Numerical Methods and Transform Calculus: Complete Guide to Laplace Transforms

Course Overview and Curricular Framework

Course Title and Affiliation:
- Course Name: Numerical Methods and Transform Calculus
- Instructor: Samala Rathan
- Institution: Indian Institute of Petroleum and Energy (IIPE), Visakhapatnam
Curriculum Syllabus:
- Laplace Transforms: Definition, linearity property, conditions for existence, shifting properties, Laplace transform of derivatives and integrals, unit step function, Dirac-delta and error function, differentiation and integration of transforms, convolution theorem, inversion, periodic functions, evaluation of integrals by Laplace transforms, solution of initial and boundary value problems.
- Fourier Series and Fourier Transforms: Fourier series representation of a function and its convergent properties, half-range series, sine and cosine series, Fourier integral representation of a function, Parseval's identity, Fourier transform, Fourier sine and cosine transforms, linearity, scaling, shifting properties, convolution theorem, applications to initial and boundary value problems.
Standard Course References:
- R. K. Jain and S. R. K. Iyengar, Advanced Engineering Mathematics, Narosa publisher.
- Erwin Kreyszig, Advanced Engineering Mathematics, Wiley publisher.
- W. Feller, An Introduction to Probability Theory and Its Applications.
- Peter V. O'Neil, Advanced Engineering Mathematics, 6th edition.

Fundamentals of the Laplace Transformation

Purpose and Operational Concept:
- The Laplace transformation is an operational tool designed to solve linear, constant-coefficient ordinary differential equations (ODEs) and partial differential equations (PDEs) under specified initial and boundary conditions.
- The operational workflow converts a calculus problem in the time domain ( $t$ ) into an algebraic problem in the complex frequency domain ( $s$ ):
- Step 1: Apply the Laplace transform to the differential equation: $\text{Initial Value Problem} \xrightarrow{\mathcal{L}} \text{Algebraic Equation}$ .
- Step 2: Formulate an algebraic subsidiary equation in terms of the transform $Y(s) = \mathcal{L}[y(t)]$ .
- Step 3: Solve the algebraic equation for $Y(s)$ .
- Step 4: Invert the algebraic solution to recover the time-domain solution: $\text{Algebraic Equation} \xrightarrow{\mathcal{L}^{-1}} \text{Solution } y(t)$ .
Core Advantages:
- Eliminates the conventional two-step procedure of finding the complementary function (general solution) followed by undetermined coefficients or variation of parameters to satisfy initial conditions. Initial values are embedded directly during the first step.
- For PDEs, applying the Laplace transform with respect to the time variable $t$ reduces the equation to an ordinary differential equation in the spatial variable.
- Readily accommodates discontinuous inputs, short-duration impulses, and periodic forcing functions without piecewise integration matching.
Mathematical Definition:
- Let $f(t)$ be a real-valued function defined for all $t \ge 0$ . The Laplace transform of $f(t)$ , denoted by $\mathcal{L}[f(t)]$ or $F(s)$ , is the improper integral: $F(s) = \mathcal{L}[f(t)] = \int_0^\infty e^{-st} f(t)\,dt$ provided the integral converges.
- The parameter $s$ is generally treated as a real or complex parameter.
- The inverse Laplace transform recovers $f(t)$ from $F(s)$ : $\mathcal{L}^{-1}[F(s)] = f(t)$
Linearity Property:
- Theorem: Let $f(t)$ and $g(t)$ be functions whose Laplace transforms exist. For any constants $\alpha$ and $\beta$ : $\mathcal{L}[\alpha f(t) + \beta g(t)] = \alpha \mathcal{L}[f(t)] + \beta \mathcal{L}[g(t)]$
- Proof: Derived directly from the linear property of definite integrals: $\mathcal{L}[\alpha f(t) + \beta g(t)] = \int_0^\infty e^{-st} [\alpha f(t) + \beta g(t)]\,dt = \alpha \int_0^\infty e^{-st} f(t)\,dt + \beta \int_0^\infty e^{-st} g(t)\,dt = \alpha \mathcal{L}[f(t)] + \beta \mathcal{L}[g(t)]$

Elementary Laplace Transforms and Inverses

Direct Derivations from the Integral Definition:
- Constant Function $f(t) = 1$ ( $t \ge 0$ ): $\mathcal{L}[1] = \int_0^\infty e^{-st}\,dt = \left[ -\frac{e^{-st}}{s} \right]_0^\infty = 0 - \left( -\frac{1}{s} \right) = \frac{1}{s}, \quad s > 0$
- Linear Function $f(t) = t$ : $\mathcal{L}[t] = \int_0^\infty e^{-st} t\,dt = \left[ -\frac{t e^{-st}}{s} - \frac{e^{-st}}{s^2} \right]_0^\infty = \frac{1}{s^2}, \quad s > 0$
- Exponential Function $f(t) = e^{at}$ : $\mathcal{L}[e^{at}] = \int_0^\infty e^{-st} e^{at}\,dt = \int_0^\infty e^{-(s-a)t}\,dt = \left[ -\frac{e^{-(s-a)t}}{s-a} \right]_0^\infty = \frac{1}{s-a}, \quad s > a$
- Hyperbolic Sine $f(t) = \sinh(\omega t)$ : $\sinh(\omega t) = \frac{e^{\omega t} - e^{-\omega t}}{2}$ $\mathcal{L}[\sinh(\omega t)] = \frac{1}{2} \left( \frac{1}{s - \omega} - \frac{1}{s + \omega} \right) = \frac{1}{2} \left( \frac{(s + \omega) - (s - \omega)}{s^2 - \omega^2} \right) = \frac{\omega}{s^2 - \omega^2}, \quad s > |\omega|$
- Hyperbolic Cosine $f(t) = \cosh(\omega t)$ : $\cosh(\omega t) = \frac{e^{\omega t} + e^{-\omega t}}{2}$ $\mathcal{L}[\cosh(\omega t)] = \frac{1}{2} \left( \frac{1}{s - \omega} + \frac{1}{s + \omega} \right) = \frac{s}{s^2 - \omega^2}, \quad s > |\omega|$
- Trigonometric Functions $f(t) = \cos(at)$ and $f(t) = \sin(at)$ :
- Utilizing Euler's identity $e^{iat} = \cos(at) + i \sin(at)$ , where $i = \sqrt{-1}$ : $\mathcal{L}[e^{iat}] = \frac{1}{s - ia} = \frac{s + ia}{(s - ia)(s + ia)} = \frac{s + ia}{s^2 + a^2} = \frac{s}{s^2 + a^2} + i \frac{a}{s^2 + a^2}$
- Separating real and imaginary parts: $\mathcal{L}[\cos(at)] = \text{Re}\left(\mathcal{L}[e^{iat}]\right) = \frac{s}{s^2 + a^2}, \quad s > 0$ $\mathcal{L}[\sin(at)] = \text{Im}\left(\mathcal{L}[e^{iat}]\right) = \frac{a}{s^2 + a^2}, \quad s > 0$
Master Transform Table:

Function $f(t)$	Laplace Transform $\mathcal{L}[f(t)] = F(s)$	Region of Convergence
$1$	$\frac{1}{s}$	$s > 0$
$t^n \quad (n = 1, 2, \dots)$	$\frac{n!}{s^{n+1}}$	$s > 0$
$t^\alpha \quad (\alpha > -1)$	$\frac{\Gamma(\alpha + 1)}{s^{\alpha + 1}}$	$s > 0$
$e^{at}$	$\frac{1}{s-a}$	$s > a$
$\sin(at)$	$\frac{a}{s^2 + a^2}$	$s > 0$
$\cos(at)$	$\frac{s}{s^2 + a^2}$	$s > 0$
$\sinh(\omega t)$	$\frac{\omega}{s^2 - \omega^2}$	$s > \|\omega\|$
$\cosh(\omega t)$	$\frac{s}{s^2 - \omega^2}$	$s > \|\omega\|$

Standard Inverse Laplace Transforms:
- $\mathcal{L}^{-1}\left[\frac{1}{s}\right] = 1$
- $\mathcal{L}^{-1}\left[\frac{n!}{s^{n+1}}\right] = t^n$
- $\mathcal{L}^{-1}\left[\frac{1}{s-a}\right] = e^{at}$
- $\mathcal{L}^{-1}\left[\frac{a}{s^2 + a^2}\right] = \sin(at)$
- $\mathcal{L}^{-1}\left[\frac{s}{s^2 - \omega^2}\right] = \cosh(\omega t)$

Existence Criteria and Asymptotic Theorems

Existence of the Laplace Transform:
- Theorem: Let $f: [0, \infty) \to \mathbb{R}$ be a function satisfying two conditions:
1. $f(t)$ is piecewise continuous on every finite interval $[0, T]$ for any $T > 0$ .
2. $f(t)$ is of exponential order as $t \to \infty$ , meaning there exist constants $M > 0$ , $a \in \mathbb{R}$ , and $T > 0$ such that: $|f(t)| \le M e^{at}, \quad \forall t \ge T$ Then the Laplace transform $\mathcal{L}[f(t)] = \int_0^\infty e^{-st} f(t)\,dt$ converges absolutely and exists for all $s > a$ .
- Proof:
- Partition the improper integral at $T$ : $\int_0^\infty e^{-st} f(t)\,dt = \int_0^T e^{-st} f(t)\,dt + \int_T^\infty e^{-st} f(t)\,dt$
- Because $f(t)$ is piecewise continuous on $[0, T]$ , the product $e^{-st} f(t)$ is piecewise continuous on the closed bounded interval, guaranteeing that $\int_0^T e^{-st} f(t)\,dt$ exists and is finite.
- Over $[T, \infty)$ , apply the exponential order bound: $\left| e^{-st} f(t) \right| = e^{-st} |f(t)| \le M e^{-st} e^{at} = M e^{-(s-a)t}$
- Evaluate the bounding integral for $s > a$ (where $s - a > 0$ ): $\int_T^\infty M e^{-(s-a)t}\,dt = M \left[ -\frac{e^{-(s-a)t}}{s-a} \right]_T^\infty = \frac{M}{s-a} e^{-(s-a)T} < \infty$
- By the direct comparison test for improper integrals, $\int_T^\infty e^{-st} f(t)\,dt$ converges absolutely. Hence, $\mathcal{L}[f(t)]$ exists for all $s > a$ .
Sufficiency versus Necessity:
- The conditions of the theorem are sufficient, not necessary. A function may fail piecewise continuity or boundedness near the origin and still possess a valid transform.
- Counterexample $f(t) = t^{-1/2}$ :
- Discontinuous at $t = 0$ ( $\lim_{t \to 0^+} t^{-1/2} = \infty$ ).
- Evaluating $\mathcal{L}[t^{-1/2}] = \int_0^\infty e^{-st} t^{-1/2}\,dt$ :
  - Substitute $t = u^2$ , which gives $dt = 2u\,du$ and $t^{-1/2} = \frac{1}{u}$ .
  - $\mathcal{L}[t^{-1/2}] = \int_0^\infty e^{-s u^2} \left(\frac{1}{u}\right) 2u\,du = 2 \int_0^\infty e^{-s u^2}\,du$ .
  - Substitute $z = u\sqrt{s}$ , so $du = \frac{dz}{\sqrt{s}}$ . Using the Gaussian integral $\int_0^\infty e^{-z^2}\,dz = \frac{\sqrt{\pi}}{2}$ : $\mathcal{L}[t^{-1/2}] = \frac{2}{\sqrt{s}} \int_0^\infty e^{-z^2}\,dz = \frac{2}{\sqrt{s}} \left(\frac{\sqrt{\pi}}{2}\right) = \sqrt{\frac{\pi}{s}}, \quad s > 0$
Transform of an Elementary Step-Like Function:
- For $f(t) = \begin{cases} 0, & 0 \le t < 2 \\ k, & t \ge 2 \end{cases}$ where $k$ is constant: $\mathcal{L}[f(t)] = \int_2^\infty k e^{-st}\,dt = -k \left[ \frac{e^{-st}}{s} \right]_2^\infty = \frac{k e^{-2s}}{s}, \quad s > 0$
Transform of Squared Trigonometric Functions:
- For $f(t) = \cos^2(at)$ , apply the power-reduction identity $\cos^2(at) = \frac{1 + \cos(2at)}{2}$ : $\mathcal{L}[\cos^2(at)] = \frac{1}{2} \mathcal{L}[1] + \frac{1}{2} \mathcal{L}[\cos(2at)] = \frac{1}{2s} + \frac{1}{2} \left( \frac{s}{s^2 + 4a^2} \right) = \frac{(s^2 + 4a^2) + s^2}{2s(s^2 + 4a^2)} = \frac{s^2 + 2a^2}{s(s^2 + 4a^2)}$
Asymptotic Behavior Theorems:
- Theorem: If $f(t)$ satisfies the existence theorem criteria with $\mathcal{L}[f(t)] = F(s)$ , then:
1. $\lim_{s \to \infty} F(s) = 0$
2. $\lim_{s \to \infty} [s F(s)]$ is bounded.
- Consequence: Functions of $s$ that do not vanish as $s \to \infty$ (such as $1$ , $s$ , $s^2$ , or $\frac{s^2}{s+1}$ ) cannot be the Laplace transforms of regular piecewise continuous functions of exponential order.

Inverse Laplace Transforms and Partial Fraction Decomposition

Inversion of Linear Combinations:
- Evaluate $\mathcal{L}^{-1}\left[ \frac{2s + 5}{s^2 + 25} \right]$ : $\mathcal{L}^{-1}\left[ \frac{2s + 5}{s^2 + 25} \right] = 2 \mathcal{L}^{-1}\left[ \frac{s}{s^2 + 25} \right] + \mathcal{L}^{-1}\left[ \frac{5}{s^2 + 25} \right] = 2 \cos(5t) + \sin(5t)$
Partial Fraction Expansion for Distinct Linear Factors:
- Evaluate $\mathcal{L}^{-1}\left[ \frac{5s^2 + 3s - 16}{(s - 1)(s - 2)(s + 3)} \right]$ :
- Decompose into partial fractions: $\frac{5s^2 + 3s - 16}{(s - 1)(s - 2)(s + 3)} = \frac{A}{s - 1} + \frac{B}{s - 2} + \frac{C}{s + 3}$
- Clear denominators: $5s^2 + 3s - 16 = A(s - 2)(s + 3) + B(s - 1)(s + 3) + C(s - 1)(s - 2)$
- Solve for undetermined coefficients:
  - Set $s = 1$ : $5(1)^2 + 3(1) - 16 = A(1 - 2)(1 + 3) \implies -8 = -4A \implies A = 2$
  - Set $s = 2$ : $5(2)^2 + 3(2) - 16 = B(2 - 1)(2 + 3) \implies 10 = 5B \implies B = 2$
  - Set $s = -3$ : $5(-3)^2 + 3(-3) - 16 = C(-3 - 1)(-3 - 2) \implies 45 - 9 - 16 = C(-4)(-5) \implies 20 = 20C \implies C = 1$
- Invert term-by-term: $F(s) = \frac{2}{s - 1} + \frac{2}{s - 2} + \frac{1}{s + 3}$ $f(t) = 2 \mathcal{L}^{-1}\left[ \frac{1}{s - 1} \right] + 2 \mathcal{L}^{-1}\left[ \frac{1}{s - 2} \right] + \mathcal{L}^{-1}\left[ \frac{1}{s + 3} \right] = 2e^t + 2e^{2t} + e^{-3t}$

Transforms of Derivatives and Solutions to Initial Value Problems

First Derivative Theorem:
- Theorem: Let $f(t)$ be continuous and of exponential order on $[0, \infty)$ , and let $f'(t)$ be piecewise continuous and of exponential order on $[0, \infty)$ . Then: $\mathcal{L}[f'(t)] = s \mathcal{L}[f(t)] - f(0) = s F(s) - f(0)$
- Proof (Continuous Case):
- Apply integration by parts with $u = e^{-st}$ and $dv = f'(t)\,dt$ (so $du = -s e^{-st}\,dt$ and $v = f(t)$ ): $\mathcal{L}[f'(t)] = \int_0^\infty e^{-st} f'(t)\,dt = \left[ e^{-st} f(t) \right]_0^\infty - \int_0^\infty (-s e^{-st}) f(t)\,dt$
- Since $f(t)$ is of exponential order, $\lim_{t \to \infty} e^{-st} f(t) = 0$ for sufficiently large $s$ : $\mathcal{L}[f'(t)] = \left( 0 - f(0) \right) + s \int_0^\infty e^{-st} f(t)\,dt = s \mathcal{L}[f(t)] - f(0)$
- Proof (Piecewise Continuous Derivative Case):
- Assume $f'(t)$ has a finite jump discontinuity at $t = T$ : $\mathcal{L}[f'(t)] = \lim_{\epsilon \to 0^+} \left( \int_0^{T - \epsilon} e^{-st} f'(t)\,dt + \int_{T + \epsilon}^\infty e^{-st} f'(t)\,dt \right)$
- Integrating by parts on each respective interval: $\mathcal{L}[f'(t)] = \lim_{\epsilon \to 0^+} \left( \left[ e^{-st} f(t) \right]_0^{T - \epsilon} + s \int_0^{T - \epsilon} e^{-st} f(t)\,dt + \left[ e^{-st} f(t) \right]_{T + \epsilon}^\infty + s \int_{T + \epsilon}^\infty e^{-st} f(t)\,dt \right)$
- Since $f(t)$ is globally continuous, $f(T^-) = f(T^+) = f(T)$ , causing boundary terms at $T$ to cancel: $e^{-sT} f(T^-) - e^{-sT} f(T^+) = 0$
- Taking $\lim_{t \to \infty} e^{-st} f(t) = 0$ leaves: $\mathcal{L}[f'(t)] = -f(0) + s \int_0^\infty e^{-st} f(t)\,dt = s \mathcal{L}[f(t)] - f(0)$
Inversion Consequence for Zero Initial Value:
- If $f(0) = 0$ , then $\mathcal{L}[f'(t)] = s F(s)$ .
- Inversely: $\mathcal{L}^{-1}[s F(s)] = f'(t) = \frac{d}{dt} \mathcal{L}^{-1}[F(s)]$ .
Higher-Order Derivatives:
- Second Derivative: $\mathcal{L}[f''(t)] = s \mathcal{L}[f'(t)] - f'(0) = s [s \mathcal{L}[f(t)] - f(0)] - f'(0) = s^2 \mathcal{L}[f(t)] - s f(0) - f'(0)$
- Third Derivative: $\mathcal{L}[f'''(t)] = s \mathcal{L}[f''(t)] - f''(0) = s^3 \mathcal{L}[f(t)] - s^2 f(0) - s f'(0) - f''(0)$
- General $n$ -th Derivative: $\mathcal{L}[f^{(n)}(t)] = s^n \mathcal{L}[f(t)] - s^{n-1} f(0) - s^{n-2} f'(0) - \dots - f^{(n-1)}(0) = s^n F(s) - \sum_{k=0}^{n-1} s^{n-1-k} f^{(k)}(0)$
- Special Case with Vanishing Initial Values:
- If $f(0) = f'(0) = \dots = f^{(n-1)}(0) = 0$ , then: $\mathcal{L}[f^{(n)}(t)] = s^n \mathcal{L}[f(t)] \iff \mathcal{L}^{-1}[s^n F(s)] = f^{(n)}(t)$
Illustrative Examples of Derivative Formulas:
- Find $\mathcal{L}[\sin^2(t)]$ :
- Let $f(t) = \sin^2(t)$ . Then $f(0) = 0$ , and $f'(t) = 2 \sin(t) \cos(t) = \sin(2t)$ .
- Applying the derivative theorem: $\mathcal{L}[f'(t)] = s \mathcal{L}[f(t)] - f(0) \implies \mathcal{L}[\sin(2t)] = s \mathcal{L}[\sin^2(t)]$ .
- Since $\mathcal{L}[\sin(2t)] = \frac{2}{s^2 + 4}$ : $\mathcal{L}[\sin^2(t)] = \frac{2}{s(s^2 + 4)}$
- Find $\mathcal{L}[t^3]$ via differentiation:
- Let $f(t) = t^3$ . Then $f(0) = 0$ , $f'(0) = 0$ , $f''(0) = 0$ , and $f'''(t) = 6$ .
- Applying the third-derivative formula: $\mathcal{L}[f'''(t)] = s^3 \mathcal{L}[f(t)] \implies \mathcal{L}[6] = s^3 \mathcal{L}[t^3]$ .
- $\frac{6}{s} = s^3 \mathcal{L}[t^3] \implies \mathcal{L}[t^3] = \frac{6}{s^4}$
- Evaluate $\mathcal{L}^{-1}\left[ \frac{s}{s^2 + 9} \right]$ via derivative inversion:
- Let $F(s) = \frac{1}{s^2 + 9}$ . Then $f(t) = \mathcal{L}^{-1}[F(s)] = \frac{1}{3} \sin(3t)$ .
- Because $f(0) = \frac{1}{3} \sin(0) = 0$ : $\mathcal{L}^{-1}[s F(s)] = f'(t) = \frac{d}{dt}\left( \frac{1}{3} \sin(3t) \right) = \cos(3t)$
Solution of Second-Order Initial Value Problems:
- Consider the differential equation $a y'' + b y' + c y = f(t)$ with $y(0) = y_0$ and $y'(0) = y'_0$ .
- Taking the transform yields: $a [s^2 Y(s) - s y(0) - y'(0)] + b [s Y(s) - y(0)] + c Y(s) = F(s)$
- Collect terms containing $Y(s)$ to form the subsidiary equation: $(a s^2 + b s + c) Y(s) = F(s) + a [s y(0) + y'(0)] + b y(0)$
- Solve algebraically for $Y(s)$ : $Y(s) = \frac{F(s) + a [s y(0) + y'(0)] + b y(0)}{a s^2 + b s + c}$
Step-by-Step Initial Value Problem Solutions:
- Example 1: Undamped Harmonic Oscillator
- Solve $y'' + 4y = 0$ with $y(0) = 1$ , $y'(0) = 6$ .
- Laplace domain: $[s^2 Y(s) - s(1) - 6] + 4 Y(s) = 0$ .
- $(s^2 + 4)Y(s) = s + 6 \implies Y(s) = \frac{s}{s^2 + 4} + \frac{6}{s^2 + 4}$ .
- Invert term-by-term: $y(t) = \cos(2t) + 3 \sin(2t)$
- Example 2: Second-Order Linear Inhomogeneous Equation
- Solve $y'' + 2y' - 3y = 3$ with $y(0) = 4$ , $y'(0) = -7$ .
- Laplace domain: $[s^2 Y(s) - 4s - (-7)] + 2[s Y(s) - 4] - 3 Y(s) = \frac{3}{s}$ .
- $(s^2 + 2s - 3)Y(s) - 4s + 7 - 8 = \frac{3}{s} \implies (s^2 + 2s - 3)Y(s) = 4s + 1 + \frac{3}{s}$ .
- Factor the characteristic polynomial: $s^2 + 2s - 3 = (s - 1)(s + 3)$ .
- Express $Y(s)$ : $Y(s) = \frac{4s + 1}{(s - 1)(s + 3)} + \frac{3}{s(s - 1)(s + 3)} = \frac{4s^2 + s + 3}{s(s - 1)(s + 3)}$
- Partial fractions expansion: $Y(s) = \frac{A}{s} + \frac{B}{s - 1} + \frac{C}{s + 3}$ .
  - $4s^2 + s + 3 = A(s - 1)(s + 3) + B s (s + 3) + C s (s - 1)$
  - At $s = 0$ : $3 = A(-1)(3) \implies A = -1$ .
  - At $s = 1$ : $8 = B(1)(4) \implies B = 2$ .
  - At $s = -3$ : $4(9) - 3 + 3 = C(-3)(-4) \implies 36 = 12C \implies C = 3$ .
- $Y(s) = -\frac{1}{s} + \frac{2}{s - 1} + \frac{3}{s + 3}$ .
- Inverting yields: $y(t) = -1 + 2e^t + 3e^{-3t}$
- Example 3: Non-Homogeneous System with Exponential Forcing
- Solve $y'' - 5y' + 4y = e^{2t}$ with $y(0) = \frac{19}{12}$ , $y'(0) = \frac{8}{3}$ .
- Laplace domain: $\left[ s^2 Y(s) - s\left(\frac{19}{12}\right) - \frac{8}{3} \right] - 5\left[ s Y(s) - \frac{19}{12} \right] + 4 Y(s) = \frac{1}{s - 2}$
- $(s^2 - 5s + 4)Y(s) - \frac{19}{12}s - \frac{32}{12} + \frac{95}{12} = \frac{1}{s - 2} \implies (s^2 - 5s + 4)Y(s) = \frac{1}{s - 2} + \frac{19s - 63}{12}$
- Factoring $s^2 - 5s + 4 = (s - 1)(s - 4)$ , we obtain: $Y(s) = \frac{1}{(s - 2)(s - 1)(s - 4)} + \frac{19s - 63}{12(s - 1)(s - 4)}$
- Partial fraction decomposition of combined $Y(s)$ : $Y(s) = -\frac{1}{2(s - 2)} + \frac{14}{9(s - 1)} + \frac{19}{36(s - 4)}$
- Taking the inverse Laplace transform: $y(t) = -\frac{1}{2}e^{2t} + \frac{14}{9}e^t + \frac{19}{36}e^{4t}$

Transform of an Integral and Integro-Differential Equations

Laplace Transform of an Integral:
- Theorem: Let $f(t)$ be piecewise continuous and of exponential order on $[0, \infty)$ . Then: $\mathcal{L}\left[ \int_0^t f(\tau)\,d\tau \right] = \frac{1}{s} \mathcal{L}[f(t)] = \frac{F(s)}{s}, \quad s > 0, \, s > \alpha$
- Proof:
- Define $\phi(t) = \int_0^t f(\tau)\,d\tau$ . By the fundamental theorem of calculus, $\phi'(t) = f(t)$ almost everywhere, and $\phi(0) = \int_0^0 f(\tau)\,d\tau = 0$ .
- Applying the transform of a derivative: $\mathcal{L}[\phi'(t)] = s \mathcal{L}[\phi(t)] - \phi(0) \implies \mathcal{L}[f(t)] = s \mathcal{L}\left[ \int_0^t f(\tau)\,d\tau \right]$
- Dividing by $s$ : $\mathcal{L}\left[ \int_0^t f(\tau)\,d\tau \right] = \frac{1}{s} \mathcal{L}[f(t)] = \frac{F(s)}{s}$
- Equivalent Inversion Formula: $\mathcal{L}^{-1}\left[ \frac{F(s)}{s} \right] = \int_0^t f(\tau)\,d\tau$
- Operational Rule: Multiplication by $\frac{1}{s}$ in the complex frequency domain corresponds to integration from $0$ to $t$ in the time domain: $\frac{1}{s} \iff \int_0^t (\cdot)\,d\tau$
Applications to Inverse Transforms:
- Single Integration Example:
- Find $\mathcal{L}^{-1}\left[ \frac{1}{s(s^2 + 9)} \right]$ .
- Let $F(s) = \frac{1}{s^2 + 9}$ , so $f(t) = \mathcal{L}^{-1}[F(s)] = \frac{1}{3}\sin(3t)$ .
- By the integral formula: $\mathcal{L}^{-1}\left[ \frac{1}{s(s^2 + 9)} \right] = \int_0^t \frac{1}{3} \sin(3\tau)\,d\tau = \left[ -\frac{\cos(3\tau)}{9} \right]_0^t = \frac{1}{9}(1 - \cos(3t))$
- Double Integration Example:
- Find $\mathcal{L}^{-1}\left[ \frac{1}{s^2(s^2 + 4)} \right]$ .
- First stage: Let $F(s) = \frac{1}{s^2 + 4}$ , then $\mathcal{L}^{-1}[F(s)] = \frac{1}{2}\sin(2t)$ . $\mathcal{L}^{-1}\left[ \frac{1}{s(s^2 + 4)} \right] = \int_0^t \frac{1}{2} \sin(2\tau)\,d\tau = \left[ -\frac{\cos(2\tau)}{4} \right]_0^t = \frac{1}{4}(1 - \cos(2t))$
- Second stage: $\mathcal{L}^{-1}\left[ \frac{1}{s^2(s^2 + 4)} \right] = \int_0^t \frac{1}{4}(1 - \cos(2\tau))\,d\tau = \frac{1}{4} \left[ \tau - \frac{\sin(2\tau)}{2} \right]_0^t = \frac{t}{4} - \frac{\sin(2t)}{8}$
- Alternative verification via partial fractions: $\frac{1}{s^2(s^2 + 4)} = \frac{1}{4} \left( \frac{1}{s^2} - \frac{1}{s^2 + 4} \right)$ $\mathcal{L}^{-1}\left[ \frac{1}{4}\left( \frac{1}{s^2} - \frac{1}{s^2 + 4} \right) \right] = \frac{1}{4} \left( t - \frac{1}{2}\sin(2t) \right) = \frac{t}{4} - \frac{\sin(2t)}{8}$
Application to Integro-Differential Equations:
- Volterra Integro-Differential Equation:
- Solve $y' + 3y + 2 \int_0^t y(\tau)\,d\tau = t$ with $y(0) = 0$ .
- Taking the Laplace transform: $[s Y(s) - y(0)] + 3 Y(s) + 2 \frac{Y(s)}{s} = \frac{1}{s^2}$
- With $y(0) = 0$ , multiply through by $s$ : $s^2 Y(s) + 3s Y(s) + 2 Y(s) = \frac{1}{s} \implies (s^2 + 3s + 2)Y(s) = \frac{1}{s}$
- Solve for $Y(s)$ : $Y(s) = \frac{1}{s(s + 1)(s + 2)}$
- Rewrite as $Y(s) = \frac{F(s)}{s}$ where $F(s) = \frac{1}{(s + 1)(s + 2)} = \frac{1}{s + 1} - \frac{1}{s + 2}$ .
- Compute $f(t) = \mathcal{L}^{-1}[F(s)] = e^{-t} - e^{-2t}$ .
- Invert using integration in time: $y(t) = \int_0^t (e^{-\tau} - e^{-2\tau})\,d\tau = \left[ -e^{-\tau} + \frac{1}{2}e^{-2\tau} \right]_0^t = \left( -e^{-t} + \frac{1}{2}e^{-2t} \right) - \left( -1 + \frac{1}{2} \right) = \frac{1}{2}e^{-2t} - e^{-t} + \frac{1}{2}$
- Series RLC Circuit Model:
- Applying Kirchhoff's voltage law to a series RLC circuit excited by an AC voltage source $E_0 \cos(\omega t)$ : $L \frac{di}{dt} + R i + \frac{1}{C} \int_0^t i(\tau)\,d\tau = E_0 \cos(\omega t)$
- Transforming to the frequency domain: $L [s I(s) - i(0)] + R I(s) + \frac{1}{C s} I(s) = \frac{E_0 s}{s^2 + \omega^2}$
- Regrouping gives the algebraic equation: $\left( L s + R + \frac{1}{C s} \right) I(s) = \frac{E_0 s}{s^2 + \omega^2} + L i(0)$

The Shifting Theorems

First Shifting Theorem (Frequency Shift):
- Theorem: Let $\mathcal{L}[f(t)] = F(s)$ for $s > \alpha$ . For any real constant $a$ : $\mathcal{L}[e^{at} f(t)] = F(s - a), \quad s > a + \alpha$
- Proof: $\mathcal{L}[e^{at} f(t)] = \int_0^\infty e^{-st} e^{at} f(t)\,dt = \int_0^\infty e^{-(s - a)t} f(t)\,dt = F(s - a)$
- Inverse Formulation: $\mathcal{L}^{-1}[F(s - a)] = e^{at} f(t) = e^{at} \mathcal{L}^{-1}[F(s)]$
Applications of the First Shifting Theorem:
- Damped Sinusoids and Polynomials:
- $\mathcal{L}[e^{at} \cos(bt)] = \frac{s - a}{(s - a)^2 + b^2}$
- $\mathcal{L}[e^{-\alpha t} \sin(\beta t)] = \frac{\beta}{(s + \alpha)^2 + \beta^2}$
- $\mathcal{L}[e^{at} t^3] = \frac{3!}{(s - a)^4} = \frac{6}{(s - a)^4}$
- $\mathcal{L}[e^{-\omega t} t^2] = \frac{2!}{(s + \omega)^3} = \frac{2}{(s + \omega)^3}$
- Quadratic Completion for Inverses:
- Invert $\frac{1}{s^2 - 4s + 8}$ : $s^2 - 4s + 8 = (s - 2)^2 + 2^2 \implies \mathcal{L}^{-1}\left[ \frac{1}{(s - 2)^2 + 2^2} \right] = \frac{1}{2} e^{2t} \sin(2t)$
- Invert $\frac{4}{s^2 - s + 2}$ : $s^2 - s + 2 = \left(s - \frac{1}{2}\right)^2 + \frac{7}{4} = \left(s - \frac{1}{2}\right)^2 + \left(\frac{\sqrt{7}}{2}\right)^2$ $\mathcal{L}^{-1}\left[ \frac{4}{\left(s - \frac{1}{2}\right)^2 + \left(\frac{\sqrt{7}}{2}\right)^2} \right] = \frac{4}{\frac{\sqrt{7}}{2}} e^{t/2} \sin\left( \frac{\sqrt{7}}{2} t \right) = \frac{8}{\sqrt{7}} e^{t/2} \sin\left( \frac{\sqrt{7}}{2} t \right)$
- Invert $\frac{3s - 1}{(s - 2)^2}$ : $\frac{3s - 1}{(s - 2)^2} = \frac{3(s - 2) + 5}{(s - 2)^2} = \frac{3}{s - 2} + \frac{5}{(s - 2)^2}$ $\mathcal{L}^{-1}\left[ \frac{3}{s - 2} + \frac{5}{(s - 2)^2} \right] = 3e^{2t} + 5t e^{2t}$
- Invert $\frac{s + 6}{s^2 + 6s + 13}$ : $s^2 + 6s + 13 = (s + 3)^2 + 4 = (s + 3)^2 + 2^2, \quad s + 6 = (s + 3) + 3$ $\frac{s + 6}{s^2 + 6s + 13} = \frac{s + 3}{(s + 3)^2 + 2^2} + \frac{3}{2} \left( \frac{2}{(s + 3)^2 + 2^2} \right)$ $\mathcal{L}^{-1}\left[ \frac{s + 6}{s^2 + 6s + 13} \right] = e^{-3t} \cos(2t) + \frac{3}{2} e^{-3t} \sin(2t)$
- Initial Value Problem with Repeated Roots:
- Solve $y'' + 4y' + 4y = 12 t^2 e^{-2t}$ with $y(0) = 2$ , $y'(0) = 1$ .
- Transformed equation: $[s^2 Y(s) - 2s - 1] + 4[s Y(s) - 2] + 4 Y(s) = \mathcal{L}[12 t^2 e^{-2t}] = \frac{12 \cdot 2!}{(s + 2)^3} = \frac{24}{(s + 2)^3}$
- $(s^2 + 4s + 4)Y(s) - 2s - 9 = \frac{24}{(s + 2)^3} \implies (s + 2)^2 Y(s) = \frac{24}{(s + 2)^3} + 2s + 9$
- Decompose into powers of $(s + 2)$ : $2s + 9 = 2(s + 2) + 5$ $Y(s) = \frac{24}{(s + 2)^5} + \frac{2(s + 2) + 5}{(s + 2)^2} = \frac{24}{(s + 2)^5} + \frac{2}{s + 2} + \frac{5}{(s + 2)^2}$
- Inverting each term using $\mathcal{L}^{-1}\left[\frac{n!}{(s+2)^{n+1}}\right] = t^n e^{-2t}$ : $\mathcal{L}^{-1}\left[ \frac{24}{(s + 2)^5} \right] = t^4 e^{-2t}, \quad \mathcal{L}^{-1}\left[ \frac{2}{s + 2} \right] = 2e^{-2t}, \quad \mathcal{L}^{-1}\left[ \frac{5}{(s + 2)^2} \right] = 5t e^{-2t}$ $y(t) = (2 + 5t + t^4)e^{-2t}$
- Initial Value Problem with Damped Oscillations:
- Solve $y'' + 4y' + 13y = e^{-t}$ with $y(0) = 0$ , $y'(0) = 2$ .
- Transformed equation: $[s^2 Y(s) - 0 - 2] + 4[s Y(s) - 0] + 13 Y(s) = \frac{1}{s + 1}$ $(s^2 + 4s + 13)Y(s) = \frac{1}{s + 1} + 2 = \frac{2s + 3}{s + 1} \implies Y(s) = \frac{2s + 3}{(s + 1)(s^2 + 4s + 13)}$
- Partial fraction expansion: $\frac{2s + 3}{(s + 1)(s^2 + 4s + 13)} = \frac{A}{s + 1} + \frac{B s + C}{s^2 + 4s + 13}$ $2s + 3 = A(s^2 + 4s + 13) + (B s + C)(s + 1)$
  - At $s = -1$ : $1 = A(1 - 4 + 13) = 10A \implies A = \frac{1}{10}$ .
  - Equating $s^2$ coefficients: $0 = A + B \implies B = -\frac{1}{10}$ .
  - At $s = 0$ : $3 = 13A + C = \frac{13}{10} + C \implies C = \frac{17}{10}$ .
- Grouping quadratic terms: $Y(s) = \frac{1}{10} \left( \frac{1}{s + 1} - \frac{s - 17}{(s + 2)^2 + 3^2} \right) = \frac{1}{10} \left( \frac{1}{s + 1} - \frac{(s + 2) - 19}{(s + 2)^2 + 3^2} \right)$ $Y(s) = \frac{1}{10} \left( \frac{1}{s + 1} - \frac{s + 2}{(s + 2)^2 + 3^2} + \frac{19}{3} \frac{3}{(s + 2)^2 + 3^2} \right)$
- Inverse transform: $y(t) = \frac{1}{10} \left( e^{-t} - e^{-2t} \cos(3t) + \frac{19}{3} e^{-2t} \sin(3t) \right)$

Discontinuous Functions and the Second Shifting Theorem

Heaviside or Unit Step Function:
- Defined mathematically as: $H(t) = u(t) = \begin{cases} 0, & t < 0 \\ 1, & t \ge 0 \end{cases}$
- For a jump discontinuity shifted to $t = a$ ( $a \ge 0$ ): $H(t - a) = u(t - a) = u_a(t) = \begin{cases} 0, & t < a \\ 1, & t \ge a \end{cases}$
- Representing Switching and Windows:
- Half-wave or activation: $f_1(t) = \sin(t) u_\pi(t) = \begin{cases} 0, & t < \pi \\ \sin(t), & t \ge \pi \end{cases}$
- Delayed activation: $f_2(t) = t^2 u_2(t) = \begin{cases} 0, & t < 2 \\ t^2, & t \ge 2 \end{cases}$
- Rectangular pulse between $t = a$ and $t = b$ ( $a < b$ ): $f_3(t) = u_a(t) - u_b(t) = \begin{cases} 0, & t < a \\ 1, & a \le t < b \\ 0, & t \ge b \end{cases}$
- Deactivating a profile: For $f(t) = \begin{cases} 5t, & 0 \le t < 2 \\ 0, & t \ge 2 \end{cases}$ , the linear ramp is turned off at $t = 2$ : $f(t) = 5t - 5t u_2(t) = 5t[1 - u_2(t)] = 5t[u_0(t) - u_2(t)]$
Second Shifting Theorem (Time Shift):
- Theorem: Let $\mathcal{L}[f(t)] = F(s)$ and let $a \ge 0$ . Then: $\mathcal{L}[f(t - a) u_a(t)] = e^{-as} F(s)$ $\mathcal{L}^{-1}[e^{-as} F(s)] = f(t - a) u_a(t)$
- Proof: $\mathcal{L}[f(t - a) u_a(t)] = \int_0^\infty e^{-st} f(t - a) u_a(t)\,dt = \int_a^\infty e^{-st} f(t - a)\,dt$
- Introduce the change of variables $\tau = t - a$ , meaning $t = \tau + a$ and $dt = d\tau$ : $\int_0^\infty e^{-s(\tau + a)} f(\tau)\,d\tau = e^{-as} \int_0^\infty e^{-s\tau} f(\tau)\,d\tau = e^{-as} F(s)$
- Transform of the Bare Unit Step: $\mathcal{L}[u_a(t)] = \mathcal{L}[1 \cdot u_a(t)] = \frac{e^{-as}}{s}$
- Alternative Direct Shifting Identity:
- When an unshifted function $f(t)$ is multiplied by $u_a(t)$ , express $f(t) = g(t - a)$ where $g(t) = f(t + a)$ : $\mathcal{L}[f(t) u_a(t)] = e^{-as} \mathcal{L}[f(t + a)]$
Worked Inversion and Transformation Problems:
- Transform of Shifting Polynomials:
- Find $\mathcal{L}[(t - 3)^2 u_3(t)]$ : Here $f(t) = t^2$ , so $\mathcal{L}[(t - 3)^2 u_3(t)] = e^{-3s} \mathcal{L}[t^2] = \frac{2 e^{-3s}}{s^3}$ .
- Find $\mathcal{L}[t^2 u_3(t)]$ : $\mathcal{L}[t^2 u_3(t)] = e^{-3s} \mathcal{L}[(t + 3)^2] = e^{-3s} \mathcal{L}[t^2 + 6t + 9] = e^{-3s} \left( \frac{2}{s^3} + \frac{6}{s^2} + \frac{9}{s} \right)$
- Inverse Transform Evaluations:
- Evaluate $\mathcal{L}^{-1}\left[ \frac{e^{-3s}}{s + 2} \right]$ :
  - Here $a = 3$ and $F(s) = \frac{1}{s + 2}$ , which gives $f(t) = e^{-2t}$ .
  - $\mathcal{L}^{-1}\left[ \frac{e^{-3s}}{s + 2} \right] = e^{-2(t - 3)} u_3(t)$
- Evaluate $\mathcal{L}^{-1}\left[ \frac{4 e^{-\pi s / 2}}{s^2 + 16} \right]$ :
  - Here $a = \frac{\pi}{2}$ and $F(s) = \frac{4}{s^2 + 16}$ , so $f(t) = \sin(4t)$ .
  - $f\left(t - \frac{\pi}{2}\right) = \sin\left(4\left(t - \frac{\pi}{2}\right)\right) = \sin(4t - 2\pi) = \sin(4t)$ .
  - $\mathcal{L}^{-1}\left[ \frac{4 e^{-\pi s / 2}}{s^2 + 16} \right] = \sin(4t) u_{\pi/2}(t)$
- Evaluate $\mathcal{L}^{-1}\left[ \frac{e^{-2s}}{s^2} \right]$ :
  - Here $a = 2$ and $F(s) = \frac{1}{s^2}$ , giving $f(t) = t$ .
  - $\mathcal{L}^{-1}\left[ \frac{e^{-2s}}{s^2} \right] = (t - 2) u_2(t)$
Differential Equations with Discontinuous Forcing:
- First-Order Switched System:
- Solve $y' + y = f(t)$ with $y(0) = 2$ , where $f(t) = \begin{cases} 0, & 0 \le t < \frac{\pi}{2} \\ \cos(t), & t \ge \frac{\pi}{2} \end{cases}$ .
- Represent forcing function: $f(t) = \cos(t) u_{\pi/2}(t)$ .
- Transform forcing input: $\mathcal{L}\left[ \cos(t) u_{\pi/2}(t) \right] = e^{-\pi s / 2} \mathcal{L}\left[ \cos\left(t + \frac{\pi}{2}\right) \right] = e^{-\pi s / 2} \mathcal{L}[-\sin(t)] = -\frac{e^{-\pi s / 2}}{s^2 + 1}$
- Laplace domain equation: $s Y(s) - 2 + Y(s) = -\frac{e^{-\pi s / 2}}{s^2 + 1} \implies (s + 1)Y(s) = 2 - \frac{e^{-\pi s / 2}}{s^2 + 1}$ $Y(s) = \frac{2}{s + 1} - \frac{e^{-\pi s / 2}}{(s + 1)(s^2 + 1)}$
- Decompose $\frac{1}{(s + 1)(s^2 + 1)}$ : $\frac{1}{(s + 1)(s^2 + 1)} = \frac{1}{2} \left( \frac{1}{s + 1} - \frac{s - 1}{s^2 + 1} \right) = \frac{1}{2} \left( \frac{1}{s + 1} - \frac{s}{s^2 + 1} + \frac{1}{s^2 + 1} \right)$
- Invert using the second shifting theorem: $y(t) = 2e^{-t} - \frac{1}{2} \left[ e^{-(t - \pi/2)} - \cos\left(t - \frac{\pi}{2}\right) + \sin\left(t - \frac{\pi}{2}\right) \right] u_{\pi/2}(t)$
- Using $\cos\left(t - \frac{\pi}{2}\right) = \sin(t)$ and $\sin\left(t - \frac{\pi}{2}\right) = -\cos(t)$ : $y(t) = 2e^{-t} - \frac{1}{2} \left[ e^{-(t - \pi/2)} - \sin(t) - \cos(t) \right] u_{\pi/2}(t)$
- Piecewise representation: $y(t) = \begin{cases} 2e^{-t}, & 0 \le t < \frac{\pi}{2} \\ 2e^{-t} - \frac{1}{2} \left( e^{-(t - \pi/2)} - \sin(t) - \cos(t) \right), & t \ge \frac{\pi}{2} \end{cases}$
Integro-Differential Equation with a Finite Duration Pulse:
- Solve $y' + 5y + 4 \int_0^t y(\tau)\,d\tau = f(t)$ with $y(0) = 2$ , where $f(t)$ is a rectangular pulse of magnitude $k$ active from $t = a$ to $t = b$ ( $a < b$ ): $f(t) = k [u_a(t) - u_b(t)]$
- Transformed equation: $s Y(s) - 2 + 5 Y(s) + \frac{4}{s} Y(s) = k \left( \frac{e^{-as}}{s} - \frac{e^{-bs}}{s} \right)$ $(s^2 + 5s + 4)Y(s) = 2s + k(e^{-as} - e^{-bs})$ $Y(s) = \frac{2s}{(s + 1)(s + 4)} + \frac{k(e^{-as} - e^{-bs})}{(s + 1)(s + 4)}$
- Partial fraction decompositions: $\frac{2s}{(s + 1)(s + 4)} = \frac{2}{3} \left( \frac{4}{s + 4} - \frac{1}{s + 1} \right)$ $\frac{1}{(s + 1)(s + 4)} = \frac{1}{3} \left( \frac{1}{s + 1} - \frac{1}{s + 4} \right)$
- Time-domain solution: $y(t) = \frac{2}{3}(4e^{-4t} - e^{-t}) + \frac{k}{3} \left[ (e^{-(t - a)} - e^{-4(t - a)}) u_a(t) - (e^{-(t - b)} - e^{-4(t - b)}) u_b(t) \right]$
- Expanded piecewise piecewise representation across three intervals: $y(t) = \begin{cases} \frac{2}{3}(4e^{-4t} - e^{-t}), & 0 \le t < a \\ \frac{2}{3}(4e^{-4t} - e^{-t}) + \frac{k}{3} [e^{-(t - a)} - e^{-4(t - a)}], & a \le t < b \\ \frac{2}{3}(4e^{-4t} - e^{-t}) + \frac{k}{3} \left( [e^{-(t - a)} - e^{-4(t - a)}] - [e^{-(t - b)} - e^{-4(t - b)}] \right), & t \ge b \end{cases}$

Impulse and the Dirac-Delta Function

Physical Concept and Generalized Distributions:
- Physical systems frequently experience forces of immense magnitude acting over negligible durations (mechanical impact, electrical surges, lightning strikes). Such phenomena are modeled by the Dirac-delta function $\delta(t)$ .
- $\delta(t)$ is not a classical function, but a generalized function (distribution), formalised by Laurent Schwartz in the Theory of Distributions.
- Physical analogy: A unit force pressed on the edge of a plate producing pressure distribution $w(x)$ , where the total integral equals the applied force: $\int_{-\infty}^\infty w(x)\,dx = 1$ .
Approximating Delta Sequences:
- Rectangular Sequence: $w_k(x) = \begin{cases} \frac{k}{2}, & |x| < \frac{1}{k} \\ 0, & |x| \ge \frac{1}{k} \end{cases}$
- Area: $\int_{-\infty}^\infty w_k(x)\,dx = \int_{-1/k}^{1/k} \frac{k}{2}\,dx = \frac{k}{2} \left( \frac{2}{k} \right) = 1$ .
- As $k \to \infty$ , height $\frac{k}{2} \to \infty$ and width $\frac{2}{k} \to 0$ , concentrating the unit area at $x = 0$ : $\delta(x) = \lim_{k \to \infty} w_k(x)$ .
- Lorentzian (Cauchy) Sequence: $w_k(x) = \frac{k}{\pi(1 + k^2 x^2)}, \quad k > 0$
- Area: $\int_{-\infty}^\infty w_k(x)\,dx = \frac{k}{\pi} \int_{-\infty}^\infty \frac{dx}{1 + k^2 x^2} = \frac{1}{\pi} \left[ \tan^{-1}(kx) \right]_{-\infty}^\infty = \frac{1}{\pi}\left( \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) \right) = 1$ .
- Narrowing Pulse Function in Time: $\delta_k(t) = \begin{cases} 0, & t < 0 \\ \frac{1}{k}, & 0 \le t < k \\ 0, & t \ge k \end{cases} = \frac{1}{k}[H(t) - H(t - k)]$
- Total integral is unity: $\int_0^\infty \delta_k(t)\,dt = \frac{1}{k} \int_0^k dt = 1$ .
- Distributional limit: $\delta(t) = \lim_{k \to 0^+} \delta_k(t)$ .
Relationship to the Heaviside Step Function:
- In the distribution sense, the derivative of the Heaviside unit step function is the Dirac-delta function: $H'(t) = \delta(t)$ $H'(t - a) = \delta(t - a)$
Filtering (Sifting) Property:
- Theorem: Let $f(t)$ be continuous and integrable on $[0, \infty)$ . Then: $\int_0^\infty f(t) \delta(t - a)\,dt = f(a)$
- Proof:
- Represent $\delta(t - a)$ by its rectangular pulse sequence $\delta_k(t - a) = \begin{cases} \frac{1}{k}, & a \le t < a + k \\ 0, & \text{otherwise} \end{cases}$ : $\int_0^\infty f(t) \delta_k(t - a)\,dt = \frac{1}{k} \int_a^{a + k} f(t)\,dt$
- By the Mean Value Theorem for definite integrals, there exists some $t_0 \in (a, a + k)$ such that: $\frac{1}{k} \int_a^{a + k} f(t)\,dt = \frac{1}{k} f(t_0) \int_a^{a + k} dt = \frac{1}{k} f(t_0) (k) = f(t_0)$
- Taking the limit as $k \to 0^+$ , the point $t_0 \to a$ . Since $f$ is continuous, $\lim_{k \to 0^+} f(t_0) = f(a)$ . Therefore: $\int_0^\infty f(t) \delta(t - a)\,dt = f(a)$
Laplace Transform of the Dirac-Delta Function:
- Using the sequence representation: $\mathcal{L}\{\delta_k(t - a)\} = \frac{1}{k} \left( \frac{e^{-as}}{s} - \frac{e^{-(a + k)s}}{s} \right) = \frac{e^{-as}}{ks} (1 - e^{-ks})$
- Taking the limit as $k \to 0^+$ via L'Hôpital's rule: $\lim_{k \to 0^+} \frac{1 - e^{-ks}}{k} = \lim_{k \to 0^+} \frac{s e^{-ks}}{1} = s$ $\mathcal{L}\{\delta(t - a)\} = \lim_{k \to 0^+} \frac{e^{-as}}{s} \left( \frac{1 - e^{-ks}}{k} \right) = \frac{e^{-as}}{s} (s) = e^{-as}$
- Alternatively, via the filtering property directly: $\mathcal{L}\{\delta(t - a)\} = \int_0^\infty e^{-st} \delta(t - a)\,dt = e^{-as}$
- Impulse at the Origin ( $a = 0$ ): $\mathcal{L}\{\delta(t)\} = e^{0} = 1 \iff \mathcal{L}^{-1}\{1\} = \delta(t)$ $\mathcal{L}\{H'(t)\} = \mathcal{L}\{\delta(t)\} = 1$
Initial Value Problem Involving Impulse Excitation:
- Solve $y'' + 2y' + 5y = \delta(t - 2)$ with $y(0) = 0$ , $y'(0) = 0$ .
- Transformed equation: $(s^2 + 2s + 5)Y(s) = e^{-2s} \implies Y(s) = \frac{e^{-2s}}{(s + 1)^2 + 4}$
- Unshifted inverse: $\mathcal{L}^{-1}\left[ \frac{1}{(s + 1)^2 + 4} \right] = \frac{1}{2} e^{-t} \sin(2t)$ .
- Applying the second shifting theorem: $y(t) = \frac{1}{2} e^{-(t - 2)} \sin(2(t - 2)) H(t - 2) = \begin{cases} 0, & 0 \le t < 2 \\ \frac{1}{2} e^{-(t - 2)} \sin(2(t - 2)), & t \ge 2 \end{cases}$
- Derivative Jump Discontinuity:
- Integrating the differential equation across an infinitesimal interval $[2 - \epsilon, 2 + \epsilon]$ : $\int_{2 - \epsilon}^{2 + \epsilon} y''\,dt + 2 \int_{2 - \epsilon}^{2 + \epsilon} y'\,dt + 5 \int_{2 - \epsilon}^{2 + \epsilon} y\,dt = \int_{2 - \epsilon}^{2 + \epsilon} \delta(t - 2)\,dt$ $[y'(2 + \epsilon) - y'(2 - \epsilon)] + 2[y(2 + \epsilon) - y(2 - \epsilon)] + 5(0) = 1$
- Because displacement $y(t)$ is continuous ( $y(2^+) - y(2^-) = 0$ ), the velocity exhibits a unit step discontinuity: $y'(2^+) - y'(2^-) = 1$

Differentiation of Laplace Transforms

Differentiation Theorem:
- Theorem: Let $f(t)$ be piecewise continuous on $[0, \infty)$ and of exponential order. If $\mathcal{L}[f(t)] = F(s)$ , then: $\mathcal{L}[t f(t)] = -F'(s) = -\frac{d}{ds}F(s)$ $\mathcal{L}[t^n f(t)] = (-1)^n F^{(n)}(s) = (-1)^n \frac{d^n}{ds^n}F(s), \quad s > \alpha$
- Proof:
- Differentiating the definition of the Laplace transform under the integral sign with respect to parameter $s$ : $F(s) = \int_0^\infty e^{-st} f(t)\,dt$ $\frac{dF}{ds} = \frac{d}{ds} \int_0^\infty e^{-st} f(t)\,dt = \int_0^\infty \frac{\partial}{\partial s}(e^{-st}) f(t)\,dt = \int_0^\infty (-t e^{-st}) f(t)\,dt = -\int_0^\infty e^{-st} [t f(t)]\,dt$ $\frac{dF}{ds} = -\mathcal{L}[t f(t)] \implies \mathcal{L}[t f(t)] = -F'(s)$
- Higher-order powers follow by mathematical induction: $\mathcal{L}[t^2 f(t)] = \mathcal{L}[t(t f(t))] = -\frac{d}{ds}\left( -F'(s) \right) = F''(s)$ $\mathcal{L}[t^n f(t)] = (-1)^n F^{(n)}(s)$
- Inverse Operational Form: $\mathcal{L}^{-1}[F^{(n)}(s)] = (-1)^n t^n f(t)$
Computational Examples of Transform Differentiation:
- Transform of $t \sin(4t)$ : $\mathcal{L}[\sin(4t)] = \frac{4}{s^2 + 16}$ $\mathcal{L}[t \sin(4t)] = -\frac{d}{ds}\left( \frac{4}{s^2 + 16} \right) = -\left( -\frac{4(2s)}{(s^2 + 16)^2} \right) = \frac{8s}{(s^2 + 16)^2}$
- Transform of $t^2 \cos(3t)$ : $\mathcal{L}[\cos(3t)] = \frac{s}{s^2 + 9}$
- First derivative: $\mathcal{L}[t \cos(3t)] = -\frac{d}{ds}\left( \frac{s}{s^2 + 9} \right) = -\frac{1(s^2 + 9) - s(2s)}{(s^2 + 9)^2} = \frac{s^2 - 9}{(s^2 + 9)^2}$
- Second derivative: $\mathcal{L}[t^2 \cos(3t)] = -\frac{d}{ds}\left( \frac{s^2 - 9}{(s^2 + 9)^2} \right) = -\frac{2s(s^2 + 9)^2 - (s^2 - 9) 2(s^2 + 9)(2s)}{(s^2 + 9)^4}$ $\mathcal{L}[t^2 \cos(3t)] = -\frac{2s(s^2 + 9) - 4s(s^2 - 9)}{(s^2 + 9)^3} = -\frac{2s^3 + 18s - 4s^3 + 36s}{(s^2 + 9)^3} = \frac{2s(s^2 - 27)}{(s^2 + 9)^3}$
- Transform of $t e^{5t}$ and $t^2 e^{-2t}$ : $\mathcal{L}[t e^{5t}] = -\frac{d}{ds}\left( \frac{1}{s - 5} \right) = \frac{1}{(s - 5)^2}, \quad s > 5$ $\mathcal{L}[t^2 e^{-2t}] = \frac{d^2}{ds^2}\left( \frac{1}{s + 2} \right) = \frac{d}{ds}\left( -\frac{1}{(s + 2)^2} \right) = \frac{2}{(s + 2)^3}$
Inversion Calculations Using Derivatives:
- Evaluate $\mathcal{L}^{-1}\left[ \frac{2(s + 1)}{(s^2 + 2s + 2)^2} \right]$ :
- Note that $s^2 + 2s + 2 = (s + 1)^2 + 1$ .
- Recognize derivative structure: $\frac{d}{ds}\left( \frac{1}{(s + 1)^2 + 1} \right) = -\frac{2(s + 1)}{[(s + 1)^2 + 1]^2} \implies \frac{2(s + 1)}{(s^2 + 2s + 2)^2} = -\frac{d}{ds}\left( \frac{1}{(s + 1)^2 + 1} \right)$
- Since $\mathcal{L}^{-1}\left[ \frac{1}{(s + 1)^2 + 1} \right] = e^{-t} \sin(t)$ , using $\mathcal{L}^{-1}[-F'(s)] = t f(t)$ : $\mathcal{L}^{-1}\left[ \frac{2(s + 1)}{(s^2 + 2s + 2)^2} \right] = t e^{-t} \sin(t)$
- Evaluate $\mathcal{L}^{-1}\left[ \frac{1}{(s + 5)^4} \right]$ :
- Differentiating $\frac{1}{s + 5}$ repeatedly: $\frac{d^3}{ds^3}\left( \frac{1}{s + 5} \right) = -\frac{6}{(s + 5)^4} \implies \frac{1}{(s + 5)^4} = -\frac{1}{6} \frac{d^3}{ds^3}\left( \frac{1}{s + 5} \right)$
- Since $\mathcal{L}^{-1}\left[ \frac{1}{s + 5} \right] = e^{-5t}$ , using $\mathcal{L}^{-1}[F'''(s)] = (-1)^3 t^3 f(t) = -t^3 f(t)$ : $\mathcal{L}^{-1}\left[ \frac{1}{(s + 5)^4} \right] = -\frac{1}{6} (-t^3 e^{-5t}) = \frac{t^3 e^{-5t}}{6}$
- Evaluate $\mathcal{L}^{-1}\left[ \frac{1}{(s^2 + 9)^2} \right]$ :
- Rewrite using the identity $\frac{1}{(s^2 + 9)^2} = \frac{(s^2 + 9) + (9 - s^2)}{18(s^2 + 9)^2} = \frac{1}{18} \left( \frac{1}{s^2 + 9} - \frac{s^2 - 9}{(s^2 + 9)^2} \right)$ .
- From prior derivations, $\mathcal{L}^{-1}\left[ \frac{1}{s^2 + 9} \right] = \frac{1}{3}\sin(3t)$ and $\mathcal{L}^{-1}\left[ \frac{s^2 - 9}{(s^2 + 9)^2} \right] = t \cos(3t)$ .
- Substituting into the expression: $\mathcal{L}^{-1}\left[ \frac{1}{(s^2 + 9)^2} \right] = \frac{1}{18} \left( \frac{1}{3} \sin(3t) - t \cos(3t) \right) = \frac{\sin(3t)}{54} - \frac{t \cos(3t)}{18}$
Differential Equations with Variable (Polynomial) Coefficients:
- Example 1: Initial Value Problem with First-Degree Polynomial Coefficients
- Solve $y'' + t y' - 2y = 6 - t$ with $y(0) = 0$ , $y'(0) = 1$ .
- Applying transform identities: $\mathcal{L}[y''] = s^2 Y(s) - s y(0) - y'(0) = s^2 Y(s) - 1$ $\mathcal{L}[t y'] = -\frac{d}{ds}\mathcal{L}[y'] = -\frac{d}{ds}(s Y(s) - y(0)) = -\frac{d}{ds}(s Y) = -(s Y' + Y)$ $\mathcal{L}[6 - t] = \frac{6}{s} - \frac{1}{s^2}$
- Substitute into the transformed differential equation: $(s^2 Y - 1) - (s Y' + Y) - 2Y = \frac{6}{s} - \frac{1}{s^2}$ $-s Y' + (s^2 - 3)Y = 1 + \frac{6}{s} - \frac{1}{s^2}$ $s Y' + (3 - s^2)Y = -1 - \frac{6}{s} + \frac{1}{s^2}$
- Divide by $s$ to establish standard first-order linear form: $Y' + \left( \frac{3}{s} - s \right) Y = -\frac{1}{s} - \frac{6}{s^2} + \frac{1}{s^3}$
- Compute the integrating factor: $I.F. = e^{\int (3/s - s)\,ds} = e^{3\ln(s) - s^2/2} = s^3 e^{-s^2/2}$
- Multiply by $I.F.$ and integrate: $\frac{d}{ds}\left( s^3 e^{-s^2/2} Y \right) = s^3 e^{-s^2/2} \left( -\frac{1}{s} - \frac{6}{s^2} + \frac{1}{s^3} \right) = e^{-s^2/2} (-s^2 - 6s + 1)$
- Evaluate $\int e^{-s^2/2} (-s^2 - 6s + 1)\,ds$ :
  - Note $\frac{d}{ds}(s e^{-s^2/2}) = e^{-s^2/2} - s^2 e^{-s^2/2} = (1 - s^2)e^{-s^2/2}$ .
  - Note $\frac{d}{ds}(6 e^{-s^2/2}) = -6s e^{-s^2/2}$ .
  - Combining yields $\frac{d}{ds}[(s + 6)e^{-s^2/2}] = (-s^2 - 6s + 1)e^{-s^2/2}$ . $s^3 e^{-s^2/2} Y = (s + 6)e^{-s^2/2} + C \implies Y(s) = \frac{s + 6}{s^3} + C \frac{e^{s^2/2}}{s^3}$
- Physical existence criterion requires $\lim_{s \to \infty} Y(s) = 0$ . Because $\lim_{s \to \infty} \frac{e^{s^2/2}}{s^3} = \infty$ , the boundary requirement demands $C = 0$ : $Y(s) = \frac{s + 6}{s^3} = \frac{1}{s^2} + \frac{6}{s^3}$
- Invert using $\mathcal{L}^{-1}\left[\frac{1}{s^2}\right] = t$ and $\mathcal{L}^{-1}\left[\frac{6}{s^3}\right] = 3t^2$ : $y(t) = t + 3t^2$
- Example 2: Initial Value Problem with Arbitrary Conditions / Vanishing Leading Coefficient
- Solve $t y'' + 2t y' + 2y = 2$ with $y(0) = 1$ , where $y'(0)$ is an unspecified constant.
- Transform term-by-term: $\mathcal{L}[t y''] = -\frac{d}{ds}[s^2 Y - s(1) - y'(0)] = -(s^2 Y' + 2s Y - 1)$ $\mathcal{L}[2t y'] = -2 \frac{d}{ds}[s Y - 1] = -2(s Y' + Y)$ $\mathcal{L}[2y] = 2Y$ $\mathcal{L}[2] = \frac{2}{s}$
- Substitute and simplify: $-(s^2 Y' + 2s Y - 1) - 2(s Y' + Y) + 2Y = \frac{2}{s}$ $-(s^2 + 2s)Y' - 2s Y + 1 = \frac{2}{s} \implies -s(s + 2)Y' - 2s Y = \frac{2}{s} - 1 = \frac{2 - s}{s}$ $Y' + \frac{2}{s + 2} Y = \frac{s - 2}{s^2 (s + 2)}$
- Integrating factor: $I.F. = e^{\int \frac{2}{s + 2}\,ds} = (s + 2)^2$ $(s + 2)^2 Y = \int \frac{(s - 2)(s + 2)^2}{s^2 (s + 2)}\,ds + C = \int \frac{s^2 - 4}{s^2}\,ds + C = \int \left( 1 - \frac{4}{s^2} \right)\,ds + C = s + \frac{4}{s} + C$ $Y(s) = \frac{s^2 + C s + 4}{s(s + 2)^2}$
- Partial fraction expansion: $\frac{s^2 + C s + 4}{s(s + 2)^2} = \frac{1}{s} + \frac{C - 4}{(s + 2)^2}$
- Taking the inverse Laplace transform: $y(t) = 1 + (C - 4) t e^{-2t}$
- Differentiating the solution: $y'(t) = (C - 4)e^{-2t} - 2(C - 4)t e^{-2t} \implies y'(0) = C - 4$ .
- Because $y'(0)$ is arbitrary, $C$ remains completely arbitrary. The problem exhibits a non-unique family of solutions because the coefficient of the highest derivative $y''$ vanishes at the initial point $t = 0$ .

Integration of Laplace Transforms

Integration Theorem:
- Theorem: Let $f(t)$ be piecewise continuous on $[0, \infty)$ and of exponential order, and suppose that $\lim_{t \to 0^+} \frac{f(t)}{t}$ exists. If $\mathcal{L}[f(t)] = F(s)$ , then: $\mathcal{L}\left[ \frac{f(t)}{t} \right] = \int_s^\infty F(s')\,ds', \quad s > \alpha$
- Proof:
- By definition, $F(s') = \int_0^\infty e^{-s't} f(t)\,dt$ . Integrating both sides with respect to $s'$ from $s$ to $\infty$ : $\int_s^\infty F(s')\,ds' = \int_s^\infty \left( \int_0^\infty e^{-s't} f(t)\,dt \right) ds'$
- Under the stated conditions, Fubini's theorem permits exchanging the order of integration: $\int_s^\infty F(s')\,ds' = \int_0^\infty f(t) \left( \int_s^\infty e^{-s't}\,ds' \right) dt$
- For $t > 0$ : $\int_s^\infty e^{-s't}\,ds' = \left[ -\frac{e^{-s't}}{t} \right]_s^\infty = 0 - \left( -\frac{e^{-st}}{t} \right) = \frac{e^{-st}}{t}$
- Substituting back into the outer integral: $\int_s^\infty F(s')\,ds' = \int_0^\infty e^{-st} \left( \frac{f(t)}{t} \right)\,dt = \mathcal{L}\left[ \frac{f(t)}{t} \right]$
- Inverse Operational Identity: $\mathcal{L}^{-1}\left[ \int_s^\infty F(s')\,ds' \right] = \frac{f(t)}{t} \iff f(t) = t \mathcal{L}^{-1}\left[ \int_s^\infty F(s')\,ds' \right]$
Worked Applications of Transform Integration:
- Transform of $\frac{\sin(\omega t)}{t}$ :
- Since $\mathcal{L}[\sin(\omega t)] = \frac{\omega}{s^2 + \omega^2}$ , apply the integration theorem: $\mathcal{L}\left[ \frac{\sin(\omega t)}{t} \right] = \int_s^\infty \frac{\omega}{(s')^2 + \omega^2}\,ds' = \left[ \tan^{-1}\left( \frac{s'}{\omega} \right) \right]_s^\infty$
- For $\omega > 0$ : $\mathcal{L}\left[ \frac{\sin(\omega t)}{t} \right] = \frac{\pi}{2} - \tan^{-1}\left( \frac{s}{\omega} \right) = \tan^{-1}\left( \frac{\omega}{s} \right)$
- Inverse Transform of Logarithmic Functions:
- Evaluate $\mathcal{L}^{-1}\left[ \ln\left( \frac{s + c}{s + d} \right) \right]$ :
  - Let $G(s) = \ln\left( \frac{s + c}{s + d} \right) = \ln(s + c) - \ln(s + d)$ .
  - Define $F(s) = -G'(s) = -\left( \frac{1}{s + c} - \frac{1}{s + d} \right) = \frac{1}{s + d} - \frac{1}{s + c}$ .
  - The inverse transform of $F(s)$ is $f(t) = \mathcal{L}^{-1}[F(s)] = e^{-dt} - e^{-ct}$ .
  - By the fundamental theorem of calculus, $G(s) = \int_s^\infty F(s')\,ds'$ .
  - Applying the inverse integration identity: $\mathcal{L}^{-1}\left[ \ln\left( \frac{s + c}{s + d} \right) \right] = \frac{f(t)}{t} = \frac{e^{-dt} - e^{-ct}}{t}$
- Inverse Transform of Repeated Rational Functions:
- Evaluate $\mathcal{L}^{-1}\left[ \frac{1}{s(s + 3)^2} \right]$ :
  - Let $G(s) = \frac{1}{s(s + 3)^2}$ . Partial fractions decomposition gives: $G(s) = \frac{1}{9} \left( \frac{1}{s} - \frac{1}{s + 3} - \frac{3}{(s + 3)^2} \right)$
  - Define $F(s) = -G'(s) = \frac{1}{9} \left( \frac{1}{s^2} - \frac{1}{(s + 3)^2} - \frac{6}{(s + 3)^3} \right)$ .
  - Compute $f(t) = \mathcal{L}^{-1}[F(s)]$ : $f(t) = \frac{1}{9} \left( t - t e^{-3t} - 3t^2 e^{-3t} \right)$
  - Applying the inverse integration formula: $\mathcal{L}^{-1}[G(s)] = \frac{f(t)}{t} = \frac{1}{9}(1 - e^{-3t} - 3t e^{-3t})$
- Inverse Transform via Indefinite Frequency Integration:
- Evaluate $\mathcal{L}^{-1}\left[ \frac{s}{(s^2 - 9)^2} \right]$ :
  - Using $f(t) = t \mathcal{L}^{-1}\left[ \int_s^\infty \frac{s'}{((s')^2 - 9)^2}\,ds' \right]$ .
  - Evaluate the inner improper integral: $\int_s^\infty \frac{s'}{((s')^2 - 9)^2}\,ds' = \left[ -\frac{1}{2((s')^2 - 9)} \right]_s^\infty = 0 - \left( -\frac{1}{2(s^2 - 9)} \right) = \frac{1}{2(s^2 - 9)}$
  - Compute the inverse transform: $f(t) = \frac{t}{2} \mathcal{L}^{-1}\left[ \frac{1}{s^2 - 9} \right] = \frac{t}{2} \left( \frac{1}{3} \sinh(3t) \right) = \frac{t}{6} \sinh(3t)$

Summary of Essential Operational Theorems

Core Operational Triad:
- Impulse Input: $\mathcal{L}\{\delta(t - a)\} = e^{-as}$
- Multiplication by Power of Time ( $t^n$ ): $\mathcal{L}[t^n f(t)] = (-1)^n F^{(n)}(s)$
- Division by Time ( $1/t$ ): $\mathcal{L}\left[ \frac{f(t)}{t} \right] = \int_s^\infty F(s')\,ds'$
Time Domain vs. Frequency Domain Duality:

Operation in Time Domain ( $t$ )	Representation in Frequency Domain ( $s$ )
Differentiation: $f'(t)$	Multiplication by $s$ : $s F(s) - f(0)$
Integration: $\int_0^t f(\tau)\,d\tau$	Division by $s$ : $\frac{F(s)}{s}$
Exponential Scaling: $e^{at} f(t)$	Frequency Translation: $F(s - a)$
Time Delay: $f(t - a) u_a(t)$	Exponential Damping: $e^{-as} F(s)$
Multiplication by $t$ : $t f(t)$	Frequency Derivative: $-F'(s)$
Division by $t$ : $\frac{f(t)}{t}$	Frequency Integration: $\int_s^\infty F(s')\,ds'$