Download COMPLEX NUMBERS, UNDETERMINED

COMPLEX NUMBERS, UNDETERMINED COEFFICIENTS, AND
LAPLACE TRANSFORMS
BORIS HASSELBLATT
C ONTENTS
1. Complex numbers, Euler’s formula
2. Homogeneous differential equations
3. Undetermined coefficients
4. Laplace transforms
5. Partial fractions
References
1
6
8
10
11
16
The purpose of these notes is to introduce complex numbers and their
use in solving ordinary differential equations. The main feature is that
trigonometric functions can be omitted from the methods even when
they arise in a given problem or its solution.
One one hand this approach is illustrated with the method of undetermined coefficients, where this approach offers a parallel development to
the one in the text [1]. On the other hand, this is done with the Laplace
transform, where the use of complex numbers replaces a great deal of
machinery needed otherwise and at the same time covers situations that
the methods in the text do not. The main effort here is to show how to
produce partial-fractions decompositions using complex numbers.
The reason the method of undetermined coefficients is revisited here
in the complex context is that fluency with this method is very helpful in
using the Laplace transform method reliably.
The exercises included here are not for credit but will help you read
this text actively.
1. C OMPLEX NUMBERS , E ULER ’ S FORMULA
We introduce the symbol i with the property
i 2 = −1
A complex number is an expression that can be written in the form a +i b
with real numbers a and b.
1
2
BORIS HASSELBLATT
1.1. Real and imaginary part, complex conjugate. If a and b are real
numbers, then a is called the real part of a + i b, and b is called the imginary part. (Note that the imaginary part is a real number!)
The expression a − i b is called the complex conjugate of a + i b. It is
sometimes denoted by a bar:
a + i b = a − i b.
Adding a complex number and its complex conjugate always gives a real
number:
a + i b + a − i b = 2a.
This is twice the real part. So, if we are given a complex number z = a +i b
in any form, we can express the real part as
ℜ(z) = real part of z =
z +z
.
2
The imaginary part can be expressed as (check!)
ℑ(z) = imaginary part of z =
z −z
.
2i
1. Exercise. Verify that z = z.
2. Exercise. Verify that any real number x satisfies x = x.
3. Exercise. Verify that a complex number z satisfying z = z is a real number.
1.2. Multiplying complex numbers. To multiply two complex numbers
just use i 2 = −1 and group terms:
(a + i b)(c + i d ) = ac + ai d + i bc + i bi d = ac − bd + i (ad + bc).
Multiplying a complex number and its complex conjugate always gives a
real number:
p
(a + i b)(a − i b) = a 2 + b 2 .
a 2 + b 2 the absolute value or modulus of a + i b:
p
|a + i b| = a 2 + b 2
p
4. Exercise. Verify that |z| = zz.
We call
5. Exercise. Verify that z 1 z 2 = z 1 z 2 .
6. Exercise. Verify that z 1 z 2 = z 1 z 2 .
COMPLEX NUMBERS, UNDETERMINED COEFFICIENTS, AND LAPLACE TRANSFORMS 3
1.3. Dividing complex numbers. To divide two complex numbers and
write the result as real part plus i ×imaginary part, multiply top and bottom of this fraction by the complex conjugate of the denominator:
a + i b a + i b c − i d (a + i b)(c − i d ) ac + bd + i (bc − ad )
=
×
=
=
.
c +id c +id c −id
c2 + d2
c2 + d2
1.4. Factoring polynomials. Factoring polynomials is no harder (or easier) when complex numbers are allowed but in this case all factors are
linear. The reason is that factors x − α are now legal even when α is complex.
7. Example. The polynomial s 2 + 1 is irreducible over the real numbers,
but we have
s 2 + 1 = (s − i )(s + i ).
1.5. Euler’s formula. Complex numbers are useful in our context because
they give Euler’s formula
e i θ = cos θ + i sin θ.
This formula is easy to remember. In case you are not sure whether to
attach the i to the cos or to the sin, just plug in θ = 0.
It’s worthwhile recalling some power series to see why this is so. We
have
∞ zn
X
1
e z = 1 + z + z 2 /2 + . . . =
n!
n=0
cos z = 1 − z 2 /2 + . . .
=
∞ (−1)n z 2n
X
(2n)!
n=0
sin z = z − z 3 /6 + . . .
=
∞ (−1)n z 2n+1
X
n=0 (2n + 1)!
With z = i θ this gives
e
iθ
∞ (i θ)n
∞ i n θn
∞ i 2n θ 2n
∞ i 2n+1 θ 2n+1
X
X
X
X
=
=
=
+
n=0 n!
n=0 n!
n=0 (2n)!
n=0 (2n + 1)!
Here we have simply summed even and odd powers separately. The reason we did this is that we can rewrite
i 2n = (i 2 )n = (−1)n
and i 2n+1 = i (i 2 )n = i (−1)n .
If we pull out the leading i s from the sum over odd powers of t , we find
eiθ =
∞ (−1)n θ 2n+1
∞ (−1)n θ 2n
X
X
+i
= cos θ + i sin θ.
(2n)!
n=0
n=0 (2n + 1)!
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BORIS HASSELBLATT
In summary, we will use the “forward” and “backward” Euler formulas
e i θ = cos θ + i sin θ
cos θ =
e i θ + e −i θ
2
sin θ =
e i θ − e −i θ i −i θ
= (e
− eiθ)
2i
2
8. Exercise. Show that e z = e z .
9. Exercise. Verify that if λ is complex and ae λt +be λt is real for all t , then
b = a via the following steps.
(1) Rewrite the given information using Exercise 2 (and Exercise 8).
(2) The deriviative of ae λt +be λt is also real—compute it and proceed
as in the previous part.
(3) Take t = 0 in the expressions obtained in both previous parts.
1.6. The complex plane. While real numbers can be geometrically represented by a number line, complex numbers can be represented by a
plane, called the complex plane. Taking the real number line as a horizontal axis, one can introduce a vertical axis with i as the unit on it. Then
a + i b can be understood as representing a point in this plane in terms of
cartesian coordinates.
The power of this representation is related to the fact that Euler’s formula represents the same plane but using polar coordinates. (Recall that
polar coordinates in the plane are of the form (r cos θ, r sin θ), where r is
the distance from the origin and θ the angle with the horizontal axis.) To
see this, we contemplate
z = e a+i b = e a e i b = e a cos b + i e a sin b.
Here e a plays the role of r in polar coordinates, and indeed,
p
p
p
|e a+i b | = |z| = z z̄ = e a+i b e a−i b = e 2a = e a .
So in the complex plane, e a+i b has distance e a from the origin and lies
in a direction relative to the horizontal axis given by the angle b, which is
called the argument of e a+i b .
One consequence is that multiplication of complex numbers can be
interpreted geometrically:
e a e i b · e c e i d = e a+i b · e c+i d = e a e c e i (b+d ) .
The absolute value of this product is e a e c , which is the product of the
absolute values of the 2 numbers. The argument is the sum of the arguments of the 2 numbers. Another way of saying this, since the argument
is an angle, is that multiplying a complex number by another rotates and
scales it.
COMPLEX NUMBERS, UNDETERMINED COEFFICIENTS, AND LAPLACE TRANSFORMS 5
1.7. Exponential shift. As in [1, p. 123], the exponential shift works for
complex exponentials (you can check that the calculation on the bottom
half of that page do not use that λ is real):
P (D)[e λt y] = e λt P (D + λ)y.
1.8. Hyperbolic functions. While we think of t as a real variable in Euler’s formula, one gets an interesting result when one plugs imaginary
numbers into cos and sin:
e i (i x) + e −i (i x) e −x + e x e x + e −x
=
=
=: cosh x.
cos i x =
2
2
2
Here, cosh x is the hyperbolic cosine function defined by this last equality.
Likewise,
e i (i x) − e −i (i x) e −x − e x
e x − e −x
=
=i
= i sinh x.
2i
2i
2
1.9. The trigonometric-identity machine. Euler’s formula produces many
trigonometric identities by using the rules of exponents.
sin i x =
10. Example. Find formulas for cos(a + b) and for sin(a + b).
We start with the law of exponents e i (a+b) = e i a e i b and rewrite every
exponential using Euler’s formula. On the left we have
e i (a+b) = cos(a + b) + i sin(a + b).
On the right we have
¡
¢¡
¢
e i a e i b = cos a + i sin a cos b + i sin b
¡
¢ ¡
¢
= cos a cos b − sin a sin b + i sin a cos b + cos a sin b .
Since these two things are the same, we get
¡
¢ ¡
¢
cos(a+b)+i sin(a+b) = cos a cos b−sin a sin b +i sin a cos b+cos a sin b .
Matching real and imaginary parts, we get
cos(a +b) = cos a cos b −sin a sin b and sin(a +b) = sin a cos b +cos a sin b.
11. Example. To get double-angle formulas, take a = b in the previous
example. Note that cos 2a then simplifies to cos2 a − sin2 a.
¡ ¢3
12. Example. To obtain triple-angle formulas use the law e 3i a = e i a
and rewrite every term using Euler’s formula, multiply out and sort terms.
(Or use Example 10 with b = 2a and use double-angle formulas along the
way.)
13. Example. Should you ever need a formula for cos(a + b + c), you can
start with the law of exponents e i (a+b+c) = e i a e i b e i c and rewrite every exponential using Euler’s formula, then multiply out and sort terms.
6
BORIS HASSELBLATT
14. Example ([1, p. 158]). Second-order differential equations often have
solutions that can be written as
c
c̄
x(t ) = ℜ(ce λt ) = e λt + e λ̄t .
2
2
If we are given initial conditions x(0) = x 0 and x 0 (0) = v 0 and write λ =
−σ + i ω, then we find that x 0 = x(0) = ℜ(ce λ0 ) = ℜc and
v 0 = x 0 (0) = ℜ(cλe λ0 ) = ℜ(cλ) = ℜ((x 0 + i ℑc) · (−σ + i ω)) = −σx 0 − ωℑc,
0
so c = ℜc + i ℑc = x 0 − i v 0 +σx
ω
The solution can be rewritten in a way that involves only a sine or a
cosine, but not both. To see how, write c = e a−i α to get
e i (ωt −α) + e −i (ωt −α)
c λt c̄ λ̄t 1 a−i α (−σ+i ω)t a+i α (−σ−i ω)t
e + e = e
e
+e
e
= e a e −σt
.
2
2
2
2
This means that
x(t ) = ℜ(ce λt ) = Ae −σt cos(ωt − α),
q
0 2
where A = e a = |c| = x 02 + ( v 0 +σx
) and
ω
cos α + i sin α = e i α = e −i α =
1 a−i α c̄
x0
v 0 + σx 0
e
= =
+i
.
a
e
A
A
ωA
2. H OMOGENEOUS DIFFERENTIAL EQUATIONS
If one uses complex numbers, then the method described in [1, Sections 2.5, 2.6] looks slightly different, and there are a few choices one can
make.
The point of [1, p. 131] is that the method of [1, Section 2.5] works even
for complex roots of the characteristic polynomial, but that some care
should be taken to make sure one ultimately writes down real solutions
only.
15. Example ([1, Example 2.6.1]). Solve (D 2 + 4D + 5)x = 0.
The characteristic polynomial has roots λ± = −2 ± i , which gives solutions e λ± t , or e λ+ t and e λ− t , or
e (−2+i )t and e (−2−i )t .
However, neither of these is a real solution2. There are different ways of
obtaining real solutions. For the purpose of finding the general solution,
the easiest is to choose, as in [1, Section 2.6] but with different notation,
the solutions
ℜ(e (−2+i )t ) and ℑ(e (−2+i )t ).
2i.e., a real-valued function that solves the differential equation.
COMPLEX NUMBERS, UNDETERMINED COEFFICIENTS, AND LAPLACE TRANSFORMS 7
(Think about why these are solutions, and why these 2 “contain the same
information” as the pair of complex solutions above.) This gives the formulas at the end of [1, Section 2.6].
16. Example ([1, Exercise 2.6.17]). Solve the initial-value problem
(5D 2 + 2D + 1)x = 0;
x(0) = 0, x 0 (0) = 1.
The roots of the characteristic polynomial are λ± = − 15 ± 25 i , giving the
“general solution” c + e λ+ t + c − e λ− t , or
1
2
1
2
c + e (− 5 + 5 i )t + c − e (− 5 − 5 i )t .
This is not quite “proper” since it involves complex functions, but unlike
in the previous example, this is also not the desired final answer anyway
since we wish to solve an initial-value problem. While we could proceed
as before and write out the general solution in real terms and then determine the correct coefficients from the initial values, we can instead
directly do so with the complex solution above.
The condition
x(0) = 0,
x 0 (0) = 1
1
2
1
2
applied to x(t ) = c + e (− 5 + 5 i )t + c − e (− 5 − 5 i )t gives
c + + c − = 0,
c + λ+ + c − λ− = 1.
Cramer’s rule gives
µ
¶
0 1
det
1 λ−
1
1
5 1
c+ =
.
= 2 =
µ
¶ =−
λ− − λ+ 2 · 5 i 2 2i
1
1
det
λ+ λ−
Now, Exercise 9 (on page 4) or an analogous computation or the first
equation tells us that c − = −c + = − 52 2i1 . Therefore the solution of the
initial-value problem is
2
2
5 1 (− 1 + 2 i )t 5 1 (− 1 − 2 i )t 5 −t /5 e 5 i t − e − 5 i t 5 −t /5
2
e 5 5 −
e 5 5 = e
·
= e
sin t .
2 2i
2 2i
2
2i
2
5
Which version works better is a matter of taste; the one presented here
leads directly to the coefficients from the complex form with a computation involving 2 equations in 2 unknowns, and it is reassuring (basic
sanity check) that the end result is real.
8
BORIS HASSELBLATT
17. Example ([1, Exercise 2.6.17] with real functions). Solve the initialvalue problem
(5D 2 + 2D + 1)x = 0;
x(0) = 0, x 0 (0) = 1.
The roots of the characteristic polynomial are λ± = − 15 ± 25 i , giving by Exercise 9 the general solution ce λ+ t + c̄e λ− t , or
1
2
1
2
ce (− 5 + 5 i )t + c̄e (− 5 − 5 i )t .
Since this is a real function by Exercise 9, it is quite proper, even though
the coefficients are complex. There is only one (complex) coefficient to
determine.
The condition
x(0) = 0,
x 0 (0) = 1
1
2
1
2
applied to x(t ) = ce (− 5 + 5 i )t + c̄e (− 5 − 5 i )t gives
c + c̄ = 0,
cλ+ + c̄ λ̄+ = 1.
The first equation (which says that c is imaginary) gives c̄ = −c, so the
second equation gives
2
1 = c · (λ+ − λ̄+ ) = c · 2i ℑλ+ = c · 2i ,
5
so
5 1
.
2 2i
As above, one then finds that the solution of the initial-value problem is
c=
5
2
x(t ) = e −t /5 sin t .
2
5
3. U NDETERMINED COEFFICIENTS
The method of undetermined coefficients works much like in [1, Section 2.7], but we use complex exponentials and complex factoring of polynomials where appropriate.
18. Example ([1, Example 2.7.5]). Solve
(N)
(D 2 − 4)x = 1 + 65e t cos 2t .
Since D 2 −4 = (D −2)(D +2), the general solution of the associated homogeneous equation
(H)
(D 2 − 4)x = 0
COMPLEX NUMBERS, UNDETERMINED COEFFICIENTS, AND LAPLACE TRANSFORMS 9
is
x = H (t ) = c 1 e 2t + c 2 e −2t .
The forcing term of (N) can be rewritten as 1 + 65
e (1+2i )t + 65
e (1−2i )t and
2
2
is therefore annihilated by
D(D − (1 + 2i ))(D − (1 − 2i )).
Therefore a particular solution x = p(t ) of (N) will also satisfy
D(D − (1 + 2i ))(D − (1 − 2i ))(D − 2)(D + 2)x = 0.
(H*)
This has characteristic polynomial r (r − (1 + 2i ))(r − (1 − 2i ))(r − 2)(r + 2),
so p(t ) has to be of the form
p(t ) = k 1 + k 2 e (1+2i )t + k 3 e (1−2i )t + k 4 e 2t + k 5 e −2t .
19. Remark. When we use this method later to predict what terms to expect in applying the Laplace transform method, then we can stop at this
point.
We get a simplified guess by omitting the terms that solve (H):
p(t ) = k 1 + k 2 e (1+2i )t + k 3 e (1−2i )t .
To determine the (as yet undetermined) coefficients k 1 , k 2 and k 3 , insert
this simplified guess into (N), written with the forcing term on the left:
1+
65 (1+2i )t 65 (1−2i )t
e
+ e
= (D 2 − 4)p(t )
2
2
= (D 2 − 4)[k 1 + k 2 e (1+2i )t + k 3 e (1−2i )t ]
= (D 2 − 4)k 1 + (D 2 − 4)k 2 e (1+2i )t
+ (D 2 − 4)k 3 e (1−2i )t
(complex exponential shift)
= −4k 1 + e (1+2i )t ([D + (1 + 2i )]2 − 4)k 2
+ e (1−2i )t ([D + (1 − 2i )]2 − 4)k 3
= −4k 1 + e (1+2i )t ((1 + 2i )2 − 4)k 2
+ e (1−2i )t ((1 − 2i )2 − 4)k 3
Since the functions 1, e (1+2i )t and e (1−2i )t are linearly independent, we
can equate coefficients of like terms on left and right:
−4k 1 = 1,
((1 + 2i )2 − 4)k 2 =
65
,
2
((1 − 2i )2 − 4)k 3 =
65
.
2
Thus, k 1 = −1/4 and
65
65
65
1
k2 =
=
=
= (−4i − 7)
2
2((1 + 2i ) − 4) 2(1 + 4i − 4 − 4) 2(4i − 7) 2
10
BORIS HASSELBLATT
and (either by a like computation or because it must be the complex conjugate3)
1
k 3 = (4i − 7).
2
This gives
1 1
1
p(t ) = − + (−4i − 7)e (1+2i )t + (4i − 7)e (1−2i )t
4 2
2
7
1 4i t 2i t
= − − e (e − e −2i t ) − e t (e 2i t + e −2i t )
4 2
2
1
= − + 4e t sin 2t − 7e t cos 2t .
4
The general solution of (N) then is the sum of H (t ) and p(t ):
x = H (t ) + p(t ) = c 1 e 2t + c 2 e −2t −
1
+ 4e t sin 2t − 7e t cos 2t .
4
This completes the example. We note that when using trigonometric
functions as in the text, one has to solve a 2 × 2 system of equations to
find k 2 and k 3 , while with complex exponentials one instead computes
the reciprocal of a complex number.
4. L APLACE TRANSFORMS
The definition of the Laplace transform in the textbook [1, page 412]
(see also [1, Example 5.2.1] and [1, Example 5.2.4]) gives
Z ∞
n λt
L [t e ] =
e −st t n e λt d t
0
Z ∞
=
t n e −(s−λ)t d t
0

1


if n = 0
¡integration by parts if n ≥ 1:¢
s −λ
Z ∞
=
n
−(s−λ)t
n
u=t , d v=e
dt

0 +
t n−1 e −(s−λ)t d t if n ≥ 1
s −λ 0

1


if n = 0

s −λ
=



 n L [t n−1 e λt ] if n ≥ 1.
s −λ
3Don’t use the “must be the complex conjugate”-shortcut if you don’t know why it’s
true!
COMPLEX NUMBERS, UNDETERMINED COEFFICIENTS, AND LAPLACE TRANSFORMS 11
Applying this recursively, we find
(1)
n!
L [t e ] =
(s − λ)n+1
n λt
L
−1
·
¸
1
1
=
t n−1 e λt
n
(s − λ)
(n − 1)!
This will go quite far when combined with partial fractions. Note that for
n = 0 this gives the Laplace transform of an exponential function and for
λ = 0 it gives the Laplace transform of a power of t .
5. PARTIAL FRACTIONS
Partial-fractions decompositions are easier with complex numbers, because irreducible quadratics are no longer a difficulty: everything reduces
to linear factors. One just has to patiently work the partial-fractions decompositions that arise and put up with complex coefficients.
20. Example ([1, Example 5.3.3]: Distinct roots). Solve the initial-value
problem
x 0 − x = 2 sin t
x(0) = 0.
Note first that the method of undetermined coefficients tells us to expect
no terms other than e t , cos t and sin t in the solution (or: no terms other
than e t , e i t and e −i t ). This was the purpose of Remark 19.
Applying L we get
hi
i
i
i
2
(s − 1)L [x] = 2L (e −i t − e i t ) =
−
=
2
s + i s − i (s − i )(s + i )
and hence
·
¸
2
−1
x =L
.
(s − 1)(s − i )(s + i )
To find x we look for the partial-fractions decomposition of L [x]:
2
A
B
C
=
+
+
(s − 1)(s + i )(s − i ) s − 1 s − i s + i
To determine A, B , and C , clear fractions and then insert the values of s
for which the original denominator is zero, that is, s = 1, ±i .
2 = A(s + i )(s − i ) + B (s − 1)(s + i ) +C (s − 1)(s − i )
For s = 1 this gives 2 = A(1 + i )(1 − i ) = A(1 − (i 2 )) = 2A, so A = 1.
For s = i this gives
2 = B (i − 1)(i + i ) = 2B (i − 1)i = 2B (i 2 − i ) = 2B (−1 − i ),
1
1 i
=− + .
so B =
−1 − i
2 2
Finally, for s = −i we get
2 = C (−i − 1)(−i − i ) = 2C (−i − 1)(−i ) = 2C ((−i )2 + i ) = 2C (−1 + i ),
12
BORIS HASSELBLATT
so C =
1
1 −1 − i
−1 − i
1 i
=
=
=
−
− .
−1 + i −1 + i −1 − i (−1)2 − i 2
2 2
21. Remark. Time-saver: B and C are complex conjugates. This is no
accident!
We thus obtain the partial-fractions decomposition
− 12 + 2i − 21 − 2i
2
1
=
+
+
(s − 1)(s 2 + 1) s − 1
s −i
s +i
Note that the two summands involving i are complex conjugates of each
other, so they sum to a real number. This is why we expect their numerators B and C to be complex conjugates.
Now we “untransform” and sort terms:
"
#
− 12 + 2i − 21 − 2i
1
−1
x =L
+
+
s −1
s −i
s +i
¡ 1 i¢
¡ 1 i¢
= e t + − + e i t + − − e −i t
2 2
2 2
1
i
= e t − (e i t + e −i t ) + (e i t − e −i t )
2
2
t
= e − cos t − sin t
We conclude that the initial-value problem
D x − x = 2 sin t
x(0) = 0.
has the unique solution
x = e t − cos t − sin t .
The form matches with our expectations from the method of undetermined coefficients (Remark 19), so one only needs to check the initial
condition to verify that this is correct: x(0) = e 0 − cos 0 − sin 0 = 1 − 1, as
required.
22. Example ([1, Example 5.6.3]: Pair of double complex roots).
h
i
s
−1
Find L
.
(s 2 + 1)2
We rewrite
s
s
=
2
2
2
(s + 1)
(s − i ) (s + i )2
and produce the partial-fractions decomposition:
s
A
B
C
D
=
+
+
+
(s − i )2 (s + i )2 s − i (s − i )2 s + i (s + i )2
becomes
s = A(s − i )(s + i )2 + B (s + i )2 +C (s + i )(s − i )2 + D(s − i )2 .
COMPLEX NUMBERS, UNDETERMINED COEFFICIENTS, AND LAPLACE TRANSFORMS 13
Set s = i to get
i
i = B (2i )2 = −4B, hence B = − .
4
Set s = −i to get
i
−i = D(−2i )2 = −4D, hence D = .
4
Before going on we consolidate the corresponding terms:
i
i
B (s + i )2 + D(s − i )2 = − (s + i )2 + (s − i )2
4
4
¢
i
i¡
= − (s 2 + 2si − 1) + (s 2 − 2si − 1) = − (4si ) = s.
4
4
Therefore we can rewrite
s = A(s − i )(s + i )2 + B (s + i )2 +C (s + i )(s − i )2 + D(s − i )2
as
0 = A(s − i )(s + i )2 +C (s + i )(s − i )2 .
This means that we can (and hence must!) take A = C = 0.
Therefore,
h
i
h
s
i /4
i /4 i
−1
L −1
=
L
−
+
(s − i )2 (s + i )2
(s − i )2 (s + i )2
i i t i −i t t h e i t − e −i t i t
= − te + te
=
= sin t .
4
4
2
2i
2
23. Example ([1, Example 5.6.5]: Application to initial-value problem).
Solve
(D 2 + 1)x = cos t ,
x(0) = x 0 (0) = 0.
Apply L to both sides to get
1
1³ 1
1 ´
s
(s 2 + 1) L [x] = L [e i t + e −i t ] =
+
=
| {z }
2
2 s −i s +i
(s − i )(s + i )
=(s−i )(s+i )
Thus, x = L −1
h
i t
s
= sin t
(s − i )2 (s + i )2
2
using the previous example.
24. Example ([1, Example 5.6.4]: Same pair of double complex roots).
h
i
1
Find L −1 2
.
(s + 1)2
1
1
=
and produce the partial-fractions deWe write 2
2
2
(s + 1)
(s − i ) (s + i )2
composition:
1
A
B
C
D
=
+
+
+
2
2
2
(s − i ) (s + i )
s − i (s − i )
s + i (s + i )2
14
BORIS HASSELBLATT
becomes
1 = A(s − i )(s + i )2 + B (s + i )2 +C (s + i )(s − i )2 + D(s − i )2 .
This can be solved for A, B , C and D by multiplying out the right-hand
side, sorting by powers of s and comparing coefficients to get 4 linear
equations in these 4 unknowns. We try sampling useful s-values first to
break the problem up.
Set s = i to get
1
1 = B (2i )2 = −4B, hence B = − .
4
Set s = −i to get
1
1 = D(−2i )2 = −4D, hence D = − .
4
Consolidate:
¢
1¡
1
B (s + i )2 + D(s − i )2 = − (s 2 + 2si − 1) + (s 2 − 2si − 1) = − (s 2 − 1).
4
2
Therefore,
1 = A(s − i )(s + i )2 + B (s + i )2 +C (s + i )(s − i )2 + D(s − i )2
1
= A(s − i )(s + i )2 +C (s + i )(s − i )2 − (s 2 − 1)
2
1 2
and, adding 2 (s − 1) to both sides,
1 2
1
(s + 1) = 1 + (s 2 − 1) = A(s − i )(s + i )2 +C (s + i )(s − i )2
2
2
= A(s 2 + 1)(s + i ) +C (s 2 + 1)(s − i ).
Dividing by the common factor s 2 + 1, this becomes
1
= A(s + i ) +C (s − i ) = (A +C )s + i (A −C ) .
|
{z
}
2
sorted by powers of s
Comparing coefficients of like powers of s we find that
1
A +C = 0 and
= i (A −C ) = 2i A,
2
so
1
1
A=
and C = − .
4i
4i
Inserting these into the partial-fractions decomposition we find
1
A
B
C
D
=
+
+
+
2
2
2
(s − i ) (s + i )
s − i (s − i )
s + i (s + i )2
1/4i
1/4
1/4i
1/4
=
−
−
−
2
s − i (s − i )
s + i (s + i )2
COMPLEX NUMBERS, UNDETERMINED COEFFICIENTS, AND LAPLACE TRANSFORMS 15
This gives us
h
i
1
1/4
1/4i
1/4 i
−1 1/4i
=
L
−
−
−
(s 2 + 1)2
s − i (s − i )2 s + i (s + i )2
1
t
1
t
= e i t − e i t − e −i t − e −i t
4i
4
4i
4
1 e i t − e −i t t e i t + e −i t
=
−
2
2i
2
2
t
1
= sin t − cos t .
2
2
25. Example (Initial-value problem producing triple complex roots). Solve
L −1
h
(D 2 + 9)2 x = −4 sin 3t ,
x(0) = x 0 (0) = x 00 (0) = 0, x 000 (0) = 1.
We first note that we should expect only the functions sin 3t , cos 3t ,
t sin 3t , t cos 3t , t 2 sin 3t , t 2 cos 3t in the ultimate solution (Remark 19).
Apply L to get
3
(s 2 + 9)2 L [x] − 1 = −4 2
,
s +9
so
=(s 2 +9)−12
(2)
z }| {
1
12
s2 − 3
L [x] = 2
−
=
(s + 9)2 (s 2 + 9)3 (s 2 + 9)3
| {z }
=(s−3i )(s+3i )
We need to decompose the right-hand side as follows:
B
C
s2 − 3
A
D
E
F
+
+
=
+
+
+
.
3
3
2
2
3
(s − 3i ) (s + 3i )
s − 3i s + 3i (s − 3i ) (s + 3i ) (s − 3i ) (s + 3i )3
Clear fractions to get
s 2 − 3 =A(s − 3i )2 (s + 3i )3 + B (s − 3i )3 (s + 3i )2
+C (s − 3i )(s + 3i )3 + D(s − 3i )3 (s + 3i )
+ E (s + 3i )3 + F (s − 3i )2 .
For s = 3i this becomes
2·6
1
−9 − 3
=
=
,
(6i )3
6 · 3 · 2 · 6i 18i
1
and F must be the complex conjugate: F = −
. We consolidate these 2
18i
terms first:
E=
3
)
(s−3i )
= (s+3i
18i − 18i
z }| {
E
F
1/18i
1/18i
s2 − 3
+
=
−
=
.
(s − 3i )3 (s + 3i )3 (s − 3i )3 (s + 3i )3 (s 2 + 9)3
3
16
BORIS HASSELBLATT
Note from equation (2) above that we have just stumbled upon the partialfractions decomposition we need, that is, that (using equation (1)) we
have
· 2
¸
·
¸
s −3
1
1
1
1 t 2 e 3i t − t 2 e −3i t
−1
−1
x =L
=
L
−
=
.
(s 2 + 9)3
18i
(s − 3i )3 (s + 3i )3
18
2i
In short, the solution to the given initial-value problem is
1
x = t 2 sin 3t .
18
By inspection (really!) one can see that this satisfies the initial condition.
This last example illustrates that we can in this way handle complex
roots of any multiplicity, not just double complex roots.
R EFERENCES
[1] Martin M. Guterman, Zbigniew H. Nitecki, Differential Equations – A First Course,
3rd ed., Saunders (1992).