Fourier series

5.5 Fourier series

Harmonic oscillator: $m y^{″} = - k y$ , $y = A \cos ω t + B \sin ω t$ .

Wave: $\frac{\partial^{2} u}{\partial t^{2}} = c^{2} \frac{\partial^{2} u}{\partial x^{2}}$ . Methods.

1.: Reducing order. $(\frac{\partial}{\partial t} - c \frac{\partial}{\partial x}) (\frac{\partial}{\partial t} + c \frac{\partial}{\partial x}) u = 0$ By means of method of characteristic line, we could get the traveling-wave solution.
2.: Separation of variables. Assuming $u (x, t) = U (t) V (x)$ , then the original PDE becomes 2 ODEs, and then we could get the standing-wave solution.

Assuming a periodic function $f$ with period $T$ , can it be expressed in the form of $f (x) = A_{0} + A_{1} \cos \frac{2 π x}{T} + B_{1} \sin \frac{2 π x}{T} + \dots + A_{n} \cos \frac{2 n π x}{T} + B_{n} \sin \frac{2 n π x}{T} + \dots$ and what’s its convergence?

Definition 5.5.1 Inner product of periodic functions. Assuming $f, g$ are periodic functions with period $2 π$ , then $⟨ f . g ⟩ = \int_{0}^{2 π} f (x) g (x) d x$

When

m, n \in N^{*}

, notice that

\begin{aligned} ⟨ \cos n x, \cos m x ⟩ & = \int_{0}^{2 π} \frac{\cos (n + m) x + \cos (n - m) x}{2} d x = {\begin{cases} 0 & n \neq m \\ π & n = m \end{cases} \\ ⟨ \sin n x, \sin m x ⟩ & = \int_{0}^{2 π} \frac{\cos (n - m) x - \cos (n + m) x}{2} d x = {\begin{cases} 0 & n \neq m \\ π & n = m \end{cases} \\ ⟨ \cos n x, \sin m x ⟩ & = 0 \end{aligned}

Assuming $f (x) = A_{0} + \sum_{n = 1}^{+ \infty} (A_{n} \cos n x + B_{n} \sin n x)$ Then we have $\int_{0}^{2 π} f (x) \cos n x d x = ⟨ f (x), \cos n x ⟩ = A_{n} ⟨ \cos n x, \cos n x ⟩ = {\begin{cases} 2 π A_{0} & n = 0 \\ π A_{n} & n \geq 1 \end{cases}$ Therefore $A_{0} = \frac{1}{2 π} \int_{0}^{2 π} f (x) d x, A_{n} = \frac{1}{π} \int_{0}^{2 π} f (x) \cos n x d x$ Similarly we have $B_{n} = \frac{1}{π} \int_{0}^{2 π} f (x) \sin n x d x$

Definition 5.5.2 Assuming $f$ is periodic with period $2 π$ , $f$ is integrable and absolutely integrable, let $A_{n} = \frac{1}{π} \int_{0}^{2 π} f (x) \cos n x d x, B_{n} = \frac{1}{π} \int_{0}^{2 π} f (x) \sin n x d x, \forall n \in N$ which is called the Euler-Fourier formula, then the Fourier series of $f$ is defined as $f (x) \sim \frac{A_{0}}{2} + \sum_{n = 1}^{+ \infty} (A_{n} \cos n x + B_{n} \sin n x)$

Example 5.5.1 $f (x) = \frac{π - x}{2}$ when $x \in (0, 2 π)$ , $f$ is periodic with period $2 π$ , seek its Fourier series.

$f$ is odd $\Rightarrow A_{n} = 0, \forall n \in N$ , and $B_{n} = \frac{1}{π} \int_{0}^{2 π} \frac{π - x}{2} \sin n x d x = \frac{1}{n}$ So its Fourier series is $\sum_{n = 1}^{+ \infty} \frac{\sin n x}{n}$ . Yet we’ve already known that $\sum_{n = 1}^{+ \infty} \frac{\sin n x}{n} = \frac{π - x}{2}, x \in (0, 2 π)$ So its Fourier series converges to itself when $x \neq 2 k π, k \in Z$ .

Example 5.5.2 $f (x) = x^{2}$ when $x \in [- π, π]$ , $f$ is periodic with period $2 π$ , seek its Fourier series.

$f$ is odd $\Rightarrow B_{n} = 0, \forall n \in N$ , and $A_{n} = \frac{2}{π} \int_{0}^{π} x^{2} \cos n x d x = \frac{4 (- 1)^{n}}{n^{2}}, \forall n \in N^{*}$ So its Fourier series is $\frac{π^{2}}{3} + \sum_{n = 1}^{+ \infty} \frac{4 (- 1)^{n}}{n^{2}} \cos n x$ . Yet we’ve already known that $\sum_{n = 1}^{+ \infty} \frac{\cos n x}{n^{2}} = \frac{π^{2}}{6} + \frac{x^{2} - 2 π x}{4}, x \in [0, 2 π]$ Therefore, $\sum_{n = 1}^{+ \infty} \frac{(- 1)^{n}}{n^{2}} \cos n x = \sum_{n = 1}^{+ \infty} \frac{\cos n (x + π)}{n^{2}} = \frac{π^{2}}{6} + \frac{(x + π)^{2} - 2 π (x + π)}{4}$ After computation we could verify that its Fourier series constantly converges to itself.