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The Ordinary Differential Equations Project Instructor’s Version

Thomas W. Judson

Section 6.1 The Laplace Transform

Consider the electrical circuit governed by the differential equation
\begin{equation*} L I'' + RI' + \frac{1}{C} I = E'(t). \end{equation*}
The voltage function, \(E'(t)\text{,}\) might have discontinuities. For example, the voltage in the circuit can be periodically turned on and off. The previous methods that we have used to solve second order linear differential equations may not apply here. However, the Laplace transform, an integral transform, gives a method of solving such equations.
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As a second example, let us consider a population of fish that is governed by exponential growth,
\begin{align*} \frac{dP}{dt} & = kP\\ P(0) & = P_0, \end{align*}
and suppose that we wish to determine the effects of seasonal fishing. In other words, harvesting will not be continuous. For example, we might only allow fishing at a constant rate \(r\) during the first half of the year,
\begin{equation*} H(t) = \begin{cases} r & 0 \leq t \leq 1/2, \\ 0 & 1/2 \lt t \lt 1, \\ r & 1 \leq t \leq 3/2, \\ 0 & 3/2 \lt t \lt 2, \\ & \vdots \end{cases} \end{equation*}
Our initial value problem now becomes
\begin{align*} \frac{dP}{dt} & = kP - H\\ P(0) & = P_0, \end{align*}
It should be clear that we need some additional tools to analyze differential equations possessing discontinuous terms.
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Given an initial value problem
\begin{align*} ay'' + by' + cy & = g(t)\\ y(0) & = y_0\\ y'(0) & = y_0', \end{align*}
the idea is to use the Laplace transform to change the differential equation into an equation that can be solved algebraically and then transform the algebraic solution back into a solution of the differential equation. Surprisingly, this method will even work when \(g\) is a discontinuous function, provided the discontinuities are not too bad.
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Subsection 6.1.1 Definition of the Laplace Transform

We shall define the Laplace transform of a function \(f(t)\) by
\begin{equation*} {\mathcal L}(f)(s)= F(s) = \int_0^\infty e^{-st} f(t) \, dt, \end{equation*}
provided this integral converges. The Laplace transform of a function has many nice properties, especially with respect to the derivative of \(f\text{.}\) However, before we investigate these properties, let us compute several Laplace transforms.
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Example 6.1.1.

  1. Let \(f(t) = k\text{,}\) where \(k\) is a constant. Then
    \begin{equation*} {\mathcal L}(f)(s) = \int_0^\infty e^{-st}k \; dt = \frac{k}{s}, \end{equation*}
    for \(s \gt 0\text{.}\)
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  2. For \(f(t) = t\text{,}\) the Laplace transform is
    \begin{equation*} {\mathcal L}(f)(s) = \int_0^\infty e^{-st} t \, dt = \frac{1}{s^2}, \end{equation*}
    for \(s \gt 0\text{.}\)
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  3. If \(f(t) = e^{at}\) with \(t > 0\text{,}\) then
    \begin{equation*} {\mathcal L}(f)(s) = \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. \end{equation*}
    Noting that \(e^{-(s -a)t} \to 0\) as \(t \to \infty\) for \(s \gt a\text{,}\) we find that
    \begin{equation*} {\mathcal L}(e^{at}) = \frac{1}{s - a} \end{equation*}
    for \(s \gt a\text{.}\)
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Example 6.1.2.

The Laplace transform of a function does not always exist, even for functions that are infinitely differentiable. For example, let \(f(t) = e^{t^2}\text{.}\) Then for any real number \(s\text{,}\) we know that
\begin{equation*} e^{t^2} e^{-st} = e^{t(t-s)} \gt 1\text{,} \end{equation*}
for \(t \gt s\) with \(t \geq 0\text{;}\) hence,
\begin{equation*} \int_0^b e^{t^2} e^{-st} \, dt = \int_0^b e^{t(t-s)} \, dt \geq b. \end{equation*}
Thus, the integral
\begin{equation*} \int_0^\infty e^{t^2} e^{-st} \, dt \end{equation*}
does not converge.
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The Laplace transform, however, does exist in many cases. In Subsection 6.1.3, we will show that the Laplace transform of a function exists provided the function does not grow too quickly and does not possess bad discontinuities.
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Subsection 6.1.2 Properties of the Laplace Transform

One of the most important properties of the Laplace transform is linearity. That is,
\begin{equation*} {\mathcal L} ( \alpha f + \beta g )(s) = \alpha {\mathcal L}(f)(s) + \beta {\mathcal L} ( g)(s), \end{equation*}
where \(\alpha, \beta \in {\mathbb R}\) provided the Laplace transforms of \(f\) and \(g\) exist. The proof of this statement follows directly from the definition of the Laplace transform and the properties of integration,
\begin{align*} {\mathcal L} ( \alpha f + \beta g )(s) \amp = \int_0^\infty e^{-st} [\alpha f(t) + \beta g(t)] \, dt\\ \amp = \lim_{b \to \infty} \int_0^b e^{-st} [\alpha f(t) + \beta g(t)] \, dt\\ \amp = \alpha \lim_{b \to \infty} \int_0^b e^{-st} f(t) \, dt + \beta \lim_{b \to \infty} \int_0^b e^{-st}\beta g(t) \, dt\\ \amp = \alpha {\mathcal L}(f)(s) + \beta {\mathcal L} ( g)(s) \end{align*}
We state the linearity of the Laplace transform as a theorem. Transforms of functions having the linearity property are called linear operators.
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Example 6.1.4.

Recall that \(\sinh at = (e^{at} - e^{-at})/2\text{.}\) Provided \(a \gt 0\text{,}\) linearity makes it very easy to compute the Laplace transform of \(\sinh at\text{.}\) By Example 6.1.1,
\begin{equation*} {\mathcal L} ( \sinh at) = \frac{1}{2} {\mathcal L} ( e^{at}) - \frac{1}{2} {\mathcal L} ( e^{-at}) = \frac{1}{2} \left( \frac{1}{s - k} - \frac{1}{s + k} \right). \end{equation*}
Thus, for \(s \gt k \gt 0\text{,}\)
\begin{equation*} {\mathcal L} ( \sinh at) = \frac{k}{s^2 - k^2}. \end{equation*}
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The Laplace transform of discontinuous functions also exist, provided the discontinuities are not too bad. We say that a function \(f\) is piecewise continuous on an interval \([a, b]\) if \(f\) satisfies the following conditions.
  1. There are a finite number of discontinuities of \(f\) on \([a, b]\text{.}\)
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  2. If \(c \in [a, b]\) is a discontinuity, then the one-sided limits
    \begin{equation*} \lim_{t \to c^-} f(t) \qquad \mbox{and} \qquad \lim_{t \to c^+} f(t) \end{equation*}
    both exist. We also assume that \(\lim_{t \to a^+} f(t)\) and \(\lim_{t \to b^-} f(t)\) exist.
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We say that a function is piecewise continuous on an interval \([0, \infty)\text{,}\) if it is piecewise continuous on the interval \([0, b]\) for all \(b \gt 0\text{.}\) The types of discontinuities that we are describing are sometimes called jump discontinuities. If a function is piecewise continuous, then the continuities are not too bad.
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Example 6.1.5.

One of the simplest piecewise continuous functions is
\begin{equation*} u(t) = \begin{cases} 0, & t \lt 0 \\ 1, & t \geq 0. \end{cases} \end{equation*}
The function \(u(t)\) has a jump discontinuity at \(t = 0\text{.}\) It follows very quickly that \({\mathcal L}(u(t)) = 1/s\) for \(s \gt 0\text{.}\)
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If we define the unit step function at \(a\) to be
\begin{equation*} u_a(t) = u(t - a) = \begin{cases} 0, & t \lt a \\ 1, & t \geq a, \end{cases} \end{equation*}
then we have a jump discontinuity at \(t = a\) (Figure 6.1.6). If \(a \gt 0\text{,}\) then
\begin{equation*} {\mathcal L}(u_a(t)) = \int_0^\infty e^{-st} u_a(t) \, dt = \int_a^\infty e^{-st} \, dt = \lim_{b \to \infty} \left[ -\frac{e^{-st}}{s} \right]_a^b = \frac{e^{-sa}}{s}, \end{equation*}
where \(s \gt 0\) and \(a \gt 0\text{.}\) A function of the form \(u_a(t) = u(t - a)\) is called a step function or Heaviside function, named for the British engineer Oliver Heaviside.
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Figure 6.1.6. Unit step function for \(a = 0\)
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Example 6.1.7.

Given a function \(f(t)\text{,}\) consider the new function
\begin{equation*} g(t) = u_a(t) f( t - a). \end{equation*}
To obtain the function \(g\) from \(f\text{,}\) we shift the graph of the \(f\) to the right by \(a\) units and let \(g(t) = 0\) for \(0 \leq t \lt a\) (see Figure 6.1.8). Let us compute the Laplace transform of \(g\text{,}\)
\begin{equation*} {\mathcal L}(g) = \int_0^\infty g(t) e^{-st} \, dt =\int_a^\infty f(t - a) e^{-st} \, dt. \end{equation*}
If we make the substitution \(w = t - a\text{,}\) then
\begin{equation*} {\mathcal L}(g) = \int_0^\infty f(w) e^{-s(w+a)} \, dt = e^{-sa} \int_0^\infty f(w) e^{-sw} \, dt = e^{-sa} {\mathcal L}(f). \end{equation*}
In other words, if \({\mathcal L}(f) = F(s)\text{,}\) then \({\mathcal L}(u_a(t) f( t- a)) = e^{-as} F(s)\text{.}\)
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B Activate menu driven exploration
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Cursor down Explore next lower level
Cursor up Explore next upper level
Cursor right Explore next element on level
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X Toggle expert mode
W Extra details if available
Space Repeat speech
M Activate step magnification
Comma Activate direct magnification
N Deactivate magnification
Z Toggle subtitles
C Cycle contrast settings
T Monochrome colours
L Toggle language (if available)
K Kill current sound
Y Stop sound output
O Start and stop sonification
P Repeat sonification output
Figure 6.1.8. \(u_a(t)f(t-a)\text{,}\) where \(a = 1\)
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Activity 6.1.1. Finding Laplace Transforms.

Find the Laplace transform for each of the functions below. Recall that \(\cosh t = (e^t + e^{-t})/2\) and \(\sinh t = (e^t - e^{-t})/2\)
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(c)
\(f(t) = 3\sin^2 \left(\sqrt{2}t\right)\)
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Subsection 6.1.3 Existence and Uniqueness of the Laplace Transform

If we are to use Laplace transforms to study differential equations, we would like to know which functions actually have Laplace transforms. Furthermore, if two functions have the same Laplace transform, we can ask if the functions must be the same. In other words, we wish to know if the Laplace transform of a function exists and is unique. We can answer both of these questions affirmatively if the function \(f(t)\) is piecewise continuous on \([0, \infty)\) and does not grow too quickly as \(t \to \infty\text{.}\) We say a function \(f(t)\) is exponentially bounded on \([0, \infty)\) if there exist constants \(M \geq 0\) and \(a\) such that
\begin{equation*} |f(t)| \leq Me^{at}, \end{equation*}
for all \(t\) in \([0, \infty)\text{.}\) In other words, the graph of \(f\) must lie between the curves \(y = Me^{at}\) and \(y = -Me^{at}\text{.}\) The following two theorems tell us that Laplace transforms exist and are unique for piecewise continuous, exponentially bounded functions on the interval \([0, \infty)\text{.}\) We leave the proofs of these theorem as exercises.
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In light of Theorem 6.1.10, it now makes sense to define the inverse Laplace transform of a function \(F(s)\text{,}\) which we will denote by \({\mathcal L}^{-1}(F(s))(t)\text{,}\) as the function \(f(t)\) whose unique Laplace transform is \(F(s)\text{.}\) Furthermore, the inverse Laplace transform is linear,
\begin{equation*} {\mathcal L}^{-1} ( \alpha F + \beta G ) = \alpha {\mathcal L}^{-1}(F) + \beta {\mathcal L}^{-1} ( G). \end{equation*}
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Subsection 6.1.4 Finding Laplace Transforms and Inverse Transforms

To use the method of Laplace transforms effectively, we need either tables or computer software such as Sage so that we can easily find Laplace transforms and inverse Laplace transforms.
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Table 6.1.11. Table of Laplace Transforms
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\(f(t) = {\mathcal L}^{-1}(F(s))\) \(F(s) = {\mathcal L}(f(t))\)
1 \(1/s\text{,}\) \(s \gt 0\)
\(e^{at}\) \(1/(s - a)\text{,}\) \(s \gt a\)
\(t^n\text{,}\) \(n \in {\mathbb N}\) \(n!/s^{n+1}\text{,}\) \(s \gt 0\)
\(t^a\text{,}\) \(a > -1\) \(\Gamma(a + 1)/s^{a + 1}\text{,}\) \(s \gt 0\)
\(\sin at\) \(a/(s^2 + a^2)\text{,}\) \(s \gt 0\)
\(\cos at\) \(s/(s^2 + a^2)\text{,}\) \(s \gt 0\)
\(\sinh at\) \(a/(s^2 - a^2)\text{,}\) \(s \gt |a|\)
\(\cosh at\) \(s/(s^2 - a^2)\text{,}\) \(s \gt |a|\)
\(e^{at} \sin bt\) \(b/[(s-a)^2 + b^2]\text{,}\) \(s \gt a\)
\(e^{at} \cos bt\) \((s-a)/[(s-a)^2 + b^2]\text{,}\) \(s \gt a\)
\(t^n e^{at}\text{,}\) \(n \in {\mathbb N}\) \(n!/(s - a)^{n + 1}\text{,}\) \(s \gt a\)
\(u_c(t)\) \(e^{-cs}/ s\text{,}\) \(s \gt 0\)
\(u_c(t) f(t - c)\) \(e^{-cs} F(s)\)
\(e^{ct} f(t)\) \(F(s - c)\)
\(f(ct)\) \((1/c) F(s/c)\text{,}\) \(c \gt 0\)
\(\delta(t - c) = \delta_c(t)\) \(e^{-cs}\)
\(f'(t)\) \(s F(s) - f(0)\)
\(f''(t)\) \(s^2 F(s) - sf(0) - f'(0)\)
We can also use Sage to find Laplace transforms and inverse transforms (see Subsection 6.1.8).
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Activity 6.1.2. Finding Inverse Transforms.

Find the inverse Laplace transform for each of the following functions.
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(c)
\(F(s) = \dfrac{2}{(s - 3)^2 + 4}\)
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(d)
\(F(s) = \dfrac{2s^2}{(s^2 + 1)(s - 1)}\)
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(e)
\(F(s) = \dfrac{2s - 13}{s(s^2 - 4s + 13)}\)
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Subsection 6.1.5 Important Lessons

  • Many initial value problems have discontinuous forcing terms.
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  • The Laplace transform of a function \(f(t)\) by
    \begin{equation*} {\mathcal L}(f)(s)= F(s) = \int_0^\infty e^{-st} f(t) \, dt, \end{equation*}
    provided the integral converges.
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  • If \(f\) is a piecewise continuous, exponentially bounded function defined \([0, \infty)\text{,}\) then the Laplace transform of \(f\text{,}\)
    \begin{equation*} {\mathcal L}(f)(s)= F(s) = \int_0^\infty e^{-st} f(t) \, dt \end{equation*}
    exists.
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  • The Laplace transform is a linear operator; i.e.,
    \begin{equation*} {\mathcal L} ( \alpha f + \beta g )(s) = \alpha {\mathcal L}(f)(s) + \beta {\mathcal L} ( g)(s), \end{equation*}
    where \(\alpha, \beta \in {\mathbb R}\) provided the Laplace transforms of \(f\) and \(g\) exist.
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  • If we know the Laplace transform \(F(s)\) of a function, we can recover the original function using the inverse Laplace transform of a function \({\mathcal L}^{-1}(F(s))(t)\text{.}\)
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  • The inverse Laplace transform is linear,
    \begin{equation*} {\mathcal L}^{-1} ( \alpha F + \beta G ) = \alpha {\mathcal L}^{-1}(F) + \beta {\mathcal L}^{-1} ( G). \end{equation*}
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  • If \({\mathcal L}(f) = F(s)\text{,}\) then \({\mathcal L}(u_c(t) f( t- c)) = e^{-sc} F(s)\text{.}\)
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  • If \(c \geq 0\text{,}\) we define the Heaviside function to be
    \begin{equation*} u_c(t) = \begin{cases} 0 & t \lt c \\ 1 & t \geq c. \end{cases} \end{equation*}
    The Laplace transform of \(u_c\) is
    \begin{equation*} {\mathcal L}(u_c(t))(s) = \frac{e^{-cs}}{s}. \end{equation*}
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Reading Questions 6.1.6 Reading Questions

1.

Does the Laplace transform of a function always exist? If so, why? If not, given an example?
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2.

What is the Heaviside function? Explain.
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Exercises 6.1.7 Exercises

Finding Laplace transforms.

Find the Laplace transform for each of the functions in Exercise Group 6.1.7.1–8. Recall that \(\cosh t = (e^t + e^{-t})/2\) and \(\sinh t = (e^t - e^{-t})/2\text{.}\)
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1.
\(f(t) = \sin at\)
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Solution.
\(\mathcal{L}(\sin(at))(s) = \dfrac{a}{s^2 + a^2}\), for \(s > 0.\)
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2.
\(f(t) = \cos at\)
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Solution.
\(\mathcal{L}(\cos(at))(s) = \dfrac{s}{s^2 + a^2}\), for \(s > 0.\)
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3.
\(f(t) = e^{at} \sin bt\)
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Solution.
Using the shift property,
\begin{equation*} \mathcal{L}(e^{at}\sin(bt))(s) = \frac{b}{(s-a)^2 + b^2}, \end{equation*}
for \(s > a\text{.}\)
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4.
\(f(t) = \cosh at\)
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Solution.
Since \(\cosh(at) = \tfrac12(e^{at}+e^{-at})\text{,}\)
\begin{equation*} \mathcal{L}(\cosh(at))(s) = \frac{s}{s^2 - a^2}, \end{equation*}
for \(s > |a|\text{.}\)
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5.
\(f(t) = 2 \cos at - 3\sinh at\)
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Solution.
By linearity,
\begin{equation*} \mathcal{L}(f)(s) = 2\frac{s}{s^2 + a^2} - 3\frac{a}{s^2 - a^2}, \end{equation*}
for \(s > |a|\text{.}\)
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6.
\(f(t) = t e^{at}\)
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Solution.
Using the shift property and \(\mathcal{L}(t)(s)=1/s^2\text{,}\)
\begin{equation*} \mathcal{L}(t e^{at})(s) = \frac{1}{(s-a)^2}, \end{equation*}
for \(s > a\text{.}\)
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7.
\(f(t) = t \sin at\)
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Solution.
Using \(\mathcal{L}(t f(t)) = -\frac{d}{ds}\mathcal{L}(f)\text{,}\)
\begin{equation*} \mathcal{L}(t\sin(at))(s) = \frac{2as}{(s^2 + a^2)^2}, \end{equation*}
for \(s > 0\text{.}\)
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8.
\(f(t) = t^2 \cos at\)
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Solution.
Using \(\mathcal{L}(t^2 f(t)) = \frac{d^2}{ds^2}\mathcal{L}(f)\text{,}\)
\begin{equation*} \mathcal{L}(t^2\cos(at))(s) = \frac{2s(s^2 - 3a^2)}{(s^2 + a^2)^3}, \end{equation*}
for \(s > 0\text{.}\)
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Finding inverse Laplace transforms.

Find the inverse Laplace transform for each of the functions in Exercise Group 6.1.7.9–16. You will find partial fraction decomposition very useful.
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9.
\(F(s) = \dfrac{1}{s(s + 1)}\)
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Solution.
Using partial fractions,
\begin{equation*} \frac{1}{s(s+1)} = \frac{1}{s} - \frac{1}{s+1}. \end{equation*}
Hence,
\begin{equation*} \mathcal{L}^{-1}\!\left(\frac{1}{s(s+1)}\right) = 1 - e^{-t}. \end{equation*}
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10.
\(F(s) = s^{-3/2}\)
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Solution.
Since
\begin{equation*} \mathcal{L}(t^{1/2})(s) = \frac{\Gamma(3/2)}{s^{3/2}}, \end{equation*}
and \(\Gamma(3/2)=\frac{\sqrt{\pi}}{2}\text{,}\) we obtain
\begin{equation*} \mathcal{L}^{-1}(s^{-3/2}) = \frac{2}{\sqrt{\pi}}\,t^{1/2}. \end{equation*}
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11.
\(F(s) = \dfrac{1}{s + 7}\)
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Solution.
\(\mathcal{L}^{-1}\!\left(\dfrac{1}{s+7}\right) = e^{-7t}.\)
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12.
\(F(s) = \dfrac{2}{s^2 + 9}\)
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Solution.
Since
\begin{equation*} \mathcal{L}(\sin(3t))(s) = \frac{3}{s^2+9}, \end{equation*}
we get
\begin{equation*} \mathcal{L}^{-1}\!\left(\frac{2}{s^2+9}\right) = \frac{2}{3}\sin(3t). \end{equation*}
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13.
\(F(s) = \dfrac{5s - 4}{2s^2 + s - 1}\)
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Solution.
Using partial fractions,
\begin{equation*} \frac{5s-4}{(2s-1)(s+1)} = \frac{3}{2s-1} + \frac{2}{s+1}. \end{equation*}
Thus,
\begin{equation*} \mathcal{L}^{-1}(F)(t) = 3e^{t/2} + 2e^{-t}. \end{equation*}
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14.
\(F(s) = \dfrac{2s^2 - s + 4}{s^3 + 4s}\)
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Solution.
Write
\begin{equation*} \frac{2s^2-s+4}{s(s^2+4)} = \frac{1}{s} + \frac{s}{s^2+4} - \frac{1}{s^2+4}. \end{equation*}
Hence,
\begin{equation*} \mathcal{L}^{-1}(F)(t) = 1 + \cos(2t) - \frac12\sin(2t). \end{equation*}
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15.
\(F(s) = \dfrac{1}{(s + 2)^3}\)
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Solution.
Since
\begin{equation*} \mathcal{L}(t^2 e^{-2t})(s) = \frac{2}{(s+2)^3}, \end{equation*}
we obtain
\begin{equation*} \mathcal{L}^{-1}\!\left(\frac{1}{(s+2)^3}\right) = \frac{t^2}{2}e^{-2t}. \end{equation*}
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16.
\(F(s) = \dfrac{1}{(s + 1)^2(s^2 - 9)}\)
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Solution.
Using partial fractions,
\begin{equation*} \frac{1}{(s+1)^2(s-3)(s+3)} = \frac{1}{32}\frac{1}{s-3} - \frac{1}{32}\frac{1}{s+3} - \frac{1}{16}\frac{1}{s+1} - \frac{1}{8}\frac{1}{(s+1)^2}. \end{equation*}
Taking inverse transforms,
\begin{equation*} \mathcal{L}^{-1}(F)(t) = \frac{1}{32}e^{3t} - \frac{1}{32}e^{-3t} - \frac{1}{16}e^{-t} - \frac{1}{8}te^{-t}. \end{equation*}
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17.

Prove Theorem 6.1.9: If \(f\) is a piecewise continuous, exponentially bounded function defined \([0, \infty)\text{,}\) then the Laplace transform of \(f\text{,}\)
\begin{equation*} {\mathcal L}(f)(s)= F(s) = \int_0^\infty e^{-st} f(t) \, dt \end{equation*}
exists.
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Solution.
Since \(f\) is exponentially bounded, there exist constants \(M > 0\text{,}\) \(a \in \mathbb{R}\text{,}\) and \(T \ge 0\) such that
\begin{equation*} |f(t)| \le M e^{at} \quad \text{for } t \ge T. \end{equation*}
Then for \(t \ge T\text{,}\)
\begin{equation*} |e^{-st} f(t)| \le M e^{-(s-a)t}. \end{equation*}
If \(s > a\text{,}\) the integral
\begin{equation*} \int_T^\infty M e^{-(s-a)t}\,dt \end{equation*}
converges. By the comparison test,
\begin{equation*} \int_T^\infty e^{-st} f(t)\,dt \end{equation*}
also converges absolutely. Since \(f\) is piecewise continuous,
\begin{equation*} \int_0^T e^{-st} f(t)\,dt \end{equation*}
is finite. Therefore,
\begin{equation*} \int_0^\infty e^{-st} f(t)\,dt \end{equation*}
exists for all \(s > a\text{.}\)
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18.

Let \(f\) be a function whose Laplace transform \({\mathcal L}(f)(s)= F(s)\) exists for \(s \gt a\text{.}\) Prove that
\begin{equation*} {\mathcal L}(e^{ct} f(t))(s)= F(s - c) \end{equation*}
for \(s \gt a + c\text{.}\)
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Solution.
By definition of the Laplace transform,
\begin{equation*} \mathcal{L}(e^{ct}f(t))(s) = \int_0^\infty e^{-st} e^{ct} f(t)\,dt. \end{equation*}
Combine the exponentials:
\begin{equation*} e^{-st} e^{ct} = e^{-(s-c)t}. \end{equation*}
Hence,
\begin{equation*} \mathcal{L}(e^{ct}f(t))(s) = \int_0^\infty e^{-(s-c)t} f(t)\,dt. \end{equation*}
By the definition of \(F\text{,}\)
\begin{equation*} \int_0^\infty e^{-(s-c)t} f(t)\,dt = F(s-c). \end{equation*}
Since \(F(s)\) exists for \(s > a\text{,}\) the expression \(F(s-c)\) exists when \(s-c > a\text{,}\) or equivalently \(s > a + c\text{.}\) Therefore,
\begin{equation*} \mathcal{L}(e^{ct} f(t))(s) = F(s - c) \end{equation*}
for all \(s > a + c\text{.}\)
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19.

Prove Theorem 6.1.10: Suppose that \(f\) and \(g\) are piecewise continuous, exponentially bounded functions defined \([0, \infty)\text{,}\) with Laplace transforms of \(F(s)\) and \(G(s)\text{,}\) respectively. If \(F(s) = G(s)\text{,}\) then \(f(t) = g(t)\text{.}\)
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Solution.
Suppose \(f\) and \(g\) are piecewise continuous and exponentially bounded on \([0,\infty)\text{,}\) and that their Laplace transforms satisfy \(F(s)=G(s)\) for all sufficiently large \(s\text{.}\) Consider the function \(h(t)=f(t)-g(t)\text{.}\) Since both \(f\) and \(g\) are piecewise continuous and exponentially bounded, so is \(h\text{.}\) By linearity of the Laplace transform,
\begin{equation*} \mathcal{L}(h)(s) = \mathcal{L}(f)(s) - \mathcal{L}(g)(s) = F(s) - G(s) = 0. \end{equation*}
The only piecewise continuous, exponentially bounded function whose Laplace transform is identically zero is the zero function. Hence,
\begin{equation*} h(t) = 0 \quad \text{for all } t \ge 0. \end{equation*}
Therefore,
\begin{equation*} f(t) = g(t) \quad \text{for all } t \ge 0. \end{equation*}
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20.

If \(f\) is a piecewise continuous, exponentially bounded function defined \([0, \infty)\text{,}\) prove that the Laplace transform of \(f\text{,}\)
\begin{equation*} {\mathcal L}(f)(s)= F(s) = \int_0^\infty e^{-st} f(t) \, dt \end{equation*}
exists.
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Solution.
Since \(f\) is exponentially bounded, there exist constants \(M > 0\text{,}\) \(a \in \mathbb{R}\text{,}\) and \(T \ge 0\) such that
\begin{equation*} |f(t)| \le M e^{at} \quad \text{for } t \ge T. \end{equation*}
For \(t \ge T\text{,}\) we have
\begin{equation*} |e^{-st} f(t)| \le M e^{-(s-a)t}. \end{equation*}
If \(s > a\text{,}\) the integral
\begin{equation*} \int_T^\infty M e^{-(s-a)t}\,dt \end{equation*}
converges. By the comparison test,
\begin{equation*} \int_T^\infty e^{-st} f(t)\,dt \end{equation*}
also converges absolutely. Since \(f\) is piecewise continuous on \([0,T]\text{,}\)
\begin{equation*} \int_0^T e^{-st} f(t)\,dt \end{equation*}
is finite. Therefore,
\begin{equation*} \int_0^\infty e^{-st} f(t)\,dt \end{equation*}
exists for all \(s > a\text{.}\) Hence the Laplace transform \(\mathcal{L}(f)(s)\) exists.
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21.

Define the gamma function to be
\begin{equation*} \Gamma(x) = \int_0^\infty e^{-t} t^{x-1} \; dt \end{equation*}
for \(x \gt 0\text{.}\) The gamma function is very useful for expressing the Laplace transform of the function \(t^\alpha\text{.}\)
  1. Show that \(\Gamma(1) = 1\text{.}\)
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  2. Prove that \(\Gamma(x + 1) = x\Gamma(x)\) and deduce that \(\Gamma(n+1) = n!\) for \(n \in {\mathbb N}\text{.}\)
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  3. Show that the Laplace transform of \(f(t) = t^\alpha\) is
    \begin{equation*} {\mathcal L}(f)(s) = \frac{\Gamma(\alpha + 1)}{s^{\alpha + 1}}. \end{equation*}
    If \(n\) is a positive integer, then \({\mathcal L}(t^n)(s) = n!/s^{n + 1}\text{.}\)
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Solution.
  1. By definition,
    \begin{equation*} \Gamma(1) = \int_0^\infty e^{-t} t^{1-1}\,dt = \int_0^\infty e^{-t}\,dt. \end{equation*}
    Since
    \begin{equation*} \int_0^\infty e^{-t}\,dt = 1, \end{equation*}
    we conclude that \(\Gamma(1) = 1\text{.}\)
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  2. For \(x > 0\text{,}\) integration by parts gives
    \begin{equation*} \Gamma(x+1) = \int_0^\infty e^{-t} t^x\,dt. \end{equation*}
    Let \(u=t^x\) and \(dv=e^{-t}dt\text{.}\) Then \(du = x t^{x-1}dt\) and \(v=-e^{-t}\text{.}\) Thus,
    \begin{equation*} \Gamma(x+1) = \Big[-t^x e^{-t}\Big]_0^\infty + x\int_0^\infty e^{-t} t^{x-1}\,dt. \end{equation*}
    The boundary term is zero, so
    \begin{equation*} \Gamma(x+1) = x\Gamma(x). \end{equation*}
    Since \(\Gamma(1)=1\text{,}\) repeated application gives
    \begin{equation*} \Gamma(n+1) = n(n-1)\cdots 1 = n! \end{equation*}
    for all \(n \in \mathbb{N}\text{.}\)
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  3. The Laplace transform of \(f(t)=t^\alpha\) is
    \begin{equation*} \mathcal{L}(f)(s) = \int_0^\infty e^{-st} t^\alpha\,dt. \end{equation*}
    Let \(u=st\text{.}\) Then \(t=u/s\) and \(dt=du/s\text{,}\) so
    \begin{equation*} \mathcal{L}(f)(s) = \int_0^\infty e^{-u} \left(\frac{u}{s}\right)^\alpha \frac{du}{s}. \end{equation*}
    This simplifies to
    \begin{equation*} \mathcal{L}(f)(s) = \frac{1}{s^{\alpha+1}} \int_0^\infty e^{-u} u^\alpha\,du. \end{equation*}
    By definition of the Gamma function,
    \begin{equation*} \int_0^\infty e^{-u} u^\alpha\,du = \Gamma(\alpha+1). \end{equation*}
    Hence,
    \begin{equation*} \mathcal{L}(t^\alpha)(s) = \frac{\Gamma(\alpha+1)}{s^{\alpha+1}}. \end{equation*}
    If \(n\) is a positive integer, then \(\Gamma(n+1)=n!\text{,}\) so
    \begin{equation*} \mathcal{L}(t^n)(s) = \frac{n!}{s^{n+1}}. \end{equation*}
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Subsection 6.1.8 Laplace Transforms with Sage

Computer algebra systems have now replaced tables of Laplace transforms just as the calculator has replaced the slide rule. It is easy to calculate Laplace transforms with Sage. For example, suppose that we wish to compute the Laplace transform of \(f(x) = t^3 e^t - \cos t\text{.}\) We can use the Sage command laplace.
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To recover the original function, we can use the Sage command inverse_laplace. Suppose that
\begin{equation*} F(s) = -\frac{s}{s^2 + 1} + \frac{6}{(s - 1)^4}. \end{equation*}
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We can even use Sage to find the Laplace transform of piecewise-defined functions. Suppose that
\begin{equation*} f(t) = \begin{cases} 1, \amp 0 \leq t \leq 3 \\ 0, \amp t \gt 3. \end{cases} \end{equation*}
The Sage command to define a piecewise-defined function is piecewise.
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Sage can be used to find the Laplace transform of a piecewise-defined function.
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Exercises 6.1.9 Exercises

1.

Use Sage to find the Laplace transform of
\begin{equation*} f(t) = \begin{cases} -t, \amp 0 \leq t \leq 1 \\ 2, \amp t \gt 1. \end{cases} \end{equation*}
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