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Math Help - Hermite polynomials.

  1. #1
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    Hermite polynomials.

    Im practicing some problems for applied mathematics, and couldnt quite solve the following:
    In differential equations of equilibrium, from u(0) = a, du/dx(0) = b, u(1) = c, du/dx(1) = d we get the Hermite polynomial u = a(x-1)(2x+1) + b(x-1)x + cx(3-2x) + dx(x-1). But how do I get the cubic which substitutes u = a(x-1)(2x+1) + b(x-1)x + cx(3-2x) + dx(x-1) when we move the right end of the interval from x=1 to x=h?
    And how do I get the new form of 2(s0) + 1(s1) = 3(u1) 3(u0) for the case described above, and given that du/dx = 0 at x = 0 (in which u are the heights and s the slopes)
    If you know the solution, or could guide me in the right direction, I would appreciate it a lot! Thanks in advance..
    Last edited by mr fantastic; April 25th 2010 at 05:34 AM. Reason: Restored deleted question.
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  2. #2
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    Hi

    Using the following base
    x(x-h)
    (x-h)^3
    x(x-h)
    x

    instead of the canonical base x^3, x, x, 1

    I can find

    u(x) = \frac{a}{h^3} (x-h)^2(2x+h) + \frac{b}{h^2}x(x-h)^2 + \frac{c}{h^3}x^2(3h-2x) + \frac{d}{h^2}x^2(x-h)
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  3. #3
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    .

    Hi running-gag,
    Thanks for taking a look at my question, but I couldn't quite figure out how to get from the canonical base to the new one. Could you show me how?
    Last edited by mr fantastic; April 25th 2010 at 05:35 AM. Reason: Restored deleted post.
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  4. #4
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    You are looking for a third degree polynomial

    To find it you can use the canonical base x^3, x^2, x^3, 1,
    set u(x) = \alpha x^3 + \beta x^2 + \gamma x + \delta
    and use the conditions u(0) = a, \frac{du}{dx}(0) = b, u(h) = c, \frac{du}{dx}(h) = d

    But you can use another base in order to simplify the calculations

    I used x(x-h)^2, (x-h)^3, x^2(x-h), x^2 also because I wanted to find a result with a form close to the one you already had for h=1

    To show that x(x-h)^2, (x-h)^3, x^2(x-h), x^2 is a base you set
    \alpha x(x-h)^2 + \beta(x-h)^3 + \gamma x^2(x-h) + \delta x^2 = 0
    and you show that this leads to \alpha= 0, \beta = 0 , \gamma = 0 and \delta =  0

    To find u(x) you set
    u(x) = \alpha x(x-h)^2 + \beta (x-h)^3 +\gamma x^2(x-h) + \delta x^2
    and you use the conditions u(0) = a, \frac{du}{dx}(0) = b, u(h) =  c, \frac{du}{dx}(h) = d
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  5. #5
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    Thanks running-gag, its clear to me that you know your math quite well! Thanks for showing me the way..
    Do you by any chance know the second part of my question? I think I should end up with 2s(0)+s(1)=(3u(1)-3u(0))/h but I'm not quite sure...
    Last edited by mr fantastic; April 25th 2010 at 05:36 AM. Reason: Restored deleted post.
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