Davidson K.R., Donsig A.P. Real Analysis with Real Applications

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13.4 Convergence in the Open Disk 433

Now we extend this argument to apply to the various partial derivatives of u.

This procedure justiﬁes term-by-term differentiation under appropriate condi-

tions on the convergence.

13.4.2. LEMMA. Suppose that u

(x, y) are C

functions on an open set R for

n ≥ 0 such that

∞

n=0

(x, y) converges uniformly to u(x, y) and

∞

n=0

∂

∂x

(x, y)

converges uniformly to v(x, y). Then

∂

∂x

u(x, y) = v(x, y).

PROOF. It is enough to verify the theorem on a small square about an arbitrary

point (x

, y

) in R. By limiting ourselves to a square, the whole line segment from

, y) to (x, y) will lie in R. We deﬁne functions on this small square by

(x, y) =

k=0

(x, y) =

k=0

, y) +

∂

∂x

k=0

(t, y) dt

and

w(x, y) = u(x

, y) +

v(t, y) dt.

Then since the integrands

∂

∂x

k=0

(t, y) converge uniformly to v(t, y), Corol-

lary 8.3.2 shows that w

(x, y) converges uniformly to w(x, y). But this limit is

u(x, y). Therefore, by the Fundamental Theorem of Calculus,

∂

∂x

u(x, y) =

∂

∂x

w(x, y)

∂

∂x

u(x

, y) +

v(t, y) dt

(x, y) = v(x, y).

We may apply this to our function u(r, θ).

13.4.3. THEOREM. Let f be an absolutely integrable function with associated

function u(r, θ) = A

∞

n=1

cosnθ + B

sinnθ. Then u satisﬁes the

heat equation ∆u(r, θ) = 0 in the open disk D.

PROOF. Let u

(r, θ) = A

cosnθ + B

sinnθ. Then

∂

∂r

(r, θ) = nA

n−1

cosnθ + nB

n−1

sinnθ

∂

∂r

(r, θ) = n(n − 1)A

n−2

cosnθ + n(n − 1)B

n−2

sinnθ

∂

∂θ

(r, θ) = −nA

sinnθ + nB

cosnθ

∂

∂θ

(r, θ) = −n

cosnθ − n

sinnθ.

We will apply the M-test to the series

∞

n=0

(r, θ) and to each series of partial

derivatives. Indeed, on the disk

for 0 ≤ R < 1, each of the preceding terms is

434 Fourier Series and Physics

uniformly bounded by 4kfk

n−2

. Apply the Ratio Test (Exercise 3.2.I) to the

series

∞

n=0

4kfk

n−2

lim

n→∞

4kfk

(n + 1)

n−1

4kfk

n−2

= R < 1.

Thus this series converges; and therefore it follows from the Weierstrass M-test

that each series of partial derivatives converges uniformly on

Therefore, by Lemma 13.4.2, the partial derivative of the sum equals the sum

of the partial derivatives. This means that

∂

∂r

u(r, θ) =

∞

n=0

∂

∂r

(r, θ)

∂

∂r

u(r, θ) =

∞

n=0

∂

∂r

(r, θ)

∂

∂θ

u(r, θ) =

∞

n=0

∂

∂θ

(r, θ)

∂

∂θ

u(r, θ) =

∞

n=0

∂

∂θ

(r, θ).

This convergence is uniform on every disk

for R < 1.

Using ∆u = u

θθ

, we deduce from Lemma 13.4.2 that

∆u(r, θ) =

∞

n=0

∆u

(r, θ)

and that this convergence is uniform on any disk

. However, we constructed the

functions r

cosnθ and r

sinnθ as solutions of ∆u = 0. Thus the right-hand side

of this equation is zero. Hence ∆u(r, θ) = 0 everywhere in the open disk D. ¥

13.4.4. DEFINITION. A function u such that ∆u = 0 is called a harmonic

function. The function u(r, θ) determined by the Fourier series of a 2π-periodic f

is called the harmonic extension of f.

Exercises for Section 13.4

A. Find the Fourier series of f (θ) = θ

for −π ≤ θ ≤ π.

(a) Show that the series for u(r, θ) converges uniformly on the closed disk D.

(b) Show that the series for u

θθ

does not converge uniformly on D.

B. Let A

and B

be the Fourier coefﬁcients of a continuous function f (θ).

(a) Show that v

(r, θ) = u(rt, θ) = A

∞

n=1

(rt)

cosnθ + B

(rt)

sinnθ

converges uniformly on the closed disk D for 0 < t < 1.

(b) Show that v

converges uniformly to u on D as t → 1

−

if and only if u extends to

a continuous function on D.

13.5 The Poisson Formula 435

C. (a) Suppose that a 2π-periodic function f is C

(i.e., f

is continuous), then the Fourier

coefﬁcients of f satisfy |A

| + |B

| ≤ Cn

−1

for some constant C.

HINT: Integrate by parts and compare with Lemma 13.3.1.

(b) Show by induction that if f is C

, then |A

|+ |B

| ≤ Cn

−k

for some constant C.

D. Suppose f ∼ A

n≥1

cosnθ + B

sinnθ is a 2π-periodic function and that

n≥1

| + |B

| < ∞. Prove that the Fourier series converges uniformly to f.

E. Suppose f ∼ A

n≥1

cosnθ + B

sinnθ is a 2π-periodic function and that

n≥1

n|A

| + n|B

| < ∞. By modifying the proof of Lemma 13.4.2, establish that

n≥1

−nA

sinnθ + nB

cosnθ converges uniformly to f

F. The Fourier coefﬁcients of a 2π-periodic function f satisfy |A

| + |B

| ≤ Cn

−k

for

an integer k ≥ 2 and some constant C. Show that f is in the class C

k−2

HINT: Use term-by-term differentiation and the M-test.

G. Show that u(r, θ) has partial derivatives

∂

j+k

∂r

∂θ

u of all orders. Hence show that u is

∞

HINT: First verify this for the functions u

(r, θ). Then show that the sequence of

partial derivatives converges uniformly on each D

H. Give necessary and sufﬁcient conditions for A

∞

n=1

cosnθ + B

sinnθ to be

the Fourier series of a C

∞

function.

HINT: Combine Exercises C and E.

13.5. The Poisson Formula

The next step is to ﬁnd a formula for u(r, θ) in terms of the boundary function

f(θ). The basic idea is to substitute the formula for each Fourier coefﬁcient and in-

terchange the order of the summation and integration. This again requires uniform

convergence.

Compute u(r, θ) for a bounded integrable function f:

u(r, θ) = A

∞

n=1

cosnθ + B

sinnθ

2π

−π

f(t) dt +

∞

n=1

−π

f(t) cosnt dt r

cosnθ+

−π

f(t) sinnt dt r

sinnθ

2π

−π

f(t) dt +

∞

n=1

−π

f(t)r

cosnt cosnθ + sin nt sinnθ

2π

−π

f(t) dt +

∞

n=1

−π

f(t)r

cosn(θ − t) dt.

436 Fourier Series and Physics

Yet again, we apply the M-test. Since

kf(t)r

cosn(θ − t)k ≤ kfk

∞

for 0 ≤ r ≤ R,

and

∞

n=0

kfk

∞

= kfk

∞

/(1 − R) < ∞, the series converges uniformly on

Therefore, by Theorem 8.3.1, we can interchange the order of the summation and

the integral

u(r, θ) =

−π

f(t)

2π

1 + 2

∞

n=1

cosn(θ − t)

−π

f(t)P (r, θ − t) dt =

−π

f(θ − u)P (r, u) du.

We have introduced the function

P (r, θ) =

2π

1 + 2

∞

n=1

cosnθ

for 0 ≤ r < 1, θ ∈ R.

Notice that the last step is a change of variables in which we make use of the fact

that both f and P are 2π-periodic in θ.

The function P (r, θ) is known as the Poisson kernel. The purpose of this

section is to develop its basic properties.

We will use the fact that exponentiation and trigonometric functions are related

using complex variables. (See Appendix 13.10.) In particular, e

= cost + i sint

and consequently

cosnt =

int

+ e

−int

and sinnt =

int

− e

−int

13.5.1. THE POISSON FORMULA.

Let f be a bounded integrable function on [−π, π]. Then for 0 ≤ r < 1 and

−π ≤ θ ≤ π, the harmonic extension of f is given by

u(r, θ) =

−π

f(θ − t)P (r, t) dt,

where P (r, t) =

2π

1 − r

1 − 2r cost + r

PROOF. From the discussion preceding the theorem, it only remains to evaluate the

series for P (r, t). Changing the cosines to complex exponentials allows us to sum

a geometric series.

2πP (r, t) = 1 + 2

∞

n=1

cosnt = 1 +

∞

n=1

int

+ e

−int

)

= 1 +

∞

n=1

(re

)

∞

n=1

(re

−it

)

13.5 The Poisson Formula 437

= 1 +

1 − re

−it

1 − re

−it

(1 − re

− re

−it

+ r

) + (re

− r

) + (re

−it

− r

)

(1 − re

)(1 − re

−it

)

1 − r

1 − r(e

+ e

−it

) + r

1 − r

1 − 2r cost + r

The Poisson kernel has a number of very nice properties. The most important

is that P (r, t) is positive and integrates to 1. Hence u(r, θ) is a weighted average

of the values f(θ − t). The function P (r, t) peaks dramatically at t = 0 when r is

close to 1, and thus eventually u(r, θ) depends mostly on the values of f(u) for u

close to θ. See Figure 13.2.

13.5.2. PROPERTIES OF THE POISSON KERNEL.

For 0 ≤ r < 1 and −π ≤ t ≤ π,

(1) P (r, t) > 0.

(2) P (r, −t) = P (r, t).

(3)

−π

P (r, t) dt = 1.

(4) P (r, t) is decreasing in t on [0, π] for ﬁxed r.

(5) For any δ > 0, lim

r→1

−

max

δ≤|t|≤π

P (r, t) = 0.

PROOF. Statements (1) and (2) are routine. For (3), take f = 1. The Fourier series

of f is A

= 1 and A

= B

= 0 for n ≥ 1. Hence u(r, θ) = 1. Plugging this into

the Poisson formula, we obtain

1 = u(r, θ) =

−π

P (r, t) dt.

For ﬁxed r, the numerator 1 − r

of P (r, t) is constant while the denominator

1 − 2r cost + r

is monotone increasing on [0, π]; whence P (r, t) is monotone

decreasing in t. This veriﬁes (4). Finally, from (2) and (4), we see that

lim

r→1

−

max

δ≤|t|≤π

P (r, t) = lim

r→1

−

P (r, δ) = lim

r→1

−

1 − r

1 − 2r cosδ + r

2(1 − cos δ)

= 0.

In the next chapter, we will develop two other integral kernels, the Dirichlet

kernel and the Fej

er kernel. The reader may want to compare their properties and

compare Figures 13.2, 14.2, and 14.4.

438 Fourier Series and Physics

–2 2

FIGURE 13.2. Graphs of P (r, t) for r = 1/2, 3/4, 9/10, and 19/20.

13.6 Poisson’s Theorem 439

Exercises for Section 13.5

A. Compute the Fourier series of f (θ) = P (s, θ) for 0 ≤ s < 1.

B. Find an explicit value of r < 1 for which

.01

−.01

P (r, t) dt > 0.999.

C. Prove that

1 − r

1 + r

≤ 2πP (r, θ) ≤

1 + r

1 − r

D. Let f be a positive continuous 2π-periodic function with harmonic extension u(r, θ).

(a) Prove that u(r, θ) ≥ 0.

(b) Prove Harnack’s inequality:

1 − r

1 + r

u(0, 0) ≤ u(r, θ) ≤

1 + r

1 − r

u(0, 0).

HINT: Use the previous exercise.

E. Suppose that f is a 2π-periodic continuous function such that L ≤ f (θ) ≤ M for

−π ≤ θ ≤ π. Let u(r, θ) be the Poisson extension of f. Show that L ≤ u(r, θ) ≤ M

for 0 ≤ r ≤ 1 and −π ≤ θ ≤ π.

F. Show that if f(θ) is absolutely integrable on [−π, π] and g

(θ) are continuous func-

tions that converge uniformly to g(θ) on [−π, π], then

lim

n→∞

−π

f(θ)g

(θ) dθ =

−π

f(θ)g(θ) dθ.

HINT: Look at the proof of Theorem 8.3.1.

G. Use the previous exercise to show that the Poisson formula is valid for absolutely

integrable functions on [−π, π].

H. Prove that

−π

P (r, θ − t)P (s, t) dt = P (rs, θ).

HINT: Use the series expansion of the Poisson kernel.

I. Let f(θ) be a continuous 2π-periodic function. Let u(r, θ) be the harmonic extension

of f. Let 0 < s < 1, and deﬁne g(θ) = u(s, θ). Prove that the harmonic extension of

g is u(rs, θ).

HINT: Use the Poisson formula twice to obtain theharmonic extension of g as a double

integral, and interchange the order of integration.

13.6. Poisson’s Theorem

Using the properties of the Poisson kernel, it is now possible to show that

u(r, θ) approaches f(θ) uniformly as r tends to 1. This means that our proposed

solution to the heat problem is continuous on the closed disk and has the desired

boundary values. This puts us very close to solving the steady-state heat problem.

It also provides a stronger reason for calling u(r, θ) an extension of f.

440 Fourier Series and Physics

13.6.1. POISSON’S THEOREM.

Let f be a continuous 2π-periodic function on [−π, π] with harmonic extension

u(r, θ). Then f

(θ) := u(r, θ) converges uniformly to f on [−π, π] as r → 1

−

Consequently, the function u(r, θ) may be deﬁned on the boundary of the closed

disk by u(1, θ) = f(θ) to obtain a continuous function on D that is harmonic on D

and agrees with f on the boundary.

PROOF. Let ε > 0 be given. We will ﬁnd an r

< 1 so that

|f(θ) − f

(θ)| ≤ ε for all r

≤ r < 1, −π ≤ θ ≤ π,

which will establish uniform convergence.

Let M = kf k

∞

and set ε

= (4πM +1)

−1

ε. By Theorem 5.5.9, f is uniformly

continuous. Therefore, there is a δ > 0 such that

|f(θ) − f(t)| ≤ ε

for all |θ − t| ≤ δ.

By property (5), there is an R < 1 such that

max

δ≤|t|≤π

P (r, t) ≤ ε

for all R ≤ r < 1.

Now, using the Poisson formula, compute

f(θ) − f

(θ) = f (θ)

−π

P (r, t) dt −

−π

f(θ − t)P (r, t) dt

−π

f(θ) − f(θ − t)

P (r, t) dt.

For R ≤ r < 1, split this integral into two pieces to estimate |f (θ) − f

(θ)|

≤

−δ

|f(θ) − f(θ − t)|P (r, t) dt +

−δ

−π

|f(θ) − f(θ − t)|P (r, t) dt

≤

−δ

P (r, t) dt +

−δ

−π

2Mε

≤ ε

+ 2π(2Mε

) = (4πM + 1)ε

= ε.

This establishes that kf − f

∞

< ε for R ≤ r < 1. Hence f

(θ) converges

uniformly to f(θ) on [−π, π].

Extend the deﬁnition of u(r, θ) to the boundary of the closed disk by setting

u(1, θ) = f(θ). The previous paragraphs show that this function is continuous on

the closed disk

D. By Theorem 13.4.3, u is harmonic on the interior of the disk. ¥

This results yields several important consequences quite easily. The ﬁrst is

the existence of a solution to the heat problem. The question of uniqueness of the

solution will be handled in the next section.

13.6.2. COROLLARY. Let f be a continuous 2π-periodic function. Then the

steady-state heat equation ∆u = 0 and u(1, θ) = f(θ) has a continuous solution.

13.6 Poisson’s Theorem 441

PROOF. By Theorem 13.4.3, the function u(r, θ) is differentiable on the open disk

D and satisﬁes ∆u = 0. By Poisson’s Theorem, u extends to be continuous on the

closed unit disk D and attains the boundary values u(1, θ) = f(θ). ¥

This next corollary shows that the Fourier series of a continuous function de-

termines the function uniquely. Continuity is essential for this result. For example,

the function that is zero except for a discontinuous point f(0) = 1 will have the

zero Fourier series but is not the zero function.

13.6.3. COROLLARY. If two continuous 2π-periodic functions on R have equal

Fourier series, then they are equal functions.

PROOF. Suppose that f and g have the same Fourier series. Since the harmonic

extension u(r, θ) is deﬁned only in terms of the Fourier coefﬁcients, this is the same

function for both f and g. Now, by Poisson’s Theorem,

f(θ) = lim

r→1

−

u(r, θ) = g(θ) for − π ≤ θ ≤ π.

Thus f = g. ¥

The ﬁnal application concerns the case in which the Fourier series itself con-

verges uniformly. This is not always the case, and the delicate question of the

convergence of the Fourier series will be dealt with later.

13.6.4. COROLLARY. Suppose that the Fourier series

f ∼ A

∞

n=1

cosnθ + B

sinnθ

of a continuous function f converges uniformly. Then the series in fact converges

uniformly to f.

PROOF. Let the uniform limit be g. By Theorem 8.2.1, g is continuous; and it is

2π-periodic as it is the limit of 2π-periodic functions. Compute the Fourier series

of g using Theorem 8.3.1:

−π

g(t) sinnt dt = lim

N→∞

−π

k=1

coskt + B

sinkt

sinnt dt

= lim

N→∞

= B

The other coefﬁcients are computed in the same way. It follows that f and g have

the same Fourier series. Thus by the previous corollary, they are equal. That is, the

Fourier series converges uniformly to f. ¥

442 Fourier Series and Physics

13.6.5. EXAMPLE. Recall Example 13.3.2. It was shown that

|θ| ∼

−

∞

k=0

cos((2k + 1)θ)

(2k + 1)

Since

∞

k=0

(2k+1)

converges (by the integral test or by comparison with

it follows from the Weierstrass M -test (8.4.7) that this series converges uniformly.

By Corollary 13.6.4, it follows that the Fourier series converges uniformly to |θ|.

Let us evaluate this at θ = 0.

0 =

−

∞

k=0

(2k + 1)

Therefore,

∞

k=0

(2k + 1)

A little manipulation yields that

∞

k=1

∞

k=0

(2k + 1)

∞

k=1

(2k)

∞

k=1

Solving, we obtain a famous identity due to Euler (by different methods):

∞

k=1

We conclude this section with an important application of Poisson’s Theorem

to approximation by trigonometric polynomials. Another proof of this fact will be

given in Theorem 14.6.4.

13.6.6. COROLLARY. Every continuous 2π-periodic function is the uniform

limit of a sequence of trigonometric polynomials.

PROOF. Let f be a continuous 2π-periodic function. Given n ∈ N, Poisson’s

Theorem (Theorem 13.6.1) shows that there is an r < 1 such that kf −f

∞

Let M = kfk

and choose K so large that

∞

k=K+1

4Mr

. Then

(θ) = A

k=1

coskθ + B

sinkθ