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8.3: Fourier Series II

( \newcommand{\kernel}{\mathrm{null}\,}\)

In this section we discuss Fourier expansions in terms of the eigenfunctions of Problems 1-4 for Section 8.1.

Fourier Cosine Series

From Exercise 8.1.20, the eigenfunctions

1,\, \cos{\pi x\over L}, \, \cos{2\pi x\over L},\dots, \, \cos{n\pi x\over L},\dots\nonumber

of the boundary value problem

\label{eq:11.3.1} y''+\lambda y=0,\quad y'(0)=0,\quad y'(L)=0

(Problem 2) are orthogonal on [0,L]. If f is integrable on [0,L] then the Fourier expansion of f in terms of these functions is called the Fourier cosine series of f on [0,L]. This series is

a_0+\sum_{n=1}^\infty a_n\cos{n\pi x\over L},\nonumber

where

a_0={\int_0^Lf(x)\,dx\over\int_0^L\,dx}={1\over L}\int_0^Lf(x)\,dx\nonumber

and

a_n={\int_0^Lf(x)\cos{n\pi x\over L}\,dx\over\int_0^L \cos^2{n\pi x\over L}\,dx}={2\over L}\int_0^Lf(x)\cos{n\pi x\over L}\,dx,\quad n=1,2,3,\dots.\nonumber

Comparing this definition with Theorem 8.2.6a shows that the Fourier cosine series of f on [0,L] is the Fourier series of the function

f_{1}(x)=\left\{\begin{array}{cc}{f(-x),}&{-L<x<0,}\\{f(x),}&{}0\leq x\leq L\end{array} \right.\nonumber

obtained by extending f over [-L,L] as an even function (Figure \PageIndex{1}).

clipboard_e78f3dad1a0ca2025b28f4735d0b46356.png
Figure \PageIndex{1}

Applying Theorem 8.2.4 to f_1 yields the next theorem.

Theorem \PageIndex{1}

If f is piecewise smooth on [0,L], then the Fourier cosine series

C(x)=a_0+\sum_{n=1}^\infty a_n\cos{n\pi x\over L}\nonumber

of f on [0,L], with

a_0={1\over L}\int_0^Lf(x)\,dx \quad \text{and} \quad a_n={2\over L}\int_0^Lf(x)\cos{n\pi x\over L}\,dx,\quad n=1,2,3,\dots,\nonumber

converges for all x in [0,L]; moreover,

C(x)=\left\{\begin{array}{cl}{f(0+)}&{if\: x=0}\\{f(x)}&{if\: 0<x<L\:and\:f\:is\:continuous\:at\:x}\\{\frac{f(x-)+f(x+)}{2}}&{if\:0<x<L\:and\:f\:is\:discontinuous\:at\:x}\\{f(L-)}&{if\:x=L}\end{array} \right. \nonumber

Example \PageIndex{1}

Find the Fourier cosine series of f(x)=x on [0,L].

The coefficients are

a_0={1\over L}\int_0^Lx\,dx=\left. {1\over L}{x^2\over2} \right|_{0}^{L}={L\over2}\nonumber

and, if n\ge1

\begin{aligned} a_n&={2\over L}\int_0^Lx\cos{n\pi x\over L}\,dx =\left. {2\over n\pi}\left[x\sin{n\pi x\over L}\right|_{0}^{L}- \int_0^L \sin{n\pi x\over L}\,dx\right]\\ &=-{2\over n\pi}\int_0^L \sin{n\pi x\over L}\,dx =\left.{2L\over n^2\pi^2}\cos{n\pi x\over L}\right|_{0}^{L} ={2L\over n^2\pi^2}[(-1)^n-1]\\ &= \left\{\begin{array}{cl} -{4L\over(2m-1)^2\pi^2}&\mbox{if $n=2m-1$},\\ 0&\mbox{if $n=2m$}. \end{array}\right.\end{aligned}\nonumber

Therefore

C(x)=\frac{L}{2}-\frac{4L}{\pi ^{2}}\sum _{n=1}^{\infty}\frac{1}{(2n-1)^{2}}\cos\frac{(2n-1)\pi x}{L}\nonumber

Theorem \PageIndex{1} implies that

C(x)=x,\quad 0\le x\le L.\nonumber

Fourier Sine Series

From Exercise 8.1.19, the eigenfunctions

\sin{\pi x\over L}, \, \sin{2\pi x\over L},\dots, \, \sin{n\pi x\over L},\dots\nonumber

of the boundary value problem

y''+\lambda y=0,\quad y(0)=0,\quad y(L)=0\nonumber

(Problem 1) are orthogonal on [0,L]. If f is integrable on [0,L] then the Fourier expansion of f in terms of these functions is called the Fourier sine series of f on [0,L]. This series is

\sum_{n=1}^\infty b_n\sin{n\pi x\over L},\nonumber

where

b_n={\int_0^Lf(x)\sin{n\pi x\over L}\,dx\over\int_0^L \sin^2{n\pi x\over L}\,dx}={2\over L}\int_0^Lf(x)\sin{n\pi x\over L}\,dx,\quad n=1,2,3,\dots.\nonumber

Comparing this definition with Theorem 11.2.6b shows that the Fourier sine series of f on [0,L] is the Fourier series of the function

f_{2}(x)=\left\{\begin{array}{cc}{-f(-x),}&{-L<x<0}\\{f(x),}&{0\leq x\leq L,}\end{array} \right.\nonumber

obtained by extending f over [-L,L] as an odd function (Figure \PageIndex{2}).

clipboard_e6ea67e089734d7f81550452a017d6229.png
Figure \PageIndex{2}

Applying Theorem 8.2.4 to f_2 yields the next theorem.

Theorem \PageIndex{2}

If f is piecewise smooth on [0,L], then the Fourier sine series

S(x)=\sum_{n=1}^\infty b_n\sin{n\pi x\over L}\nonumber

of f on [0,L], with

b_n={2\over L}\int_0^Lf(x)\sin{n\pi x\over L}\,dx,\nonumber

converges for all x in [0,L]; moreover,

S(x)=\left\{\begin{array}{cl}{0}&{if\:x=0}\\{f(x)}&{if\:0<x<L\:and\:f\:is\:continuous\:at\:x}\\{\frac{f(x-)+f(x+)}{2}}&{if\:0<x<L\:and\:f\:is\:discontinuous\:at\:x}\\{0}&{if\:x=L}\end{array} \right.\nonumber

Example \PageIndex{2}

Find the Fourier sine series of f(x)=x on [0,L].

Solution

The coefficients are

\begin{aligned} b_n&={2\over L}\int_0^Lx\sin{n\pi x\over L}\,dx =\left.-{2\over n\pi}\left[x\cos{n\pi x\over L}\right|_{0}^{L}- \int_0^L \cos{n\pi x\over L}\,dx\right]\\ &=\left. (-1)^{n+1}{2L\over n\pi}+{2L\over n^2\pi^2}\sin{n\pi x\over L}\right|_{0}^{L} =(-1)^{n+1}{2L\over n\pi}.\end{aligned}\nonumber

Therefore

S(x)=-{2L\over\pi}\sum_{n=1}^\infty{(-1)^n\over n} \sin{n\pi x\over L}.\nonumber

Theorem \PageIndex{2} implies that

S(x)= \left\{\begin{array}{cl} x,&0\le x< L,\\0,& x=L. \end{array}\right.\nonumber

Mixed Fourier Cosine Series

From Exercise 8.1.22, the eigenfunctions

\cos{\pi x\over 2L}, \, \cos{3\pi x\over 2L},\dots, \, \cos{(2n-1)\pi x\over 2L},\dots\nonumber

of the boundary value problem

\label{eq:8.3.2} y''+\lambda y=0,\quad y'(0)=0,\quad y(L)=0

(Problem 4) are orthogonal on [0,L]. If f is integrable on [0,L] then the Fourier expansion of f in terms of these functions is

\sum_{n=1}^\infty c_n\cos{(2n-1)\pi x\over2L},\nonumber

where

c_n={\int_0^Lf(x)\cos{(2n-1)\pi x\over2L}\,dx\over\int_0^L \cos^2{(2n-1)\pi x\over L}\,dx}={2\over L}\int_0^Lf(x)\cos{(2n-1)\pi x\over2L}\,dx.\nonumber

We’ll call this expansion the mixed Fourier cosine series of f on [0,L], because the boundary conditions of ( Equation \ref{eq:8.3.2}) are “mixed” in that they require y to be zero at one boundary point and y' to be zero at the other. By contrast, the “ordinary” Fourier cosine series is associated with ( Equation \ref{eq:8.3.1}), where the boundary conditions require that y' be zero at both endpoints.

It can be shown (Exercise 8.3.57) that the mixed Fourier cosine series of f on [0,L] is simply the restriction to [0,L] of the Fourier cosine series of

f_3(x)= \left\{\begin{array}{cl} f(x),&0\le x\le L,\\-f(2L-x),&L< x\le 2L \end{array}\right.\nonumber

on [0,2L] (Figure \PageIndex{3}).

clipboard_e422102b4804e75637bcf91a5e99550b1.png
Figure \PageIndex{3}

Applying Theorem \PageIndex{1} with f replaced by f_3 and L replaced by 2L yields the next theorem.

Theorem \PageIndex{3}

If f is piecewise smooth on [0,L], then the mixed Fourier cosine series

C_M(x)=\sum_{n=1}^\infty c_n\cos{(2n-1)\pi x\over2L}\nonumber

of f on [0,L], with

c_n={2\over L}\int_0^Lf(x)\cos{(2n-1)\pi x\over2L}\,dx,\nonumber

converges for all x in [0,L]; moreover,

C_{M}(x)=\left\{\begin{array}{cl}{f(0+)}&{if\:x=0}\\{f(x)}&{if\:0<x<L\:and\:f\:is\:continuous\:at\:x}\\{\frac{f(x-)+f(x+)}{2}}&{if\:0<x<L\:and\:f\:is\:discontinuous\:at\:x}\\{0}&{if\:x=L}\end{array} \right.\nonumber

Example \PageIndex{3}

Find the mixed Fourier cosine series of f(x)=x-L on [0,L].

Solution

The coefficients are

\begin{aligned} c_n&={2\over L}\int_0^L(x-L)\cos{(2n-1)\pi x\over2L}\,dx\\ &=\left.{4\over(2n-1)\pi}\left[(x-L)\sin{(2n-1)\pi x\over2L}\right|_{0}^{L}-\int_0^L \sin{(2n-1)\pi x\over2L}\,dx\right]\\ &=\left.{8L\over(2n-1)^2\pi^2} \cos{(2n-1)\pi x\over2L}\right|_{0}^{L} =-{8L\over(2n-1)^2\pi^2}.\end{aligned}\nonumber

Therefore

C_M(x)=-{8L\over\pi^2}\sum_{n=1}^\infty{1\over(2n-1)^2} \cos{(2n-1)\pi x\over2L}.\nonumber

Theorem \PageIndex{3} implies that

C_M(x)= x-L,\quad 0\le x\le L.\nonumber

Mixed Fourier Sine Series

From Exercise 8.1.21, the eigenfunctions

\sin{\pi x\over 2L}, \, \sin{3\pi x\over 2L},\dots, \, \sin{(2n-1)\pi x\over 2L},\dots\nonumber

of the boundary value problem

y''+\lambda y=0,\quad y(0)=0,\quad y'(L)=0\nonumber

(Problem 3) are orthogonal on [0,L]. If f is integrable on [0,L], then the Fourier expansion of f in terms of these functions is

\sum_{n=1}^\infty d_n\sin{(2n-1)\pi x\over2L},\nonumber

where

d_n={\int_0^Lf(x)\sin{(2n-1)\pi x\over2L}\,dx\over\int_0^L \sin^2{(2n-1)\pi x\over2L}\,dx}={2\over L}\int_0^Lf(x)\sin{(2n-1)\pi x\over2L}\,dx.\nonumber

We’ll call this expansion the mixed Fourier sine series of f on [0,L].

It can be shown (Exercise 8.3.58) that the mixed Fourier sine series of f on [0,L] is simply the restriction to [0,L] of the Fourier sine series of

f_4(x)= \left\{\begin{array}{cl} f(x),&0\le x\le L,\\f(2L-x),&L< x\le 2L, \end{array}\right.\nonumber

on [0,2L] (Figure \PageIndex{4}).

clipboard_e7ade36fe4d1c38e1c87341d756f57d5b.png
Figure \PageIndex{4}

Applying Theorem \PageIndex{2} with f replaced by f_4 and L replaced by 2L yields the next theorem.

Theorem \PageIndex{4}

If f is piecewise smooth on [0,L], then the mixed Fourier sine series

S_M(x)=\sum_{n=1}^\infty d_n\sin{(2n-1)\pi x\over2L}\nonumber

of f on [0,L], with

d_n={2\over L}\int_0^Lf(x)\sin{(2n-1)\pi x\over2L}\,dx,\nonumber

converges for all x in [0,L]; moreover,

S{M}(x)=\left\{\begin{array}{cl}{0}&{if\:x=0}\\{f(x)}&{if\:0<x<L\:and\:f\:is\:continuous\:at\:x}\\{\frac{f(x-)+f(x+)}{2}}&{if\:0<x<L\:and\:f\:is\:discontinuous\:at\:x}\\{f(L-)}&{if\:x=L}\end{array} \right.\nonumber

Example \PageIndex{4}

Find the mixed Fourier sine series of f(x)=x on [0,L].

Solution

The coefficients are

\begin{aligned} d_n&={2\over L}\int_0^Lx\sin{(2n-1)\pi x\over2L}\,dx\\ &=\left.-{4\over(2n-1)\pi}\left[x\cos{(2n-1)\pi x\over2L}\right|_{0}^{L}- \int_0^L \cos{(2n-1)\pi x\over2L}\,dx\right]\\ &={4\over(2n-1)\pi} \int_0^L \cos{(2n-1)\pi x\over2L}\,dx\\ &=\left.{8L\over(2n-1)^2\pi^2}\sin{(2n-1)\pi x\over2L}\right|_{0}^{L}=(-1)^{n+1}{8L\over(2n-1)^2\pi^2}.\end{aligned}\nonumber

Therefore

S_M(x)=-{8L\over\pi^2}\sum_{n=1}^\infty{(-1)^n\over(2n-1)^2} \sin{(2n-1)\pi x\over2L}.\nonumber

Theorem \PageIndex{4} implies that

S_M(x)=x,\quad 0\le x\le L.\nonumber

A Useful Observation

In applications involving expansions in terms of the eigenfunctions of Problems 1-4, the functions being expanded are often polynomials that satisfy the boundary conditions of the problem under consideration. In this case the next theorem presents an efficient way to obtain the coefficients in the expansion.

Theorem \PageIndex{5}

  1. If f'(0)=f'(L)=0, f'' is continuous, and f''' is piecewise continuous on [0,L], then \label{eq:11.3.3} f(x)=a_0+\sum_{n=1}^\infty a_n\cos{n\pi x\over L}, \quad 0\le x\le L, with \label{eq:11.3.4} a_0={1\over L}\int_0^L f(x)\,dx \quad \text{and} \quad a_n= {2L^2\over n^3\pi^3}\int_0^L f'''(x)\sin{n\pi x\over L}\,dx, \quad n\ge1. Now suppose f' is continuous and f'' is piecewise continuous on [0,L].
  2. If f(0)=f(L)=0, then f(x)=\sum_{n=1}^\infty b_n\sin{n\pi x\over L}, \quad 0\le x\le L,\nonumber with \label{eq:11.3.5} b_n=-{2L\over n^2\pi^2}\int_0^L f''(x)\sin{n\pi x\over L}\,dx.
  3. If f'(0)=f(L)=0, then f(x)= \sum_{n=1}^\infty c_n\cos{(2n-1)\pi x\over2L}, \quad 0\le x\le L,\nonumber with \label{eq:11.3.6} c_n=-{8L\over(2n-1)^2\pi^2}\int_0^L f''(x)\cos{(2n-1)\pi x\over2L} \,dx.
  4. If f(0)=f'(L)=0, then f(x)= \sum_{n=1}^\infty d_n\sin{(2n-1)\pi x\over2L}, \quad 0\le x\le L,\nonumber with \label{eq:11.3.7} d_n=-{8L\over(2n-1)^2\pi^2}\int_0^L f''(x)\sin{(2n-1)\pi x\over2L} \,dx.
Proof

We'll prove (a) and leave the rest to you (Exercises 8.3.35, 8.3.42, and 8.3.50). Since f is continuous on [0,L], Theorem \PageIndex{1} implies ( Equation \ref{eq:11.3.3}) with a_0, a_1, a_2,... as defined in Theorem \PageIndex{1}. We already know that a_0 is as in ( Equation \ref{eq:11.3.4}). If n\ge1, integrating twice by parts yields

\begin{aligned} a_n&= {2\over L}\int_0^L f(x)\cos{n\pi x\over L}\,dx\\ &=\left.{2\over n\pi}\left[f(x)\sin{n\pi x\over L}\right|_{0}^{L} -\int_0^Lf'(x)\sin{n\pi x\over L}\,dx\right]\\ &=-{2\over n\pi} \int_0^Lf'(x)\sin{n\pi x\over L}\,dx \mbox{ (since $\sin0=\sin n\pi=0$)}\\ &=\left.{2L\over n^2\pi^2}\left[f'(x)\cos{n\pi x\over L}\right|_{0}^{L} -\int_0^Lf''(x)\cos{n\pi x\over L}\right]\,dx\\ &= -{2L\over n^2\pi^2}\int_0^Lf''(x)\cos{n\pi x\over L}\,dx \mbox{ (since $f'(0)=f'(L)=0$)}\\ &=\left.-{2L^2\over n^3\pi^3}\left[f''(x)\sin{n\pi x\over L}\right|_{0}^{L} -\int_0^Lf'''(x)\sin{n\pi x\over L}\,dx\right]\\ &= {2L^2\over n^3\pi^3}\int_0^Lf'''(x)\sin{n\pi x\over L}\,dx \mbox{ (since $\sin0=\sin n\pi=0$).}\end{aligned}\nonumber

(By an argument similar to one used in the proof of Theorem 8.3.1, the last integration by parts is legitimate in the case where f''' is undefined at finitely many points in [0,L], so long as it is piecewise continuous on [0,L].) This completes the proof.

Example \PageIndex{5}

Find the Fourier cosine expansion of f(x)=x^2(3L-2x) on [0,L].

Solution

Here

a_0={1\over L}\int_0^L(3Lx^2-2x^3)\,dx=\left.{1\over L}\left(Lx^3-{x^4\over2} \right)\right|_{0}^{L}={L^3\over2}\nonumber

and

a_n={2\over L}\int_0^L(3Lx^2-2x^3)\cos{n\pi x\over L}\,dx,\quad n\ge1.\nonumber

Evaluating this integral directly is laborious. However, since f'(x)=6Lx-6x^2, we see that f'(0)=f'(L)=0. Since f'''(x)=-12, we see from ( Equation \ref{eq:11.3.4}) that if n\ge1 then

\begin{aligned} a_n&=-{24L^2\over n^3\pi^3}\int_0^L\sin{n\pi x\over L}\,dx =\left.{24L^3\over n^4\pi^4}\cos{n\pi x\over L}\right|_{0}^{L}={24L^3\over n^4\pi^4}[(-1)^n-1]\\ &= \left\{\begin{array}{cl} -{48L^3\over(2m-1)^4\pi^4}&\mbox{if $n=2m-1$},\\ 0&\mbox{if $n=2m$.} \end{array}\right.\end{aligned}\nonumber

Therefore

C(x)={L^3\over2}-{48L^3\over\pi^4}\sum_{n=1}^\infty{1\over (2n-1)^4}\cos{(2n-1)\pi x\over L}.\nonumber

Example \PageIndex{6}

Find the Fourier sine expansion of f(x)=x(x^2-3Lx+2L^2) on [0,L].

Solution

Since f(0)=f(L)=0 and f''(x)=6(x-L), we see from ( Equation \ref{eq:11.3.5}) that

\begin{aligned} b_n&=- {12L\over n^2\pi^2}\int_0^L(x-L)\sin{n\pi x\over L}\,dx\\ &=\left.{12L^2\over n^3\pi^3}\left[(x-L)\cos{n\pi x\over L}\right|_{0}^{L} -\int_0^L\cos{n\pi x\over L}\,dx\right]\\ &=\left.{12L^2\over n^3\pi^3}\left[L-\frac{L}{n\pi}\sin\frac{n\pi x}{L}\right|_{0}^{L}\right] ={12L^3\over n^3\pi^3}.\end{aligned}\nonumber

Therefore

S(x)=\frac{12L^{3}}{\pi ^{3}}\sum_{n=1}^{\infty}\frac{1}{n^{3}}\sin\frac{n\pi x}{L}\nonumber

Example \PageIndex{7}

Find the mixed Fourier cosine expansion of f(x)=3x^3-4Lx^2+L^3 on [0,L].

Solution

Since f'(0)=f(L)=0 and f''(x)=2(9x-4L), we see from ( Equation \ref{eq:11.3.6}) that

\begin{aligned} c_n&= -{16L\over(2n-1)^2\pi^2} \int_0^L(9x-4L)\cos{(2n-1)\pi x\over2L}\,dx\\ &=\left.-{32L^2\over(2n-1)^3\pi^3}\left[(9x-4L)\sin{(2n-1)\pi x\over2L} \right|_{0}^{L}-9\int_0^L\sin{(2n-1)\pi x\over2L}\right]\,dx\\ &=\left.-{32L^2\over(2n-1)^3\pi^3} \left[(-1)^{n+1}5L+{18L\over(2n-1)\pi}\cos{(2n-1)\pi x\over2L} \right|_{0}^{L}\right] \\ &={32L^3\over(2n-1)^3\pi^3} \left[(-1)^n5+{18\over(2n-1)\pi}\right].\end{aligned}\nonumber

Therefore

C_{M}(x)=\frac{32L^{3}}{\pi ^{3}}\sum_{n=1}^{\infty}\frac{1}{(2n-1)^{3}}\left[(-1)^{n}5+\frac{18}{(2n-1)\pi } \right]\cos\frac{(2n-1)\pi x}{2L}\nonumber

Example \PageIndex{8}

Find the mixed Fourier sine expansion of

f(x)=x(2x^2-9Lx+12L^2)\nonumber

on [0,L].

Solution

Since f(0)=f'(L)=0, and f''(x)=6(2x-3L), we see from ( Equation \ref{eq:11.3.7}) that

\begin{aligned} d_{n}&=-\frac{48L}{(2n-1)^{2}\pi ^{2}}\int_{0}^{L}(2x-3L)\sin\frac{(2n-1)\pi x}{2L}dx \\ &=\left.\frac{96L^{2}}{(2n-1)^{3}\pi ^{3}}\left[(2x-3L)\cos\frac{(2n-1)\pi x}{2L}\right|_{0}^{L} - 2\int_{0}^{L}\cos\frac{(2n-1)\pi x}{2L}dx \right] \\ &=\left. \frac{96L^{2}}{(2n-1)^{3}\pi ^{3}}\left[3L-\frac{4L}{(2n-1)\pi }\sin\frac{(2n-1)\pi x}{2L}\right|_{0}^{L} \right] \\&=\frac{96L^{3}}{(2n-1)^{3}\pi ^{3}}\left[3+(-1)^{n}\frac{4}{(2n-1)\pi } \right] \end{aligned}\nonumber

Therefore

S_{M}(x)=\frac{96L^{3}}{\pi ^{3}}\sum_{n=1}^{\infty}\frac{1}{(2n-1)^{3}}\left[3+(-1)^{n}\frac{4}{(2n-1)\pi } \right]\sin\frac{(2n-1)\pi x}{2L}\nonumber


This page titled 8.3: Fourier Series II is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Zoya Kravets.

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