Notes on the Laplace transform

David Pierce

November 2, 2006

Our aim is to develop a method for solving initial-value problems

\begin{matrix}y^{(n)} + a_{1}y^{(n−1)} + ⋯ + a_{n−1}y′ + a_{n}y = g, \\ y(t_{0}) = y_{0}, y′(t_{0}) = y′_{0}, …, y^{(n−1)}(t_{0}) = y_{0}^{(n−1)}, \end{matrix}  (1)

where the coefficients a_{i} are constant, although the forcing function g need not be constant. To give the conditions that the forcing function will have to satisfy, we need some definitions:

A function is piecewise continuous on an interval [a, b] if the interval has a partition

a = t_{0} < t_{1} < ⋯ < t_{n−1} < t_{n} = b  (2)

such that the function is continuous on each sub-interval (t_{k}, t_{k+1}) and has finite limits at the endpoints of these intervals. The value of the function at the points t_{k} is irrelevant. Indeed, we may say that f is piecewise continuous on [a, b] even if it is undefined at the points t_{k}. Theorem and definition. Suppose f is piecewise continuous on [0, b] for each positive b, and

lim_{t→∞} e^{−at}f(t) = 0  (3)

for some a. Then f has a Laplace transform, namely the function F on (a, ∞) given by

F (s) = ∫_{0}^{∞} e^{−st}f(t) d t.  (4)

Proof. By (3), there is some (positive) R such that |e^{−at}f(t)| < 1 when R < t. But then f is piecewise continuous on [0, R], so ∫_{0}^{R} e^{−st}f(t) d t exists. Also, when R < t, we have

|e^{−st}f(t)| ⩽ e^{(a−s)t};  (5)

and when a < s, the integral ∫_{R}^{∞} e^{(a−s)t} d t converges; so ∫_{R}^{∞} e^{−st}f(t) d t converges. Therefore the integral in (4) converges when a < s.

The property expressed by equation (3) is that f has exponential order. If f has the Laplace transform F, then we may write

F = ℒ {f} .  (6)

Often F is expressed in terms of its values F (s), and f in terms of f(t); then we may write

F (s) = ℒ {f(t)} .  (7)

The solution of problems (1) by means of Laplace transforms involves three essential lemmas: Lemma 1. The operation f ↦ ℒ {f} is linear, that is,

ℒ {c_{0}f_{0} + c_{1}f_{1}} = c_{0} · ℒ {f_{0}} + c_{1} · ℒ {f_{1}} .  (8)

Proof. Multiplication by e^{−st} and integration are linear.

Remembering the proof of the following may be an aid in remembering the lemma itself: Lemma 2. If the derivatives f^{(k)} are of exponential order and are continuous on [0, ∞) when k < n, and if f^{(n)} is piecewise continuous on [0, ∞), then ℒ {f^{(n)}} exists and is given by

ℒ {f^{(n)}} = s^{n} · ℒ {f} − (s^{n−1} · f(0) + s^{n−2} · f′(0) + ⋯ + f^{(n−1)}(0)). (9)

In particular,

ℒ {f′} = s · ℒ {f} − f(0),  (10)
ℒ {f′′} = s^{2} · ℒ {f} − s · f(0) − f′(0).  (11)

Proof. Use integration by parts and induction. Equation (9) holds trivially when n = 0. By integration by parts,

ℒ {f^{(m+1)}} = ∫_{0}^{∞} e^{−st}f^{(m)}(t) d t  (12)
= [e^{−st}f^{(m)}(t)]_{0}^{∞} + s · ∫_{0}^{∞} e^{−st}f^{(m)}(t) d t  (13)
= −f^{(m)}(0) + s · ℒ {f^{(m)}} ,  (14)

assuming f^{(m)} has exponential order. Hence, if (9) holds when n = m, then it holds when n = m + 1 (under the stated conditions).

Lemma 3. Suppose f and g are piecewise continuous on each interval [0, a]. Then ℒ {f} = ℒ {g} if and only if f(t) = g(t) for all but finitely many t in each interval [0, a].

Proof. Part of this is easy, since the integral of a function is not changed if the value of the function is changed at finitely many points. We shall not try to prove the hard part.

Lemma 3 means that, if ℒ {f} = F, then we can write

ℒ^{−1} {F } = f :  (15)

in a word, the operation f ↦ ℒ {f} is invertible.

To simplify matters, let us say we want to solve the intermediate-value problem

y′′ + ay′ + by = g, y(0) = y_{0}, y′(0) = y′_{0}.  (16)

(What is important here is that the differential equation is linear with constant coefficients.) By Lemma 1 we get

ℒ {y′′} + a · ℒ {y′} + b · ℒ {y} = ℒ {g} .  (17)

Writing Y for ℒ {y}, by Lemma 2 we get

s^{2} · Y − s · y_{0} − y′_{0} + as · Y − ay_{0} + b · Y = ℒ {g} ,  (18)

which we can solve for Y:

Y = \frac{y_{0}s + ay_{0} + y′_{0}}{s^{2} + as + b} + \frac{ℒ \{g\}}{s^{2} + as + b}.  (19)

So, to be able to solve (16) by means of Laplace transforms, we need:

1. to be able to find ℒ {g} easily;

2. to be able to recover y as ℒ^{−1} {Y }.

If g is a reasonable function, then (because of Lemma 3) we can accomplish these tasks by knowing some standard Laplace transforms.

First, we find the transforms of polynomials: Theorem 1. For all non-negative integers n,

ℒ \{t^{n}\} = \frac{n!}{s^{n+1}} .  (20)

Proof. Just compute:

ℒ \{1\} = ∫_{0}^{∞} e^{−st} d t = [\frac{e^{−st}}{−s} ]_{0}^{∞} = \frac{1}{s};  (21)
ℒ {t^{k+1}} = ∫_{0}^{∞} e^{−st}t^{k+1} d t  (22)
= [\frac{e^{−st}t^{k+1}}{−s} ]_{0}^{∞} + \frac{k + 1}{s} ∫_{0}^{∞} e^{−st}t^{k} d t  (23)
= \frac{k + 1}{s} · ℒ \{t^{k}\};  (24)

Now the claim follows by induction.

If we know that ℒ {f} = F, then we can recover the function whose Laplace transform is F (s − c): Theorem 2.

ℒ {e^{ct}f(t)} = F (s − c).  (25)

Proof. Apply definition (4) twice:

ℒ {e^{ct}f(t)} = ∫_{0}^{∞} e^{−(s−c)t}f(t) d t,  (26)

which is F (s − c).

Theorem 3.

ℒ \{cosh ct\} = \frac{s}{s^{2} − c^{2}} ,  ℒ \{cos ct\} = \frac{s}{s^{2} + c^{2}} ,
ℒ \{sinh ct\} = \frac{c}{s^{2} − c^{2}} ,  ℒ \{sin ct\} = \frac{c}{s^{2} + c^{2}} .

Proof. By Theorem 2 and (21),

ℒ \{e^{ct}\} = \frac{1}{s − c}.  (27)

Hence, by using the definitions

cosh t = \frac{e^{t} + e^{−t}}{2} , cos t = cosh it,
sinh t = \frac{e^{t} − e^{−t}}{2} , sin t = \frac{sinh it}{i},

(28)

we can compute the desired transforms.

An alternative proof of Theorem 3 uses the following rule. Theorem 4.

ℒ \{f(ct)\} = \frac{1}{c} · F (\frac{s}{c}) .  (29)

Proof. The computation is

ℒ \{f(ct)\} = ∫_{0}^{∞} e^{−st}f(ct) d t = \frac{1}{c} ∫_{0}^{∞} e^{−su/c}f(u) d u,  (30)

which is (1/c) · F (s/c).

A special case of (27) is

ℒ \{exp\} = \frac{1}{s − 1}.  (31)

Then, from (28), we get

ℒ \{cosh\} = \frac{s}{s^{2} − 1},  ℒ \{cos\} = \frac{s}{s^{2} + 1},
ℒ \{sinh\} = \frac{1}{s^{2} − 1},  ℒ \{sin\} = \frac{1}{s^{2} + 1}.

Now Theorem (3) follows. Theorem 5. If F is a rational function, then ℒ^{−1} {F } exists and is a linear combination of e^{ct} and e^{ct} cos bt and e^{ct} sin bt (for various b and c).

Proof. If F is a rational function, then we can write F (s) as a sum of ‘partial’ fractions of the forms

\frac{A}{s − a} and \frac{As + B}{s^{2} + as + b},  (32)

where a^{2} < 4b in the latter. Hence we can write F as a linear combination of fractions of the forms

\frac{1}{s − c} and \frac{s − c}{(s − c)^{2} + b^{2}} and \frac{b}{(s − c)^{2} + b^{2}} .  (33)

These are respectively the transforms of

e^{ct} and e^{ct} cos bt and e^{ct} sin bt;  (34)

so, by Lemma 1, we can compute ℒ^{−1} {F }.

Now we can use Laplace transforms to solve (16) (and higher-order problems) in case g is a linear combination of polynomial and exponential and trigonometric functions. The solution does not require solving a homogeneous equation first, or finding a general solution before using the initial conditions.

Piecewise-continuous forcing functions

If g is merely piecewise continuous, will still may be able to compute ℒ {g} by expressing g in terms of step-functions. Let u_{c} be the function on (−∞, ∞) given by

u_{c}(t) = \begin{cases}0, & if t < c; \\ 1, & if c ⩽ t. \end{cases}  (35)

If, for example, g is given by

g(t) = \begin{cases}g_{0}(t), & if  t ⩽ t_{1}; \\ g_{1}(t), & if t_{1} < t < t_{2}; \\ g_{2}(t), & if t_{2} ⩽ t; \end{cases}  (36)

then

g(t) = g_{0}(t) + u_{t_{1}}(t) · (g_{1}(t) − g_{0}(t)) + u_{t_{2}}(t) · (g_{2}(t) − g_{1}(t)),  (37)

unless t = t_{1}; this one difference won’t affect computation of the Laplace transform, by Lemma 3. We can perform this computation by means of the following. Theorem 6.

ℒ {u_{c}(t) · f(t − c)} = e^{−cs}F (s).  (38)

Proof. Compute:

ℒ {u_{c}(t) · f(t − c)} = ∫_{c}^{∞} e^{−st}f(t − c) d t = ∫_{0}^{∞} e^{−s(u+c)}f(u) d u,  (39)

which is e^{−cs}F (s).

Impulse functions

We have encountered no function whose Laplace transform is a non-zero constant. In calculus, when we note that 1/x is not the derivative of a rational function, we define ln x = ∫_{1}^{x} d t/t. Similarly, we look for a function δ (the Dirac delta) satisfying

ℒ {δ} = 1  (40)

and—so that Theorem 6 is satisfied—

ℒ {δ(t − c)} = e^{−cs}.  (41)

These properties are achieved if δ(t) = 0 when t ≠ 0, but ∫_{−∞}^{∞} δ = 1. There is no such function, literally; but the equation (19) still makes sense if g is a multiple of δ(t − c), and then we can find y by means of Theorem 6.

Convolution

Theorem and definition. If ℒ {f} = F and ℒ {g} = G, then

F · G = ℒ {f ∗ g} ,  (42)

where the convolution f ∗ g is defined by

f ∗ g(t) = ∫_{0}^{t} f(t − x) · g(x) d x.  (43)

In particular, Equation (19) has the solution y = ϕ + g ∗ h, where

ℒ \{ϕ\} = \frac{y_{0}s + ay_{0} + y′_{0}}{s^{2} + as + b} and ℒ \{h\} = \frac{1}{s^{2} + as + b}.  (44)

Note that ϕ and h are independent of g; also ϕ = 0 if y_{0} = 0 and y′_{0} = 0. Once ϕ and h are found, then solutions to (19) can be obtained by integration.