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Mathematics

MATH 251

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Second Order Linear Partial Differential Equations
Part II
Fourier series; Euler▯Fourier formulas; Fourier Convergence Theorem;
Even and odd functions; Cosine and Sine Series Extensions; Particular
solution of the heat conduction equation
Fourier Series
Suppose f is a periodic function with a period T = 2L. Then ther
series representation of f is a trigonometric series (that is, it is an infinite
series consists of sine and cosine terms) of the form
a ∞ nπ x nπ x
f (x)= 0+ ∑ a nos +b nin
2 n=1 L L
Where the coefficients are given by the Euler▯Fourier formulas:
L
a = 1 f (x)cosmπ x dx
m L ∫ L , m = 0, 1, 2, 3, …
−L
1 L nπ x
bn= ∫ f( )sin dx, n = 1, 2, 3, …
L −L L
The coefficients a’s are called the Fourier cosine coefficients (i0cluding a ,
the constant term, which is in reality the 0▯th cosine term), and b’s are called
the Fourier sine coefficients.
© 2008, 2012 Zachary S Tseng E▯2 ▯ 1 Note 1: Thus, every periodic function can be decomposed into a sum of one
or more cosine and/or sine terms of selected frequencies determined solely
by that of the original function. Conversely, by superimposing cosines and/
or sines of a certain selected set of frequencies we can reconstruct any
periodic function.
Note 2: If f is piecewise continuous, then the definite integrals in the Euler▯
Fourier formulas always exist (i.e. even in the cases where they are improper
integrals, the integrals will converge). On the other hand, f needs not to be
piecewise continuous to have a Fourier series. It just needs to be periodic.
However, if f is not piecewise continuous, then there is no guarantee that we
could find its Fourier coefficients, because some of the integrals used to
compute them could be improper integrals which aredivergent.
Note 3: Even though that the “=” sign is usually used to equate a periodic
function and its Fourier series, we need to be a little careful. The function f
and its Fourier series “representation” are only equal to each other if, and
whenever, f is continuous. Hence, if f is continuous for −∞ < x < ∞, then f is
exactly equal to its Fourier series; but if f is piecewise continuous, then it
disagrees with its Fourier series at every discontinuity. (See the Fourier
Convergence Theorem below for what happens to the Fourier series at a
discontinuity of f .)
Note 4: Recall that a function f is said to be periodic if there exists a positive
number T, such that f (x + T) = f(x), for all x in its domain. In such a case
the number T is called a period of f. A period is not unique, since if f(x + T)
= f(x), then f(x + 2T) = f(x) and f(x + 3T) = f(x) and so on. That is, every
integer▯multiple of a period is again another period. The smallest such T is
called the fundamental period of the given function f. A special case is the
constant functions. Every constant function is clearly a periodic function,
with an arbitrary period. It, however, has no fundamental period, because its
period can be an arbitrarily small real number. The Fourier series
representation defined above is unique for each function with a fixed period
T = 2L. However, since a periodic function has infinitely many (non▯
fundamental) periods, it can have many different Fourier series by using
different values of L in the definition above. The difference, however, is
really in a technical sense. After simplification they would look the same.
© 2008, 2012 Zachary S Tseng E▯2 ▯ 2 Therefore, technically at least, a Fourier series of a periodic function
depends both on the function as well as its chosen period.
Note 5: The definite integrals in the Euler▯Fourier formulas can be found be
integrating over any interval of length 2L. However, from −L to L is the
convention, and is often the most convenient interval to use.
Note 6: Since the Fourier coefficients are calculated by definite integrals,
which are insensitive to the value of the function at finitely many points.
Consequently, piecewise continuous functions of the same period that differ
from each other at finitely many points (notably, at isolated discontinuities)
per period will have the same Fourier series.
Note 7: The constant term in the Fourier series, which has expression
a0 1 1 L 1 L
= ⋅ ∫ f (x)cos(0)dx= ∫ f (x)dx ,
2 2 L −L 2L −L
is just the average or mean value of f(x) on the interval [−L, L]. Since f is
periodic, this average value is the same for everyperiod of f. Therefore, the
constant term in a Fourier series represents the average value of the function
f over its entire domain.
© 2008, 2012 Zachary S Tseng E▯2 ▯ 3 Example: Find a Fourier series for f(x) = x, −2 < x < 2, f(x + 4) = f(x).
First note that T = 2L = 4, hence L = 2.
The constant term is one half of:
L 2 2
1 mπ x 1 1 x2 1
a0 = ∫ f x)cos dx = ∫ xdx = = (2− 2) = 0
L−L L 2 −2 2 2 −2 2
The rest of the cosine coefficients, for n = 1, 2, 3, …, are
L 2
1 nπ x 1 nπx
an= L ∫ f (x)cos L dx = 2 ∫xcos 2 dx
−L −2
2 2
= 1 x sinn x − 2 sinn x dx
2 n π 2 −2 nπ −2 2
1 2x n x 4 n x 2
= sin + 2 2 cos
2 π 2 n π 2 −2
1 4 4
= 2 0 + n2π 2cos(nπ ) 0 + n2π 2cos(−nπ) = 0
Hence, there is no nonzero cosine coefficient for this function. That is,
its Fourier series contains no cosine terms at all. (We shall see the
significance of this fact a little later.)
© 2008, 2012 Zachary S Tseng E▯2 ▯ 4 The sine coefficients, for n = 1, 2, 3, …, are
L 2
b = 1 f (x)sinnπ xdx = 1 xsinnπ x dx
n L −L L 2 −2 2
1 − 2x nπx 2 −2 2 nπx
= cos − ∫ cos dx
2 n π 2 −2 nπ −2 2
2
1 − 2x nπx 4 nπx
= cos + 2 2 sin
2 n π 2 nπ 2 −2
= 1 − 4cos(nπ )−0 − 4 cos(−nπ )−0
2 nπ n π
= − 2(cos(nπ) +os(nπ) = − 4)cos(nπ)
nπ nπ
4 , n = odd n+1
= nπ = (−1) 4
−4 nπ .
nπ , n =even
4 ∞ ( 1)n+1 nπ x
Therefore,( )= ∑ sin .
π n=1 n 2
© 2008, 2012 Zachary S Tseng E▯2 ▯ 5 Figure : the graph of the partial sum of the first 30 terms of the
Fourier series
∞ n+1
f (x) =4 (−1) sin nπ x
π n=1 n 2 .
Compare it against the graph of the actual function the series
represents the function f(x) = x, −2 < x < 2, f(x + 4) = f(x), seen
earlier.
© 2008, 2012 Zachary S Tseng E▯2 ▯ 6 Example: Find a Fourier series for f(x) = x, 0 < x < 4, f(x + 4) = f(x). How
will it be different from the series above?
4 2 4
1 1 x 1
a0= 2 ∫ xdx = 2 2 = 2 (8−0) = 4
0 0
For n = 1, 2, 3, … :
4
a = 1 xcos nπx dx
n 2 ∫ 2
0
1 2x n x 4 n x 4
= sin + 2 2cos
2 π 2 n π 2 0
= 1 0+ 4 cos(2nπ) − 0+ 4 cos(0)= 0
2 n π 2 n π 2
1 4 nπx
bn= ∫ xsin dx
2 0 2
4
= 1 − 2x cosn x + 4 sinn x
2 nπ 2 n2π2 2
0
1 − 8 − 4
= cos(2n π 0− − (0 0 =
2 nπ nπ
Consequently,
a0 ∞ nπx nπ x − 4 ∞ 1 nπx
f( )= 2 + ∑ ancos L +bn sin L =2+ π ∑ nsin 2 .
n=1 n=1
© 2008, 2012 Zachary S Tseng E▯2 ▯ 7 Example: Find a Fourier series for f(x) = | x|, −2 < x < 2, f(x + 4) = f(x).
8 ∞ 1 (2n −1) x
Answer : f( ) 1− ∑ cos
π 2 n=12 n −1)2 2
Example: Find a Fourier series for
f (x)= − 2, −1≤ x < 0
2, 0 ≤ x <1 , f(x + 2) = f(x).
8 ∞ 1
Answer: f (x) = ∑ sin((2n −1)π x)
π n=1(2n −1)
© 2008, 2012 Zachary S Tseng E▯2 ▯ 8 Comment: Just because a Fourier series could have infinitely many (nonzero)
terms does not mean that it will always have that many terms. If a periodic
function f can be expressed by finitely many terms normally found in a
Fourier series, then the expression must be the Fourier series of f. (This is
analogous to the fact that the Maclaurin series of any polynomial function is
just the polynomial itself, which is a sum of finitely many powers of x.)
Example: The Fourier series (period 2π) representing f (x) = 5 + cos(4x) −
sin(5x) is just f(x) = 5 + cos(4x) − sin(5x).
Example: The Fourier series (period 2π) representing f (x) = 6cos(x)sin(x) is
not exactly itself as given, since the product cos(x)sin(x) is not a term in a
Fourier series representation. However, we can use the double▯angle
formula of sine to obtain the result: 6cos(x)sin(x) = 3sin(2x).
Consequently, the Fourier series is f(x) = 3sin(2x).
© 2008, 2012 Zachary S Tseng E▯2 ▯ 9 The Fourier Convergence Theorem
Here is a theorem that states a sufficient condition for the convergence of a
given Fourier series. It also tells us to what value does the Fourier series
converge to at each point on the real line.
Theorem: Suppose f and f ′ are piecewise continuous on the interval
−L ≤ x ≤ L. Further, suppose that f is defined elsewhere so that it is periodic
with period 2L. Then f has a Fourier series as stated previously whose
coefficients are given by the Euler▯Fourier formulas. The Fourier series
converge to f (x) at all points where f is continuous, and to
lim f (x) lim f (x) /2
x→c − x→c+
at every point c where f is discontinuous.
Comment: As seen before, the fact that f is piecewise continuous guarantees
that the Fourier coefficients can be found. The condition that f ′ is also
piecewise continuous is a sufficient condition to guarantee that the series
thusly found will be convergent everywhere on the real line. As well, recall
that, suppose f is continuous at c, then by definition f(c) equals both one▯
sided limits of f(x) as x approaches c. Therefore, the second part of the
theorem could be even more succinctly stated as that the Fourier series
representing f will always converge to
x→c − (x)+ lix→c+(x) /2
at every point c (and not just at discontinuities of f ).
A consequence of this theorem is that the Fourier series of f will “fill in” any
removable discontinuity the original function might have. A Fourier series
will not have any removable▯type discontinuity.
© 2008, 2012 Zachary S Tseng E▯2 ▯ 10 Example: Let us revisit the earlier calculation of the Fourier series
representing f(x) = x, −2 < x < 2, f(x + 4) = f(x).
The Fourier series, as we have found, is
∞ n+1
f (x) = 4 (−1) sinnπ x
π n=1 n 2 .
The following figures are the graphs of various finite n▯th partial sums of the
series above.
n = 3
n = 10
© 2008, 2012 Zachary S Tseng E▯2 ▯ 11 n = 20
n = 30
n = 50
© 2008, 2012 Zachary S Tseng E▯2 ▯ 12 Note that superimposed sinusoidal curves take on the general shape of the
piecewise continuous periodic function f (x) almost immediately. As well,
for the parts of the curve where f(x) is continuous (where the Fourier
Convergence Theorem predicts a perfect match) the composite curve of the
Fourier series converges rapidly to that of f (x), as predicted. The
convergence is not as rapidly near the jump discontinuities. Indeed, for all
but the lowest partial sums of the Fourier series, the curve seems to
“overshoot” that of f (x) near each jump discontinuity by a noticeable margin.
Further more, this discrepancy does not fade away for any finitely larger n.
That is, the convergence of a Fourier series, while predictable, is not uniform.
(That is a small price we pay for approximating a piecewise continuous
periodic function by sinusoidal curves. It can be done, but the Fourier series
does not converge uniformly to the actual function.)
This behavior is known as the Gibbs Phenomenon. It further states that the
partial sums of a Fourier series will overshoot a jump discontinuity by an
amount approximately equal to 9% of the jump. That is, near each jump
discontinuity, the overshoot amounts to about
0.09 lim f +x)− lim f (x)− ,
x →c x →c
for large n. Further, this overshoot does not go away for any finitely large n.
© 2008, 2012 Zachary S Tseng E▯2 ▯ 13 Question: Sketch the graph of the Fourier series of
f(x) = x, −2 < x < 2, f(x + 4) = f(x).
We have seen a few graphs of its partial sums. But what will the graph of
the actual Fourier series look like?
Example: Sketch the graph of the Fourier series of
f(x) = | x|, −2 < x < 2, f(x + 4) = f(x).
Example: Sketch the graph of the Fourier series of
− 2, −1≤ x < 0
f (x)= , f(x + 2) = f(x).
2, 0 ≤ x <1
© 2008, 2012 Zachary S Tseng E▯2 ▯ 14 Even and Odd Functions
Recall that an even function is any function f such that
f(−x) = f(x), for all x in its domain.
2 4 6 −2 −4
Examples: cos(x), sec(x), any constant function, x , x , x , …

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