In this chapter, we will discuss the Laplace transform 1
Just like the Laplace equation and the Laplacian, the Laplace transform is also named after Pierre-Simon, marquis de Laplace (1749–1827).
. The Laplace transform is a very efficient method to solve certain ODE or PDE problems. The transform takes a differential equation and turns it into an algebraic equation. If the algebraic equation can be solved, applying the inverse transform gives us our desired solution. The Laplace transform also has applications in the analysis of electrical circuits, NMR spectroscopy, signal processing, and elsewhere. Finally, understanding the Laplace transform will also help with understanding the related Fourier transform, which, however, requires more understanding of complex numbers. We will not cover the Fourier transform.
The Laplace transform also gives a lot of insight into the nature of the equations we are dealing with. It can be seen as converting between the time and the frequency domain. For example, take the standard equation
\begin{equation*}
m x''(t) + c x'(t) + k x(t) = f(t) .
\end{equation*}
We can think of \(t\) as time and \(f(t)\) as incoming signal. The Laplace transform will convert the equation from a differential equation in time to an algebraic (no derivatives) equation, where the new independent variable \(s\) is the frequency. 2
Really, it is the “frequency” in terms of complex numbers, but we digress.
We can think of the Laplace transform as a black box. It eats functions and spits out functions in a new variable. We write \(\mathcal{L} \bigl\{ f(t) \bigr\} = F(s)\) for the Laplace transform of \(f(t)\text{.}\) It is common to write lower case letters for functions in the time domain and upper case letters for functions in the frequency domain. We use the same letter to denote that one function is the Laplace transform of the other. For example, \(F(s)\) is the Laplace transform of \(f(t)\text{.}\) Let us define the transform.
We note that we are only considering \(t \geq 0\) in the transform. Of course, if we think of \(t\) as time, there is no problem, we are generally interested in finding out what will happen in the future (the Laplace transform is one place where it is safe to ignore the past). Let us compute some simple transforms.
A common function is the unit step function, which is sometimes called the Heaviside function 3
The function is named after the English mathematician, engineer, and physicist Oliver Heaviside (1850–1925). Only by coincidence is the function “heavy” on “one side.”
Let us find the Laplace transform of \(u(t-a)\text{,}\) where \(a \geq 0\) is some constant. That is, it is the function that is \(0\) for \(t < a\) and \(1\) for \(t \geq a\text{.}\)
By applying similar procedures, we can compute the transforms of many elementary functions. Many basic transforms are listed in Table 6.1 (see also Appendix B).
Verify the theorem. That is, show that \(\mathcal{L} \bigl\{ A f(t) + B g(t) \bigr\} =
A \mathcal{L} \bigl\{ f(t) \bigr\} +
B \mathcal{L} \bigl\{ g(t) \bigr\}\text{.}\)
Moreover, not all functions have a Laplace transform. For example, the function \(\frac{1}{t}\) does not have a Laplace transform as the integral diverges for all \(s\text{.}\) Similarly, \(\tan t\) or \(e^{t^2}\) do not have Laplace transforms.
When does the Laplace transform exist? A function \(f(t)\) is of exponential order as \(t\) goes to infinity if
\begin{equation*}
\lvert f(t) \rvert \leq M e^{ct} ,
\end{equation*}
for some constants \(M\) and \(c\text{,}\) for sufficiently large \(t\) (say for all \(t > t_0\) for some \(t_0\)). The simplest way to check this condition is to try and compute
Use L’Hopital’s rule from calculus to show that a polynomial is of exponential order. Hint: Note that a sum of two exponential-order functions is also of exponential order. Then show that \(t^n\) is of exponential order for any \(n\text{.}\)
Let \(f(t)\) be continuous on the interval \([0,\infty)\) and of exponential order for a certain constant \(c\text{.}\) Then \(F(s) = \mathcal{L} \bigl\{ f(t) \bigr\}\) is defined for all \(s > c\text{.}\)
The existence is not difficult to see. Let \(f(t)\) be of exponential order, that is, \(\lvert f(t) \rvert \leq M e^{ct}\) for all \(t > 0\) (for simplicity \(t_0 = 0\)). Let \(s > c\text{,}\) or in other words \((s-c) > 0\text{.}\) By the comparison theorem from calculus, the improper integral defining \(\mathcal{L} \bigl\{ f(t) \bigr\}\) exists because the following integral exists
\begin{equation*}
\int_0^\infty e^{-st} ( M e^{ct} ) \,dt
=
M \int_0^\infty e^{-(s-c)t} \,dt = M \left[ \frac{e^{-(s-c)t}}{-(s-c)}
\right]_{t=0}^\infty = \frac{M}{s-c} .
\end{equation*}
The transform also exists for some other functions that are not of exponential order, but that will not be relevant to us. Before dealing with uniqueness, we note that for functions of exponential order, their Laplace transform decays at infinity:
Let \(f(t)\) and \(g(t)\) be continuous and of exponential order. Suppose that there exists a constant \(C\text{,}\) such that \(F(s) = G(s)\) for all \(s > C\text{.}\) Then \(f(t) = g(t)\) for all \(t \geq 0\text{.}\)
Both theorems hold for piecewise continuous functions as well. Recall that piecewise continuous means that the function is continuous except perhaps at a discrete set of points, where it has jump discontinuities like the Heaviside function. Uniqueness, however, does not “see” values at the discontinuities. So we can only conclude that \(f(t) = g(t)\) outside of discontinuities. For example, the unit step function is sometimes defined using \(u(0) = \nicefrac{1}{2}\text{.}\) This new step function, however, has the exact same Laplace transform as the one we defined earlier, where \(u(0) = 1\text{.}\)
As we said, the Laplace transform will allow us to convert a differential equation into an algebraic equation. Once we solve the algebraic equation in the frequency domain, we will want to get back to the time domain, as that is what we are interested in. Given a function \(F(s)\text{,}\) we wish to find a function \(f(t)\) such that \(\mathcal{L} \bigl\{ f(t) \bigr\} = F(s)\text{.}\)Theorem 6.1.3 says that the solution \(f(t)\) is unique. So we can without fear make the following definition.
There is an integral formula for the inverse, but it is not as simple as the transform itself—it requires complex numbers and path integrals. For us it will suffice to compute the inverse using Table 6.1.
Putting the right-hand side over a common denominator and equating the numerators we get \(A(s^2+1) + s(Bs+C) = s^2+s+1\text{.}\) Expanding and equating coefficients, we obtain \(A+B = 1\text{,}\)\(C=1\text{,}\)\(A=1\text{,}\) and thus \(B=0\text{.}\) In other words,
The shifting property can be used, for example, when the denominator is a more complicated quadratic that may come up in the method of partial fractions. We complete the square and write such quadratics as \({(s+a)}^2+b\) and then use the shifting property.
where \(F(s)\) and \(G(s)\) are polynomials. If \(\frac{F(s)}{G(s)}\) is the Laplace transform of an exponential-order function, it goes to zero as \(s \to \infty\text{,}\) and so the degree of \(F(s)\) must be smaller than that of \(G(s)\text{.}\) Such rational functions are called proper rational functions and we can always apply the method of partial fractions without polynomial division. Of course, we still need to be able to factor the denominator into linear and quadratic terms, which involves finding the roots of the denominator.
