Partial Differential Equations

PDE notes: Wave Equation

June 1, 2022June 1, 2022Michael LiuLeave a comment

2.1. Introduction

The one-dimensional wave equation for small displacement of a perfectly elastic string of length ${l}$ with no frictional forces and no restoring forces is

$\displaystyle \rho_0 u_{tt} = T_0 u_{xx} \ \ \ \ \ (88)$

$\displaystyle u_{tt} = c^2 u_{xx} \ \ \ \ \ (89)$

for ${t\geq0}$ , ${0\leq x \leq l}$ , where ${c^2=\frac{T_0}{\rho_0}}$ . We assume both tension ${T_0}$ and linear density ${\rho_0}$ to be constant. Tension ${T_0}$ has units of ${\frac{mass}{time^2}}$ and linear density has units of ${\frac{mass}{length^2}}$ , so ${c=\sqrt{\frac{T_0}{\rho_0}}}$ has the units of ${\frac{length}{time}}$ , which is also the unit of speed. Thus, ${c}$ turns out to be the velocity of wave propagation along the string. The one-dimensional wave equation models sound waves, water waves, vibrations in solids, and longitudinal or torsional vibrations in a rod, among other things.

Since the PDE in (89) contains the second time derivative, two initial conditions are required. The initial conditions usually take the follwing form:

$\displaystyle \mathrm{Initial \hspace{0.2cm} displacement:} \hspace{0.5cm} u(x,0)= f(x) \ \ \ \ \ (90)$

$\displaystyle \mathrm{Initial \hspace{0.2cm} velocity:} \hspace{0.5cm} v(x,0)=u_t(x,0)= g(x) \ \ \ \ \ (91)$

Typical boundary conditions are of the same form as thus given in the discussion of one-dimensional heat equation. For homogeneous Dirichlet conditions,

$\displaystyle u(0,t)=0=u(l,t) \ \ \ \ \ (92)$

for ${t \geq 0}$ . The end ends of the vibrating strings are fixed.

For homogeneous Neumann conditions,

$\displaystyle u_x(0,t)=0=u_x(l,t) \ \ \ \ \ (93)$

for ${t \geq 0}$ . These conditions are be achieved, for example, by attaching the ends of the string to a frictionless sleeve that moves vertically.

2.2. Solution by separation of variables

We now solve the one-dimensional wave equation with homogeneous Dirichlet boundary conditions. The following problem is defined for ${0 \leq x \leq L}$ and ${t \geq 0}$ :

$\displaystyle \mathrm{PDE: } \hspace{0.5cm} u_{tt}=c^2u_{xx} \ \ \ \ \ (94)$

$\displaystyle \mathrm{BC1:} \hspace{0.5cm} u(0,t)=0 \ \ \ \ \ (95)$

$\displaystyle \mathrm{BC2:}\hspace{0.5cm} u(L,t)=0 \ \ \ \ \ (96)$

$\displaystyle \mathrm{IC1:}\hspace{0.5cm} u(x,0) = f(x) \ \ \ \ \ (97)$

$\displaystyle \mathrm{IC2:}\hspace{0.5cm} u_x(x,0) = g(x) \ \ \ \ \ (98)$

Solution Since both the PDE and the boundary conditions are linear and homogeneous, the method of separation of variables is attempted. We look for special product solutions of the form:

$\displaystyle u(x,t)=\phi(x)h(t) \ \ \ \ \ (99)$

Substitute (99) into (94) yields

$\displaystyle \phi(x)h''(t)=c^2h(t)\phi''(x) \ \ \ \ \ (100)$

Divide both sides by ${\phi(x)h(t)c^2}$ separates the variables:

$\displaystyle \frac{1}{h(t)c^2}h''(t)= \frac{1}{\phi(x)}\phi''(x)=-\lambda \ \ \ \ \ (101)$

where ${\lambda}$ is the separation constant. This implies that

$\displaystyle h''=-\lambda hc^2 \ \ \ \ \ (102)$

$\displaystyle \phi''=-\lambda \phi \ \ \ \ \ (103)$

We consider (103) first since it has a complete set of boundary conditions. Letting ${\phi=e^{rx}}$ , then ${\phi''=r^2e^{rx}}$ . So ${r=\pm\sqrt{-\lambda}}$ . Using results from the heat equation, we know that ${\lambda < 0}$ and ${\lambda=0}$ yields trivial solutions. For ${\lambda > 0}$ , the general solution is

$\displaystyle \phi(x) = c_1\mathrm{cos}(\sqrt{\lambda}x)+c_2\mathrm{sin}(\sqrt{\lambda}x) \ \ \ \ \ (104)$

Apply ${\phi(0)=0}$ and we arrive at ${c_1=0}$ . Consider ${\phi(L)=0}$ , we arrive at ${c_2\mathrm{sin}(\sqrt{\lambda}L)=0}$ . If ${c_2=0}$ , we get a trivial solution. Therefore, we get a nontrivial solution if and only if

$\displaystyle \mathrm{sin}(\sqrt{\lambda}L)=0 \ \ \ \ \ (105)$

This means that

$\displaystyle \sqrt{\lambda}L= n\pi \ \ \ \ \ (106)$

The eigenvalues are

$\displaystyle \lambda= (\frac{n\pi}{L})^2, n=1,2,3... \ \ \ \ \ (107)$

The eigenfunctions corresponding to the eigenvalues are

$\displaystyle \phi(x)=c_2 \mathrm{sin}(\frac{n\pi x}{L}), n=1,2,3... \ \ \ \ \ (108)$

The time-dependent part of the solution is

$\displaystyle h(t) = c_3\mathrm{cos}(\frac{cn\pi t}{L})+c_4\mathrm{sin}(\frac{cn\pi t}{L}) \ \ \ \ \ (109)$

for ${n=1,2,3...}$ . Thus, for ${n\geq 0}$ , each of the functions

$\displaystyle u_n(x,t)=\phi_n(x)h_n(t) = \mathrm{sin}(\frac{n\pi x}{L})(A_n\mathrm{cos}(\frac{cn\pi t}{L})+B_n\mathrm{sin}(\frac{cn\pi t}{L})) \ \ \ \ \ (110)$

is a solution to the PDE and satisfies the boundary condition. By superposition principle, we can solve the initial value problem by considering a linear combinations of all product solutions:

$\displaystyle u(x,t) = \sum^{\infty}_{0}\mathrm{sin}(\frac{n\pi x}{L})(A_n\mathrm{cos}(\frac{cn\pi t}{L})+B_n\mathrm{sin}(\frac{cn\pi t}{L})) \ \ \ \ \ (111)$

The initial conditions in (97) and (98) are satisfied if,

$\displaystyle f(x)=\sum^{\infty}_{0}A_n\mathrm{sin}(\frac{n\pi x}{L}) \ \ \ \ \ (112)$

$\displaystyle g(x)=\sum^{\infty}_{0}B_n\frac{cn\pi}{L}\mathrm{sin}(\frac{n\pi x}{L}) \ \ \ \ \ (113)$

We can consider the fact that ${\mathrm{sin}(\frac{n\pi x}{L})}$ satisfies the following orthogonality relation:

$\displaystyle \int_{0}^{L} \mathrm{sin}(\frac{n \pi x}{L})\mathrm{sin}(\frac{m \pi x}{L}) \,dx = \left\{ \begin{array}{lr} 0 & n \neq m \\ \frac{L}{2} & n=m \\ \end{array} \right. \ \ \ \ \ (114)$

Multiply (112) by ${\mathrm{sin}(\frac{m \pi x}{L})}$ and integrating from ${0}$ to ${L}$ yields

$\displaystyle \int_{0}^{L} f(x)\mathrm{sin}(\frac{m \pi x}{L})\, dx = A_{n}\int_{0}^{L}\mathrm{sin^2}(\frac{m\pi x}{L})\, dx \ \ \ \ \ (115)$

Solving for ${A_n}$ yields,

$\displaystyle A_n=\frac{2}{L} \int_{0}^{L}f(x)\mathrm{sin}(\frac{n \pi x}{L})\, dx \ \ \ \ \ (116)$

Multiply (113) by ${\mathrm{sin}(\frac{m \pi x}{L})}$ and integrating from ${0}$ to ${L}$ yields

$\displaystyle \int_{0}^{L} f(x)\mathrm{sin}(\frac{m \pi x}{L})\, dx = B_n\frac{cn\pi}{L}\int_{0}^{L}\mathrm{sin^2}(\frac{m\pi x}{L})\, dx \ \ \ \ \ (117)$

Solving for ${B_n}$ yields,

$\displaystyle B_n\frac{cn\pi}{L}=\frac{2}{L} \int_{0}^{L}f(x)\mathrm{sin}(\frac{n \pi x}{L})\, dx \ \ \ \ \ (118)$

Therefore, the PDE with homogeneous Dirichlet boundary conditions has a simple explicit solution.

${\square}$

The product solutions are also called the normal modes of vibration. The coefficients of ${t}$ inside the product solutions, namely ${\frac{n\pi c}{L}, n=1,2,3...}$ , are called the frequencies. The fundamental mode of the string has a frequency of ${\frac{\pi c}{L}}$ . The ${n}$ th overtone is just the ${n}$ th integral multiple of the fundamental.

We now consider the motion of a vibrating string governed by the homogeneous Neumann boundary conditions for the wave equation:

$\displaystyle \mathrm{PDE: } \hspace{0.5cm} u_{tt}=c^2u_{xx} \ \ \ \ \ (119)$

$\displaystyle \mathrm{BC1:} \hspace{0.5cm} u_x(0,t)=0 \ \ \ \ \ (120)$

$\displaystyle \mathrm{BC2:}\hspace{0.5cm} u_x(L,t)=0 \ \ \ \ \ (121)$

$\displaystyle \mathrm{IC1:}\hspace{0.5cm} u(x,0) = f(x) \ \ \ \ \ (122)$

$\displaystyle \mathrm{IC2:}\hspace{0.5cm} u_x(x,0) = g(x) \ \ \ \ \ (123)$

Solution Again, we use the method of separation of variables. We look for production solutions of the form:

$\displaystyle u(x,t)=\phi(x)h(t) \ \ \ \ \ (124)$

with

$\displaystyle h''=-\lambda hc^2 \ \ \ \ \ (125)$

$\displaystyle \phi''=-\lambda \phi \ \ \ \ \ (126)$

The general solution for (126) is

$\displaystyle \phi(x) = c_1\mathrm{cos}(\sqrt{\lambda}x)+c_2\mathrm{sin}(\sqrt{\lambda}x) \ \ \ \ \ (127)$

We also need

$\displaystyle \phi'(x) = \sqrt{\lambda}(c_2\mathrm{cos}(\sqrt{\lambda}x)-c_1\mathrm{sin}(\sqrt{\lambda}x)) \ \ \ \ \ (128)$

Apply ${\phi'(0)=0}$ and we arrive at ${c_2=0}$ . Apply ${\phi'(L)=0}$ and we arrive at ${\sqrt{\lambda} c_1\mathrm{sin}(\sqrt{\lambda}x)=0}$ . Since ${c_1=0}$ yields a trivial solution, we get a nontrivial solution if and only if

$\displaystyle \mathrm{sin}(\sqrt{\lambda}L)=0 \ \ \ \ \ (129)$

This means that

$\displaystyle \sqrt{\lambda}L= n\pi \ \ \ \ \ (130)$

The eigenvalues are

$\displaystyle \lambda= (\frac{n\pi}{L})^2, n=1,2,3... \ \ \ \ \ (131)$

The eigenfunctions corresponding to the eigenvalues are

$\displaystyle \phi(x)=c_1 \mathrm{cos}(\frac{n\pi x}{L}), n=1,2,3... \ \ \ \ \ (132)$

The time-dependent part of the solution is

$\displaystyle h(t) = c_3\mathrm{cos}(\frac{cn\pi t}{L})+c_4\mathrm{sin}(\frac{cn\pi t}{L}) \ \ \ \ \ (133)$

for ${n=0,1,2,3...}$ . Thus, for ${n\geq 0}$ , each of the functions

$\displaystyle u_n(x,t)=\phi_n(x)h_n(t) = \mathrm{cos}(\frac{n\pi x}{L})(A_n\mathrm{cos}(\frac{cn\pi t}{L})+B_n\mathrm{sin}(\frac{cn\pi t}{L})) \ \ \ \ \ (134)$

is a solution to the PDE and satisfies the boundary condition. By superposition principle, we can solve the initial value problem by considering a linear combinations of all product solutions:

$\displaystyle u(x,t) = A_0+\sum^{\infty}_{1}\mathrm{cos}(\frac{n\pi x}{L})(A_n\mathrm{cos}(\frac{cn\pi t}{L})+B_n\mathrm{sin}(\frac{cn\pi t}{L})) \ \ \ \ \ (135)$

The initial conditions in (97) and (98) are satisfied if,

$\displaystyle f(x)=A_0+\sum^{\infty}_{1}A_n\mathrm{cos}(\frac{n\pi x}{L}) \ \ \ \ \ (136)$

$\displaystyle g(x)=A_0+\sum^{\infty}_{1}B_n\frac{cn\pi}{L}\mathrm{cos}(\frac{n\pi x}{L}) \ \ \ \ \ (137)$

for ${0<x<L}$ . We use the fact that ${\mathrm{cos}(\frac{n\pi x}{L})}$ satisfies the following orthogonality relation:

$\displaystyle \int_{0}^{L} \mathrm{cos}(\frac{n \pi x}{L})\mathrm{cos}(\frac{m \pi x}{L}) \,dx = \left\{ \begin{array}{lr} 0 & n \neq m \\ \frac{L}{2} & n=m\neq 0 \\ L & n=m=0 \end{array} \right. \ \ \ \ \ (138)$

Multiply (136) by ${\mathrm{cos}(\frac{m \pi x}{L})}$ and integrating from ${0}$ to ${L}$ yields

$\displaystyle \int_{0}^{L} f(x)\mathrm{cos}(\frac{m \pi x}{L})\mathrm{dx} = A_{n}\int_{0}^{L}\mathrm{cos^2}(\frac{m\pi x}{L})\mathrm{dx} \ \ \ \ \ (139)$

Solving for ${A_n}$ yields,

$\displaystyle A_0= \frac{1}{L} \int_{0}^{L}f(x) \mathrm{dx} \ \ \ \ \ (140)$

$\displaystyle A_n=\frac{2}{L} \int_{0}^{L}f(x)\mathrm{cos}(\frac{n \pi x}{L})\mathrm{dx}, n\geq 1 \ \ \ \ \ (141)$

Multiply (137) by ${\mathrm{cos}(\frac{m \pi x}{L})}$ and integrating from ${0}$ to ${L}$ yields

$\displaystyle B_n\frac{cn\pi}{L}=\frac{2}{L} \int_{0}^{L}f(x)\mathrm{cos}(\frac{n \pi x}{L})\mathrm{dx}, n\geq 1 \ \ \ \ \ (142)$

${\square}$

Notes on Fourier analysis (part 2)

May 7, 2022May 7, 2022Michael LiuLeave a comment

1.3. Uniform convergence

Continuity is not necessarily preserved under pointwise limits. For example, Suppose that ${f_n:[0,1]\rightarrow \mathbb{R}}$ is defined by ${f_n(x)=x^n}$ . If ${0<x<1}$ , then ${x^n\rightarrow0}$ as ${n\rightarrow\infty}$ . If ${x=1}$ , then ${x^n\rightarrow1}$ as ${n\rightarrow\infty}$ . Thus, ${f_n}$ is continuous on ${[0,1]}$ but the pointwise limit ${f}$ is not (it is discontinuous at 1). On the other hand, uniform convergence of continuous functions does guarantee continuity.

Definition 3.1 Uniform convergence The series of continuous function ${\sum^{\infty}_{n=1} f_n}$ converges uniformly on the closed interval ${[a,b]}$ if the following two conditions are satisfied: First, the series converges to the sum ${S(x)}$ for each ${x}$ value in the interval; Second, given any number ${\epsilon >0}$ , there exists a ${N\in\mathbb{N}}$ such that for all ${m\geq N}$ the partial sums ${S_m=\sum^{m}_{n=1} f_n}$ satisfy ${\mid S(x)-S_m(x) \mid\leq\epsilon}$ for every ${x}$ value in the interval.

The following theorem gives an easily applied condition for determining uniform convergence. It is especially useful in its applicaion to Fourier series.

Theorem 3.2 Weierstrass’ ${M}$ -Test If for all points ${x}$ in ${[a,b]}$ we have ${\mid f_n(x)\mid \leq M_n }$ from a certain ${n=k}$ on, and the series of positive numbers ${\sum^{\infty}_{n=k}M_n}$ converges, then the series ${\sum^{\infty}_{n=1}f_n}$ converges uniformly on ${[a,b]}$ .

Proof For each value ${x}$ in the interval ${[a,b]}$ , since ${\mid f_n(x)\mid \leq M_n }$ and ${\sum^{\infty}_{n=k}M_n}$ converges, then by comparison test, the series ${\sum^{\infty}_{n=k}f_n}$ converges absolutely. Therefore, the series ${\sum^{\infty}_{n=k}f_n}$ converges absolutely for every ${x}$ on the interval ${[a,b]}$ .

Given some ${\epsilon>0}$ , then there exist an ${ \in \mathbb{N}}$ such that ${m>N}$ implies ${\sum^{\infty}_{n=m+1}M_n < \epsilon}$ .

By Triangle inequality, we have for each ${x}$ in ${[a,b]}$

$\displaystyle \lvert S(x)-S_m(x)\rvert \leq \lvert\sum^{\infty}_{n=m+1}f_n(x)\rvert \leq\sum^{\infty}_{n=m+1}\lvert f_n(x)\rvert \leq\sum^{\infty}_{n=m+1}M_n < \epsilon \ \ \ \ \ (62)$

Thus, for all ${m\geq\mathbb{N}}$ ,

$\displaystyle \mid S(x)-S_m(x) \mid < \epsilon \ \ \ \ \ (63)$

for every ${x}$ in ${[a,b]}$ .

${\square}$

In the case of Fourier series, we have the following estimates

$\displaystyle \hspace{0.5cm} \lvert a_n\mathrm{cos}n\theta\rvert\leq \lvert a_n \rvert \hspace{0.5cm} \lvert b_n\mathrm{sin}n\theta\rvert\leq \lvert b_n \rvert \hspace{0.5cm} \lvert c_ne^{in\theta}\rvert\leq \lvert c_n \rvert \ \ \ \ \ (64)$

Hence the Weierstrass’s ${M}$ -test will apply to a Fourier series in trigonometric form if ${\sum^{\infty}_0\lvert a_n \rvert<\infty}$ and ${\sum^{\infty}_1\lvert b_n \rvert<\infty}$ , and to Fourier series in exponential form if ${\sum^{\infty}_{-\infty}\lvert c_n \rvert<\infty}$ .

Another theorem that we will use for the proof of uniform convergence is the Cauchy-Schwarz Inequality.

Theorem 3.3 Cauchy-Schwarz Inequality Let ${ {a_n}^N_{n=1}}$ and ${ {b_n}^N_{n=1}}$ be two finite sets of real numbers. Then,

$\displaystyle (\sum^{N}_{n=1}a_nb_n)^2\leq (\sum^{N}_{n=1}a_n^2)(\sum^{N}_{n=1}b_n^2) \ \ \ \ \ (65)$

When expressed in vector form, the Cauchy-Schwarz inequality says that the dot product of two vectors is bounded by the product of their norms.

Proof Expanding out the brackets and collecting identical terms, we have

$\displaystyle \begin{aligned} \sum^{N}_{i=1}\sum^{N}_{j=1}(a_ib_j-a_jb_i)^2&=\sum^{N}_{i=1}a_i^2\sum^{N}_{j=1}b_j^2+\sum^{N}_{j=1}a_j^2\sum^{N}_{i=1}b_i^2-2\sum^{N}_{i=1}a_ib_i\sum^{N}_{j=1}a_jb_j\\ &= 2(\sum^{N}_{i=1}a_i^2)(\sum^{N}_{i=1}b_i^2)-2 (\sum^{N}_{i=1}a_ib_i)^2 \end{aligned} \ \ \ \ \ (66)$

The left-hand side of the equation is greater than or equal to zero since it is a sum of the squares of real numbers. Thus,

$\displaystyle (\sum^{N}_{i=1}a_i^2)(\sum^{N}_{i=1}b_i^2)\geq(\sum^{N}_{i=1}a_ib_i)^2 \ \ \ \ \ (67)$

${\square}$

The graph of periodic functions that are both continuous and piecewise smooth is a smooth curve except that is can have “corners” where the derivatives jump. The fundamental theorem of calculus,

$\displaystyle f(b)-f(a)=\int^b_a f'(\theta) \mathrm{d}\theta \ \ \ \ \ (68)$

applies to functions ${f}$ that are continuous and piecewise smooth, even though ${f'}$ is defined at the “corners”. To see this, let ${f}$ to be differentiable except at point ${c\in(a,b)}$ , we have

$\displaystyle \begin{aligned} \int^b_a f'(\theta) \mathrm{d}\theta &= \int^b_c f'(\theta) \mathrm{d}\theta+\int^c_a f'(\theta) \mathrm{d}\theta\\ &= (f(b)-f(c))+(f(c)-f(a))\\ &= f(b)-f(a) \end{aligned} \ \ \ \ \ (69)$

We introduce the following theorem as a preliminary step towards the main convergence theorem.

Theorem 3.4 Suppose ${f}$ is ${2\pi}$ -periodic, continuous, and piecewise smooth. Let ${a_n}$ , ${b_n}$ , and ${c_n}$ be the Fourier coefficients of ${f}$ defined in (2.5) and (2.6), and let ${a_n'}$ , ${b_n'}$ , and ${c_n'}$ be the corresponding Fourier coefficients of ${f'}$ . Then

$\displaystyle \hspace{0.5cm} a_n'=nb_n, \hspace{0.5cm} b_n'=-na_n, \hspace{0.5cm} c'_n=inc_n \ \ \ \ \ (70)$

Proof This is a simple matter of integration by parts.

${\square}$

From theorem 3.4 we obtain the following results on differentiation and integration of Fourier series.

Theorem 3.5 Suppose ${f}$ is ${2\pi}$ -periodic, continuous, and piecewise smooth, and suppose also that ${f'}$ is piecewise smooth. If

$\displaystyle \frac{1}{2}a_0+\sum_{n=1}^{\infty}(a_n\mathrm{cos}(n\theta)+b_n\mathrm{sin}(n\theta))=\sum_{-\infty}^{\infty}c_ne^{in\theta} \ \ \ \ \ (71)$

is the Fourier series of ${f(\theta)}$ , then ${f'(\theta)}$ is the sum of the derived series

$\displaystyle \frac{1}{2}a_0+\sum_{n=1}^{\infty}(nb_n\mathrm{cos}(n\theta)-na_n\mathrm{sin}(n\theta))=\sum_{-\infty}^{\infty}inc_ne^{in\theta} \ \ \ \ \ (72)$

for all ${\theta}$ at which ${f'(\theta)}$ exists. At the exceptional points ${f'}$ has jumps, the series converges to ${\frac{1}{2}[f'(\theta_-)+f'(\theta_+)]}$ .

Proof Since ${f'}$ is piecewise smooth, by Theorem 2.6, it is the sum of it Fourier series at every point. By theorem 3.4, the coefficients of ${e^{in\theta}}$ , ${\mathrm{sin}(n\theta)}$ , ${\mathrm{cos}(n\theta)}$ in this series are ${inc_n}$ , ${-na_n}$ , and ${nb_n}$ respectively. Thus theorem 3.5 follows.

${\square}$

Theorem 3.6 Suppose ${f}$ is ${2\pi-}$ periodic and piecewise continuous, with Fourier coefficients ${a_n}$ , ${b_n}$ , ${c_n}$ , and let ${F(\theta)=\int^\theta_{0}f(\phi)\mathrm{d}\phi}$ . If ${c_0(=\frac{1}{2}a_0)=0}$ , then for all ${\theta}$ we have

$\displaystyle F(\theta)=\frac{1}{2}A_0+\sum_{n=1}^{\infty}(\frac{a_n}{n}\mathrm{sin}(n\theta)-\frac{b_n}{n}\mathrm{cos}(n\theta))=C_0+\sum_{n \neq 0}\frac{c_n}{in}e^{in\theta} \ \ \ \ \ (73)$

where the constant term is the mean value of ${F}$ on ${[-\pi,\pi]}$ :

$\displaystyle C_0=\frac{1}{2}A_0=\frac{1}{2\pi}\int^{\pi}_{-\pi}F(\theta)\mathrm{d}\theta \ \ \ \ \ (74)$

The series on the right of (72) is the series obtained by formally integrating the Fourier series of ${f}$ term by term, whether the latter series actually converges or not. If ${c_0 \neq 0}$ , the sum of the series on the right of (72) is ${F(\theta)-c_0\theta}$ .

Proof ${F}$ is continuous and piecewise smooth, being the integral of a piecewise continuous function. Moreover, if ${c_0=0}$ , ${F}$ is ${2\pi-}$ periodic, for

$\displaystyle F(\theta+2\pi)-F(\theta)=\int^{\theta+2\pi}_{\theta} f(\phi)\mathrm{d}\phi=\int^{\pi}_{-\pi} f(\phi)\mathrm{d}\phi=2\pi c_0=0 \ \ \ \ \ (75)$

Hence, by theorem 2.6, ${F(\theta)}$ is the sum of its Fourier series at every ${\theta}$ . But by theorem 3.4 applied to F, the Fourier coefficients ${A_n, B_n}$ , and ${C_n}$ of ${F}$ are related to those of ${f}$ by

$\displaystyle \hspace{0.5cm} A_n'=-\frac{b_n}{n}, \hspace{0.5cm} B_n'=\frac{a_n}{n}, \hspace{0.5cm}C_n=\frac{c_n}{in} \hspace{0.5cm} (n\neq0). \ \ \ \ \ (76)$

The formula (74) for the constant ${C_0}$ or ${\frac{1}{2}A_0}$ is just the usual formula for the zeroth Fourier coefficient of ${F}$ . If ${c\neq0}$ , these arguments can be applied to the function ${f(\theta)-c_0}$ rather than ${f(\theta)}$ , yielding the final assertion.

${\square}$

We will now prove a theorem on uniform convergence of Fourier series.

Theorem 3.7 If ${f}$ is ${2\pi-}$ periodic, continuous, and piecewise smooth, then the Fourier series of ${f}$ converges to ${f}$ absolutely and uniformly on ${\mathbb{R}}$ .

Proof Let ${c_n'}$ denote the Fourier coefficients of ${f'}$ . By theorem 3.4 we know that ${c_n=(in)^-1c_n'}$ for ${n \neq 0}$ , and by (29) (Bessel’s inequality) applied to ${f'}$ ,

$\displaystyle \sum^{\infty}_{-\infty}|c_n'|^2\leq \frac{1}{2\pi}\int^{\pi}_{-\pi}|f'(\theta)|^2\mathrm{d}\theta<\infty \ \ \ \ \ (77)$

Hence, by the Cauchy-Schwartz inequality,

$\displaystyle \sum^{\infty}_{-\infty}|c_n|=\lvert c_0\rvert+\sum_{n\neq0}\lvert \frac{c_n'}{in} \rvert\leq \lvert c_0\rvert+(\sum_{n\neq 0}\frac{1}{n^2})^{\frac{1}{2}}(\sum_{n\neq 0}\lvert c_n'\rvert^2)^{\frac{1}{2}} <\infty \ \ \ \ \ (78)$

since ${\sum_{n\neq 0}(\frac{1}{n^2})=2\sum_{n=1}^{\infty}(\frac{1}{n^2})<\infty}$ . This completes the proof.

${\square}$

Notes on Fourier analysis

May 3, 2022May 25, 2022Michael LiuLeave a comment

1.1. Definition

When solving heat equations with linear and homogeneous boundary conditions, we obtained product solutions in terms of infinite series of sines and cosines. Joseph Fourier developed his ideas on the convergence of these trigonometric series while studying heat flow.

We now study Fourier series of a periodic function. Suppose ${f(\theta)}$ is a function defined on the real line such that ${f(\theta+2\pi)=f(\theta)}$ for all ${\theta}$ , then we say that ${f(\theta)}$ is ${2\pi}$ -periodic. We shall assume that ${f}$ is Riemann integrable on every bounded interval; this will be the case if ${f}$ is bounded and is continuous except at finitely many points in each bounded interval. We allow ${f}$ is be complex-valued (we will see why later). We wish to obtain the following series expansion:

$\displaystyle f(\theta)=\frac{1}{2}a_0+\sum_{n=1}^{\infty}(a_n\mathrm{cos}(n\theta)+b_n\mathrm{cos}(n\theta)) \ \ \ \ \ (1)$

The series expansion of ${f(\theta)}$ can also be written in terms of complex exponential functions ${e^{i\theta}}$ . We use the following properties:

$\displaystyle \mathrm{cos}(\theta)=\frac{e^{i\theta}+e^{-i\theta}}{2} \ \ \ \ \ (2)$

$\displaystyle \mathrm{sin}(\theta)=\frac{e^{i\theta}-e^{-i\theta}}{2i} \ \ \ \ \ (3)$

$\displaystyle e^{i\theta}= \mathrm{cos}(\theta)+i\mathrm{sin}(\theta) \ \ \ \ \ (4)$

Equation (1) can be rewritten as

$\displaystyle f(\theta)=\sum_{-\infty}^{\infty}c_ne^{in\theta} \ \ \ \ \ (5)$

where

$\displaystyle c_0=\frac{1}{2}a_0 {\hspace{0.5cm}} c_n=\frac{1}{2}(a_n-ib_n) {\hspace{0.5cm}} c_{-n}=\frac{1}{2}(a_n+ib_n) {\hspace{0.5cm}} n=1,2,3... \ \ \ \ \ (6)$

We can consider (5) as the sum of two infinity series, one going from ${n=0}$ to ${+\infty}$ and one going from ${n=-1}$ to ${-\infty}$ .

Alternatively, we can express the coefficients in the following way:

$\displaystyle a_0=2c_0 {\hspace{0.5cm}} a_n=c_n+c_{-n} {\hspace{0.5cm}} b_n=i(c_n-c_{-n}){\hspace{0.5cm}} {\hspace{0.5cm}} n=1,2,3... \ \ \ \ \ (7)$

How can the coefficient ${c_n}$ be expressed in terms of ${f}$ ? Let’s first assume that the series is term by term integrable, we multiply equation (5) by ${e^{-ik\theta}}$ , ${k}$ is an integer, and integrate both sides from ${-\pi}$ to ${\pi}$ . We obtain

$\displaystyle \int^{\pi}_{-\pi}f(\theta)e^{-ik\theta}\mathrm{d}\theta = \sum_{-\infty}^{\infty}c_n \int^{\pi}_{-\pi}e^{i(n-k)\theta}\mathrm{d}\theta \ \ \ \ \ (8)$

Note that

$\displaystyle \int^{\pi}_{-\pi}e^{i(n-k)\theta}\mathrm{d}\theta=\frac{1}{{i(n-k)}}e^{i(n-k)\theta}\Big|^\pi_{-\pi}=0 {\hspace{0.5cm}} n\neq k \ \ \ \ \ (9)$

$\displaystyle \int^{\pi}_{-\pi}{d}\theta = 2\pi {\hspace{0.5cm}} n= k \ \ \ \ \ (10)$

Hence, the only non-zero term after the integration is the term ${n=k}$ , and thus we have

$\displaystyle \int^{\pi}_{-\pi}f(\theta)e^{-ik\theta}\mathrm{d}\theta=c_k2\pi \ \ \ \ \ (11)$

Relabeling the term ${c_k}$ as ${c_n}$ we get

$\displaystyle c_n=\frac{1}{2\pi}\int^{\pi}_{-\pi}f(\theta)e^{-in\theta}\mathrm{d}\theta \ \ \ \ \ (12)$

This means that we can also find ${a_0}$ , ${a_n}$ and ${b_n}$ easily

$\displaystyle a_0=2c_0=\frac{1}{\pi}\int^{\pi}_{-\pi}f(\theta)\mathrm{d}\theta \ \ \ \ \ (13)$

$\displaystyle a_n=c_n+c_{-n}= \frac{1}{2\pi}\int^{\pi}_{-\pi}f(\theta)(e^{-in\theta}+e^{in\theta})\mathrm{d}\theta = \frac{1}{\pi}\int^{\pi}_{-\pi}f(\theta)\mathrm{cos}(n\theta)\mathrm{d}\theta \ \ \ \ \ (14)$

$\displaystyle b_n=i(c_n-c_{-n})=\frac{i}{2\pi}\int^{\pi}_{-\pi}f(\theta)(e^{-in\theta}-e^{in\theta})\mathrm{d}\theta = \frac{1}{\pi}\int^{\pi}_{-\pi}f(\theta)\mathrm{sin}(n\theta)\mathrm{d}\theta \ \ \ \ \ (15)$

for ${n=1,2,3...}$

To recapitulate: if ${f}$ has a series expansion of the form in (1), and if the series converges decently such that term by term integration is permissible, then the coefficients ${a_n, b_n}$ and ${c_n}$ are given by (14), (15), (12) respectively. But now if ${f}$ is any Riemann-integrable periodic functions, the integrals in (14), (15), and (12) make perfectly good sense, and we use them to define the coefficients ${a_n, b_n}$ and ${c_n}$ . We are able to give the formal definition as follows:

Definition 1.1 Suppose ${f}$ is periodic with period ${2\pi}$ and integrable over ${[-\pi,\pi]}$ . The numbers ${a_n}$ and ${b_n}$ defined in (14) and (15), or the number ${c_n}$ defined in (12), are called the Fourier coefficients of ${f}$ , and the corresponding series

$\displaystyle \frac{1}{2}a_0+\sum_{n=1}^{\infty}(a_n\mathrm{cos}(n\theta)+b_n\mathrm{sin}(n\theta)) {\hspace{0.5cm}} \mathrm{or} {\hspace{0.5cm}} \sum_{-\infty}^{\infty}c_ne^{in\theta} \ \ \ \ \ (16)$

are called the Fourier series of ${f}$ .

We also present the following Lemmas:

Lemma 1.2 With reference to the formulas (14) and (15),

if f is even,

$\displaystyle a_n=\frac{2}{\pi}\int^{\pi}_{0}f(\theta)\mathrm{cos}(n\theta)\mathrm{d}\theta {\hspace{0.5cm}}and {\hspace{0.5cm}} b_n=0 \ \ \ \ \ (17)$

if f is odd,

$\displaystyle b_n=\frac{2}{\pi}\int^{\pi}_{0}f(\theta)\mathrm{sin}(n\theta)\mathrm{d}\theta {\hspace{0.5cm}}and {\hspace{0.5cm}} a_n=0 \ \ \ \ \ (18)$

For Fourier series of a ${2\pi}$ -periodic function, the constant term of series is

$\displaystyle c_0=\frac{1}{2}a_0=\frac{1}{2\pi}\int^{\pi}_{-\pi}f(\theta)\mathrm{d}\theta \ \ \ \ \ (19)$

which is equivalent to the mean value of ${f}$ in the interval ${[-\pi,\pi]}$ . Thus,

Lemma 1.3 The constant term in the Fourier series of a ${2\pi}$ -periodic function ${f}$ is the mean value of ${f}$ on an interval of length ${2\pi}$ .

We now turn to the discussion of periodic functions with an arbitrary period ${2L}$ . Similarly, we obtain the following definition of the Fourier series.

Definition 1.4 For a function ${f}$ with period ${2L}$ the Fourier coefficients of ${f}$ are

$\displaystyle a_n=\frac{1}{L}\int^{L}_{-L}f(x)\mathrm{cos}(\frac{n\pi x}{L})\mathrm{d}x \ \ \ \ \ (20)$

$\displaystyle b_n=\frac{1}{L}\int^{L}_{-L}f(x)\mathrm{sin}(\frac{n\pi x}{L})\mathrm{d}x \ \ \ \ \ (21)$

$\displaystyle c_0=\frac{1}{2L}\int^{L}_{-L}f(x)e^{-i(\frac{n\pi x}{L})}\mathrm{d}x \ \ \ \ \ (22)$

and the corresponding series

$\displaystyle \frac{1}{2}a_0+\sum_{n=1}^{\infty}(a_n\mathrm{cos}(n\frac{\pi x}{L})+b_n\mathrm{sin}(\frac{n\pi x}{L})) {\hspace{0.5cm}} \mathrm{or} {\hspace{0.5cm}} \sum_{-\infty}^{\infty}c_ne^{i\frac{n\pi x}{L}} \ \ \ \ \ (23)$

are called the Fourier series of function ${f}$ .

Figure 1. Graph of partial sums (n=10, 30, 80) of the Fourier series of a square wave.

Definition 1.5 Let ${f}$ be a function defined on the interval ${[-L,L]}$ . The periodic extension of ${f}$ is that function, denoted by ${f_p}$ , which satisfies for ${p=2L}$

$\displaystyle f_p(x+kp)=f(x) \ \ \ \ \ (24)$

for each integer k.

For an even function ${f}$ over ${[-L,L]}$ , the Fourier series for ${f}$ has the form

$\displaystyle f(x)=\frac{1}{2}a_0+\sum_{n=1}^{\infty}a_n\mathrm{cos}(n\frac{\pi x}{L}) \hspace{0.5cm} a_n=\frac{2}{L}\int^{L}_{0}f(x)\mathrm{cos}(\frac{n\pi x}{L})\mathrm{d}x \ \ \ \ \ (25)$

Formula (25) is called the cosine series of ${f}$ over ${[0,L]}$ .

For an odd function ${f}$ over ${[-L,L]}$ , the Fourier series for ${f}$ has the form

$\displaystyle f(x)=\frac{1}{2}a_0+\sum_{n=1}^{\infty}b_n\mathrm{sin}(n\frac{\pi x}{L}) \hspace{0.5cm} a_n=\frac{2}{L}\int^{L}_{0}f(x)\mathrm{sin}(\frac{n\pi x}{L})\mathrm{d}x \ \ \ \ \ (26)$

Formula (26) is called the sine series of ${f}$ over ${[0,L]}$ .

Definition 1.6 Let the function ${f}$ be defined on ${[0,L]}$ . The even extension ${f_e}$ and the odd extension ${f_o}$ of ${f}$ are the following functions

$\displaystyle f_e(x) \left\{ \begin{array}{lr} f(x) \\ f(-x) \\ \end{array} \right. \ \ \ \ \ (27)$

$\displaystyle f_o(x) \left\{ \begin{array}{lr} f(x) \\ -f(-x) \\ \end{array} \right. \ \ \ \ \ (28)$

${\square}$

1.2. A convergence theorem

Now that we have formally introduced the Fourier series. However, one important question remains unresolved: How do we know whether Fourier series converges for all function ${f(x)}$ ? Proving the convergence of the Fourier series is not a simple matter. Firstly, let’s derive an estimate of the Fourier coefficients that will be needed to establish convergence.

Theorem 2.1 (Bessel’s Inequality) If ${f}$ is ${2\pi}$ -periodic and Riemann integrable on ${[-\pi,\pi]}$ , and the Fourier coefficients are defined by (16), then

$\displaystyle \sum^{\infty}_{-\infty}|c_n|^2\leq \frac{1}{2\pi}\int^{\pi}_{-\pi}|f(\theta)|^2\mathrm{d}\theta \ \ \ \ \ (29)$

Proof We use the following property of complex numbers,

$\displaystyle \mid z \mid^2=z\overline{z} \ \ \ \ \ (30)$

and let

$\displaystyle \begin{aligned} \mid f(\theta)-\sum^{N}_{-N}c_ne^{in\theta}. \mid^2 & =( f(\theta)-\sum^{N}_{-N}c_ne^{in\theta})( \overline{f(\theta)}-\sum^{N}_{-N}\overline{c_n}e^{-in\theta}) \\ & = \mid f(\theta) \mid^2 -\sum^{N}_{-N} [f(\theta)\overline{c_n}e^{-in\theta}+\overline{f(\theta)}c_ne^{in\theta}]+\sum^{N}_{-N}c_n\overline{c_n}\\ \end{aligned} \ \ \ \ \ (31)$

Divide both sides of (31) by ${2\pi}$ and integrate from ${-\pi}$ to ${\pi}$ , we obtain the following using formula (12),

$\displaystyle \begin{aligned} \frac{1}{2\pi}\int^{\pi}_{-\pi}\mid f(\theta)-\sum^{N}_{-N}c_ne^{in\theta}. \mid^2 \mathrm{d}\theta & = \frac{1}{2\pi}\int^{\pi}_{-\pi} \mid f(\theta) \mid^2\mathrm{d}\theta -\sum^{N}_{-N} \overline{c_n}c_n+c_n\overline{c_n}+\sum^{N}_{-N}c_n\overline{c_n}\\ & = \frac{1}{2\pi}\int^{\pi}_{-\pi} \mid f(\theta) \mid^2\mathrm{d}\theta -\sum^{N}_{-N} \overline{c_n}c_n\\ & = \frac{1}{2\pi}\int^{\pi}_{-\pi} \mid f(\theta) \mid^2 \mathrm{d}\theta-\sum^{N}_{-N}\mid c_n\mid^2\\ \end{aligned} \ \ \ \ \ (32)$

It is obvious that the left-hand-side of (32) is nonnegative, thus

$\displaystyle \frac{1}{2\pi}\int^{\pi}_{-\pi} \mid f(\theta) \mid^2 \mathrm{d}\theta-\sum^{N}_{-N} \mid c_n\mid^2 \geq 0 \ \ \ \ \ (33)$

Letting ${n\rightarrow \infty}$ , we have

$\displaystyle \frac{1}{2\pi}\int^{\pi}_{-\pi} \mid f(\theta) \mid^2 \mathrm{d}\theta \geq \sum^{\infty}_{-\infty}\mid c_n\mid^2 \ \ \ \ \ (34)$

${\square}$

Bessel’s inequality can also be stated in terms of ${a_n}$ and ${b_n}$ defined by (16). By equation (7), for ${n\geq1}$ we have

$\displaystyle \begin{aligned} \mid a_n \mid^2 + \mid b_n \mid^2 & = a_n\overline{a_n}+b_n\overline{b_n}\\ & = (c_n+c_{-n})(\overline{c_n}+\overline{c_{-n}})+i(c_n-c_{-n})(-i)(\overline{c_n}-\overline{c_{-n}})\\ & = 2c_n\overline{c_n}+2c_{-n}\overline{c_{-n}} \end{aligned} \ \ \ \ \ (35)$

so that

$\displaystyle \mid a_0 \mid^2=4\mid c_0 \mid^2, {\hspace{0.5cm}} \mid a_n \mid^2+\mid b_n \mid^2 = 2(\mid c_n \mid^2 + \mid c_{-n} \mid^2) {\hspace{0.3cm}} \mathrm{for} {\hspace{0.3cm}} n\geq1 \ \ \ \ \ (36)$

Therefore,

$\displaystyle \frac{1}{4} \mid a_0 \mid^2+\frac{1}{2}\sum^{\infty}_{n=1} (\mid a_n \mid^2+\mid b_n \mid^2)= \sum^{\infty}_{-\infty} \overline\mid c_n\mid^2 \leq \frac{1}{2\pi}\int^{\pi}_{-\pi} \mid f(\theta) \mid^2 \mathrm{d}\theta \ \ \ \ \ (37)$

The significance of (37) is that the series ${\sum \mid a_0 \mid^2}$ , ${\sum \mid b_0 \mid^2}$ and ${\sum \mid c_0 \mid^2}$ are all convergent. As a consequence, we obtain the following result.

Corollary 2.2 The Fourier coefficients ${a_n}$ , ${b_n}$ , and ${c_n}$ all tend to ${0}$ as ${n\rightarrow\infty}$ (and as ${n\rightarrow-\infty}$ in the case of ${c_n}$ ).

Proof ${\mid a_n \mid^2}$ , ${\mid b_n \mid^2}$ and ${\mid c_n \mid^2}$ are the ${n}$ th term of convergent series, so they all tend to ${0}$ as ${n\rightarrow\infty}$ ; hence so do ${a_n}$ , ${b_n}$ , and ${c_n}$ .

In fact, corollary 2.2 is a special case of the Riemann-Lebesgue lemma. Next, we define the class of functions with which we shall be working.

Definition 2.3 Suppose ${-\infty<a<b<\infty}$ , we say that a function ${f}$ on the closed interval ${[a,b]}$ is piecewise continuous provided that (i) f is continous on ${[a,b]}$ except perhaps at finitely many points ${x_1,...,x_k}$ ;(ii) at each of the points ${x_1,...,x_k}$ the left-hand side and right-hand side limits of ${f}$ ,

$\displaystyle f(x_j-)=\lim_{h\rightarrow0, h>0}(x_j-h) \ \ \ \ \ (38)$

and

$\displaystyle f(x_j+)=\lim_{h\rightarrow0, h>0}(x_j+h) \ \ \ \ \ (39)$

exists. (If ${a}$ (or ${b}$ ) are one of the exception points ${x_j}$ , then we require only the left-hand (or the right-hand) limit to exist). Thus, we say that ${f}$ is piecewise continuous on the ${[a,b]}$ if f is continuous there except for finitely many finite jump continuities.

Figure 2. A piecewise continuous function.

Definition 2.4 A function ${f}$ on the closed interval ${[a,b]}$ is piecewise smooth if ${f}$ and its first derivative ${f'}$ are both piecewise continuous.

Pictorially, ${f}$ is piecewise smooth if its graph is a smooth curve except for finitely many jumps (where ${f}$ is discontinuous) and corners (where ${f'}$ is discontinuous). We do not allow infinite discontinuities or sharp cusps.

Figure 3. A piecewise smooth function (left) and a function that is not piecewise smooth (right).

A very useful tool in Fourier analysis is the Dirichlet’s kernel for the ${n}$ th partial sum

$\displaystyle S_N^f(\theta)=\frac{1}{2}a_0+\sum_{n=1}^{N}(a_n\mathrm{cos}(n\theta)+b_n\mathrm{cos}(n\theta))=\sum_{-N}^{N}c_ne^{in\theta} \ \ \ \ \ (40)$

of the Fourier series for ${f}$ . Substitute ${c_n}$ as defined in (16) into (31), we get

$\displaystyle S_N^f(\theta)=\sum_{-N}^{N}\frac{1}{2\pi}\int^{\pi}_{-\pi}f(\psi)\mathrm e^{in(\theta-\psi)} {d}\psi \ \ \ \ \ (41)$

Note that we changed the variable of integration from ${\theta}$ to ${\psi}$ for later convenience. We obtain the following

$\displaystyle S_N^f(\theta)= \frac{1}{2\pi}\sum_{-N}^{N}\int^{\pi}_{-\pi}f(\psi) e^{in(\psi-\theta)} \mathrm{d}\psi \ \ \ \ \ (42)$

by replacing ${n}$ with ${-n}$ . The sum is not affected since it ranges from ${-N}$ to ${N}$ . Now, letting ${\phi=\psi-\theta}$ , we have

$\displaystyle S_N^f(\theta)= \frac{1}{2\pi}\sum_{-N}^{N}\int^{\pi}_{-\pi}f(\phi+\theta) e^{in\phi} \mathrm{d}\phi\\ \ \ \ \ \ (43)$

We can write (34) as

$\displaystyle S_N^f(\theta)=\int^{\pi}_{-\pi}f(\phi+\theta)D_N(\phi)\mathrm{d}\phi \hspace{0.5cm} \mathrm{where} \hspace{0.5cm} D_N(\phi)= \frac{1}{2\pi}\sum_{-N}^{N}e^{in\phi} \ \ \ \ \ (44)$

The function ${D_N(\phi)}$ is called the ${N}$ th Dirichlet kernel. We introduce the following properties of the Dirichlet kernal:

${D_N(\phi)=\frac{1}{2\pi}+\frac{1}{\pi}\sum^{N}_{n=1}\mathrm{cos}(n\theta)}$
${D_N(\phi)=\frac{1}{2\pi}\frac{\mathrm{sin}(N+\frac{1}{2})\phi}{\mathrm{sin}\frac{1}{2}(\phi)}}$

Proof For (1), we use the property ${e^{ix}+e^{-ix}=\mathrm{cos}(x)}$ to obtain

$\displaystyle \begin{aligned} D_N(\phi)&=\frac{1}{2\pi}\sum^{N}_{-N}e^{in\theta}\\ &= \frac{1}{2\pi}(\sum^{N}_{1}[e^{in\theta}+e^{-in\theta}]+e^{i(0)\theta})\\ &= \frac{1}{\pi}\sum^{N}_{1}\mathrm{cos}(n\theta)+\frac{1}{2\pi} \end{aligned} \ \ \ \ \ (45)$

For (2), we recognize that ${D_N(\phi)}$ is the sum of a finite geometric series:

$\displaystyle D_N(\phi)= \frac{1}{2\pi}e^{-iN\theta}(1+e^{i\theta}+...+e^{i(2N+1)\theta}) \ \ \ \ \ (46)$

Since ${\sum^n_1r^{k}=\frac{r^{n+1}-1}{r-1}}$ , we have

$\displaystyle D_N(\phi)= \frac{1}{2\pi}e^{-iN\phi}\frac{e^{i(2N+1)\phi}-1}{e^{i\phi}-1}=\frac{1}{2\pi}\frac{e^{i(N+1)\phi}-e^{-iN\phi}}{e^{i\phi}-1} \ \ \ \ \ (47)$

Multiplying top and bottom by ${e^{-\frac{i\theta}{2}}}$ ,

$\displaystyle D_N(\phi)=\frac{1}{2\pi}\frac{e^{i(N+\frac{1}{2})\phi}-e^{-i(N-\frac{1}{2})\phi}}{e^{\frac{i\phi}{2}}-e^{-\frac{i\theta}{2}}} \ \ \ \ \ (48)$

using the property ${e^{ix}-e^{-ix}=\mathrm{sin}(x)}$ , we get

$\displaystyle D_N(\phi)=\frac{1}{2\pi}\frac{\mathrm{sin}(N+\frac{1}{2})\phi}{\mathrm{sin}\frac{1}{2}(\phi)} \ \ \ \ \ (49)$

${\square}$

Figure 4. Graph of the Dirichlet kernel (solid) with N=25 on the interval from -π to π .

Lemma 2.5 For any ${N}$ ,

$\displaystyle \int^{0}_{-\pi}D_N(\theta)d\theta = \int^{\pi}_{0}D_N(\theta)d\theta=\frac{1}{2} \ \ \ \ \ (50)$

Proof We integrate both sides of property (45) from ${0}$ to ${\pi}$ ,

$\displaystyle \begin{aligned} \int^{\pi}_{0}D_N(\theta)d\theta&= \int^{\pi}_{0} \frac{1}{\pi}\sum^{N}_{1}\mathrm{cos}(n\theta)+\frac{1}{2\pi}d\theta\\ &= \bigg[\frac{1}{\pi}\sum^{N}_{1}\frac{1}{n}\mathrm{sin}(n\theta)+\frac{\theta}{2\pi}\bigg]^\pi_0\\ &= \frac{1}{2} \end{aligned} \ \ \ \ \ (51)$

${\square}$

We are now ready to present the main convergence theorem. It says that the Fourier series of a function ${f}$ that is piecewise smooth on every bounded interval ${[a,b]}$ converges pointwise to ${f}$ , provided that we redefine ${f}$ at its points of discontinuities to be the average of its left and right hand limits.

Theorem 2.6 If ${f}$ is ${2\pi}$ periodic and piecewise smooth on ${\mathbb{R}}$ , and ${S^{f}_N}$ is defined by (40), then

$\displaystyle \lim_{n\rightarrow\infty} S^{f}_N(\theta)=\frac{1}{2}[f(\theta-)+f(\theta+)] \ \ \ \ \ (52)$

for every ${theta}$ . In particular, ${\lim_{N \rightarrow \infty} S^{f}_N(\theta) = f(\theta)}$ for every ${\theta}$ at which ${f}$ is continuous.

Proof By formula (50), we have

$\displaystyle \frac{1}{2}(f(\theta-))= \int^{0}_{-\pi}f(\theta-)D_N(\phi)d\phi \ \ \ \ \ (53)$

$\displaystyle \frac{1}{2}(f(\theta+))= \int^{\pi}_{0}f(\theta+)D_N(\phi)d\phi \ \ \ \ \ (54)$

and hence by equation (44)

$\displaystyle S^{f}_N(\theta)-\frac{1}{2}[f(\theta-)+f(\theta+)] \ \ \ \ \ (55)$

$\displaystyle \begin{aligned} &=\int^{\pi}_{0}f(\phi+\theta)D_N(\phi)d\phi-\int^{\pi}_{0}f(\theta+)D_N(\phi)d\phi\\ &+\int^{0}_{-\pi}f(\phi+\theta)D_N(\phi)d\phi-\int^{0}_{-\pi}f(\theta-)D_N(\phi)d\phi\\ &=\int^{\pi}_{0}(f(\phi+\theta)-f(\theta+))D_N(\phi)d\phi+\int^{0}_{-\pi}(f(\phi+\theta)-f(\theta-))D_N(\phi)d\phi \end{aligned} \ \ \ \ \ (56)$

We wish to show that for each fixed ${\theta}$ , equation (56) tends to 0 as ${N\rightarrow\infty}$ . By (47) we can write it as

$\displaystyle\frac{1}{2\pi} \int^{\pi}_{-\pi}g(\phi)(e^{i(N+1)\phi}-e^{-iN\phi})d\phi \ \ \ \ \ (57)$

where

$\displaystyle g(\phi)= \frac{f(\phi+\theta)-f(\theta+)}{e^{i\phi}-1}, -\pi<\phi<0 \ \ \ \ \ (58)$

$\displaystyle g(\phi)= \frac{f(\phi+\theta)-f(\theta-)}{e^{i\phi}-1}, 0<\phi<\pi \ \ \ \ \ (59)$

We see that ${g(\phi)}$ is a well-behaved function except when near ${\phi=0}$ . However, since ${\lim_{\phi\rightarrow0} f(\phi+\theta)-f(\theta-)=0}$ and ${\lim_{\phi\rightarrow0} e^{i\phi}-1=0}$ , by L’Hôpital’s rule,

$\displaystyle \lim_{\phi\rightarrow0} g(\phi) = \lim_{\phi\rightarrow0} \frac{f(\phi+\theta)-f(\theta-)}{e^{i\phi}-1} = \lim_{\phi\rightarrow0}\frac{f'(\phi+\theta)}{ie^{i\phi}}=\frac{f'(\theta)}{i} \ \ \ \ \ (60)$

Likewise, ${g(\phi)}$ approaches ${\frac{f'(\theta)}{i}}$ as ${\phi}$ approaches ${0}$ from the right. Thus, ${g(\phi)}$ is piecewise continuous on ${[-\pi,\pi]}$ . By Bessel’s inequality, the Fourier coefficient

$\displaystyle c_n=\frac{1}{2\pi}\int^{\pi}_{-\pi}g(\phi)e^{-in\theta}\mathrm{d}\phi \ \ \ \ \ (61)$

tends to 0 as ${n\rightarrow\pm\infty}$ . However, we know that formula (57) is just ${C_{-(N+1)}-C_N}$ . Thus, equation (55) tends to ${0}$ as ${n\rightarrow\pm\infty}$ . This is what we needed to show.

${\square}$

Theorem 2.6 says that the Fourier series of a ${2\pi}$ -periodic piecewise smooth function ${f}$ converges to ${f}$ everywhere, provided that ${f}$ is (re)defined at each of its points of discontinuity to be the average of its left- and right-hand limits there. With this understanding, we have the following uniqueness theorem.

Corollary 2.7 If ${f}$ and ${g}$ are 2 ${\pi}$ -periodic and piecewise smooth, and ${f}$ and ${g}$ have the same Fourier coefficients, then ${f=g}$ .

Proof ${f}$ and ${g}$ are both the sums of the same Fourier series.

${\square}$

April 28, 2022April 28, 2022Michael LiuLeave a comment

Even, odd, and periodic functions

The concepts of oddness, evenness, and periodicity are closely related to the Fourier series.

Definition A function ${\phi(x)}$ that is defined for ${-\infty<x<\infty}$ is called periodic if there is a number ${p>0}$ such that

$\displaystyle \phi(x)=\phi(x+p) \hspace{0.5cm} \mathrm{for\hspace{3pt}all}\hspace{3pt} x \ \ \ \ \ (1)$

p is called a period of ${\phi(X)}$ . A periodic function has the following properties:

If ${\phi(x)}$ has period ${p}$ , then ${\phi(x+np)=\phi(x)}$ for all ${x}$ and for all integers ${n}$ .
The sum of two functions of period ${p}$ has period ${p}$ .
If ${\phi(x)}$ has period ${p}$ , then ${\int_{a}^{a+p}\phi(x)}$ does not depend on ${a}$ .

For a function defined on the interval ${-l<x<l}$ , its periodic extension is

$\displaystyle \phi_{per}(x)=\phi(x-2lm)\hspace{3pt} \mathrm{for}\hspace{3pt} -l+2lm<x<l+2lm \ \ \ \ \ (2)$

for all integers ${m}$ . This definition does not specify what the periodic extension is at the endpoints ${x = l + 2lm}$ . In fact, the function ${\phi(x)}$ may have jump discontinuities at the endpoints if the the one-sided limits ${\phi(l^-)}$ and ${\phi(-l^+)}$ both exist but not equal.

An even function is a function the satisfy the equation

$\displaystyle \phi(x)=\phi(-x) \ \ \ \ \ (3)$

This mean that the graph ${y=\phi(x)}$ is symmetric with respect to the y axis.

An odd function is a function the satisfy the equation

$\displaystyle \phi(-x)=-\phi(x) \ \ \ \ \ (4)$

This mean that the graph ${y=\phi(x)}$ is symmetric with respect to the origin.

A monomial ${x^n}$ is an even function if ${n}$ is even and is an odd function if n is odd. The functions cos ${x}$ , cosh ${x}$ , and any function of ${x^2}$ are even functions. The functions sin ${x}$ , tan ${x}$ , and sinh ${x}$ are odd functions. In fact, the products of functions follow the usual rules: even × even = even, odd × odd = even, odd × even = odd. The sum of two odd functions is again odd, and the sum of two evens is even.

But the sum of an even and an odd function can be anything. Proof: Let ${f(x)}$ be any function at all defined on ${(-l,l)}$ . Let ${\phi(x)=\frac{1}{2}[\phi(x)+\phi(-x)]}$ and ${\psi(x)=\frac{1}{2}[\psi(x)-\psi(-x)]}$ . Then we easily check that ${f(x)=\phi(x)+\psi(x)}$ , that ${\phi(x)}$ is even and that ${\psi(x)}$ is odd. The functions ${\phi}$ and ${\psi}$ are called the even and odd parts of ${f}$ , respectively. If ${p(x)}$ is any polynomial, its even part is the sum of its even terms, and its odd part is the sum of its odd terms.

Integration and differentiation change the parity (evenness or oddness) of a function. That is, if ${\phi(x)}$ is even, then both ${\frac{d\phi}{dx}}$ and ${\int_0^x\phi(s)ds}$ are odd. If ${\phi(x)}$ is odd, then its derivative and integral are even.

The graph of an odd function ${\phi(x)}$ must pass through the origin since ${\phi(0) = 0}$ follows directly from (4) by putting ${x = 0}$ . The graph of an even function ${\phi(x)}$ must cross the ${y}$ axis horizontally, ${\phi'(x)=0}$ , since the derivative is odd (provided the derivative exists).

The concepts of oddness, evenness, and periodicity have the following relationships with the boundary conditions:

${u(0,t)=u(l,t)=0}$ : Dirichlet BCs corresponding to the odd extension
${u_x(0,t)=u_x(l,t)=0}$ : Nuemann BCs corresponding to the even extension
${u(l,t)=u_(-l,t), u_x(l,t)=u_x=(-l,t)}$ : Periodic BCs corresponding to the periodic extension

PDE notes: heat equation

April 26, 2022June 1, 2022Michael LiuLeave a comment

1. Heat equation

1.1. Introduction

The heat equation corresponding to no sources and constant thermal properties is given as

$\displaystyle u_t = ku_{xx} \ \ \ \ \ (1)$

Equation (1) describes how heat energy spreads out. Other physical quantities besides temperature smooth out in much the same manner, satisfying the same partial differential equation (1). For this reason, (1) is also called the diffusion equation.

Since the heat equation in (1) has one time derivative, we must be given one initial condition (IC) (usually at t = 0), the initial temperature. It is possible that the initial temperature is not constant, but depends on x. Thus, we must be given the initial temperature distribution,

$\displaystyle u(x,0)= f(x) \ \ \ \ \ (2)$

We also need to know that happens at the two boundaries, ${x = 0}$ and ${x = L}$ . Without knowing this information, we cannot predict the future. Two conditions are needed corresponding to the second spatial derivatives present in (1), usually one condition at each end. We call them boundary conditions. We deal with mainly four types of boundary conditions: Dirichlet, Neumann, Robin, and Periodic.

1.2. Separation of Variables (SoV)

When the PDE and the BCs are linear and homogeneous, we use a technique called the method of separation of variables to find the analytic solutions.

We introduce the heat operator,

$\displaystyle L(u)=u_t-ku_{xx} \ \ \ \ \ (3)$

The heat operator is a linear operator that satisfies the linearity property ${L(c_1u_1+c_2u_2)=c_1L(u_1)+c_2L(u_2)}$ .

A linear equation for the unknown ${u}$ is of the form

$\displaystyle L(u)=f \ \ \ \ \ (4)$

where ${L}$ is a linear operator and ${f}$ is known. The heat equation is a linear PDE:

$\displaystyle L(u)=u_t-ku_{xx}=f(x,t) \ \ \ \ \ (5)$

${f(x,t)}$ is a forcing term. If ${f=0}$ , then ${L(u)=0}$ and the PDE becomes

$\displaystyle u_t-ku_{xx}=0 \ \ \ \ \ (6)$

Equation (6) is a linear homogeneous equation.

We now propose to study heat equation with zero temperatures at finite ends:

$\displaystyle \mathrm{PDE: } \hspace{0.5cm} u_t=ku_{xx} \ \ \ \ \ (7)$

$\displaystyle \mathrm{BC1:} \hspace{0.5cm} u(0,t)=0 \ \ \ \ \ (8)$

$\displaystyle \mathrm{BC2:}\hspace{0.5cm} u(L,t)=0 \ \ \ \ \ (9)$

$\displaystyle \mathrm{IC:}\hspace{0.5cm} u(x,0) = f(x) \ \ \ \ \ (10)$

defined for ${0\leq x \leq L}$ and ${t \geq 0 }$ . This problem consists of linear homogeneous partial differential equations with linear homogeneous boundary conditions.

Solution. We attempt to determine the product solution in the form of

$\displaystyle u(x,t)=G(t)\phi(x) \ \ \ \ \ (11)$

First, we substitute equation (11) into equation (7),

$\displaystyle \phi(x)G'(t)=kG(t)\phi^{''}(x) \ \ \ \ \ (12)$

We can “separate variables” by divide both sides of equation (12) by ${kG(t)\phi(x)}$ :

$\displaystyle \frac{1}{kG}G'=\frac{1}{\phi}\phi^{''} \ \ \ \ \ (13)$

Now the variables are “separated” in a sense that the left hand side is only a function of ${t}$ and the right hand side is only a function of ${x}$ . Next, we can claim that both sides of (13) must equal the same constant:

$\displaystyle \frac{1}{kG}G'=\frac{1}{\phi}\phi^{''}=-\lambda \ \ \ \ \ (14)$

where ${\lambda}$ is a constant known as the separation constant.

Equation (14) yields 2 ordinary differential equations,

$\displaystyle G'=-\lambda kG \ \ \ \ \ (15)$

and

$\displaystyle \phi''=-\lambda\phi \ \ \ \ \ (16)$

We now see that the advantage of the product method is that it transform a PDE, which we do not have to solve, into two ODEs. The boundary conditions impose two conditions on the ${x}$ -dependent ODE. The time-dependent ODE, equation (15), has no additional conditions. We can find a general solution for equation (15) quite easily,

$\displaystyle G(t)=ce^{-\lambda kt} \ \ \ \ \ (17)$

where ${c}$ is an arbitrary multiplicative constant. There can be three cases for ${\lambda}$ : ${\lambda>0}$ , ${\lambda=0}$ , ${\lambda<0}$ . We will determine all allowable values of ${\lambda}$ next.

The ODE in (16) has two boundary conditions:

$\displaystyle \phi(0)=0 \ \ \ \ \ (18)$

$\displaystyle \phi(L)=0 \ \ \ \ \ (19)$

We call this a boundary value problem for ordinary differential equations. We note that ${\phi(x)=0}$ satisfy the ODE (16) and both BCs (18) and (19) regardless of the value of the separation constant. Thus, ${\phi(x)=0}$ is a trivial solution of our PDE. It corresponds to ${u(x,t)\equiv0}$ since ${u(x,t)=\phi(x)G(t)}$ . However, we want to see if there are other nontrivial solutions. We will show that there are certain values of ${\lambda}$ , called eigenvalues of the boundary value problem, for which there are nontrivial solutions, ${\phi(x)}$ . The non-trivial solutions are called eigenfunctions corresponding to the eigenvalues ${\lambda}$

Now we solve equation (16). For linear and homogeneous second-order ODE, two independent solutions are usually obtained in the form of ${\phi=e^{rx}}$ . Substituting this result into equation (16) yields

$\displaystyle \phi''=r^2e^{rx} \ \ \ \ \ (20)$

and this implies that

$\displaystyle r^2=-\lambda \ \ \ \ \ (21)$

We consider the three cases of ${\lambda}$ separately.

If ${\lambda<0}$ , the roots of the characteristic polynomials are ${r=\pm\sqrt{-\lambda}}$ , so the solutions are ${e^{\sqrt{-\lambda}x}}$ and ${e^{-\sqrt{-\lambda}x}}$ . The general solution is

$\displaystyle \phi(x)=c_1e^{\sqrt{-\lambda}x}+c_2e^{-\sqrt{-\lambda}x} \ \ \ \ \ (22)$

We use the hyperbolic functions,

$\displaystyle \phi(x)=c_3\mathrm{cosh}(\sqrt{-\lambda}x)+c_4\mathrm{sinh}(\sqrt{-\lambda}x) \ \ \ \ \ (23)$

We consider the BC from (18). Since ${\mathrm{cosh}(0)=1}$ , ${\phi(0)=0}$ implies that ${c_3=0}$ . So we have ${\phi(x)=c_4\mathrm{sinh}(\sqrt{-\lambda}x)}$ . Now we consider ${\phi(L)=0}$ , which implies that ${c_4\mathrm{sinh}(\sqrt{-\lambda}L)=0}$ . Since ${\sqrt{-\lambda}L>0}$ , ${\mathrm{sinh}(\sqrt{-\lambda}L)>0}$ . It follows that ${c_4=0}$ . This implies that ${\phi(x)=0}$ . The only solution of (16) for ${\lambda<0}$ that solves the homogeneous boundary condition is the trivial solution. Thus there are no negative eigenvalues.

We now consider the case of ${\lambda=0}$ . The general solution is

$\displaystyle \phi(x)=c_1+c_2x \ \ \ \ \ (24)$

Apply ${\phi(0)=0}$ and we arrive at ${c_1=0}$ . Apply ${\phi(L)=0}$ , we have ${c_2x=0}$ which implies that ${c_2=0}$ . This implies that ${\phi(x)=0}$ . The only solution of (16) for ${\lambda=0}$ that solves the homogeneous boundary condition is the trivial solution. Thus ${\lambda=0}$ is not an eigenvalue for this problem.

We now consider the case of ${\lambda>0}$ . The roots of the characteristic polynomials are ${r=\pm\sqrt{\lambda}i}$ , so the solutions are ${e^{\sqrt{\lambda}ix}}$ and ${e^{-\sqrt{\lambda}ix}}$ . The general solution is

$\displaystyle \phi(x)=c_1e^{\sqrt{\lambda}ix}+c_2e^{-\sqrt{\lambda}ix} \ \ \ \ \ (25)$

Since ${\mathrm{cos}(\sqrt{\lambda}x)}$ and ${\mathrm{sin}(\sqrt{\lambda}x)}$ are each linear combinations of ${e^{\sqrt{\lambda}ix}}$ and ${e^{-\sqrt{\lambda}ix}}$ , the general solution can also be expresed in the form of

$\displaystyle \phi(x)=c_3\mathrm{cos}(\sqrt{\lambda}x)+c_4\mathrm{sin}(\sqrt{\lambda}x) \ \ \ \ \ (26)$

For the BC ${\phi(0)=0}$ , we have ${\phi(0)=c_3\mathrm{cos}(\sqrt{\lambda}x)}$ which implies that ${c_3=0}$ . Consider the BC ${\phi(L)=0}$ , we have ${c_4\mathrm{sin}(\sqrt{\lambda}x)=0}$ . If we let ${c_4=0}$ , then we arrive at ${u(x,t)=0}$ , a trivial solution. To find the nontrivial solutions, the eigenvalue ${\lambda}$ much satisfy

$\displaystyle \mathrm{sin}(\sqrt{\lambda}L)=0 \ \ \ \ \ (27)$

${\sqrt{\lambda}x}$ must be the zero of the sine function. Thus,

$\displaystyle \sqrt{\lambda}L=n\pi \ \ \ \ \ (28)$

The eigenvalues are

$\displaystyle \lambda=(\frac{n\pi}{L})^2, n=0,1,2,3... \ \ \ \ \ (29)$

The eigenfunctions corresponding to the eigenvalues are

$\displaystyle \phi(x)=c_4\mathrm{sin}(\frac{n\pi x}{L}), n=0,1,2,3... \ \ \ \ \ (30)$

Now we know that the product solution consists of ${\phi(x)=\\c_4\mathrm{sin}(\frac{n\pi x}{L})}$ and ${G(t)=e^{-(\frac{n\pi}{L})^2 kt}}$ where we determined from the boundary conditions ${\phi(0)=0}$ and ${\phi(L)=0}$ . We call this type of boundary conditions the Dirichlet boundary conditions. Thus, the product solutions of the heat equation with homogeneous Dirichlet boundary conditions are

$\displaystyle u(x,t)= B\mathrm{sin}(\frac{n\pi x}{L})e^{-(\frac{n\pi}{L})^2 kt}, n=0,1,2,3... \ \ \ \ \ (31)$

where B is an arbitrary constant.

1.3. Initial value problems

The principle of superposition can be extended to show that if ${u_1 , u_2 , u_3 , . . . , u_M}$ are solutions of a linear homogeneous problem, then any linear combination of these is also a solution, ${\sum_{n=1}^{M} c_nu_n}$ ,where ${c_n}$ are arbitrary constants. Since we know from the method of separation of variables that ${\mathrm{sin}(\frac{n\pi x}{L})e^{-(\frac{n\pi}{L})^2 kt}}$ is a solution of the heat equation (solving zero boundary conditions) for all positive ${n}$ , it follows that any linear combination of these solutions is also a solution of the linear homogeneous heat equation. Thus,

$\displaystyle u(t,x)=\sum_{n=1}^{M} B_n\mathrm{sin}(\frac{n\pi x}{L})e^{-(\frac{n\pi}{L})^2 kt} \ \ \ \ \ (32)$

For example, we wish to solve the following initial value problem:

$\displaystyle \mathrm{PDE: } \hspace{0.5cm} u_t = ku_{xx} \ \ \ \ \ (33)$

$\displaystyle \mathrm{BC1:} \hspace{0.5cm}u(0,t)=0 \ \ \ \ \ (34)$

$\displaystyle \mathrm{BC2:}\hspace{0.5cm}u(L,t)=0 \ \ \ \ \ (35)$

$\displaystyle \mathrm{IC:}\hspace{0.5cm} u(x,0) = 3\mathrm{sin}(\frac{4\pi x}{L})+7\mathrm{sin}(\frac{9\pi x}{L}) \ \ \ \ \ (36)$

Solution. We can let ${u_1(x,0) = 3\mathrm{sin}(\frac{4\pi x}{L})}$ and ${u_2(x,0) = 7\mathrm{sin}(\frac{9\pi x}{L})}$ such that ${u(x,0)=u_1(x,0)+u_2(x,0)}$ . We can use the product solutions from (31) and we see that the product solution satisfy the initial conditions ${B\mathrm{sin}(\frac{n\pi x}{L})}$ . By picking ${B=3}$ and ${n=4}$ , we have satisfied the initial condition ${u_1(x,0)}$ . The solution we get is

$\displaystyle u_1(t,x)=3\mathrm{sin}(\frac{4\pi x}{L})e^{-(\frac{4\pi}{L})^2 kt} \ \ \ \ \ (37)$

By picking ${B=7}$ and ${n=9}$ , we have satisfied the initial condition ${u_2(x,0)}$ . The solution we get is

$\displaystyle u_1(t,x)=7\mathrm{sin}(\frac{9\pi x}{L})e^{-(\frac{9\pi}{L})^2 kt} \ \ \ \ \ (38)$

By superposition principle we know that the product solution of our problem is a linear combinations of ${u_1(x,t)}$ and ${u_2(x,t)}$ . Thus,

$\displaystyle u(t,x)=3\mathrm{sin}(\frac{4\pi x}{L})e^{-(\frac{4\pi}{L})^2 kt}+ 7\mathrm{sin}(\frac{9\pi x}{L})e^{-(\frac{9\pi}{L})^2 kt} \ \ \ \ \ (39)$

${\square}$

The IC in the previous problem is a finite sum of the sine functions. What should we do in the situation that the IC is not a finite linear combinations of the sine functions? Consider the following problem:

$\displaystyle \mathrm{PDE: } \hspace{0.5cm} u_t = ku_{xx} \ \ \ \ \ (40)$

$\displaystyle \mathrm{BC1:} \hspace{0.5cm}u(0,t)=0 \ \ \ \ \ (41)$

$\displaystyle \mathrm{BC2:}\hspace{0.5cm}u(L,t)=0 \ \ \ \ \ (42)$

$\displaystyle \mathrm{IC:}\hspace{0.5cm} u(x,0) = 100 \ \ \ \ \ (43)$

Solution. From equation (32) we know that the solution is

$\displaystyle u(t,x)=\sum_{n=1}^{M} B_n\mathrm{sin}(\frac{n\pi x}{L})e^{-(\frac{n\pi}{L})^2 kt} \ \ \ \ \ (44)$

We want to apply the IC to our problem. First, from the previous problem, note that we can solve the heat equation if initially

$\displaystyle u(x,0)= f(x) = \sum_{n=1}^{M} B_n\mathrm{sin}(\frac{n\pi x}{L}) \ \ \ \ \ (45)$

Thus, we use the theory of Fourier series and claim that “any” (with restrictions) initial condition ${f(x)}$ can be written as an infinite linear combination of ${\mathrm{sin}(\frac{n\pi x}{L})}$ , known as a type of Fourier series:

$\displaystyle f(x) = \sum_{n=1}^{\infty} B_n\mathrm{sin}(\frac{n\pi x}{L}) \ \ \ \ \ (46)$

We want to determine the coefficient ${B_n}$ in equation (46). To do that, we use the fact the ${\mathrm{sin}(\frac{n\pi x}{L})}$ satisfies the following orthogonality relation:

$\displaystyle \int_{0}^{L} \mathrm{sin}(\frac{n \pi x}{L})\mathrm{sin}(\frac{m \pi x}{L}) \,dx = \left\{ \begin{array}{lr} 0 & n \neq m \\ \frac{L}{2} & n=m \\ \end{array} \right. \ \ \ \ \ (47)$

Multiply (46) by ${\mathrm{sin}(\frac{m \pi x}{L})}$ and integrating from ${0}$ to ${L}$ yields

$\displaystyle \int_{0}^{L} f(x)\mathrm{sin}(\frac{m \pi x}{L})\, dx = B_{m}\int_{0}^{L}\mathrm{sin^2}(\frac{m\pi x}{L})\, dx \ \ \ \ \ (48)$

Solving for ${A_m}$ yields,

$\displaystyle B_m=\frac{2}{L} \int_{0}^{L}f(x)\mathrm{sin}(\frac{m \pi x}{L})\, dx \ \ \ \ \ (49)$

Now, we calculate the coefficient ${B_n}$ from equation (49) for ${f(x)=100}$ ,

$\displaystyle B_n=\frac{2}{L} \int_{0}^{L}100\mathrm{sin}(\frac{n \pi x}{L})\, dx \ \ \ \ \ (50)$

Solving for ${B_n}$ yields,

$\displaystyle \left\{ \begin{array}{lr} 0 & n \hspace{5pt} even\\ \frac{400}{n\pi} & n\hspace{5pt} odd \\ \end{array} \right. \ \ \ \ \ (51)$

Each succeeding term in the series is much smaller than the first. We can then approximate the infinite series by only the first term:

$\displaystyle u(t,x)\approx \frac{400}{\pi}\mathrm{sin}(\frac{\pi x}{L})e^{-(\frac{\pi}{L})^2 kt} \ \ \ \ \ (52)$

${\square}$

Let us summarize the method of separation of variables as it appears for the heat equation with homogeneous Dirichlet boundary conditions:

$\displaystyle \mathrm{PDE: } \hspace{0.5cm} u_t = ku_{xx} \ \ \ \ \ (53)$

$\displaystyle \mathrm{BC1:} \hspace{0.5cm}u(0,t)=0 \ \ \ \ \ (54)$

$\displaystyle \mathrm{BC2:}\hspace{0.5cm}u(L,t)=0 \ \ \ \ \ (55)$

$\displaystyle \mathrm{IC:}\hspace{0.5cm} u(x,0) = f(x) \ \ \ \ \ (56)$

Make sure that you have a linear and homogeneous PDE with linear and homogeneous BC.
Temporarily ignore the nonzero IC.
Separate variables (determine differential equations implied by the assumption of product solutions) and introduce a separation constant.
Determine separation constants as the eigenvalues of a boundary value problem.
Solve other differential equations. Record all product solutions of the PDE obtainable by this method.
Apply the principle of superposition (for a linear combination of all product solutions).
Attempt to satisfy the initial condition.
Determine coefficients using the orthogonality of the eigenfunctions.

1.4. Other Boundary Value Problems

The following problem is defined for ${0\leq x \leq L}$ and ${t \geq 0 }$ .

$\displaystyle \mathrm{PDE: } \hspace{0.5cm} u_t = ku_{xx} \ \ \ \ \ (57)$

$\displaystyle \mathrm{BC1:} \hspace{0.5cm}u_x(0,t)=0 \ \ \ \ \ (58)$

$\displaystyle \mathrm{BC2:}\hspace{0.5cm}u_x(L,t)=0 \ \ \ \ \ (59)$

$\displaystyle \mathrm{IC:}\hspace{0.5cm} u(x,0) = f(x) \ \ \ \ \ (60)$

This is a heat conduction problem in a one-dimensional rod with constant thermal properties and no sources. The ends of the rod are insulated. Both the PDE and BCs are linear and homogeneous. We call this type of BC the Neumann boundary condition. We apply SoV.

Solution. The assumed product solutions are:

$\displaystyle u(x,t)=\phi(x)G(t) \ \ \ \ \ (61)$

Substitute into the PDE,

$\displaystyle \phi G'= kG\phi^{''}= -\lambda \ \ \ \ \ (62)$

where ${\lambda}$ is the separation constant. This implies that

$\displaystyle G^{'}=-\lambda kG \ \ \ \ \ (63)$

$\displaystyle \phi^{''}=-\lambda\phi \ \ \ \ \ (64)$

Solving (64),

$\displaystyle G(t)=e^{-\lambda kt} \ \ \ \ \ (65)$

The insulated BCs imply that the separated solutions must satisfy

$\displaystyle \phi^{'}(0)=0 \ \ \ \ \ (66)$

$\displaystyle \phi^{'}(L)=0 \ \ \ \ \ (67)$

We determine the separation constant ${\lambda}$ by finding those ${\lambda}$ for which nontrivial solutions exist.

For ${\lambda <0}$ , the general solution is

$\displaystyle \phi(x)=c_1e^{\sqrt{-\lambda}x}+c_2e^{-\sqrt{-\lambda}x} \ \ \ \ \ (68)$

We also need

$\displaystyle \phi^{'}(x)=\sqrt{-\lambda}(c_1e^{\sqrt{-\lambda}x}-c_2e^{-\sqrt{-\lambda}x}) \ \ \ \ \ (69)$

The BC ${\phi^{'}(0)=0}$ implies that ${c_1=c_2}$ , and ${\phi^{'}(L)=0}$ means that

$\displaystyle \sqrt{-\lambda}(c_1e^{\sqrt{-\lambda}L}-c_2e^{-\sqrt{-\lambda}L})=2\sqrt{-\lambda}c_1\mathrm{sinh}(\sqrt{-\lambda} L)=0 \ \ \ \ \ (70)$

Since ${\sqrt{-\lambda} L >0}$ , ${\mathrm{sinh}(\sqrt{-\lambda} L)>0}$ . This implies that ${c_1=0}$ . Since ${\phi(x)=0}$ is a trivial solution, there is no negative eigenvalues.

For ${\lambda=0}$ , the solution is

$\displaystyle \phi(x)=c_1+c_2x \ \ \ \ \ (71)$

and

$\displaystyle \phi^{'}(x)=c_2 \ \ \ \ \ (72)$

Both BCs give ${c_2=0}$ . Thus, there are non trivial solutions of the BVP for ${\lambda=0}$ , namely,

$\displaystyle \phi(x)=c_1 \ \ \ \ \ (73)$

The time dependent part gives ${G(t)=e^{-\lambda kt}=1}$ . Thus, the resulting product solution of the PDE is

$\displaystyle u(x,t)=A \ \ \ \ \ (74)$

where ${A}$ is any constant.

For ${\lambda>0}$ , the solutions is in the form

$\displaystyle \phi(x)=e^{\pm\sqrt{\lambda}ix} \ \ \ \ \ (75)$

which gives

$\displaystyle \phi(x)=c_1\mathrm{cos}(\sqrt{\lambda} x)+c_2\mathrm{sin}(\sqrt{\lambda} x) \ \ \ \ \ (76)$

and

$\displaystyle \phi^{'}(x)=\sqrt{\lambda}(c_2\mathrm{cos}(\sqrt{\lambda} x)-c_1\mathrm{sin}(\sqrt{\lambda} x)) \ \ \ \ \ (77)$

The BC ${\phi^{'}(0)=0}$ implies that ${c_2\sqrt{\lambda}=0}$ , so ${c_2=0}$ . The BC ${\phi^{'}(L)=0}$ gives

$\displaystyle -\sqrt{\lambda}c_1\mathrm{sin}(\sqrt{\lambda}L)=0 \ \ \ \ \ (78)$

For nontrivial solutions, ${c_1\neq0}$ . This gives us ${\mathrm{sin}(\sqrt{\lambda}L)=0}$ .

$\displaystyle \lambda= (\frac{n\pi}{L})^2, n=0,1,2,3... \ \ \ \ \ (79)$

However, the corresponding eigenfunctions are cosines,

$\displaystyle \phi(x)=c_1\mathrm{cos}(\frac{n\pi x}{L}), n=0,1,2,3... \ \ \ \ \ (80)$

The product solution of the PDE is

$\displaystyle u(t,x)=Ae^{-(\frac{n\pi}{L})^2kt}\mathrm{cos}(\frac{n\pi x}{L}), n=0,1,2,3... \ \ \ \ \ (81)$

where A is a arbitrary multiplicative constant.

In order to satisfy the IC, we use the principle of superposition so that,

$\displaystyle u(t,x)=A_0+\sum_{n=1}^{\infty}A_{n}e^{-(\frac{n\pi}{L})^2kt}\mathrm{cos}(\frac{n\pi x}{L}) \ \ \ \ \ (82)$

The IE ${u(x,0)=f(x)}$ is satisfied if

$\displaystyle f(x)=A_0+\sum_{n=1}^{\infty}A_{n}\mathrm{cos}(\frac{n\pi x}{L}) \ \ \ \ \ (83)$

for ${0\leq x \leq L}$ .

To complete the solution, we need to determine the arbitrary coefficient ${A_0}$ and ${A_n}$ . We use the fact that ${\mathrm{cos}(\frac{n\pi x}{L})}$ satisfies the following orthogonality relation: