[PDF] Solutions to Practice Problems for Final Examination

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Solutions to Practice Problems for Final Examination

Question 1.Given the function

f(x) =x,-π < x < π find the Fourier series forfand use Dirichlet"s convergence theorem to show that n=1(-1)n-1sinna n=a2 for 0< a < π.

Solution:Sincef(x) is an odd function on the interval [-π,π],the Fourier series off(x) is given by

f(x)≂∞? n=1b nsinnx where b n=2 0 xsinnxdx 2 -1nxcosnx????π0-1n? 0 cosnxdx? =-2 n(-1)n and b n=2(-1)n-1 n forn≥1.

Therefore

f(x)≂2∞? n=1(-1)n-1sinnx n,

and from Dirichlet"s convergence theorem, sincef(x) is continuous for-π < x < π,the Fourier series

converges tof(x) for-π < x < π,that is, x= 2∞? n=1(-1)n-1sinnx n for-π < x < π,in particular, choosingx=a,we get n=1(-1)n-1sinna n=a2 for 0< a < π.

Question 2.Let 0< a < π,given the function

f(x) =???1

2aif|x|< a

0 ifx?(-π,π],and|x|> a

find the Fourier series forfand use Dirichlet"s convergence theorem to show that n=1sinna n=12(π-a) for 0< a < π.

Solution:Sincef(x) is an even function of the interval [-π,π],the Fourier series off(x) is given by

f(x)≂a0+∞? n=1a ncosnx where a 0=1 0 f(x)dx=1π? a

012adx=12π,

and a n=2 0 f(x)cosnxdx 2 a

012acosnxdx

1

πa?

a 0 cosnxdx 1

πa·1nsinnx????a0

1

πa·sinnan,

that is, a n=1

πa·sinnan

forn≥1,and f(x)≂1

2π+1πa∞

n=1sinnacosnxn for-π < x < π.

Sincef(x) is continuous on the interval-π < x < πthe Fourier series converges tof(x) for-π < x < π,

that is, f(x) =1

2π+1πa∞

n=1sinnacosnxn for-π < x < π,in particular, whenx= 0,we have 1

2a=12π+1πa∞

n=1sinnan, so that n=1sinna n=12(π-a) for 0< a < π. Question 3.Consider the regular Sturm-Liouville problem (xφ?)?+λ21

φ(1) = 0

φ(2) = 0

(a) The general solution to the differential equation is

φ(x) =Acos(λlnx) +Bsin(λlnx).

Find the eigenvaluesλ2nand the corresponding eigenfunctionsφnfor this problem.

(b) Show directly, by integration, that eigenfunctions corresponding to distinct eigenvalues are orthogonal.

(c) Use the Rayleigh quotient to estimate the smallest eigenvalue of this regular Sturm-Liouville problem.

Note:From part (a), the first eigenvalue and eigenfunction are 2

1=?π

ln2?

2≈20.5423 andφ1(x) = sin?πlnxln2?

Try to find a reasonable estimate.

Solution:

(a) If

φ(x) =Acos(λlnx) +Bsin(λlnx)

for 1< x <2,then ?(x) =-λA xsin(λlnx) +λBxcos(λlnx), so that xφ ?(x) =-λAsin(λlnx) +λBcos(λlnx), and (xφ?(x))?=-λ2A xcos(λlnx)-λ2Bxsin(λlnx).

Therefore

(xφ?(x))?+λ21 xφ(x) = 0 for 1< x <2,andφ(x) is a solution to the differential equation. In order to satisfy the first boundary conditionφ(1) = 0,we need

φ(1) =Acos0 +Bsin0 =A= 0,

and the solution is now

φ(x) =Bsin(λlnx)

for 1< x <2. In order to satisfy the second boundary conditionφ(2) = 0,we need

φ(2) =Bsin(λln2) = 0,

and ifB= 0 we get the trivial solution. Therefore we have a nontrivial solution to the boundary value problem if and only if sin(λln2) = 0, that is, if and only ifλln2 =nπfor some integern. The eigenvalues and eigenfunctions for this boundary valueproblem are given by 2 n=?nπ ln2?

2andφn(x) = sin?nπlnxln2?

,1< x <2 forn≥1.

(b) From the differential equation, eigenfunctions corresponding to distinct eigenvalues will be orthogonal

on the interval [1,2] with respect to the weight functionσ(x) =1 x. To show this directly, suppose thatmandnare positive integers withm?=n,then 2 1 m(x)φn(x)1 xdx=? 2 1 sin?mπlnxln2? sin?nπlnxln2? 1xdx = ln2 1 0 sin(mπt)sin(nπt)dt(t= lnx/ln2) = 0 ifm?=n. (c) Letu(x) be a test function satisfying only the boundary conditions u(1) = 0 andu(2) = 0, the simplest such function is the quadratic u(x) = (2-x)(x-1) =-x2+ 3x-2 withu?(x) =-2x+ 3.

The Rayleigh quotient for this function is

R(u) =-p(x)u(x)u?(x)????21+?

2

1?p(x)u?(x)2-q(x)u(x)2?dx

?2 1 u(x)2σ(x)dx, wherep(x) =x, q(x) = 0,andσ(x) =1 x.

ComputingR(u),we have

R(u) =?

2 1 xu?(x)2dx ?2 1 u(x)2σ(x)dx 2 1 x(2x-3)2dx ?2

1?(2-x)2(x-1)2/x?dx

1/2 -11/4 + 4ln2 and sinceλ1is the minimum Rayleigh quotient over all such test functions, then -11/4 + 4ln2≈23. Question 4.Find the solution of theexterior Dirichlet problem for a disk, that is find a bounded solution to the problem: 1 r∂∂r? r∂u∂r? +1r2∂

2u∂θ2= 0, a < r <∞,-π < θ < π

u(r,π) =u(r,-π)a < r <∞ ∂u ∂θ(r,π) =∂u∂θ(r,-π)a < r <∞ u(a,θ) =f(θ)-π < θ < π.

Solution:A solution to Laplace"s equation in polar coordinates whichsatisfies the periodicity conditions

is given by u(r,θ) =A0+B0logr+∞? n=1? rn?Ancosnθ+Bnsinnθ?+1 rn?Cncosnθ+Dnsinnθ??, and in order to satisfy the boundedness condition we needB0=An=Bn= 0,forn= 1,2,3,...,so that u(r,θ) =A0+∞? n=11 rn?Cncosnθ+Dnsinnθ?.

Now, whenr=awe have

f(θ) =u(a,θ) =A0+∞? n=11 an?Cncosnθ+Dnsinnθ?, where A 0=1

2π?

-πf(φ)dφ, C n=an -πf(φ) cosnφdφ, D n=an -πf(φ) sinnφdφ forn= 1,2,3....

Therefore

u(r,θ) =1

2π?

-πf(φ)dφ+1π∞ n=1? ar? n?π -πf(φ)?cosnφcosnθ+ sinnφsinnθ?dφ, that is, u(r,θ) =1

2π?

-πf(φ)?

1 + 2∞?

n=1? ar? ncosn(θ-φ)? dφ. We can actually sum the series to get a much simpler expression foru(r,θ).Letz=a rei(θ-φ),then z n=?a r? nein(θ-φ)=?ar? n[cosn(θ-φ) +isinn(θ-φ)], and

1 + 2∞?

n=1? a r? ncosn(θ-φ) = Re?

1 + 2∞?

n=1z n?

Since|z|=ar<1,then

1 + 2 n=1? a r? ncosn(θ-φ) = Re?

1 +2z1-z?

= Re?1 +z1-z? =r2-a2a2-2arcos(θ-φ) +r2. The solution to the exterior Dirichlet problem for the disk is therefore u(r,θ) =1

2π?

-π(r2-a2)f(φ)a2-2arcos(θ-φ) +r2dφ, fora < r <∞,-π < θ < π. Question 5.Find all functionsφfor whichu(x,t) =φ(x+ct) is a solution of the heat equation 2u ∂x2=1k∂u∂t wherekandcare constants. Solution:Ifu(x,t) =φ(x+ct) is a solution to the heat equation 2u ∂x2=1k∂u∂t, letξ=x+ct,then from the chain rule we have ∂u 2u ∂x2=ddξ? dφdξ? ∂ξ∂x=d2φdξ2, ∂u Therefore,φsatisfies the ordinary differential equation d 2φ dξ2-ckdφdξ= 0, and the solution is given by

φ(ξ) =A+B ec

kξ, that is, u(x,t) =A+B ec k(x+ct) whereAandBare arbitrary constants.

Question 6.Consider torsional oscillations of a homogeneous cylindrical shaft. Ifω(x,t) is the angular

displacement at timetof the cross section atx,then 2ω ∂t2=a2∂2ω∂x20< x < L, t >0.

Solve this problem if

ω(x,0) =f(x) 0< x < L

∂t(x,0) = 0 0< x < L, and the ends of the shaft are fixed elastically: ∂x(0,t)-αω(0,t) = 0t >0 ∂x(L,t) +αω(L,t) = 0t >0 withαa positive constant.

Solution:Since the partial differential equation is linear and homogeneous and the boundary conditions

are linear and homogeneous, we can use separation of variables. Assuming a solution of the form and separating variables, we have two ordinary differentialequations: ?(0)-αφ(0) = 0 ?(L) +αφ(L) = 0 We use the Rayleigh quotient to show thatλ >0 for all eigenvaluesλ.

Letλbe an eigenvalue of the Sturm Liouville problem, and letφ(x) be the corresponding eigenfunction,

then -p(x)φ(x)φ?(x)????L0=-φ(L)φ?(L) +φ(0)φ?(0) =α(φ(0)2+φ(L)2)>0,

λ=α(φ(0)2+φ(L)2) +?

L 0

φ?(x)2dx

?L 0

φ(x)2dx≥0

Note that ifλ= 0,then

?φ(0)2+φ(L)2?+? L 0

φ?(x)2dx= 0

implies that ?φ(0)2+φ(L)2?= 0 and? L 0

φ?(x)2dx= 0.

Sinceα >0,this implies thatφ(0) = 0 andφ(L) = 0; and sinceφ?is continuous on [0,L],thatφ?(x) = 0 for

eigenvalue, and all of the eigenvaluesλof this Sturm-Liouville problem satisfyλ >0.

Ifλ >0,thenλ=μ2,whereμ?= 0,and the differential equation isφ??+μ2φ= 0 with general solution

φ(x) =Acosμx+Bsinμxandφ?(x) =-μAsinμx+μBcosμx

From the first boundary condition

?(0)-αφ(0) =μB-αA= 0, andA=μB

α,and the solution is now

φ(x) =B(μcosμx+αsinμx).

From the second boundary condition

?(L) +αφ(L) =B?-μ2sinμL+αμcosμL+αμcosμL+α2sinμL?= 0, that is,

B?(α2-μ2)sinμL+ 2αμcosμL?= 0,

and the boundary value problem has a nontrivial solution if and only if tanμL=2αμ

μ2-α2,

that is, if and only if tan

λL=2α⎷λ

λ-α2.

In order to determine the eigenvalues we sketch the graphs ofthe functions f(μ) = tanμLandg(μ) =2αμ

μ2-α2

forμ >0.

Note that forμ >0,we have

g(μ) =2αμ

μ2-α2=α?1μ+α+1μ-α?

so that g ?(μ) =-α?1 (μ+α)2+1(μ-α)2? <0

andgis decreasing on the interval (0,α) and on the interval (α,∞) and the lineμ=αis a vertical asymptote

to the graph. The graphs ofgandfare shown below. 2

μ0 y

y = y= tan αL

μ2_μα

2

From the figure it is clear that there are an infinite number of distinct solutionsμnto the equation

tanμL=2αμ

μ2-α2,

and the eigenvalues areλn=μ2n,forn≥1,while the corresponding eigenfunctions are forn≥1. The corresponding solutions to the time equation are G n(t) =ancosμnat+bnsinμnat, t≥0 and from the superposition principle, the function

ω(x,t) =∞?

n=1φ n(x)·Gn(t) =∞? satisfies the partial differential equation and the boundaryconditions.

Since the spatial problem is a regular Sturm-Liouville problem, then the eigenfunctions are orthogonal on

the interval [0,L],and we use this fact to satisfy the initial conditions

ω(x,0) =f(x) =∞?

n=1a nφn(x) and∂ω ∂t(x,0) =∞? n=1b nμnφn(x) = 0, and the generalized Fourier coefficients are given by a n=? L 0 f(x)φn(x)dx ?L 0 n(x)2dxandbn= 0 forn≥1.

Therefore the solution is

ω(x,t) =∞?

n=1a where a n=? L 0 f(x)φn(x)dx ?L 0 n(x)2dx forn≥1.

Question 7.Use D"Alembert"s solution of the wave equation to solve the initial value - boundary value

problem: 2u ∂x2=1c2∂

2u∂t2- ∞< x <∞, t >0

quotesdbs_dbs42.pdfusesText_42

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