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Rayleigh-Ritz Approximation (Obtaining the constants)

  1. Sep 18, 2011 #1
    1. The problem statement, all variables and given/known data
    Consider the differential equation

    [tex]-\frac{d^2u}{dx^2}=cos\pi x[/tex] for 0<x<1

    subject to the boundary condition

    u(0)=0, u(1)=0

    Determine a three-parameter solution, with trigonometric functions the least-squares method

    Given: [tex]u=\phi_0 + c_1\phi_1 + c_2\phi_2 + c_3\phi_3[/tex]

    [tex]\phi_0=0, \phi_1=sin\pi x, \phi_2=sin2\pi x, \phi_3=sin3\pi x[/tex]

    2. Relevant equations
    1st and 2nd derivatives of u with respect to x
    Derivative of R with respect to ci

    [tex]0=\int_0^1R\frac{\partial R}{\partial c_i}dx[/tex]

    3. The attempt at a solution

    [tex]u=c_1sin\pi x + c_2sin2\pi x + c_3sin3\pi x=\Sigma_{i=1}^n c_isin(i\pi x)[/tex]

    [tex]\frac{d^2u}{dx^2}=-\Sigma_{i=1}^n c_i(i\pi)^2sin(i\pi x)[/tex]

    and

    [tex]\frac{\partial R}{\partial c_i}=(i\pi)^2sin(i\pi x)[/tex]

    so then,

    [tex]R=-\frac{d^2u}{dx^2}-cos(\pi x)=\Sigma_{i=1}^n c_i(i\pi)^2sin(i\pi x)-cos(\pi x)[/tex]
    [tex]0=\int_0^1R\frac{\partial R}{\partial c_i}dx=\int_0^1\left[\Sigma_{i=1}^n c_i(i\pi)^2sin(i\pi x)-cos(\pi x)\right]\left[(i\pi)^2sin(i\pi x)\right]dx[/tex]

    The term,
    [tex](i\pi)^2[/tex] is dropped out because it's just a multiplicative factor.

    So then I arrive here, at which point I'm stuck
    [tex]0=\int_0^1\left[\Sigma_{i=1}^n c_i(i\pi)^2sin(i\pi x)-cos(\pi x)\right]sin(i\pi x)dx[/tex]

    I know that I need to somehow get [tex]c_1, c_2, c_3[/tex]
    If I integrate the above integral across the summation, I'll end up with one equation with the three constants.

    I would think that I need three equations. Do I perform the summation from i=1,2,3 and then integrate, which would give me one equation, or do I integrate and then do the summation? If I remember correctly I can take the summation out before the integral. Is this correct?
     
  2. jcsd
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