Hassani S. Mathematical Methods: For Students of Physics and Related Fields
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29.2 Fourier Transform and Green’s Functions 711
29.2.3 Green’s Function for the Wave Equation
The wave equation, which we write as
1
c
2
∂
2
Ψ
∂t
2
−∇
2
Ψ=0, (29.43)
with c the speed of the wave, is a PDE in 4 variables. As in the case of the
heat equation, we let the fourth variable have 0 as subscript. Then
p(k
j
)=−
k
2
0
c
2
+ k
2
1
+ k
2
2
+ k
2
3
≡−
k
2
0
c
2
+ k
2
,
and the Green’s function can be written as
G(x − x
)=−
1
(2π)
4
#
d
4
k
e
ik
0
(x
0
−x
0
)+ik·(r−r
)
k
2
0
/c
2
− k
2
= −
c
2
(2π)
4
#
d
3
ke
ik·(r−r
)
#
∞
−∞
dk
0
e
ik
0
t
k
2
0
− c
2
k
2
, (29.44)
where we substituted t for x
0
and assumed x
0
= t
=0.
Let us concentrate on the k
0
integration and use the calculus of residues
to calculate it. The integrand has two poles k
0
= ±ck on the real axis, and
depending on how these poles are handled, different Green’s functions are
obtained. One way to handle the poles is to move them up slightly, i.e.,
give them an infinitesimal positive imaginary part. If t>0, the contour of
integration should be in the UHP with zero contribution from the large circle
there. If t<0, the contour of integration should be in the LHP for which
the integral vanishes because there are no poles inside the contour. Thus,
denoting the integrand by f ,wehave
#
∞
−∞
dk
0
e
ik
0
t
k
2
0
− c
2
k
2
=2πi [Res(f (ck)) + Res(f(−ck))] .
But
Res(f(ck)) = lim
k
0
→ck
(
(k
0
− ck)
e
ik
0
t
k
2
0
− c
2
k
2
)
= lim
k
0
→ck
(
e
ik
0
t
k
0
+ ck
)
=
e
ickt
2ck
.
Similarly, Res(f(ck)) = −e
−ickt
/2ck,andthek
0
integral gives
#
∞
−∞
dk
0
e
ik
0
t
k
2
0
− c
2
k
2
=2πi
e
ickt
2ck
−
e
−ickt
2ck
= −2π
sin ckt
ck
.
Substituting this in (29.44) yields
G(x −x
)=
c
(2π)
3
#
d
3
ke
ik·(r−r
)
sin ckt
k
,
712 Integral Transforms
which, through a by-now-familiar routine in k-space spherical integration
yields
G(r −r
; t)=
2c
(2π)
2
|r − r
|
#
∞
0
dk sin(k|r −r
|)sinckt
=
c
(2π)
2
|r − r
|
#
∞
−∞
dk
e
ik|r−r
|
− e
−ik|r−r
|
2i
e
ickt
− e
−ickt
2i
.
Multiply the exponentials and note that
∞
−∞
e
−ix
dx =
∞
−∞
e
ix
dx to obtain
G(r − r
; t)=−
c
4(2π)
2
|r − r
|
2
#
∞
−∞
dk
*
e
ick(t+|r−r
|/c)
− e
ick(t−|r−r
|/c)
+
= −
1
2(2π)
2
|r − r
|
[2πδ(t + |r −r
|/c) − 2πδ(t −|r −r
|/c)] .
(29.45)
The first delta function vanishes because t>0. Therefore, the final form of
Retarded Green’s
function for wave
equation
the Green’s function for the wave equation is
G
ret
(r − r
; t)=
δ(t −|r − r
|/c)
4π|r −r
|
. (29.46)
The subscript “ret” on the Green’s function stands for retarded.Asthe
argument of the delta function implies, G
ret
(r − r
; t) is zero unless t = |r −
r
|/c, i.e., unless the wave has had time to move from the source point r
to the
observation point r. The signal is “retarded” by this amount of time. Had we
given the poles of the k
0
integral of (29.44) an infinitesimal negative imaginary
part and chosen t to be negative, the first delta function of (29.45) would have
survived and we would have obtained the advanced Green’s function:
Advanced Green’s
function for wave
equation
G
adv
(r − r
; t)=
δ(t + |r −r
|/c)
4π|r −r
|
. (29.47)
29.3 The Laplace Transform
In the previous section, the power of the Fourier transform was illustrated by
formalism and application. Fourier transform is by far the most important of
all the transforms used in mathematical analysis. Another transform which is
widely used in solving ordinary differential equations is the Laplace transform,
the subject of this section.
Let f(t) be a sufficiently well-behaved function. The Laplace transform
of f is another function L[f]whosevalueats is given by
Laplace transform
defined
L[f](s)=
#
∞
0
e
−st
f(t) dt. (29.48)
29.3 The Laplace Transform 713
s could be complex, although it is usually taken to be real. To assure the
convergence of the integral, s must have a positive real part. The left-hand
side of (29.48) is usually denoted by F (s). It is also common to write it (less
precisely) as L[f(t)] with the letter s understood!
Example 29.3.1.
The Laplace transform of the unit function—the function whose
valueeverywhereis1—evaluatedats can easily be shown to be 1/s. The Laplace
transform of e
iωt
can be readily calculated as well:
F (s)=
#
∞
0
e
−st
e
iωt
dt =
#
∞
0
e
(−s+iω)t
dt =
e
(−s+iω)t
−s + iω
∞
0
=
1
s − iω
.
The Laplace transforms of sin ωt and cos ωt can now be evaluated:
L[cos ωt]=Re
L[e
iωt
]
=Re
1
s − iω
=Re
s + iω
s
2
+ ω
2
=
s
s
2
+ ω
2
, (29.49)
and
L[sin ωt]=Im
L[e
iωt
]
=Im
1
s − iω
=Im
s + iω
s
2
+ ω
2
=
ω
s
2
+ ω
2
. (29.50)
The Laplace transform of the step function θ(t −a) is very useful in applications
[see Section 5.1.3 for the definition of the step function].
L[θ(t −a)] =
#
∞
0
e
−st
θ(t − a) dt =
#
∞
a
e
−st
dt =
e
−as
s
.
The lower limit of integration was changed because θ(t −a)iszerofort<a(and it
isequalto1fort>a).
Knowing L[1], we can find the Laplace transform of any power of t because
L[t
n
]=
#
∞
0
t
n
e
−st
dt =(−1)
n
d
n
ds
n
#
∞
0
e
−st
dt.
Since L[1](s)=1/s,wehave
L[t
n
]=(−1)
n
d
n
ds
n
1
s
=
n!
s
n+1
. (29.51)
What if n in the above equation is not an integer? Let’s evaluate L[t
ν
]directly.
L[t
ν
]=
#
∞
0
t
ν
e
−st
dt =
#
∞
0
u
s
ν
e
−u
1
s
du =
1
s
ν+1
#
∞
0
u
ν
e
−u
du =
Γ(ν +1)
s
ν+1
,
(29.52)
where Γ is the gamma function introduced in Section 11.1.1. Note that if ν = n,we
regain (29.51) because Γ(n +1)=n!.
29.3.1 Properties of Laplace Transform
In a typical application, one obtains the Laplace transform of a function from,
say a differential equation, and inverts it to find the actual function. This is
what was done in the case of the Fourier transform, and indeed in any other
transform used. While the formula for inverting a Fourier transform [see
714 Integral Transforms
Equation (29.7)] is nice and symmetric, that for the Laplace transform is not
as nice. Furthermore, Fourier transform adapts itself very naturally to partial
differential equations as demonstrated in the previous section. However, the
adaptation of Laplace transform to PDEs is not so natural. That is why the
Fourier transform techniques are much more powerful—both formally and for
calculations—than the Laplace transform.
Because of this drawback, one has to rely on some formal properties of the
Laplace transform and its inverse—as well as a lot of examples—to be able to
reconstruct the original function.
Linearity
One such property is the linearity of the Laplace transform and it inverse:
L[af + bg]=aL[f]+bL[g], L
−1
[af + bg]=aL
−1
[f]+bL
−1
[g]. (29.53)
First shift property
Another is the first shift property.IfF (s) is the Laplace transform of f(t),
then F(s −a) is the Laplace transform of e
at
f(t). This can easily be verified:
F (s −a)=
#
∞
0
e
−(s−a)t
f(t) dt =
#
∞
0
e
−st
e
at
f(t)
!
dt = L[e
at
f(t)]. (29.54)
A more useful way of writing this equation is
L
−1
[F (s − a)] = e
at
L
−1
[F (s)]. (29.55)
Second shift property
The second shift property involves the step function:
L[θ(t − a)f(t −a)] = e
−as
L[f](s). (29.56)
This is because
L[θ(t−a)f(t−a)] =
#
∞
a
e
−st
f(t−a) dt =
#
∞
0
e
−s(τ+a)
f(τ) dτ = e
−as
L[f](s),
where in the second equality we changed the variable of integration to τ = t−a.
Denoting by F (s) the Laplace transform of f(t), we write (29.56) as
L
−1
[e
−as
F (s)] = θ(t −a)f(t −a)=
0
f(t − a)ifa>0
0ifa<0.
(29.57)
Example 29.3.2.
Since L[t
n
]=n!/s
n+1
, using the first shift property, we get
L[t
n
e
at
]=
n!
(s − a)
n+1
. (29.58)
29.3 The Laplace Transform 715
In particular, if n =0,wehaveL[e
at
]=1/(s − a). From this, and the linearity
property, we can find the Laplace transforms of the hyperbolic sine:
L[sinh γt]=L[
1
2
e
γt
− e
−γt
!
]=
1
2
L[e
γt
] − L[e
−γt
]
!
=
1
2
1
s − γ
−
1
s + γ
=
γ
s
2
− γ
2
(29.59)
and hyperbolic cosine:
L[cosh γt]=L[
1
2
e
γt
+ e
−γt
!
]=
1
2
L[e
γt
]+L[e
−γt
]
!
=
1
2
1
s − γ
+
1
s + γ
=
s
s
2
− γ
2
. (29.60)
With our accumulated knowledge of the Laplace transform, let’s see if we can find
the inverse transform of 1/(s
2
+2as + b
2
). Complete the square in the denominator
and consider three cases: b>a, b<a,andb = 0. First assume b>aand define
ω
2
= b
2
− a
2
.Then
L
−1
1
(s + a)
2
+ b
2
− a
2
=
1
ω
L
−1
ω
(s + a)
2
+ ω
2
=
e
−at
ω
L
−1
ω
s
2
+ ω
2
=
e
−at
ω
sin ωt
where we used (29.55) and (29.50). Substituting for ω,weget
L
−1
1
s
2
+2as + b
2
=
e
−at
√
b
2
− a
2
sin(
b
2
− a
2
t),a<b.
For b<adefine γ
2
= a
2
− b
2
.Then
L
−1
1
(s + a)
2
+ b
2
− a
2
=
1
γ
L
−1
γ
(s + a)
2
− γ
2
=
e
−at
γ
L
−1
γ
s
2
− γ
2
=
e
−at
γ
sinh γt
where we used (29.55) and (29.59). Substituting for γ,weget
L
−1
1
s
2
+2as + b
2
=
e
−at
√
a
2
− b
2
sinh(
a
2
− b
2
t),a>b.
If b = a, then the denominator is a complete square and
L
−1
1
(s + a)
2
= e
−at
t
by (29.58).
Similarly, we can show that
L
−1
s
s
2
+2as + b
2
=e
−at
cos(
b
2
− a
2
t)
−
a
√
b
2
− a
2
e
−at
sin(
b
2
− a
2
t),b > a
L
−1
s
s
2
+2as + b
2
=e
−at
cosh(
a
2
− b
2
t)
−
a
√
a
2
− b
2
e
−at
sinh(
a
2
− b
2
t),b<a
We shall use the formulas derived in this example in solving differential
equations.
716 Integral Transforms
Periodic functions
Although Fourier series are better suited for periodic functions, Laplace trans-
form of periodic functions is also of interest. If f (t) is periodic of period T ,
i.e., if f(t + T )=f(t), then
L[f(t)] =
#
T
0
e
−st
f(t) dt
call this F
1
(s)
+
#
∞
T
e
−st
f(t) dt = F
1
(s)+
#
∞
0
e
−s(u+T )
f(u + T ) du
= F
1
(s)+e
−sT
#
∞
0
e
−su
f(u) du = F
1
(s)+e
−sT
=L[f (t)]
#
∞
0
e
−st
f(t) dt .
We thus have L[f(t)] = F
1
(s)+e
−sT
L[f(t)], which upon solving for L[f(t)]
yields
L[f(t)] =
1
1 − e
−sT
F
1
(s). (29.61)
Example 29.3.3.
The Laplace transform of the square wave function of Example
10.6.1 defined by
V (t)=
0
V
0
if 0 ≤ t ≤ T,
0ifT<t≤ 2T,
can be readily found. We simply note that the period is 2T and F
1
(s)is
F
1
(s)=
#
2T
0
e
−st
V (t) dt = V
0
#
2T
T
e
−st
dt =
V
0
s
(e
−sT
− e
−2sT
).
Substituting this in (29.61), we obtain
L[V (t)] =
1
1 − e
−2sT
V
0
s
(e
−sT
− e
−2sT
)
=
V
0
e
−sT
(1 − e
−sT
)
s(1 − e
−sT
)(1 + e
−sT
)
=
V
0
e
−sT
s(1 + e
−sT
)
=
V
0
s(1 + e
sT
)
.
Convolution
The convolution of two functions is defined as
(f ∗ g)(t)=
#
t
0
f(u)g(t − u)du.
Let v = t −u and change the variable of integration to v.Then
(f ∗g)(t)=
#
0
t
f(t − v)g(v)(−dv)=
#
t
0
g(v)f(t −v)dv =(g ∗f)(t),
showing that convolution is commutative. Commutativity is only one of the
following properties of convolution:
29.3 The Laplace Transform 717
1. c(f ∗g)=cf ∗ g = f ∗ cg, c a constant;
2. f ∗g = g ∗ f (commutative property);
3. f ∗(g ∗ h)=(f ∗g) ∗h (associative property);
4. f ∗(g + h)=f ∗ g + f ∗ h (distributive property).
The notion of convolution is useful for Laplace transform because one can
show the following:
Box 29.3.1. The Laplace transform of the convolution of two functions
is the product of the Laplace transforms of the two functions:
L[f ∗ g](s)=L[f ](s) ·L[g](s)or(f ∗g)(t)=L
−1
.
L[f](s) · L[g](s)
/
Example 29.3.4. Suppose we want to find the inverse transform of s/(s
2
+ a
2
)
2
.
We can write this as [see Equations (29.49) and (29.50)]
s
(s
2
+ a
2
)
2
=
1
a
s
s
2
+ a
2
·
a
s
2
+ a
2
=
1
a
L[cos at] · L[sin at].
Therefore,
L
−1
s
(s
2
+ a
2
)
2
=
1
a
cos at ∗ sin at =
1
a
#
t
0
cos au sin a(t − u)du =
1
2a
t sin at.
Similarly
L
−1
s
2
(s
2
+ a
2
)
2
= L
−1
.
L[cos at] · L[cos at]
/
=cosat ∗ cos at
=
#
t
0
cos au cos a(t − u)du =
1
2
t cos at +
1
2a
sin at.
29.3.2 Derivative and Integral of the Laplace Transform
By differentiating the integral of a Laplace transform, one can easily obtain
the formulas
d
n
ds
n
F (s)=L[(−1)
n
t
n
f(t)] or L
−1
*
F
(n)
(s)
+
=(−1)
n
t
n
L
−1
[F (s)],
(29.62)
where F
(n)
denotes the nth derivative of F .Forn = 1, this formula leads to
the following useful relation:
L
−1
[F (s)] = −
1
t
L
−1
[F
(s)] , (29.63)
because sometimes it is easier to find the inverse Laplace transform of the
derivative of a function than the function itself.
718 Integral Transforms
Example 29.3.5. It is not easy to find the inverse transform of F (s) = ln[(s +
a)/(s + b)] directly. But the inverse transform of
F
(s)=
d
ds
[ln(s + a) − ln(s + b)] =
1
s + a
−
1
s + b
is much easier to find. In fact,
L
−1
.
F
(s)
/
= L
−1
1
s + a
− L
−1
1
s + b
= e
−at
− e
−bt
,
by (29.58) with n = 0. Therefore, according to (29.63)
L
−1
ln
s + a
s + b
=
e
−bt
− e
−at
t
.
One can also find the primitive (antiderivative, indefinite integral) of F (s).
Recall that the indefinite integral of a function can be written as a definite
integral with one of its limits being a variable [see Equation (3.18)]. Therefore,
let’s write the indefinite integral of F (s)as−
∞
s
F (u) du.Thisintegralcan
be easily evaluated:
#
∞
s
F (u) du =
#
∞
s
du
#
∞
0
e
−ut
f(t) dt =
#
∞
0
f(t) dt
#
∞
s
e
−ut
du
=
#
∞
0
f(t) dt
−
e
−ut
t
u=∞
u=s
=
#
∞
0
e
−st
f(t)
t
dt.
This can be written as
#
∞
s
L[f](u) du = L
f(t)
t
. (29.64)
Example 29.3.6.
Let’s use (29.64) to find the Laplace transform of sin ωt/t.From
(29.50), we have
L
sin ωt
t
=
#
∞
s
ω
u
2
+ ω
2
du =tan
−1
u
ω
∞
s
=
π
2
− tan
−1
s
ω
=tan
−1
ω
s
(see Problem 29.17 for the last equality). Similarly,
L
sinh γt
t
=
#
∞
s
γ
u
2
− γ
2
du =
1
2
lim
x→∞
#
x
s
1
u − γ
−
1
u + γ
du
=
1
2
lim
x→∞
ln
x − γ
x + γ
− ln
s − γ
s + γ
=
1
2
ln
s + γ
s − γ
.
29.3.3 Laplace Transform and Differential Equations
Certain differential equations with appropriate boundary conditions or initial
values can be nicely solved by Laplace transform techniques. For the appli-
cation of Laplace transform to differential equations, we need to know the
transform of the derivative of a function. Using integration by parts, we have
#
∞
0
e
−st
f
(t) dt = e
−st
f(t)
∞
0
+ s
#
∞
0
e
−st
f(t) dt = −f(0) + sL[f](s).
29.3 The Laplace Transform 719
Therefore,
L[f
](s)=sL[f](s) −f(0). (29.65)
This can be iterated to give
L[f
](s)=sL[f
](s) − f
(0) = s[sL[f](s) −f(0)] −f
(0),
or
L[f
](s)=s
2
L[f](s) − sf(0) − f
(0). (29.66)
We can continue iterating the formula, but since most differential equations
encountered in applications are of second order, we stop at the second deriva-
tive.
To solve a differential equation, take the Laplace transform of both sides
and use (29.65) and (29.66). Solve for L[f](s) and take the inverse transform
to find the solution. Let’s look at a specific example. Consider a mass m
attached to a spring of spring constant k. The differential equation of motion
of this system is
m¨x + kx =0 or ¨x + ω
2
0
x =0,ω
0
=
k
m
.
Taking the Laplace transform of both sides gives
L[¨x](s)+ω
2
0
L[x](s)=0.
Using (29.66), this becomes
s
2
L[x](s) − sx(0) − ˙x(0) + ω
2
0
L[x](s)=0,
or, letting x
0
= x(0) and ˙x
0
=˙x(0), we get
(s
2
+ ω
2
0
)L[x](s)=sx
0
+˙x
0
or L[x](s)=
x
0
s +˙x
0
s
2
+ ω
2
0
,
and from (29.49) and (29.50) we obtain
x(t)=x
0
L
−1
s
s
2
+ ω
2
0
+
˙x
0
ω
0
L
−1
ω
0
s
2
+ ω
2
0
= x
0
cos ω
0
t +
˙x
0
ω
0
sin ω
0
t.
Note how the initial values are automatically included in the solution.
A more general problem has a damping term as well as a driving force.
This leads to a differential equation of the form
¨x + γ ˙x + ω
2
0
x = f(t),ω
0
=
k
m
. (29.67)
To solve this, once again take the Laplace transform of both sides:
L[¨x](s)+γL[˙x](s)+ω
2
0
L[x](s)=L[f](s),
720 Integral Transforms
and use (29.65) and (29.66) to get
s
2
L[x](s) − x
0
s − ˙x
0
+ γ{sL[x](s) − x
0
}+ ω
2
0
L[x](s)=L[f](s),
where x
0
= x(0) and ˙x
0
=˙x(0). Therefore,
(s
2
+ γs + ω
2
0
)L[x](s)=L[f](s)+x
0
s +˙x
0
+ γx
0
,
which yields
L[x](s)=
L[f](s)+x
0
s +˙x
0
+ γx
0
s
2
+ γs + ω
2
0
.
The solution can be obtained by inversion once we know L[f ](s). Symbolically,
we write
x(t)=L
−1
L[f](s)
s
2
+ γs + ω
2
0
+ x
0
L
−1
s
s
2
+ γs + ω
2
0
+(˙x
0
+ γx
0
)L
−1
1
s
2
+ γs + ω
2
0
. (29.68)
We consider only the case of a damped harmonic oscillator, i.e., that
ω
0
>γ/2. The second and third inversions are given in Example 29.3.2
with a = γ/2andb = ω
0
. Then, with Ω ≡
ω
2
0
− (γ/2)
2
,wehave
L
−1
s
s
2
+ γs + ω
2
0
= e
−γt/2
cos Ωt −
γ
2Ω
e
−γt/2
sin Ωt,
L
−1
1
s
2
+ γs + ω
2
0
=
e
−γt/2
Ω
sin Ωt. (29.69)
Substituting these in (29.68), we obtain
x(t)=e
−γt/2
x
0
cos Ωt +
˙x
0
+ x
0
γ/2
Ω
sin Ωt
+ L
−1
L[f](s)
s
2
+ γs + ω
2
0
.
(29.70)
Let us denote the last term of this equation by Φ(t) and evaluate the equation
at t = 0 to obtain x(0) = x
0
+ Φ(0) implying that Φ(0) = 0. Similarly,
differentiating the equation and evaluating the result at t =0yields ˙x(0) =
˙x
0
+
˙
Φ(0) implying that
˙
Φ(0) = 0. This is an interesting result since f(t)
is quite arbitrary! The following example looks at a specific instance of this
result.
Example 29.3.7.
As an example of the general formula (29.70), let’s consider a
damped harmonic oscillator driven by a sinusoidal source f(t)=A sin ω
0
t operating
at the natural frequency of the oscillator as given in (29.67). Then by (29.50)
L[f](s)=L[A sin ω
0
t]=
Aω
0
s
2
+ ω
2
0
,
and the last term of (29.70) becomes
L
−1
Aω
0
(s
2
+ ω
2
0
)(s
2
+ γs + ω
2
0
)
.