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Note: 1.5 – 2 lectures, §10.1 in [EP], §6.1 and parts of §6.2 in [BD]

In this chapter we will discuss the Laplace transform^{1}. The Laplace transform turns out to be a very efficient method to solve certain ODE problems. In particular, the transform can take a differential equation and turn 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

We can think of as time and 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 is the frequency.

We can think of the Laplace transform as a black box. It eats functions and spits out functions in a new variable. We write for the Laplace transform of . 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 is the Laplace transform of . Let us define the transform.

We note that we are only considering in the transform. Of course, if we think of as time there is no problem, we are generally interested in finding out what will happen in the future (Laplace transform is one place where it is safe to ignore the past). Let us compute some simple transforms.

The limit (the improper integral) only exists if . So is only defined for .

Example 6.1.4: A common function is the unit step function, which is sometimes called the Heaviside function^{2}. This function is generally given as

Let us find the Laplace transform of , where is some constant. That is, the function that is 0 for and 1 for .

where of course (and as we said before).

By applying similar procedures we can compute the transforms of many elementary functions. Many basic transforms are listed in Table 6.1.

Exercise 6.1.1: Verify Table 6.1.

Since the transform is defined by an integral. We can use the linearity properties of the integral. For example, suppose is a constant, then

So we can “pull out” a constant out of the transform. Similarly we have linearity. Since linearity is very important we state it as a theorem.

Theorem 6.1.1 (Linearity of the Laplace transform). Suppose that , , and are constants, then

and in particular

These rules together with Table 6.1 make it easy to find the Laplace transform of a whole lot of functions already. But be careful. It is a common mistake to think that the Laplace transform of a product is the product of the transforms. In general

It must also be noted that not all functions have a Laplace transform. For example, the function does not have a Laplace transform as the integral diverges for all . Similarly, or do not have Laplace transforms.

Let us consider when does the Laplace transform exist in more detail. First let us consider functions of exponential order. The function is of exponential order as goes to infinity if

for some constants and , for sufficiently large (say for all for some ). The simplest way to check this condition is to try and compute

If the limit exists and is finite (usually zero), then is of exponential order.

Exercise 6.1.3: 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 is of exponential order for any .

For an exponential order function we have existence and uniqueness of the Laplace transform.

Theorem 6.1.2 (Existence). Let be continuous and of exponential order for a certain constant . Then is defined for all .

The existence is not difficult to see. Let be of exponential order, that is for all (for simplicity ). Let , or in other words . By the comparison theorem from calculus, the improper integral defining exists if the following integral exists

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, let us note that for exponential order functions we obtain that their Laplace transform decays at infinity:

Theorem 6.1.3 (Uniqueness). Let and be continuous and of exponential order. Suppose that there exists a constant , such that for all . Then for all .

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 outside of discontinuities. For example, the unit step function is sometimes defined using . This new step function, however, has the exact same Laplace transform as the one we defined earlier where .

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. If we have a function , to be able to find such that , we need to first know if such a function is unique. It turns out we are in luck by Theorem 6.1.3. So we can without fear make the following definition.

If for some function . We define the inverse Laplace transform as

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.

As the Laplace transform is linear, the inverse Laplace transform is also linear. That is,

Of course, we also have . Let us demonstrate how linearity can be used.

Example 6.1.6: Take . Find the inverse Laplace transform.

First we use the method of partial fractions to write in a form where we can use Table 6.1. We factor the denominator as and write

Putting the right hand side over a common denominator and equating the numerators we get . Expanding and equating coefficients we obtain , , , and thus . In other words,

By linearity of the inverse Laplace transform we get

Another useful property is the so-called shifting property or the first shifting property

where is the Laplace transform of .

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 and then use the shifting property.

First we complete the square to make the denominator . Next we find

Putting it all together with the shifting property, we find

In general, we want to be able to apply the Laplace transform to rational functions, that is functions of the form

where and are polynomials. Since normally, for the functions that we are considering, the Laplace transform goes to zero as , it is not hard to see that the degree of must be smaller than that of . Such rational functions are called proper rational functions and we can always apply the method of partial fractions. Of course this means we need to be able to factor the denominator into linear and quadratic terms, which involves finding the roots of the denominator.

^{1}Just like the Laplace equation and the Laplacian, the Laplace transform is also named after Pierre-Simon, marquis de Laplace (1749 – 1827).

^{2}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.”