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Math Help - Special Matrices and Applications

  1. #1
    Newbie Erich's Avatar
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    Special Matrices and Applications

    Find the 5 by 5 matrix A_{0}(h = \frac{1}{6}) that approximates
     \frac{-d^2u}{ dx^2} = f(x), \frac{du}{dx}(0) = \frac{du}{dx}(1) = 0

    replacing these boundary conditions by u_{0} = u_{1} and u_{6}=u_{5}. Check that your A_{0} times the constant vector (C, C, C, C, C), yields zero; A_{0} is singular. Analogously, if u(x) is a solution of the continuous problem, then so is u(x)+c.

    Here's my work:
     \left[\begin{array}{ccccc}\frac{\Delta U}{\Delta x}  | x_{1}\\\\\frac{\Delta U}{\Delta x}  | x_{2}\\\\\frac{\Delta U}{\Delta x}  | x_{3}\\\\.\\\\.\\\\\frac{\Delta U}{\Delta x}  | x_{n}\end{array}\right] =  \frac{U_{i+1} - U_{i}}{x_{i+1}-x_{i}}= \frac{U_{i+1}-U_{i}}{h} where \ h = \frac{1}{n+1}
     f_{1} = f(x_{1})
    f_{2}=f(x_{2})
    u_{1} = u(x_{1})
    u_{2}=u(x_{2})

     \frac{\Delta U}{\Delta x} = \frac{U_{1+1}-U_{1}}{h}=\frac{U_{2}-U_{1}}{h}

    Follow the pattern until you get to  \frac{U_{6}-U_{5}}{h} because  U_{6} = U_{5}; \frac{U_{6}-U_{5}}{h} = 0

    Therefore let Forward U be labeled F, the matrix is:
     F_{u}=\frac{1}{h}\left[\begin{array}{ccccc}-1&1&0&0&0\\0&-1&1&0&0\\0&0&-1&1&0\\0&0&0&-1&1\\0&0&0&0&0\end{array}\right]\left[\begin{array}{c}u_{1}\\u_{2}\\u_{3}\\u_{4}\\u_{5}\  end{array}\right]



     \frac{\Delta U}{\Delta x} = \frac{U_{i}-U_{i-1}}{h}

    Follow the pattern as before, but going backwards, and since [tex] U_{1}=U_{0}=0, you get to  \frac{U_{6}-U_{5}}{h}  because  U_{6} = U_{5}; \frac{U_{6}-U_{5}}{h} = 0; the \ resulting \frac{\Delta U}{\Delta x} = \frac{U_{1}-U_{1-1}}{h}=0
    Let B= Backwards of U
     \left[\begin{array}{ccccc}\frac{\Delta U}{\Delta x}  |   x_{1}\\\\\frac{\Delta U}{\Delta x}  | x_{2}\\\\\frac{\Delta U}{\Delta x}    | x_{3}\\\\.\\\\.\\\\\frac{\Delta U}{\Delta x}  |   x_{n}\end{array}\right] = B_{u} = \frac{1}{h}\left[\begin{array}{ccccc}0&0&0&0&0\\-1&1&0&0&0\\0&-1&1&0&0\\0&0&-1&1&0\\0&0&0&-1&1\end{array}\right]\left[\begin{array}{c}u_{1}\\u_{2}\\u_{3}\\u_{4}\\u_{5}\  end{array}\right]

     \frac{-d^2u}{ dx^2} = \frac{(\frac{\Delta U}{\Delta x})_{F} - (\frac{\Delta U}{\Delta x})_B}{h^2}
    Factor out U to get
    \frac{F_{u} - B_{u}}{h^2}
    Now because that one was for \frac{d^2u}{ dx^2}, do I have to do the exact opposite?
    \frac{B_{u} - F_{u}}{h^2}? For the first element, I keep getting -1 - 0 for the 1st element in the 5 by 5 matrix, yet the book gets 1, and I have no idea why...









    *I worked on this problem from 3 in the afternoon, til 7 at night, and I still don't have a clue how the author got the answer out of the book. The way my professor showed the class how to solve a similar problem is different from what the book did, and that's the examples I'll use to show what I've done so far. I had 3 professors help me, and they were also confused, which means they were unable to answer the question.
    Last edited by Erich; February 18th 2011 at 11:22 AM. Reason: Left out a h^2
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