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CHAPTER 4
FOURIER SERIES AND INTEGRALS
4.1 FOURIER SERIES FOR PERIODIC FUNCTIONS
This section explains three Fourier series: sines, cosines, and exponentials eikx. Square waves (1 or 0 or −1) are great examples, with delta functions in the derivative. We look at a spike, a step function, and a ramp—and smoother functions too.
Start with sin x. It has period 2π since sin(x + 2π) = sin x. It is an odd function since sin(−x) = − sin x, and it vanishes at x = 0 and x = π. Every function sin nx has those three properties, and Fourier looked at infinite combinations of the sines:
Fourier sine series S(x) = b 1 sin x + b 2 sin 2x + b 3 sin 3x + · · · =
∑^ ∞
n=
bn sin nx (1)
If the numbers b 1 , b 2 ,... drop off quickly enough (we are foreshadowing the im- portance of the decay rate) then the sum S(x) will inherit all three properties:
Periodic S(x + 2π) = S(x) Odd S(−x) = −S(x) S(0) = S(π) = 0
200 years ago, Fourier startled the mathematicians in France by suggesting that any function S(x) with those properties could be expressed as an infinite series of sines. This idea started an enormous development of Fourier series. Our first step is to compute from S(x) the number bk that multiplies sin kx.
Suppose S(x) =
bn sin nx. Multiply both sides by sin kx. Integrate from 0 to π: ∫ (^) π
0
S(x) sin kx dx =
∫ (^) π
0
b 1 sin x sin kx dx + · · · +
∫ (^) π
0
bk sin kx sin kx dx + · · · (2)
On the right side, all integrals are zero except the highlighted one with n = k. This property of “orthogonality” will dominate the whole chapter. The sines make 90 ◦^ angles in function space, when their inner products are integrals from 0 to π:
Orthogonality
∫ (^) π
0
sin nx sin kx dx = 0 if n = k. (3)
318 Chapter 4 Fourier Series and Integrals
Zero comes quickly if we integrate
cos mx dx =
[ (^) sin mx m
]π 0 = 0^ −^ 0. So we use this:
Product of sines sin nx sin kx =
cos(n − k)x −
cos(n + k)x. (4)
Integrating cos mx with m = n − k and m = n + k proves orthogonality of the sines.
The exception is when n = k. Then we are integrating (sin kx)^2 = 12 − 12 cos 2kx: ∫ (^) π
0
sin kx sin kx dx =
∫ (^) π
0
dx −
∫ (^) π
0
cos 2kx dx =
π 2
The highlighted term in equation (2) is bkπ/ 2. Multiply both sides of (2) by 2/π:
Sine coefficients S(−x) = −S(x)
bk =
π
∫ (^) π
0
S(x) sin kx dx =
π
∫ (^) π
−π
S(x) sin kx dx. (6)
Notice that S(x) sin kx is even (equal integrals from −π to 0 and from 0 to π).
I will go immediately to the most important example of a Fourier sine series. S(x) is an odd square wave with SW (x) = 1 for 0 < x < π. It is drawn in Figure 4.1 as an odd function (with period 2π) that vanishes at x = 0 and x = π.
(^) x
SW (x) = 1
−π 0 π^2 π
Figure 4.1: The odd square wave with SW (x + 2π) = SW (x) = {1 or 0 or − 1 }.
Example 1 Find the Fourier sine coefficients bk of the square wave SW (x).
Solution For k = 1, 2 ,... use the first formula (6) with S(x) = 1 between 0 and π:
bk =
π
∫ (^) π
0
sin kx dx =
π
[
− cos kx k
]π
0
π
The even-numbered coefficients b 2 k are all zero because cos 2kπ = cos 0 = 1. The odd-numbered coefficients bk = 4/πk decrease at the rate 1/k. We will see that same 1 /k decay rate for all functions formed from smooth pieces and jumps.
Put those coefficients 4/πk and zero into the Fourier sine series for SW (x):
Square wave SW (x) =
π
[
sin x 1
sin 3x 3
sin 5x 5
sin 7x 7
]
Figure 4.2 graphs this sum after one term, then two terms, and then five terms. You can see the all-important Gibbs phenomenon appearing as these “partial sums”
320 Chapter 4 Fourier Series and Integrals
Every cosine has period 2π. Figure 4.3 shows two even functions, the repeating ramp RR(x) and the up-down train UD(x) of delta functions. That sawtooth ramp RR is the integral of the square wave. The delta functions in UD give the derivative of the square wave. (For sines, the integral and derivative are cosines.) RR and UD will be valuable examples, one smoother than SW , one less smooth.
First we find formulas for the cosine coefficients a 0 and ak. The constant term a 0 is the average value of the function C(x):
a 0 = Average a 0 =
π
∫ (^) π
0
C(x) dx =
2 π
∫ (^) π
−π
C(x) dx. (11)
I just integrated every term in the cosine series (10) from 0 to π. On the right side, the integral of a 0 is a 0 π (divide both sides by π). All other integrals are zero:
∫ (^) π
0
cos nx dx =
[
sin nx n
]π
0
In words, the constant function 1 is orthogonal to cos nx over the interval [0, π].
The other cosine coefficients ak come from the orthogonality of cosines. As with sines, we multiply both sides of (10) by cos kx and integrate from 0 to π:
∫ (^) π
0
C(x) cos kx dx =
∫ (^) π
0
a 0 cos kx dx+
∫ (^) π
0
a 1 cos x cos kx dx+··+
∫ (^) π
0
ak(cos kx)^2 dx+··
You know what is coming. On the right side, only the highlighted term can be nonzero. Problem 4.1.1 proves this by an identity for cos nx cos kx—now (4) has a plus sign. The bold nonzero term is akπ/ 2 and we multiply both sides by 2/π:
Cosine coefficients C(−x) = C(x) ak^ =
π
∫ (^) π
0
C(x) cos kx dx =
π
∫ (^) π
−π
C(x) cos kx dx. (13)
Again the integral over a full period from −π to π (also 0 to 2π) is just doubled.
(^) x −π 0 π^2 π
RR(x) = |x|
Repeating Ramp RR(x) Integral of Square Wave
−π x 0 π^2 π
− 2 δ(x + π)
2 δ(x)
− 2 δ(x − π)
2 δ(x − 2 π)
Up-down U D(x)
Figure 4.3: The repeating ramp RR and the up-down UD (periodic spikes) are even. The derivative of RR is the odd square wave SW. The derivative of SW is U D.
4.1 Fourier Series for Periodic Functions 321
Example 2 Find the cosine coefficients of the ramp RR(x) and the up-down UD(x).
Solution The simplest way is to start with the sine series for the square wave:
SW (x) =
π
[
sin x 1
sin 3x 3
sin 5x 5
sin 7x 7
]
Take the derivative of every term to produce cosines in the up-down delta function:
Up-down series UD(x) =
π
[cos x + cos 3x + cos 5x + cos 7x + · · · ]. (14)
Those coefficients don’t decay at all. The terms in the series don’t approach zero, so officially the series cannot converge. Nevertheless it is somehow correct and important. Unofficially this sum of cosines has all 1 ’s at x = 0 and all − 1 ’s at x = π. Then +∞ and −∞ are consistent with 2 δ(x) and − 2 δ(x − π). The true way to recognize δ(x) is by the test
δ(x)f (x) dx = f (0) and Example 3 will do this. For the repeating ramp, we integrate the square wave series for SW (x) and add the average ramp height a 0 = π/ 2 , halfway from 0 to π:
Ramp series RR(x) =
π 2
π 4
[
cos x 12
cos 3x 32
cos 5x 52
cos 7x 72
]
The constant of integration is a 0. Those coefficients ak drop off like 1 /k^2. They could be computed directly from formula (13) using
x cos kx dx, but this requires an integration by parts (or a table of integrals or an appeal to Mathematica or Maple). It was much easier to integrate every sine separately in SW (x), which makes clear the crucial point: Each “degree of smoothness” in the function is reflected in a faster decay rate of its Fourier coefficients ak and bk.
No decay Delta functions (with spikes) 1 /k decay Step functions (with jumps) 1 /k^2 decay Ramp functions (with corners) 1 /k^4 decay Spline functions (jumps in f ′′′) rk^ decay with r < 1 Analytic functions like 1/(2 − cos x)
Each integration divides the kth coefficient by k. So the decay rate has an extra 1 /k. The “Riemann-Lebesgue lemma” says that ak and bk approach zero for any continuous function (in fact whenever
|f (x)|dx is finite). Analytic functions achieve a new level of smoothness—they can be differentiated forever. Their Fourier series and Taylor series in Chapter 5 converge exponentially fast.
The poles of 1/(2 − cos x) will be complex solutions of cos x = 2. Its Fourier series converges quickly because rk^ decays faster than any power 1/kp. Analytic functions are ideal for computations—the Gibbs phenomenon will never appear.
Now we go back to δ(x) for what could be the most important example of all.
4.1 Fourier Series for Periodic Functions 323
−π 0 π
δ 5 (x)
δ 10 (x)
height 11/ 2 π
height 21/ 2 π
height − 1 / 2 π
height 1/ 2 π
Figure 4.4: The sums δN (x) = (1 + 2 cos x + · · · + 2 cos Nx)/ 2 π try to approach δ(x).
Complete Series: Sines and Cosines
Over the half-period [0, π], the sines are not orthogonal to all the cosines. In fact the integral of sin x times 1 is not zero. So for functions F (x) that are not odd or even, we move to the complete series (sines plus cosines) on the full interval. Since our functions are periodic, that “full interval” can be [−π, π] or [0, 2 π]:
Complete Fourier series F (x) = a 0 +
∑^ ∞
n=
an cos nx +
∑^ ∞
n=
bn sin nx. (19)
On every “2π interval” all sines and cosines are mutually orthogonal. We find the Fourier coefficients ak and bk in the usual way: Multiply (19) by 1 and cos kx and sin kx, and integrate both sides from −π to π:
a 0 =
2 π
∫ (^) π
−π
F (x) dx ak =
π
∫ (^) π
−π
F (x) cos kx dx bk =
π
∫ (^) π
−π
F (x) sin kx dx. (20)
Orthogonality kills off infinitely many integrals and leaves only the one we want.
Another approach is to split F (x) = C(x) + S(x) into an even part and an odd part. Then we can use the earlier cosine and sine formulas. The two parts are
C(x) = Feven(x) =
F (x) + F (−x) 2
S(x) = Fodd(x) =
F (x) − F (−x) 2
The even part gives the a’s and the odd part gives the b’s. Test on a short square pulse from x = 0 to x = h—this one-sided function is not odd or even.
324 Chapter 4 Fourier Series and Integrals
Example 4 Find the a’s and b’s if F (x) = square pulse =
1 for 0 < x < h 0 for h < x < 2 π
Solution The integrals for a 0 and ak and bk stop at x = h where F (x) drops to zero. The coefficients decay like 1 /k because of the jump at x = 0 and the drop at x = h:
Coefficients of square pulse a 0 =
2 π
∫ (^) h
0
1 dx =
h 2 π
= average
ak =
π
∫ (^) h
0
cos kx dx =
sin kh πk
bk =
π
∫ (^) h
0
sin kx dx =
1 − cos kh πk
If we divide F (x) by h, its graph is a tall thin rectangle: height (^) h^1 , base h, and area = 1.
When h approaches zero, F (x)/h is squeezed into a very thin interval. The tall rectangle approaches (weakly) the delta function δ(x). The average height is area/ 2 π = 1 / 2 π. Its other coefficients ak/h and bk/h approach 1 /π and 0 , already known for δ(x):
F (x) h
→ δ(x)
ak h
π
sin kh kh
π
and
bk h
1 − cos kh πkh
→ 0 as h → 0. (23)
When the function has a jump, its Fourier series picks the halfway point. This example would converge to F (0) = 12 and F (h) = 12 , halfway up and halfway down.
The Fourier series converges to F (x) at each point where the function is smooth. This is a highly developed theory, and Carleson won the 2006 Abel Prize by proving convergence for every∫ x except a set of measure zero. If the function has finite energy |F (x)|^2 dx, he showed that the Fourier series converges “almost everywhere.”
Energy in Function = Energy in Coefficients
There is an extremely important equation (the energy identity) that comes from integrating (F (x))^2. When we square the Fourier series of F (x), and integrate from −π to π, all the “cross terms” drop out. The only nonzero integrals come from 1^2 and cos^2 kx and sin^2 kx, multiplied by a^20 and a^2 k and b^2 k:
Energy in F (x) =
∫ (^) π −π(a^0 +^
ak cos kx +
bk sin kx)^2 dx ∫ (^) π −π(F^ (x))
(^2) dx = 2πa 2 0 +^ π(a
2 1 +^ b
2 1 +^ a
2 2 +^ b
2 2 +^ · · ·^ ).^
The energy in F (x) equals the energy in the coefficients. The left side is like the length squared of a vector, except the vector is a function. The right side comes from an infinitely long vector of a’s and b’s. The lengths are equal, which says that the Fourier transform from function to vector is like an orthogonal matrix. Normalized by constants
2 π and
π, we have an orthonormal basis in function space. What is this function space? It is like ordinary 3-dimensional space, except the “vectors” are functions. Their length ‖f ‖ comes from integrating instead of adding: ‖f ‖^2 =
|f (x)|^2 dx. These functions fill Hilbert space. The rules of geometry hold:
326 Chapter 4 Fourier Series and Integrals
Notice that c 0 = a 0 is still the average of F (x), because e^0 = 1. The orthogonality of einx^ and eikx^ is checked by integrating, as always. But the complex inner product (F, G) takes the complex conjugate G of G. Before integrating, change eikx^ to e−ikx:
Complex inner product Orthogonality of einx^ and eikx
(F, G) =
∫ (^) π
−π
F (x)G(x) dx
∫ (^) π
−π
ei(n−k)xdx =
[
ei(n−k)x i(n − k)
]π
−π
Example 5 Add the complex series for 1 /(2 − eix) and 1 /(2 − e−ix). These geometric series have exponentially fast decay from 1 / 2 k. The functions are analytic. ( 1 2
eix 4
e^2 ix 8
e−ix 4
e−^2 ix 8
cos x 2
cos 2x 4
cos 3x 8
When we add those functions, we get a real analytic function:
1 2 − eix^
2 − e−ix^
(2 − e−ix) + (2 − eix) (2 − eix)(2 − e−ix)
4 − 2 cos x 5 − 4 cos x
This ratio is the infinitely smooth function whose cosine coefficients are 1 / 2 k.
Example 6 Find ck for the 2 π-periodic shifted pulse F (x) =
1 for s ≤ x ≤ s + h 0 elsewhere in [−π, π]
Solution The integrals (26) from −π to π become integrals from s to s + h:
ck =
2 π
∫ (^) s + h
s
1 · e−ikx^ dx =
2 π
[
e−ikx −ik
]s + h
s
= e−iks
1 − e−ikh 2 πik
Notice above all the simple effect of the shift by s. It “modulates” each ck by e−iks. The energy is unchanged, the integral of |F |^2 just shifts, and all |e−iks| = 1:
Shift F (x) to F (x − s) ←→ Multiply ck by e−iks. (30)
Example 7 Centered pulse with shift s = −h/ 2. The square pulse becomes centered around x = 0. This even function equals 1 on the interval from −h/ 2 to h/ 2 :
Centered by s = − h 2 ck = eikh/^2
1 − e−ikh 2 πik
2 π
sin(kh/2) k/ 2
Divide by h for a tall pulse. The ratio of sin(kh/2) to kh/ 2 is the sinc function:
Tall pulse
Fcentered h
2 π
∑^ ∞
−∞
sinc
kh 2
eikx^ =
1 /h for − h/ 2 ≤ x ≤ h/ 2 0 elsewhere in [−π, π]
That division by h produces area = 1. Every coefficient approaches (^21) π as h → 0. The Fourier series for the tall thin pulse again approaches the Fourier series for δ(x).
4.1 Fourier Series for Periodic Functions 327
Hilbert space can contain vectors c = (c 0 , c 1 , c− 1 , c 2 , c− 2 , · · · ) instead of functions F (x). The length of c is 2π
|ck|^2 =
|F |^2 dx. The function space is often denoted by L^2 and the vector space is ^2. The energy identity is trivial (but deep). Integrating the Fourier series for F (x) times F (x), orthogonality kills every cnck for n = k. This leaves the ckck = |ck|^2 :
∫ (^) π
−π
|F (x)|^2 dx =
∫ (^) π
−π
cneinx)(
cke−ikx)dx = 2π(|c 0 |^2 + |c 1 |^2 + |c− 1 |^2 + ··). (31)
This is Plancherel’s identity: The energy in x-space equals the energy in k-space.
Finally I want to emphasize the three big rules for operating on F (x) =
ckeikx:
- The derivative
dF dx
has Fourier coefficients ikck (energy moves to high k).
- The integral of F (x) has Fourier coefficients
ck ik
, k = 0 (faster decay).
- The shift to F (x−s) has Fourier coefficients e−iksck (no change in energy).
Application: Laplace’s Equation in a Circle
Our first application is to Laplace’s equation. The idea is to construct u(x, y) as an infinite series, choosing its coefficients to match u 0 (x, y) along the boundary. Every- thing depends on the shape of the boundary, and we take a circle of radius 1.
Begin with the simple solutions 1, r cos θ, r sin θ, r^2 cos 2θ, r^2 sin 2θ, ... to Laplace’s equation. Combinations of these special solutions give all solutions in the circle:
u(r, θ) = a 0 + a 1 r cos θ + b 1 r sin θ + a 2 r^2 cos 2θ + b 2 r^2 sin 2θ + · · · (32)
It remains to choose the constants ak and bk to make u = u 0 on the boundary. For a circle u 0 (θ) is periodic, since θ and θ + 2π give the same point:
Set r = 1 u 0 (θ) = a 0 + a 1 cos θ + b 1 sin θ + a 2 cos 2θ + b 2 sin 2θ + · · · (33)
This is exactly the Fourier series for u 0. The constants ak and bk must be the Fourier coefficients of u 0 (θ). Thus the problem is completely solved, if an infinite series (32) is acceptable as the solution.
Example 8 Point source u 0 = δ(θ) at θ = 0 The whole boundary is held at u 0 = 0, except for the source at x = 1, y = 0. Find the temperature u(r, θ) inside.
Fourier series for δ u 0 (θ) =
2 π
π
(cos θ + cos 2θ + cos 3θ + · · · ) =
2 π
∑^ ∞
−∞
einθ
4.1 Fourier Series for Periodic Functions 329
WORKED EXAMPLE
A hot metal bar is moved into a freezer (zero temperature). The sides of the bar are coated so that heat only escapes at the ends. What is the temperature u(x, t) along the bar at time t? It will approach u = 0 as all the heat leaves the bar.
Solution The heat equation is ut = uxx. At t = 0 the whole bar is at a constant temperature, say u = 1. The ends of the bar are at zero temperature for all time t > 0. This is an initial-boundary value problem:
Heat equation ut = uxx with u(x, 0) = 1 and u(0, t) = u(π, t) = 0. (40)
Those zero boundary conditions suggest a sine series. Its coefficients depend on t:
Series solution of the heat equation u(x, t) =
∑^ ∞
1
bn(t) sin nx. (41)
The form of the solution shows separation of variables. In a comment below, we look for products A(x) B(t) that solve the heat equation and the boundary conditions. What we reach is exactly A(x) = sin nx and the series solution (41).
Two steps remain. First, choose each bn(t) sin nx to satisfy the heat equation:
Substitute into ut = uxx b (^) n′(t) sin nx = −n^2 bn(t) sin nx bn(t) = e−n
(^2) t bn(0).
Notice b (^) n′ = −n^2 bn. Now determine each bn(0) from the initial condition u(x, 0) = 1 on (0, π). Those numbers are the Fourier sine coefficients of SW (x) in equation (38):
Box function/square wave
∑^ ∞
1
bn(0) sin nx = 1 bn(0) =
πn
for odd n
This completes the series solution of the initial-boundary value problem:
Bar temperature u(x, t) =
odd n
πn
e−n
(^2) t sin nx. (42)
For large n (high frequencies) the decay of e−n
(^2) t is very fast. The dominant term (4/π)e−t^ sin x for large times will come from n = 1. This is typical of the heat equation and all diffusion, that the solution (the temperature profile) becomes very smooth as t increases.
Numerical difficulty I regret any bad news in such a beautiful solution. To compute u(x, t), we would probably truncate the series in (42) to N terms. When that finite series is graphed on the website, serious bumps appear in uN (x, t). You ask if there is a physical reason but there isn’t. The solution should have maximum temperature at the midpoint x = π/2, and decay smoothly to zero at the ends of the bar.
330 Chapter 4 Fourier Series and Integrals
Those unphysical bumps are precisely the Gibbs phenomenon. The initial u(x, 0) is 1 on (0, π) but its odd reflection is −1 on (−π, 0). That jump has produced the slow 4/πn decay of the coefficients, with Gibbs oscillations near x = 0 and x = π. The sine series for u(x, t) is not a success numerically. Would finite differences help?
Separation of variables We found bn(t) as the coefficient of an eigenfunction sin nx. Another good approach is to put u = A(x) B(t) directly into ut = uxx:
Separation A(x) B ′(t) = A ′′(x) B(t) requires
A ′′(x) A(x)
B ′(t) B(t)
= constant. (43)
A ′′/A is constant in space, B ′/B is constant in time, and they are equal:
A ′′ A
= −λ gives A = sin
λ x and cos
λ x
B ′
B
= −λ gives B = e−λt
The products AB = e−λt^ sin
λ x and e−λt^ cos
λ x solve the heat equation for any number λ. But the boundary condition u(0, t) = 0 eliminates the cosines. Then u(π, t) = 0 requires λ = n^2 = 1, 4 , 9 ,... to have sin
λ π = 0. Separation of variables has recovered the functions in the series solution (42).
Finally u(x, 0) = 1 determines the numbers 4/πn for odd n. We find zero for even n because sin nx has n/2 positive loops and n/2 negative loops. For odd n, the extra positive loop is a fraction 1/n of all loops, giving slow decay of the coefficients.
Heat bath (the opposite problem) The solution on the website is 1 − u(x, t), because it solves a different problem. The bar is initially frozen at U(x, 0) =
- It is placed into a heat bath at the fixed temperature U = 1 (or U = T 0 ). The new unknown is U and its boundary conditions are no longer zero.
The heat equation and its boundary conditions are solved first by UB (x, t). In this example UB ≡ 1 is constant. Then the difference V = U − UB has zero boundary values, and its initial values are V = −1. Now the eigenfunction method (or sepa- ration of variables) solves for V. (The series in (42) is multiplied by −1 to account for V (x, 0) = −1.) Adding back UB solves the heat bath problem: U = UB + V = 1 − u(x, t).
Here UB ≡ 1 is the steady state solution at t = ∞, and V is the transient solution. The transient starts at V = −1 and decays quickly to V = 0.
Heat bath at one end The website problem is different in another way too. The Dirichlet condition u(π, t) = 1 is replaced by the Neumann condition u ′(1, t) = 0. Only the left end is in the heat bath. Heat flows down the metal bar and out at the far end, now located at x = 1. How does the solution change for fixed-free?
Again UB = 1 is a steady state. The boundary conditions apply to V = 1 − UB :
Fixed-free eigenfunctions V^ (0) = 0^ and^ V^
′(1) = 0 lead to A(x) = sin
n +
πx. (44)
332 Chapter 4 Fourier Series and Integrals
5 Plot the first three partial sums and the function itself:
x(π − x) =
π
sin x 1
sin 3x 27
sin 5x 125
, 0 < x < π.
Why is 1/k^3 the decay rate for this function? What is the second derivative?
6 What constant function is closest in the least square sense to f = cos^2 x? What multiple of cos x is closest to f = cos^3 x?
7 Sketch the 2π-periodic half wave with f (x) = sin x for 0 < x < π and f (x) = 0 for −π < x < 0. Find its Fourier series.
8 (a) Find the lengths of the vectors u = (1, 12 , 14 , 18 ,.. .) and v = (1, 13 , 19 ,.. .) in Hilbert space and test the Schwarz inequality |uTv|^2 ≤ (uTu)(vTv). (b) For the functions f = 1 + 12 eix^ + 14 e^2 ix^ + · · · and g = 1 + 13 eix^ + 19 e^2 ix^ + · · · use part (a) to find the numerical value of each term in ∣ ∣∣ ∣
∫ (^) π
−π
f (x) g(x) dx
2 ≤
∫ (^) π
−π
|f (x)|^2 dx
∫ (^) π
−π
|g(x)|^2 dx.
Substitute for f and g and use orthogonality (or Parseval).
9 Find the solution to Laplace’s equation with u 0 = θ on the boundary. Why is this the imaginary part of 2(z − z^2 /2 + z^3 / 3 · · · ) = 2 log(1 + z)? Confirm that on the unit circle z = eiθ, the imaginary part of 2 log(1 + z) agrees with θ.
10 If the boundary condition for Laplace’s equation is u 0 = 1 for 0 < θ < π and u 0 = 0 for −π < θ < 0, find the Fourier series solution u(r, θ) inside the unit circle. What is u at the origin?
11 With boundary values u 0 (θ) = 1 + 12 eiθ^ + 14 e^2 iθ^ + · · · , what is the Fourier series solution to Laplace’s equation in the circle? Sum the series.
12 (a) Verify that the fraction in Poisson’s formula satisfies Laplace’s equation.
(b) What is the response u(r, θ) to an impulse at the point (0, 1), at the angle ϕ = π/2? (c) If u 0 (ϕ) = 1 in the quarter-circle 0 < ϕ < π/2 and u 0 = 0 elsewhere, show that at points on the horizontal axis (and especially at the origin)
u(r, 0) =
2 π
tan−^1
1 − r^2 − 2 r
by using
∫ dϕ b + c cos ϕ
b^2 − c^2
tan−^1
b^2 − c^2 sin ϕ c + b cos ϕ
4.1 Fourier Series for Periodic Functions 333
13 When the centered square pulse in Example 7 has width h = π, find
(a) its energy
|F (x)|^2 dx by direct integration (b) its Fourier coefficients ck as specific numbers (c) the sum in the energy identity (31) or (24)
If h = 2π, why is c 0 = 1 the only nonzero coefficient? What is F (x)?
14 In Example 5, F (x) = 1+(cos x)/2+· · ·+(cos nx)/ 2 n^ +· · · is infinitely smooth:
(a) If you take 10 derivatives, what is the Fourier series of d^10 F/dx^10? (b) Does that series still converge quickly? Compare n^10 with 2n^ for n^1024.
15 (A touch of complex analysis) The analytic function in Example 5 blows up when 4 cos x = 5. This cannot happen for real x, but equation (28) shows blowup if eix^ = 2 or 12. In that case we have poles at x = ±i log 2. Why are there also poles at all the complex numbers x = ±i log 2 + 2πn?
16 (A second touch) Change 2’s to 3’s so that equation (28) has 1/(3 − eix) + 1 /(3 − e−ix). Complete that equation to find the function that gives fast decay at the rate 1/ 3 k.
17 (For complex professors only) Change those 2’s and 3’s to 1’s:
1 1 − eix^
1 − e−ix^
(1 − e−ix) + (1 − eix) (1 − eix)(1 − e−ix)
2 − eix^ − e−ix 2 − eix^ − e−ix^
A constant! What happened to the pole at eix^ = 1? Where is the dangerous series (1 + eix^ + · · · ) + (1 + e−ix^ + · · · ) = 2 + 2 cos x + · · · involving δ(x)?
18 Following the Worked Example, solve the heat equation ut = uxx from a point source u(x, 0) = δ(x) with free boundary conditions u ′(π, t) = u ′(−π, t) = 0. Use the infinite cosine series for δ(x) with time decay factors bn(t).
