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splitringtail
09-26-2008, 10:42 PM
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
10-12-2008, 03:44 PM
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
04-15-2009, 05:23 AM