Powers D.L. Boundary Value Problems and Partial Differential Equations
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368 Chapter 6 Laplace Transform
b.
f (t) =
0, 0 < t < a,
1, a < t < b,
0, b < t;
c. f (t) =
t, 0 < t < a,
a, a < t.
4. The Heaviside step function is defined by the formula
H
a
(t) =
1, t > a,
0, t < a.
Assuming a ≥ 0, show that the Laplace transform of H
a
is
L
H
a
(t)
=
e
−as
s
.
5. Use completion of square and the shifting theorem to find the inverse trans-
form of
a.
1
s
2
+2s
,
b.
s +1
s
2
+2s + 2
,
c.
1
s
2
+2as + b
2
, b > a.
6. Find the Laplace transform of the square-wave function
f (t) =
1, 0 < x < a,
0, a < x < 2a,
f (x +2a) = f (x).
Hint: Break up the integral as shown in the following, evaluate the integrals,
and add up a geometric series:
F(s) =
∞
n=0
2(n+1)a
2na
f (t)e
−st
dt.
7. Use any method to find the inverse transform of the following.
a.
1
(s −a)(s −b)
;
c.
s
2
(s
2
+ω
2
)
2
;
b.
s
(s
2
−a
2
)
2
;
d.
1
(s −a)
3
;
e.
1 −e
−s
s
.
8. Use any theorem or formula to find the transform of the following.
a.
1 −cos(ωt)
t
;
c. t
2
e
−at
;
b.
t
0
sin(at
)
t
dt
;
d. t cos(ωt);
e. sinh(at) sin(ωt).
9. Find the inverse transform of these functions of s by any method.
6.2 Partial Fractions and Convolutions 369
a.
1
(s
2
+ω
2
)
2
;
c.
s
2
(s
2
+ω
2
)
2
;
b.
s
(s
2
+ω
2
)
2
;
d.
s
3
(s
2
+ω
2
)
2
.
6.2 Partial Fractions and Convolutions
Because of the formula for the transform of derivatives, the Laplace transform
finds important application to linear differential equations with constant co-
efficients, subject to initial conditions. In order to solve the simple problem
u
+au = 0, u(0) = 1,
we transform the entire equation, obtaining
L(u
) +aL(u) = 0
or
sU −1 +aU =0,
where U =
L(u). The derivative has been “transformed out,” and U is deter-
mined by simple algebra to be
U(s) =
1
s +a
.
By consulting Table 2 we find that u(t) = e
−at
.
Equationsofhigherordercanbesolvedinthesameway.Whentrans-
formed, the problem
u
+ω
2
u =0, u(0) =1, u
(0) =0
becomes
s
2
U −s ·1 −0 +ω
2
U = 0.
Note how both initial conditions have been incorporated into this one equa-
tion. Now we solve the transformed equation algebraically to find
U(s) =
s
s
2
+ω
2
,
the transform of cos(ωt) = u(t).
370 Chapter 6 Laplace Transform
In general we may outline our procedure as follows:
Original
problem
L
Solution of
original problem
Transformed
problem
Solution of
transformed
problem
L
−1
In the step marked L
−1
, we must compute the function of t to which the solu-
tion of the transformed problem corresponds. This is the difficult part of the
process. The key property of the inverse transform is its linearity, as expressed
by
L
−1
c
1
F
1
(s) +c
2
F
2
(s)
=c
1
L
−1
F
1
(s)
+c
2
L
−1
F
2
(s)
.
This property allows us to break down a complicated transform into a sum of
simple ones.
A simple mass–spring–damper system leads to the initial value problem
u
+au
+ω
2
u =0, u(0) = u
0
, u
(0) =u
1
,
whose transform is
s
2
U −su
0
−u
1
+a(sU −u
0
) +ω
2
U = 0.
Determination of U gives it as the ratio of two polynomials:
U(s) =
su
0
+(u
1
+au
0
)
s
2
+as + ω
2
.
Although this expression is not in Table 2, it can be worked around to a func-
tion of s + a/2, whose inverse transform is available. The shift theorem then
gives u(t).Thereis,however,abetterway.
The inversion of a rational function of s (that is, the ratio of two polynomi-
als) can be accomplished by the technique of “partial fractions.” Suppose we
wishtocomputetheinversetransformof
U(s) =
cs +d
s
2
+as + b
.
The denominator has two roots, r
1
and r
2
, which we assume for the moment
to be distinct. Thus
s
2
+as + b = (s −r
1
)(s −r
2
),
U(s) =
cs +d
(s −r
1
)(s −r
2
)
.
6.2 Partial Fractions and Convolutions 371
The rules of elementary algebra suggest that U can be written as a sum,
cs +d
(s −r
1
)(s −r
2
)
=
A
1
s −r
1
+
A
2
s −r
2
, (1)
for some choice of A
1
and A
2
. Indeed, by finding the common denominator
form for the right-hand side and matching powers of s in the numerator, we
obtain
cs +d
(s −r
1
)(s −r
2
)
=
A
1
(s −r
2
) +A
2
(s −r
1
)
(s −r
1
)(s −r
2
)
,
c = A
1
+A
2
, d =−A
1
r
2
−A
2
r
1
.
When A
1
and A
2
are determined, the inverse transform of the right-hand side
of Eq. (1) is easily found:
L
−1
A
1
s −r
1
+
A
2
s −r
2
=A
1
exp(r
1
t) +A
2
exp(r
2
t).
For a specific example, suppose that
U(s) =
s +4
s
2
+3s + 2
.
The roots of the denominator are r
1
=−1andr
2
=−2. Thus
s +4
s
2
+3s + 2
=
A
1
s +1
+
A
2
s +2
=
(A
1
+A
2
)s +(2A
1
+A
2
)
(s +1)(s +2)
.
We fi nd A
1
=3, A
2
=−2. Hence
L
−1
s +4
s
2
+3s + 2
=
L
−1
3
s +1
−
2
s +2
=3e
−t
−2e
−2t
.
A little calculus takes us much further. Suppose that U has the form
U(s) =
q(s)
p(s)
,
where p and q are polynomials and the degree of q is less than the degree of p.
Assume that p has distinct roots r
1
,...,r
k
:
p(s) =(s −r
1
)(s −r
2
) ···(s −r
k
).
We t r y to w ri te U in the fraction form
U(s) =
A
1
s −r
1
+
A
2
s −r
2
+···+
A
k
s −r
k
=
q(s)
p(s)
.
372 Chapter 6 Laplace Transform
The algebraic determination of the A’s is very tedious, but notice that
(s −r
1
)q(s)
p(s)
=A
1
+A
2
s −r
1
s −r
2
+···+A
k
s −r
1
s −r
k
.
If s is set equal to r
1
, the right-hand side is just A
1
. The left-hand side becomes
0/0, but L’Hôpital’s rule gives
lim
s→r
1
(s −r
1
)q(s)
p(s)
= lim
s→r
1
(s −r
1
)q
(s) +q(s)
p
(s)
=
q(r
1
)
p
(r
1
)
.
Therefore A
1
and all the other A’s are given by
A
i
=
q(r
i
)
p
(r
i
)
.
Consequently, our rational function takes the form
q(s)
p(s)
=
q(r
1
)
p
(r
1
)
1
s −r
1
+···+
q(r
k
)
p
(r
k
)
1
s −r
k
.
From this point we can easily obtain the inverse transform, as expressed in the
conclusion of the following theorem.
Theorem 1. Let p and q be polynomials, q of lower degree than p, and let p have
only simple roots, r
1
,r
2
,...,r
k
.Then
L
−1
q(s)
p(s)
=
q(r
1
)
p
(r
1
)
exp(r
1
t) +···+
q(r
k
)
p
(r
k
)
exp(r
k
t). (2)
(Equation (2) is known as Heaviside’s formula.)
Let us apply the theorem to the example in which q(s) = s + 4, p(s) = s
2
+
3s +2, p
(s) =2s +3. Then
L
−1
s +4
s
2
+3s + 2
=
−1 +4
2(−1) +3
e
−t
+
−2 +4
2(−2) +3
e
−2t
=3e
−t
−2e
−2t
.
In nonhomogeneous differential equations also, the Laplace transform is a
useful tool. To solve the problem
u
+au = f (t), u(0) = u
0
,
we again transform the entire equation, obtaining
sU −u
0
+aU = F(s),
U(s) =
u
0
s +a
+
1
s +a
F(s).
6.2 Partial Fractions and Convolutions 373
The first term in this expression is recognized as the transform of u
0
e
−at
.If
F(s) is a rational function, partial fractions may be used to invert the second
term. However, we can identify that term by solving the problem another way.
For example,
e
at
(u
+au) = e
at
f (t),
ue
at
= e
at
f (t),
ue
at
=
t
0
e
at
f
t
dt
+c,
u(t) =
t
0
e
−a(t−t
)
f
t
dt
+ce
−at
.
The initial condition requires that c = u
0
. On comparing the two results, we
see that
L
t
0
e
−a(t−t
)
f
t
dt
=
1
s +a
F(s).
Thus the transform of the combination of e
−at
and f (t) on the left is the prod-
uct of the transforms of e
−at
and f (t). This simple result can be generalized in
the following way.
Theorem 2. If g(t) and f (t) have Laplace transforms G(s) and F(s),respectively,
then
L
t
0
g
t −t
f
t
dt
=G(s)F(s). (3)
(This is known as the convolution theorem.)
The integral on the left is called the convolution of g and f , written
g(t) ∗ f (t) =
t
0
g
t −t
f
t
dt
.
It can be shown that the convolution follows these rules:
g ∗f = f ∗ g, (4a)
f ∗( g ∗ h) = ( f ∗g) ∗h, (4b)
f ∗( g +h) = f ∗g + f ∗ h. (4c)
The convolution theorem provides an important device for inverting
Laplace transforms, which we shall apply to find the general solution of the
nonhomogeneous problem
u
−au = f (t), u(0) = u
0
, u
(0) =u
1
.
374 Chapter 6 Laplace Transform
The transformed equation is readily solved, yielding
U(s) =
su
0
+u
1
s
2
−a
+
1
s
2
−a
F(s).
Because 1/(s
2
−a) is the transform of sinh(
√
at)/
√
a, we easily determine that
u is
u(t) = u
0
cosh
√
at
+
u
1
√
a
sinh
√
at
+
t
0
sinh(
√
a(t −t
))
√
a
f
t
dt
. (5)
A slightly different problem occurs if the mass in a spring–mass system is
struck while the system is in motion. The mathematical model of the system
might be
u
+ω
2
u =f (t), u(0) = u
0
, u
(0) =u
1
,
where f (t) = F
0
for t
0
< t < t
1
and f (t) = 0 for other values. The transform of
u is
U(s) =
su
0
+u
1
s
2
+ω
2
+
1
s
2
+ω
2
F(s).
The inverse transform of U(s) is then
u(t) = u
0
cos(ωt) +u
1
sin(ωt)
ω
+
t
0
sin(ω(t −t
))
ω
f
t
dt
.
The convolution in this case is easy to calculate:
t
0
sin(ω(t −t
))
ω
f
t
dt
=
0, t < t
0
,
F
0
1 −cos(ω(t −t
0
))
ω
2
, t
0
< t < t
1
,
F
0
cos(ω(t − t
1
)) −cos(ω(t −t
0
))
ω
2
, t
1
< t.
EXERCISES
1. Solve these initial value problems.
a. u
−2u = 0, u(0) = 1;
b. u
+2u = 0, u(0) = 1;
c. u
+4u
+3u = 0, u(0) = 1, u
(0) =0;
d. u
+9u = 0, u(0) = 0, u
(0) =1.
6.2 Partial Fractions and Convolutions 375
2. Solve the initial value problem
u
+2au
+u = 0, u(0) = u
0
, u
(0) =u
1
in these three cases: 0 < a < 1, a = 1, a > 1.
3. Solve these nonhomogeneous problems with zero initial conditions.
a. u
+au = 1;
c. u
+4u = sin(t);
e. u
+2u
=1 −e
−t
;
b. u
+u = t;
d. u
+4u = sin(2t);
f. u
−u = 1.
4. Complete the square in the denominator and use the shift theorem
[F(s −a) =
L(e
at
f (t))]toinvert
U(s) =
su
0
+(u
1
+2au
0
)
s
2
+2as + ω
2
.
There are three cases, corresponding to
ω
2
−a
2
> 0, =0,<0.
5. Use partial fractions to invert the following transforms.
a.
1
s
2
−4
;
c.
(s +3)
s(s
2
+2)
;
b.
1
s
2
+4
;
d.
4
s(s +1)
.
6. Prove properties (4a) and (4c) of the convolution.
7. Compute the convolution f ∗g for
a. f (t) = 1, g(t) = sin(t);
b. f (t) = e
t
, g(t) = cos(ωt);
c. f (t) = t, g(t) =sin(t).
8. Demonstrate the following properties of convolution either directly or by
using Laplace transform.
a. 1 ∗ f
(t) = f (t) −f (0);
b. (t ∗ f (t))
=f (t);
c. ( f ∗g)
=f
∗g =f ∗g
,iff (0) =g(0) = 0.
376 Chapter 6 Laplace Transform
6.3 Partial Differential Equations
In applying the Laplace transform to partial differential equations, we treat
variables other than t as parameters. Thus, the transform of a function u(x, t)
is defined by
L
u(x, t)
=
∞
0
e
−st
u(x, t) dt = U(x, s).
For instance, we easily find the transforms
L
e
−at
sin(πx)
=
1
s +a
sin(πx),
L
sin(x +t)
=
s sin(x) +cos(x)
s
2
+1
.
The transform U naturally is a function not only of s but also of the “untrans-
formed” variable x. We assume that derivatives or integrals with respect to the
untransformed variable pass through the transform
L
∂u
∂x
=
∞
0
∂u(x, t)
∂x
e
−st
dt
=
∂
∂x
∞
0
u(x, t)e
−st
dt =
∂
∂x
U(x, s)
.
Ifwewishtofocusontheroleofx as a variable and keep s in the background
as a parameter, we might use the symbol for the ordinary derivative:
L
∂u
∂x
=
dU
dx
.
The rule for transforming a derivative with respect to t can be found, as
before, with integration by parts:
L
∂u
∂t
=s
L
u(x, t)
−u(x, 0).
If the Laplace transform is applied to a boundary value–initial value prob-
lem in x and t, all time derivatives disappear, leaving an ordinary differential
equation in x. We shall illustrate this technique with some trivial examples.
Incidentally,weassumefromhereonthatproblemshavebeenprepared(for
example, by dimensional analysis) so as to eliminate as many parameters as
possible.
6.3 Partial Differential Equations 377
Example 1.
∂
2
u
∂x
2
=
∂u
∂t
, 0 < x < 1, 0 < t,
u(0, t) = 1, u(1, t) = 1, 0 < t,
u(x, 0) = 1 +sin(π x), 0 < x < 1.
The partial differential equation and the boundary conditions (that is, every-
thing that is valid for t > 0) are transformed, while the initial condition is
incorporated by the transform
d
2
U
dx
2
=sU −
1 +sin(π x )
, 0 < x < 1,
U(0, s) =
1
s
, U(1, s) =
1
s
.
This boundary value problem is solved to obtain
U(x, s) =
1
s
+
sin(πx)
s +π
2
.
We dire ct our at tent ion now to U as a function of s. Because sin(π x) is a con-
stant with respect to s,tablesmaybeusedtofind
u(x, t) = 1 + sin(πx) exp
−π
2
t
.
Example 2.
∂
2
u
∂x
2
=
∂
2
u
∂t
2
, 0 < x < 1, 0 < t,
u(0, t) = 0, u(1, t) = 0, 0 < t,
u(x, 0) = sin(π x), 0 < x < 1,
∂u
∂t
(x, 0) =−sin(π x), 0 < x < 1.
Under transformation the problem becomes
d
2
U
dx
2
=s
2
U −s sin(πx) +sin(π x), 0 < x < 1,
U(0, s) = 0, U(1, s) = 0.
The function U is found to be
U(x, s) =
s −1
s
2
+π
2
sin(πx),