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Math 555: Differential Equations

6.1 Laplace Transforms

6.1.1 Integral Transforms

In many physics and engineering applications, systems are often subject to nonhomogeneous influence or forcing terms. A common technique to studying differential equations in these cases involves using an integral transform the changes the problem into a form that closely matches the nature of forcing terms or the differential equation itself. The two most common are the Fourier transform , which is useful for oscillating systems, and the Laplace transform , that handles discontinuous and "impulse" style forcing terms particularly well.

6.1.2 Improper Integrals

The Laplace transform

$$ \mathcal{L}\left\{f(t)\right\} = F(s) = \int_0^\infty e^{-st} f(t)\, dt $$
is defined using an improper integral, so we shall review some integral properties and compute some of these transforms for practice.

The improper integral of $f(t)$ from $a$ to $\infty$ is interpreted as

$$ \int_a^\infty f(t)\, dt = \lim_{A\rightarrow\infty} \int_a^A f(t)\, dt. $$
This integral is said to converge if its value exists and is finite and diverge otherwise.

Example 6.1.1

Determine if the improper integral $\int_a^\infty \frac{dt}{t}$ converges or diverges.

Solution

By the definition,

$$ \int_a^\infty \frac{dt}{t} = \lim_{A\rightarrow\infty} \int_a^A \frac{dt}{t} = \lim_{A\rightarrow\infty} \ln A = \infty. $$
The integral diverges.

Example 6.1.2

Assuming that $s > 0$, determine if the improper integral $\int_0^\infty e^{-st}\, dt$ converges or diverges.

Solution

We have

$$\begin{aligned} \int_0^\infty e^{-st}\, dt &= \lim_{A\rightarrow\infty} \int_0^A e^{-st}\, dt \\ \\ &= \lim_{A\rightarrow\infty} \left. \dfrac{e^{-st}}{-s}\right\vert_{\ 0}^{\ A} \\ \\ &= \lim_{A\rightarrow\infty} \dfrac{e^{-sA} - e^{-s\cdot 0}}{-s} \\ \\ &= \dfrac{0 - 1}{-s} \\ \\ &= \dfrac{1}{s}. \end{aligned} $$ This is convergent.

Exercise 6.1.1

Find all values of $p\in\mathbb{R}$ for which

$$ \int_1^\infty t^{-p}\, dt $$
converges. For which values of $p$ does it diverge?

Check Your Work It converges for $p\gt 1$. The result should be reminiscent of the integral test from Calculus II that proves that the series
$$ \sum_{n=0}^\infty \dfrac{1}{n^p} $$
converges for $p\gt 1$.

Follow Along If $p = 1$, then we have Example 6.1.1 . So we assume that $p\neq 1$ and

$$ \int_1^\infty t^{-p}\, dt = \lim_{A\rightarrow\infty} \int_1^A t^{-p}\, dt = \lim_{A\rightarrow\infty} \left.\dfrac{t^{1-p}}{1-p}\right\vert_{\ 1}^{\ A} = \lim_{A\rightarrow\infty} \dfrac{A^{1-p} -1}{1-p} $$
For this limit to be finite as $A\rightarrow\infty$, we need $1-p \lt 0$. So the condition is that $p\gt 1$.

6.1.3 Integrals of Piecewise Functions

An important theoretical note about integration is that we allow the operation to be performed on functions that are piecewise continuous on an interval $[a,b]$. A function is piecewise continuous on $[a,b]$ if there exists a partition of the interval using finitely many points

$$ a = t_0\lt t_1\lt t_2\lt\ldots\lt t_n=b $$
so that

  1. the function $f(t)$ is continuous on each open subinterval $(t_{j-1},t_j)$
  2. $f(t)$ approaches a finite limit as $t\rightarrow t_{j-1}^+$ and $t\rightarrow t_{j}^-$ (the limits at the endpoints approached from within the subinterval are finite).

Integrating piecewise continuous functions is straightforward, they are just the sum of the integrals on all of the subintervals:

$$ \int_a^b f(t)\, dt = \int_a^{t_1} f(t)\, dt + \int_{t_1}^{t_2} f(t)\, dt + \ldots + \int_{t_{n-1}}^{b} f(t)\, dt $$

The most relevant feature of these types of functions are the presence of jump discontinuities . A piecewise continuous function with multiple jump discontinuities

Theorem 6.1.1

The improper integral of a function bounded above by a convergent improper integral converges and the improper integral of a function bounded below by a divergent improper integral diverges.
If $f$ is piecewise continuous for $t\geq a$, $|f(t)|\leq g(t)$ when $t\geq M$ for a positive constant $M$, and $\int_M^\infty g(t)\,dt$ converges, then $\int_M^\infty f(t)\,dt$ also converges.
Likewise, if $f(t) \geq g(t)\geq 0$ for $t\geq M$ and $\int_M^\infty g(t)\,dt$ diverges, then $\int_M^\infty f(t)\,dt$ also diverges.

Since piecewise continuous functions may be integrated without issue, these discontinuities are "smoothed out" and made easier to work with under the correct conditions. This is one of the chief benefits of employing a tool like the Laplace transform, which we are ready to discuss.

6.1.4 Laplace Transform

As noted above the Laplace transform is defined as

$$ \mathcal{L}\left\{f(t)\right\} = F(s) = \int_0^\infty e^{-st} f(t)\, dt $$
and is an example of an integral transform . In general, integral transforms look like

$$ F(s) = \int_a^b K(s,t)f(t)\,dt $$
where $K(s,t)$ is a function called the kernel of the transformation. The limits of integration $a$ and $b$ may be $-\infty$ and $\infty$ in some cases, such as we see with the Laplace transform. The purpose of the integral transform is to take a function $f$ and transform it into another function $F$, often called the transform of $f$.

Theorem 6.1.2

Existence Conditions for the Laplace Transform of a Function
If

  1. $f$ is a piecewise continuous function on the interval $t\in [0,A]$ for any $A\gt 0$
  2. there exist real constants $K$, $a$, and $M$ with $K,M\gt 0$ such that

    $$ |f(t)\leq Ke^{at}\text{ when } t\geq M $$

then the Laplace transform $\mathcal\{f(t)\} = F(s)$ exists for $s\gt a$.

The procedure for using the Laplace transform is

  1. Apply $\mathcal{L}$ to the initial value problem for an unknown function $f$ in the $t$-domain to transform it into a simpler (algebraic) problem for $F$ in the $s$-domain.
  2. Determine $F$ using algebra.
  3. "Invert the transform," meaning find $f$ using its transform $F$.

We will employ this procedure beginning next section. For now, we will compute the transforms of a few elementary functions to obtain some familiarity.

Exercise 6.1.2

Find $\mathcal{L}\{1\}$.

Check Your Work $$\mathcal{L}\{1\} = \dfrac{1}{s},\quad s \gt 0. $$

Hint Example 6.1.2.

Example 6.1.4

Find $\mathcal{L}\{e^{at}\}$.

Solution


$$\begin{aligned} \mathcal{L}\{e^{at}\} &= \int_0^\infty e^{-st} e^{at}\, dt = \int_0^\infty e^{-(s-a)t}\, dt, \\ \\ &= \dfrac{1}{s-a},\quad s\gt a. \end{aligned} $$

Example 6.1.5

Find the Laplace transform of the pictured function Unit pulse function

Solution

First, we use the picture to write down the piecewise representation of the function

$$ f(t) = \left\{\begin{matrix} 1, & t\in [0,1) \\ k, & t = 1 \\ 0, & t\in (1,\infty). \end{matrix} \right. $$
The value of $k$ does not matter for the integral, since there is no "width" to integrate at exactly the point $t = 1$. The Laplace transform looks like

$$ \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t)\, dt = \int_0^1 e^{-st} \, dt = \left.\dfrac{e^{-st}}{s}\right\vert_{\ 0}^{\ 1} = \dfrac{1 - e^{-s}}{s},\quad s\gt 0. $$

Exercise 6.1.3

Find $\mathcal{L}\{\cos(at)\}$.

Check Your Work $$ \mathcal{L}\{\cos(at)\} = \dfrac{s}{s^2 + a^2},\quad s \gt 0. $$

Video Solution

6.1.5 Properties of the Laplace Transform

The Laplace transform operator is linear , so it behaves nicely with scalar multiplication and addition:

$$ \mathcal{L}\{\alpha f(t)\} = \alpha\mathcal{L}\{f(t)\},\qquad \mathcal{L}\{f(t) + g(t)\} = \mathcal{L}\{f(t)\} + \mathcal{L}\{g(t)\}. $$

Exercise 6.1.4

Using the linearity property of the Laplace transform, find $\mathcal{L}\{ 3e^{-3t} + 2\cos(5t)\}$.

Check Your Work

$$ \mathcal{L}\{ 3e^{-3t} + 2\cos(5t)\} = \dfrac{3}{s + 3} + \dfrac{2s}{s^2 + 25} $$

6.1.6 Table of Laplace Transforms

In practice, these transforms are not constantly computed as they are needed, but instead found using a reference. Having a reference table is especially useful for when we need to invert the transforms, since that computation cannot be done directly in most cases. The table also provides a guide of algebraic forms for us to try to manipulate our problem into to be able to invert $F(s)$ using the table.

Laplace Transform Table

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