Laplace Transform

Laplace Transform

The Laplace transform can be used to solve differential equations. Be- sides being a different and efficient alternative to variation of parame- ters and undetermined coefficients, the Laplace method is particularly advantageous for input terms that are piecewise-defined, periodic or im- pulsive.

The direct Laplace transform or the Laplace integral of a function f (t) defined for 0 ≤ t < ∞ is the ordinary calculus integration problem

Z ∞ 0

f (t)e

^−st

dt,

succinctly denoted L(f (t)) in science and engineering literature. The L–notation recognizes that integration always proceeds over t = 0 to t = ∞ and that the integral involves an integrator e

^−st

dt instead of the usual dt. These minor differences distinguish Laplace integrals from the ordinary integrals found on the inside covers of calculus texts.

7.1 Introduction to the Laplace Method

The foundation of Laplace theory is Lerch’s cancellation law

R∞

0

y(t)e

^−st

dt =

^R₀^∞

f (t)e

^−st

dt implies y(t) = f (t), or

L(y(t) = L(f (t)) implies y(t) = f (t).

(1)

In differential equation applications, y(t) is the sought-after unknown while f (t) is an explicit expression taken from integral tables.

Below, we illustrate Laplace’s method by solving the initial value prob- lem

y

⁰

= −1, y(0) = 0.

The method obtains a relation L(y(t)) = L(−t), whence Lerch’s cancel- lation law implies the solution is y(t) = −t.

The Laplace method is advertised as a table lookup method, in which

the solution y(t) to a differential equation is found by looking up the

answer in a special integral table.

7.1 Introduction to the Laplace Method 247

Laplace Integral. The integral

^R₀^∞

g(t)e

^−st

dt is called the Laplace integral of the function g(t). It is defined by lim

_{N →∞}^R₀^N

g(t)e

^−st

dt and depends on variable s. The ideas will be illustrated for g(t) = 1, g(t) = t and g(t) = t

²

, producing the integral formulas in Table 1.

R∞

0

(1)e

^−st

dt = −(1/s)e

^−st

t=∞

t=0

Laplace integral of g(t) = 1.

= 1/s Assumed s > 0.

R∞

0

(t)e

^−st

dt =

^R₀^∞

−

_ds^d

(e

^−st

)dt Laplace integral of g(t) = t.

= −

_ds^{d R}₀^∞

(1)e

^−st

dt Use

^R _ds^d

F (t, s)dt =

_ds^{d R}

F (t, s)dt.

= −

_ds^d

(1/s) Use L(1) = 1/s.

= 1/s

²

Differentiate.

R∞

0

(t

²

)e

^−st

dt =

^R₀^∞

−

_ds^d

(te

^−st

)dt Laplace integral of g(t) = t

²

.

= −

_ds^{d R}₀^∞

(t)e

^−st

dt

= −

_ds^d

(1/s

²

) Use L(t) = 1/s

²

.

= 2/s

³

Table 1. The Laplace integral R∞

0 g(t)e^−stdt for g(t) = 1, t and t².

R∞

0

(1)e

^−st

dt = 1 s

R∞

0

(t)e

^−st

dt = 1 s

²

R∞

0

(t

²

)e

^−st

dt = 2 s

³

In summary, L(t

ⁿ

) = n!

s

¹⁺ⁿ

An Illustration. The ideas of the Laplace method will be illus- trated for the solution y(t) = −t of the problem y

⁰

= −1, y(0) = 0. The method, entirely different from variation of parameters or undetermined coefficients, uses basic calculus and college algebra; see Table 2.

Table 2. Laplace method details for the illustration y⁰= −1, y(0) = 0.

y

⁰

(t)e

^−st

= −e

^−st

Multiply y

⁰

= −1 by e

^−st

.

R∞

0

y

⁰

(t)e

^−st

dt =

^R₀^∞

−e

^−st

dt Integrate t = 0 to t = ∞.

R∞

0

y

⁰

(t)e

^−st

dt = −1/s Use Table 1.

s

^R₀^∞

y(t)e

^−st

dt − y(0) = −1/s Integrate by parts on the left.

R∞

0

y(t)e

^−st

dt = −1/s

²

Use y(0) = 0 and divide.

R∞

0

y(t)e

^−st

dt =

^R₀^∞

(−t)e

^−st

dt Use Table 1.

y(t) = −t Apply Lerch’s cancellation law.

In Lerch’s law, the formal rule of erasing the integral signs is valid pro- vided the integrals are equal for large s and certain conditions hold on y and f – see Theorem 2. The illustration in Table 2 shows that Laplace theory requires an in-depth study of a special integral table, a table which is a true extension of the usual table found on the inside covers of calculus books. Some entries for the special integral table appear in Table 1 and also in section 7.2, Table 4.

The L-notation for the direct Laplace transform produces briefer details, as witnessed by the translation of Table 2 into Table 3 below. The reader is advised to move from Laplace integral notation to the L–notation as soon as possible, in order to clarify the ideas of the transform method.

Table 3. Laplace method L-notation details for y⁰ = −1, y(0) = 0 translated from Table 2.

L(y

⁰

(t)) = L(−1) Apply L across y

⁰

= −1, or multiply y

⁰

=

−1 by e

^−st

, integrate t = 0 to t = ∞.

L(y

⁰

(t)) = −1/s Use Table 1.

sL(y(t)) − y(0) = −1/s Integrate by parts on the left.

L(y(t)) = −1/s

²

Use y(0) = 0 and divide.

L(y(t)) = L(−t) Apply Table 1.

y(t) = −t Invoke Lerch’s cancellation law.

Some Transform Rules. The formal properties of calculus integrals plus the integration by parts formula used in Tables 2 and 3 leads to these rules for the Laplace transform:

L(f (t) + g(t)) = L(f (t)) + L(g(t)) The integral of a sum is the sum of the integrals.

L(cf (t)) = cL(f (t)) Constants c pass through the integral sign.

L(y

⁰

(t)) = sL(y(t)) − y(0) The t-derivative rule, or inte- gration by parts. See Theo- rem 3.

L(y(t)) = L(f (t)) implies y(t) = f (t) Lerch’s cancellation law. See Theorem 2.

1 Example (Laplace method) Solve by Laplace’s method the initial value problem y

⁰

= 5 − 2t, y(0) = 1.

Solution:

Laplace’s method is outlined in Tables 2 and 3. The L-notation of Table 3 will be used to find the solution y(t) = 1 + 5t − t².

7.1 Introduction to the Laplace Method 249

L(y⁰(t)) = L(5 − 2t) Apply L across y⁰ = 5 − 2t.

L(y⁰(t)) = 5 s− 2

s² Use Table 1.

sL(y(t)) − y(0) = 5 s − 2

s² Apply the t-derivative rule, page 248.

L(y(t)) = 1 s+ 5

s² − 2

s³ Use y(0) = 1 and divide.

L(y(t)) = L(1) + 5L(t) − L(t²) Apply Table 1, backwards.

= L(1 + 5t − t²) Linearity, page 248.

y(t) = 1 + 5t − t² Invoke Lerch’s cancellation law.

2 Example (Laplace method) Solve by Laplace’s method the initial value problem y

⁰⁰

= 10, y(0) = y

⁰

(0) = 0.

Solution:

The L-notation of Table 3 will be used to find the solution y(t) = 5t². L(y⁰⁰(t)) = L(10) Apply L across y⁰⁰= 10.

sL(y⁰(t)) − y⁰(0) = L(10) Apply the t-derivative rule to y⁰, that is, replace y by y⁰ on page 248.

s[sL(y(t)) − y(0)] − y⁰(0) = L(10) Repeat the t-derivative rule, on y.

s²L(y(t)) = L(10) Use y(0) = y⁰(0) = 0.

L(y(t)) = 10

s³ Use Table 1. Then divide.

L(y(t)) = L(5t²) Apply Table 1, backwards.

y(t) = 5t² Invoke Lerch’s cancellation law.

Existence of the Transform. The Laplace integral

^R₀^∞

e

^−st

f (t) dt is known to exist in the sense of the improper integral definition

¹

Z ∞

0

g(t)dt = lim

N →∞

Z N 0

g(t)dt

provided f (t) belongs to a class of functions known in the literature as functions of exponential order. For this class of functions the relation

t→∞

lim f (t)

e

^at

= 0 (2)

is required to hold for some real number a, or equivalently, for some constants M and α,

|f (t)| ≤ M e

^αt

. (3)

In addition, f (t) is required to be piecewise continuous on each finite subinterval of 0 ≤ t < ∞, a term defined as follows.

1An advanced calculus background is assumed for the Laplace transform existence proof. Applications of Laplace theory require only a calculus background.

Definition 1 (piecewise continuous)

A function f (t) is piecewise continuous on a finite interval [a, b] pro- vided there exists a partition a = t

₀

< · · · < t

_n

= b of the interval [a, b]

and functions f

1

, f

2

, . . . , f

n

continuous on (−∞, ∞) such that for t not a partition point

f (t) =







f

₁

(t) t

₀

< t < t

₁

,

.. . .. .

f

_n

(t) t

_n−1

< t < t

_n

. (4)

The values of f at partition points are undecided by equation (4). In particular, equation (4) implies that f (t) has one-sided limits at each point of a < t < b and appropriate one-sided limits at the endpoints.

Therefore, f has at worst a jump discontinuity at each partition point.

3 Example (Exponential order) Show that f (t) = e

^t

cos t + t is of expo- nential order, that is, show that f (t) is piecewise continuous and find α > 0 such that lim

t→∞

f (t)/e

^αt

= 0.

Solution:

Already, f (t) is continuous, hence piecewise continuous. From L’Hospital’s rule in calculus, limt→∞p(t)/e^αt = 0 for any polynomial p and any α > 0. Choose α = 2, then

t→∞lim f (t)

e^2t = lim

t→∞

cos t e^t + lim

t→∞

t e^2t = 0.

Theorem 1 (Existence of L(f ))

Let f (t) be piecewise continuous on every finite interval in t ≥ 0 and satisfy

|f (t)| ≤ M e

^αt

for some constants M and α. Then L(f (t)) exists for s > α and lim

_s→∞

L(f (t)) = 0.

Proof:

It has to be shown that the Laplace integral of f is finite for s > α.

Advanced calculus implies that it is sufficient to show that the integrand is ab- solutely bounded above by an integrable function g(t). Take g(t) = M e^−(s−α)t. Then g(t) ≥ 0. Furthermore, g is integrable, because

Z ∞ 0

g(t)dt = M s − α.

Inequality |f(t)| ≤ Me^αt implies the absolute value of the Laplace transform integrand f (t)e^−st is estimated by

f (t)e^−st

≤ Me^αte^−st= g(t).

The limit statement follows from |L(f(t))| ≤R∞

0 g(t)dt = M

s − α, because the right side of this inequality has limit zero at s = ∞. The proof is complete.

7.1 Introduction to the Laplace Method 251 Theorem 2 (Lerch)

If f

₁

(t) and f

₂

(t) are continuous, of exponential order and

^R₀^∞

f

₁

(t)e

^−st

dt =

R∞

0

f

₂

(t)e

^−st

dt for all s > s

₀

, then f

₁

(t) = f

₂

(t) for t ≥ 0.

Proof: See Widder [?].

Theorem 3 (t-Derivative Rule) If f (t) is continuous, lim

t→∞

f (t)e

^−st

= 0 for all large values of s and f

⁰

(t) is piecewise continuous, then L(f

⁰

(t)) exists for all large s and L(f

⁰

(t)) = sL(f (t)) − f (0).

Proof: See page 276.

Exercises 7.1

Laplace method

. Solve the given initial value problem using Laplace’s method.

1. y⁰= −2, y(0) = 0.

2. y⁰= 1, y(0) = 0.

3. y⁰= −t, y(0) = 0.

4. y⁰= t, y(0) = 0.

5. y⁰= 1 − t, y(0) = 0.

6. y⁰= 1 + t, y(0) = 0.

7. y⁰= 3 − 2t, y(0) = 0.

8. y⁰= 3 + 2t, y(0) = 0.

9. y⁰⁰= −2, y(0) = y⁰(0) = 0.

10. y⁰⁰= 1, y(0) = y⁰(0) = 0.

11. y⁰⁰= 1 − t, y(0) = y⁰(0) = 0.

12. y⁰⁰= 1 + t, y(0) = y⁰(0) = 0.

13. y⁰⁰= 3 − 2t, y(0) = y⁰(0) = 0.

14. y⁰⁰= 3 + 2t, y(0) = y⁰(0) = 0.

Exponential order

. Show that f (t) is of exponential order, by finding a constant α ≥ 0 in each case such that

t→∞lim f (t)

e^αt = 0.

15. f (t) = 1 + t 16. f (t) = e^tsin(t) 17. f (t) =PN

n=0cnxⁿ, for any choice of the constants c0, . . . , cN.

18. f (t) = PN

n=1cnsin(nt), for any choice of the constants c1, . . . , cN.

Existence of transforms

. Let f (t) = te^t2sin(e^t2). Establish these results.

19. The function f (t) is not of expo- nential order.

20. The Laplace integral of f (t), R∞

0 f (t)e^−stdt, converges for all s > 0.

Jump Magnitude

. For f piecewise continuous, define the jump at t by J(t) = lim

h→0+f (t + h) − lim

h→0+f (t − h).

Compute J(t) for the following f . 21. f (t) = 1 for t ≥ 0, else f(t) = 0 22. f (t) = 1 for t ≥ 1/2, else f(t) = 0 23. f (t) = t/|t| for t 6= 0, f(0) = 0 24. f (t) = sin t/| sin t| for t 6= nπ,

f (nπ) = (−1)ⁿ

Taylor series

. The series relation L(P∞

n=0cntⁿ) = P∞

n=0cnL(tⁿ) often holds, in which case the result L(tⁿ) = n!s⁻¹⁻ⁿ can be employed to find a series representation of the Laplace transform. Use this idea on the fol- lowing to find a series formula for L(f(t)).

25. f (t) = e^2t=P∞

n=0(2t)ⁿ/n!

26. f (t) = e^−t=P∞

n=0(−t)ⁿ/n!

7.2 Laplace Integral Table

The objective in developing a table of Laplace integrals, e.g., Tables 4 and 5, is to keep the table size small. Table manipulation rules appear- ing in Table 6, page 257, effectively increase the table size manyfold, making it possible to solve typical differential equations from electrical and mechanical problems. The combination of Laplace tables plus the table manipulation rules is called the Laplace transform calculus.

Table 4 is considered to be a table of minimum size to be memorized.

Table 5 adds a number of special-use entries. For instance, the Heaviside entry in Table 5 is memorized, but usually not the others.

Derivations are postponed to page 270. The theory of the gamma func- tion Γ(x) appears below on page 255.

Table 4. A minimal Laplace integral table with L-notation

R∞

0 (tⁿ)e^−stdt = n!

s¹⁺ⁿ L(tⁿ) = n!

s¹⁺ⁿ R∞

0 (e^at)e^−stdt = 1

s − a L(e^at) = 1

s − a R∞

0 (cos bt)e^−stdt = s

s²+ b² L(cos bt) = s

s²+ b² R∞

0 (sin bt)e^−stdt = b

s²+ b² L(sin bt) = b

s²+ b²

Table 5. Laplace integral table extension

L(H(t − a)) = e^−as

s (a ≥ 0) Heaviside unit step, defined by H(t) =

 1 for t ≥ 0, 0 otherwise.

L(δ(t − a)) = e^−as Dirac delta, δ(t) = dH(t).

Special usage rules apply.

L(floor(t/a)) = e^−as

s(1 − e^−as) Staircase function,

floor(x) = greatest integer ≤ x.

L(sqw(t/a)) = 1

stanh(as/2) Square wave,

sqw(x) = (−1)floor^(x). L(a trw(t/a)) = 1

s²tanh(as/2) Triangular wave, trw(x) =Rx

0 sqw(r)dr.

L(t^α) =Γ(1 + α)

s^1+α Generalized power function,

Γ(1 + α) =R∞

0 e^−xx^αdx.

L(t^−1/2) =r π

s Because Γ(1/2) =√

π.

7.2 Laplace Integral Table 253

4 Example (Laplace transform) Let f (t) = t(t − 1) − sin 2t + e

^3t

. Compute L(f (t)) using the basic Laplace table and transform linearity properties.

Solution:

L(f(t)) = L(t²− 5t − sin 2t + e^3t) Expand t(t − 5).

= L(t²) − 5L(t) − L(sin 2t) + L(e^3t) Linearity applied.

= 2 s³ − 5

s² − 2

s²+ 4 + 1

s − 3 Table lookup.

5 Example (Inverse Laplace transform) Use the basic Laplace table back- wards plus transform linearity properties to solve for f (t) in the equation

L(f (t)) = s

s

²

+ 16 + 2

s − 3 + s + 1 s

³

. Solution:

L(f(t)) = s

s²+ 16+ 2 1 s − 3+ 1

s²+1 2

2

s³ Convert to table entries.

= L(cos 4t) + 2L(e^3t) + L(t) + ¹₂L(t²) Laplace table (backwards).

= L(cos 4t + 2e^3t+ t +¹₂t²) Linearity applied.

f (t) = cos 4t + 2e^3t+ t +¹₂t² Lerch’s cancellation law.

6 Example (Heaviside) Find the Laplace transform of f (t) in Figure 1.

1

1 3 5

5

Figure 1. A piecewise defined function f (t) on 0 ≤ t < ∞: f(t) = 0 except for 1 ≤ t < 2 and 3 ≤ t < 4.

Solution:

The details require the use of the Heaviside function formula H(t − a) − H(t − b) =

 1 a ≤ t < b, 0 otherwise.

The formula for f (t):

f (t) =







1 1 ≤ t < 2, 5 3 ≤ t < 4, 0 otherwise

=

 1 1 ≤ t < 2, 0 otherwise + 5

 1 3 ≤ t < 4, 0 otherwise Then f (t) = f1(t) + 5f2(t) where f1(t) = H(t − 1) − H(t − 2) and f2(t) = H(t − 3) − H(t − 4). The extended table gives

L(f(t)) = L(f¹(t)) + 5L(f²(t)) Linearity.

= L(H(t − 1)) − L(H(t − 2)) + 5L(f2(t)) Substitute for f1.

=e^−s− e^−2s

s + 5L(f²(t)) Extended table used.

=e^−s− e^−2s+ 5e^−3s− 5e^−4s

s Similarly for f2.

7 Example (Dirac delta) A machine shop tool that repeatedly hammers a die is modeled by the Dirac impulse model f (t) =

^PN_n=1

δ(t − n). Show that L(f (t)) =

^PN_n=1

e

^−ns

.

Solution:

L(f(t)) = L PN

n=1δ(t − n)

=PN

n=1L(δ(t − n)) Linearity.

=PN

n=1e^−ns Extended Laplace table.

8 Example (Square wave) A periodic camshaft force f (t) applied to a me- chanical system has the idealized graph shown in Figure 2. Show that f (t) = 1 + sqw(t) and L(f (t)) =

¹_s

(1 + tanh(s/2)).

0 2

1 3

Figure 2. A periodic force f (t) applied to a mechanical system.

Solution:

1 + sqw(t) =

 1 + 1 2n ≤ t < 2n + 1, n = 0, 1, . . ., 1 − 1 2n + 1 ≤ t < 2n + 2, n = 0, 1, . . .,

=

 2 2n ≤ t < 2n + 1, n = 0, 1, . . ., 0 otherwise,

= f (t).

By the extended Laplace table, L(f(t)) = L(1) + L(sqw(t)) = 1

s+tanh(s/2)

s .

9 Example (Sawtooth wave) Express the P -periodic sawtooth wave repre- sented in Figure 3 as f (t) = ct/P − c floor(t/P ) and obtain the formula

L(f (t)) = c

P s

²

− ce

^{−P s}

s − se

^{−P s}

.

0 c

P 4P

Figure 3. A P -periodic sawtooth wave f (t) of height c > 0.

7.2 Laplace Integral Table 255

Solution:

The representation originates from geometry, because the periodic function f can be viewed as derived from ct/P by subtracting the correct con- stant from each of intervals [P, 2P ], [2P, 3P ], etc.

The technique used to verify the identity is to define g(t) = ct/P − c floor(t/P ) and then show that g is P -periodic and f (t) = g(t) on 0 ≤ t < P . Two P - periodic functions equal on the base interval 0 ≤ t < P have to be identical, hence the representation follows.

The fine details: for 0 ≤ t < P , floor(t/P ) = 0 and floor(t/P + k) = k. Hence g(t + kP ) = ct/P + ck − c floor(k) = ct/P = g(t), which implies that g is P -periodic and g(t) = f (t) for 0 ≤ t < P .

L(f(t)) = c

PL(t) − cL(floor(t/P )) Linearity.

= c

P s²− ce^{−P s}

s − se^{−P s} Basic and extended table applied.

10 Example (Triangular wave) Express the triangular wave f of Figure 4 in terms of the square wave sqw and obtain L(f (t)) = 5

πs

²

tanh(πs/2).

0 5

2π

Figure 4. A 2π-periodic triangular wave f (t) of height 5.

Solution:

The representation of f in terms of sqw is f (t) = 5Rt/π

0 sqw(x)dx.

Details: A 2-periodic triangular wave of height 1 is obtained by integrating the square wave of period 2. A wave of height c and period 2 is given by c trw(t) = cRt

0sqw(x)dx. Then f (t) = c trw(2t/P ) = cR2t/P

0 sqw(x)dx where c = 5 and P = 2π.

Laplace transform details: Use the extended Laplace table as follows.

L(f(t)) = 5

πL(π trw(t/π)) = 5

πs²tanh(πs/2).

Gamma Function. In mathematical physics, the Gamma func- tion or the generalized factorial function is given by the identity

Γ(x) =

Z ∞

0

e

^−t

t

^x−1

dt, x > 0.

(1)

This function is tabulated and available in computer languages like For- tran, C and C++. It is also available in computer algebra systems and numerical laboratories. Some useful properties of Γ(x):

Γ(1 + x) = xΓ(x) (2)

Γ(1 + n) = n! for integers n ≥ 1.

(3)

Details for relations (2) and (3):

Start with R∞

0 e^−tdt = 1, which gives Γ(1) = 1. Use this identity and successively relation (2) to obtain relation (3).

To prove identity (2), integration by parts is applied, as follows:

Γ(1 + x) =R∞

0 e^−tt^xdt Definition.

= −t^xe^−t|^t=∞t=0 +R∞

0 e^−txt^x−1dt Use u = t^x, dv = e^−tdt.

= xR∞

0 e^−tt^x−1dt Boundary terms are zero for x > 0.

= xΓ(x).

Exercises 7.2

Laplace transform

. Compute L(f(t)) using the basic Laplace table and the linearity properties of the transform. Do not use the direct Laplace transform!

1. L(2t) 2. L(4t)

3. L(1 + 2t + t²) 4. L(t²− 3t + 10) 5. L(sin 2t) 6. L(cos 2t) 7. L(e^2t) 8. L(e^−2t) 9. L(t + sin 2t) 10. L(t − cos 2t) 11. L(t + e^2t) 12. L(t − 3e^−2t) 13. L((t + 1)²) 14. L((t + 2)²) 15. L(t(t + 1)) 16. L((t + 1)(t + 2)) 17. L(P10

n=0tⁿ/n!) 18. L(P10

n=0tⁿ⁺¹/n!) 19. L(P10

n=1sin nt) 20. L(P10

n=0cos nt)

Inverse Laplace transform

. Solve the given equation for the function f (t). Use the basic table and linearity properties of the Laplace transform.

21. L(f(t)) = s⁻² 22. L(f(t)) = 4s⁻²

23. L(f(t)) = 1/s + 2/s²+ 3/s³ 24. L(f(t)) = 1/s³+ 1/s 25. L(f(t)) = 2/(s²+ 4) 26. L(f(t)) = s/(s²+ 4) 27. L(f(t)) = 1/(s − 3) 28. L(f(t)) = 1/(s + 3) 29. L(f(t)) = 1/s + s/(s²+ 4) 30. L(f(t)) = 2/s − 2/(s²+ 4) 31. L(f(t)) = 1/s + 1/(s − 3) 32. L(f(t)) = 1/s − 3/(s − 2) 33. L(f(t)) = (2 + s)²/s³ 34. L(f(t)) = (s + 1)/s² 35. L(f(t)) = s(1/s²+ 2/s³) 36. L(f(t)) = (s + 1)(s − 1)/s³ 37. L(f(t)) =P10

n=0n!/s¹⁺ⁿ 38. L(f(t)) =P10

n=0n!/s²⁺ⁿ 39. L(f(t)) =P10

n=1

n s²+ n² 40. L(f(t)) =P10

n=0

s s²+ n²

7.3 Laplace Transform Rules 257

7.3 Laplace Transform Rules

In Table 6, the basic table manipulation rules are summarized. Full statements and proofs of the rules appear in section 7.7, page 275.

The rules are applied here to several key examples. Partial fraction expansions do not appear here, but in section 7.4, in connection with Heaviside’s coverup method.

Table 6. Laplace transform rules

L(f(t) + g(t)) = L(f(t)) + L(g(t)) Linearity.

The Laplace of a sum is the sum of the Laplaces.

L(cf(t)) = cL(f(t)) Linearity.

Constants move through the L-symbol.

L(y⁰(t)) = sL(y(t)) − y(0) The t-derivative rule.

Derivatives L(y⁰) are replaced in transformed equations.

L Rt

0g(x)dx

= 1

sL(g(t)) The t-integral rule.

L(tf(t)) = −d

dsL(f(t)) The s-differentiation rule.

Multiplyingf by t applies −d/ds to the transform of f .

L(e^atf (t)) = L(f(t))|s→(s−a) First shifting rule.

Multiplyingf by e^atreplacess by s − a.

L(f(t − a)H(t − a)) = e^−asL(f(t)), L(g(t)H(t − a)) = e^−asL(g(t + a))

Second shifting rule.

First and second forms.

L(f(t)) = RP

0 f (t)e^−stdt

1 − e^{−P s} Rule for P -periodic functions.

Assumed here isf (t + P ) = f (t).

L(f(t))L(g(t)) = L((f ∗ g)(t)) Convolution rule.

Define(f ∗ g)(t) =R^t

0f (x)g(t − x)dx.

11 Example (Harmonic oscillator) Solve by Laplace’s method the initial value problem x

⁰⁰

+ x = 0, x(0) = 0, x

⁰

(0) = 1.

Solution:

The solution is x(t) = sin t. The details:

L(x⁰⁰) + L(x) = L(0) Apply L across the equation.

sL(x⁰) − x⁰(0) + L(x) = 0 Use the t-derivative rule.

s[sL(x) − x(0)] − x⁰(0) + L(x) = 0 Use again the t-derivative rule.

(s²+ 1)L(x) = 1 Use x(0) = 0, x⁰(0) = 1.

L(x) = 1

s²+ 1 Divide.

= L(sin t) Basic Laplace table.

x(t) = sin t Invoke Lerch’s cancellation law.

12 Example (s-differentiation rule) Show the steps for L(t

²

e

^5t

) = 2 (s − 5)

³

. Solution:

L(t²e^5t) =



−d ds

 

−d ds



L(e^5t) Apply s-differentiation.

= (−1)² d ds

d ds

 1 s − 5



Basic Laplace table.

= d ds

 −1

(s − 5)²



Calculus power rule.

= 2

(s − 5)³ Identity verified.

13 Example (First shifting rule) Show the steps for L(t

²

e

^−3t

) = 2 (s + 3)

³

. Solution:

L(t²e^−3t) = L(t²)

s→s−(−3) First shifting rule.

=

 2 s²⁺¹

  

_{s→s−(−3)}

Basic Laplace table.

= 2

(s + 3)³ Identity verified.

14 Example (Second shifting rule) Show the steps for L(sin t H(t − π)) = e

^−πs

s

²

+ 1 . Solution:

The second shifting rule is applied as follows.

L(sin t H(t − π)) = L(g(t)H(t − a) Choose g(t) = sin t, a = π.

= e^−asL(g(t + a) Second form, second shifting theorem.

= e^−πsL(sin(t + π)) Substitute a = π.

= e^−πsL(− sin t) Sum rule sin(a + b) = sin a cos b + sin b cos a plus sin π = 0, cos π = −1.

= e^−πs −1

s²+ 1 Basic Laplace table. Identity verified.

15 Example (Trigonometric formulas) Show the steps used to obtain these Laplace identities:

(a) L(t cos at) = s

²

− a

²

(s

²

+ a

²

)

²

(c) L(t

²

cos at) = 2(s

³

− 3sa

²

) (s

²

+ a

²

)

³

(b) L(t sin at) = 2sa

(s

²

+ a

²

)

²

(d) L(t

²

sin at) = 6s

²

a − a

³

(s

²

+ a

²

)

³

7.3 Laplace Transform Rules 259

Solution:

The details for (a):

L(t cos at) = −(d/ds)L(cos at) Use s-differentiation.

= − d ds

 s

s²+ a²



Basic Laplace table.

= s²− a²

(s²+ a²)² Calculus quotient rule.

The details for (c):

L(t²cos at) = −(d/ds)L((−t) cos at) Use s-differentiation.

= d ds



− s²− a² (s²+ a²)²



Result of (a).

=2s³− 6sa²)

(s²+ a²)³ Calculus quotient rule.

The similar details for (b) and (d) are left as exercises.

16 Example (Exponentials) Show the steps used to obtain these Laplace identities:

(a) L(e

^at

cos bt) = s − a

(s − a)

²

+ b

²

(c) L(te

^at

cos bt) = (s − a)

²

− b

²

((s − a)

²

+ b

²

)

²

(b) L(e

^at

sin bt) = b

(s − a)

²

+ b

²

(d) L(te

^at

sin bt) = 2b(s − a) ((s − a)

²

+ b

²

)

²

Solution:

Details for (a):

L(e^atcos bt) = L(cos bt)|s→s−a First shifting rule.

=

 s

s²+ b²

   _s→s−a

Basic Laplace table.

= s − a

(s − a)²+ b² Verified (a).

Details for (c):

L(te^atcos bt) = L(t cos bt)|s→s−a First shifting rule.

=



−d

dsL(cos bt)

   _s→s−a

Apply s-differentiation.

=



−d ds

 s

s²+ b²



  _s→s−a

Basic Laplace table.

=

 s²− b² (s²+ b²)²

   s→s−a

Calculus quotient rule.

= (s − a)²− b²

((s − a)²+ b²)² Verified (c).

Left as exercises are (b) and (d).

17 Example (Hyperbolic functions) Establish these Laplace transform facts about cosh u = (e

^u

+ e

^−u

)/2 and sinh u = (e

^u

− e

^−u

)/2.

(a) L(cosh at) = s

s

²

− a

²

(c) L(t cosh at) = s

²

+ a

²

(s

²

− a

²

)

²

(b) L(sinh at) = a

s

²

− a

²

(d) L(t sinh at) = 2as

(s

²

− a

²

)

²

Solution:

The details for (a):

L(cosh at) = ¹2(L(e^at) + L(e^−at)) Definition plus linearity of L.

=1 2

 1

s − a+ 1 s + a



Basic Laplace table.

= s

s²− a² Identity (a) verified.

The details for (d):

L(t sinh at) = − d ds

 a

s²− a²



Apply the s-differentiation rule.

= a(2s)

(s²− a²)² Calculus power rule; (d) verified.

Left as exercises are (b) and (c).

18 Example (s-differentiation) Solve L(f (t)) = 2s

(s

²

+ 1)

²

for f (t).

Solution:

The solution is f (t) = t sin t. The details:

L(f(t)) = 2s (s²+ 1)²

= −d ds

 1

s²+ 1



Calculus power rule (uⁿ)⁰= nuⁿ⁻¹u⁰.

= − d

ds(L(sin t)) Basic Laplace table.

= L(t sin t) Apply the s-differentiation rule.

f (t) = t sin t Lerch’s cancellation law.

19 Example (First shift rule) Solve L(f (t)) = s + 2

2

²

+ 2s + 2 for f (t).

Solution:

The answer is f (t) = e^−tcos t + e^−tsin t. The details:

L(f(t)) = s + 2

s²+ 2s + 2 Signal for this method: the denom- inator has complex roots.

= s + 2

(s + 1)²+ 1 Complete the square, denominator.

7.3 Laplace Transform Rules 261

= S + 1

S²+ 1 Substitute S for s + 1.

= S

S²+ 1+ 1

S²+ 1 Split into Laplace table entries.

= L(cos t) + L(sin t)|s→S=s+1 Basic Laplace table.

= L(e^−tcos t) + L(e^−tsin t) First shift rule.

f (t) = e^−tcos t + e^−tsin t Invoke Lerch’s cancellation law.

20 Example (Damped oscillator) Solve by Laplace’s method the initial value problem x

⁰⁰

+ 2x

⁰

+ 2x = 0, x(0) = 1, x

⁰

(0) = −1.

Solution:

The solution is x(t) = e^−tcos t. The details:

L(x⁰⁰) + 2L(x⁰) + 2L(x) = L(0) Apply L across the equation.

sL(x⁰) − x⁰(0) + 2L(x⁰) + 2L(x) = 0 The t-derivative rule on x⁰. s[sL(x) − x(0)] − x⁰(0)

+2[L(x) − x(0)] + 2L(x) = 0

The t-derivative rule on x.

(s²+ 2s + 2)L(x) = 1 + s Use x(0) = 1, x⁰(0) = −1.

L(x) = s + 1

s²+ 2s + 2 Divide.

= s + 1

(s + 1)²+ 1 Complete the square in the de- nominator.

= L(cos t)|s→s+1 Basic Laplace table.

= L(e^−tcos t) First shifting rule.

x(t) = e^−tcos t Invoke Lerch’s cancellation law.

21 Example (Rectified sine wave) Compute the Laplace transform of the rectified sine wave f (t) = | sin ωt| and show it can be expressed in the form

L(| sin ωt|) = ω coth

^πs_2ω

s

²

+ ω

²

.

Solution:

The periodic function formula will be applied with period P = 2π/ω. The calculation reduces to the evaluation of J =RP

0 f (t)e^−stdt. Because sin ωt ≤ 0 on π/ω ≤ t ≤ 2π/ω, integral J can be written as J = J1+ J2, where

J1= Z π/ω

0

sin ωt e^−stdt, J2= Z 2π/ω

π/ω − sin ωt e^−stdt.

Integral tables give the result Z

sin ωt e^−stdt = −ωe^−stcos(ωt)

s²+ ω² −se^−stsin(ωt) s²+ ω² . Then

J1= ω(e^−π∗s/ω+ 1)

s²+ ω² , J2= ω(e^−2πs/ω+ e^−πs/ω) s²+ ω² ,

J = ω(e^−πs/ω+ 1)² s²+ ω² .

The remaining challenge is to write the answer for L(f(t)) in terms of coth.

The details:

L(f(t)) = J

1 − e^{−P s} Periodic function formula.

= J

(1 − e^{−P s/2})(1 + e^{−P s/2}) Apply 1 − x²= (1 − x)(1 + x), x = e^{−P s/2}.

= ω(1 + e^{−P s/2})

(1 − e^{−P s/2})(s²+ ω²) Cancel factor 1 + e^{−P s/2}.

= e^{P s/4}+ e^{−P s/4} e^{P s/4}− e^{−P s/4}

ω

s²+ ω² Factor out e^{−P s/4}, then cancel.

= 2 cosh(P s/4) 2 sinh(P s/4)

ω

s²+ ω² Apply cosh, sinh identities.

=ω coth(P s/4)

s²+ ω² Use coth u = cosh u/ sinh u.

= ω coth _2ω^πs

s²+ ω² Identity verified.

22 Example (Half–wave rectification) Compute the Laplace transform of the half–wave rectification of sin ωt, denoted g(t), in which the negative cycles of sin ωt have been canceled to create g(t). Show in particular that

L(g(t)) = 1 2

ω s

²

+ ω

²



1 + coth



πs 2ω



Solution:

The half–wave rectification of sin ωt is g(t) = (sin ωt + | sin ωt|)/2.

Therefore, the basic Laplace table plus the result of Example 21 give L(2g(t)) = L(sin ωt) + L(| sin ωt|)

= ω

s²+ ω² +ω cosh(πs/(2ω)) s²+ ω²

= ω

s²+ ω²(1 + cosh(πs/(2ω)) Dividing by 2 produces the identity.

23 Example (Shifting rules) Solve L(f (t)) = e

^−3s

s + 1

s

²

+ 2s + 2 for f (t).

Solution:

The answer is f (t) = e^3−tcos(t − 3)H(t − 3). The details:

L(f(t)) = e^−3s s + 1

(s + 1)²+ 1 Complete the square.

= e^−3s S

S²+ 1 Replace s + 1 by S.

= e^−3S+3(L(cos t))|s→S=s+1 Basic Laplace table.

7.3 Laplace Transform Rules 263

= e³ e^−3sL(cos t)


_s→S=s+1 Regroup factor e^−3S.

= e³(L(cos(t − 3)H(t − 3)))|s→S=s+1 Second shifting rule.

= e³L(e^−tcos(t − 3)H(t − 3)) First shifting rule.

f (t) = e^3−tcos(t − 3)H(t − 3) Lerch’s cancellation law.

24 Example () Solve L(f (t) = s + 7

s

²

+ 4s + 8 for f (t).

Solution:

The answer is f (t) = e^−2t(cos 2t +⁵₂sin 2t). The details:

L(f(t)) = s + 7

(s + 2)²+ 4 Complete the square.

= S + 5

S²+ 4 Replace s + 2 by S.

= S

S²+ 4+5 2

2

S²+ 4 Split into table entries.

= s

s²+ 4+5 2

2 s²+ 4

  

s→S=s+2

Prepare for shifting rule.

= L(cos 2t) +⁵2L(sin 2t)

_s→S=s+2 Basic Laplace table.

= L(e^−2t(cos 2t +⁵₂sin 2t)) First shifting rule.

f (t) = e^−2t(cos 2t +⁵₂sin 2t) Lerch’s cancellation law.

7.4 Heaviside’s Method

This practical method was popularized by the English electrical engineer Oliver Heaviside (1850–1925). A typical application of the method is to solve

2s

(s + 1)(s

²

+ 1) = L(f (t))

for the t-expression f (t) = −e

^−t

+ cos t + sin t. The details in Heaviside’s method involve a sequence of easy-to-learn college algebra steps.

More precisely, Heaviside’s method systematically converts a polyno- mial quotient

a

₀

+ a

₁

s + · · · + a

_n

s

ⁿ

b

₀

+ b

₁

s + · · · + b

m

s

^m

(1)

into the form L(f (t)) for some expression f (t). It is assumed that a

₀

, .., a

_n

, b

₀

, . . . , b

_m

are constants and the polynomial quotient (1) has limit zero at s = ∞.

Partial Fraction Theory

In college algebra, it is shown that a rational function (1) can be ex- pressed as the sum of terms of the form

A (s − s

₀

)

^k

(2)

where A is a real or complex constant and (s − s

₀

)

^k

divides the denomi- nator in (1). In particular, s

0

is a root of the denominator in (1).

Assume fraction (1) has real coefficients. If s

₀

in (2) is real, then A is real. If s

0

= α + iβ in (2) is complex, then (s − s

0

)

^k

also appears, where s

₀

= α − iβ is the complex conjugate of s

₀

. The corresponding terms in (2) turn out to be complex conjugates of one another, which can be combined in terms of real numbers B and C as

A

(s − s

₀

)

^k

+ A

(s − s

₀

)

^k

= B + C s ((s − α)

²

+ β

²

)

^k

. (3)

Simple Roots. Assume that (1) has real coefficients and the denomi- nator of the fraction (1) has distinct real roots s

₁

, . . . , s

_N

and distinct complex roots α

₁

+ iβ

₁

, . . . , α

M

+ iβ

M

. The partial fraction expansion of (1) is a sum given in terms of real constants A

_p

, B

_q

, C

_q

by

a

₀

+ a

₁

s + · · · + a

_n

s

ⁿ

b

₀

+ b

₁

s + · · · + b

m

s

^m

=

N

X

p=1

A

_p

s − s

p

+

M

X

q=1

B

_q

+ C

_q

(s − α

_q

)

(s − α

q

)

²

+ β

_q²

.

(4)

7.4 Heaviside’s Method 265

Multiple Roots. Assume (1) has real coefficients and the denomi- nator of the fraction (1) has possibly multiple roots. Let N

_p

be the multiplicity of real root s

_p

and let M

_q

be the multiplicity of complex root α

q

+ iβ

q

, 1 ≤ p ≤ N , 1 ≤ q ≤ M . The partial fraction expansion of (1) is given in terms of real constants A

_p,k

, B

_q,k

, C

_q,k

by

N

X

p=1

X

1≤k≤Np

A

_p,k

(s − s

_p

)

^k

+

M

X

q=1

X

1≤k≤Mq

B

_q,k

+ C

_q,k

(s − α

q

) ((s − α

_q

)

²

+ β

_q²

)

^k

. (5)

Heaviside’s Coverup Method

The method applies only to the case of distinct roots of the denominator in (1). Extensions to multiple-root cases can be made; see page 266.

To illustrate Oliver Heaviside’s ideas, consider the problem details 2s + 1

s(s − 1)(s + 1) = A s + B

s − 1 + C s + 1 (6)

= L(A) + L(Be

^t

) + L(Ce

^−t

)

= L(A + Be

^t

+ Ce

^−t

)

The first line (6) uses college algebra partial fractions. The second and third lines use the Laplace integral table and properties of L.

Heaviside’s mysterious method. Oliver Heaviside proposed to find in (6) the constant C =

¹₂

by a cover–up method:

2s + 1 s(s − 1)

    s+1 =0

= C

.

The instructions are to cover–up the matching factors (s + 1) on the left and right with box , then evaluate on the left at the root s which makes the contents of the box zero. The other terms on the right are replaced by zero.

To justify Heaviside’s cover–up method, multiply (6) by the denominator s + 1 of partial fraction C/(s + 1):

(2s + 1) (s + 1)

s(s − 1) (s + 1) = A (s + 1)

s + B (s + 1)

s − 1 + C (s + 1) (s + 1) . Set (s + 1) = 0 in the display. Cancellations left and right plus annihi- lation of two terms on the right gives Heaviside’s prescription

2s + 1 s(s − 1)

   s+1=0

= C.

The factor (s + 1) in (6) is by no means special: the same procedure applies to find A and B. The method works for denominators with simple roots, that is, no repeated roots are allowed.

Extension to Multiple Roots. An extension of Heaviside’s method is possible for the case of repeated roots. The basic idea is to factor–out the repeats. To illustrate, consider the partial fraction expansion details

R = 1

(s + 1)

²

(s + 2) A sample rational function having repeated roots.

= 1

s + 1



1

(s + 1)(s + 2)



Factor–out the repeats.

= 1

s + 1



1

s + 1 + −1 s + 2



Apply the cover–up method to the simple root fraction.

= 1

(s + 1)

²

+ −1

(s + 1)(s + 2) Multiply.

= 1

(s + 1)

²

+ −1

s + 1 + 1

s + 2 Apply the cover–up method to the last fraction on the right.

Terms with only one root in the denominator are already partial frac- tions. Thus the work centers on expansion of quotients in which the denominator has two or more roots.

Special Methods. Heaviside’s method has a useful extension for the case of roots of multiplicity two. To illustrate, consider these details:

R = 1

(s + 1)

²

(s + 2) A fraction with multiple roots.

= A

s + 1 + B

(s + 1)

²

+ C

s + 2 See equation (5).

= A

s + 1 + 1

(s + 1)

²

+ 1

s + 2 Find B and C by Heaviside’s cover–

up method.

= −1

s + 1 + 1

(s + 1)

²

+ 1

s + 2 Multiply by s+1. Set s = ∞. Then 0 = A + 1.

The illustration works for one root of multiplicity two, because s = ∞ will resolve the coefficient not found by the cover–up method.

In general, if the denominator in (1) has a root s

₀

of multiplicity k, then the partial fraction expansion contains terms

A

₁

s − s

₀

+ A

₂

(s − s

₀

)

²

+ · · · + A

_k

(s − s

₀

)

^k

.

Heaviside’s cover–up method directly finds A

_k

, but not A

₁

to A

_k−1

.

7.5 Heaviside Step and Dirac Delta 267

7.5 Heaviside Step and Dirac Delta

Heaviside Function. The unit step function or Heaviside func- tion is defined by

H(x) =

(

1 for x ≥ 0, 0 for x < 0.

The most often–used formula involving the Heaviside function is the characteristic function of the interval a ≤ t < b, given by

H(t − a) − H(t − b) =

(

1 a ≤ t < b, 0 t < a, t ≥ b.

(1)

To illustrate, a square wave sqw(t) = (−1)floor

^(t)

can be written in the series form

∞

X

n=0

(−1)

ⁿ

(H(t − n) − H(t − n − 1)).

Dirac Delta. A precise mathematical definition of the Dirac delta, denoted δ, is not possible to give here. Following its inventor P. Dirac, the definition should be

δ(t) = dH(t).

The latter is nonsensical, because the unit step does not have a cal- culus derivative at t = 0. However, dH(t) could have the meaning of a Riemann-Stieltjes integrator, which restrains dH(t) to have meaning only under an integral sign. It is in this sense that the Dirac delta δ is defined.

What do we mean by the differential equation x

⁰⁰

+ 16x = 5δ(t − t

₀

)?

The equation x

⁰⁰

+ 16x = f (t) represents a spring-mass system without damping having Hooke’s constant 16, subject to external force f (t). In a mechanical context, the Dirac delta term 5δ(t − t

₀

) is an idealization of a hammer-hit at time t = t

₀

> 0 with impulse 5.

More precisely, the forcing term f (t) can be formally written as a Riemann- Stieltjes integrator 5dH(t−t

₀

) where H is Heaviside’s unit step function.

The Dirac delta or “derivative of the Heaviside unit step,” nonsensical as it may appear, is realized in applications via the two-sided or central difference quotient

H(t + h) − H(t − h)

2h ≈ dH(t).

Therefore, the force f (t) in the idealization 5δ(t − t

₀

) is given for h > 0 very small by the approximation

f (t) ≈ 5 H(t − t

₀

+ h) − H(t − t

₀

− h)

2h .

The impulse

²

of the approximated force over a large interval [a, b] is computed from

Z b a

f (t)dt ≈ 5

Z h

−h

H(t − t

₀

+ h) − H(t − t

₀

− h)

2h dt = 5,

due to the integrand being 1/(2h) on |t − t

₀

| < h and otherwise 0.

Modeling Impulses. One argument for the Dirac delta idealization is that an infinity of choices exist for modeling an impulse. There are in addition to the central difference quotient two other popular difference quotients, the forward quotient (H(t + h) − H(t))/h and the backward quotient (H(t) − H(t − h))/h (h > 0 assumed). In reality, h is unknown in any application, and the impulsive force of a hammer hit is hardly constant, as is supposed by this naive modeling.

The modeling logic often applied for the Dirac delta is that the external force f (t) is used in the model in a limited manner, in which only the momentum p = mv is important. More precisely, only the change in momentum or impulse is important,

^R_a^b

f (t)dt = ∆p = mv(b) − mv(a).

The precise force f (t) is replaced during the modeling by a simplistic piecewise-defined force that has exactly the same impulse ∆p. The re- placement is justified by arguing that if only the impulse is important, and not the actual details of the force, then both models should give similar results.

Function or Operator? The work of physics Nobel prize winner P.

Dirac (1902–1984) proceeded for about 20 years before the mathematical community developed a sound mathematical theory for his impulsive force representations. A systematic theory was developed in 1936 by the soviet mathematician S. Sobolev. The French mathematician L.

Schwartz further developed the theory in 1945. He observed that the idealization is not a function but an operator or linear functional, in particular, δ maps or associates to each function φ(t) its value at t = 0, in short, δ(φ) = φ(0). This fact was observed early on by Dirac and others, during the replacement of simplistic forces by δ. In Laplace theory, there is a natural encounter with the ideas, because L(f (t)) routinely appears on the right of the equation after transformation. This term, in the case

2Momentum is defined to be mass times velocity. If the force f is given by Newton’s law as f (t) = _dt^d(mv(t)) and v(t) is velocity, thenRb

af(t)dt = mv(b) − mv(a) is the net momentum or impulse.

7.5 Heaviside Step and Dirac Delta 269

of an impulsive force f (t) = c(H(t−t

₀

−h)−H(t−t

₀

+h))/(2h), evaluates for t

₀

> 0 and t

₀

− h > 0 as follows:

L(f (t)) =

Z ∞

0

c

2h (H(t − t

0

− h) − H(t − t

0

+ h))e

^−st

dt

=

Z t0+h t0−h

c 2h e

^−st

dt

= ce

^−st0

e

^sh

− e

^−sh

2sh

!

The factor e

^sh

− e

^−sh

2sh is approximately 1 for h > 0 small, because of L’Hospital’s rule. The immediate conclusion is that we should replace the impulsive force f by an equivalent one f

^∗

such that

L(f

^∗

(t)) = ce

^−st0

. Well, there is no such function f

^∗

!

The apparent mathematical flaw in this idea was resolved by the work of L. Schwartz on distributions. In short, there is a solid foundation for introducing f

^∗

, but unfortunately the mathematics involved is not elementary nor especially accessible to those readers whose background is just calculus.

Practising engineers and scientists might be able to ignore the vast lit-

erature on distributions, citing the example of physicist P. Dirac, who

succeeded in applying impulsive force ideas without the distribution the-

ory developed by S. Sobolev and L. Schwartz. This will not be the case

for those who wish to read current literature on partial differential equa-

tions, because the work on distributions has forever changed the required

background for that topic.