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Homework Statement
We are suppose to find the Eigenfunctions and Eigenvalues of the following
on the interval 0<x<a;
[tex]U''[x] = - k U'[x][/tex]
for the following case
[tex]a)\ U(0) = 0\ and\ U(a) = 0.[/tex]
[tex]b)\ U(0) = 0\ and\ U'(a) = 0.[/tex]
[tex]c)\ U'(0)= 0\ and\ U'(a) = 0.[/tex]
[tex]d)\ U(0)+a U'(0)=0\ and\ U(a)-a U'(a)=0.[/tex]
Homework Equations
[tex]Professor's\ Statement:\ The\ operator\ determines\ the\ eigenfunction\ and[/tex]
[tex]the\ boundary\ conditions/initial\ conditions\ determine\ the\ eigenvalues.[/tex]
[tex]U_{n}[x]=A_{n}\ Cos[ \sqrt{k_{n}}x] + B_{n}\ Sin[\sqrt{k_{n}}x][/tex]
[tex]U_{n}[x]=C_{n}\ Exp [i \sqrt{k_{n}}x] + D_{n}\ Exp [-i \sqrt{k_{n}}x][/tex]
[tex]R\ Exp [i \sqrt{k_{n}}x]=R\ (Cos[ \sqrt{k_{n}}x] + Sin[\sqrt{k_{n}}x])[/tex]
[tex]Initial Conditions: (a)\ to\ (d)[/tex]
The Attempt at a Solution
This is where I get confused, because I want to say the eigenfunctions are
[tex]U_{n}[x]=C_{n}\ Exp [i \sqrt{k_{n}}x] + D_{n}\ Exp [-i \sqrt{k_{n}}x][/tex]
since this is the Fourier Transform and I like exponentials in general. From (a),
[tex]U_{n}(0) = 0\ \Rightarrow\ C_{n}=-D_{n}[/tex]
[tex]U_{n}(a) = 0\ \Rightarrow\ -D_{n}\ Exp [i \sqrt{k_{n}}a] + D_{n}\ Exp [-i \sqrt{k_{n}}a]=0[/tex]
[tex]D_{n}\ Exp [i \sqrt{k_{n}}a]= D_{n}\ Exp [-i\sqrt{k_{n}}a][/tex]
[tex]Exp [i \sqrt{k_{n}}a]= Exp [-i \sqrt{k_{n}}a][/tex]
[tex]i\sqrt{k_{n}}a= -i \sqrt{k_{n}}a \Rightarrow 1=-1[/tex]
For the other eigenfunction, I can find the eigenvalues.
[tex]U_{n}[x]=A_{n}\ Cos[ \sqrt{k_{n}}x] + B_{n}\ Sin[\sqrt{k_{n}}x][/tex]
Again from (a),
[tex]U_{n}(0) = 0\ \Rightarrow\ A_{n} + 0 = 0[/tex]
[tex]U_{n}(0) = 0\ \Rightarrow\ B_{n} Sin[\sqrt{k_{n}}a] = 0[/tex]
[tex]\Rightarrow\ \sqrt{k_{n}} = \frac{2 \pin}{a}\ where\ n=1,2,3,...[/tex]
[tex]\Rightarrow\ k_{n}=(\frac{2 \pin}{a})^{2}[/tex]
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 too but no luck and it is a lot of work to present here.