AB2.8: Transforms of Derivatives and Integrals. Differential Equations

THEOREM 5.2.1 Suppose that f (t) is continuous for all t ≥ 0 satisfies the condition |f (t)| ≤ M ekt

(t ≥ 0)

for some constants k and M . and has a derivative f 0 (t) that is piecewise continuous on any finite interval in the range t ≥ 0. Then the Laplace transform of the derivative f 0 (t) exists when s > k, and L (f 0 ) = sL (f ) − f (0), s > k.

PROOF We first consider the case when f 0 (t) is continuous for all t ≥ 0. Then, by definitions and integration by parts, we obtain the desired result L (f 0 ) =

Z ∞ 0

∞

e−st f 0 (t)dt == e−st f (t) + s 0

Z ∞ 0

e−st f (t)dt = 0 − f (0) + sL (f ), s > k.

By applying the theorem to the second derivative, we obtain L (f 00 ) = sL (f 0 ) − f 0 (0) = s[sL (f ) − f (0)] − f 0 (0) = s2 L (f ) − sf (0) − f 0 (0). Similarly, L (f 000 ) = s3 L (f ) − s2 f (0) − sf 0 (0) − f 00 (0). Using the mathematical induction, we obtain the Laplace transforms for the general case of the derivatives f (n) of an arbitrary order n = 1, 2, . . . L (f (n) ) = sn L (f ) − sn−1 f (0) − sn−2 f 0 (0) − . . . − sf (n−2) (0) − f (n−1) (0).

EXAMPLE 1 Laplace transform of f (t) = t2 Let f (t) = t2 , t ≥ 0. Find F (s). Solution. We have f 0 (t) = 2t,

f 00 (t) = 2,

so that f (0) = 0,

f 0 (0) = 0,

f 00 (0) = 2.

Therefore, L (f 00 ) = L (2) =

2 = s2 L (f ), s

Resolving the latter equality with respect to s, we obtain the result L (t2 ) =

2 . s3

EXAMPLE 2 Laplace transform of cos ωt Derive the Laplace transform of f (t) = cos ωt. Solution. We have f 00 (t) = −ω 2 cos ωt = −ω 2 f (t), and f (0) = 1,

f 0 (0) = 0.

Therefore, L (f 00 ) = −ω 2 L (f ) = s2 L (f ) − sf (0) − f 0 (0) = s2 L (f ) − s. Resolving the latter equality with respect to L, we obtain the result L (f ) =

s . s2 + ω 2

EXAMPLE 2’ Laplace transform of sin ωt Derive the Laplace transform of f (t) = sin ωt. Solution. We have f 0 (t) = ω cos ωt,

f 00 (t) = −ω 2 sin ωt = −ω 2 f (t),

so that f (0) = 0,

f 0 (0) = ω.

Therefore, L (f 00 ) = −ω 2 L (f ) = s2 L (f ) − sf (0) − f 0 (0) = s2 L (f ) − ω. Resolving the latter equality with respect to L, we obtain the result L (f ) =

s2

ω . + ω2

EXAMPLE 3 Find the Laplace transform of f (t) = sin2 t. Solution. We have f 0 (t) = 2 sin t cos t = sin 2t,

f 00 (t) = 2 cos 2t,

so that f (0) = 0,

f 0 (0) = 0,

f 00 (0) = 2.

Therefore, L (f 00 ) = 2L (cos 2t) =

2s = s2 L (f ) − sf (0) − f 0 (0) = s2 L (f ). +4

s2

Resolving the equality 2s = s2 L (f ) +4

s2 with respect to s, we obtain the result

2 . + 4)

L (sin2 t) =

s(s2

EXAMPLE 4 Find the Laplace transform of f (t) = t sin ωt. Solution. We have f (0) = 0 and f 0 (t) = sin ωt + ωt cos ωt,

f 0 (0) = 0;

f 00 (t) = 2ω cos ωt − ω 2 t sin ωt = 2ω cos ωt − ω 2 f (t), so that L (f 00 ) = 2ωL (cos ωt) − ω 2 L (f ) = s2 L (f ) − sf (0) − f 0 (0) = s2 L (f ). Resolving the equality

s − ω 2 L (f ) = s2 L (f ) 2 +ω with respect to L, we obtain the result 2ω

s2

L (f ) = L (t sin ωt) =

(s2

2ωs . + ω 2 )2

Differential Equations. Initial value problems Begin with the initial value problem y 00 + ay 0 + by = r(t),

y(0) = K0 ,

y 0 (0) = K1

(a, b = const).

(1)

Describe Laplace’s method for solving this nitial value problem. First step. Writing Y = L, R = L (r) we apply the Laplace transform to (1) and obtain L (y 00 ) + aL (y 0 ) + bL (y) = R or [s2 Y − sy(0) − y 0 (0)] + a[sY − y(0)] + bY = R. This is called the subsidiary equation and reduces to (s2 + as + b)Y = (s + a)y(0) + y 0 (0) + R(s). Second step. Solve the subsidiary equation using the transfer function Q = Q(s) =

s2

1 + as + b

(2)

to obtain Y (s) = [(s + a)y(0) + y 0 (0)]Q(s) + R(s)Q(s). If y(0) = y 0 (0) = 0 this is simply Y = RQ and Q is the quotient Q=

Y L(output) = R L(input)

Third step. Determine y(t) = L−1 (Y ) by rearranging (3) algebraically to a sum of elementary transforms.

EXAMPLE 5 Solve the initial value problem y 00 − y = t,

y(0) = 1,

y 0 (0) = 1.

Solution. In this case, we have, in terms of (1), a = 0,

b = −1,

r(t) = t,

K0 = K1 = 1.

Apply Laplace’s method for solving this nitial value problem. First step. Write 1 Y = L, R = L (r) = L (t) = 2 s 00 and apply the Laplace transform to y − y = t to obtain L (y 00 ) − L (y) = R and then the subsidiary equation [s2 Y − sy(0) − y 0 (0)] − Y = or

1 . s2

1 . s2 Second step. Solve the subsidiary equation using the transfer function (s2 − 1)Y = s + 1 +

Q = Q(s) =

s2

1 −1

to obtain Y (s) = [s + 1]Q(s) +

s+1 1 1 1 1 1 Q(s) = 2 + 2 2 = + 2 − 2. 2 s s − 1 s (s − 1) s−1 s −1 s

Third step. Determine −1

y(t) = L

−1

(Y ) = L













1 1 1 + L−1 2 − L−1 2 = et + sinh t − t. s−1 s −1 s

(3)

Laplace transform of the integral of a function

THEOREM 5.2.2 Let F (s) be the Laplace transform of f (t). If f (t) is piecewise continuous on any finite interval in the range t ≥ 0 and satisfies the condition |f (t)| ≤ M ekt

(t ≥ 0)

for some constants k and M , then L

Z t

f (τ )dτ

0

or

Z t



1 = F (s) (s > 0, s > k), s

f (τ )dτ = L

−1

0





1 F (s) s

PROOF If f (t) is piecewise continuous and satisfies the condition |f (t)| ≤ M ekt for k > 0 and M > 0, the integral g(t) =

Z t

(t ≥ 0) f (τ )dτ

0

is continuous and from the above inequality, we obtain Z t Z t M kt M kt |g(t)| ≤ (e − 1) ≤ e (k > 0), |f (τ )|dτ ≤ M ekτ dτ = k k 0 0 which means that g(t) also satisfies this inequality. Also g 0 (t) = f (t); consequently, g 0 (t) is piecewise continuous on each finite interval in the range t ≥ 0 and L [f (t)] = L [g 0 (t)] = sL (g) − g(0)

(s > k).

Here, g(0) = 0 and F (s) = L (f ) = sL (g) and finally L (g) = L

Z t

f (τ )dτ

0



1 = F (s). s

EXAMPLE 7 Let L (f ) = Find f (t). Solution. We have L so that L

−1

"

−1



s(s2

1 . + ω2)



1 1 = sin ωt 2 2 s +ω ω

# 1Zt 1 Zt 1 = sin ωτ dτ = (1 − cos ωt). s(s2 + ω 2 ) ω 0 ω2 0

PROBLEM 5.2.1 Solve the initial value problem y 0 + 3y = 10 sin t,

y(0) = 0.

Solution. Apply Laplace’s method for solving this nitial value problem. First step. Write 10 Y = L, R = L (10 sin t) = 2 s +1 0 and apply the Laplace transform to y + 3y = 10 sin t to obtain L (y 0 ) + 3L (y) =

10 +1

s2

and then the subsidiary equation [sY − y(0)] + 3Y = or

10 . +1

s2

10 . +1 Second step. Solve the subsidiary equation using the transfer function (s + 3)Y =

s2

Q = Q(s) = to obtain Y (s) =

1 s+3

1 10 10 A Bs + C Q(s) = 2 = + 2 . +1 s +1s+3 s+3 s +1

s2

Find A, B, C by equating 10 = A(s2 + 1) + (Bs + C)s + 3, which yields A = 1,

B = −1,

C = 3.

Third step. Determine the desired solution −1

y(t) = L

−1

(Y ) = L













1 s 1 − L−1 2 + 3L−1 2 = e−3t − cos t + 3 sin t. s+3 s +1 s +1

PROBLEM 5.2.5 Solve the initial value problem y 00 + ay 0 − 2a2 y = 0,

y(0) = 6,

y 0 (0) = 0

(a = const).

Solution. In this case, we have, in terms of (1), b = −2a2 ,

r(t) = 0,

K0 = 6,

K1 = 0.

Apply Laplace’s method for solving this nitial value problem.

First step. Write Y = L(y),

R = L (r) = 0

and apply the Laplace transform to y 00 + ay 0 − 2a2 y = 0 to obtain L (y 00 ) + aL (y 0 ) − 2a2 L (y) = 0 and then the subsidiary equation [s2 Y − sy(0) − y 0 (0)] + a[sY − y(0)] − 2a2 Y = 0. or (s2 + as − 2a2 )Y = 6(s + a). Second step. Solve the subsidiary equation using the transfer function Q = Q(s) =

s2

1 + as − 2a2

to obtain Y (s) = 6(s + a)Q(s) = 6 Third step. Determine y(t) = L−1 (Y ) = 6 6

s2

s2

s+a + as − 2a2

s a +6 2 = 2 + as − 2a s + as − 2a2

s a +6 = 2 2 2 (s + − (a/2) − 2a (s + a/2) − (a/2)2 − 2a2 s a 6 +6 = 2 2 2 (s + a/2) − (3a/2) (s + a/2) − (3a/2)2 a/2)2

6e−(a/2)t cosh (3a/2)t + 4e−(a/2)t sinh (3a/2)t = 2e−2at + 4eat .

PROBLEM 5.2.11 Find F (s) = L (cos2 t). Solution. Set f = cos2 t,

cos2 t = 1 − sin2 t,

f (0) = 1,

f 0 = −2 cos t sin t = −2 sin 2t.

Now

2 = sL (f ) − 1. +4 Resolving the latter equality with respect to L yields L (− sin 2t) = − 

s2



1 2 1 L (f ) = 1− 2 = s s +4 s

!

s2 + 2 . s2 + 4

PROBLEM 5.2.12a Find the Laplace transform of f (t) = t cos ωt. Solution. We have f (0) = 0 and f 0 (t) = cos ωt − ωt sin ωt,

f 0 (0) = 1;

f 00 (t) = −2ω sin ωt − ω 2 t cos ωt = −2ω sin ωt − ω 2 f (t),

so that

L (f 00 ) = −2ωL (sin ωt) − ω 2 L (f ) = s2 L (f ) − sf (0) − f 0 (0) = s2 L (f ) − 1. Resolving the equality 2ω 2 − ω 2 L (f ) = s2 L (f ) − 1 2 2 s +ω with respect to L, we obtain the result −

L (f ) = L (t cos ωt) =

s2 − ω 2 . (s2 + ω 2 )2

PROBLEM 5.2.12b Prove that −1

L

!

1 2 (s + ω 2 )2

Solution. We have

=

1 (sin ωt − ωt cos ωt). 2ω 3

ω(s2 − ω 2 ) , (s2 + ω 2 )2 ω , L (sin ωt) = 2 s + ω2

−L (ωt cos ωt) = −

so that

ω ω(s2 − ω 2 ) 2ω 3 L (sin ωt − ωt cos ωt) = 2 − 2 = 2 , s + ω2 (s + ω 2 )2 (s + ω 2 )2 which yields the desired formula.

PROBLEM 5.2.13 Find f (t) for F (s) = L (f ) = Solution. We have

L

Z t

and f (t) =

−4τ

e

0

Z t 0

1 + 4s

1 1 1 = L (e−4t ), ss+4 s

F (s) = so that

s2

dτ



1 = L (e−4t ), s

1 e−4τ dτ = (1 − e−4t ). 4