The Laplace Transform

In this chapter we study a specific integral transform, the Laplace transform, and its applications to differential equations. Its power comes from a single trick: differentiation in $t$ becomes multiplication by $s$ . A differential equation in $t$ therefore turns into an algebraic equation in $s$ , which is solved by ordinary algebra and then transformed back.

What Is an Integral Transform?

A powerful method for solving certain ordinary and partial differential equations is the use of integral transforms. In general, a linear integral transform of a function $f (t)$ is an expression of the form

\mathcal{T}\left\{f(t)\right\} = \int_a^b K(s,t)\,f(t)\,dt \tag{i}

where $K (s, t)$ , a given function of two variables, is called the kernel of the transform. The result of the integral is a function of $s$ , say $F (s)$ . Form (i) is called an integral transform because it transforms a function in $t$ -space into another function in $s$ -space by integration.

Different choices of kernel and limits give different transforms:

If $K (s, t) = e^{- s t}$ , $a = 0$ , and $b = + \infty$ , the transform is the Laplace transform. This is sometimes called the unilateral Laplace transform; taking $a = - \infty$ gives the bilateral version. Only the unilateral transform is considered in this book.
If $K (s, t) = \frac{1}{\sqrt{2 π}} e^{- i s t}$ , with $a = - \infty$ and $b = + \infty$ , the transform is the Fourier transform.

Many other integral transforms are defined the same way, each suited to a different class of problems.

Why the Method Works

The central identity, proved in Section 7.5, is

\mathscr{L}\left\{f'(t)\right\} = sF(s) - f(0) .

Differentiation in the $t$ world becomes multiplication by $s$ in the $s$ world, with the initial value appearing automatically. Applying this twice turns a second-order equation into a linear algebraic equation for $F (s)$ .

The resulting three-step procedure is unlike anything in the earlier chapters:

Transform. Apply $ℒ$ to both sides of the differential equation. Initial conditions are absorbed in this step rather than fitted at the end.
Solve algebraically. Rearrange to get $F (s)$ explicitly. This is ordinary algebra with no calculus involved.
Invert. Find $f (t) = ℒ^{- 1} {F (s)}$ , usually by partial fractions and a table.

What This Chapter Covers

Section	What you will learn
7.1 Definition of the Laplace Transform	The defining integral, linearity, transforms of $e^{a t}$ , $\sin k t$ , $\cos k t$ , $t^{n}$ , and $t^{α}$ via the gamma function.
7.2 Piecewise Continuous Functions, Exponential Order	Which functions are guaranteed to have a transform, and why the conditions are sufficient but not necessary.
7.3 The Inverse Laplace Transform	Recovering $f (t)$ from $F (s)$ , uniqueness up to isolated points, and partial fractions.
7.4 Further Properties of Laplace Transforms	Shifting in $s$ , multiplication by $t$ , division by $t$ , and transforms of periodic functions.
7.5 Laplace Transform of Derivatives and Integrals	The identities that make the whole method work, and solving initial value problems with them.
7.6 The Unit Step Function. Time Shifting Theorem	Handling forcing terms that switch on at a given moment.
7.7 Dirac Delta Function	Modeling an instantaneous impulse such as a hammer blow or a voltage spike.

What the Laplace Transform Adds

The earlier chapters could already solve constant-coefficient equations, so it is fair to ask what is gained.

Initial conditions are built in. No general solution with arbitrary constants is ever formed. The data enters at the transform step and the answer emerges already fitted.
Discontinuous forcing is routine. A voltage switched on at $t = 3$ defeats undetermined coefficients but is handled cleanly by the unit step function of Section 7.6.
Impulses become legitimate. A hammer blow delivering finite momentum in zero time is described by the Dirac delta of Section 7.7, which has no classical function representation at all.
Systems become algebra. A coupled system transforms into a system of linear algebraic equations in $s$ , solvable by elimination.
It is the engineer's language. Transfer functions, poles, and frequency response are all statements about $F (s)$ , and control theory is written in these terms.

Notation

The original function depends on $t$ and its Laplace transform is a function of $s$ . We denote the original function by a lowercase letter and its transform by the same letter in capital: $ℒ {f (t)} = F (s)$ , $ℒ {y (t)} = Y (s)$ . This convention is used throughout and saves a great deal of writing.

Frequently Asked Questions

What is the Laplace transform, in one sentence?

It is the operation that sends a function $f (t)$ defined for $t > 0$ to the function $F (s) = \int_{0}^{\infty} e^{- s t} f (t) d t$ , converting problems about derivatives into problems about algebra.

Why does the integral start at zero?

Because the transform is built for initial value problems, where the story begins at $t = 0$ and behavior before then is irrelevant. This is also why the transform cannot see what $f$ does for $t < 0$ , and why the theory pairs so naturally with the unit step function.

Does every function have a Laplace transform?

No. The integral must converge, which fails for functions growing too fast, such as $e^{t^{2}}$ . Section 7.2 gives sufficient conditions: piecewise continuity together with exponential order guarantee existence. Those conditions are not necessary, and some functions violating them still have transforms.

Is the Laplace transform reversible?

Yes, essentially uniquely. Two continuous functions with the same transform are identical, and in general two functions with the same transform can differ only at isolated points, which does not matter for solving differential equations. Section 7.3 makes this precise.

When should I reach for the Laplace transform instead of the earlier methods?

Whenever the forcing function is discontinuous, impulsive, or periodic, and whenever you want the answer already fitted to given initial conditions. For a homogeneous equation with no initial data, the characteristic equation of Chapter 4 is quicker.