8.1 Introduction to the Laplace Transform

Definition of the Laplace Transform

To define the Laplace transform, we first recall the definition of an improper integral. If \(g\) is integrable over the interval \([a,T]\) for every \(T>a\), then the improper integral of \(g\) over \([a,\infty)\) is defined as

We say that the improper integral converges if the limit in (8.1.1) exists; otherwise, we say that the improper integral diverges or does not exist. Here’s the definition of the Laplace transform of a function \(f\).

It is important to keep in mind that the variable of integration in (8.1.2) is \(t\), while \(s\) is a parameter independent of \(t\). We use \(t\) as the independent variable for \(f\) because in applications the Laplace transform is usually applied to functions of time.

The Laplace transform can be viewed as an operator \({\cal L}\) that transforms the function \(f=f(t)\) into the function \(F=F(s)\). Thus, (8.1.2) can be expressed as

The functions \(f\) and \(F\) form a transform pair, which we’ll sometimes denote by

It can be shown that if \(F(s)\) is defined for \(s=s_0\) then it’s defined for all \(s>s_0\) (Exercise 14(b)).

Computation of Some Simple Laplace Transforms

Solution From (8.1.2) with \(f(t)=1\),

If \(s\ne 0\) then

Therefore

If \(s=0\) the integrand reduces to the constant \(1\), and

Therefore \(F(0)\) is undefined, and

This result can be written in operator notation as

or as the transform pair

Solution From (8.1.2) with \(f(t)=t\),

If \(s\ne0\), integrating by parts yields

If \(s=0\), the integral in (8.1.5) becomes

Therefore \(F(0)\) is undefined and

This result can also be written as

or as the transform pair

Solution From (8.1.2) with \(f(t)=e^{at}\),

Combining the exponentials yields

However, we know from Example 8.1.1 that

Replacing \(s\) by \(s-a\) here shows that

This can also be written as

Solution Define

and

If \(s>0\), integrating (8.1.6) by parts yields

so

If \(s>0\), integrating (8.1.7) by parts yields

so

Now substitute from (8.1.8) into this to obtain

Solving this for \(G(s)\) yields

This and (8.1.8) imply that

Tables of Laplace transforms

Extensive tables of Laplace transforms have been compiled and are commonly used in applications. The brief table of Laplace transforms in the Appendix will be adequate for our purposes.

Solution The table includes the transform pair

Setting \(n=3\) and \(a=4\) here yields

We’ll sometimes write Laplace transforms of specific functions without explicitly stating how they are obtained. In such cases you should refer to the table of Laplace transforms.

Linearity of the Laplace Transform

The next theorem presents an important property of the Laplace transform.

Proof We give the proof for the case where \(n=2\). If \(s>s_0\) then

Solution By definition,

Therefore

where the first transform on the right is defined for \(s>b\) and the second for \(s>-b\); hence, both are defined for \(s>|b|\). Simplifying the last expression in (8.1.9) yields

The First Shifting Theorem

The next theorem enables us to start with known transform pairs and derive others. (For other results of this kind, see Exercises 6 and 13.)

Proof. Replacing \(s\) by \(s-a\) in (8.1.10) yields

if \(s-a>s_0\); that is, if \(s>s_0+a\). However, (8.1.11) can be rewritten as

which implies the conclusion.

Solution In the following table the known transform pairs are listed on the left and the required transform pairs listed on the right are obtained by applying Theorem 8.1.3.

Existence of Laplace Transforms

Not every function has a Laplace transform. For example, it can be shown (Exercise 3) that

for every real number \(s\). Hence, the function \(f(t)=e^{t^2}\) does not have a Laplace transform.

Our next objective is to establish conditions that ensure the existence of the Laplace transform of a function. We first review some relevant definitions from calculus.

Recall that a limit

exists if and only if the one-sided limits

both exist and are equal; in this case,

Recall also that \(f\) is continuous at a point \(t_0\) in an open interval \((a,b)\) if and only if

which is equivalent to

For simplicity, we define

so (8.1.12) can be expressed as

If \(f(t_0+)\) and \(f(t_0-)\) have finite but distinct values, we say that \(f\) has a jump discontinuity at \(t_0\), and

is called the jump in \(f\) at \(t_0\) (Figure 8.1.1).

If \(f(t_0+)\) and \(f(t_0-)\) are finite and equal, but either \(f\) isn’t defined at \(t_0\) or it’s defined but

we say that \(f\) has a removable discontinuity at \(t_0\) (Figure 8.1.2). This terminolgy is appropriate since a function \(f\) with a removable discontinuity at \(t_0\) can be made continuous at \(t_0\) by defining (or redefining)

Figure 8.1.3 shows the graph of a typical piecewise continuous function.

It is shown in calculus that if a function is piecewise continuous on a finite closed interval then it’s integrable on that interval. But if \(f\) is piecewise continuous on \([0,\infty)\), then so is \(e^{-st}f (t)\), and therefore

exists for every \(T>0\). However, piecewise continuity alone does not guarantee that the improper integral

converges for \(s\) in some interval \((s_0,\infty)\). For example, we noted earlier that (8.1.13) diverges for all \(s\) if \(f(t)=e^{t^2}\). Stated informally, this occurs because \(e^{t^2}\) increases too rapidly as \(t\to\infty\). The next definition provides a constraint on the growth of a function that guarantees convergence of its Laplace transform for \(s\) in some interval \((s_0,\infty)\) .

The next theorem gives useful sufficient conditions for a function \(f\) to have a Laplace transform. The proof is sketched in Exercise 10.

Solution Since \(f\) is defined by different formulas on \([0,1)\) and \([1,\infty)\), we write

Since

and

it follows that

This is consistent with Theorem 8.1.6, since

and therefore \(f\) is of exponential order \(s_0=-1\).

8.1 Exercises

1. Find the Laplace transforms of the following functions by evaluating the integral \(F(s)=\int_0^\infty e^{-st} f(t)\,dt\).

   (a) \(t\)	(b) \(te^{-t}\)	(c) \(\sinh bt\)
   (d) \(e^{2t}-3e^t\)	(e) \(t^2\)

   Show answer

   (a) \(\dst{1\over s^2}\) (b) \(\dst{1\over(s+1)^2}\) (c) \(\dst{b\over s^2-b^2}\) (d) \(\dst{-2s+5\over(s-1)(s-2)}\) (e) \(\dst{2\over s^3}\)

2. Use the table of Laplace transforms to find the Laplace transforms of the following functions.

   (a) \(\cosh t\sin t\)	(b) \(\sin^2t\)	(c) \(\cos^2 2t\)
   (d) \(\cosh^2 t\)	(e) \(t\sinh 2t\)	(f) \(\sin t\cos t\)
   (g) \(\dst{\sin\left(t+{\pi\over 4}\right)}\)	(h) \(\cos 2t -\cos 3t\)	(i) \(\sin 2t +\cos 4t\)

   Show answer

   (a) \(\dst{{s^2+2\over\left[(s-1)^2+1\right] \left[(s+1)^2 +1\right]}}\) (b) \(\dst{2\over s(s^2+4)}\) (c) \(\dst{s^2+8\over s(s^2+16)}\) (d) \(\dst{s^2-2\over s(s^2-4)}\)

      (e) \(\dst{4s\over(s^2-4)^2}\) (f) \(\dst{1\over s^2+4}\) (g) \(\dst{{1\over\sqrt{2}}\,{s+1\over s^2+1}}\) (h) \(\dst{5s\over(s^2+4)(s^2+9)}\) (i) \(\dst{{s^3+2s^2+4s+32\over (s^2+4)(s^2+16)}}\)

3. Show that

   \[ \int_0^\infty e^{-st}e^{t^2} dt=\infty \]

   for every real number \(s\).

4. Graph the following piecewise continuous functions and evaluate \(f(t+)\), \(f(t-)\), and \(f(t)\) at each point of discontinuity.

   (a) \(f(t)=\left\{\begin{array}{cl} -t, & 0\le t<2,\\ t-4, & 2\le t<3,\\ 1, & t\ge 3.\end{array}\right.\)	(b) \(f(t)=\left\{\begin{array}{cl} t^2+2, & 0 \le t<1,\\4, & t=1,\\ t, & t> 1.\end{array}\right.\)
   (c) \(f(t)=\left\{\begin{array}{rl} \sin t, & 0\le t<\pi/ 2,\\ 2\sin t, &\pi/ 2 \le t<\pi,\\ \cos t, & t\ge\pi.\end{array}\right.\)	(d) \(f(t)=\left\{\begin{array}{cl}t, & 0\le t<1,\\ 2, & t=1,\\ 2-t, & 1 \le t<2,\\ 3, & t=2,\\ 6, & t> 2.\end{array}\right.\)

   Show answer

   (a) \(\dst{f(3-)=-1,\,f(3)=f(3+)=1}\) (b) \(\dst{f(1-)=3,\,f(1)=4,\,f(1+)=1}\)

      (c) \(\dst{f\left({\pi\over2}-\right)=1,\, f\left({\pi\over2}\right)=f\left({\pi\over2}+\right)=2,\, f(\pi-)=0,\,f(\pi)=f(\pi+)=-1}\)

      (d) \(\dst{f(1-)=1,\,f(1)=2,\,f(1+)=1,\,f(2-)=0,\,f(2)=3,\,f(2+)=6}\)

5. Find the Laplace transform:

   (a) \(f(t)=\left\{\begin{array}{rl} e^{-t}, & 0\le t<1,\\ e^{-2t}, & t\ge 1.\end{array}\right.\)	(b) \(f(t)=\left\{\begin{array}{rl} 1, & 0\le t< 4,\\ t, & t\ge 4.\end{array}\right.\)
   (c) \(f(t)=\left\{\begin{array}{rl} t, & 0\le t<1,\\ 1, & t\ge 1.\end{array}\right.\)	(d) \(f(t)=\left\{\begin{array}{rl} te^t, & 0\le t<1,\\\phantom{t} e^t, & t\ge 1.\end{array}\right.\)

   Show answer

   (a) \(\dst{{1 -e^{-(s+1)}\over s+1}+ {e^{-(s+2)}\over s+2}}\) (b) \(\dst{{1\over s}+e^{-4s}\left({1\over s^2}+{3\over s}\right)}\) (c) \(\dst{1-e^{-s}\over s^2}\) (d) \(\dst{1-e^{-(s-1)}\over(s-1)^2}\)

6. Prove that if \(f(t)\leftrightarrow F(s)\) then \(t^kf(t)\leftrightarrow (-1)^kF^{(k)}(s)\). Hint: Assume that it’s permissible to differentiate the integral \(\int_0^\infty e^{-st}f(t)\,dt\) with respect to \(s\) under the integral sign.

7. Use the known Laplace transforms

   \[ {\cal L}(e^{\lambda t}\sin\omega t)={\omega\over(s-\lambda)^2+\omega^2} \quad\mbox{and }\quad {\cal L}(e^{\lambda t}\cos\omega t)={s-\lambda\over(s-\lambda)^2+\omega^2} \]

   and the result of Exercise 6 to find \({\cal L}(te^{\lambda t}\cos\omega t)\) and \({\cal L}(te^{\lambda t}\sin\omega t)\).

   Show answer

   \(\dst{{\cal L}(e^{\lambda t}\cos\omega t)={(s-\lambda)^2 - \omega^2\over\left((s-\lambda)^2+\omega^2\right)^2}\; {\cal L}(e^{\lambda t}\sin\omega t)= {2\omega(s-\lambda)\over\left((s-\lambda)^2+\omega^2\right)^2}}\)

8. Use the known Laplace transform \({\cal L}(1)=1/s\) and the result of Exercise 6 to show that

   \[ {\cal L}(t^n)={n!\over s^{n+1}},\quad n=\mbox{ integer}. \]

9. 1. Show that if \(\lim_{t\to\infty} e^{-s_0t} f(t)\) exists and is finite then \(f\) is of exponential order \(s_0\).

   2. Show that if \(f\) is of exponential order \(s_0\) then \(\lim_{t \to\infty} e^{-st} f(t)=0\) for all \(s>s_0\).

   3. Show that if \(f\) is of exponential order \(s_0\) and \(g(t)=f(t+\tau)\) where \(\tau>0\), then \(g\) is also of exponential order \(s_0\).

10. Recall the next theorem from calculus.

    Theorem A. Let \(g\) be integrable on \([0,T]\) for every \(T>0.\) Suppose there’s a function \(w\) defined on some interval \([\tau,\infty)\) (with \(\tau\ge 0\)) such that \(|g(t)|\le w(t)\) for \(t\ge\tau\) and \(\int^\infty_\tau w(t)\,dt\) converges. Then \(\int_0^\infty g(t)\,dt\) converges.

    Use Theorem A to show that if \(f\) is piecewise continuous on \([0,\infty)\) and of exponential order \(s_0\), then \(f\) has a Laplace transform \(F(s)\) defined for \(s>s_0\).

11. Prove: If \(f\) is piecewise continuous and of exponential order then \(\lim_{s\to\infty}F(s)~=~0\).

12. Prove: If \(f\) is continuous on \([0,\infty)\) and of exponential order \(s_0>0\), then

    \[ {\cal L}\left(\int^t_0 f(\tau)\,d\tau\right)={1\over s} {\cal L} (f), \quad s>s_0. \]

    Hint: Use integration by parts to evaluate the transform on the left.

13. Suppose \(f\) is piecewise continuous and of exponential order, and that \(\lim_{t\to 0+} f(t)/t\) exists. Show that

    \[ {\cal L}\left({f(t)\over t}\right)=\int^\infty_s F(r)\,dr. \]

    Hint: Use the results of Exercises \(6\) and \(11\).

14. Suppose \(f\) is piecewise continuous on \([0,\infty)\).

    1. Prove: If the integral \(g(t)=\int^t_0 e^{-s_0\tau} f(\tau)\,d\tau\) satisfies the inequality \(|g(t)|\le M\; (t\ge 0)\), then \(f\) has a Laplace transform \(F(s)\) defined for \(s>s_0\).

       Hint

       Use integration by parts to show that

       \[ \int_0^T e^{-st}f(t)\,dt = e^{-(s-s_0)T}g(T) +(s-s_0)\int_0^Te^{-(s-s_0)t}g(t)\,dt. \]

    2. Show that if \({\cal L}(f)\) exists for \(s=s_0\) then it exists for \(s>s_0\). Show that the function

       \[ f(t)=te^{t^2}\cos(e^{t^2}) \]

       has a Laplace transform defined for \(s>0\), even though \(f\) isn’t of exponential order.

    3. Show that the function

       \[ f(t)=te^{t^2}\cos(e^{t^2}) \]

       has a Laplace transform defined for \(s>0\), even though \(f\) isn’t of exponential order.

15. Use the table of Laplace transforms and the result of Exercise 13 to find the Laplace transforms of the following functions.

    (a) \(\dst{{\sin\omega t\over t}\quad(\omega>0)}\)	(b) \(\dst{{\cos\omega t-1\over t}}\quad (\omega>0)\)	(c) \(\dst{{e^{at}-e^{bt}\over t}}\)
    (d) \(\dst{{\cosh t-1\over t}}\)	(e) \(\dst{{\sinh^2 t\over t}}\)

    Show answer

    (a) \(\dst{\tan^{-1} {\omega\over s},\quad s>0}\) (b) \(\dst{{1\over 2}\ln{s^2\over s^2+\omega^2},\quad s>0}\) (c) \(\dst{{\ln{s-b\over s-a},\quad s>\max (a,b)}}\)

       (d) \(\dst{{1\over 2}\ln{s^2\over s^2-1},\quad s>1}\) (e) \(\dst{{1\over 4}\ln{s^2\over s^2-4},\quad s>2}\)

16. The gamma function is defined by

    \[ \Gamma (\alpha)=\int_0^\infty x^{\alpha-1}e^{-x}\,dx, \]

    which can be shown to converge if \(\alpha>0\).

    1. Use integration by parts to show that

       \[ \Gamma (\alpha+1)=\alpha\Gamma (\alpha),\quad\alpha>0. \]

    2. Show that \(\Gamma(n+1)=n!\) if \(n=1\), \(2\), \(3\),….

    3. From (b) and the table of Laplace transforms,

       \[ {\cal L}(t^\alpha)={\Gamma (\alpha+1)\over s^{\alpha+1}},\quad s>0, \]

       if \(\alpha\) is a nonnegative integer. Show that this formula is valid for any \(\alpha>-1\). Hint: Change the variable of integration in the integral for \(\Gamma (\alpha+1)\).

17. Suppose \(f\) is continuous on \([0, T]\) and \(f(t+T)=f(t)\) for all \(t\ge 0\). (We say in this case that \(f\) is periodic with period \(T\).)

    1. Conclude from Theorem 8.1.6 that the Laplace transform of \(f\) is defined for \(s>0\). Hint: Since \(f\) is continuous on \([0,T]\) and periodic with period \(T\), it’s bounded on \([0,\infty)\).

    2. (b) Show that

       \[ F(s)={1\over 1-e^{-sT}}\int_0^T e^{-st}f(t)\,dt,\quad s>0. \]

       Hint

       Write

       \[ F(s)=\sum^\infty_{n=0}\int^{(n+1)T}_{nT}e^{-st} f(t)\,dt. \]

       Then show that

       \[ \int^{(n+1)T}_{nT} e^{-st}f(t)\,dt=e^{-nsT}\int_0^T e^{-st}f(t)\,dt, \]

       and recall the formula for the sum of a geometric series.

18. Use the formula given in Exercise 17(b) to find the Laplace transforms of the given periodic functions:

    1. \(\dst{f(t)=\left\{\begin{array}{cl} t, & 0\le t<1,\\ 2-t, & 1\le t<2,\end{array}\right.\hskip30pt f(t+2)=f(t), \quad t\ge 0}\)

    2. \(\dst{f(t)=\left\{\begin{array}{rl}1, & 0\le t<{1\over 2},\\ -1, & {1\over 2}\le t<1,\end{array}\right. \hskip30pt f(t+1)=f(t),\quad t\ge 0}\)

    3. \(f(t)=|\sin t|\)

    4. \(\dst{f(t)=\left\{\begin{array}{cl}\sin t, & 0\le t< \pi, \\ 0, &\pi\le t<2\pi,\end{array}\right.\hskip30pt f(t+2\pi)=f(t)}\)

    Show answer

    (a) \(\dst{1\over s^2}\tanh{s\over 2}\)  (b) \(\dst{1\over s}\tanh{s\over 4}\)  (c) \(\dst{{1\over s^2+1}\coth {\pi s\over 2}}\)  (d) \(\dst{1\over(s^2+1)(1-e^{-\pi s})}\)