is always real.

Then :

A.Only (I) is true
B.Only (II) is true
C.Both (I) and (II) are true
D.Both (I) and (II) are false

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

Both (I) and (II) are true.

For a real symmetric kernel, \(K(t, s) = K(s, t)\), so

\[ \begin{aligned} \langle K\phi, \psi \rangle &= \int\!\!\int K(t, s)\phi(s)\psi(t)\, ds\, dt \\ &= \langle \phi, K\psi \rangle \end{aligned} \]

which makes K self-adjoint (I). For a self-adjoint operator, \(\langle K\phi, \phi \rangle = \langle \phi, K\phi \rangle\), which is \(\overline{\langle K\phi, \phi \rangle}\), so the inner product is real (II). This is why symmetric kernels have real characteristic values and orthogonal eigenfunctions.

Since both statements hold, the other options are wrong.

Discuss in Forum

18.

Solution of the Fredholm integral equation

\[ \int_{0}^{\infty} \cos \lambda t\, u(t)\, dt = e^{-\lambda} \]

is :

A.\(\dfrac{2}{\pi}\dfrac{1}{(1 + t^{2})}\)
B.\(\dfrac{2}{\pi}\dfrac{1}{(1 - t^{2})}\)
C.\(\dfrac{2}{\pi}\dfrac{1}{(1 + t)^{2}}\)
D.\(\dfrac{2}{\pi}\dfrac{1}{(1 - t)^{2}}\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

The solution is \(u(t) = \dfrac{2}{\pi}\dfrac{1}{1 + t^{2}}\).

The equation says the Fourier cosine transform of \(u\) is \(e^{-\lambda}\). Inverting:

\[ \begin{aligned} u(t) &= \frac{2}{\pi}\int_{0}^{\infty} e^{-\lambda}\cos\lambda t\, d\lambda \\ &= \frac{2}{\pi}\cdot\frac{1}{1 + t^{2}} \end{aligned} \]

using \(\int_{0}^{\infty} e^{-\lambda}\cos\lambda t\, d\lambda = 1/(1 + t^{2})\). Check: \(\int_{0}^{\infty} \frac{\cos\lambda t}{1 + t^{2}}\, dt\) equals \(\frac{\pi}{2}e^{-\lambda}\), so the factor \(2/\pi\) is right. The options with \(1 - t^{2}\) or \((1 \pm t)^{2}\) do not come from this integral.

Discuss in Forum

19.

The solution of the homogeneous integral equation

\[ x(t) = \lambda \int_{0}^{1} t s\, x(s)\, ds, \]

for \(\lambda \pm 0\), is :

A.\(t^{2}\)
B.\(1 + t^{2}\)
C.\(t\)
D.\(1 + t\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The solution is \(x(t) = t\), up to a constant multiple. (The stem prints '\(\lambda \pm 0\)' where \(\lambda \neq 0\) is meant.)

The kernel \(ts\) is separable. Pulling \(t\) out of the integral:

\[ x(t) = \lambda t \int_{0}^{1} s\, x(s)\, ds = c\,\lambda t \]

so every solution is a multiple of \(t\). Substituting \(x = t\) back gives \(\int_{0}^{1} s^{2}\, ds = 1/3\), so a nonzero solution exists exactly for the eigenvalue \(\lambda = 3\), and the eigenfunction is \(t\).

The forms \(t^{2}\), \(1 + t\) and \(1 + t^{2}\) cannot be solutions, because the right-hand side of the equation is always a multiple of \(t\).

Discuss in Forum

20.

The kernel of the linear integral equation

\[ x(t) = \lambda \int_{0}^{1} (t^{2} - s^{2})\, x(s)\, ds \]

is :

A.Separable
B.Symmetric
C.Convolution
D.Singular

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The kernel \(t^{2} - s^{2}\) is separable.

A separable (degenerate) kernel is a finite sum of products \(a_{i}(t) b_{i}(s)\). Here \(t^{2} - s^{2} = t^{2} \cdot 1 + 1 \cdot (-s^{2})\), a sum of two such products, so the integral equation reduces to a 2 × 2 linear system.

It is not symmetric: swapping \(t\) and \(s\) gives \(s^{2} - t^{2}\), the negative, so the kernel is skew-symmetric. It is not a convolution kernel, which would be a function of \(t - s\) alone, and it is bounded and continuous on the square, so it is not singular.

Linear Integral Equations Questions and Answers

Topics

Other Subjects

Tags: Linear Integral Equations Questions and Answers

About Us

Linear Integral Equations Questions and Answers

Topics

Other Subjects

Tags: Linear Integral Equations Questions and Answers

About Us

Stay informed about all important Dates & Notifications