Instructions: The figures in the margin indicate full marks for the questions. Attempt all questions including internal choices as per the guide below.
(a) What happens to Fourier series expansion of a periodic function, if the function is even in nature?
Answer: If a periodic function f(x) is even in nature, i.e., f(-x) = f(x), the sine coefficients (bn) in the Fourier series become zero. The series consists only of the constant term and cosine terms.
f(x) = a0/2 + Σ [an cos(nπx/l)]
(b) What are Dirichlet's conditions for a Fourier series expansion?
Answer: For a function f(x) to be expanded in a Fourier series in the interval (c, c+2l), it must satisfy:
(c) How will you change the function f(x) in the interval (-π, π) to (-l, +l) in Fourier series?
Answer: To change the interval, we use a change of variable. Let x be the variable in (-π, π) and z be the variable in (-l, l). We use the transformation:
z = (l/π) * x or x = (πz/l)
The Fourier series then uses arguments (nπz/l) instead of nx.
Find the series of sines and cosines of multiples of x which represents f(x) in the interval -π < x < π where f(x) = 0 for -π < x ≤ 0 and f(x) = πx/4 for 0 < x ≤ π. Hence deduce π²/8 = 1 + 1/3² + 1/5² + ...
Answer:
1. Calculation of Coefficients:
a0 = (1/π) ∫ f(x) dx from -π to π = (1/π) [0 + ∫(πx/4)dx from 0 to π] = (1/π) * (π/4) * [x²/2] from 0 to π = π²/8.
an = (1/π) ∫ f(x) cos(nx) dx = (1/π) * (π/4) ∫ x cos(nx) dx = (1/4) [x sin(nx)/n + cos(nx)/n²] from 0 to π.
an = (1/4n²) [cos(nπ) - 1]. If n is even, an = 0. If n is odd, an = -1/2n².
bn = (1/π) ∫ f(x) sin(nx) dx = (1/4) ∫ x sin(nx) dx = (1/4) [-x cos(nx)/n + sin(nx)/n²] from 0 to π.
bn = (1/4) [-π cos(nπ)/n] = -π/4n for odd n, and π/4n for even n.
2. Deduction:
Set x = 0 (point of discontinuity). f(0) = [f(0-) + f(0+)]/2 = [0 + 0]/2 = 0. This doesn't help directly. Set x = π. f(π) = π²/4. The series at π converges to [f(π) + f(-π)]/2 = [π²/4 + 0]/2 = π²/8. Substituting into the series leads to the required summation.
(a) Obtain a Fourier expression for f(x) = x³ for -π < x < π.
Answer: Since f(x) = x³ is an odd function [f(-x) = -x³], a0 = 0 and an = 0. We only find bn:
bn = (2/π) ∫ x³ sin(nx) dx from 0 to π.
Using integration by parts repeatedly:
bn = (2/π) [(-x³ cos(nx)/n) + (3x² sin(nx)/n²) + (6x cos(nx)/n³) - (6 sin(nx)/n⁴)] from 0 to π.
bn = (2/π) [(-π³ cos(nπ)/n) + (6π cos(nπ)/n³)] = 2 * (-1)^n * [(6/n³) - (π²/n)].
(b) Obtain the complex form of the Fourier series of f(x) = 0 for -π ≤ x ≤ 0 and f(x) = 1 for 0 ≤ x ≤ π.
Answer: The complex Fourier series is f(x) = Σ cn e^(inx).
cn = (1/2π) ∫ f(x) e^(-inx) dx from -π to π.
cn = (1/2π) [0 + ∫ 1 * e^(-inx) dx from 0 to π] = (1/2π) [e^(-inx) / (-in)] from 0 to π.
cn = (1/2πin) [1 - e^(-inπ)].
If n is even (n ≠ 0), cn = 0. If n is odd, cn = 1/(inπ).
For n = 0, c0 = (1/2π) ∫ 1 dx = 1/2.
(a) Explain the terms 'ordinary point' and 'singular point' with reference to ordinary differential equation.
Answer: For an ODE y'' + P(x)y' + Q(x)y = 0:
Ordinary Point: A point x = a is ordinary if both P(x) and Q(x) are analytic at x = a.
Singular Point: If either P(x) or Q(x) is not analytic at x = a, it is a singular point.
(b) Check whether x = 0 is ordinary or singular for: (i) y'' + xy' + (x²+2)y = 0 (ii) (2+x)y'' + (1+x)y' + 3y = 0.
Answer:
(i) P(x) = x, Q(x) = x²+2. Both analytic at x=0. Ordinary Point.
(ii) P(x) = (1+x)/(2+x), Q(x) = 3/(2+x). Both analytic at x=0. Ordinary Point.
(c) State Hermite and Laguerre differential equation of second order.
Answer:
Hermite: y'' - 2xy' + 2ny = 0
Laguerre: x y'' + (1-x)y' + ny = 0
(a) Obtain a series solution in powers of x for 2x(1-x)y'' + (1-x)y' + 3y = 0 using Frobenius method.
Answer: Let y = Σ ak x^(m+k). Substitute y, y', y'' into the equation.
The indicial equation is found by equating the coefficient of the lowest power of x (x^(m-1)) to zero:
2m(m-1) + m = 0 => 2m² - m = 0 => m(2m - 1) = 0. Roots: m = 0, 1/2.
Recurrence relation: ak+1 = ak * [(2(m+k)(m+k+1) - 2(m+k) - 3) / (2(m+k+1)(m+k) + (m+k+1))].
Solve for k=0, 1, 2... for both roots of m to get the general solution y = A(y1) + B(y2).
(b) Applying Frobenius method to solve Hermite differential equation y'' - 2xy' + 2ny = 0.
Answer: x=0 is an ordinary point. Let y = Σ ak x^k. Substitute into the ODE.
Σ k(k-1)ak x^(k-2) - 2Σ k ak x^k + 2nΣ ak x^k = 0.
Shift index: Σ (k+2)(k+1)ak+2 x^k - Σ 2(k-n)ak x^k = 0.
Recurrence relation: ak+2 = [2(k-n) / (k+1)(k+2)] ak.
This gives two independent series (even and odd powers). If n is an integer, one series terminates to become a Hermite Polynomial.
(a) Write the Rodrigue's formula for Legendre polynomial.
Pn(x) = (1 / (2^n * n!)) * (d^n / dx^n) (x² - 1)^n
(b) What is the orthogonality condition of the Legendre polynomial?
∫ Pn(x) Pm(x) dx from -1 to 1 = 0 if n ≠ m, and 2/(2n+1) if n = m.
(c) Show that Pn(-x) = (-1)^n Pn(x).
Answer: From the generating function (1 - 2xt + t²)^(-1/2) = Σ Pn(x) t^n. Replace x with -x: (1 + 2xt + t²)^(-1/2) = Σ Pn(-x) t^n. Also, replacing t with -t in the original gives (1 + 2xt + t²)^(-1/2) = Σ Pn(x) (-t)^n = Σ (-1)^n Pn(x) t^n. Comparing coefficients of t^n: Pn(-x) = (-1)^n Pn(x).
(a) Write down the conditions for Laplace transform to exist.
Answer: 1. f(t) must be piecewise continuous in every finite interval. 2. f(t) must be of exponential order (i.e., |f(t)| ≤ M e^(at) for some constants M and a).
(b) Explain two properties of Laplace transform.
Answer: 1. Linearity: L{af(t) + bg(t)} = aL{f(t)} + bL{g(t)}. 2. First Shifting Theorem: If L{f(t)} = F(s), then L{e^(at) f(t)} = F(s-a).
Find the Laplace transform of e^(at) sin(bt).
Answer: We know L{sin(bt)} = b / (s² + b²).
Using the first shifting theorem: L{e^(at) f(t)} = F(s-a).
Therefore, L{e^(at) sin(bt)} = b / ((s-a)² + b²).
(a) What do you mean by inverse Laplace transform?
Answer: If L{f(t)} = F(s), then f(t) is called the inverse Laplace transform of F(s), written as f(t) = L⁻¹{F(s)}.
(b) State convolution theorem for inverse Laplace transform.
L⁻¹{F(s)G(s)} = ∫ f(u) g(t-u) du from 0 to t.
(c) Find the inverse Laplace transform of 6 / (s² + 36).
Answer: We know L{sin(at)} = a / (s² + a²). Here a = 6.
Therefore, L⁻¹{6 / (s² + 6²)} = sin(6t).