Advanced Engineering Mathematics: One-Dimensional Wave Equation, Assignments of Mathematics

Information on the one-dimensional wave equation, its solution using separation of variables, and traveling wave solutions. It includes boundary conditions, initial conditions, and the d'alembert solution. From a university mathematics course, math 348, and is related to partial differential equations and wave propagation.

Typology: Assignments

Pre 2010

Uploaded on 08/16/2009

koofers-user-7je
koofers-user-7je 🇺🇸

10 documents

1 / 2

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
MATH 348 - Advanced Engineering Mathematics April 2, 2008
Homework 8, Spring 2008 Due: April 11, 2008
partial differential equations - the one-dimensional wave equation
Consider the one-dimensional wave equation,
2u
∂t2=c22u
∂x2,(1)
x(0, L), t (0,), c2=T
ρ.(2)
Equations (1)-(2) model the time-evolution of the displacement, u=u(x, t), of an elastic medium in one-dimension.
The object, of length L, is assumed to have a homogenous lateral tension T, and linear density ρ. That is, T, ρ R+.
Define,
f(x) = x, 0< x L
x+ 2L, L < x < 2L(3)
1. Consider the one-dimensional wave equation, (1)-(2), with the boundary conditions1,
ux(0, t)=0, u(2L, t) = 0,(4)
and initial conditions
u(x, 0) = f(x), ut(x, 0) = g(x) (5)
(a) Assume that the solution to (1)-(2) is such that u(x, t) = F(x)G(t) and use separation of variables to find
the general solution to (1)-(2), which satisfies (4)-(5). 2
(b) Solve for the unknown constants assuming (3) and zero initial velocity for all points on the object.
2. Consider the one-dimensional wave equation, (1)-(2), with the boundary conditions3,
ux(0, t)=0, ux(2L, t)=0,(6)
and initial conditions
u(x, 0) = f(x), ut(x, 0) = g(x) (7)
(a) Assume that the solution to (1)-(2) is such that u(x, t) = F(x)G(t) and use separation of variables to find
the general solution to (1)-(2), which satisfies (6)-(7). 4 5
(b) Let L= 1 and solve for the unknown constants assuming (3) and zero initial velocity for all points on the
object.
3. Many applications consider traveling wave solutions, f(x, t) = f(xct), of the sinusoidal form, f(x, t) =
Acos(kx ωt). Assume that u(x, t) = Aei(kxω t)is a solution to the following wave-like equation:6
utt uxx +u= 0.(8)
Show that the phase velocity, cp=ω
k, of the traveling wave solutions to (8) is given by cp=±1 + k2.7
1These boundary conditions imply that the object must have zero curvature at the left endpoint and zero displacement at the right
endpoint.
2It is important to notice that the solution to the spatial portion of the problem is the same as the heat problem of hw5 prob1.
3These boundary conditions imply that the object must have zero curvature at each endpoint.
4It is important to notice that the solution to the spatial portion of the problem is the same as the heat problem of hw5 prob2.
5Remember that in this case we have nontrivial solutions for k0= 0. You should find that G0(t) = C1+C2t.
6Here we choose to work with complex exponential functions since calculation of derivatives is less clumsy than trigonometric functions.
Notice that the real-part of uis equal to f.
7This implies that different waves which solve (8) travel at different velocities.
1
pf2

Partial preview of the text

Download Advanced Engineering Mathematics: One-Dimensional Wave Equation and more Assignments Mathematics in PDF only on Docsity!

MATH 348 - Advanced Engineering Mathematics April 2, 2008 Homework 8, Spring 2008 Due: April 11, 2008

partial differential equations - the one-dimensional wave equation

Consider the one-dimensional wave equation,

∂^2 u ∂t^2

= c^2 ∂

(^2) u ∂x^2

x ∈ (0, L) , t ∈ (0, ∞) , c^2 = T ρ

Equations (1)-(2) model the time-evolution of the displacement, u = u(x, t), of an elastic medium in one-dimension. The object, of length L, is assumed to have a homogenous lateral tension T , and linear density ρ. That is, T, ρ ∈ R+. Define,

f (x) =

x, 0 < x ≤ L −x + 2L, L < x < 2 L (3)

  1. Consider the one-dimensional wave equation, (1)-(2), with the boundary conditions^1 ,

ux(0, t) = 0, u(2L, t) = 0, (4)

and initial conditions u(x, 0) = f (x), ut(x, 0) = g(x) (5)

(a) Assume that the solution to (1)-(2) is such that u(x, t) = F (x)G(t) and use separation of variables to find the general solution to (1)-(2), which satisfies (4)-(5). 2 (b) Solve for the unknown constants assuming (3) and zero initial velocity for all points on the object.

  1. Consider the one-dimensional wave equation, (1)-(2), with the boundary conditions^3 ,

ux(0, t) = 0, ux(2L, t) = 0, (6)

and initial conditions u(x, 0) = f (x), ut(x, 0) = g(x) (7) (a) Assume that the solution to (1)-(2) is such that u(x, t) = F (x)G(t) and use separation of variables to find the general solution to (1)-(2), which satisfies (6)-(7). 4 5 (b) Let L = 1 and solve for the unknown constants assuming (3) and zero initial velocity for all points on the object.

  1. Many applications consider traveling wave solutions, f (x, t) = f (x − ct), of the sinusoidal form, f (x, t) = A cos(kx − ωt). Assume that u(x, t) = Aei(kx−ωt)^ is a solution to the following wave-like equation:^6

utt − uxx + u = 0. (8)

Show that the phase velocity, cp = ω k

, of the traveling wave solutions to (8) is given by cp = ±

1 + k−^2.^7

(^1) These boundary conditions imply that the object must have zero curvature at the left endpoint and zero displacement at the right endpoint. 2

3 It is important to notice that the solution to the spatial portion of the problem is the same as the heat problem of hw5 prob1. 4 These boundary conditions imply that the object must have zero curvature at each endpoint. 5 It is important to notice that the solution to the spatial portion of the problem is the same as the heat problem of hw5 prob2. 6 Remember that in this case we have nontrivial solutions for^ k^0 = 0. You should find that^ G^0 (t) =^ C^1 +^ C^2 t. Here we choose to work with complex exponential functions since calculation of derivatives is less clumsy than trigonometric functions. Notice that the 7 real-part of u is equal to f. This implies that different waves which solve (8) travel at different velocities.

1

  1. Show that by direct substitution that the function u(x, t) given by,

u(x, t) =

2 [u^0 (x^ −^ ct) +^ u^0 (x^ +^ ct)] +

2 c

∫ (^) x+ct

x−ct

v 0 (y)dy, (9)

is a solution to the one-dimensional wave equation where u 0 and v 0 are the initial displacement and velocity of the elastic string, respectively. 8

  1. Consider the non-homogenous one-dimensional wave equation,

∂^2 u ∂t^2 =^ c

2 ∂^2 u ∂x^2 +^ F^ (x, t)^ ,^ (10) x ∈ (0, L) , t ∈ (0, ∞) , c^2 =

T

ρ.^ (11) with boundary conditions and initial conditions,

u(0, t) = u(L, t) = 0, (12) u(x, 0) = ut(x, 0) = 0. (13)

Letting F (x, t) = A sin(ωt) gives the following Fourier Series Representation of the forcing function F ,

F (x, t) =

∑^ ∞

n=

fn(t) sin

( (^) nπx L

where fn(t) =

2 A

nπ (1^ −^ (−1)

n) sin(ωt). (15)

(a) Show that substitution of (14)-(15) into (10) gives the ODE,

G′′ n +

( (^) cnπ L

Gn =^2 nπA (1 − (−1)n) sin(ωt). (16)

(b) The solution to (16) is given by,

Gn(t) = Bn cos

( (^) cnπ L x

  • B n∗ sin

( (^) cnπ L x

  • Gpn(t), (17)

where Bn, B∗ n ∈ R and Gpn(t) is the particular solution to (16). i. If ω 6 = cnπ/L then what would the choice for Gpn(t) be, assuming you were solving for Gpn(t) using the method of undetermined coefficients? do not solve for these coefficients ii. If ω = cnπ/L then what would the choice for Gpn(t) be, assuming you were solving for Gpn(t) using the method of undetermined coefficients? do not solve for these coefficients iii. For the latter case what is lim t→∞ u(x, t)? iv. What does this limit imply physically?

(^8) This is called the D’Alembert solution to the wave equation.