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Homework Help: A simple 2nd order ODE using Green's functions

  1. Dec 10, 2009 #1
    1. The problem statement, all variables and given/known data

    Solve [tex] \frac{d^2 u(x)}{dx^2}=x(1-x) [/tex] subject to the homogeneous boundary conditions u(0)=0=u(1), using Green's functions.

    2. Relevant equations

    Green's functions...

    3. The attempt at a solution

    There are three parts:

    i - homogeneous eqn:

    [tex] \frac{d^2 u(x)}{dx^2}=0 [/tex] which has the obvious solutions a, bx. Given the BCs these can be modified to be u(x) = x or u(x) = 1-x; the first satisfies the first BC and the second satisfies the second.

    ii - construction of Green's function:

    [tex] G(x,z) = \left\{ \begin{array}{rcl}
    c_1(z) x, \hspace{3mm} 0 < z < x \\
    c_2(z) (1-x), \hspace{3mm} x < z < 1
    \end{array}\right [/tex]

    iii - application of continuity requirements

    We demand that:

    [tex] c_1(z) z - c_2(z) (1-z) = 0 [/tex]
    and
    [tex] c_1(z) + c_2(z) = -1 [/tex]

    which tells us after some algebra that

    [tex] c_1(z) = -(1+z), c_2(z) = z[/tex]. Thus

    [tex] G(x,z) = \left\{ \begin{array}{rcl}
    -(1+z) x, 0 < z < x \\
    z (1-x), x < z < 1
    \end{array}\right [/tex].

    So we should just be able to solve for u(x) problem-free:

    [tex] u(x) = \int_0^1 G(x,z) z(1-z) dz = \int_0^x -(1+z) x z(1-z) dz + \int_x^1 z (1-x) z(1-z) dz = (11/12)x^4-x^2-(1/12)x-(1/4)x^5+1/12-(1/3)x^3 [/tex]

    But this u(x) does not satisfy the original ODE!!!!

    What am I doing wrong? I suspect that I approached part i wrong. It seemed suspicious - I thought that both my two homogeneous solns had to individually satisfy BOTH BC's, and each only satisfies one - is that where I'm going wrong? :confused:
     
  2. jcsd
  3. Dec 11, 2009 #2

    HallsofIvy

    User Avatar
    Science Advisor

    You have these reversed. Since the first satisfies G(0,z)= 0, you must have 0< x< z. Since the second satisfies G(1,z)= 0, it must be for z< x< 1.

     
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