Chapter 4

Laplace Transform

It’s time to stop guessing solutions and find a systematic way of finding solutions to non homogeneous linear ODEs. We define the Laplace transform of a function f in the following way. Definition 4.1. The Laplace transform of f (t), written F (s), is given by (4.1)

L(f ) =

∞

e−st f (t) dt.

0

That is, the Laplace transform acts on a function, f (t), integrates the t out, and creates function of s, which we denote F (s). Before we see why this is useful, we might want to know if the integral in the definition exists - it is after all an improper integral. We need another definition. Definition 4.2. We say a function f (t) is exponentially bounded or of exponential order if there exists non negative numbers a, k and M such that (4.2)

|f (t)| ≤ keat

t ≥ M.

We want to allow for interesting functions forcing an ODE. Definition 4.3. A function f is said to be piecewise continuous on a bounded interval if it has a finite number of discontinuities and the left and right limits exist at each discontinuity. It is said to be piecewise continuous on [0, ∞] if it is piecewise continuous on every bounded subinterval I ⊂ [0, ∞]. And finally, Definition 4.4. A function f is said to be piecewise smooth if f and its derivative are piecewise continuous. Now we can say when the Laplace transform of a function exists. Theorem 4.5. Suppose f is piecewise continuous on [0, ∞] and exponentially bounded. Then L(f ) = F (s) exists for all s > a. 37

38

4. Laplace Transform

We will not prove the theorem. However, the examples below will show why it is reasonable. Example 4.6. Compute L(1). Solution. Let’s apply Theorem 4.5 to see what to expect. The function f (t) = 1 is exponentially bounded. Indeed, we can choose k = 1, a = M = 0 in (4.2). Theorem 4.5 then implies the Laplace transform exists for all s > a = 0. To find L(1) we compute it using the definition L(1) =

∞

0

1 e−st 1 dt = lim − e−st A→∞ s

A 0

= lim

A→∞

1 (1 − e−As ). s

The limit of the exponential is zero as long as s > 0. Therefore, 1 s>0 L(1) = , s as expected. Example 4.7. Compute L(eat ). Solution. Again we apply Theorem 4.5 to see what to expect. The function f (t) = eat is exponentially bounded if we choose k = 1, a = a, M = 1 in (4.2). Theorem 4.5 then implies the Laplace transform exists for all s > a. To find the L(eat ) we compute it using the definition ∞

L(eat ) =

∞

e−st eat dt =

e(a−s)t dt

0

0

1 1 (a−s)t A e (1 − eA(a−s) ). = lim = lim − A→∞ s − a A→∞ a−s 0 We need the exponent to be negative for the limit to exist. This is the case provided s > a. Therefore, 1 L(eat ) = s > a. s−a The next example is more difficult, but it shows the usefulness of the Laplace transform. Example 4.8. Compute L(f ′ (t)), where f is a piecewise smooth exponentially bounded function. Solution. We are told there exists k, a, and M such that |f (t)| ≤ keat for all t ≥ M . We use the definition of Laplace transform and integrate by parts L(f ′ (t))

∞

=

A

e−st f ′ (t) dt = lim

A→∞

0

=

lim

A→∞

e−st f (t)

A 0

e−st f ′ (t) dt 0

A

se−st f (t) dt

+ 0

A

=

lim

A→∞

e−sA f (A) − f (0) +

se−st f (t) dt . 0

39

4. Laplace Transform

Since f is exponentially bounded lim |e−sA f (A)| ≤ lim |e−sA keaA | = lim |ke(a−s)A | = 0

A→∞

A→∞

A→∞

provided s > a. Returning to our calculation A

L(f ′ (t))

=

−f (0) +

lim

A→∞ ∞

e−st f (t) dt − f (0)

=

s

=

sL(f (t)) − f (0).

0

se−st f (t) dt 0

This may look more useful if f is replaced with y. Then we have the formula L(y ′ ) = sL(y) − y(0).

That is, the Laplace transform turns a derivative y ′ into an algebraic expression! Quite useful for solving ODEs! Indeed, we can easily compute L(y ′′ ). Using the formula for L(y ′ ) we find L(y ′′ ) = sL(y ′ ) − y ′ (0) = s2 L(y) − sy(0) − y ′ (0).

Table 1 below provides the Laplace transform for many common functions. Example 4.9. Find the solution to y ′′ + 3y ′ + 2y = 0, y(0) = 1, y ′ (0) = 0 by using Laplace transforms. Solution. We just take the Laplace transform of both sides. The left side of the ODE gives L(y ′′ + 3y ′ + 2y) = L(y ′′ ) + 3L(y ′ ) + 2L(y).

It distributes this way because integrals behave this way, and the Laplace transform is an integral. Obviously L(0) = 0. Applying our formulas s2 L(y) − sy(0) − y ′ (0) + 3 sL(y) − y(0) + 2L(y) = 0. Next we use the initial data and rearrange to find (s2 + 3s + 2)L(y) = s + 3, that is, s+3 . + 3s + 2 The Laplace transform turned an ODE into an algebraic problem. The only catch is that we need to find a function whose Laplace transform is (s + 3)/(s2 + 3s + 2). Table 1 does not seem to help since this is not in the table. We have to modify the expression to use Table 1. We have using partial fractions s+3 s+3 A B = = + . 2 s + 3s + 2 (s + 1)(s + 2) s+1 s+2 L(y) =

s2

40

4. Laplace Transform

1.

f (t) = L−1 (F (s)) 1

2.

eat

1 s−a

s>a

3.

tn

n! sn+1

s>0

4.

tn eat

n! (s−a)n+1

s>a

5.

sin at

a s2 +a2

s>0

6.

cos at

s s2 +a2

s>0

7.

eat sin bt

b (s−a)2 +b2

s>a

8.

eat cos bt

s−a (s−a)2 +b2

s>a

9.

eat f (t)

F (s − a) s > a

10.

uc (t)f (t)

e−cs L(f (t + c))

11.

uc (t)f (t − c)

12.

δ(t − c)

F (s) = L(f (t)) 1 s s>0

e−cs L(f (t)) e−cs

t

13. 0

f (t − τ )g(τ ) dt

F (s)G(s)

14.

y ′ (t)

sL(y) − y(0)

15.

y ′′ (t)

s2 L(y) − sy(0) − y ′ (0)

Table 1. Laplace and inverse Laplace transforms for common functions.

This implies s + 3 = A(s + 2) + B(s + 1), and A = −1 and B = 2. Therefore, L(y) =

2 1 − . s+1 s+2

We may now use Table 1 to find y(t) = 2e−t − e−2t . No guessing involved!

41

4. Laplace Transform

Example 4.10. Find the inverse Laplace transform of 2s + 1 F (s) = 2 , s + 4s + 5 Solution. Just a matter of making it look like ones in the table. When the denominator does not factor, we complete the square. Recall to do this we take half the coefficient in front of the middle term, s, square it, then add and subtract it. Here s2 + 4s + 5 = s2 + 4s + 4 − 4 + 5 = (s + 2)2 + 1. Then 2s + 1 (s + 2) 1 =2 −3 . 2 + 4s + 5 (s + 2) + 1 (s + 2)2 + 1 This is readily found in Table 1 and F (s) =

s2

f (t) = 2e−2t cos t − 3e−2t sin t

Homework 4.1

Find the inverse Laplace transform of the following. 1. F (s) = 2. F (s) = 3. F (s) = 4. F (s) = 5. F (s) =

3 s2 +4 4 (s−1)3

2s−3 s2 −4 2s+1 s2 −2s+2

6. F (s) = 7. F (s) =

2 s2 +3s−4 3s s2 −s−6 2s+2 s2 +2s+5

8s2 −4s+12 s(s2 +4) 1−2s s2 +4s+5 2s−3 s2 +2s+10

8. F (s) = 9. F (s) = 10. F (s) =

Use the Laplace transform to solve the following ODEs 11. y ′′ − y ′ − 6y = 0; 12. y ′′ − 2y ′ + 2y = 0;

y(0) = 1, y ′ (0) = −1 y(0) = 0, y ′ (0) = 1

13. y ′′ + 2y ′ + 5y = 0; y(0) = 2, y ′ (0) = −1 14. y IV − y ′ = 0; y(0) = 1, y ′ (0) = 1, y ′′ (0) = 1, y ′′′ (0) = 0 15. y ′′ − 2y ′ + 2y = cos t; 1. f (t) =

3 2

y(0) = 1, y ′ (0) = 0

sin 2t

2. f (t) = 2t2 et 3. f (t) = 25 et − 25 e−4t

Answers 9. f (t) = −2e−2t cos t + 5e−2t sin t

10. f (t) = 2e−t cos 3t − 53 e−t sin 3t 11. y(t) = 15 (e3t + 4e−2t )

4. f (t) = 59 e3t − 56 e−2t

12. y(t) = et sin t

5. f (t) = 2e−t cos 2t

13. y(t) = 2e−t cos 2t + 12 e−t sin 2t

6. f (t) =??

14. y(t) =

7. f (t) = 2et cos t + 3et sin t

15. y(t) = 2et sin t)

8. f (t) = 3 + 5 cos 2t − 2 sin 2t

et +e−t 2 1 (cos t 5

− 2 sin t + 4et cos t −

42

4. Laplace Transform

4.2. Heaviside Function We can force ODEs with more interesting functions now that we have a more non guessing method for solving ODEs. Indeed, consider the Heaviside function given by 0 1

uc (t) =

(4.3)

t 0. We can think of the Heaviside function as a switch. It turns on at t = c. Thus the function 1 − uc (t) is a switch that turns off at t = c.

0.5

0

0.5

0

0.5

1 t

1.5

2

0

0

0.5

1 t

1.5

Figure 1. Left: Heaviside function with c = 1. Right: graph of 1 − u1 (t).

Example 4.11. Write the function f (t) =

0 sin t

t < 2π t ≥ 2π

in terms of Heaviside functions. Solution. The function f starts out as the function 0 until t = 2π. At t = 2π zero turns off and sin t turns on. Thus f (t) = 0 − u2π (t)0 + u2π (t) sin t = u2π (t) sin t.

Keep in mind, uc (t) is a switch that turns on at t = c. A graph of f is given below.

Example 4.12. Write the function   0, 0 < t < 1, f (t) = 1, 1 ≤ t < 2,  0, t ≥ 2, in terms of Heaviside functions.

2

43

4.2. Heaviside Function

1

f(t)

0.5

0

-0.5

-1 0

10 t

5

20

15

Figure 2. Graph of f (t) = u2π (t) sin t in Example 4.11.

Solution. The function f starts out as the function 0 until t = 1. At t = 1 zero turns off and 1 turns on. At t = 2 one turns off 0 turns on. Thus f (t) =

0 − u1 (t)0 + u1 (t)1 − u2 (t)1 + u2 (t)0

=

u1 (t) − u2 (t).

A graph of this function is given below.

1.5

f(t)

1

0.5

0

0

0.5

1

1.5 t

2

3

2.5

Figure 3. Graph of f (t) = u1 (t) − u2 (t) in Example 4.12.

We need to compute the Laplace transform of the Heaviside function. In particular, the above examples imply we need to find the Laplace transform of functions in the form uc (t)f (t). Using the definition of Laplace transform, we find L(uc (t)f (t))

∞

=

∞

∞

e−s(r+c) f (r + c) dr = e−cs

0

=

e−st f (t) dt

c

0

=

∞

e−st uc (t)f (t) dt =

e−cs L(f (t + c)).

0

e−sr f (r + c) dr

44

4. Laplace Transform

Example 4.13. Compute the Laplace transform of u2π (t) sin t. Solution. Using 10. in Table 1 (just derived above) L(u2π (t) sin t) = =

e−2πs L(sin(t + 2π)) e−2πs . e−2πs L(sin t) = 2 s +1

Example 4.14. Compute the Laplace transform of u2 (t)(tt + t + 1). Solution. Using 10. in Table 1 L(u2 (t)(t2 + t + 1) = =

=

e−2s L((t + 2)2 + (t + 2) + 1))

e−2s L(t2 + 5t + 7) 2 5 7 e−2s + 2+ s3 s s

.

Example 4.15. Compute the Laplace transform of  t, 0 ≤ t < 3,  (t − 3), 3 ≤ t < 4, f (t) =  1, t≥4. Solution. Here f (t) = t − u3 (t)t + u3 (t)(t − 3) − u4 (t)(t − 3) + u4 (t).

The Laplace transform is F (s)

= =

e−4s 1 −3s −3s −4s − e L((t + 3)) + e L((t + 3) − 3) − e L((t + 4) − 3) + s2 s −3s −4s e e 1 1 3 1 1 + 2 − e−4s + − e−3s + + . s2 s2 s s s2 s s

Example 4.16. Compute the inverse Laplace transform of G(s) =

(2s − 3)e−s . s2 + 2s + 10

Solution. We need to make this look like 11 in Table 1. We may write G(s) as G(s)

= =

3e−s 5 (s + 1)e−s − 2 2 (s + 1) + 3 3 (s + 1)2 + 32 e−s F1 (s) + e−s F2 (s).

2

Next we find the inverse of F1 (s) and F2 (s) from the table. Here 5 f2 (t) = e−t sin 3t. f1 (t) = 2e−t cos 3t, 3 The inverse Laplace transform is g(t) = =

u1 (t)f1 (t − 1) + u1 (t)f2 (t)

5 u1 (t) 2e−(t−1) cos(3(t − 1)) − e−(t−1) sin(3(t − 1)) . 3

45

4.2. Heaviside Function

Example 4.17. Find the solution to the ODE y ′′ + y =

1 − t, 0,

0 ≤ t < 1, t≥1.

y(0) = 1,

y ′ (0) = 0 .

Solution. The right side is the same as (1 − t) − u1 (t))(1 − t). Taking the Laplace transform of both sides, 1 1 (s2 L(y) − s) + L(y) = − 2 − e−s L(1 − (t + 1)), s s or s 1 1 e−s L(y) = 2 + − 2 2 + 2 2 , 2 s + 1 s(s + 1) s (s + 1) s (s + 1) Set 1 s 1 = − 2 F (s) := 2 s(s + 1) s (s + 1) 1 1 1 = 2− 2 . G(s) := 2 2 s (s + 1) s (s + 1) Then s + F (s) − G(s) + e−s G(s). L(y) = 2 s +1 It follows that y(t) = cos(t) + f (t) − g(t) + u1 (t)g(t − 1). Here f (t) g(t) Hence,

= 1 − cos t = t − sin t.

y(t) = 1 − t + sin t + u1 (t)(t − 1 − sin(t − 1)).

Homework 4.2

Find the Laplace transform of the following.   0 0