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27-06 — Laplace Transforms

Phase: Ordinary Differential Equations | Subject: 27-06 Prerequisites: 27-05-systems-of-odes.md, 05-02-integration-techniques.md Next subject: 27-07-series-solutions.md

Learning Objectives

By the end of this subject, you will be able to:

  1. Define the Laplace transform and compute transforms of elementary functions
  2. Apply the key properties: linearity, shifting, derivatives, and convolution
  3. Use Laplace transforms to solve linear ODEs with initial conditions
  4. Compute inverse Laplace transforms using partial fractions and tables
  5. Handle discontinuous forcing functions via the unit step function

Core Content

Definition of the Laplace Transform

The Laplace transform converts a function $f(t)$ (time domain) to $F(s)$ (complex frequency domain):

$$\mathcal{L}{f(t)} = F(s) = \int_0^\infty e^{-st} f(t) \, dt$$

where $s$ is a complex variable. The transform exists for $f(t)$ that are piecewise continuous and of exponential order (i.e., $|f(t)| \leq Me^{ct}$ for some $M, c$).

Key insight: Differentiation in the time domain becomes multiplication in the $s$-domain — turning ODEs into algebraic equations.

Table of Basic Transforms

$f(t)$ $F(s) = \mathcal{L}{f(t)}$ Region of Convergence
$1$ $\frac{1}{s}$ $\operatorname{Re}(s) > 0$
$t^n$ $\frac{n!}{s^{n+1}}$ $\operatorname{Re}(s) > 0$
$e^{at}$ $\frac{1}{s-a}$ $\operatorname{Re}(s) > a$
$\sin(\omega t)$ $\frac{\omega}{s^2 + \omega^2}$ $\operatorname{Re}(s) > 0$
$\cos(\omega t)$ $\frac{s}{s^2 + \omega^2}$ $\operatorname{Re}(s) > 0$
$\sinh(\omega t)$ $\frac{\omega}{s^2 - \omega^2}$ $\operatorname{Re}(s) >
$\cosh(\omega t)$ $\frac{s}{s^2 - \omega^2}$ $\operatorname{Re}(s) >
$t e^{at}$ $\frac{1}{(s-a)^2}$ $\operatorname{Re}(s) > a$
$e^{at}\sin(\omega t)$ $\frac{\omega}{(s-a)^2 + \omega^2}$ $\operatorname{Re}(s) > a$
$e^{at}\cos(\omega t)$ $\frac{s-a}{(s-a)^2 + \omega^2}$ $\operatorname{Re}(s) > a$
$\delta(t-a)$ (Dirac delta) $e^{-as}$ all $s$

Key Properties

1. Linearity: $$\mathcal{L}{a f(t) + b g(t)} = a F(s) + b G(s)$$

2. First Shifting Theorem (s-shift): $$\mathcal{L}{e^{at} f(t)} = F(s - a)$$

3. Second Shifting Theorem (t-shift): If $u_c(t)$ is the unit step at $t=c$: $$\mathcal{L}{u_c(t) f(t-c)} = e^{-cs} F(s)$$

4. Transform of Derivatives: $$\mathcal{L}{f'(t)} = sF(s) - f(0)$$ $$\mathcal{L}{f''(t)} = s^2 F(s) - s f(0) - f'(0)$$ $$\mathcal{L}{f^{(n)}(t)} = s^n F(s) - s^{n-1}f(0) - s^{n-2}f'(0) - \cdots - f^{(n-1)}(0)$$

⚠️ CRITICAL: The derivative property is what makes Laplace transforms so powerful for IVPs — initial conditions are automatically incorporated. But you MUST use the correct initial values at $t=0$ (or $t=0^+$ for problems with impulses).

5. Transform of Integrals: $$\mathcal{L}\left{\int_0^t f(\tau) \, d\tau\right} = \frac{F(s)}{s}$$

6. Convolution Theorem: $$(f * g)(t) = \int_0^t f(\tau) g(t-\tau) \, d\tau$$ $$\mathcal{L}{(f * g)(t)} = F(s) G(s)$$

This is crucial: convolution in time = multiplication in $s$-domain.

Solving ODEs with Laplace Transforms

The method in four steps:

  1. Take the Laplace transform of both sides of the ODE
  2. Use the derivative property to express $\mathcal{L}{y}$ in terms of $s$ and initial conditions
  3. Solve the resulting algebraic equation for $Y(s) = \mathcal{L}{y(t)}$
  4. Take the inverse Laplace transform to recover $y(t)$

Example: Solve $y'' + 4y = 0$, $y(0) = 1$, $y'(0) = 0$.

  1. $\mathcal{L}{y''} + 4\mathcal{L}{y} = 0$
  2. $[s^2 Y(s) - s y(0) - y'(0)] + 4Y(s) = 0$
  3. $s^2 Y(s) - s(1) - 0 + 4Y(s) = 0 \Rightarrow Y(s)(s^2 + 4) = s \Rightarrow Y(s) = \frac{s}{s^2+4}$
  4. $y(t) = \mathcal{L}^{-1}\left{\frac{s}{s^2+4}\right} = \cos(2t)$

Inverse Laplace Transform

Computed using: - Table lookup: match $F(s)$ to known forms - Partial fractions: decompose rational $F(s)$ into simpler terms - Completing the square: for quadratic denominators with no real roots - Convolution: when $F(s) = G(s)H(s)$

Discontinuous and Impulse Forcing

Unit step function (Heaviside):

$$u_c(t) = \begin{cases} 0, & t < c \ 1, & t \geq c \end{cases}$$

$$\mathcal{L}{u_c(t)} = \frac{e^{-cs}}{s}$$

Dirac delta function $\delta(t-a)$: an idealized impulse of unit area at $t=a$. $$\mathcal{L}{\delta(t-a)} = e^{-as}$$

This models instantaneous forces (hammer blows, electrical spikes).

Applications

Key Terms

Worked Examples

Example 1: Basic Transform

Compute $\mathcal{L}{t^2 e^{3t}}$.

Solution: Method 1 — using the first shifting theorem: $\mathcal{L}{t^2} = \frac{2!}{s^3} = \frac{2}{s^3}$. Then $\mathcal{L}{e^{3t} t^2} = \frac{2}{(s-3)^3}$.

Method 2 — from the table: $\mathcal{L}{t^n e^{at}} = \frac{n!}{(s-a)^{n+1}}$, with $n=2, a=3$ gives $\frac{2}{(s-3)^3}$.

Click for answer $\mathcal{L}\{t^2 e^{3t}\} = \frac{2}{(s-3)^3}$, valid for $\operatorname{Re}(s) > 3$.

Example 2: Solve an IVP

Solve $y'' - 3y' + 2y = e^{4t}$, with $y(0) = 1$, $y'(0) = 0$.

Solution:

  1. Take Laplace transform: $[s^2 Y - s(1) - 0] - 3[sY - 1] + 2Y = \frac{1}{s-4}$

  2. Simplify: $s^2 Y - s - 3sY + 3 + 2Y = \frac{1}{s-4}$ $Y(s^2 - 3s + 2) = s - 3 + \frac{1}{s-4}$

  3. Solve for $Y$: $Y = \frac{s-3}{s^2-3s+2} + \frac{1}{(s-4)(s^2-3s+2)}$ Factor: $s^2-3s+2 = (s-1)(s-2)$

  4. Partial fractions on first term: $\frac{s-3}{(s-1)(s-2)} = \frac{2}{s-1} - \frac{1}{s-2}$ Partial fractions on second term: $\frac{1}{(s-4)(s-1)(s-2)} = \frac{1/6}{s-4} - \frac{1/3}{s-1} + \frac{1/2}{s-2}$

  5. Combine: $Y = \left(\frac{2}{s-1} - \frac{1}{s-2}\right) + \left(\frac{1/6}{s-4} - \frac{1/3}{s-1} + \frac{1/2}{s-2}\right)$ $Y = \frac{5/3}{s-1} - \frac{1/2}{s-2} + \frac{1/6}{s-4}$

  6. Inverse: $y(t) = \frac{5}{3}e^t - \frac{1}{2}e^{2t} + \frac{1}{6}e^{4t}$

Verify: $y(0) = \frac{5}{3} - \frac{1}{2} + \frac{1}{6} = \frac{10-3+1}{6} = \frac{8}{6} = \frac{4}{3}$... wait, that doesn't match $y(0)=1$. Let's recheck.

Actually, let's recompute more carefully: $Y[(s-1)(s-2)] = s-3 + \frac{1}{s-4}$ $Y = \frac{s-3}{(s-1)(s-2)} + \frac{1}{(s-4)(s-1)(s-2)}$

$\frac{s-3}{(s-1)(s-2)} = \frac{A}{s-1} + \frac{B}{s-2}$. Multiply: $s-3 = A(s-2) + B(s-1)$. At $s=1$: $1-3 = A(1-2) \Rightarrow -2 = -A \Rightarrow A=2$. At $s=2$: $2-3 = B(2-1) \Rightarrow -1 = B$. So $\frac{s-3}{(s-1)(s-2)} = \frac{2}{s-1} - \frac{1}{s-2}$.

$\frac{1}{(s-4)(s-1)(s-2)} = \frac{A}{s-4} + \frac{B}{s-1} + \frac{C}{s-2}$. $1 = A(s-1)(s-2) + B(s-4)(s-2) + C(s-4)(s-1)$. $s=4$: $1 = A(3)(2) = 6A \Rightarrow A = 1/6$. $s=1$: $1 = B(-3)(-1) = 3B \Rightarrow B = 1/3$. $s=2$: $1 = C(-2)(1) = -2C \Rightarrow C = -1/2$.

$Y = \frac{2}{s-1} - \frac{1}{s-2} + \frac{1/6}{s-4} + \frac{1/3}{s-1} - \frac{1/2}{s-2}$ $Y = \frac{7/3}{s-1} - \frac{3/2}{s-2} + \frac{1/6}{s-4}$

$y(t) = \frac{7}{3}e^t - \frac{3}{2}e^{2t} + \frac{1}{6}e^{4t}$

Check $y(0) = \frac{7}{3} - \frac{3}{2} + \frac{1}{6} = \frac{14-9+1}{6} = \frac{6}{6} = 1$. ✓

Click for answer $y(t) = \frac{7}{3}e^t - \frac{3}{2}e^{2t} + \frac{1}{6}e^{4t}$

Example 3: Step Function Forcing

Solve $y'' + y = u_\pi(t)$, with $y(0)=0, y'(0)=0$, where $u_\pi$ is the unit step at $t=\pi$.

Solution:

  1. Transform: $s^2 Y + Y = \frac{e^{-\pi s}}{s}$
  2. $Y = \frac{e^{-\pi s}}{s(s^2+1)}$
  3. Partial fractions: $\frac{1}{s(s^2+1)} = \frac{1}{s} - \frac{s}{s^2+1}$
  4. So $Y = e^{-\pi s}\left(\frac{1}{s} - \frac{s}{s^2+1}\right)$

  5. By the second shifting theorem: $\mathcal{L}^{-1}{e^{-\pi s} \cdot \frac{1}{s}} = u_\pi(t) \cdot 1$ $\mathcal{L}^{-1}{e^{-\pi s} \cdot \frac{s}{s^2+1}} = u_\pi(t) \cos(t-\pi) = u_\pi(t)(-\cos t)$ since $\cos(t-\pi) = -\cos t$.

  6. $y(t) = u_\pi(t)[1 - (-\cos t)] = u_\pi(t)[1 + \cos t]$

Interpretation: $y = 0$ for $t < \pi$, then $y = 1 + \cos t$ for $t \geq \pi$.

Click for answer $y(t) = \begin{cases} 0, & t < \pi \\ 1 + \cos t, & t \geq \pi \end{cases}$ The system is at rest until $t=\pi$, when a unit step force is applied.

Quiz

Q1: What does the concept of Laplace transform primarily refer to in this subject?

A) The definition and application of Laplace transform B) A historical anecdote about Laplace transform C) A computational error related to Laplace transform D) A visual representation of Laplace transform

Correct: A)

Q2: Which of the following is the key formula discussed in this subject?

A) The inverse operation of the formula in question B) An unrelated formula from a different topic C) \mathcal{L}\{f(t)\} = F(s) = \int_0^\infty e^{-st} f(t) \, dt D) A simplified version of \mathcal{L}\{f(t)\} = F(s) ...

Correct: C)

Q3: What is the primary purpose of Dirac delta function?

A) It is primarily a historical notation system B) It replaces all other methods in this domain C) It is used to dirac delta function in mathematical analysis D) It is used only in advanced research contexts

Correct: C)

Q4: Which statement about Inverse Laplace Transform is TRUE?

A) Inverse Laplace Transform is not related to this subject B) Inverse Laplace Transform is an advanced topic beyond this subject's scope C) Inverse Laplace Transform is a fundamental concept covered in this subject D) Inverse Laplace Transform is mentioned only as a historical footnote

Correct: C)

Q5: Based on the worked examples in this subject, what is the correct result?

A) A different result from a common mistake B) An unrelated numerical value C) 0$. D) The inverse of the correct answer

Correct: C)

Q6: How are Inverse Laplace Transform and Partial Fraction Decomposition related?

A) Inverse Laplace Transform is the inverse of Partial Fraction Decomposition B) Inverse Laplace Transform and Partial Fraction Decomposition are closely related concepts C) Inverse Laplace Transform and Partial Fraction Decomposition are completely unrelated topics D) Inverse Laplace Transform is a special case of Partial Fraction Decomposition

Correct: B)

Q7: What is a common pitfall when working with First Shifting Theorem?

A) First Shifting Theorem is always computed the same way in all contexts B) The main error with First Shifting Theorem is using it when it is not needed C) First Shifting Theorem has no common misconceptions D) A common mistake is confusing First Shifting Theorem with a similar concept

Correct: D)

Q8: When should you apply Second Shifting Theorem?

A) Avoid Second Shifting Theorem unless explicitly instructed B) Apply Second Shifting Theorem to solve problems in this subject's domain C) Second Shifting Theorem is not practically useful D) Use Second Shifting Theorem only in pure mathematics contexts

Correct: B)

Practice Problems

  1. Find $\mathcal{L}{e^{-2t} \cos(3t)}$.

    Click for answer Using first shifting theorem: $\mathcal{L}\{\cos(3t)\} = \frac{s}{s^2+9}$, so $\mathcal{L}\{e^{-2t}\cos(3t)\} = \frac{s+2}{(s+2)^2+9}$.

  2. Find $\mathcal{L}^{-1}\left{\frac{1}{s^2 + 4s + 13}\right}$.

    Click for answer Complete the square: $s^2 + 4s + 13 = (s+2)^2 + 9$. So $F(s) = \frac{1}{(s+2)^2+9} = \frac{1}{3} \cdot \frac{3}{(s+2)^2+9}$. Inverse: $\frac{1}{3}e^{-2t}\sin(3t)$.

  3. Solve $y'' + 4y' + 4y = 0$, $y(0)=1$, $y'(0)=2$.

    Click for answer $s^2Y - s(1) - 2 + 4(sY-1) + 4Y = 0$ → $Y(s^2+4s+4) = s+6$ → $Y = \frac{s+6}{(s+2)^2} = \frac{1}{s+2} + \frac{4}{(s+2)^2}$ → $y(t) = e^{-2t} + 4t e^{-2t}$.

  4. Express $f(t) = \begin{cases} 0, & t<2 \ t-2, & t\geq 2 \end{cases}$ using unit step and find its Laplace transform.

    Click for answer $f(t) = u_2(t)(t-2)$. $\mathcal{L}\{u_2(t)(t-2)\} = e^{-2s}\mathcal{L}\{t\} = \frac{e^{-2s}}{s^2}$.

  5. Use convolution to find $\mathcal{L}^{-1}\left{\frac{1}{(s^2+1)^2}\right}$.

    Click for answer $\frac{1}{(s^2+1)^2} = \frac{1}{s^2+1} \cdot \frac{1}{s^2+1}$. Convolution: $\sin t * \sin t = \int_0^t \sin(\tau)\sin(t-\tau) d\tau$. Using trig identity: $= \frac{1}{2}[\sin t - t\cos t]$.

Summary

Key takeaways:

Pitfalls

Next Steps

Next up: 27-07-series-solutions.md — power series methods for ODEs when closed-form solutions don't exist.