10.4 Derivatives, Integrals, and Products of Transforms

The Laplace transform of the (initially unknown) solution of a differential equation is sometimes recognizable as the product of the transforms of two known functions. For example, when we transform the initial value problem

x″+x=cost;x(0)=x′(0)=0,
x′′+x=cost;x(0)=x'(0)=0,

we get

X(s)=s(s2+1)2=ss2+1⋅1s2+1=L{cos t}⋅L{sin t}.
X(s)=s(s2+1)2=ss2+1⋅1s2+1=L{cos t}⋅L{sin t}.

This strongly suggests that there ought to be a way of combining the two functions sin tsin t and cos tcos t to obtain a function x(t) whose transform is the product of their transforms. But obviously x(t) is not simply the product of cos tcos t and sin tsin t, because

L{costsint}=L{12sin 2t}=1s2+4≠s(s2+1)2.
L{costsint}=L{12sin 2t}=1s2+4≠s(s2+1)2.

Thus L{cos tsin t}≠L{cos t}·L{sin t}L{cos tsin t}≠L{cos t}⋅L{sin t}.

Theorem 1 of this section will tell us that the function

h(t)=∫t0f(τ)g(t−τ)dτ
h(t)=∫t0f(τ)g(t−τ)dτ
(1)

has the desired property that

L{h(t)}=H(s)=F(s)⋅G(s)⋅
L{h(t)}=H(s)=F(s)⋅G(s)⋅
(2)

The new function of t defined as the integral in (1) depends only on f and g and is called the convolution of f and g. It is denoted by f∗g,f∗g, the idea being that it is a new type of product of f and g, so tailored that its transform is the product of the transforms of f and g.

We will also write f(t)∗g(t)f(t)∗g(t) when convenient. In terms of the convolution product, Theorem 1 of this section says that

L{f∗g}=L{f}⋅L{g}.
L{f∗g}=L{f}⋅L{g}.

If we make the substitution u=t−τu=t−τ in the integral in (3), we see that

f(t)∗g(t)=∫t0f(τ)g(t−τ)dτ=∫0tf(t−u)g(u)(−du)=∫t0g(u)f(t−u)du=g(t)∗f(t).
f(t)∗g(t)==∫t0f(τ)g(t−τ)dτ=∫0tf(t−u)g(u)(−du)∫t0g(u)f(t−u)du=g(t)∗f(t).

Thus the convolution is commutative: f∗g=g∗ff∗g=g∗f.

Example 1

The convolution of cos tcos t and sin tsin t is

(cost)∗(sint)=∫t0cosτsin(t−τ)dτ.
(cost)∗(sint)=∫t0cosτsin(t−τ)dτ.

We apply the trigonometric identity

cosAsinB=12[sin(A+B)−sin(A−B)]
cosAsinB=12[sin(A+B)−sin(A−B)]

to obtain

(cost)∗(sint)=∫t012[sint−sin(2τ−t)]dτ=12[τsint+12cos(2τ−t)]tτ=0;
(cost)∗(sint)==∫t012[sint−sin(2τ−t)]dτ12[τsint+12cos(2τ−t)]tτ=0;

that is,

(cost)∗(sint)=12tsint.
(cost)∗(sint)=12tsint.

And we recall from Example 5 of Section 10.2 that the Laplace transform of 12tsin t12tsin t is indeed s/(s2+1)2.s/(s2+1)2.

Theorem 1 is proved at the end of this section.

Thus we can find the inverse transform of the product F(s)·G(s),F(s)⋅G(s), provided that we can evaluate the integral

L−1{F(s)⋅G(s)}=∫t0f(τ)g(t−τ)dτ.
L−1{F(s)⋅G(s)}=∫t0f(τ)g(t−τ)dτ.
(5′)

Example 2 illustrates the fact that convolution often provides a convenient alternative to the use of partial fractions for finding inverse transforms.

Example 2

With f(t)=sin 2tf(t)=sin 2t and g(t)=et,g(t)=et, convolution yields

L−1{2(s−1)(s2+4)}=(sin 2t)∗et=∫t0et−τsin 2τdτ=et∫t0e−τsin 2τdτ=et[e−τ5(−sin 2τ−2 cos 2τ)]t0,
L−1{2(s−1)(s2+4)}==(sin 2t)∗et=∫t0et−τsin 2τdτet∫t0e−τsin 2τdτ=et[e−τ5(−sin 2τ−2 cos 2τ)]t0,

so

L−1{2(s−1)(s2+4)}=25et−15sin 2t−25cos 2t.
L−1{2(s−1)(s2+4)}=25et−15sin 2t−25cos 2t.

Differentiation of Transforms

According to Theorem 1 of Section 10.2, if f(0)=0f(0)=0 then differentiation of f(t) corresponds to multiplication of its transform by s. Theorem 2, proved at the end of this section, tells us that differentiation of the transform F(s) corresponds to multiplication of the original function f(t) by −t−t.

Example 3

Find L{t2 sin kt}L{t2 sin kt}.

Solution

Equation (8) gives

L{t2sinkt}=(−1)2d2ds2(ks2+k2)=dds[−2ks(s2+k2)2]=6ks2−2k3(s2+k2)3.
L{t2sinkt}==(−1)2d2ds2(ks2+k2)dds[−2ks(s2+k2)2]=6ks2−2k3(s2+k2)3.
(9)

The form of the differentiation property in Eq. (7) is often helpful in finding an inverse transform when the derivative of the transform is easier to work with than the transform itself.

Example 4

Find L−1{tan−1(1/s)}L−1{tan−1(1/s)}.

Solution

The derivative of tan−1(1/s)tan−1(1/s) is a simple rational function, so we apply Eq. (7):

L−1{tan−11s}=−1tL−1{ddstan−11s}=−1tL−1{−1/s21+(1/s)2}=−1tL−1{−1s2+1}=−1t(−sint).
L−1{tan−11s}===−1tL−1{ddstan−11s}−1tL−1{−1/s21+(1/s)2}−1tL−1{−1s2+1}=−1t(−sint).

Therefore,

L−1{tan−11s}=sintt.
L−1{tan−11s}=sintt.

Equation (8) can be applied to transform a linear differential equation having polynomial, rather than constant, coefficients. The result will be a differential equation involving the transform; whether this procedure leads to success depends, of course, on whether we can solve the new equation more readily than the old one.

Example 5

Let x(t) be the solution of Bessel’s equation of order zero,

tx″+x′+tx=0,
tx′′+x'+tx=0,

such that x(0)=1x(0)=1 and x′(0)=0.x'(0)=0. This solution of Bessel’s equation is customarily denoted by J0(t).J0(t). Because

L{x′(t)}=sX(s)−1andL{x″(t)}=s2X(s)−s,
L{x'(t)}=sX(s)−1andL{x′′(t)}=s2X(s)−s,

and because x and x″x′′ are each multiplied by t, application of Eq. (6) yields the transformed equation

−dds[s2X(s)−s]+[sX(s)−1]−dds[X(s)]=0.
−dds[s2X(s)−s]+[sX(s)−1]−dds[X(s)]=0.

The result of differentiation and simplification is the differential equation

(s2+1)X′(s)+sX(s)=0.
(s2+1)X'(s)+sX(s)=0.

This equation is separable—

X′(s)X(s)=−ss2+1;
X'(s)X(s)=−ss2+1;

its general solution is

X(s)=C√s2+1.
X(s)=Cs2+1−−−−−√.

In Problem 39 we outline the argument that C=1.C=1. Because X(s)=L{J0(t)},X(s)=L{J0(t)}, it follows that

L{J0(t)}=1√s2+1.
L{J0(t)}=1s2+1−−−−−√.
(10)

Integration of Transforms

Differentiation of F(s) corresponds to multiplication of f(t) by t (together with a change of sign). It is therefore natural to expect that integration of F(s) will correspond to division of f(t) by t. Theorem 3, proved at the end of this section, confirms this, provided that the resulting quotient f(t)/tf(t)/t remains well behaved as t→0t→0 from the right; that is, provided that

limt→0+f(t)t exists and is finite.
limt→0+f(t)t exists and is finite.
(11)

Example 6

Find L{(sinht)/t}L{(sinht)/t}.

Solution

We first verify that the condition in (11) holds:

limt→0sinhtt=limt→0et−e−t2t=limt→0et+e−t2=1,
limt→0sinhtt=limt→0et−e−t2t=limt→0et+e−t2=1,

with the aid of l’Hôpital’s rule. Then Eq. (12), with f(t)=sinht,f(t)=sinht, yields

L{sinhtt}=∫∞sL{sinht}dσ=∫∞sdσσ2−1=12∫∞s(1σ−1−1σ+1)dσ=12[lnσ−1σ+1]∞s.
L{sinhtt}==∫∞sL{sinht}dσ=∫∞sdσσ2−112∫∞s(1σ−1−1σ+1)dσ=12[lnσ−1σ+1]∞s.

Therefore,

L{sinhtt}=12lns+1s−1,
L{sinhtt}=12lns+1s−1,

because ln 1=0ln 1=0.

The form of the integration property in Eq. (13) is often helpful in finding an inverse transform when the indefinite integral of the transform is easier to handle than the transform itself.

Example 7

Find L−1{2s/(s2−1)2}L−1{2s/(s2−1)2}.

Solution

We could use partial fractions, but it is much simpler to apply Eq. (13). This gives

L−1{2s(s2−1)2}=tL−1{∫∞s2σ(σ2−1)2dσ}=tL−1{[−1σ2−1]∞s}=tL−1{1s2−1},
L−1{2s(s2−1)2}==tL−1{∫∞s2σ(σ2−1)2dσ}tL−1{[−1σ2−1]∞s}=tL−1{1s2−1},

and therefore

L−1{2s(s2−1)2}=tsinht.
L−1{2s(s2−1)2}=tsinht.

* Proofs of Theorems

Proof of Theorem 1:

The transforms F(s) and G(s) exist when s>cs>c by Theorem 2 of Section 10.1. For any τ>0τ>0 the definition of the Laplace transform gives

G(s)=∫∞0e−sug(u)du=∫∞τe−s(t−τ)g(t−τ)dt(u=t−τ),
G(s)=∫∞0e−sug(u)du=∫∞τe−s(t−τ)g(t−τ)dt(u=t−τ),

and therefore

G(s)=esτ∫∞0e−stg(t−τ)dt,
G(s)=esτ∫∞0e−stg(t−τ)dt,

because we may define f(t) and g(t) to be zero for t<0.t<0. Then

F(s)G(s)=G(s)∫∞0e−sτf(τ)dτ=∫∞0e−sτf(τ)G(s)dτ=∫∞0e−sτf(τ)(esτ∫∞0e−sτg(t−τ)dt)dτ=∫∞0(∫∞0e−stf(τ)g(t−τ)dt)dτ.
F(s)G(s)===G(s)∫∞0e−sτf(τ)dτ=∫∞0e−sτf(τ)G(s)dτ∫∞0e−sτf(τ)(esτ∫∞0e−sτg(t−τ)dt)dτ∫∞0(∫∞0e−stf(τ)g(t−τ)dt)dτ.

Now our hypotheses on f and g imply that the order of integration may be reversed. (The proof of this requires a discussion of uniform convergence of improper integrals, and can be found in Chapter 2 of Churchill’s Operational Mathematics, 3rd ed. (New York: McGraw-Hill, 1972).) Hence

F(s)G(s)=∫∞0(∫∞0e−stf(τ)g(t−τ)dτ)dt=∫∞0e−st(∫t0f(τ)g(t−τ)dτ)dt=∫∞0e−st[f(t)∗g(t)]dt,
F(s)G(s)===∫∞0(∫∞0e−stf(τ)g(t−τ)dτ)dt∫∞0e−st(∫t0f(τ)g(t−τ)dτ)dt∫∞0e−st[f(t)∗g(t)]dt,

and therefore,

F(s)G(s)=L{f(t)∗g(t)}.
F(s)G(s)=L{f(t)∗g(t)}.

We replace the upper limit of the inner integral with t because g(t−τ)=0g(t−τ)=0 whenever τ>t.τ>t. This completes the proof of Theorem 1.

Proof of Theorem 2:

Because

F(s)=∫∞0e−stf(t)dt,
F(s)=∫∞0e−stf(t)dt,

differentiation under the integral sign yields

F(s)=dds∫∞0e−stf(t)dt=∫∞0dds[e−stf(t)] dt=∫∞0e−st[−tf(t)] dt;
F(s)==dds∫∞0e−stf(t)dt∫∞0dds[e−stf(t)] dt=∫∞0e−st[−tf(t)] dt;

thus

F′(s)=L{−tf(t)},
F'(s)=L{−tf(t)},

which is Eq. (6). We obtain Eq. (7) by applying L−1L−1 and then dividing by −t.−t. The validity of differentiation under the integral sign depends on uniform convergence of the resulting integral; this is discussed in Chapter 2 of the book by Churchill just mentioned.

Proof of Theorem 3:

By definition,

F(σ)=∫∞0e−σtf(t) dt.
F(σ)=∫∞0e−σtf(t) dt.

So integration of F(σ)F(σ) from s to +∞+∞ gives

∫∞sF(σ) dσ=∫∞s(∫∞0e−σtf(t) dt)dσ.
∫∞sF(σ) dσ=∫∞s(∫∞0e−σtf(t) dt)dσ.

Under the hypotheses of the theorem, the order of integration may be reversed (see Churchill’s book once again); it follows that

∫∞sF(σ)dσ=∫∞0(∫∞se−σtf(t)dσ)dt=∫∞0[e−σt−t]∞σ=sf(t)dt=∫∞0e−stf(t)tdt=L{f(t)t}.
∫∞sF(σ)dσ===∫∞0(∫∞se−σtf(t)dσ)dt∫∞0[e−σt−t]∞σ=sf(t)dt∫∞0e−stf(t)tdt=L{f(t)t}.

This verifies Eq. (12), and Eq. (13) follows upon first applying L−1L−1 and then multiplying by t.

10.4 Problems

Find the convolution f(t)∗g(t)f(t)∗g(t) in Problems 1 through 6.

  1. f(t)=t, g(t)≡1f(t)=t, g(t)≡1

     

  2. f(t)=t, g(t)=eatf(t)=t, g(t)=eat

     

  3. f(t)=t, g(t)=sintf(t)=t, g(t)=sint

     

  4. f(t)=t2, g(t)=costf(t)=t2, g(t)=cost

     

  5. f(t)=g(t)=eatf(t)=g(t)=eat

     

  6. f(t)=eat, g(t)=ebt (a≠b)f(t)=eat, g(t)=ebt (a≠b)

Apply the convolution theorem to find the inverse Laplace transforms of the functions in Problems 7 through 14.

  1. F(s)=1s(s−3)F(s)=1s(s−3)

     

  2. F(s)=1s(s2+4)F(s)=1s(s2+4)

     

  3. F(s)=1(s2+9)2F(s)=1(s2+9)2

     

  4. F(s)=1s2(s2+k2)F(s)=1s2(s2+k2)

     

  5. F(s)=s2(s2+4)2F(s)=s2(s2+4)2

     

  6. F(s)=1s(s2+4s+5)F(s)=1s(s2+4s+5)

     

  7. F(s)=s(s−3)(s2+1)F(s)=s(s−3)(s2+1)

     

  8. F(s)=ss4+5s2+4F(s)=ss4+5s2+4

In Problems 15 through 22, apply either Theorem 2 or Theorem 3 to find the Laplace transform of f(t).

  1. f(t)=tsin 3tf(t)=tsin 3t

     

  2. f(t)=t2cos 2tf(t)=t2cos 2t

     

  3. f(t)=te2tcos 3tf(t)=te2tcos 3t

     

  4. f(t)=te−tsin2tf(t)=te−tsin2t

     

  5. f(t)=sinttf(t)=sintt

     

  6. f(t)=1−cos 2ttf(t)=1−cos 2tt

     

  7. f(t)=e3t−1tf(t)=e3t−1t

     

  8. f(t)=et−e−ttf(t)=et−e−tt

Find the inverse transforms of the functions in Problems 23 through 28.

  1. F(s)=lns−2s+2F(s)=lns−2s+2

     

  2. F(s)=lns2+1s2+4F(s)=lns2+1s2+4

     

  3. F(s)=lns2+1(s+2)(s−3)F(s)=lns2+1(s+2)(s−3)

     

  4. F(s)=tan−13s+2F(s)=tan−13s+2

     

  5. F(s)=ln(1+1s2)F(s)=ln(1+1s2)

     

  6. F(s)=s(s2+1)3F(s)=s(s2+1)3

In Problems 29 through 34, transform the given differential equation to find a nontrivial solution such that x(0)=0x(0)=0.

  1. tx″+(t−2)x′+x=0tx′′+(t−2)x'+x=0

     

  2. tx″+(3t−1)x′+3x=0tx′′+(3t−1)x'+3x=0

     

  3. tx″−(4t+1)x′+2(2t+1)x=0tx′′−(4t+1)x'+2(2t+1)x=0

     

  4. tx″+2(t−1)x′−2x=0tx′′+2(t−1)x'−2x=0

     

  5. tx″−2x′+tx=0tx′′−2x'+tx=0

     

  6. tx″+(4t−2)x′+(13t−4)x=0tx′′+(4t−2)x'+(13t−4)x=0

     

  7. Apply the convolution theorem to show that

    L−1{1(s−1)√s}=2et√π∫√t0e−u2 du=eterf√t.
    L−1{1(s−1)s√}=2etπ−−√∫t√0e−u2 du=eterft√.

    (Suggestion: Substitute u=√tu=t√.)

In Problems 36 through 38, apply the convolution theorem to derive the indicated solution x(t) of the given differential equation with initial conditions x(0)=x′(0)=0x(0)=x'(0)=0.

  1. x″+4x=f(t); x(t)=12∫t0f(t−τ)sin 2τ dτx′′+4x=f(t); x(t)=12∫t0f(t−τ)sin 2τ dτ

     

  2. x″+2x′+x=f(t); x(t)=∫t0τe−τf(t−τ) dτx′′+2x'+x=f(t); x(t)=∫t0τe−τf(t−τ) dτ

     

  3. x″+4x′+13x=f(t);x(t)=13∫t0f(t−τ)e−2τ sin 3τ dτx′′+4x'+13x=f(t);x(t)=13∫t0f(t−τ)e−2τ sin 3τ dτ

Termwise Inverse Transformation of Series

In Chapter 2 of Churchill’s Operational Mathematics, the following theorem is proved. Suppose that f(t) is continuous for t≧0,t≧0, that f(t) is of exponential order as t→+∞,t→+∞, and that

F(s)=∞∑n=0ansn+k+1
F(s)=∑n=0∞ansn+k+1

where 0≦k<10≦k<1 and the series converges absolutely for s>c.s>c. Then

f(t)=∞∑n=0antn+kΓ(n+k+1).
f(t)=∑n=0∞antn+kΓ(n+k+1).
Apply this result in Problems 39 through 41.

  1. In Example 5 it was shown that

    L{J0(t)}=C√s2+1=Cs(1+1s2)−1/2.
    L{J0(t)}=Cs2+1−−−−−√=Cs(1+1s2)−1/2.

    Expand with the aid of the binomial series and then compute the inverse transformation term by term to obtain

    J0(t)=C∞∑n=0(−1)nt2n22n(n!)2.
    J0(t)=C∑n=0∞(−1)nt2n22n(n!)2.

    Finally, note that J0(0)=1J0(0)=1 implies that C=1C=1.

  2. Expand the function F(s)=s−1/2e−1/sF(s)=s−1/2e−1/s in powers of s−1s−1 to show that

    L−1{1√se−1/s}=1√πtcos 2√t.
    L−1{1s√e−1/s}=1πt−−√cos 2t√.

  3. Show that

    L−1{1se−1/s}=J0(2√t).
    L−1{1se−1/s}=J0(2t√).
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