9.7: The Laplace Transform
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Up to this point we have only explored Fourier exponential transforms as one type of integral transform. The Fourier transform is useful on infinite domains. However, students are often introduced to another integral transform, called the Laplace transform, in their introductory differential equations class. These transforms are defined over semi-infinite domains and are useful for solving initial value problems for ordinary differential equations.
The Laplace transform is named after Pierre-Simon de Laplace ( \(1749-1827\) ). Laplace made major contributions, especially to celestial mechanics, tidal analysis, and probability.
Integral transform on \([a, b]\) with respect to the integral kernel, \(K(x, k)\).
The Fourier and Laplace transforms are examples of a broader class of transforms known as integral transforms. For a function \(f(x)\) defined on an interval \((a, b)\), we define the integral transform \[F(k)=\int_{a}^{b} K(x, k) f(x) d x,\nonumber \] where \(K(x, k)\) is a specified kernel of the transform. Looking at the Fourier transform, we see that the interval is stretched over the entire real axis and the kernel is of the form, \(K(x, k)=e^{i k x}\). In Table \(\PageIndex{1}\) we show several types of integral transforms.
| Laplace Transform | \(F(s)=\int_{0}^{\infty} e^{-s x} f(x) d x\) |
|---|---|
| Fourier Transform | \(F(k)=\int_{-\infty}^{\infty} e^{i k x} f(x) d x\) |
| Fourier Cosine Transform | \(F(k)=\int_{0}^{\infty} \cos (k x) f(x) d x\) |
| Fourier Sine Transform | \(F(k)=\int_{0}^{\infty} \sin (k x) f(x) d x\) |
| Mellin Transform | \(F(k)=\int_{0}^{\infty} x^{k-1} f(x) d x\) |
| Hankel Transform | \(F(k)=\int_{0}^{\infty} x J_{n}(k x) f(x) d x\) |
It should be noted that these integral transforms inherit the linearity of integration. Namely. let \(h(x)=\alpha f(x)+\beta g(x)\), where \(\alpha\) and \(\beta\) are constants. Then, \[\begin{align} H(k) &=\int_{a}^{b} K(x, k) h(x) d x,\nonumber \\ &=\int_{a}^{b} K(x, k)(\alpha f(x)+\beta g(x)) d x,\nonumber \\ &=\alpha \int_{a}^{b} K(x, k) f(x) d x+\beta \int_{a}^{b} K(x, k) g(x) d x,\nonumber \\ &=\alpha F(x)+\beta G(x) .\label{eq:1} \end{align}\] Therefore, we have shown linearity of the integral transforms. We have seen the linearity property used for Fourier transforms and we will use linearity in the study of Laplace transforms.
The Laplace transform of \(f, F=\mathcal{L}[f]\).
We now turn to Laplace transforms. The Laplace transform of a function \(f(t)\) is defined as \[F(s)=\mathcal{L}[f](s)=\int_{0}^{\infty} f(t) e^{-s t} d t, \quad s>0 .\label{eq:2}\] This is an improper integral and one needs \[\lim _{t \rightarrow \infty} f(t) e^{-s t}=0\nonumber \] to guarantee convergence.
Laplace transforms also have proven useful in engineering for solving circuit problems and doing systems analysis. In Figure \(\PageIndex{1}\) it is shown that a signal \(x(t)\) is provided as input to a linear system, indicated by \(h(t)\). One is interested in the system output, \(y(t)\), which is given by a convolution of the input and system functions. By considering the transforms of \(x(t)\) and \(h(t)\), the transform of the output is given as a product of the Laplace transforms in the s-domain. In order to obtain the output, one needs to compute a convolution product for Laplace transforms similar to the convolution operation we had seen for Fourier transforms earlier in the chapter. Of course, for us to do this in practice, we have to know how to compute Laplace transforms.
Properties and Examples of Laplace Transforms
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 \(\PageIndex{2}\). Combining some of these simple Laplace transforms with the properties of the Laplace transform, as shown in Table \(\PageIndex{3}\), 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.
We begin with some simple transforms. These are found by simply using the definition of the Laplace transform.
Show that \(\mathcal{L}[1]=\frac{1}{\varsigma}\).
Solution
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 \[\begin{align} \mathcal{L}[1] &=\int_{0}^{\infty} e^{-s t} d t\nonumber \\ &=\lim _{a \rightarrow \infty} \int_{0}^{a} e^{-s t} d t\nonumber \\ &=\lim _{a \rightarrow \infty}\left(-\frac{1}{s} e^{-s t}\right)_{0}^{a}\nonumber \\ &=\lim _{a \rightarrow \infty}\left(-\frac{1}{s} e^{-s a}+\frac{1}{s}\right)=\frac{1}{s} .\label{eq:3} \end{align}\]
Thus, we have found that the Laplace transform of 1 is \(\frac{1}{\frac{1}{S}}\). This result can be extended to any constant \(c\), using the linearity of the transform, \(\mathcal{L}[c]=c \mathcal{L}[1]\). Therefore, \[\mathcal{L}[c]=\frac{c}{s}. \nonumber \]
Show that \(\mathcal{L}\left[e^{a t}\right]=\frac{1}{s-a}\), for \(s>a\).
Solution
For this example, we can easily compute the transform. Again, we only need to compute the integral of an exponential function. \[\begin{align} \mathcal{L}\left[e^{a t}\right] &=\int_{0}^{\infty} e^{a t} e^{-s t} d t\nonumber \\ &=\int_{0}^{\infty} e^{(a-s) t} d t\nonumber \\ &=\left(\frac{1}{a-s} e^{(a-s) t}\right)_{0}^{\infty}\nonumber \\ &=\lim _{t \rightarrow \infty} \frac{1}{a-s} e^{(a-s) t}-\frac{1}{a-s}=\frac{1}{s-a} .\label{eq:4} \end{align}\]
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 \(a, s>\operatorname{Re}(a) .1\)
Show that \(\mathcal{L}[\cos a t]=\frac{s}{s^{2}+a^{2}}\) and \(\mathcal{L}[\sin a t]=\frac{a}{s^{2}+a^{2}}\).
Solution
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 \(e^{\text {iat }}=\cos a t+i \sin\) at. 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 \[\mathcal{L}\left[e^{i a t}\right]=\int_{0}^{\infty} e^{i a t} e^{-s t} d t=\int_{0}^{\infty} e^{-(s-i a) t} d t=\frac{1}{s-i a} .\nonumber \] Note that we could easily have used the result for the transform of an exponential, which was already proven. In this case \(s>\operatorname{Re}(\mathrm{ia})=0\).
We now extract the real and imaginary parts of the result using the complex conjugate of the denominator: \[\frac{1}{s-i a}=\frac{1}{s-i a} \frac{s+i a}{s+i a}=\frac{s+i a}{s^{2}+a^{2}} .\nonumber \] Reading off the real and imaginary parts, we find the sought transforms, \[\begin{align} \mathcal{L}[\cos a t] &=\frac{s}{s^{2}+a^{2}}\nonumber \\ \mathcal{L}[\sin a t] &=\frac{a}{s^{2}+a^{2}} .\label{eq:5} \end{align}\]
Show that \(\mathcal{L}[t]=\frac{1}{s^{2}}\).
Solution
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: \[\begin{align} \int_{0}^{\infty} t e^{-s t} d t &=-\left.t \frac{1}{s} e^{-s t}\right|_{0} ^{\infty}+\frac{1}{s} \int_{0}^{\infty} e^{-s t} d t\nonumber \\ &=\frac{1}{s^{2}} .\label{eq:6} \end{align}\]
Show that \(\mathcal{L}\left[t^{n}\right]=\frac{n !}{s^{n}+1}\) for nonnegative integer \(n\).
Solution
We have seen the \(n=0\) and \(n=1\) cases: \(\mathcal{L}[1]=\frac{1}{s}\) and \(\mathcal{L}[t]=\frac{1}{s^{2}}\). 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:\(^{1}\) \[\begin{align} \int_{0}^{\infty} t^{n} e^{-s t} d t &=-\left.t^{n} \frac{1}{s} e^{-s t}\right|_{0} ^{\infty}+\frac{n}{s} \int_{0}^{\infty} t^{-n} e^{-s t} d t\nonumber \\ &=\frac{n}{s} \int_{0}^{\infty} t^{-n} e^{-s t} d t .\label{eq:7} \end{align}\]
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 \[\mathcal{L}\left[t^{n}\right]=\frac{n}{s} \mathcal{L}\left[t^{n-1}\right] .\nonumber \]
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 \(I_{0}=\mathcal{L}[1]=\frac{1}{s}\), we have the following: \[I_{n}=\frac{n}{s} I_{n-1}, \quad I_{0}=\frac{1}{s} .\label{eq:8}\] 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 \[I_{n-1}=\frac{n-1}{s} I_{n-2} \text {. }\nonumber \]
Repeating the process, we find \[\begin{align} I_{n} &=\frac{n}{s} I_{n-1}\nonumber \\ &=\frac{n}{s}\left(\frac{n-1}{s} I_{n-2}\right)\nonumber \\ &=\frac{n(n-1)}{s^{2}} I_{n-2}\nonumber \\ &=\frac{n(n-1)(n-2)}{s^{3}} I_{n-3}\label{eq:9} \end{align}\]
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 figure out how many steps will get us to the known initial value. A list of iterates is easily written out: \[\begin{align} I_{n} &=\frac{n}{s} I_{n-1}\nonumber \\ &=\frac{n(n-1)}{s^{2}} I_{n-2}\nonumber \\ &=\frac{n(n-1)(n-2)}{s^{3}} I_{n-3}\nonumber \\ &=\ldots \nonumber \\ &=\frac{n(n-1)(n-2) \ldots(n-k+1)n(n-k)}{s^{k}} I_{n-k} .\label{eq:10} \end{align}\]
Since we know \(I_{0}=\frac{1}{s}\), we choose to stop at \(k=n\) obtaining \[I_{n}=\frac{n(n-1)(n-2) \ldots(2)(1)}{s^{n}} I_{0}=\frac{n !}{s^{n+1}} .\nonumber \] Therefore, we have shown that \(\mathcal{L}\left[t^{n}\right]=\frac{n !}{s^{n+1}}\).
Such iterative techniques are useful in obtaining a variety of integrals, such as \(I_{n}=\int_{-\infty}^{\infty} x^{2 n} e^{-x^{2}} d x .\)
This integral can just as easily be done using differentiation. We note that \[\left(-\frac{d}{d s}\right)^{n} \int_{0}^{\infty} e^{-s t} d t=\int_{0}^{\infty} t^{n} e^{-s t} d t .\nonumber\] Since \[\begin{aligned} &\qquad \int_{0}^{\infty} e^{-s t} d t=\frac{1}{s}, \\ &\int_{0}^{\infty} t^{n} e^{-s t} d t=\left(-\frac{d}{d s}\right)^{n} \frac{1}{s}=\frac{n !}{s^{n+1}} . \end{aligned}\]
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 5.4. Recall that the Gamma function is the generalization of the factorial function and is defined as \[\Gamma(x)=\int_{0}^{\infty} t^{x-1} e^{-t} d t .\label{eq:11}\] 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 \[\Gamma(x)=(x-1) ! \text {. }\nonumber \] Thus, the Gamma function can be viewed as a generalization of the factorial and we have shown that \[\mathcal{L}\left[t^{p}\right]=\frac{\Gamma(p+1)}{s^{p+1}}\nonumber \] for \(p>-1\).
Now we are ready to introduce additional properties of the Laplace transform in Table \(\PageIndex{3}\). We have already discussed the first property, which is a consequence of linearity of the integral transforms. We will prove the other properties in this and the following sections.
| Laplace Transform Properties |
|---|
| \(\mathcal{L}[a f(t)+b g(t)]=a F(s)+b G(s)\) |
| \(\mathcal{L}[t f(t)]=-\frac{d}{d s} F(s)\) |
| \(\mathcal{L}\left[\frac{d f}{d t}\right]=s F(s)-f(0)\) |
| \(\mathcal{L}\left[\frac{d^{2} f}{d t^{2}}\right]=s^{2} F(s)-s f(0)-f^{\prime}(0)\) |
| \(\mathcal{L}\left[e^{a t} f(t)\right]=F(s-a)\) |
| \(\mathcal{L}[H(t-a) f(t-a)]=e^{-a s} F(s)\) |
| \(\mathcal{L}[(f * g)(t)]=\mathcal{L}\left[\int_{0}^{t} f(t-u) g(u) d u\right]=F(s) G(s)\) |
Show that \(\mathcal{L}\left[\frac{d f}{d t}\right]=s F(s)-f(0)\).
Solution
We have to compute \[\mathcal{L}\left[\frac{d f}{d t}\right]=\int_{0}^{\infty} \frac{d f}{d t} e^{-s t} d t .\nonumber \] 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 \[\begin{align} \mathcal{L}\left[\frac{d f}{d t}\right] &=\int_{0}^{\infty} \frac{d f}{d t} e^{-s t} d t\nonumber \\ &=\left.f(t) e^{-s t}\right|_{0} ^{\infty}+s \int_{0}^{\infty} f(t) e^{-s t} d t\nonumber \\ &=-f(0)+s F(s) .\label{eq:12} \end{align}\] Here we have assumed that \(f(t) e^{-\text {st }}\) vanishes for large \(t\).
The final result is that \[\mathcal{L}\left[\frac{d f}{d t}\right]=s F(s)-f(0) .\nonumber \]
Show that \(\mathcal{L}\left[\frac{d^{2} f}{d t^{2}}\right]=s^{2} F(s)-s f(0)-f^{\prime}(0)\).
Solution
We can compute this Laplace transform using two integrations by parts, or we could make use of the last result. Letting \(g(t)=\frac{d f(t)}{d t}\), we have \[\mathcal{L}\left[\frac{d^{2} f}{d t^{2}}\right]=\mathcal{L}\left[\frac{d g}{d t}\right]=s G(s)-g(0)=s G(s)-f^{\prime}(0) .\nonumber \] But, \[G(s)=\mathcal{L}\left[\frac{d f}{d t}\right]=s F(s)-f(0)\nonumber \] So, \[\begin{align} \mathcal{L}\left[\frac{d^{2} f}{d t^{2}}\right] &=s G(s)-f^{\prime}(0)\nonumber \\ &=s[s F(s)-f(0)]-f^{\prime}(0)\nonumber \\ &=s^{2} F(s)-s f(0)-f^{\prime}(0) .\label{eq:13} \end{align}\]
We will return to the other properties in Table \(\PageIndex{3}\) after looking at a few applications.