Laplace Transforms

Contents

L.T. of Elementary Functions

Multiplication by tn & Division by t

L.T. of Integrals and Special Functions

Evaluation of Integrals using L.T

Inverse Laplace Transforms

Convolution Theorem

Application to Differential Equations

1M Questions

Definition: Let \( f(t) \) be a function of \( t \) defined for all \( t \ge 0 \). Then the Laplace transform of \( f(t) \), denoted by \( L\{ f(t) \} \), is defined by: \[ L\{ f(t) \} = \int_{0}^{\infty} e^{-st} f(t) \, dt \quad \text{(1)} \] provided that the integral exists. Here, \( s \) is a parameter that may be either real or complex.

\( L\{ f(t) \} \) is a function of \( s \) and is briefly written as \( F(s) \), i.e., \[ L\{ f(t) \} = F(s). \] The symbol \( L \), which transforms \( f(t) \) into \( F(s) \), is called the Laplace transform operator.

Sufficient Conditions for the Existence of Laplace Transform of \( f(t) \)

The Laplace transform of \( f(t) \) exists if the improper integral on the right-hand side of (1) converges (has a finite value) when the following sufficient conditions are satisfied:

  1. \( f(t) \) is piecewise continuous, i.e., \( f(t) \) is continuous in every sub-interval and has a finite limit at the end points of each of these sub-intervals.
  2. \( f(t) \) is of exponential order \( a \), i.e., there exist constants \( M \) and \( a \) such that: \[ |f(t)| < M e^{at}, \] i.e., \[ \lim_{t \to \infty} e^{-at} f(t) = \text{a finite quantity} \]

For example, \( f(t) = t^2 \), \( \sin(at) \), \( e^{-at} \), etc., are all of exponential order and continuous. However, \( e^{t^2} \) is not of exponential order, and its Laplace transform does not exist.

Linearity Property

If \( c_1 \) and \( c_2 \) are constants, and \( f \) and \( g \) are functions of \( t \), then: \[ L\{ c_1 f(t) \pm c_2 g(t) \} = c_1 L\{ f(t) \} \pm c_2 L\{ g(t) \} \]

Laplace Transforms of Elementary Functions

The direct application of the definition gives the following formulae:

1) \( L\{ 1 \} = \frac{1}{s}, \quad s > 0 \).

Proof: By the definition of Laplace transform, we have: \[ L\{ f(t) \} = \int_{0}^{\infty} e^{-st} f(t) \, dt \] For \( f(t) = 1 \): \[ L\{ 1 \} = \int_{0}^{\infty} e^{-st} \cdot 1 \, dt = \left[ \frac{e^{-st}}{-s} \right]_{0}^{\infty} = -\frac{1}{s} \left( 0 - e^{0} \right) \quad \left( \because \lim_{t \to \infty} e^{-st} = 0 \right). \] Simplifying: \[ L\{ 1 \} = \frac{1}{s}, \quad s > 0. \]

2) Laplace Transform of \( t^n \):

\[ L\{ t^n \} = \frac{n!}{s^{n+1}}, \quad n = 0, 1, 2, \dots \]

Proof: By the definition of Laplace transform, \[ L\{ f(t) \} = \int_{0}^{\infty} e^{-st} f(t) \, dt \] For \( f(t) = t^n \): \[ L\{ t^n \} = \int_{0}^{\infty} e^{-st} t^n \, dt \] Using integration by parts, let \( u = t^n \) and \( dv = e^{-st} dt \). Then: \[ du = n t^{n-1} dt, \quad v = \frac{e^{-st}}{-s} \] Applying integration by parts: \[ L\{ t^n \} = \left[ t^n \cdot \frac{e^{-st}}{-s} \right]_0^\infty - \int_{0}^{\infty} n t^{n-1} \cdot \frac{e^{-st}}{-s} \, dt \] Simplifying: \[ L\{ t^n \} = \frac{n}{s} \int_{0}^{\infty} t^{n-1} e^{-st} \, dt = \frac{n}{s} L\{ t^{n-1} \} \] By repeatedly applying this result \( n \) times: \[ L\{ t^n \} = \frac{n}{s} \cdot \frac{n-1}{s} \cdot \frac{n-2}{s} \cdots \frac{1}{s} L\{ 1 \} \] Since \( L\{ 1 \} = \frac{1}{s} \), we get: \[ L\{ t^n \} = \frac{n!}{s^{n+1}} \] For example,

  • if n=1 then \( L\{ t \} = \frac{1}{s^2} \)
  • if n=2 then \( L\{ t^2 \} = \frac{2}{s^3} \)
  • if n=3 then \( L\{ t^3 \} = \frac{6}{s^4} \)

3) Laplace Transform of \( e^{at} \):

\[ L\{ e^{at} \} = \frac{1}{s - a}, \quad s > a. \]

Proof: By the definition of Laplace transform, \[ L\{ e^{at} \} = \int_{0}^{\infty} e^{-st} e^{at} \, dt = \int_{0}^{\infty} e^{-t(s - a)} \, dt \] Evaluating the integral: \[ L\{ e^{at} \} = \left[ \frac{e^{-t(s - a)}}{-(s - a)} \right]_0^\infty = -\frac{1}{s - a} \left[ 0 - 1 \right]. \] Simplifying: \[ L\{ e^{at} \} = \frac{1}{s - a}, \quad s > a. \] Similarly, \[ L\{ e^{-at} \} = \frac{1}{s + a}, \quad s > -a. \]

4) Laplace Transform of \( \cos(at) \) and \( \sin(at) \):

\[ L\{ \cos(at) \} = \frac{s}{s^2 + a^2}, \quad L\{ \sin(at) \} = \frac{a}{s^2 + a^2} \]

Proof: Using Euler's formula, \( e^{iat} = \cos(at) + i \sin(at) \), and the Laplace transform of \( e^{iat} \): \[ L\{ e^{iat} \} = \frac{1}{s - ia} \] Rationalizing the denominator: \[ L\{ e^{iat} \} = \frac{s + ia}{(s - ia)(s + ia)} = \frac{s + ia}{s^2 + a^2} \] \[ L\{ \cos at + i \sin at \} = \frac{s}{s^2 + a^2}+i\frac{a}{s^2 + a^2} \] Equating real and imaginary parts: \[ L\{ \cos(at) \} = \frac{s}{s^2 + a^2}, \quad L\{ \sin(at) \} = \frac{a}{s^2 + a^2} \]

5) Laplace Transform of \( \cosh(at) \) and \( \sinh(at) \):

\[ L\{ \cosh(at) \} = \frac{s}{s^2 - a^2}, \quad L\{ \sinh(at) \} = \frac{a}{s^2 - a^2} \]

Proof: Using the definitions of hyperbolic functions: \[ \cosh(at) = \frac{e^{at} + e^{-at}}{2}, \quad \sinh(at) = \frac{e^{at} - e^{-at}}{2} \] Taking the Laplace transform of \( \cosh(at) \): \[ L\{ \cosh(at) \} = \frac{1}{2} \left[ L\{ e^{at} \} + L\{ e^{-at} \} \right] = \frac{1}{2} \left[ \frac{1}{s - a} + \frac{1}{s + a} \right]. \] Simplifying: \[ L\{ \cosh(at) \} = \frac{s}{s^2 - a^2} \] Similarly, for \( \sinh(at) \): \[ L\{ \sinh(at) \} = \frac{a}{s^2 - a^2} \]

First Shifting Theorem:

If \( L\{ f(t) \} = F(s) \), then: \[ L\{ e^{at} f(t) \} = F(s - a). \] Note: \( L\{ e^{-at} f(t) \} = F(s + a) \).

Change of Scale Property:

If \( L\{ f(t) \} = F(s) \), then: \[ L\{ f(at) \} = \frac{1}{a} F\left( \frac{s}{a} \right), \quad a > 0. \]

Application of First Shifting Theorem:

By the first shifting theorem: \[ L\{ e^{at} f(t) \} = \left[ L\{ f(t) \} \right]_{s \to (s - a)} = \left[ F(s) \right]_{s \to (s - a)} \]

Using this, we derive the following results:

  1. \[ L\{ e^{at} t^n \} = \left[ L\{ t^n \} \right]_{s \to (s - a)} = \left[ \frac{n!}{s^{n+1}} \right]_{s \to (s - a)} = \frac{n!}{(s - a)^{n+1}} \]
  2. \[ L\{ e^{at} \cos(bt) \} = \left[ L\{ \cos(bt) \} \right]_{s \to (s - a)} = \left[ \frac{s}{s^2 + b^2} \right]_{s \to (s - a)} = \frac{s - a}{(s - a)^2 + b^2} \]
  3. \[ L\{ e^{at} \sin(bt) \} = \left[ L\{ \sin(bt) \} \right]_{s \to (s - a)} = \left[ \frac{b}{s^2 + b^2} \right]_{s \to (s - a)} = \frac{b}{(s - a)^2 + b^2} \]
  4. \[ L\{ e^{at} \cosh(bt) \} = \left[ L\{ \cosh(bt) \} \right]_{s \to (s - a)} = \left[ \frac{s}{s^2 - b^2} \right]_{s \to (s - a)} = \frac{s - a}{(s - a)^2 - b^2} \]
  5. \[ L\{ e^{at} \sinh(bt) \} = \left[ L\{ \sinh(bt) \} \right]_{s \to (s - a)} = \left[ \frac{b}{s^2 - b^2} \right]_{s \to (s - a)} = \frac{b}{(s - a)^2 - b^2} \]

Example 1: Find the Laplace transforms of the following:

  1. \( 2t^3 + \cos(4t) + \sin(4t) + e^{-2t} \)

    2. Solution: \[ L\{ 2t^3 + \cos(4t) + \sin(4t) + e^{-2t} \} = 2L\{ t^3 \} + L\{ \cos(4t) \} + L\{ \sin(4t) \} + L\{ e^{-2t} \} \] Substituting the known Laplace transforms: \[ = 2 \cdot \frac{3!}{s^{3+1}} + \frac{s}{s^2 + 16} + \frac{4}{s^2 + 16} + \frac{1}{s + 2} \] Simplifying: \[ = \frac{12}{s^4} + \frac{s + 4}{s^2 + 16} + \frac{1}{s + 2} \]

  2. \( \sin(2t) \cos(3t) \)

    3. Solution: Using the trigonometric identity \( \sin(A) \cos(B) = \frac{1}{2} [\sin(A + B) + \sin(A - B)] \): \[ L\{ \sin(2t) \cos(3t) \} = L\left\{ \frac{\sin(5t) + \sin(-t)}{2} \right\} = \frac{1}{2} L\{ \sin(5t) - \sin(t) \} \] Substituting the known Laplace transforms: \[ = \frac{1}{2} \left[ \frac{5}{s^2 + 25} - \frac{1}{s^2 + 1} \right] \] Simplifying: \[ = \frac{2(s^2 - 5)}{(s^2 + 25)(s^2 + 1)} \]

  3. \( \cosh^3(2t) \)

    4. Solution: Using the identity \( \cosh^3(\theta) = \frac{\cosh(3\theta) + 3\cosh(\theta)}{4} \): \[ L\{ \cosh^3(2t) \} = L\left\{ \frac{\cosh(6t) + 3\cosh(2t)}{4} \right\} \] Substituting the known Laplace transforms: \[ = \frac{1}{4} \left[ \frac{s}{s^2 - 36} + \frac{3s}{s^2 - 4} \right]. \] Simplifying: \[ = \frac{s(s^2 - 28)}{(s^2 - 36)(s^2 - 4)} \]

  4. \( 3\cosh(2t) - 4\sinh(5t) \)

    5. Solution: \[ L\{ 3\cosh(2t) - 4\sinh(5t) \} = 3L\{ \cosh(2t) \} - 4L\{ \sinh(5t) \} \] Substituting the known Laplace transforms: \[ = 3 \cdot \frac{s}{s^2 - 4} - 4 \cdot \frac{5}{s^2 - 25} \] Simplifying: \[ = \frac{3s - 20}{s^2 - 25} \]

Example 2: Find the Laplace transforms of the following:

  1. \( t^2 e^{2t} \)

    2. Solution: Using the first shifting theorem: \[ L\{ t^2 e^{2t} \} = \left[ L\{ t^2 \} \right]_{s \to (s - 2)} = \left[ \frac{2!}{s^{2+1}} \right]_{s \to (s - 2)} = \frac{2}{(s - 2)^3} \]

  2. \( e^{2t} (3t^5 - \cos(4t)) \)

    3. Solution: Using the first shifting theorem: \[ L\{ e^{2t} (3t^5 - \cos(4t)) \} = \left[ L\{ 3t^5 - \cos(4t) \} \right]_{s \to (s - 2)} \] Substituting the known Laplace transforms: \[ = 3 \cdot \frac{5!}{s^6} - \frac{s}{s^2 + 16} \] Applying the shift: \[ = \frac{360}{(s - 2)^6} - \frac{s - 2}{(s - 2)^2 + 16} \]

  3. \( e^{-t} \sin^2(t) \)

    4. Solution: Using the identity \( \sin^2(t) = \frac{1 - \cos(2t)}{2} \): \[ L\{ e^{-t} \sin^2(t) \} = \left[ L\{ \sin^2(t) \} \right]_{s \to (s + 1)} \] Substituting the known Laplace transforms: \[ = \frac{1}{2} \left[ \frac{1}{s} - \frac{s}{s^2 + 4} \right]_{s \to (s + 1)} \] \[ = \frac{1}{2} \left[ \frac{1}{s+1} - \frac{s+1}{(s+1)^2 + 4} \right] \] Simplifying: \[ = \frac{2}{(s + 1)(s^2 + 2s + 5)} \]

Example 3: Find \( L\{ e^{-3t} \cosh(3t) \} \) using the change of scale property

Solution: Let \( f(t) = e^{-t} \cosh(t) \). Then: \[ L\{ f(t) \} = L\{ e^{-t} \cosh(t) \} = L\{ \cosh(t) \}_{s \to (s + 1)} = \left[ \frac{s}{s^2 - 1} \right]_{s \to (s + 1)} = \frac{s + 1}{s^2 + 2s} = F(s). \] By the change of scale property: \[ L\{ f(3t) \} = \frac{1}{3} F\left( \frac{s}{3} \right). \] Substituting \( F\left( \frac{s}{3} \right) \): \[ L\{ f(3t) \} = \frac{1}{3} \cdot \frac{\left( \frac{s}{3} + 1 \right)}{\left( \frac{s}{3} \right)^2 + 2 \left( \frac{s}{3} \right)} \] Simplifying: \[ L\{ e^{-3t} \cosh(3t) \} = \frac{s + 3}{s^2 + 6s} \]

Example 4: Find \( L\{ g(t) \} \), where

\[ g(t) = \begin{cases} \sin(t), & 0 < t < \pi, \\ 0, & t > \pi. \end{cases} \]

Solution: By the definition of the Laplace transform: \[ L\{ g(t) \} = \int_{0}^{\infty} e^{-st} g(t) \, dt = \int_{0}^{\pi} e^{-st} \sin(t) \, dt + \int_{\pi}^{\infty} e^{-st} \cdot 0 \, dt \] Evaluating the integral: \[ L\{ g(t) \} = \int_{0}^{\pi} e^{-st} \sin(t) \, dt \] Using integration by parts: \[ L\{ g(t) \} = \left[ \frac{e^{-st}}{s^2 + 1} (-s \sin(t) - \cos(t)) \right]_{0}^{\pi} \] Substituting the limits: \[ L\{ g(t) \} = \frac{e^{-s\pi}}{s^2 + 1} (-s \sin(\pi) - \cos(\pi)) - \frac{1}{s^2 + 1} (-s \sin(0) - \cos(0)). \] Simplifying: \[ L\{ g(t) \} = \frac{e^{-s\pi}}{s^2 + 1} (0 + 1) - \frac{1}{s^2 + 1} (0 - 1) = \frac{1 + e^{-s\pi}}{s^2 + 1} \]

If \( L\{ f(t) \} = F(s) \), then for: \[ g(t) = \begin{cases} f(t - a), & t > a, \\ 0, & t < a, \end{cases} \] The Laplace transform of \( g(t) \) is: \[ L\{ g(t) \} = e^{-as} F(s) = e^{-as} \left( \frac{1 + e^{-s\pi}}{s^2 + 1} \right) \]

Example 5: Find \( L\{ g(t) \} \), where

\[ g(t) = \begin{cases} \cos\left(t - \frac{2\pi}{3}\right), & t > \frac{2\pi}{3}, \\ 0, & t < \frac{2\pi}{3} \end{cases} \]

Solution: By the second shifting theorem: \[ L\{ g(t) \} = e^{-as} F(s), \] where \( a = \frac{2\pi}{3} \) and \( F(s) = L\{ \cos(t) \} = \frac{s}{s^2 + 1} \). Substituting: \[ L\{ g(t) \} = e^{-\frac{2\pi s}{3}} \cdot \frac{s}{s^2 + 1} \]

Model Problems

Find the Laplace transform of the following functions:

  1. \[ 3\cosh(5t) - 4\sinh(5t) \]
  2. \[ \sinh(3t) \cos^2(t) \]
  3. \[ e^{2t} + 4t^3 - 2\sin(3t) + 3\cos(3t) \]
  4. \[ \sin(2t) \sin(3t) \]
  5. \[ \cos^3(3t) \]
  6. \[ \sin^3(3t) \]
  7. \[ \cos(wt + \theta) \]
  8. \[ (\sin t - \cos t)^2 \]
  9. \[ e^{-3t} (2\cos(5t) - 3\sin(5t)) \]
  10. \[ e^{-2t} \cos^2(t) \]
  11. \[ \cosh(at) \sinh(at) \]
  12. \[ e^{t} \left( \cos(2t) + \frac{1}{2} \sinh(2t) \right) \]
  13. \[ f(t) = \begin{cases} 1, & 0 < t < 1, \\ t, & 1 < t < 2, \\ 0, & t > 2. \end{cases} \]
  14. \[ f(t) = |t - 1| + |t + 1|, \quad t \ge 0 \]
  15. Find \( L\{ f(3t) \} \), if \[ L\{ f(t) \} = \frac{9s^2 - 12s + 15}{(s - 1)^3} \]
  16. Find \( L\left\{ \frac{\sin(at)}{t} \right\} \), given that \[ L\left\{ \frac{\sin(t)}{t} \right\} = \tan^{-1}\left( \frac{1}{s} \right) \]
  17. If \( L\{ f(t) \} = \frac{1}{s} e^{-\frac{1}{s}} \), then prove that: \[ L\{ e^{-t} f(3t) \} = \frac{1}{s + 1} e^{-\frac{3}{s + 1}} \]

Answers

  1. \[ \frac{3s - 20}{s^2 - 25} \]
  2. \[ \frac{3}{2} \left( \frac{1}{S^2 - 9} + \frac{S^2 - 13}{S^4 - 10S^2 + 169} \right) \]
  3. \[ \frac{1}{s - 2} + \frac{24}{S^4} + \frac{3s - 6}{S^2 + 9} \]
  4. \[ \frac{12s}{(s^2 + 1)(s^2 + 25)} \]
  5. \[ \frac{s(s^2 + 63)}{(s^2 + 9)(s^2 + 81)} \]
  6. \[ \frac{48}{(s^2 - 4)(s^2 - 36)} \]
  7. \[ \frac{s \cos \theta - w \sin \theta}{s^2 + w^2} \]
  8. \[ \frac{s^2 - 4s + 4}{s(s^2 + 4)} \]
  9. \[ \frac{2s - 9}{s^2 + 6s + 34} \]
  10. \[ \frac{1}{2} \left[ \frac{1}{s - 2} + \frac{s - 2}{S^2 - 4s + 7} \right] \]
  11. \[ \frac{1}{2} \left[ \frac{b}{(s - 2)^2 - b^2} + \frac{b}{(s + 2)^2 - b^2} \right] \]
  12. \[ \frac{s - 1}{s^2 - 2s + 3} + \frac{1}{s^2 - 2s - 5} \]
  13. \[ \frac{1 - 2e^{-2s}}{s} + \frac{e^{-s} - e^{-2s}}{s^2} \]
  14. \[ \frac{2}{s^2} (s + e^{-s}) \]
  15. \[ \frac{9(s^2 - 4s + 15)}{(s - 3)^2} \]
  16. \[ \tan^{-1} \left( \frac{a}{s} \right) \]

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