Laplace Transforms: Solving Differential Equations

Laplace Transform and Application

• The Laplace transform is named after French mathematician Laplace, who studied it in 1782
• Laplace transforms are integral transforms

Introduction

• Laplace transforms solve differential equations with boundary values
• Laplace transform method relies on algebra to solve linear differential equations
• This is a powerful tool for dealing with linear differential equations with discontinuous forcing functions

Objectives

• Laplace transforms are useful for solving differential equations with boundary values
• Different properties of Laplace and Inverse Laplace transforms facilitate solving physics problems
• The unit will teach how to use the Laplace transform for differential equations

Laplace Transform

• The Laplace transform of a function f(t) is defined as F(s) = integral from 0 to infinity of e^(-st) * f(t) dt
• This definition holds for all positive values of t where the integral exists
• The Laplace transform is denoted as L[f(t)] = F(s) = integral from 0 to infinity of e^(-st) * f(t) dt

Linearity of the Laplace Transform

• The Laplace transform is a linear operation
• Given functions f(t) and g(t) with existing transforms, and constants a and b, the transform of af(t) + bg(t) exists: L[af(t) + bg(t)] = aL[f(t)] + bL[g(t)]
• Integration is a linear operation

Change of Scale Property

• If the Laplace transform of f(t) is F(s), then L[f(at)] = (1/a) * F(s/a)

First Shifting Theorem

• If F(s) is the Laplace transform of f(t), then L[e^(at) f(t)] = F(s − a)

Second Shifting Theorem (Heaviside's Shifting Theorem)

• If L[f(t)] = F(s) and g(t) = f(t − a) for t > a, and g(t) = 0 for 0 < t < a, then L[g(t)] = e^(-as) F(s)

Laplace Transform of the Derivative of f(t)

• If L[f(t)] = F(s) and f'(t) is the derivative, then L[f'(t)] = sL[f(t)] – f(0)

Laplace Transform of the Derivative of Order n

• L[f^(n)(t)] = s^n L[f(t)] – s^(n-1) f(0) – s^(n-2) f'(0) – s^(n-3) f''(0) – … – f^(n-1)(0)

Laplace Transform of the Integral of f(t)

• If L[f(t)] = F(s), then L[integral of f(t) from 0 to t] = (1/s) * F(s)

Laplace Transform of Some Important Functions

• L(1) = 1/s
• L(e^(at)) = 1/(s-a), where s > a
• L(sin at) = a / (s^2 + a^2)
• L(cos at) = s / (s^2 + a^2)
• L(sinh at) = a / (s^2 - a^2)
• L(cosh at) = s / (s^2 - a^2)
• L(t^n) = n! / s^(n+1), where n and s are positive

Laplace Transform Of t. f(t)

• L[t^n * f(t)] = (-1)^n * (d^n / ds^n) * F(s)

Unit Step Function

• The unit step function is defined as u(t − a) = 0 when t < a, and u(t − a) = 1 when t ≥ a, where a ≥ 0

Laplace Transform of Unit Step Function

• L[u(t − a)] = e^(-as) / s

Periodic Functions

• For a periodic function f(t) with period T where f(t + T) = f(t), then L[f(t)] = (integral from 0 to T of e^(-st) * f(t) dt) / (1 - e^(-sT))

Some Important Formulae of Laplace Transform

• f(t) = e^(at), then F(s) = 1/(s-a)
• f(t) = sin at, then F(s) = a/(s^2 + a^2)
• f(t) = cos at, then F(s) = s/(s^2 + a^2)
• f(t) = sinh at, then F(s) = a/(s^2 - a^2)
• f(t) = cosh at, then F(s) = s/(s^2 - a^2)
• f(t) = t^n, then F(s) = n!/s^(n+1)
• f(t) = e^(bt) sin at, then F(s) = a/((s-b)^2 + a^2)
• f(t) = e^(bt) cos at, then F(s) = (s-b) / ((s-b)^2 + a^2)
• f(t) = (t/2a)sin at, then F(s) = s / (s^2 + a^2)^2
• f(t) = t cos at, then F(s) = (s^2 - a^2) / (s^2 + a^2)^2

Inverse Laplace Transform

• If F(s) is the Laplace Transform of a function f(t), then f(t) is the Inverse Laplace Transform: f(t) = L^(-1)[F(s)]
• The Inverse Laplace Transform is useful for solving differential equations

Some Important Formulae of Inverse Laplace Transform

• F(s) = 1/(s-a) then f(t) = e^(at)
• F(s) = a/(s^2 + a^2) then f(t) = sin at
• F(s) = s/(s^2 + a^2) then f(t) = cos at
• F(s) = a/(s^2 - a^2) then f(t) = sinh at
• F(s) = s/(s^2 - a^2) then f(t) = cosh at
• F(s) = n!/s^(n+1) then f(t) = t^n
• F(s) = = a/((s-b)^2 + a^2) then f(t) = e^(bt)sin at
• F(s) = (s-b)/((s-b)^2 + a^2) then f(t) = e^(bt) cos at
• F(s) = = s/(s^2 + a^2)^2 then f(t) = (t/2a)sin at
• F(s) = (s^2 - a^2)/(s^2 + a^2)^2 then f(t) = t cos at
• F(s) = = 1/s then f(t) = 1

Multiplication by s

• L^(-1)[sF(s)] = d/dt f(t) + f(0)δ(t)

Division by s (Multiplication By 1/s)

• L^(-1)[F(s) / s] = integral from 0 to t of [L^(-1) |F(s)|] dt = integral from 0 to t of f(t) dt

First Shifting Property

• If the inverse Laplace transform of F(s) is f(t), then L^(-1)F(s + a) = e^(-at) L^(-1)[F(s)]

Second Shifting Property

• L^(-1)[e^(-as) F(s)] = f(t − a)u(t – a)

Inverse Laplace Transforms of Derivatives

• L^(-1) [d/ds F(s)] = -tL^(-1)[F(s)] = −tf(t)

Inverse Laplace Transform Of Integrals

• L^(-1)[integral from s to ∞ of F(s)ds] = f(t)/t = 1/t L^(-1)[F(s)]

Inverse Laplace Transform by Partial Fraction Method

• Uses partial fractions to simplify the function before applying inverse Laplace transform

Solution of Differential Equations by Laplace Transforms

• Ordinary linear differential equations with constant coefficients can be solved using Laplace Transformation
• Steps:
• Set up the subsidiary equation by transforming the given differential equation
• Solve the subsidiary equation by algebra using the transfer function
• Find the inverse transform