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Numerical PDE's

  1. Mar 28, 2008 #1
    1. The problem statement, all variables and given/known data

    A = [ b c ... 0000000000000000000 ]
    [ c b c ... .000000000000000 0]
    [ ... ]
    [ 000000000000000000 c b c ]
    [ 000000000000000000 b c ]
    where a,b are real. This matrix is tridigonal and symmetric.

    I need to show that this matrix has e-values lamda_i = b +2cos((i * pi)/(N+1))
    and e-vectors x_i = [sin ((i* pi)/(N+1), sin ((2*i*pi)/(N+1)), ...., sin((N*i*pi)/(N+1))]

    2. Relevant equations



    3. The attempt at a solution

    I could find the deterministic equation to find the e-values but i don't see how that gives rise to trigonometric functions.
     
  2. jcsd
  3. Mar 29, 2008 #2

    HallsofIvy

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    The matrix you've shown is not symmetric. The right bottom corner is
    [tex]\left[\begin{array}{cc}b & c \\ b & c\end{array}\right][/itex]
    which is not symmetric.
    Do you mean for the bottom row to end with "c b" rather than "b c"?

    For problems like this it is always a good idea to look at simple cases- If n= 3 this is
    [tex]\left[\begin{array}{ccc}b & c & 0 \\ c & b & c \\ 0 & b & c\end{array}\right][/tex]
    where I have altered the matrix as I suggested to make it symmetric.

    The characteristic equation is [itex](b-\lambda)^3- 2c^2(b- \lambda)= (b- \lambda)((b-\lambda^2- 2c^20)= 0[/itex] which has the obvious solutions [itex]\lambda= b[/itex], [itex]\lambda= b+ c\sqrt{2}[/itex], and [itex]\lambda= b- c\sqrt{2}[/itex]. Of course, here N= 3 so N+1= 4 and [itex]\pi/(N+1)= \pi/4[/itex].
    According to your formula, [itex]\lambda_1= b+ 2cos(\pi/4)= b+ \sqrt{2}[/itex], [itex]/lamba_2= b+ 2cos(2\pi/4)= b[/itex] and [itex]\lambda_3= b+ 2cos(3\pi/4)= b- \sqrt{2}[/itex]
    (Your formula seems to be missing a "c".)
     
  4. Mar 29, 2008 #3

    tiny-tim

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    … (N+1)-th roots of -1 …

    Hi Nusc! :smile:

    (I haven't looked at the matrix aspect of this, but …)

    cos(nπ/(N+1)) and sin(nNπ/(N+1)) aren't really trigonometric - they're part of:
    [tex]e^{i\pi\left(\frac{n}{N+1}
    \right)}\,,\,=\,\cos\left(\frac{n\pi}{N+1}
    \right)\,+\,i\sin\left(\frac{n\pi}{N+1}
    \right)\,,[/tex]​
    which is one of the (N+1)-th roots of -1. :smile:
     
  5. Mar 29, 2008 #4
    For the tridiagonal matrix we have, b elements are along the diagonal. c elements are on the upper and lower diagonal.

    I could not write the matrix using latex
     
  6. Mar 29, 2008 #5
    That matrix is incorrect. How would i find the deterministic equation for nxn matrix in this case? Would doing that suffice to prove that the matrix always has e-values \lambda_i = b +2ccos(ipi/(n+1)) ?
     
    Last edited: Mar 29, 2008
  7. Mar 29, 2008 #6

    tiny-tim

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    … wrong answer …

    The expected answer is wrongly stated. :mad:

    It should have a factor 2c.cos, not 2cos.

    Then the eigenvalues are b + 2c.cos(nπ/(N+1)), for n = 1 … N;
    and an eigenvector for the nth eigenvalue is:
    (sin(nπ/(N+1)), sin(2nπ/(N+1)), … , sin(Nnπ/(N+1))).​

    Hint: sin((m-1)nπ/(N+1)) + sin((m+1)nπ/(N+1)) = … ? :smile:

    (And note that sin(0nπ/(N+1)) and sin((N+1)nπ/(N+1)) are both 0.)
     
    Last edited: Mar 30, 2008
  8. Mar 29, 2008 #7
    So the actual characteristic equation for a 3x3 matrix in this case is
    [itex]
    (b-\lambda)^3- 2c^2(b- \lambda) +c^2 = 0
    [/itex]
    What tricks can i use to solve for lamdba
     
  9. Mar 30, 2008 #8

    tiny-tim

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    … recurrence relation …

    Hi Nusc! :smile:

    Hint: if the characteristic equation of the n x n matrix is Pn = 0, find a recurrence relation expressing the polynomial Pn in terms of Pn-1 and Pn-2.

    Then either solve that recurrence relation by normal methods, or (since the question gives us the solution) define ∆ by: lambda = b + 2c.cos∆.

    You should find that Pn = cos(n∆).

    And then … ? :smile:
     
  10. Mar 30, 2008 #9
    I'm not familiar with recurrence relations and so I don't know how that process works.
     
  11. Mar 30, 2008 #10

    tiny-tim

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    ok, I'll rephrase it without recurrence relations …

    Look for constants p and q which do not depend on n and for which there is a formula [tex]P_n\,=\,pP_{n-1}\,+\,qP_{n-2}\,.[/tex]

    (Do it by starting with the b - lambda in the top left corner, and then starting again with the two c's next to it.)

    Then put Pn = cos(n∆) into the formula, and check that that works. :smile:
     
  12. Mar 31, 2008 #11
    I feel like I should know this elementary concept. lol sadly i do not
     
  13. Mar 31, 2008 #12

    tiny-tim

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    … finding the determinant …

    Right: the object is to find the determinant (which we're calling Pn) of the tridiagonl n x n matrix with b - µ along the diagonal and c along the off-diagonals.

    Do you know how the determinant is defined? Basically, you multiply n terms, chosen so that there's one in each of the n rows and one in each of the n columns, and then add all such possible products (some of them multiplied by minus-one).

    So start with the b - µ in the top left corner - what is the sum of all the possible products with that b - µ? Can it be written (b - µ)Pm, for some m?

    Then start again, with the two c's next to the top left corner. Can you have one without the other in any product? If not, why not? Then do the same as before … can the result, or part of the result, be written cPm, or (c^2)Pm, for some m?

    Has that left anything out?

    How have you got on so far … ? :smile:
     
  14. Mar 31, 2008 #13
    P_n = (b-lambda)P_n-1 - c^2 P_n-2
    Are you sure Pn = Cos(n*nabla) ?

    (b-lambda)P_n-1 = c^2 P_n-1

    which gives

    lambda = b - c^2 Pn-2/Pn-1

    HOw does that equal b + 2 c.cos(n*pi/(N+1) ???
     
    Last edited: Mar 31, 2008
  15. Apr 1, 2008 #14

    tiny-tim

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    … cos((n+1)∆) + cos((n-1)∆) = 2.cos∆.cos(n∆) …

    Hi Nusc! :smile:
    Excellent! :smile:
    erm … where does that come from? :confused:
    (Actually, its a Delta … a nabla is an upside-down ∆ … ∆ is a standard letter in all civilsed fonts, so if you copy-and-paste from this post, it should work. :smile:)

    Yup. Just define ∆ by lambda = b + 2c.cos∆, which gives you:
    P_n = 2c.cos∆P_n-1 - c^2 P_n-2​

    (or P_n+1 = 2c.cos∆P_n - c^2 P_n-1, which is probably slightly neater.)

    and plug P_n = 2.(c^n).cos(n∆) into it, and divide by 2.c^n.

    (:redface: oops! … I left out the 2.c^n before … :redface:)​

    In other words, use standard trig formulas to check that:

    :smile: cos((n+1)∆) + cos((n-1)∆) = 2.cos∆.cos(n∆). :smile:

    and combine that with the obvious P_1 = b - lambda = 2.c.cos∆ and P_2 = (b - lambda)^2 - c^2 = (c^2)(4(cos∆)^2 - 1) = (c^2)(2(cos∆)^2 - 2(sin∆)^2) = 2.(c^2).cos(2∆) to confirm that we have the right factor and the right starting-point for n.​

    And finally … since the characteristic equation is now 2.(c^n).cos(n∆) = 0 … what are the eigenvalues (the lambdas) … ? :smile:

    (oh … and the value for the angle in my post #6 was wrong … :redface:)
     
  16. Apr 1, 2008 #15
     
  17. Apr 1, 2008 #16
    Where did this come from
    P_n = 2.(c^n).cos(n∆) into it, and divide by 2.c^n.
    ?
     
  18. Apr 1, 2008 #17
    And you forgot a negative sign it shojld be pn = -2ccos(del*n(
     
  19. Apr 1, 2008 #18

    tiny-tim

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    Hi nusc! :smile:
    ah … I said that the characteristic equation of the matrix is Pn = 0.

    Pn is a polynomial in lambda, and we find all the lambdas by finding all n solutions of Pn = 0.

    But first we have to find Pn … and it isn't zero when we find it! :smile:

    So we found it by induction (or recurrence).
    Well, that's the official answer (more or less), so I just plugged it in to confirm that it was the right answer (and, incidentally found that I'd got the wrong factor originally!). :smile:

    We could have solved for P_n using standard recurrence relation methods, of course, but you said you hadn't done that, so I just went straight to the answer.
    ooh … you're right … well spotted! :redface:

    Soo … since the eigenvalues (the lambdas) must satisfy cos(n∆) = 0 … what are they … ? :smile:
     
  20. Apr 1, 2008 #19
    I'm lost:

    [tex]
    P_n\,=\,-2c*ccos(m*pi/(N+1))P_{n-1}\,-\,c^2P_{n-2}\,.
    [/tex]

    That does not give you:
    [tex]
    P_n = -2c^n \cos(n*del)
    [/tex]
     
  21. Apr 2, 2008 #20

    tiny-tim

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    … sorry … !

    :mad: ah … yes … I made a stupid trig mistake in P_2 ! … :mad:

    I wrote:
    P_2 = (b - lambda)^2 - c^2 = (c^2)(4(cos∆)^2 - 1) = (c^2)(2(cos∆)^2 - 2(sin∆)^2) = 2.(c^2).cos(2∆),​
    which is rubbish.

    4(cos∆)^2 - 1 isn't 2(cos∆)^2 - 2(sin∆)^2, it's 2(cos∆)^2 + cos(2∆).

    So P_2 should have been (c^2)(2.(cos∆)^2 + cos(2∆)) = (c^2).sin(3∆)/sin∆.

    My original P_1 = 2c.cos∆ was correct, but it can also be written P_1 = c.sin(2∆)/sin∆.

    So the middle of my post #14 (it's just too late for me to edit it) should read:

    P_n = -2c.cos∆P_n-1 - c^2 P_n-2.

    So plug P_n = ((-c)^n).sin((n+1)∆)/sin∆ into it, and multiply by sin∆/c^n.

    In other words, use standard trig formulas to check that:

    sin((n+1)∆) + sin((n-1)∆) = 2.cos∆.sin(n∆),​

    and combine that with:
    P_1 = b - lambda = -2.c.cos∆ = -sin(2∆)/sin∆,

    and P_2 = (b - lambda)^2 - c^2 = (c^2)(4(cos∆)^2 - 1) = (c^2)(2.(cos∆)^2 + cos(2∆)) = (c^2).sin(3∆)/sin∆,​

    to confirm that P_n = ((-c)^n).sin((n+1)∆)/sin∆ gives us the right factor, and also the right starting-point for n.


    :redface: Sorry! :redface:

    So … back to the plot … since … with a little help from you … the correct characteristic equation is now ((-c)^n).sin((n+1)∆)/sin∆ = 0 … what are the eigenvalues (the lambdas) … ? :smile:
     
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