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PDE help, Green's Function

  1. Mar 30, 2009 #1
    Hey Guys;

    I'm solving PDE's with the use of Green's function where all the boundary conditions are homogeneous.
    However, how do you solve ones in which we have non-homogeneous b.c's.

    In case it helps, the particular PDE I'm looking at is:

    [tex]y'' = -x^2[/tex]

    [tex]y(0) + y'(0) = 4, y'(1)= 2 [/tex]

    Thanks for any help.
     
  2. jcsd
  3. Mar 30, 2009 #2
    That is an ODE, the general solution is

    [tex]y = c_1 + c_2x - \frac{1}{12}x^4 [/tex]

    The equations are:

    [tex]y'(1) = c_2 - \frac{1}{3}(1)^3 = 1[/tex]

    Therefore we have c_2 = 4/3. Therefore the second equation reads:

    [tex]c_1 + c_2 = 4[/tex]

    Therefore we have c _1 = 8/3.

    Doing this problem with the method of Green functions, we begin with the piecewise solution c_1 + c_2x, x < x' , and b_1 + b_2 x, x > x'. The first boundary condition says that c_1 + c_2 = 4, and the second one says b_2 = 1. Continuity at x' yields c_1 + (4-c_1)x' = b_1 + x'. Finally, the jump condition on the first derivative yields y'(x')_right - y'(x')_left = 1 implies 1 - c_2 = 1, and so c_2 is zero, at which point I'm stuck too.
     
  4. Apr 3, 2009 #3

    Try making the substitution:
    Y(x) = y(x) - 2x - 2.

    Then
    Y(0) = y(0) - 2 , Y'(0) = y'(0) - 2 and Y'(1) = y'(1) - 2.
    These will make your bc homogeneous.
     
  5. Apr 4, 2009 #4
    Thanks, that seems to work perfectly.
     
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