Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 5: Fourier series and PDEs

42 problems · 0 figures

Exercise 4.1.101 §4.1 source ↗

Consider a spinning string of length 2 and linear density 0.1 and tension 3. Find the smallest angular velocity when the string pops out.

Answer

ω=π152

Exercise 4.1.102 §4.1 source ↗

Suppose x′′+λx=0 and x(0)=1⁠, x(1)=1⁠. Find all λ for which there is more than one solution. Also find the corresponding solutions (only for the eigenvalues).

Answer

λk=4k2π2 for k=1,2,3,… xk= cos⁡(2kπt)+ Bsin⁡(2kπt) (for any B⁠)

Exercise 4.1.103 §4.1 source ↗

Suppose x′′+x=0 and x(0)=0⁠, x′(π)=1⁠. Find all the solution(s) if any exist.

Answer

x(t)=−sin⁡(t)

Exercise 4.1.104 §4.1 source ↗

Consider x′+λx=0 and x(0)=0⁠, x(1)=0⁠. Why does it not have any eigenvalues? Why does every first-order equation with two endpoint conditions such as the one above have no eigenvalues?

Answer

General solution is x=Ce−λt⁠. Since x(0)=0 then C=0⁠, and so x(t)=0⁠. Therefore, the solution is always identically zero. One condition is always enough to guarantee a unique solution for a first-order equation.

Exercise 4.1.105 §4.1 source ↗

(challenging) Suppose x′′′+λx=0 and x(0)=0⁠, x′(0)=0⁠, x(1)=0⁠. Suppose that λ>0⁠. Find an equation that all such eigenvalues must satisfy. Hint: Note that −λ3 is a root of r3+λ=0⁠.

Answer

33e−32λ3− 33cos⁡(3λ32)+ sin⁡(3λ32)= 0

Exercise 4.2.101 §4.2 source ↗

Suppose f(t) is defined on [−π,π] as f(t)=sin⁡(t)⁠. Extend periodically and compute the Fourier series.

Answer

sin⁡(t)

Exercise 4.2.102 §4.2 source ↗

Suppose f(t) is defined on (−π,π] as f(t)=sin⁡(πt)⁠. Extend periodically and compute the Fourier series.

Answer

∑n=1∞2n(−1)nsin⁡(π2)π(π2−n2)sin⁡(nt)

Exercise 4.2.103 §4.2 source ↗

Suppose f(t) is defined on (−π,π] as f(t)=sin2⁡(t)⁠. Extend periodically and compute the Fourier series.

Answer

12−12cos⁡(2t)

Exercise 4.2.104 §4.2 source ↗

Suppose f(t) is defined on (−π,π] as f(t)=t4⁠. Extend periodically and compute the Fourier series.

Answer

π45+ ∑n=1∞(−1)n(8π2n2−48)n4cos⁡(nt)

Exercise 4.3.101 §4.3 source ↗

Let

f(t)=t2for −2<t≤2

extended periodically.

a) Compute the Fourier series for f(t)⁠.

b) Write out the series explicitly up to the 3rd harmonic.

Answer

a) 86+ ∑n=1∞16(−1)nπ2n2cos⁡(nπ2t) b) 86− 16π2cos⁡(π2t)+ 4π2cos⁡(πt)− 169π2cos⁡(3π2t)+ ⋯

Exercise 4.3.102 §4.3 source ↗

Let

f(t)=tfor −λ<t≤λ (for some λ>0)

extended periodically.

a) Compute the Fourier series for f(t)⁠.

b) Write out the series explicitly up to the 3rd harmonic.

Answer

a) ∑n=1∞(−1)n+12λnπsin⁡(nπλt) b) 2λπsin⁡(πλt)− λπsin⁡(2πλt)+ 2λ3πsin⁡(3πλt)− ⋯

Exercise 4.3.103 §4.3 source ↗

Let

f(t)=12+∑n=1∞1n(n2+1)sin⁡(nπt).

Compute f′(t)⁠.

Answer

f′(t)= ∑n=1∞πn2+1cos⁡(nπt)

Exercise 4.3.104 §4.3 source ↗

Let

f(t)=12+∑n=1∞1n3cos⁡(nt).

a) Find the antiderivative.

b) Is the antiderivative periodic?

Answer

a) F(t)= t2+ C+ ∑n=1∞1n4sin⁡(nt) b) no

Exercise 4.3.105 §4.3 source ↗

Let

f(t)=t/2for −π<t<π

extended periodically.

a) Compute the Fourier series for f(t)⁠.

b) Plug in t=π/2 to find a series representation for π/4⁠.

c) Using the first 4 terms of the result from part b) approximate π/4⁠.

Answer

a) ∑n=1∞(−1)n+1nsin⁡(nt) b) f is continuous at t=π/2 so the Fourier series converges to f(π/2)=π/4⁠. Obtain π/4= ∑n=1∞(−1)n+12n−1= 1− 1/3+ 1/5− 1/7+ ⋯⁠. c) Using the first 4 terms get 76/105≈0.72 (quite a bad approximation, you would have to take about 50 terms to start to get to within 0.01 of π/4⁠).

Exercise 4.3.106 §4.3 source ↗

Let

f(t)={0if −2<t≤0,2if −0<t≤2,

extended periodically. Suppose F(t) is the function given by the Fourier series of f⁠. Without computing the Fourier series evaluate

a) F(0)

b) F(−1)

c) F(1)

d) F(−2)

e) F(4)

f) F(−9)

Answer

a) F(0)=1⁠, b) F(−1)=0⁠, c) F(1)=2⁠, d) F(−2)=1⁠, e) F(4)=1⁠, f) F(−9)=0

Exercise 4.4.101 §4.4 source ↗

Let f(t)=t/3 on 0≤t<3⁠.

a) Find the Fourier series of the even periodic extension.

b) Find the Fourier series of the odd periodic extension.

Answer

a) 1/2+ ∑n=1n odd∞−4π2n2cos⁡(nπ3t) b) ∑n=1∞2(−1)n+1πnsin⁡(nπ3t)

Exercise 4.4.102 §4.4 source ↗

Let f(t)=cos⁡(2t) on 0≤t<π⁠.

a) Find the Fourier series of the even periodic extension.

b) Find the Fourier series of the odd periodic extension.

Answer

a) cos⁡(2t) b) ∑n=1n odd∞4nπn2−4πsin⁡(nt)

Exercise 4.4.103 §4.4 source ↗

Let f(t) be defined on 0≤t<1⁠. Consider the average of the two extensions g(t)=Fodd(t)+Feven(t)2⁠.

a) What is g(t) if 0≤t<1 (Justify!)

b) What is g(t) if −1<t<0 (Justify!)

Answer

a) f(t) b) 0

Exercise 4.4.104 §4.4 source ↗

Let f(t)=∑n=1∞1n2sin⁡(nt)⁠. Solve x′′−x=f(t) for the Dirichlet conditions x(0)=0 and x(π)=0⁠.

Answer

∑n=1∞−1n2(1+n2)sin⁡(nt)

Exercise 4.4.105 §4.4 source ↗

(challenging) Let f(t)=t+∑n=1∞12nsin⁡(nt)⁠. Solve x′′+πx=f(t) for the Dirichlet conditions x(0)=0 and x(π)=1⁠. Hint: Note that tπ satisfies the given Dirichlet conditions.

Answer

tπ+ ∑n=1∞12n(π−n2)sin⁡(nt)

Exercise 4.5.101 §4.5 source ↗

Let F(t)= sin⁡(2πt)+ 0.1cos⁡(10πt)⁠. Find the steady periodic solution to x′′+2x=F(t)⁠. Express your solution as a Fourier series.

Answer

x= 12−4π2sin⁡(2πt)+ 0.12−100π2cos⁡(10πt)

Exercise 4.5.102 §4.5 source ↗

Let F(t)=∑n=1∞e−ncos⁡(2nt)⁠. Find the steady periodic solution to x′′+3x=F(t)⁠. Express your solution as a Fourier series.

Answer

x= ∑n=1∞e−n3−(2n)2cos⁡(2nt)

Exercise 4.5.103 §4.5 source ↗

Let F(t)=|t| for −1≤t≤1 extended periodically. Find the steady periodic solution to x′′+3x=F(t)⁠. Express your solution as a series.

Answer

x= 123+ ∑n=1n odd∞−4n2π2(3−n2π2)cos⁡(nπt)

Exercise 4.5.104 §4.5 source ↗

Let F(t)=|t| for −1≤t≤1 extended periodically. Find the steady periodic solution to x′′+π2x=F(t)⁠. Express your solution as a series.

Answer

x= 12π2− 2π3tsin⁡(πt)+ ∑n=3n odd∞−4n2π4(1−n2)cos⁡(nπt)

Exercise 4.6.101 §4.6 source ↗

Find a series solution of

ut=3uxxfor 0<x<π and t>0,u(0,t)=u(π,t)=0for t>0,u(x,0)=5sin⁡(x)+2sin⁡(5x)for 0<x<π.

Answer

u(x,t)= 5sin⁡(x)e−3t+ 2sin⁡(5x)e−75t

Exercise 4.6.102 §4.6 source ↗

Find a series solution of

ut=0.1uxxfor 0<x<π and t>0,ux(0,t)=ux(π,t)=0for t>0,u(x,0)=1+2cos⁡(x)for 0<x<π.

Answer

u(x,t)=1+2cos⁡(x)e−0.1t

Exercise 4.6.103 §4.6 source ↗

Use separation of variables to find a nontrivial solution to uxt=uxx⁠.

Answer

u(x,t)=eλteλx for some λ

Exercise 4.6.104 §4.6 source ↗

Use a variation on separation of variables to find a nontrivial solution to ux+ut=u⁠. Hint: Try u(x,t)=X(x)+T(t)⁠.

Answer

u(x,t)=Aex+Bet

Exercise 4.6.105 §4.6 source ↗

Suppose that the temperature on the wire is fixed at 0 at the ends, L=1⁠, k=1⁠, and u(x,0)=100sin⁡(2πx)⁠.

a) What is the temperature at x=1/2 at any time?

b) What is the maximum and the minimum temperature on the wire at t=0⁠?

c) At what time is the maximum temperature on the wire exactly one half of the initial maximum at t=0⁠?

Answer

a) 0⁠, b) minimum −100⁠, maximum 100⁠, c) t=ln⁡24π2⁠.

Exercise 4.7.101 §4.7 source ↗

Solve

ytt=yxxfor 0<x<π,t>0,y(0,t)=y(π,t)=0for t>0,y(x,0)=sin⁡(x)for 0<x<π,yt(x,0)=sin⁡(x)for 0<x<π.

Answer

y(x,t)= sin⁡(x)(sin⁡(t)+ cos⁡(t))

Exercise 4.7.102 §4.7 source ↗

Solve

ytt=25yxxfor 0<x<2 and t>0,y(0,t)=y(2,t)=0for t>0,y(x,0)=0for 0<x<2,yt(x,0)=sin⁡(πx)+0.1sin⁡(2πx)for 0<x<2.

Answer

y(x,t)= 15πsin⁡(πx)sin⁡(5πt)+ 1100πsin⁡(2πx)sin⁡(10πt)

Exercise 4.7.103 §4.7 source ↗

Solve

ytt=2yxxfor 0<x<π and t>0,y(0,t)=y(π,t)=0for t>0,y(x,0)=xfor 0<x<π,yt(x,0)=0for 0<x<π.

Answer

y(x,t)= ∑n=1∞2(−1)n+1nsin⁡(nx)cos⁡(n2t)

Exercise 4.7.104 §4.7 source ↗

What happens when a=0⁠? Find a solution to ytt=0⁠, y(0,t)=y(π,t)=0⁠, y(x,0)=sin⁡(2x)⁠, yt(x,0)=sin⁡(x)⁠.

Answer

y(x,t)= sin⁡(2x)+ tsin⁡(x)

Exercise 4.8.101 §4.8 source ↗

Using the d’Alembert solution solve ytt=9yxx⁠, 0<x<1⁠, t>0⁠, y(0,t)=y(1,t)=0⁠, y(x,0)=sin⁡(2πx)⁠, and yt(x,0)=sin⁡(3πx)⁠.

Answer

y(x,t)= sin⁡(2π(x−3t))+sin⁡(2π(3t+x))2+ cos⁡(3π(x−3t))−cos⁡(3π(3t+x))18π

Exercise 4.8.102 §4.8 source ↗

Take ytt=4yxx⁠, 0<x<1⁠, t>0⁠, y(0,t)=y(1,t)=0⁠, y(x,0)=x−x2⁠, and yt(x,0)=0⁠. Using the d’Alembert solution find the solution at

a) t=0.1⁠,

b) t=1/2⁠,

c) t=1⁠.

You may have to split your answer up by cases.

Answer

a) y(x,0.1)= {x−x2−0.04if 0.2≤x≤0.80.6xif x≤0.20.6−0.6xif x≥0.8
b) y(x,1/2)=−x+x2 c) y(x,1)=x−x2

Exercise 4.8.103 §4.8 source ↗

Take ytt=100yxx⁠, 0<x<4⁠, t>0⁠, y(0,t)=y(4,t)=0⁠, y(x,0)=F(x)⁠, and yt(x,0)=0⁠. Suppose that F(0)=0⁠, F(1)=2⁠, F(2)=3⁠, F(3)=1⁠. Using the d’Alembert solution find

a) y(1,1)⁠,

b) y(4,3)⁠,

c) y(3,9)⁠.

Answer

a) y(1,1)=−1/2 b) y(4,3)=0 c) y(3,9)=1/2

Exercise 4.9.101 §4.9 source ↗

In the region described by 0<x<1 and 0<y<1⁠, solve the problem

Δu=0,u(x,0)=∑n=1∞1n2sin⁡(nπx),u(x,1)=0,u(0,y)=0,u(1,y)=0.

Answer

u(x,y)= ∑n=1∞1n2sin⁡(nπx)(sinh⁡(nπ(1−y))sinh⁡(nπ))

Exercise 4.9.102 §4.9 source ↗

In the region described by 0<x<1 and 0<y<2⁠, solve the problem

Δu=0,u(x,0)=0.1sin⁡(πx),u(x,2)=0,u(0,y)=0,u(1,y)=0.

Answer

u(x,y)= 0.1sin⁡(πx)(sinh⁡(π(2−y))sinh⁡(2π))

Exercise 4.10.101 §4.10 source ↗

Using series solve Δu=0⁠, u(1,θ)= 1+ ∑n=1∞1n2sin⁡(nθ)⁠.

Answer

u=1+∑n=1∞1n2rnsin⁡(nθ)

Exercise 4.10.102 §4.10 source ↗

Using the series solution find the solution to Δu=0⁠, u(1,θ)=1−cos⁡(θ)⁠. Express the solution in Cartesian coordinates (that is, using x and y⁠).

Answer

u=1−x

Exercise 4.10.103 §4.10 source ↗

a) Try and guess a solution to Δu=−1⁠, u(1,θ)=0⁠. Hint: Try a solution that only depends on r⁠. Also, don’t worry about the boundary condition at first.

b) Now solve Δu=−1⁠, u(1,θ)=sin⁡(2θ) using superposition.

Answer

a) u=−14r2+14 b) u=−14r2+14+r2sin⁡(2θ)

Exercise 4.10.104 §4.10 source ↗

(challenging) Derive the Poisson kernel solution if the region is a circle of radius ρ rather than 1. That is, solve Δu=0⁠, u(ρ,θ)=g(θ)⁠.

Answer

u(r,θ)= 12π ∫−ππρ2−r2ρ2−2rρcos⁡(θ−α)+r2g(α)dα

Loading chapter 6…

Exercises and answers are adapted from Notes on Diffy Qs: Differential Equations for Engineers by Jiří Lebl, which is free to read and download, and are used under the CC BY-NC-SA 4.0 licence. They are taken from the book’s LaTeX source and set for the web, so they may differ slightly from the printed page. This adaptation is released under the same licence and is neither affiliated with nor endorsed by Lebl or Jiří Lebl.