import sympy as sym
from sympy.abc import s,t
# Just run this code cell. Do Not Edit.
# Define the Laplace transform function
def L(f):
return sym.laplace_transform(f, t, s)
# Define the inverse Laplace transform function
def invL(F):
return sym.inverse_laplace_transform(F, s, t)4.3: Inverse Laplace Transforms
Reading: Notes on Diffy Q’s Section 6.1 Part B
Inverse Operations
When analyzing a differential equation, it is important to read and interpret the differential equation in practical terms when the variables represent physical quantities. For example, let \(s(t)\) denotes the position (measured in cm. from its initial position) of a particle \(t\) seconds from now. If the particle’s position is modeled by the differential equation
\[s' - 200\sin{t} = 0 \qquad s(0) = 0,\]
then we know the particle’s velocity is oscillating over time.
- Initially the particle is at rest (\(s'(0)=0\)).
- Then the particle moves to the right (\(s'(t) > 0\)).
- The particle stops at \(t=\frac{\pi}{2}\) seconds, and turns around.
- The particle then moves to the left (when \(s'(t) < 0\)) for a bit before turning around at \(t=\pi\) seconds
- And so on.
Let \(y=s(t)\) denote the position of a particle at time \(t\). We can find its velocity by applying the derivative, which maps a function to its rate of change.
In the case of the differential equation above, we would like to go in the opposite direction. Given a formula for the derivative \(s'\), we would like to find its corresponding position \(s(t)\). To solve the differential equation, we can:
- Apply the inverse operation (integration) to both sides.
\[ \int (s'(t) - 200\sin{t}) \, dt = \int 0 \, dt \quad \longrightarrow \quad s(t) + 200 \cos{t} = C\]
- Then solve for \(s(t)\) using the initial position and algebra.
\[s(t) = -200 \cos{t} + 200\]
Similar to the derivative, the Laplace transform is a mapping of one function \(f(t)\) to another function, its Laplace transform \(\color{dodgerblue}{F(s)= \mathscr{L} \left\{ f \right\}}\). In Worksheet 23 (the next worksheet), we will explore how Laplace transforms help us solve differential equations. Before focusing our efforts back on differential equations, in this notebook we will practice transforming functions in the opposite direction with the inverse Laplace transform.
Reviewing the Laplace Transform
So far we have explored Laplace transforms in Worksheet 20 and some useful properties of Laplace transforms in Worksheet 21. The main results of these investigations are summarized in Appendix A and Appendix B.
Before defining the inverse Laplace transform and applying this concept to solve differential equations, let’s first refresh ourselves of key properties of the Laplace transform that will be particularly useful when solving differential equations.
For example, Property 1 and Property 2 tell us that the Laplace transform is a linear operator,
\[\color{dodgerblue}{\mathscr{L} \left\{ c_1 f_1(t) + c_2 f_2(t) \right\} = c_1 \mathscr{L} \left\{ f_1(t) \right\} + c_2 \mathscr{L} \left\{ f_2(t) \right\}}.\]
Question 1:
Describe properties 3 and 5 in practical terms.
Solution to Question 1:
Property 3 : Multiplying \(f(t)\) by \(e^{at}\) and then taking the Laplace transform has what kind of effect on \(\mathscr{L} \left\{ f(t) \right\}\)?
Explain in practicel terms. For example, \(\mathscr{L} \left\{ e^{at} f(t) \right\}\) can be described as a transformation (shift, compression, and/or reflection) of \(F(s)=\mathscr{L} \left\{ f(t) \right\}\).
Property 5 : Multiplying \(f(t)\) by \(t^n\) and then taking the Laplace transform has what kind of effect on \(\mathscr{L} \left\{ f(t) \right\}\)?
Explain in practicel terms. For example, \(\mathscr{L} \left\{ t^n f(t) \right\}\) is equal to applying what operation(s) to \(F(s)=\mathscr{L} \left\{ f(t) \right\}\).
Solving Differential Equations with Laplace Transforms
One useful problem solving technique in mathematics is to:
- transform a difficult problem into an easier one,
- solve the easier problem, and
- then use its solution to obtain a solution of the original problem.
Let’s apply this technique to differential equations by applying the following steps:
Step 1: Apply the Laplace transform of both sides of the differential equation.
- This transforms the differential equation to an algebraic equation.
Step 2: Rearrange and group like terms to solve for \(\mathscr{L}\{y(t)\}=Y(s)\).
- Solving an algebraic equation is generally easier than a differential equation.
Step 3: Take the inverse Laplace transform and solve for \(y(t) = \mathscr{L}^{-1}\{Y(x) \}\).
- Undo the first step by applying the inverse Laplace transform to obtain a solution to the original differential equation.
Before we focus our attention on the inverse Laplace transform in Step 3, let’s first demonstrate how this three step process can be put into practice with the example in Question 2 below.
Question 2:
Solve \(y''-y=-t\) with \(y(0)=0\) and \(y'(0)=1\).
Note
In Step 3, we will apply the inverse Laplace transform to \(Y(s)\) in order to identify \(y(t) = \mathscr{L}^{-1} \{ Y(s) \}\).
Step 1: Apply the Laplace Transform to Both Sides
Using the properties, apply the Laplace transform to both sides:
\[ \mathscr{L} \{ y'' -y \} = \mathscr{L} \{ -t \} .\]
Solution to Question 2 Step 1:
Step 2: Solve for \(Y(s) = \mathscr{L} \{ y(t) \}\)
Using your result from Step 1, solve for \(\mathscr{L} \{ y(t) \}=Y(s)\).
Solution to Question 2 Step 2:
Step 3: Apply the Inverse Laplace Transform
Use the table of common Laplace transforms to identify a function \(y(t)\) that has \(\mathscr{L} \{ y(t) \}=Y(s)\).
Solution to Question 2 Step 3:
The Inverse Laplace Transform
Given \(F(s)\), if there is a function \(f(t)\) that is continuous on \(\lbrack 0 , \infty )\) and satisfies \(\mathscr{L} \{ f \} = F(s)\), then we say \(f(t)\) is the inverse Laplace transform of \(F(s)\) which is denoted by
\[\color{dodgerblue}{\large f(t) = \mathscr{L}^{-1} \{ F(s) \} }.\]
Question 3:
In parts (a)-(d), find the inverse Laplace transform of the function \(F(s)\).
Tip
First decide whether \(f(t)\) is of the form \(t^n\), \(\cos{(bt)}\), \(\sin{(bt)}\), or \(e^{at}\).
Question 3a:
\(\displaystyle F(s) = \frac{1}{s^2}\)
Solution to Question 3a:
Question 3b:
\(\displaystyle F(s) = \frac{2}{s^2+4}\)
Solution to Question 3b:
Question 3c:
\(\displaystyle F(s) = \frac{4s}{s^2+9}\)
Solution to Question 3c:
Question 3d:
\(\displaystyle F(s) = \frac{2}{s+6}\)
Solution to Question 3d:
Inverse Laplace Transforms with SymPy
We can use the sympy.inverse_laplace_transform(F, s, t) function from the SymPy library to check our work.
- We use
sas the variable in the resulting transform \(F(s)\). - We use
tas the variable in \(f(t)\). - The function will return the inverse Laplace transform (if it exists) \(f(t)\).
Below are a couple of additional resources for help with Laplace transforms with SymPy.
- For more examples with
sympy.inverse_laplace_transform(), here’s a good resource. - For official documentation, see SymPy Doc.
- For a review of Laplace transforms using
sympy.laplace_transform(), see the section on Computing Laplace Transforms in Python from Worksheet 20.
Question 4:
Using the inverse Laplace transform function invL(F) to verify your answers in Question 3. Below are the formulas from parts (a)-(d) from Question 3
\(\displaystyle F(s) = \frac{1}{s^2}\)
\(\displaystyle F(s) = \frac{2}{s^2+4}\)
\(\displaystyle F(s) = \frac{4s}{s^2+9}\)
\(\displaystyle F(s) = \frac{2}{s+6}\)
Solution to Question 4:
In each of the four code cells below, replace the ?? in the code cell below with an appropriate expression to verify your each of your four answers to each part of Question 3. Be sure you have run the previous code cell to define invL(F) before trying to run the code cells below.
Solution to Question 4a:
#############################################################
# STUDENT TO DO:
# Replace the ?? with an approrpriate expression
#############################################################
F = ?? # define your function with respect to s
invL(F)Solution to Question 4b:
#############################################################
# STUDENT TO DO:
# Replace the ?? with an approrpriate expression
#############################################################
F = ?? # define your function with respect to s
invL(F)Solution to Question 4c:
#############################################################
# STUDENT TO DO:
# Replace the ?? with an approrpriate expression
#############################################################
F = ?? # define your function with respect to s
invL(F)Solution to Question 4d:
#############################################################
# STUDENT TO DO:
# Replace the ?? with an approrpriate expression
#############################################################
F = ?? # define your function with respect to s
invL(F)What’s the Deal with \(\theta(t)\)?
The Heaviside function is denoted by \(\theta(t)\) in the output above. The Heaviside function is defined by
\[\color{dodgerblue}{\large \theta(t) = \left\{ \begin{array}{ll} 0, & t < 0 \\ 1, & t \geq 0 \end{array} \right.}\]
Run the code cell below to plot the graph of the Heaviside funtion.
sym.plot(sym.Heaviside(t));Heaviside Function and Inverse Laplace Transforms
The Heaviside function can be thought of as a signal that switches a function on by default at time \(t=0\). \(\theta(t)\) is useful when expressing inverse Laplace transforms since we only consider the (forward) Laplace transform of \(f(t)\) on the domain \(t \geq 0\).
For conciseness, we typically write \(\mathscr{L}^{-1} \left\{ \frac{2}{s+6} \right\} = 2e^{-6t}\) which has domain \((-\infty, \infty)\).
- However, we only consider \(\mathscr{L} \left\{ 2e^{-6t} \right\}\) on the interval \(t \leq 0\).
- Thus, \(f(t) = \mathscr{L}^{-1} \left\{ F(s) \right\}\) should only be considered over the interval \(t \geq 0\).
- We can mulitply by the Heaviside function to “turn the function on” at \(t=0\).
\[\color{dodgerblue}{\large f(t) \theta(t) = \left\{ \begin{array}{ll} 0, & t < 0 \\ 2e^{-6t}, & t \geq 0 \end{array} \right.}\]
For conciseness, when typing or writing inverse Laplace transform we simply write \(\mathbf{f(t) = 2e^{-6t}}\), and do not also write that the function is \(0\) when \(t <0\).
Run the code cell below to plot the graph of \(y=2e^{-6t}\theta(t)\).
# Modified from
# https://dynamics-and-control.readthedocs.io/en/latest/1_Dynamics/3_Linear_systems/Laplace%20transforms.html
a = sym.symbols('a')
F = 2/(s + 6)
f = 2*sym.exp(-6*t)
p = sym.plot(f.subs({a: 0}), invL(F).subs({a: 0}),
xlim=(-0.5, 1), ylim=(0, 3), show=False)
p[1].line_color = 'red'
p.show()Question 5:
Find the inverse Laplace transform of \(\displaystyle F(s) = \frac{s+2}{s^2+4s+11}\) by answering the questions below.
Question 5a:
Complete the square for the expression in the denominator of \(F(s)\) to find possible values for \(a\) and \(b\) such that \(s^2+4s+11=(s-a)^2+b\).
Solution to Question 5a:
Question 5b:
Use the table of common Laplace transforms to identify \(\mathscr{L}^{-1} \{ F(s)\}\).
Solution to Question 5b:
Question 6:
Using the inverse Laplace transform function invL(F) to verify your answer in Question 5.
Solution to Question 6:
Replace the ?? in the code cell below with an appropriate expression to check your previous answer.
#############################################################
# STUDENT TO DO:
# Replace the ?? with an approrpriate expression
#############################################################
F = ?? # define your function with respect to s
invL(F)Question 7:
Find the inverse Laplace transform of \(\displaystyle F(s) = \frac{5s-10}{s^2-3s-4}\).
Solution to Question 7:
Question 8:
Find the inverse Laplace transform of \(\displaystyle F(s) = \frac{3s-15}{2s^2-4s+10}\).
Solution to Question 8:
Question 9:
Find the inverse Laplace transform of \(\displaystyle F(s) = \frac{-5s-36}{(s+2)(s^2+9)}\).
Solution to Question 9:
Question 10:
Using the inverse Laplace transform function invL(F) to verify your answers in Question 7, Question 8, and Question 9.
Solution to Question 10:
Replace the ?? in each of the three code cells below with an appropriate expression to check your previous answers.
Solution to Question 10a: Checking Solution to Question 7
Find the inverse Laplace transform of \(\displaystyle F(s) = \frac{5s-10}{s^2-3s-4}\).
#############################################################
# STUDENT TO DO:
# Replace the ?? with an approrpriate expression
#############################################################
F = ?? # define your function with respect to s
invL(F)Solution to Question 10b: Checking Solution to Question 8
Find the inverse Laplace transform of \(\displaystyle F(s) = \frac{3s-15}{2s^2-4s+10}\).
#############################################################
# STUDENT TO DO:
# Replace the ?? with an approrpriate expression
#############################################################
F = ?? # define your function with respect to s
invL(F)Solution to Question 10c: Checking Solution to Question 9
Find the inverse Laplace transform of \(\displaystyle F(s) = \frac{-5s-36}{(s+2)(s^2+9)}\).
#############################################################
# STUDENT TO DO:
# Replace the ?? with an approrpriate expression
#############################################################
F = ?? # define your function with respect to s
invL(F)Appendix A: Common Laplace Transforms
| \(\displaystyle \large f(t)\) | \(\displaystyle \large F(s) = \mathscr{L} \left\{ f(t) \right\}\) |
|---|---|
| \(\displaystyle \large f(t)=1\) | \(\displaystyle \large F(s)=\frac{1}{s}, \ s > 0\) |
| \(\displaystyle\large f(t)=e^{at}\) | \(\displaystyle \large F(s) = \frac{1}{s-a}, \ s > a\) |
| \(\displaystyle \large f(t)=t^n, \ n=1,2, \ldots\) | \(\displaystyle \large F(s) = \frac{n!}{s^{n+1}}, \ s > 0\) |
| \(\displaystyle \large f(t)=\sin{(bt)}\) | \(\displaystyle \large F(s) = \frac{b}{s^2+b^2}, \ s > 0\) |
| \(\displaystyle \large f(t)=\cos{(bt)}\) | \(\displaystyle \large F(s) = \frac{s}{s^2+b^2}, \ s > 0\) |
| \(\displaystyle \large e^{at}t^n, \ n=1,2, \ldots\) | \(\displaystyle \large F(s) = \frac{n!}{(s-a)^{n+1}}, \ s > a\) |
| \(\displaystyle \large e^{at}\sin{(bt)}\) | \(\displaystyle \large F(s) = \frac{b}{(s-a)^2+b^2}, \ s > a\) |
| \(\displaystyle \large e^{at}\cos{(bt)}\) | \(\displaystyle \large F(s) = \frac{s-a}{(s-a)^2+b^2}, \ s > a\) |
Appendix B: Properties of Laplace Transforms
1. \(\color{dodgerblue}{\mathscr{L} \left\{ cf(t) \right\} = c \mathscr{L} \left\{ f(t) \right\}}\), where \(c\) is a constant.
2. \(\color{dodgerblue}{\mathscr{L} \left\{ f_1(t) + f_2(t) \right\} = \mathscr{L} \left\{ f_1(t) \right\} + \mathscr{L} \left\{ f_2(t)\right\}}\).
3. If \(F(s) = \mathscr{L} \left\{ f(t) \right\}\) exists for all \(s > \alpha\), then \(\color{dodgerblue}{\displaystyle \mathscr{L} \left\{ e^{at} f(t) \right\} = F(s-a)}\) for all \(s>\alpha + a\).
4. If \(F(s) =\mathscr{L} \left\{ f(t) \right\}\) exists for all \(s > \alpha\), then for all \(s>\alpha\),
\[\color{dodgerblue}{\mathscr{L} \left\{ f^{(n)}(t) \right\} = s^n \mathscr{L} \{ f(t) \}-s^{n-1} f(0)- s^{n-2} f'(0) - \ldots - f^{(n-1)}(0)}.\]
5. If \(F(s) =\mathscr{L} \left\{ f(t) \right\}\) exists for all \(s > \alpha\), then
\[\color{dodgerblue}{\mathscr{L} \left\{ t^n f(t) \right\} = (-1)^n \frac{d^nF}{ds^n} \mbox{ for all } s > \alpha}.\]
Creative Commons License Information
Exploring Differential Equations by Adam Spiegler is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
Based on a work at https://github.com/CU-Denver-MathStats-OER/ODEs and original content created by Rasmussen, C., Keene, K. A., Dunmyre, J., & Fortune, N. (2018). Inquiry oriented differential equations: Course materials. Available at https://iode.sdsu.edu.