Derivatives and the Laplace Transform

Home , Laplace transform

Section 7.2 Solving initial value problems with the Laplace transform In this section, we will see how to use the Laplace transform to solve initial value problems involving linear equations with constant coeﬃcients, i.e. problems of the form

(n) ′ (n−1) any + ... + a1y + a0y = f(t), y(0) = a0, . . . , y (0) = an−1.

However, before we do this, we must think about one more interesting property of Laplace transforms–the way that they interact with derivatives.

To understand Laplace transforms of derivatives, we should begin by recalling the types of functions that have Laplace transforms:

Theorem 2, Section 7.1. Suppose that f(t) is deﬁned for t ≥ 0, is piecewise continuous, and that there are constants M, T , and c so that

f(t) −M ≤ ≤ M for all t ≥ T. (1) ect Then L{f(t)} exists for all s > c.

The theorem says that if

1. f(t) is piecewise continuous, and

2. f(t) does not grow too fast with respect to the exponential function ect for some constant c (f(t) is of exponential order), then we are guaranteed that f(t) has a Laplace transform F (s). There is one interesting side note here: if f(t) is of exponential order, we know that the quantity

f(t) ect does not get too large as t → ∞; alternatively, we interpret this comment as saying that f(t) and ect grow at roughly the same rate. Because of this, if we increase c at all, then the denominator of the fraction will automatically grow faster than f(t). As a result,

f(t) lim = 0, s > c. t→∞ est

Derivatives and the Laplace Transform

If f(t) is a function that is continuous and of exponential order, then we know that its Laplace transform F (s) exists, and we also know that

f(t) lim = 0, s > c t→∞ est

1 Section 7.2 for some number c. We would like to understand the Laplace transform L{f ′(t)} of f ′(t) (assuming that the transform exists). We should go back to the deﬁnition of the transform: ∫ ∞ L{f ′(t)} = e−stf ′(t) dt. 0 Let’s make the calculation, using integration by parts with

u = e−st dv = f ′(t) dt du = −se−st dt v = f(t).

∫ ∞ L{f ′(t)} = e−stf ′(t) dt 0 ( ∫ ) b = lim e−stf ′(t) dt b→∞ ( 0 ∫ ) b = lim e−stf(t) b + s e−stf(t) dt b→∞ 0 ( ) 0 ( ∫ ) b = lim e−stf(t) b + lim s e−stf(t) dt →∞ 0 →∞ b ( ) b 0

= lim e−stf(t) b + sF (s) b→∞ 0 = lim (e−sbf(b) − f(0)) + sF (s) b→∞ f(b) − = lim ( − f(0)) + sF (s) b→∞ e sb = −f(0) + sF (s) = sF (s) − f(0) since f(b) lim − = 0 for s > c. b→∞ e sb In other words, if we know F (s) = L{f(t)} and f(0), then we can easily ﬁnd the Laplace transform L{f ′(t)} of f ′(t) using the formula

L{f ′(t)} = sF (s) − f(0).

This idea is stated more carefully in the following theorem: Theorem 1. Suppose that f(t) is continuous for t ≥ 0, that f ′(t) exists at all but a ﬁnite number of points on every bounded subinterval of [0, ∞), and that f(t) is of exponential order so that there are constants M, T , and c so that f(t) −M ≤ ≤ M for all t ≥ T. (2) ect

2 Section 7.2 Then L{f ′(t)} exists for all s > c, and is given by

L{f ′(t)} = sF (s) − f(0).

1 Example. Given that the Laplace transform of f(t) = sin t is F (s) = s2+1 , verify the formula for the Laplace transform of cos t. Since f ′(t) = cos t, the theorem says that the transform of cos t is given by

L{ } − cos t = sF((s) f)(0) 1 = s − 0 s2 + 1 s = , s2 + 1 s > 0.

Let’s use the Laplace transform of f ′(t) to ﬁnd the Laplace transform of f ′′(t). Using the formula from the theorem, we know that

L{f ′′(t)} = sL{f ′(t)} − f ′(0) = s(sF (s) − f(0)) − f ′(0) = s2F (s) − sf(0) − f ′(0).

Thus we begin to see the development of a remarkable property of Laplace transforms: transforms rewrite all of the derivatives of f(t) as algebraic quantities involving the transform F (s) of f(t). The corollary below makes this idea precise:

Corollary. If each of the functions f, f ′,··· , and f (n−1) are continuous on [0, ∞), each of their derivatives exists at all but a ﬁnite number of points on every bounded subinterval of [0, ∞), and each is of exponential order for the same values of M and c, then L{f (n)(t)} exists for s > c and is given in terms of the Laplace transform F (s) of f(t) by

L{f (n)(t)} = snF (s) − sn−1f(0) − · · · − sf (n−2)(0) − f (n−1)(0). (3)

Notice that the corollary does not apply to functions that are piecewise continuous, but not actually continuous; we will need to handle such functions separately.

3 Section 7.2 Solving initial value problems using Laplace transforms

With the tools that we have developed in this and in the previous section, we now have the ability to solve initial value problems involving linear equations with constant coeﬃcients. To solve the initial value problem

(n) ′ (n−1) any + ... + a1y + a0y = f(t), y(0) = a0, ··· , y (0) = an−1 using Laplace transforms, follow the steps below: 1. Apply the Laplace transform to the entire equation. 2. The resulting equation will involve only functions of s and the Laplace transform Y (s) of y(t). Use algebraic methods to solve for Y (s). 3. Since the inverse Laplace transform of Y (s) is unique–it must be y(t)–we can apply the inverse Laplace transform to the entire equation to return to an equation for y(t).

Example. Use Laplace transforms to solve the initial value problem y′′ + y = sin 2t, y(0) = 2, y′(0) = 1. Let’s start by applying the Laplace transform to both sides of the equation: y′′ + y = sin 2t 2 (s2Y (s) − sy(0) − y′(0)) + Y (s) = s2 + 4 2 (s2Y (s) − 2s − 1) + Y (s) = s2 + 4 2 s2Y (s) − 2s − 1 + Y (s) = s2 + 4

The transformed version of our equation is thus 2 s2Y (s) − 2s − 1 + Y (s) = . s2 + 4 Notice that the equation is algebraic–there are no derivatives left–and involves only the variable s and the unknown function Y (s). Let’s solve the equation for Y (s): 2 s2Y (s) − 2s − 1 + Y (s) = s2 + 4 2 Y (s)(s2 + 1) − 2s − 1 = s2 + 4 2 Y (s)(s2 + 1) = + 2s + 1 s2 + 4 2 2s 1 Y (s) = + + . (s2 + 1)(s2 + 4) s2 + 1 s2 + 1

4 Section 7.2 We now have a formula for Y (s), the Laplace transform of the unknown function y(t). As we saw in the previous section, Y (s) is the Laplace transform of precisely one function y(t). So if we can ﬁnd the inverse Laplace transform of the equation 2 2s 1 Y (s) = + + , (s2 + 1)(s2 + 4) s2 + 1 s2 + 1 we will have a formula for y(t)! We know that

y(t) = L−1{Y (s)} 2 2s 1 = L−1{ + + } (s2 + 1)(s2 + 4) s2 + 1 s2 + 1 2 2s 1 = L−1{ } + L−1{ } + L−1{ }. (s2 + 1)(s2 + 4) s2 + 1 s2 + 1

Notice that we can ﬁnd the inverse transforms of the second two fractions using a table of transforms: 2s s L−1{ } = 2L−1{ } = 2 cos t s2 + 1 s2 + 1 and 1 L−1{ } = sin t. s2 + 1 However, computing L−1{ 2 } (s2 + 1)(s2 + 4) is a bit trickier–this fraction doesn’t show up on our table of transforms. There are many possible techniques for handling such a situation; the most natural one is to try to decompose the fraction using partial fractions. If the resulting fractions have forms that appear on our table of transforms, then we’ll be able to compute the answer. The denominator of the fraction 2 (s2 + 1)(s2 + 4) is factored as far as possible, into irreducible quadratics. Thus the form of the decomposed fraction must be 2 B s + C B s + C = 1 1 + 2 2 . (s2 + 1)(s2 + 4) s2 + 1 s2 + 4 Let’s add the fractions: B s + C B s + C (B s + C )(s2 + 4) + (B s + C )(s2 + 1) 1 1 + 2 2 = 1 1 2 2 s2 + 1 s2 + 4 (s2 + 1)(s2 + 4) B s3 + C s2 + 4B s + 4C + B s3 + C s2 + B s + C = 1 1 1 1 2 2 2 2 (s2 + 1)(s2 + 4)

Since 2 B s + C B s + C B s3 + C s2 + 4B s + 4C + B s3 + C s2 + B s + C = 1 1 + 2 2 = 1 1 1 1 2 2 2 2 , (s2 + 1)(s2 + 4) s2 + 1 s2 + 4 (s2 + 1)(s2 + 4)

5 Section 7.2 we know that 3 2 3 2 B1s + C1s + 4B1s + 4C1 + B2s + C2s + B2s + C2 = 2. We can determine the constants by equating factors:

3 s B1 + B2 = 0

2 s C1 + C2 = 0

1 s 4B1 + B2 = 0

0 s 4C1 + C2 = 2 Using the second and fourth equations, we have

4C1 + C2 − C1 − C2 = 2 − 0

3C1 = 2 2 C = , 1 3 − − 2 so that C2 = C1 = 3 . Using the ﬁrst and third equations, we have

4B1 + B2 − B1 − B2 = 0 − 0

3B1 = 0

B1 = 0, so that B2 = −B1 = 0. Thus our original fraction decomposes as 2 2/3 −2/3 = + . (s2 + 1)(s2 + 4) s2 + 1 s2 + 4 Both of these fractions have easily computable Laplace transforms: 2/3 2 1 2 L−1{ } = L−1{ } = sin t s2 + 1 3 s2 + 1 3 and −2/3 1 2 1 L−1{ } = − L−1{ } = − sin 2t. s2 + 4 3 s2 + 4 3 Returning to the original problem, we have y(t) = L−1{Y (s)} 2 2s 1 = L−1{ + + } (s2 + 1)(s2 + 4) s2 + 1 s2 + 1 2 2s 1 = L−1{ } + L−1{ } + L−1{ } (s2 + 1)(s2 + 4) s2 + 1 s2 + 1 2/3 −2/3 2s 1 = L−1{ } + L−1{ } + L−1{ } + L−1{ } s2 + 1 s2 + 4 s2 + 1 s2 + 1 2 1 = sin t − sin 2t + 2 cos t + sin t 3 3 5 1 = sin t − sin 2t + 2 cos t. 3 3

6 Section 7.2 With this conclusion, we have solved the original initial value problem: 5 1 y(t) = sin t − sin 2t + 2 cos t. 3 3

Example. Use Laplace transforms to solve the initial value problem y′′ + 3y′ + 2y = t, y(1) = 0, y′(1) = 2. At first glance, this seems to be a poor choice of problem to apply Laplace transforms to; we need to know y(0) and y′(0) in order to be able to use the method! Fortunately, this difficulty is not insurmountable; we can get around the issue by defining a new function v(t): set v(t) = y(t + 1), so that v(0) = y(1) = 0, and v′(0) = y′(1) = 2. We need to rewrite the equation accordingly: since y′′(t) + 3y′(t) + 2y(t) = t, we know that y′′(t + 1) + 3y′(t + 1) + 2y(t + 1) = t + 1, so that our initial value problem becomes v′′ + 3v′ + 2v = t + 1, v(0) = 0, v′(0) = 2. Applying Laplace transforms to both sides of the equation, we have L{v′′ + 3v′ + 2v} = L{t + 1} 1 1 s2V (s) − sv(0) − v′(0) + 3(sV (s) − v(0)) + 2V (s) = + s2 s 1 1 s2V (s) − 2 + 3sV (s) + 2V (s) = + s2 s 1 1 s2V (s) + 3sV (s) + 2V (s) = + + 2 s2 s 1 s 2s2 V (s)(s2 + 3s + 2) = + + s2 s2 s2 1 + s + 2s2 V (s)(s2 + 3s + 2) = s2 1 + s + 2s2 V (s) = . s2(s2 + 3s + 2)

Now that we have a formula for V (s), we can apply the inverse Laplace transform to the entire equation to get a formula for v(t). Again, we don’t see the function 1 + s + 2s2 s2(s2 + 3s + 2)

7 Section 7.2 on the list of Laplace transforms, so it will be necessary to rewrite the fraction as a sum of component fractions. The denominator factors completely as

s2(s2 + 3s + 2) = (s − 0)2(s + 1)(s + 2), so the expanded fraction should have form

1 + s + 2s2 A B C D = + + + . s2(s2 + 3s + 2) s s2 s + 1 s + 2

Adding the four component fractions, we get

A B C D As(s2 + 3s + 2) + B(s2 + 3s + 2) + Cs2(s + 2) + Ds2(s + 1) + + + = s s2 s + 1 s + 2 s2(s + 1)(s + 2) As3 + 3As2 + 2As + Bs2 + 3Bs + 2B + Cs3 + 2Cs2 + Ds3 + Ds2 = s2(s + 1)(s + 2) As3 + Cs3 + Ds3 + 3As2 + Bs2 + 2Cs2 + Ds2 + 2As + 3Bs + 2B = s2(s + 1)(s + 2) s3(A + C + D) + s2(3A + B + 2C + D) + s(2A + 3B) + 2B = . s2(s + 1)(s + 2)

Equating coeﬃcients, we have s3 A + C + D = 0

s2 3A + B + 2C + D = 2

s1 2A + 3B = 1

s0 2B = 1

so that B = 1/2. Using the third equation, we have 3 2A + = 1 2 1 A = − . 4 This the ﬁrst equation becomes 1 C + D = , 4 while the second becomes 9 2C + D = . 4 Thus 9 1 2C + D − C − D = − 4 4 C = 2,

8 Section 7.2 so that 1 7 D = − 2 = − . 4 4 Thus our original equation 1 + s + 2s2 V (s) = s2(s2 + 3s + 2) can be rewritten as −1/4 1/2 2 −7/4 V (s) = + + + ; s s2 s + 1 s + 2 each of the component fractions above has a Laplace transform, so

v(t) = L−1{V (s)} −1/4 1/2 2 −7/4 = L−1{ + + + } s s2 s + 1 s + 2 −1/4 1/2 2 −7/4 = L−1{ } + L−1{ } + L−1{ } + L−1{ } s s2 s + 1 s + 2 1 1 1 1 1 7 1 = − L−1{ } + L−1{ } + 2L−1{ } − L−1{ } 4 s 2 s2 s + 1 4 s + 2 1 1 7 = − + t + 2e−t − e−2t. 4 2 4

Thus 1 1 7 v(t) = − + t + 2e−t − e−2t, 4 2 4 and since v(t) = y(t + 1), we think of v(t − 1) = y(t). Thus the function y(t) is given by 1 1 7 y(t) = − + (t − 1) + 2e−t+1 − e−2t+2. 4 2 4

Now that we have seen an example of the use of Laplace transforms to solve initial value problems, we should make a few comments about the advantages of using Laplace transforms over the methods of Chapter 3 when attempting to solve linear equations with constant coeﬃcients:

• The method of Laplace transforms does not require us to ﬁnd yc and yp separately.

• There is no need to make guesses, as we had to when we were looking for yp. • The method also does not require us to determine the appropriate constants based on the initial values; it automatically outputs the desired particular solution to the initial value problem.

• The method completely rewrites the diﬀerential equation as an algebraic one, which is automatically simpler to solve.

• Higher order equations may be simpler to solve using Laplace transforms, since we don’t have to ﬁnd the roots of a characteristic equation (usually a diﬃcult problem if the degree of the equation is higher than 2).

9 Section 7.2 Of course, the method has its downside: it may be quite diﬃcult to actually ﬁnd the desired inverse Laplace transform. We will consider this problem carefully over the next few sections, and see some appropriate techniques for handling it.

Alternate Method for Finding Laplace Transforms

We will often come across functions whose Laplace transforms are not easily accessible on a table; in such cases, we would like to have alternate methods for ﬁnding the transform, as opposed to reverting to the deﬁnition ∫ ∞ F (s) = e−stf(t) dt. (4) 0 In such cases, it will be helpful to recall that, if f and its derivatives are continuous, then

L{f (n)(t)} = snF (s) − sn−1f(0) − ... − sf (n−2)(0) − f (n−1)(0).

For suﬃciently nice functions, we can use this identity as an alternative to ﬁnding transforms via (4). Notice particular that, if f(0) = 0, then

L{f ′(t)} = sL{f(t)}; in other words, there is a sense in which the transform of a derivative is just a multiple (by s) of the transform of the original function. We will see momentarily that the “reverse” is also true: the transform of an integral is a multiple by 1/s of the transform of the original function.

Example. Find L{tet} without using the deﬁnition of the Laplace Transform. Setting f(t) = tet, we know that f ′(t) = et + tet = et + f(t). On one hand, we know from (3) that

L{f ′(t)} = sL{f(t)} − f(0) = sL{f(t)} since f(0) = 0. On the other hand, we can use the formula for f ′(t) to see that

L{f ′(t)} = L{et} + L{f(t)} 1 = + L{f(t)}. s − 1

10 Section 7.2 Now we have two formulas for L{f ′(t)}; equating them, we see that

1 sL{f(t)} = + L{f(t)} s − 1 1 sL{f(t)} − L{f(t)} = s − 1 1 L{f(t)}(s − 1) = s − 1 1 L{f(t)} = . (s − 1)2

Thus we have 1 L{tet} = . (s − 1)2

Example. Find L{t sinh t}. Recall that et − e−t sinh t = ; 2 setting et − e−t f(t) = t , 2 we know that et − e−t f(t) = t 2 et − e−t et + e−t f ′(t) = + t 2 2 et − e−t f ′′(t) = et + e−t + t 2 = et + e−t + f(t),

so that f(0) = 0 and f ′(0) = 0. Notice that f(t) shows up in the formula for f ′′(t); we can use this to our advantage. On one hand, we know that L{f ′′(t)} = s2L{f(t)} − sf(0) − f ′(0) = s2L{f(t)}. But we can calculate the transform of f ′′ another way, using the formula for f ′′: f ′′(t) = et + e−t + f(t) L{f ′′(t)} = L{et} + L{e−t} + L{f(t)} 1 1 = + + L{f(t)}. s − 1 s + 1

11 Section 7.2 Now we have two diﬀerent formulas for L{f ′′(t)}; thus we know that 1 1 s2L{f(t)} = + + L{f(t)}. s − 1 s + 1 Since the only unknown in this equation is L{f(t)}, we can solve it to ﬁnd the desired transform: 1 1 s2L{f(t)} = + + L{f(t)} s − 1 s + 1 1 1 s2L{f(t)} − L{f(t)} = + s − 1 s + 1 s + 1 + s − 1 L{f(t)}(s2 − 1) = (s − 1)(s + 1) 2s L{f(t)}(s2 − 1) = s2 − 1 2s L{f(t)} = . (s2 − 1)2

Thus the Laplace transform of f(t) = t sinh t is 2s L{t sinh t} = . (s2 − 1)2

The trick we used in the previous examples to calculate the Laplace transform worked speciﬁcally because f(t) shows up in the formula for f ′(t) or for f ′′(t); such a phenomenon is common for exponential functions, trig functions, and hyperbolic trig functions, so the above technique has many possible applications.

Finding Inverse Laplace Transforms

Earlier, we noted that, when f(0) = 0, the Laplace transform of a derivatives is just a multiple by s of the Laplace transform of the original function; it seems reasonable to guess that the Laplace transform of the integral of a function is a multiple by 1/s of the transform of the original function. The theorem below veriﬁes this idea:

Theorem 2. If f(t) is piecewise continuous for t ≥ 0 and satisﬁes (2), then { ∫ } t 1 F (s) L f(τ) dτ = L{f(t)} = 0 s s for s > c; equivalently, { } ∫ F (s) t L−1 = f(τ) dτ. s 0

12 Section 7.2 As indicated above, the difficulty in solving a differential equation by using Laplace transforms is dictated by how hard it is to find the inverse Laplace transform of the resulting function. Theorem 2 gives us an additional method for finding certain inverse transforms.

Example. Find the inverse Laplace transform y(t) of 1 Y (s) = . s(s2 + 1)

If we recognize the function Y (s) as 1 1 Y (s) = F (s) where F (s) = , s s2 + 1 then we can use Theorem 2 to help us ﬁnd the inverse transform of Y (s). Clearly

L−1{F (s)} = f(t) = sin t, so ∫ t L−1{Y (s)} = sin τ dτ 0 t

= − cos τ 0 = cos 0 − cos t = 1 − cos t.

We conclude that y(t) = 1 − cos t, which is easily veriﬁed using the transform:

L{y(t)} = L{1 − cos t} 1 s = − s s2 + 1 s2 + 1 − s2 = s(s2 + 1) 1 = s(s2 + 1) = Y (s).