Strum-Liouville Problem

[splitringtail]

We are suppose to find the Eigenfunctions and Eigenvalues of the following
on the interval 0<x<a;

U''[x] == - k U'[x]

for the following cases

a) U (0) = 0 and U(a) = 0.
b) U (0) = 0 and U'(a) = 0 .
c) U'(0) = 0 and U'(a) = 0 .
d) U(0)+a U'(0)=0 and U(a)-a U'(a)=0

This is where I get confused. The professor says the operator determines
eigenfunction and the boundary conditions/initial conditions determine the
eigenvalues.

Now I am having a problem with that b/c right away to me I would say the
eigenfunction is

Un[x] = Cn Exp [i Sqt(kn)x] + Dn Exp [-i Sqt(kn)x]

since I really like exponentials and this is the Fourier transform.
However, this gives me

U(0) = 0 --> Cn=-Dn

U(a) = 0 --> -Dn Exp [i Sqt(kn)a] + Dn Exp [-i Sqt(kn)a]==0

Dn Exp [i Sqt(kn)a]== Dn Exp [-i Sqt(kn)a]

Exp [i Sqt(kn)a]== Exp [-i Sqt(kn)a]

i Sqt(kn)a== -i Sqt(kn)a --> 1==-1 ; ??????


However, I could say the eigenfunction is

Un[x] = An Sin[Sqt(kn)x] + Bn Cos[Sqt(kn)x]

works a lot better and I can find eigenvalues for this problem, which is by the
way kn=(2*pi*n/a)^2. I really didn't feel like writing all the algebra out.

So, should my professor added that both the operator and B.C./I.C.
determine the eigenfunction or am I in the wrong to say that these
eigenfunctions are pretty much the same. I tried converting between the two
in such away I would get a constant times function of x, but it gets messy.
Because of the converting, I am assuming I am in the wrong.

[HallsofIvy]

We are suppose to find the Eigenfunctions and Eigenvalues of the following
on the interval 0<x<a;

U''[x] == - k U'[x]

for the following cases

a) U (0) = 0 and U(a) = 0.
b) U (0) = 0 and U'(a) = 0 .
c) U'(0) = 0 and U'(a) = 0 .
d) U(0)+a U'(0)=0 and U(a)-a U'(a)=0

This is where I get confused. The professor says the operator determines
eigenfunction and the boundary conditions/initial conditions determine the
eigenvalues.

Now I am having a problem with that b/c right away to me I would say the
eigenfunction is

Un[x] = Cn Exp [i Sqt(kn)x] + Dn Exp [-i Sqt(kn)x]

since I really like exponentials and this is the Fourier transform.
I see no "Fourier transform" here. I myself would prefer to write the functions in terms of sine and cosine: Un(x)= Cn cos(Sqt(kn) x)+ Dn sin((kn) x).

However, this gives me

U(0) = 0 --> Cn=-Dn

U(a) = 0 --> -Dn Exp [i Sqt(kn)a] + Dn Exp [-i Sqt(kn)a]==0

Dn Exp [i Sqt(kn)a]== Dn Exp [-i Sqt(kn)a]

Exp [i Sqt(kn)a]== Exp [-i Sqt(kn)a]

i Sqt(kn)a== -i Sqt(kn)a --> 1==-1 ; ??????
No, the complex exponential is NOT one-to-one. Instead, multiply both sides by Exp[i Sqt(kn)a] to get (Exp[i Sqt(kn)a])^2= 1, Now, either Exp[i Sqt(kn)a]= 1 or Exp[i Sqt(kn)a]= -1. The first is true for Sqt(kn)a any multiple of 2pi (any even multiple of pi) and the second for Sqt(kn)a any odd multiple of pi so your equation is satisifed for Sqt(kn)a= npi. kn= (npi/a)^2

However, I could say the eigenfunction is

Un[x] = An Sin[Sqt(kn)x] + Bn Cos[Sqt(kn)x]

works a lot better and I can find eigenvalues for this problem, which is by the
way kn=(2*pi*n/a)^2. I really didn't feel like writing all the algebra out.

So, should my professor added that both the operator and B.C./I.C.
determine the eigenfunction or am I in the wrong to say that these
eigenfunctions are pretty much the same. I tried converting between the two
in such away I would get a constant times function of x, but it gets messy.
Because of the converting, I am assuming I am in the wrong.[/QUOTE]
A exp(ix)+ B exp(-ix) and Ccos(x)+ Dsin(x) are the same. And it is determined by the operator. Both you and your professor are really saying the same thing: the eigefunctions are determined by the operator, the eigenvalues by the boundary conditions.

The "conversion" is not all that messy: e^ix= cos(x)+ i sin(x) and, since cosine is an even function and sine an odd function, e^(-ix)= cos(x)- i sin(x). Ae^(ix)+ Be^(-ix)= A(cos(x)+ i sin(x))+ B(cos(x)- i sin(x))= (A+ B)cos(x)+ i(A-B)sin(x). C= A+ B, D= i(A- B).

Since these problems all involve real numbers I would recommend using sine and cosine from the start.

[hdxbk666276]

日本政府14日表示，强烈敦促朝鲜重返六方会谈。俄罗斯外交部新闻局发表声明说，俄罗斯对朝鲜退出六方会谈 表示遗憾，认为朝鲜的这一决定“明显不利于”达到各方在解决朝核问题方面提出的目标。　　此刻身为“六方会 谈”的主导者中国，面对朝鲜的抵制该如何处置，这将考验中国外交智慧。　　多年来，六方会谈在推进半岛无核 化进程，为有关国家提供沟通和建立信任的平台，以及共同探讨构筑东北亚和平安全机制方面发挥了积极作用，取 得了举世瞩目的成就。可以说，“六方会谈”平台的搭建也是相关各国一个外交形象上的工程，大家自然都不愿意 看到它如此毁掉。　　外交部发言人姜瑜14日下午就朝鲜退出六方会谈一事作出回应：中方希望各方着眼大局长 远，保持冷静和克制，共同维护六方会谈进程，因为这有利于地区和平发展，应以和平方式实行半岛无核化。　　 说到底，还是要国际社会共同努力，以各种方式，使朝鲜半岛无核化重新回到六方会谈的框架内。朝鲜半岛无核化 ，符合包括中国在内的国际社会的共同利益，这一点必须坚持，不可能妥协。现在的关键是，除朝鲜之外的其他相 关五方必须团结一致，采取一致行动，像以前一样，争取重开六方会谈。只要能谈，就总能够找出问题的解决方案 。 香港六合彩 (htttp://www.hk333888.cn)