Dirichlet Convergence Theorem

MATH 467 Partial Differential Equations

J. Robert Buchanan

Department of Mathematics

Spring 2016

Objectives

In this lesson we will sketch proofs of ▶ the Dirichlet Convergence Theorem for Fourier series, and ▶ the theorem regarding the uniform convergence of Fourier

series.

Dirichlet Convergence Theorem

Theorem

Assume that f (x) is a piecewise smooth function on the interval [−L, L] extended to (−∞, ∞) periodically with period 2L. Then the Fourier series of f (x) converges for all x to the value

\frac{1}{2} (f (x+) + f (x−)).

Necessary Lemma

In order to prove Dirichlet’s convergence theorem for Fourier series we will need the following lemma.

Lemma

For any fixed L > 0 and piecewise continuous function g(x) on interval [0, L],

lim_{N→∞}∫_{0}^{L}g(t) sin \frac{(2N + 1)πt}{2L} dt = 0.

Proof

Let g(x) be piecewise continuous on [0, L].

∫_{0}^{L}g(t) sin \frac{(2N + 1)πt}{L} dt

= ∫_{0}^{L}[g(t) sin \frac{πt}{L}]cos \frac{2Nπt}{L} dt + ∫_{0}^{L}[g(t) cos \frac{πt}{L}]sin \frac{2Nπt}{L} dt

Consider the even 2L-periodic extension of g(t) sin(πt/L) and the 2Nth coefficient in the Fourier cosine series for this extension.

a_{2N} = \frac{2}{L} ∫_{0}^{L}g(t) sin \frac{πt}{L} cos \frac{2Nπt}{L} dt
\frac{L}{2}a_{2N} = ∫_{0}^{L}g(t) sin \frac{πt}{L} cos \frac{2Nπt}{L} dt
lim_{N→∞} \frac{L}{2}a_{2N} = lim_{N→∞}∫_{0}^{L}g(t) sin \frac{πt}{L} cos \frac{2Nπt}{L} dt = 0

In a similar manner the other integral converges to 0 as N → ∞.

Proof

Let g(x) be piecewise continuous on [0, L].

∫_{0}^{L}g(t) sin \frac{(2N + 1)πt}{L} dt

= ∫_{0}^{L}[g(t) sin \frac{πt}{L}]cos \frac{2Nπt}{L} dt + ∫_{0}^{L}[g(t) cos \frac{πt}{L}]sin \frac{2Nπt}{L} dt

Consider the even 2L-periodic extension of g(t) sin(πt/L) and the 2Nth coefficient in the Fourier cosine series for this extension.

a_{2N} = \frac{2}{L} ∫_{0}^{L}g(t) sin \frac{πt}{L} cos \frac{2Nπt}{L} dt
\frac{L}{2}a_{2N} = ∫_{0}^{L}g(t) sin \frac{πt}{L} cos \frac{2Nπt}{L} dt
lim_{N→∞} \frac{L}{2}a_{2N} = lim_{N→∞}∫_{0}^{L}g(t) sin \frac{πt}{L} cos \frac{2Nπt}{L} dt = 0

In a similar manner the other integral converges to 0 as N → ∞.

Nth Partial Sum Revisited

S_{N}(x) = \frac{a_{0}}{2} + ∑_{n=1}^{N}(a_{n} cos \frac{nπx}{L} + b_{n} sin \frac{nπx}{L})

= \frac{1}{2L} ∫_{−L}^{L}f (t) dt + \frac{1}{L} ∑_{n=1}^{N}[∫_{−L}^{L}f (t) cos \frac{nπt}{L} dt]cos \frac{nπx}{L}

+ \frac{1}{L} ∑_{n=1}^{N}[∫_{−L}^{L}f (t) sin \frac{nπt}{L} dt]sin \frac{nπx}{L}

= \frac{1}{2L} ∫_{−L}^{L}f (t) dt

+ \frac{1}{L} ∑_{n=1}^{N}∫_{−L}^{L}f (t) [cos \frac{nπt}{L} cos \frac{nπx}{L} + sin \frac{nπt}{L} sin \frac{nπx}{L}] dt

= \frac{1}{2L} ∫_{−L}^{L}f (t) dt + \frac{1}{L} ∑_{n=1}^{N}∫_{−L}^{L}f (t) cos \frac{nπ(t − x)}{L}dt

Dirichlet Kernel

S_{N}(x) = \frac{1}{2L} ∫_{−L}^{L}f (t) dt + \frac{1}{L} ∑_{n=1}^{N}∫_{−L}^{L}f (t) cos \frac{nπ(t − x)}{L}dt

= \frac{1}{L} ∫_{−L}^{L}f (t) [\frac{1}{2} + ∑_{n=1}^{N}cos \frac{nπ(t − x)}{L}] dt

= \frac{1}{L} ∫_{−L}^{L}f (t)D_{N}(t − x) dt

where

D_{N}(z) = \frac{1}{2} + ∑_{n=1}^{N}cos \frac{nπz}{L}

is called the Dirchlet Kernel.

Closed Form of Dirichlet Kernel (1 of 3)

D_{N}(z) = \frac{1}{2} + ∑_{n=1}^{N}cos \frac{nπz}{L}

D_{N}(z) sin \frac{πz}{L} = \frac{1}{2} sin \frac{πz}{L} + ∑_{n=1}^{N}cos \frac{nπz}{L} sin \frac{πz}{L}

= \frac{1}{2} sin \frac{πz}{L} + \frac{1}{2} ∑_{n=1}^{N}[sin \frac{(n + 1)πz}{L} − sin \frac{(n − 1)πz}{L} ]

according to a product-to-sum formula. The summation telescopes to produce

D_{N}(z) sin \frac{πz}{L} = \frac{1}{2} [sin \frac{(N + 1)πz}{L} + sin \frac{Nπz}{L} ] .

Closed Form of Dirichlet Kernel (2 of 3)

D_{N}(z) sin \frac{πz}{L}

= \frac{1}{2} [sin \frac{(N + 1)πz}{L} + sin \frac{Nπz}{L} ]
= \frac{1}{2} [sin \frac{2(N + 1)πz}{2L} + sin \frac{2Nπz}{2L} ]
= \frac{1}{2} [sin (\frac{(2N + 1)πz}{2L} + \frac{πz}{2L}) + sin (\frac{(2N + 1)πz}{2L} − \frac{πz}{2L})]
= sin \frac{(2N + 1)πz}{2L} cos \frac{πz}{2L}

using a product-to-sum formula.

= sin \frac{(2N+1)πz}{2L} cos \frac{πz}{2L} D_{N}(z) = sin \frac{(2N+1)πz}{2L} cos \frac{πz}{2L} ; sin \frac{πz}{L} = sin \frac{(2N+1)πz}{2L} cos \frac{πz}{2L} _{\overline{sin \frac{2πz}{2L}}} = sin \frac{(2N+1)πz}{2L} cos \frac{πz}{2L} _{\overline{2 sin \frac{πz}{2L} cos \frac{πz}{2L}}} D_{N}(z) = \frac{sin \frac{(2N+1πz}{2L}}{2 sin \frac{πz}{2L}} using the double-angle formula for the sine.

Fourier Series Uniform Convergence

Theorem

Assume that f (x) is continuous on [−L, L] with f (−L) = f (L) and f′(x) is piecewise continuous on [−L, L], then the Fourier series of f (x) converges uniformly to f (x) on [−L, L].

Homework

▶ Read Section x.y ▶ Exercises: