The Laplace Transform: Definition, Properties, and Solving IVPs

A linear constant-coefficient equation can be solved by guessing a form for the particular solution and fitting the arbitrary constants to the initial data at the end. The Laplace transform reorganizes that work: it converts the whole initial value problem into a single algebraic equation, folds the initial conditions in from the start, and treats forced and unforced equations identically. It adds one inversion step at the end, run against a table.

The idea is an integral transform: multiply the unknown function by a fixed kernel $K (s, t)$ and integrate out the time variable, producing a new function of $s$ .¹

F (s) = α \int β K (s, t) f (t) d t .

Different kernels give different transforms. The Laplace transform takes $K (s, t) = e^{- s t}$ on $[0, \infty)$ . That kernel suits linear constant-coefficient differential equations because the derivatives of $e^{- s t}$ reproduce $e^{- s t}$ up to a factor of $s$ , so integration by parts trades a derivative in $t$ for a multiplication by $s$ .

The definition

The integral runs to infinity, so convergence is not automatic. Two examples fix the boundary cases. For the constant function $f (t) = 1$ ,

L {1} = 0 \int \infty e^{- s t} d t = A \to \infty lim [- \frac{e ^{- s t}}{s}]_{0}^{A} = \frac{1}{s}, s > 0,

and the limit exists only when $s > 0$ ; for $s \leq 0$ the integrand does not decay and the integral diverges. For the exponential $f (t) = e^{a t}$ ,

L {e^{a t}} = 0 \int \infty e^{- (s - a) t} d t = \frac{1}{s - a}, s > a .

Each transform comes with a half-line of validity $s > a$ : the transform is a function, but it is only defined where its defining integral converges.

Existence: piecewise continuity and exponential order

Two structural conditions on $f$ guarantee the integral converges. The first controls behavior on finite intervals; the second controls growth at infinity.

A function of exponential order stays under an envelope

K e^{a t}

; multiplying by

e^{- s t}

with

s > a

forces the integrand to decay, so the defining integral converges.

The proof splits $\int_{0}^{\infty}$ at $M$ . The piece $\int_{0}^{M}$ exists because $f$ is piecewise continuous. On the tail, $∣ e^{- s t} f (t) ∣ \leq K e^{(a - s) t}$ , and $\int_{M}^{\infty} e^{(a - s) t} d t$ converges whenever $a - s < 0$ , i.e. $s > a$ . A comparison with the decaying exponential does the rest. Every function treated here except the Dirac delta satisfies these hypotheses.

Linearity and the transform table

The integral is linear in its integrand, so the transform is a linear operator.

Linearity plus a few integrals fills a table. The transform of $t^{n}$ follows by induction (or from the gamma function), and $sinh, cosh$ come from writing them as sums of exponentials. The trigonometric entries take two integrations by parts, which fold the transform back onto itself.³

$f (t)$	$F (s) = L {f (t)}$	valid for
$1$	$1/ s$	$s > 0$
$e^{a t}$	$1/ (s - a)$	$s > a$
$t^{n}$ , $n$ a positive integer	$n! / s^{n + 1}$	$s > 0$
$t^{p}$ , $p > - 1$	$Γ (p + 1) / s^{p + 1}$	$s > 0$
$sin (a t)$	$a / (s^{2} + a^{2})$	$s > 0$
$cos (a t)$	$s / (s^{2} + a^{2})$	$s > 0$
$sinh (a t)$	$a / (s^{2} - a^{2})$	$s > ∣ a ∣$
$cosh (a t)$	$s / (s^{2} - a^{2})$	$s > ∣ a ∣$
$e^{a t} sin (b t)$	$b / ((s - a)^{2} + b^{2})$	$s > a$
$e^{a t} cos (b t)$	$(s - a) / ((s - a)^{2} + b^{2})$	$s > a$
$t^{n} e^{a t}$	$n! / (s - a)^{n + 1}$	$s > a$

Because a transform corresponds to essentially one function, reading the table right to left inverts a transform. Two piecewise-continuous functions can differ at isolated points and share a transform, but among continuous functions the correspondence is one-to-one, so the inverse transform $L^{- 1}$ is well defined on continuous functions, and linear: $L^{- 1} {F_{1} + \dots + F_{n}} = L^{- 1} {F_{1}} + \dots + L^{- 1} {F_{n}}$ .

The frequency-shift rule

Multiplying $f$ by $e^{c t}$ in the $t$ -domain slides its transform in the $s$ -domain. This one rule generates the last three table rows and handles quadratic denominators.

The proof is one line: $\int_{0}^{\infty} e^{- s t} e^{c t} f (t) d t = \int_{0}^{\infty} e^{- (s - c) t} f (t) d t = F (s - c)$ . Applied to $L {sin (b t)} = b / (s^{2} + b^{2})$ , it gives $L {e^{a t} sin (b t)} = b / ((s - a)^{2} + b^{2})$ directly.

The first shifting rule slides the transform: multiplying

f

e^{c t}

moves the graph of

F (s)

a distance

c

to the right in the

s

-domain, taking

F (s)

F (s - c)

Read backwards, the rule inverts transforms whose denominators do not factor over the reals: complete the square.

Transforms of derivatives

The property that turns calculus into algebra relates the transform of $f^{'}$ to the transform of $f$ .

Integration by parts gives it: $\int_{0}^{A} e^{- s t} f^{'} (t) d t = e^{- s A} f (A) - f (0) + s \int_{0}^{A} e^{- s t} f (t) d t$ , and $e^{- s A} f (A) \to 0$ as $A \to \infty$ because $s > a$ .⁴ Applying the rule to $f^{'}$ in place of $f$ compounds it:

L {f^{''} (t)} = s L {f^{'} (t)} - f^{'} (0) = s^{2} L {f (t)} - s f (0) - f^{'} (0) .

Each derivative pulls out one more power of $s$ and one initial value.

The polynomial in $s$ that multiplies $F (s)$ reproduces the characteristic polynomial of the differential operator, and the trailing terms carry the initial data.

Solving initial value problems

Transforming a constant-coefficient linear equation replaces each $L {y^{(k)}}$ by its expression in $Y (s) = L {y}$ and the initial values. The differential equation becomes an algebraic equation for $Y (s)$ ; solving it, decomposing into table entries, and inverting recovers $y (t)$ .

The transform method routes around the differential equation: transform to the s-domain, solve algebraically for

Y (s)

, and invert. The direct route (dashed) is the classical method.

The transform method as a fixed procedure: four steps from the initial value problem to the solution, with the table used at both the transform and the inversion step.

Written as a procedure, the method never varies:

Algorithm:

\textsc{SolveIVP}

— solve

a y'' + b y' + c y = g(t)

with

y(0)=y_0,\ y'(0)=y_0'

1
transform both sides: $a\,\mathcal{L}\{y''\} + b\,\mathcal{L}\{y'\} + c\,\mathcal{L}\{y\} = G(s)$
2
substitute $\mathcal{L}\{y'\} = sY - y_0$ and $\mathcal{L}\{y''\} = s^2 Y - s y_0 - y_0'$
3
collect terms and solve the algebraic equation for $Y(s)$
4
decompose $Y(s)$ into partial fractions matching table entries
5
invert term by term: $y(t) \gets \mathcal{L}^{-1}\{Y(s)\}$
6
return $y(t)$

The transform $Y (s)$ is found by solving an algebraic equation, not a differential one. The initial conditions enter at the transform step, so no arbitrary constants remain to fit at the end. A nonhomogeneous equation is handled exactly like a homogeneous one, with $G (s)$ on the right rather than zero. And the method extends to any order $n$ unchanged, using the $n$ th-derivative rule. The one genuine difficulty is the last line: recovering $y (t)$ from $Y (s)$ , the inversion problem.

Partial fractions and inversion

Inversion works by decomposing $Y (s)$ into a sum of terms that appear in the table. When $Y (s) = P (s) / Q (s)$ is a rational function with $de g P < de g Q$ and $Q$ has distinct real roots, the decomposition is a sum of simple fractions.

Partial-fraction inversion: a proper rational

Y (s) = P (s) / Q (s)

with simple real roots splits into terms

c_{k} / (s + p_{k})

, each inverting to an exponential

c_{k} e^{r_{k} t}

by the table.

The homogeneous worked example above ran exactly this split, with $Q (s) = (s - 2) (s + 1)$ and residues $1/3$ and $2/3$ . When the denominator carries an irreducible quadratic factor, the corresponding numerator is linear.

For a denominator $Q (s)$ of degree $n$ with distinct roots $r_{1}, \dots, r_{n}$ , the coefficient of $1/ (s - r_{k})$ is $P (r_{k}) / Q^{'} (r_{k})$ , and the inverse transform is $\sum_{k} \frac{P ( r _{k} )}{Q ^{'} ( r _{k} )} e^{r_{k} t}$ — a compact residue formula.

Differentiating and integrating the transform

Differentiating $F (s)$ under the integral sign multiplies the original function by $- t$ , which produces transforms of $t^{n} f (t)$ without new integration.

For example, from $L {sin (a t)} = a / (s^{2} + a^{2})$ ,

L {t sin (a t)} = - \frac{d}{d s} (\frac{a}{s ^{2} + a ^{2}}) = \frac{2 a s}{( s ^{2} + a ^{2} ) ^{2}} .

Because $t$ appears as a first-power coefficient in some variable-coefficient equations, this rule occasionally converts such an equation into a first-order equation for $Y (s)$ ; that is how the transform of the Bessel function $J_{0}$ can be found.⁵ Integrating the transform runs the other way: if $f (t) / t$ has a limit at $0$ , then $L {f (t) / t} = \int_{s}^{\infty} F (σ) d σ$ , and letting $s \to 0$ evaluates $\int_{0}^{\infty} f (t) / t d t$ . For $f = sin t$ this gives the value $π /2$ for the Dirichlet integral.

Benefits and costs of the method

The transform is a change of representation, with definite benefits and costs.

Differentiation becomes multiplication. Each $d / d t$ becomes a factor of $s$ , and the differential equation becomes algebraic.
Initial data is built in. The values $y (0), y^{'} (0), \dots$ enter at the transform step, so the general-solution-then-fit-constants detour disappears.
Forcing is uniform. Nonhomogeneous and homogeneous equations transform the same way; discontinuous and impulsive forcing are no harder in the $s$ -domain.
Roots still matter. The denominator of $Y (s)$ is the characteristic polynomial, so factoring it — the same obstruction as before — is unavoidable.
Inversion is the real work. Going from $Y (s)$ back to $y (t)$ relies on the table, partial fractions, the shift rules, and the convolution theorem.

Simmons, Differential Equations with Applications and Historical Notes, §48 — Introduction: the Laplace transform as an integral transform with kernel $e^{- pt}$ , converting differentiation into multiplication. ↩
Boyce, Elementary Differential Equations and Boundary Value Problems, §6.1, Theorems 6.1.1–6.1.2 — the comparison test for improper integrals and existence of the transform for functions piecewise continuous and of exponential order, valid for $s > a$ . ↩
Simmons, §49 — A Few Remarks on the Theory; and Boyce, §6.1, Examples 4–8 and Table 6.2.1 — the elementary transforms of $1$ , $e^{a t}$ , $t^{n}$ , $sin (a t)$ , $cos (a t)$ , and their combinations, assembled by linearity. ↩
Boyce, §6.2, Theorem 6.2.1 and Corollary 6.2.2 — $L {f^{'}} = s F (s) - f (0)$ and its $n$ -fold extension, the identity that reduces a constant-coefficient IVP to an algebraic equation whose denominator is the characteristic polynomial. ↩
Simmons, §51 — Derivatives and Integrals of Laplace Transforms: $L {- t f (t)} = F^{'} (s)$ and $L {f (t) / t} = \int_{s}^{\infty} F (σ) d σ$ , with the Bessel-equation and Dirichlet-integral applications. ↩