MA 18.03, R05
FOURIER SERIES, GENERALIZED FUNCTIONS, LAPLACE
TRANSFORM
1
Fourier Series
1.1
General facts
1. A (generalized) function f (t) of period 2L has a Fourier series of the form
f (t) ∼
a0
πt
2πt
πt
2πt
+ a1 cos
+ a2 cos
+ · · · + b1 sin
+ b2 sin
+ ···
2
L
L
L
L
or
∞
f (t) ∼
∞
a0 X
nπt X
nπt
+
an cos
+
bn sin
2
L
L
n=1
n=1
2. The coefficients an and bn can be found by the following formulas.
an
bn
=
=
1
L
Z
1
L
Z
L
f (t) cos
nπt
dt
L
f (t) sin
nπt
dt
L
−L
L
−L
3. The Fourier series it convergent at every point t for which both the right limit f (t+) and left limit
f (t−) exist and are finite. The sum of the Fourier series at t is the average of these two limits.
4. The terms in the Fourier series of a function f (t) must have the same symmetries as f (t) itself.
For instance,
• an odd function will only have a sine functions in its Fourier series (no constant);
• an even function will only have a cosine functions in its Fourier series and the constant term;
• the Fourier series of a function that is odd about L/2 will only have cos (2n+1)πt
and sin 2nπt
L
L ;
(2n+1)πt
• the Fourier series of a function that is even about L/2 will only have cos 2nπt
.
L and sin
L
It might be useful to recall that
• even · even = even
• even · odd = odd
• odd · odd = even
5. We differentiate Fourier series term by term. So given f (t) ∼
we get
f 0 (t) ∼
∞
X
n=1
∞
∞
a0 X
nπt X
nπt
+
an cos
+
bn sin
,
2 n=1
L
L
n=1
∞
bn
nπ
nπt X
nπ
nπt
cos
+
−an
sin
L
L
L
L
n=1
6. Integration is also done term by term, but the result is a Fourier series only of the series we
integrated does not have a non-zero constant term.
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MA 18.03, R05
1.2
How do we compute Fourier series?
Directly from the definition using the formulas for an and bn .
Reducing to a known Fourier series (or to a Fourier series given on the exam) by differentiating,
or integrating, or translating, or taking a dilation, etc. . . and maybe employing some trig identities.
Trig identities are especially useful for computing Fourier series of, say, sin2 t. Try to remember that
sin2 A =
1 − cos 2A
2
sin(A ± B) = sin A cos B ± cos A sin B;
sin A sin B =
cos(A − B) − cos(A + B)
;
2
sin A cos B =
cos2 A =
1 + cos 2A
2
cos(A ± B) = cos A cos B ∓ sin A sin B;
cos A cos B =
cos(A − B) + cos(A + B)
;
2
sin(A + B) − sin(A − B)
;
2
and the special values
sin
(2n + 1)π
= (−1)n ;
2
π
π
1
sin = cos = √ ;
4
4
2
1.3
cos nπ = (−1)n ;
π
1
π
sin = cos = ;
6
3
2
√
3
π
π
sin = cos =
.
3
6
2
Use in ODE
Consider the linear ODE with constant coefficients
cn y (n) + . . . + c1 y 0 + c0 y = f (t),
cn 6= 0,
f (t) periodic with period 2L.
We can make use of Fourier series to find a particular solution yp of this ODE, as follows. (But the
general solution is still y = yp + yh .)
1. First write down the Fourier series of f (t).
2. Replace f (t) by its Fourier series in the RHS of the ODE. It becomes something of the form
∞
cn y (n) + . . . + c1 y 0 + c0 y =
∞
a0 X
nπt X
nπt
+
+
.
an cos
bn sin
2
L
L
n=1
n=1
3. For each term that appears in the Fourier series find a particular solution of the corresponding
equation (use ERF or ERF’ or ERF” etc. . . ).
a0
cn y (n) + . . . + c1 y 0 + c0 y =
; get y0
2
cn y (n) + . . . + c1 y 0 + c0 y
=
an cos
nπt
L
;
get
y1,n
cn y (n) + . . . + c1 y 0 + c0 y
=
bn sin
nπt
L
;
get
y2,n
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MA 18.03, R05
4. A particular solution to the original ODE is the sum of all these pieces
yp = y0 +
∞
X
n=1
2
y1,n +
∞
X
y2,n .
n=1
Step and Delta
2.1
Definitions
(
1
The step function u(t) =
0
(
1
ua (t) = u(t − a) =
0
t>0
t<0
t>a
t<a
δ(t) = u0 (t) is 0 for all t 6= 0 and has a spike of size 1 at t = 0
δa (t) = δ(t − a) = u0a (t) is 0 for all t 6= a and has a spike of size 1 at t = a
2.2
Properties
0
1. u = δ
2. u = u0
3. δ = δ0
(
0
t<a
4. f (t)ua (t) =
f (t) t > a


t<a
0
5. f (t)(ua (t) − ub (t)) = f (t) a < t < b


0
t>b
6. f (t)δa (t) = f (a)δa (t) is 0 for all t 6= a and has a spike of size f (a) at t = a
Z
7.
b
3
c


a<b
0
f (t)δa (t)dt = f (a) b < a < c


0
c<a
Unit response and weight function
p(D) = an Dn + · · · + a1 D + a0 I
The weight function or unit impulse response of the operator p(D) is the unique function w(t)
with the property that
p(D)w = δ(t).
1. Solve the homogeneous equation p(D)x = 0 for t > 0 with the initial conditions (given by
matching singularities) x(n−1) = 1/an and x(0+) = x0 (0+) = x(n−2) (0+) = 0.
2. Take the solution you found and multiply by the step function u(t). This is your w(t).
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MA 18.03, R05
The unit step response of the operator p(D) is the unique function v(t) with the property that
p(D)v = u(t).
1. Solve the inhomogeneous equation p(D)x = 1 for t > 0 with the initial conditions (given by
matching singularities) x(0+) = x0 (0+) = x(n−1) (0+) = 0.
2. Take the solution you found and multiply by the step function u(t). This is your v(t).
Properties
1. v 0 (t) = w(t)
2. The solution to p(D)x = δ(t − a) = δa (t) is w(t − a).
3. The solution to p(D)x = u(t − a) = ua (t) is v(t − a).
4
Convolution and Green’s Formula
The convolution of two functions f and g is a function (f ∗ g) given by
Z x
f (u)g(x − u) du.
(f ∗ g)(x) =
0
Properties
1. f ∗ g = g ∗ f j
2. f ∗ (ag1 + bg2 ) = af ∗ g1 + bf ∗ g2
3. f ∗ (g ∗ h) = (f ∗ g) ∗ h
Green’s formula gives a way of finding the solution yp to the IVP
an y (n) + . . . + a1 y 0 + a0 y = f (t),
y(0) = y 0 (0) = . . . = y (n−1) (0) = 0
in terms of the convolution of the forcing function f (t) with weight function w(t) associated to the
operator p(D) = an Dn + · · · + a1 D + a0 I. Namely it tell us that the solution to the IVP is
Z t
Z t
yp (t) = (f ∗ w)(t) =
f (t − τ )w(τ )dτ =
f (τ )w(t − τ )dτ.
0
0
Note: This also gives a nice way of finding a particular solution yp to the equation
an y (n) + . . . + a1 y 0 + a0 y = f (t).
As always, the general solution is y = yp + y + h. If you have an IVP with any other initial conditions
than rest, you can
1. use Green’s formula to find yp ;
2. solve the associated homogeneous equation to find yh ;
3. write down the general solution y = yp + y + h;
4. use the initial condition to solve for the undetermined constants.
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MA 18.03, R05
5
Laplace Transform
If the function f (t) is continuous at t = 0, there is no ambiguity about the way we define its Laplace
transform:
Z ∞
L(f )(s) =
e−st f (t)dt = F (s).
0
When we have to deal with discontinuities or generalized functions, it makes sense to consider
Z ∞
L+ (f )(s) =
e−st f (t)dt that ignores the (possible) discontinuity at 0 of f (t);
0+
Z
∞
e−st f (t)dt
L− (f )(s) =
that detects if f (t) has a discontinuity at 0.
0−
5.1
General facts
L
1. af (t) + bg(t) −→ aF (s) + bG(s)
L
2. f (t) ∗ g(t) −→ F (s)G(s)
L
3. eat f (t) −→ F (s − a)
L
4. tf (t) −→ −F 0 (s)
L
5. tn f (t) −→ (−1)n F (n) (s)
L
6. u(t − a)f (t − a) −→ e−as F (s) for a ≥ 0
L
7. f 0 (t) −→ sF (s) − f (0) (use 0+ for L+ )
L
8. f 00 (t) −→ s2 F (s) − sf (0) − f 0 (0) (use 0+ for L+ )
L
9. f (n) (t) −→ sn F (s) − sn−1 f (0) − . . . − f (n−1) (0) (use 0+ for L+ )
10. L− (f (n) ) = sn F (s)
11. Basic transforms
L(u) = L(1) =
1
s
L(cos at) =
L(tn ) =
s
s2 + a2
L− (δ(t)) = 1
n!
sn+1
L(sin at) =
L(eat ) =
a
s2 + a2
L(δa (t)) = e−as (a > 0)
5
1
s−a
MA 18.03, R05
5.2
Use in ODE
Finding solutions of ODEs: take he Laplace transforms of both sides of an ODE, one obtains an
(algeraic) equation for the Laplace transform. Solve that and take the inverse Laplace transform.
That gives you a solution of the original ODE.
Finding the weight function of an operator p(D):
L(w(t)) =
1
p(s)
Pole diagram: the rightmost pole(s) of the Laplace transform F (s) of a function f (t) tells us how f (t)
behaves for very large t (as t → ∞).
• If the right most pole is at s = a, then, for large t, f (t) ∼ eat
• If the right most poles are at s = a ± ib, then, for large t, f (t) ∼ eat cos bt
• In particular, f (t) → 0 as t → ∞ if and only if the poles of F (s) have real part negative.
More precisely, if F (s) has simple poles at a, and α ± iβ, i.e. if
F (s) =
∗
,
(s − a)((s − α)2 + β 2 )
then F (s) will be written as a sum of simple fractions
F (s) =
A
B
C
A
B0s + C 0
.
+
+
=
+
s − a s − α − iβ
s − α + iβ
s − a (s − α)2 + β 2
Therefore,
f (t) = Aeat + Be(α+iβ)t + Ce(α−iβ)t
or, equivalently,
f (t) = A00 eat + B 00 eαt cos βt + C 00 eαt sin βt.
The dominant term as t gets very large will be the one with the largest coefficient on the exponential,
which is to say the one corresponding to the rightmost pole.
6
Transfer function of p(D)
W (s) =
1
= L(w(t))
p(s)
It has the property that a particular solution to p(D)x = ert is given by xp = W (r)ert (if p(r) 6= 0.)
Note that this is an exponential solution.
6