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Egienfunctions and Egienvalues

  1. Sep 26, 2008 #1
    1. The problem statement, all variables and given/known data

    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]

    2. Relevant 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]

    3. 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.
     
  2. jcsd
  3. Sep 27, 2008 #2
    Are you sure it's not [tex]U''(x) = - k U(x)[/tex]??
     
  4. Sep 27, 2008 #3

    HallsofIvy

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    Neither
    [tex]e^{ix\sqrt{k_n}}[/tex] and [tex]e^{-ix\sqrt{k_n}}[/tex]
    nor
    [tex]cos(x\sqrt{k_n}})[/tex] and [tex]sin(x\sqrt{k_n}}[/tex]
    are eigenfunctions of the equation you give.

    They are eigenfunctions of U"= -U.

    Of course, because eix= cos(x)+ i sin(x), writing the functions as complex exponential or as sine and cosine functions is a matter of choice.

    "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 don't know whether you professor needed to have added that or not but it is certainly true that the eigenfunctions for any problem are determined by both the operator and the B.C./I.C.
     
  5. Sep 27, 2008 #4
    I mean to say,

    [tex]U''[x]=-kU[x][/tex]

    I apologize for the misunderstanding, but the work above is for the ODE in question. Nor am I comfortable with Latex. I really don't have time read up on Latex, when I am trying to learn programing and software relevant to my research.
     
  6. Sep 27, 2008 #5
    This is not right
    sin(what)=0 ?
     
  7. Sep 27, 2008 #6
    [tex] \Rightarrow\ \sqrt{k_{n}} = \frac{2 \pi n}{a}\ where\ n=1,2,3,...[/tex]

    Sorry, that I am not really good at this Latex stuff. I know should know it.
     
  8. Sep 28, 2008 #7

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    LaTex has nothing to do with that: the difference between U"= -kU' and U"= -kU is that ' on the right.

    The fact that k= 4/a2 does NOT give sin(sqrt{k}a)= 0 has nothing to do with LaTex as well. sin(x)= 0 when x= n pi. Then sqrt{k}a= n pi gives k= (n pi/a)2, not just (2pi/a)2.
     
  9. Sep 28, 2008 #8
    I do know they are different operators, but I have to keep checking back and forth to see if I got the Latex working or not. Nor have I figured how to break lines in LaTex. I have tired a few commands I have found on the web, but they don't appear to work here. I have restored to putting the "tex" and "/tex" on each line. It gets pretty cumbersome for my point of veiw and things slip my attention.

    My question is mainly about why can't I use Exp[] instead of Sin[]/Cos[]. Or at least get two different answers, which are the same. I didn't want to just sit that and just say it. I wanted to show the work that I put thought into this.
     
    Last edited: Sep 28, 2008
  10. Sep 28, 2008 #9

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You can if you are happy with complex exponentials. Many people, especially engineers, don't like to use complex numbers where they know everything will come out to be real numbers in the end.

    It is clearer that C cos(kx)+ D sin(kx), with C and D real constants, will always give real values than it is with A eikx+ B e-ikx, with A and B complex numbers.
     
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