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
we get
This strongly suggests that there ought to be a way of combining the two functions sin t
Thus L{cos tsin t}≠L{cos t}·L{sin t}
Theorem 1 of this section will tell us that the function
has the desired property that
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,
We will also write f(t)∗g(t)
If we make the substitution u=t−τ
Thus the convolution is commutative: f∗g=g∗f
The convolution of cos t
We apply the trigonometric identity
to obtain
that is,
And we recall from Example 5 of Section 10.2 that the Laplace transform of 12tsin t
Theorem 1 is proved at the end of this section.
Thus we can find the inverse transform of the product F(s)·G(s),
Example 2 illustrates the fact that convolution often provides a convenient alternative to the use of partial fractions for finding inverse transforms.
With f(t)=sin 2t
so
According to Theorem 1 of Section 10.2, if f(0)=0
Find L{t2 sin kt}
Equation (8) gives
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.
Find L−1{tan−1(1/s)}
The derivative of tan−1(1/s)
Therefore,
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.
Let x(t) be the solution of Bessel’s equation of order zero,
such that x(0)=1
and because x and x″
The result of differentiation and simplification is the differential equation
This equation is separable—
its general solution is
In Problem 39 we outline the argument that C=1.
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)/t
Find L{(sinht)/t}
We first verify that the condition in (11) holds:
with the aid of l’Hôpital’s rule. Then Eq. (12), with f(t)=sinht,
Therefore,
because ln 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.
Find L−1{2s/(s2−1)2}
We could use partial fractions, but it is much simpler to apply Eq. (13). This gives
and therefore
The transforms F(s) and G(s) exist when s>c
and therefore
because we may define f(t) and g(t) to be zero for t<0.
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
and therefore,
We replace the upper limit of the inner integral with t because g(t−τ)=0
Because
differentiation under the integral sign yields
thus
which is Eq. (6). We obtain Eq. (7) by applying L−1
By definition,
So integration of F(σ)
Under the hypotheses of the theorem, the order of integration may be reversed (see Churchill’s book once again); it follows that
This verifies Eq. (12), and Eq. (13) follows upon first applying L−1
Find the convolution f(t)∗g(t)
f(t)=t, g(t)≡1
f(t)=t, g(t)=eat
f(t)=t, g(t)=sint
f(t)=t2, g(t)=cost
f(t)=g(t)=eat
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.
F(s)=1s(s−3)
F(s)=1s(s2+4)
F(s)=1(s2+9)2
F(s)=1s2(s2+k2)
F(s)=s2(s2+4)2
F(s)=1s(s2+4s+5)
F(s)=s(s−3)(s2+1)
F(s)=ss4+5s2+4
In Problems 15 through 22, apply either Theorem 2 or Theorem 3 to find the Laplace transform of f(t).
f(t)=tsin 3t
f(t)=t2cos 2t
f(t)=te2tcos 3t
f(t)=te−tsin2t
f(t)=sintt
f(t)=1−cos 2tt
f(t)=e3t−1t
f(t)=et−e−tt
Find the inverse transforms of the functions in Problems 23 through 28.
F(s)=lns−2s+2
F(s)=lns2+1s2+4
F(s)=lns2+1(s+2)(s−3)
F(s)=tan−13s+2
F(s)=ln(1+1s2)
F(s)=s(s2+1)3
In Problems 29 through 34, transform the given differential equation to find a nontrivial solution such that x(0)=0
tx″+(t−2)x′+x=0
tx″+(3t−1)x′+3x=0
tx″−(4t+1)x′+2(2t+1)x=0
tx″+2(t−1)x′−2x=0
tx″−2x′+tx=0
tx″+(4t−2)x′+(13t−4)x=0
Apply the convolution theorem to show that
(Suggestion: Substitute u=√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)=0
x″+4x=f(t); x(t)=12∫t0f(t−τ)sin 2τ dτ
x″+2x′+x=f(t); x(t)=∫t0τe−τf(t−τ) dτ
x″+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,
where 0≦k<1
In Example 5 it was shown that
Expand with the aid of the binomial series and then compute the inverse transformation term by term to obtain
Finally, note that J0(0)=1
Expand the function F(s)=s−1/2e−1/s
Show that
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