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Eigenfunction again!

  1. May 31, 2010 #1
    [urgent] Eigenfunction again!

    Find the eigenvalue and eigenfunction of the following
    [tex]\int[/tex]cos(2X)+cox(2t)y(t) dt = ky , k=eigenvalue , intergrate from 0 to Pi

    thx, urgent

    this is the original question a.jpg
    Last edited: May 31, 2010
  2. jcsd
  3. May 31, 2010 #2
    Re: [urgent] Eigenfunction again!

    Your equation makes no sense

    [tex]\int_{0}^{\pi} cos(2t)y(t)dt[/tex]

    Is a number, yet you compare it to a function ( y(t)).


    [tex]\int_{0}^{\pi} cos(2x)y(t)dt[/tex]

    is an expression of x, and again you compare it to an expression of t ( ky(t))

    Are you sure you wrote it correctly?
  4. May 31, 2010 #3
    Re: [urgent] Eigenfunction again!

    I have uploaded the original version of question, please check!!
  5. May 31, 2010 #4
    Re: [urgent] Eigenfunction again!

    Can anyone please help !?
  6. Jun 1, 2010 #5


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    Re: [urgent] Eigenfunction again!

    What is found in your attachment is
    [tex]\int_0^\pi (cos(2x)+ cos(2t))y(t)dt= ky(x)[/tex]
    not quite what you originally wrote.

    We can write that as
    [tex]cos(2x)\int_0^\pi y(t)dt+ \int_0^\pi cos(2t)y(t)dt= ky(x)[/tex]

    As elibj123 pointed out, both of those integrals are NUMBERS that, of course, depend on y(x).

    Let [tex]Y_1= \int_0^\pi y(t)dt[/tex]
    and [tex]Y_2= \int_0^\pi cos(2t)y(t)[/tex].

    Now the equation reads simply [itex]ky(t)= Y_1 cos(2x)+ Y_2[/itex] and we can solve for y(t) just by finding the two numbers [itex]Y_1[/itex] and [itex]Y_2[/itex].

    Multiply both sides of [itex]ky(x)= Y_1 cos(2x)+ Y_2[/itex] by cos(2x) and integrate from 0 to [itex]\pi[/itex]:
    [tex]k\int_0^\pi cos(2x)y(x)dx= Y_1 \int_0^\pi cos^2(2x) dx+ Y_2\int_0^\pi cos(2x)dx[/tex]

    Of course, [itex]\int_0^\pi cos(2x)y(x)dx[/itex] is the same as [itex]\int_0^\pi cos(2t)y(t)dt[/itex] which we have called [itex]Y_2[/itex]. The other two integrals do not involve y(x) and so can be integrated. Rather than do them for you I am going to write [itex]\int_0^\pi cos^2(2x)dx= A[/itex] and [itex]\int_0^\pi cos(2x)= B[/itex] (although that second one ought to be obvious!).

    Now, our equation is [itex]kY_2= AY_1+ BY_2[/itex] or [itex]AY_1+ (B- k)Y_2= 0[/itex].

    If we simply integrate [itex]ky(x)= Y_1 cos(2x)+ Y_2[/itex] itself from 0 to [itex]\pi[/itex] we get
    [tex]k\int_0^\pi y(x)dx= Y_1 \int_0^\pi cos(2x)dx+ Y_2\int_0^\pi dx= BY_1+ \pi Y_2[/tex]
    That is the same as [itex]kY_1= BY_1+ \pi Y_2[/itex] or [itex](B-k)Y_1+ \pi Y_2[/itex].

    That is, we can solve for y(x) by solving the pair of numerical equations
    [itex](B- k)Y_1+ \pi Y_2= 0[/itex] and [itex]AY_1+ (B- k)Y_2= 0[/itex]
    where A and B are given by the integrals above.

    Of course, like any eigenvalue equation, those are satified by the "trivial" solution [itex]Y_1= Y_2= 0[/itex]. Eigenvalues are values of k for which there exist non-trivial solutions. Non-trivial solutions for homogeneous systems of equations occur when the determinant of the coefficient matrix, here [itex](B- k)^2- A\pi[/itex], is 0.
    Last edited by a moderator: Jun 1, 2010
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