The Inverse Laplace Transform

The Laplace transform is of little use unless it can be undone. Given $F (s)$ , the inverse Laplace transform recovers the function $f (t)$ that produced it. Lerch's theorem guarantees the answer is essentially unique: two functions with the same transform can differ only at their points of discontinuity, which never matters in practice.

Quick Reference

Item	Statement
Notation	If $ℒ {f (t)} = F (s)$ , then $ℒ^{- 1} {F (s)} = f (t)$
Lerch's theorem	Same transform $⟹$ same function, except possibly at points of discontinuity
Consequence	Restricted to continuous functions, the inverse is unique
Linearity	$ℒ^{- 1} {a F + b G} = a ℒ^{- 1} {F} + b ℒ^{- 1} {G}$
Scaling	$ℒ^{- 1} {F (a s)} = \frac{1}{a} f (\frac{t}{a})$
Shift in $s$	$ℒ^{- 1} {F (s - a)} = e^{a t} f (t)$
Shift in $t$	$ℒ^{- 1} {e^{- a s} F (s)} = u (t - a) f (t - a)$
Derivative	\mathscr{L}^{-1}\left\{sF(s) - f(0)\right\} = f'(t)
Complex inversion integral	$ℒ^{- 1} {F (s)} = \frac{1}{2 π i} \int_{α - i \infty}^{α + i \infty} F (s) e^{s t} d s$
Practical methods	Tables, transform theorems, partial fractions, convolution

Definition and the Uniqueness Question

If $ℒ {f (t)} = F (s)$ , we write

ℒ^{- 1} {F (s)} = f (t)

and call $f (t)$ the inverse Laplace transform of $F (s)$ . The symbol $ℒ^{- 1}$ is the inverse Laplace transform operator.

For example, since $ℒ {\sin t} = \frac{1}{s^{2} + 1}$ ,

ℒ^{- 1} {\frac{1}{s^{2} + 1}} = \sin t .

A question immediately arises: is there another function $f (t) \neq \sin t$ whose transform is also $\frac{1}{s^{2} + 1}$ ? The next example settles the shape of the answer.

Find $ℒ {f (t)}$ , given

f(t) = \left\{ \begin{aligned} &0 && \text{if } t < 0 \\ &t && \text{if } 0 \leq t < 1 \\ &0.5 && \text{if } t = 1 \\ &0 && \text{if } t > 1 \end{aligned} \right.

Solution

Since $f$ vanishes outside $[0, 1)$ , the integral collapses to a finite one:

ℒ {f (t)} = \int_{0}^{\infty} f (t) e^{- s t} d t = \int_{0}^{1} t e^{- s t} d t .

Integration by parts with $u = t$ and $d v = e^{- s t} d t$ , so $v = - \frac{1}{s} e^{- s t}$ , gives

\begin{aligned} \int t\,e^{-st}\,dt &= -\frac{1}{s}te^{-st} - \int\left(-\frac{1}{s}e^{-st}\right)dt \\ &= -\frac{1}{s}te^{-st} - \frac{1}{s^{2}}e^{-st} \\ &= -\frac{st+1}{s^{2}}e^{-st} \end{aligned}

Therefore

\begin{aligned} \mathscr{L}\left\{f(t)\right\} &= \left[-\frac{st+1}{s^{2}}e^{-st}\right]_{t=0}^{t=1} \\ &= -\frac{1+s}{s^{2}}e^{-s} + \frac{1}{s^{2}} \\ &= \frac{1 - (1+s)e^{-s}}{s^{2}} . \end{aligned}

The point of the example. The value $f (1) = 0.5$ never entered the computation. Changing it, or leaving $f (1)$ undefined entirely, would give exactly the same transform. Defining

g(t) = \left\{ \begin{aligned} &0 && \text{if } t < 0 \\ &t && \text{if } 0 \leq t < 1 \\ &0.25 && \text{if } t = 1 \\ &0 && \text{if } t > 1 \end{aligned} \right.

gives $ℒ {f (t)} = ℒ {g (t)}$ . An integral cannot detect the value of its integrand at a single point.

Graph of a function equal to t on the interval from 0 to 1, with an open circle at the point (1,1), an isolated solid dot at (1, 0.5), and equal to zero for t greater than 1. — The function $f (t)$ . The isolated dot at $(1, 0.5)$ is the only place where $f$ differs from the function taking the value $0$ there, and the Laplace transform is blind to that difference.

So there are infinitely many functions, differing only in their values at isolated points, with the same Laplace transform. Consequently $ℒ {f} = ℒ {g}$ does not force $f = g$ everywhere. The next theorem says it does force them to be almost equal.

Lerch's Theorem. Suppose $f (t)$ and $g (t)$ are piecewise continuous and of exponential order as $t \to \infty$ . If $ℒ {f (t)} = ℒ {g (t)}$ for all $s > a$ , where $a$ is some constant, then

f (t) = g (t) for all t > 0

except possibly at the points of discontinuity.

It follows that if we restrict attention to functions continuous on $[0, \infty)$ , the inverse Laplace transform is unique, and we may speak of the inverse Laplace transform of $F (s)$ .

This is exactly the right amount of uniqueness for the applications. Solutions of differential equations are continuous, so the ambiguity never touches them. And when a discontinuous forcing produces a solution with a corner or a jump, the value assigned at that single instant is physically meaningless anyway.

Why the Inverse Matters

Solving differential equations is a primary application of the Laplace transform. The transform converts a differential equation in the unknown $y (t)$ into an equation in $Y (s)$ , which is then solved for $Y (s)$ . If the differential equation is linear with constant coefficients, the transformed equation is purely algebraic. The remaining task is to recover $y (t)$ from $Y (s)$ , that is, to carry out the inverse transform.

Flow diagram: the original differential equation in y(t) is transformed to an equation in Y(s), which is solved algebraically for Y(s), which is inverted back to the solution y(t). A dashed downward arrow labeled Difficult to Solve marks the direct route. — The strategy of the Laplace transform. The direct route down the left side, solving the differential equation as it stands, is hard. The route around the outside, transform then solve then invert, replaces one hard problem with two easy ones and an algebra step.

The Laplace transform would be of little value if we could not carry out the inverse transform and obtain $y (t)$ from $Y (s)$ .

How to Find the Inverse Laplace Transform

It can be shown that the inverse Laplace transform of $F (s)$ is given by the complex inversion integral, also called the Bromwich integral,

ℒ^{- 1} {F (s)} = \frac{1}{2 π i} \int_{α - i \infty}^{α + i \infty} F (s) e^{s t} d s,

where $α$ is a sufficiently large positive real number such that $F (s)$ is defined for $Re (s) \geq α$ , and $s$ is the complex variable $α + i ω$ . Evaluating such an integral requires basic complex analysis, so we shall make no use of it in this book.

Several other methods are available and are what we actually use:

Use of Laplace transform tables, read from right to left.
Use of theorems on the inverse transform, listed below.
Use of partial fractions, the workhorse for rational $F (s)$ .
Use of convolution integrals.

The Inverse Transform Is Linear

Proof of linearity

Let $ℒ {f (t)} = F (s)$ and $ℒ {g (t)} = G (s)$ . Since $ℒ$ is linear,

\begin{aligned} \mathscr{L}\left\{af(t) + bg(t)\right\} &= a\mathscr{L}\left\{f(t)\right\} + b\mathscr{L}\left\{g(t)\right\} \\ &= aF(s) + bG(s) . \end{aligned}

Applying $ℒ^{- 1}$ to both sides,

ℒ^{- 1} {a F (s) + b G (s)} = a f (t) + b g (t) .

But $f (t) = ℒ^{- 1} {F (s)}$ and $g (t) = ℒ^{- 1} {G (s)}$ , so

ℒ^{- 1} {a F (s) + b G (s)} = a ℒ^{- 1} {F (s)} + b ℒ^{- 1} {G (s)} .

Hence $ℒ^{- 1}$ is a linear operator.

Linearity is what makes partial fractions useful: break $F (s)$ into a sum of simple pieces, invert each piece from the table, and add.

Properties of the Inverse Transform

Each of the following is the direct counterpart of a property of the forward transform.

Let $ℒ^{- 1} {F (s)} = f (t)$ . Then for $a > 0$ ,

$ℒ^{- 1} {F (a s)} = \frac{1}{a} f (\frac{t}{a})$
$ℒ^{- 1} {F (s - a)} = e^{a t} f (t)$
$ℒ^{- 1} {e^{- a s} F (s)} = u (t - a) f (t - a)$
\mathscr{L}^{-1}\left\{sF(s) - f(0)\right\} = f'(t)

where

u

denotes the unit step function of Section 7.6.

Property 2 is the most immediately useful: a shift of the variable $s$ corresponds to multiplication by an exponential in $t$ . It is what turns $\frac{1}{(s - 3)^{2}}$ into $t e^{3 t}$ at a glance. Property 3 is its mirror image, and it is the reason the unit step function is unavoidable once shifted exponentials appear.

Partial Fractions in Practice

When $F (s)$ is a rational function, decomposing it into simple fractions reduces the problem to table lookups.

Find $ℒ^{- 1} {\frac{s + 7}{s^{2} - s - 6}}$ .

Solution

Factor the denominator as $(s - 3) (s + 2)$ and write

\frac{s + 7}{(s - 3) (s + 2)} = \frac{A}{s - 3} + \frac{B}{s + 2} .

Multiplying through by the denominator, $s + 7 = A (s + 2) + B (s - 3)$ .

Setting $s = 3$ gives $10 = 5 A$ , so $A = 2$ . Setting $s = - 2$ gives $5 = - 5 B$ , so $B = - 1$ .

Therefore

ℒ^{- 1} {\frac{s + 7}{s^{2} - s - 6}} = 2 ℒ^{- 1} {\frac{1}{s - 3}} - ℒ^{- 1} {\frac{1}{s + 2}} = 2 e^{3 t} - e^{- 2 t} .

Find $ℒ^{- 1} {\frac{1}{s^{2} + 4 s + 13}}$ .

Solution

The denominator has no real roots, so complete the square instead of factoring:

s^{2} + 4 s + 13 = (s + 2)^{2} + 9 .

Now

\frac{1}{(s + 2)^{2} + 9} = \frac{1}{3} \cdot \frac{3}{(s + 2)^{2} + 3^{2}} .

Since $ℒ^{- 1} {\frac{3}{s^{2} + 9}} = \sin 3 t$ , property 2 with $a = - 2$ gives

ℒ^{- 1} {\frac{1}{s^{2} + 4 s + 13}} = \frac{1}{3} e^{- 2 t} \sin 3 t .

Completing the square is the standard move whenever an irreducible quadratic appears. The shifted variable $s + 2$ signals the factor $e^{- 2 t}$ , and what remains is a plain sine or cosine.

Find $ℒ^{- 1} {\frac{2 s + 3}{s^{2} (s + 1)}}$ .

Solution

A repeated linear factor needs one term for each power:

\frac{2 s + 3}{s^{2} (s + 1)} = \frac{A}{s} + \frac{B}{s^{2}} + \frac{C}{s + 1} .

Multiplying by $s^{2} (s + 1)$ , $2 s + 3 = A s (s + 1) + B (s + 1) + C s^{2}$ .

Setting $s = 0$ gives $3 = B$ . Setting $s = - 1$ gives $1 = C$ . Comparing coefficients of $s^{2}$ gives $0 = A + C$ , so $A = - 1$ .

Therefore

ℒ^{- 1} {\frac{2 s + 3}{s^{2} (s + 1)}} = - 1 + 3 t + e^{- t},

using $ℒ^{- 1} {1 / s} = 1$ , $ℒ^{- 1} {1 / s^{2}} = t$ , and $ℒ^{- 1} {1 / (s + 1)} = e^{- t}$ .

Exercises

Find $ℒ^{- 1} {\frac{3}{s^{4}}}$ .

Solution

Since $ℒ {t^{3}} = \frac{3!}{s^{4}} = \frac{6}{s^{4}}$ , we have $ℒ^{- 1} {\frac{6}{s^{4}}} = t^{3}$ , so

ℒ^{- 1} {\frac{3}{s^{4}}} = \frac{t^{3}}{2} .

Find $ℒ^{- 1} {\frac{s - 1}{(s - 1)^{2} + 4}}$ .

Solution

The pattern $\frac{s}{s^{2} + 4}$ inverts to $\cos 2 t$ , and the shift $s \to s - 1$ multiplies by $e^{t}$ :

ℒ^{- 1} {\frac{s - 1}{(s - 1)^{2} + 4}} = e^{t} \cos 2 t .

Find $ℒ^{- 1} {\frac{5 s - 2}{s^{2} + s - 6}}$ .

Solution

Factoring, $s^{2} + s - 6 = (s + 3) (s - 2)$ , so

\frac{5 s - 2}{(s + 3) (s - 2)} = \frac{A}{s + 3} + \frac{B}{s - 2} .

Setting $s = - 3$ : $- 17 = - 5 A$ , so $A = \frac{17}{5}$ . Setting $s = 2$ : $8 = 5 B$ , so $B = \frac{8}{5}$ .

ℒ^{- 1} {\frac{5 s - 2}{s^{2} + s - 6}} = \frac{17}{5} e^{- 3 t} + \frac{8}{5} e^{2 t} .

Frequently Asked Questions

Is the inverse Laplace transform unique?

Essentially yes. By Lerch's theorem, two piecewise continuous functions of exponential order with the same transform agree everywhere except possibly at their points of discontinuity. Among continuous functions the inverse is genuinely unique, and since solutions of differential equations are continuous, the ambiguity never causes trouble.

Why can two different functions have the same transform?

Because the transform is an integral, and changing a function at finitely many isolated points does not change any integral. The worked example shows a function whose value at $t = 1$ can be anything at all without affecting $F (s)$ .

Do I need the complex inversion integral?

No. It is the theoretical definition, and it requires contour integration to evaluate. In practice, every inverse in this book is obtained by algebraic manipulation (usually partial fractions) followed by table lookup.

What is the general strategy for inverting a rational

F (s)

Factor the denominator, decompose into partial fractions, and invert each piece. Distinct linear factors give exponentials, repeated linear factors give $t^{n} e^{a t}$ , and irreducible quadratics give sines and cosines after completing the square. Linearity then assembles the answer.

What if the denominator does not factor over the reals?

Complete the square. Writing $s^{2} + 4 s + 13$ as $(s + 2)^{2} + 9$ puts $F$ into the form $F_{0} (s + 2)$ , and the shift property supplies the factor $e^{- 2 t}$ multiplying a sine or cosine. The pair of complex roots always yields a damped oscillation.

Why does an

e^{- a s}

factor call for a step function?

Because $e^{- a s} F (s)$ is the transform of a delayed version of $f$ : the original signal, shifted to start at $t = a$ and identically zero before that. The unit step $u (t - a)$ is what enforces the "zero before $a$ " part. Section 7.6 develops this fully.

Quick Reference

Definition and the Uniqueness Question

Why the Inverse Matters

How to Find the Inverse Laplace Transform

The Inverse Transform Is Linear

Properties of the Inverse Transform

Partial Fractions in Practice

Exercises

Frequently Asked Questions

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