A successful strategy for many problems is to transform them into simpler ones that can be solved more easily. For example, some problems in rectangular co- ordinates are better understood and handled in polar coordinates, so we make the usual coordinate transformationx=rcosθ and y =rsinθ. After solving the problem in polar coordinates, we can return to rectangular coordinates by the inverse transformation r =
x2+y2, θ = arctany
x. A similar technique holds true for many differential equations using integral transform meth- ods. In this chapter we introduce the Laplace transformation which has the
effect of turning a differential equation with state functionu(t) into an algebra problem for an associated transformed functionU(s); we can easily solve the algebra problem forU(s) and then return tou(t) via an inverse transformation.
The technique is applicable to both homogeneous and nonhomogeneous linear differential equations with constant coefficients, and it is a standard method for engineers and applied mathematicians. It is particularly useful for differential equations that contain piecewise continuous forcing functions or functions that act as an impulse. The transform goes back to the late 1700s and is named for the great French mathematician and scientist Pierre de Laplace, although the basic integral goes back earlier to L. Euler. The English engineer O. Heaviside developed much of the operational calculus for transform methods in the early 1900s.
Let u = u(t) be a given function defined on 0 ≤ t < ∞. The Laplace transformofu(t) is the functionU(s) defined by
U(s) = ∞
0
u(t)e−stdt, (4.1)
provided the improper integral exists. The integrand is a function oft and s, and we integrate ont, leaving a function ofs. Often we represent the Laplace transform in function notation,
L[u(t)](s) =U(s) or just L[u] =U(s).
Lrepresents a function-like operation, called an operator or transform, whose domain and range are sets of functions;Ltakes a functionu(t) and transforms it into a new functionU(s) (see figure 4.1). In the context of Laplace transfor- mations,tanduare called thetime domainvariables, andsandU are called the transform domainvariables. In summary, the Laplace transform maps functionsu(t) to functionsU(s) and is somewhat like mappings we consider in calculus, such asy=f(x) =x2, which maps numbersxto numbersy.
We can compute the Laplace transform of many common functions directly from the definition (4.1).
Example 4.1
Letu(t) =eat. Then U(s) =
∞
0
eate−stdt= ∞
0
e(a−s)tdt= 1
a−se(a−s)t|t=∞t=0 = 1
s−a, s > a.
In different notation,L[eat] = s−a1 .Observe that this transform exists only for s > a(otherwise the integral does not exist). Sometimes we indicate the values ofsfor which the transformed functionU(s) is defined.
U(s) u(t)
Laplace Transform u
t
s U
Figure 4.1 The Laplace transform as a machine that transforms functions u(t) to functionsU(s).
Example 4.2
Letu(t) =t. Then, using integration by parts, U(s) =
∞
0
te−stdt=
te−st
−s t=∞
t=0
−1 s
∞
0
1ãe−stdt= 1
s2, s >0.
Example 4.3
The unit switching functionha(t) is defined byha(t) = 0 ift < aandha(t) = 1 if t ≥ a. The switch is off if t < a, and it is on when t ≥ a. Therefore the functionha(t) is a step function where the step from 0 to 1 occurs at t =a.
The switching function is also called the Heaviside function. The Laplace transform ofha(t) is
L[ha(t)] = ∞
0
ha(t)e−stdt
= a
0
ha(t)e−stdt+ ∞
a
ha(t)e−stdt
= a
0
0ãe−stdt+ ∞
a
1ãe−stdt
= −1
se−st|t=∞t=a = 1
se−as, s >0.
Example 4.4
The Heaviside function is useful for expressing multi-lined functions in a single formula. For example, let
f(t) =
⎧⎪
⎪⎨
⎪⎪
⎩
12, 0≤t <2 t−1, 2≤t≤3 5−t2, 3< t≤6
0, t >6
(The reader should plot this function). This can be written in one line as f(t) =1
2h0(t) + (t−1−1
2)h2(t) + (5−t2−(t−1))h3(t)−(5−t2)h6(t).
The first term switches on the function 1/2 att= 0; the second term switches off 1/2 and switches ont−1 at timet= 2; the third term switches offt−1 and switches on 5−t2 at t = 3; finally, the last term switches off 5−t2 att = 6.
Later we show how to find Laplace transforms of such functions.
As you may have already concluded, calculating Laplace transforms may be tedious business. Fortunately, generations of mathematicians, scientists, and engineers have computed the Laplace transforms of many, many functions, and the results have been catalogued in tables and in software systems. Some of the tables are extensive, but here we require only a short table, which is given at the end of the chapter. The table lists a function u(t) in the first column, and its transform U(s), or Lu, in the second. The various functions in the first column are discussed in the sequel. Computer algebra systems also have commands that calculate the Laplace transform (see Appendix B).
Therefore, givenu(t), the Laplace transformU(s) can be computed by the definition, given in formula (4.1). We can also think of the opposite problem:
given U(s), find a functionu(t) whose Laplace transform is U(s). This is the inverse problem. Unfortunately, there is no elementary formula that we can write down that computes u(t) in terms of U(s) (there is a formula, but it involves a contour integration in the complex plane). In elementary treatments we are satisfied with using tables. For example, if U(s) = s−21 ,then the table givesu(t) =e2tas the function that hasU(s) as its transform. When we think of it this way, we sayu(t) =e2tis the “inverse transform” ofU(s) =s−21 ,and we write
e2t=L−1 1
s−2
. In general we use the notation
U =L(u), u=L−1[U].
We think of L as an operator (transform) and L−1 as the inverse operation (inverse transform). The functionsu(t) andU(s) form a transform pair, and they are listed together in two columns of a table. Computer algebra systems also supply inverse transforms.
One question that should be addressed concerns theexistence of the trans- form. That is, which functions have Laplace transforms? Clearly if a function grows too quickly as t gets large, then the improper integral will not exist and there will be no transform. There are two conditions that guarantee ex- istence, and these are reasonable conditions for most problems in science and engineering. First, we require thatu(t) not grow too fast; a way of stating this mathematically is to require that there exist constantsM >0 andαfor which
|u(t)| ≤M eαt
is valid for allt > t0,wheret0is some value of time. That is, beyond the value t0 the function is bounded above and below by an exponential function. Such functions are said to be ofexponential order. Second, we require thatu(t) be piecewise continuouson 0≤t <∞.In other words, the interval 0≤t <∞ can be divided into intervals on which u is continuous, and at any point of discontinuityuhas finite left and right limits, except possibly att= +∞. One can prove that if uis piecewise continuous on 0≤t < ∞and of exponential order, then the Laplace transformU(s) exists for alls > α.
What makes the Laplace transform so useful for differential equations is that it turns derivative operations in the time domain into multiplication operations in the transform domain. The following theorem gives the crucial operational formulas stating how the derivatives transform.
Theorem 4.5
Letu(t) be a function and U(s) its transform. Then
L[u] = sU(s)−u(0), (4.2)
L[u] = s2U(s)−su(0)−u(0). (4.3) Proof. These facts are easily proved using integration by parts. We have
L[u] = ∞
0
u(t)e−stdt=
u(t)e−stt=∞
t=0 − ∞
0 −su(t)e−stdt
= −u(0) +sU(s), s >0.
The second operational formula (4.3) is derived using two successive integration by parts, and we leave the calculation to the reader.
These formulas allow us to transform a differential equation with unknown u(t) into an algebraic problem with unknownU(s). We solve forU(s) and then findu(t) using the inverse transformu=L−1[U].We elaborate on this method in the next section.
Before tackling the solution of differential equations, we present additional important and useful properties.
(a) (Linearity) The Laplace transform is a linear operation; that is, the Laplace transform of a sum of two functions is the sum of the Laplace transforms of each, and the Laplace transform of a constant times a func- tion is the constant times the transform of the function. We can express these rules in symbols by a single formula:
L[c1u+c2v] =c1L[u] +c2L[v]. (4.4) Here, uandv are functions andc1 andc2are any constants. Similarly, the inverse Laplace transform is a linear operation:
L−1[c1u+c2v] =c1L−1[u] +c2L−1[v]. (4.5) (b) (Shift Property) The Laplace transform of a function times an exponen-
tial,u(t)eat, shifts the transform of U; that is,
L[u(t)eat] =U(s−a). (4.6) (c) (Switching Property) The Laplace transform of a function that switches
on att=ais given by
L[ha(t)u(t−a)] =U(s)e−as. (4.7) Proofs of some of these relations follow directly from the definition of the Laplace transform, and they are requested in the Exercises.
EXERCISES
1. Use the definition of the Laplace transform to compute the transform of the square pulse functionu(t) = 1, 1≤t≤2;u(t) = 0, otherwise.
2. Derive the operational formula (4.3).
3. Sketch the graphs of sint,sin(t−π/2),andhπ/2(t) sin(t−π/2). Find the Laplace transform of each.
4. Find the Laplace transform oft2e−3t.
5. FindL[sinhkt] andL[coshkt] using the fact thatL ekt
= s−k1 .
6. FindL
e−3t+ 4 sinkt
using the table. FindL
e−3tsin 2t
using the shift property (4.6).
7. Using the switching property (4.7), find the Laplace transform of the func- tion
u(t) =
0 t <2 e−t, t >2.
8. From the definition (4.1), find L 1/√
t
using the integral substitution st=r2 and then noting∞
0 exp(−r2)dr=√ π/2.
9. Does the functionu(t) =et2have a Laplace transform? What aboutu(t) = 1/t? Comment.
10. Derive the operational formulas (4.6) and (4.7).
11. Plot thesquare-wave function f(t) =
∞ n=0
(−1)nhn(t)
on the interval t >0 and find its transformF(s). (Hint: use the geometric series 1 +x+x2+ã ã ã= 1−x1 .)
12. Show that
L t
0
u(r)dr
= U(s) s . 13. Derive the formulas
L[tu(t)] =−U(s), L−1[U(s)] =−tu(t).
Use these formulas to find the inverse transform of arctana
s. 14. Show that
L u(t)
t
= ∞
s
U(r)dr, and use the result to find
L sinht
t
. 15. Show that
L[f(t)ha(t)] =e−asL[f(t+a)], and use this formula to computeL[t2h1(t)].
16. Find the Laplace transform of the function in Example 4.4.
ODE in u(t) Algebraic equation in U(s)
solve u(t) U(s)
L
L-1
Figure 4.2 A DE for an unknown functionu(t) is transformed to an algebraic equation for its transform U(s). The algebraic problem is solved for U(s) in the transform domain, and the solution is returned to the original time domain via the inverse transform.
17. Thegamma functionis defined by Γ(x) =
∞
0
e−ttx−1dt, x >−1.
a) Show thatΓ(n+1) =nΓ(n) andΓ(n+1) =n! for nonnegative integers n. Show thatΓ(12) =√
π.
b) Show that L[ta] = Γ(a+1)
sa+1 , s >0.