5.2: Properties and Examples of Laplace Transforms
- Page ID
- 91073
IT IS TYPICAL THAT ONE MAKES USE of Laplace transforms by referring to a Table of transform pairs. A sample of such pairs is given in Table \(\PageIndex{1}\). Combining some of these simple Laplace transforms with the properties of the Laplace transform, as shown in Table \(\PageIndex{2}\), we can deal with many applications of the Laplace transform. We will first prove a few of the given Laplace transforms and show how they can be used to obtain new transform pairs. In the next section we will show how these transforms can be used to sum infinite series and to solve initial value problems for ordinary differential equations.
| \(f(t)\) | \(F(s)\) | \(f(t)\) | \(F(s)\) |
|---|---|---|---|
| \(c\) | \(\dfrac{c}{s}\) | \(e^{a t}\) | \(\dfrac{1}{s-a}, s>a\) |
| \(t^{n}\) | \(\dfrac{n !}{s^{n+1}}, s>0\) | \(t^{n} e^{a t}\) | \(\dfrac{n !}{(s-a)^{n+1}}\) |
| \(\sin \omega t\) | \(\dfrac{\omega}{s^{2}+\omega^{2}}\) | \(e^{a t} \sin \omega t\) | \(\dfrac{\omega}{(s-a)^{2}+\omega^{2}}\) |
| \(\cos \omega t\) | \(\dfrac{s}{s^{2}+\omega^{2}}\) | \(e^{a t} \cos \omega t\) | \(\dfrac{s-a}{(s-a)^{2}+\omega^{2}}\) |
| \(t \sin \omega t\) | \(\dfrac{2 \omega s}{\left(s^{2}+\omega^{2}\right)^{2}}\) | \(t \cos \omega t\) | \(\dfrac{s^{2}-\omega^{2}}{\left(s^{2}+\omega^{2}\right)^{2}}\) |
| \(\sinh a t\) | \(\dfrac{a}{s^{2}-a^{2}}\) | \(\cosh a t\) | \(\dfrac{s}{s^{2}-a^{2}}\) |
| \(H(t-a)\) | \(\dfrac{e^{-a s}}{s}, s>0\) | \(\delta(t-a)\) | \(e^{-a s}, \quad a \geq 0, s>0\) |
We begin with some simple transforms. These are found by simply using the definition of the Laplace transform.
Show that \(\mathcal{L}[1]=\dfrac{1}{s}\).
For this example, we insert \(f(t)=1\) into the definition of the Laplace transform:
\[\mathcal{L}[1]=\int_{0}^{\infty} e^{-s t} d t \nonumber \]
This is an improper integral and the computation is understood by introducing an upper limit of \(a\) and then letting \(a \rightarrow \infty\). We will not always write this limit, but it will be understood that this is how one computes such improper integrals. Proceeding with the computation, we have
\[ \begin{aligned} \mathcal{L}[1] &=\int_{0}^{\infty} e^{-s t} d t \\ &=\lim _{a \rightarrow \infty} \int_{0}^{a} e^{-s t} d t \\ &=\lim _{a \rightarrow \infty}\left(-\dfrac{1}{s} e^{-s t}\right)_{0}^{a} \\ &=\lim _{a \rightarrow \infty}\left(-\dfrac{1}{s} e^{-s a}+\dfrac{1}{s}\right)=\dfrac{1}{s} \end{aligned} \label{5.3} \]
Thus, we have found that the Laplace transform of 1 is \(\dfrac{1}{s}\). This result can be extended to any constant \(c\), using the linearity of the transform, \(\mathcal{L}[c]=c \mathcal{L}[1] .\) Therefore,
\[\mathcal{L}[c]=\dfrac{c}{s} \nonumber \]
Show that \(\mathcal{L}\left[e^{a t}\right]=\dfrac{1}{s-a}\), for \(s>a\)
For this example, we can easily compute the transform. Again, we only need to compute the integral of an exponential function.
\[ \begin{aligned} \mathcal{L}\left[e^{a t}\right] &=\int_{0}^{\infty} e^{a t} e^{-s t} d t \\ &=\int_{0}^{\infty} e^{(a-s) t} d t \\ &=\left(\dfrac{1}{a-s} e^{(a-s) t}\right)_{0}^{\infty} \\ &=\lim _{t \rightarrow \infty} \dfrac{1}{a-s} e^{(a-s) t}-\dfrac{1}{a-s}=\dfrac{1}{s-a} \end{aligned} \label{5.4} \]
Note that the last limit was computed as \(\lim _{t \rightarrow \infty} e^{(a-s) t}=0 .\) This is only true if \(a-s<0\), or \(s>a .\) [Actually, \(a\) could be complex. In this case we would only need \(s\) to be greater than the real part of \(a\), \(s>\operatorname{Re}(a) .]\)
Show that \(\mathcal{L}[\cos a t]=\dfrac{s}{s^{2}+a^{2}}\) and \(\mathcal{L}[\sin a t]=\dfrac{a}{s^{2}+a^{2}} .\)
For these examples, we could again insert the trigonometric functions directly into the transform and integrate. For example,
\[\mathcal{L}[\cos a t]=\int_{0}^{\infty} e^{-s t} \cos a t d t \nonumber \]
Recall how one evaluates integrals involving the product of a trigonometric function and the exponential function. One integrates by parts two times and then obtains an integral of the original unknown integral. Rearranging the resulting integral expressions, one arrives at the desired result. However, there is a much simpler way to compute these transforms.
Recall that \(e^{i a t}=\cos a t+i \sin a t .\) Making use of the linearity of the Laplace transform, we have
\[\mathcal{L}\left[e^{i a t}\right]=\mathcal{L}[\cos a t]+i \mathcal{L}[\sin a t]\nonumber \]
Thus, transforming this complex exponential will simultaneously provide the Laplace transforms for the sine and cosine functions!
The transform is simply computed as
\[\mathcal{L}\left[e^{i a t}\right]=\int_{0}^{\infty} e^{i a t} e^{-s t} d t=\int_{0}^{\infty} e^{-(s-i a) t} d t=\dfrac{1}{s-i a} \nonumber \]
Note that we could easily have used the result for the transform of an exponential, which was already proven. In this case, \(s>\operatorname{Re}(i a)=0\).
We now extract the real and imaginary parts of the result using the complex conjugate of the denominator:
\[\dfrac{1}{s-i a}=\dfrac{1}{s-i a} \dfrac{s+i a}{s+i a}=\dfrac{s+i a}{s^{2}+a^{2}}\nonumber \]
Reading off the real and imaginary parts, we find the sought-after transforms,
\[ \begin{aligned} \mathcal{L}[\cos a t] &=\dfrac{s}{s^{2}+a^{2}} \\ \mathcal{L}[\sin a t] &=\dfrac{a}{s^{2}+a^{2}} \end{aligned} \label{5.5} \]
Show that \(\mathcal{L}[t]=\dfrac{1}{s^{2}}\).For this example we evaluate
\[\mathcal{L}[t]=\int_{0}^{\infty} t e^{-s t} d t \nonumber \]
This integral can be evaluated using the method of integration by parts:
\[ \begin{aligned} \int_{0}^{\infty} t e^{-s t} d t &=-\left.t \dfrac{1}{s} e^{-s t}\right|_{0} ^{\infty}+\dfrac{1}{s} \int_{0}^{\infty} e^{-s t} d t \\ &=\dfrac{1}{s^{2}} \end{aligned} \label{5.6} \]
Show that \(\mathcal{L}\left[t^{n}\right]=\dfrac{n !}{s^{n+1}}\) for nonnegative integer \(n\).
We have seen the \(n=0\) and \(n=1\) cases: \(\mathcal{L}[1]=\dfrac{1}{s}\) and \(\mathcal{L}[t]=\dfrac{1}{s^{2}}\). We now generalize these results to nonnegative integer powers, \(n>1\), of \(t\). We consider the integral
\[\mathcal{L}\left[t^{n}\right]=\int_{0}^{\infty} t^{n} e^{-s t} d t \nonumber \]
Following the previous example, we again integrate by parts: \(^{1}\)
\[\begin{aligned} \int_{0}^{\infty} t^{n} e^{-s t} d t &=-\left.t^{n} \dfrac{1}{s} e^{-s t}\right|_{0} ^{\infty}+\dfrac{n}{s} \int_{0}^{\infty} t^{-n} e^{-s t} d t \\ &=\dfrac{n}{s} \int_{0}^{\infty} t^{-n} e^{-s t} d t \end{aligned} \label{5.7} \]
- 1
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This integral can just as easily be done using differentiation. We note that \
\[\left(-\dfrac{d}{d s}\right)^{n} \int_{0}^{\infty} e^{-s t} d t=\int_{0}^{\infty} t^{n} e^{-s t} d t.\nonumber \]
Since
\[\int_{0}^{\infty} e^{-s t} d t=\dfrac{1}{s}\nonumber \]
,
\[\int_{0}^{\infty} t^{n} e^{-s t} d t=\left(-\dfrac{d}{d s}\right)^{n} \dfrac{1}{s}=\dfrac{n !}{s^{n+1}}\nonumber \]
We could continue to integrate by parts until the final integral is computed. However, look at the integral that resulted after one integration by parts. It is just the Laplace transform of \(t^{n-1} .\) So, we can write the result as
\[\mathcal{L}\left[t^{n}\right]=\dfrac{n}{s} \mathcal{L}\left[t^{n-1}\right] .\nonumber \]
We compute \(\int_{0}^{\infty} t^{n} e^{-s t} d t\) by turning it into an initial value problem for a first-order difference equation and finding the solution using an iterative method.
This is an example of a recursive definition of a sequence. In this case, we have a sequence of integrals. Denoting
\[I_{n}=\mathcal{L}\left[t^{n}\right]=\int_{0}^{\infty} t^{n} e^{-s t} d t \nonumber \]
and noting that \(I_{0}=\mathcal{L}[1]=\dfrac{1}{s}\), we have the following:
\[I_{n}=\dfrac{n}{s} I_{n-1}, \quad I_{0}=\dfrac{1}{s} \nonumber \]
This is also what is called a difference equation. It is a first-order difference equation with an "initial condition," \(I_{0}\). The next step is to solve this difference equation. Finding the solution of this first-order difference equation is easy to do using simple iteration. Note that replacing \(n\) with \(n-1\), we have
\[I_{n-1}=\dfrac{n-1}{s} I_{n-2}. \nonumber \]
Repeating the process, we find
\[ \begin{aligned} I_{n} &=\dfrac{n}{s} I_{n-1} \\ &=\dfrac{n}{s}\left(\dfrac{n-1}{s} I_{n-2}\right) \\ &=\dfrac{n(n-1)}{s^{2}} I_{n-2} \\ &=\dfrac{n(n-1)(n-2)}{s^{3}} I_{n-3} \end{aligned} \label{5.9} \]
We can repeat this process until we get to \(I_{0}\), which we know. We have to carefully count the number of iterations. We do this by iterating \(k\) times and then figuring out how many steps will get us to the known initial value. A list of iterates is easily written out:
\[ \begin{aligned} I_{n} &=\dfrac{n}{s} I_{n-1} \\ &=\dfrac{n(n-1)}{s^{2}} I_{n-2} \\ &=\dfrac{n(n-1)(n-2)}{s^{3}} I_{n-3} \\ &=\cdots \\ &=\dfrac{n(n-1)(n-2) \ldots(n-k+1)}{s^{k}} I_{n-k} \end{aligned} \label{5.10} \]
Since we know \(I_{0}=\dfrac{1}{s}\), we choose to stop at \(k=n\) obtaining
\[I_{n}=\dfrac{n(n-1)(n-2) \ldots(2)(1)}{s^{n}} I_{0}=\dfrac{n !}{s^{n+1}}\nonumber \]
Therefore, we have shown that \(\mathcal{L}\left[t^{n}\right]=\dfrac{n !}{s^{n+1}}\).
Such iterative techniques are useful in obtaining a variety of integrals, such as \(I_{n}=\int_{-\infty}^{\infty} x^{2 n} e^{-x^{2}} d x\).
As a final note, one can extend this result to cases when \(n\) is not an integer. To do this, we use the Gamma function, which was discussed in Section 4.7. Recall that the Gamma function is the generalization of the factorial function and is defined as
\[\Gamma(x)=\int_{0}^{\infty} t^{x-1} e^{-t} d t \nonumber \]
Note the similarity to the Laplace transform of \(t^{x-1}\) :
\[\mathcal{L}\left[t^{x-1}\right]=\int_{0}^{\infty} t^{x-1} e^{-s t} d t \nonumber \]
For \(x-1\) an integer and \(s=1\), we have that
\[\Gamma(x)=(x-1) !\nonumber \]
Thus, the Gamma function can be viewed as a generalization of the factorial and we have shown that
\[\mathcal{L}\left[t^{p}\right]=\dfrac{\Gamma(p+1)}{s^{p+1}}\nonumber \]
for \(p>-1\).
Now we are ready to introduce additional properties of the Laplace transform in Table \(\PageIndex{2}\). We have already discussed the first property, which is a consequence of the linearity of integral transforms. We will prove the other properties in this and the following sections.
| Laplace Transform Properties |
| \(\mathcal{L}[a f(t)+b g(t)]=a F(s)+b G(s)\) \(\mathcal{L}[t f(t)]=-\dfrac{d}{d s} F(s)\) \(\mathcal{L}\left[\dfrac{d f}{d t}\right]=s F(s)-f(0)\) \(\mathcal{L}\left[\dfrac{d^{2} f}{d t^{2}}\right]=s^{2} F(s)-s f(0)-f^{\prime}(0)\) \(\mathcal{L}\left[e^{a t} f(t)\right]=F(s-a)\) \(\mathcal{L}[H(t-a) f(t-a)]=e^{-a s} F(s)\) \(\mathcal{L}[(f * g)(t)]=\mathcal{L}\left[\int_{0}^{t} f(t-u) g(u) d u\right]=F(s) G(s)\) |
Show that \(\mathcal{L}\left[\dfrac{d f}{d t}\right]=s F(s)-f(0)\)
We have to compute
\[\mathcal{L}\left[\dfrac{d f}{d t}\right]=\int_{0}^{\infty} \dfrac{d f}{d t} e^{-s t} d t \nonumber \]
We can move the derivative off \(f\) by integrating by parts. This is similar to what we had done when finding the Fourier transform of the derivative of a function. Letting \(u=e^{-s t}\) and \(v=f(t)\), we have
\[ \begin{aligned} \mathcal{L}\left[\dfrac{d f}{d t}\right] &=\int_{0}^{\infty} \dfrac{d f}{d t} e^{-s t} d t \\ &=\left.f(t) e^{-s t}\right|_{0} ^{\infty}+s \int_{0}^{\infty} f(t) e^{-s t} d t \\ &=-f(0)+s F(s) . \end{aligned}\label{5.12} \]
Here we have assumed that \(f(t) e^{-s t}\) vanishes for large \(t\).
The final result is that
\[\mathcal{L}\left[\dfrac{d f}{d t}\right]=s F(s)-f(0) \nonumber \]
Show that \(\mathcal{L}\left[\dfrac{d^{2} f}{d t^{2}}\right]=s^{2} F(s)-s f(0)-f^{\prime}(0) .\)
We can compute this Laplace transform using two integrations by parts, or we could make use of the last result. Letting \(g(t)=\dfrac{d f(t)}{d t}\), we have
\[\mathcal{L}\left[\dfrac{d^{2} f}{d t^{2}}\right]=\mathcal{L}\left[\dfrac{d g}{d t}\right]=s G(s)-g(0)=s G(s)-f^{\prime}(0)\nonumber \]
But,
\[G(s)=\mathcal{L}\left[\dfrac{d f}{d t}\right]=s F(s)-f(0) \nonumber \]
So,
\[ \begin{aligned} \mathcal{L}\left[\dfrac{d^{2} f}{d t^{2}}\right] &=s G(s)-f^{\prime}(0) \\ &=s[s F(s)-f(0)]-f^{\prime}(0) \\ &=s^{2} F(s)-s f(0)-f^{\prime}(0) \end{aligned} \label{5.13} \]
We will return to the other properties in Table \(5.3\) after looking at a few applications.