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Non linear 2nd order ode not able to solve

  1. Jul 6, 2011 #1
    u'u''-k1u=-a*cos(hy)-b
    where,u'=du/dy;
    and and a,b,k1 are constants
    conditions
    u(-H)=0;
    u'(0)=0;
    where 2H is height of the channel where the liquid is flowing
    please help any suggestions are welcome
    i couldn't find the analytical soln
    numerical soln also am havin a dead end so plz
     
  2. jcsd
  3. Jul 6, 2011 #2

    MathematicalPhysicist

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    Gold Member

    u'u''-k1u=-a*cos(hy)-b

    (u')^2 u'' - k1 uu' =(-a cos(hy) -b)u'

    1/3((u')^3)' - k1/2 ((u)^2)' = (-a cos(hy) -b) u'

    Integrate both sides wrt y, where the RHS is:
    [tex]\int_{-H}^{y} (-a \cdot cos(hy) -b) u' = (-a \cdot cos(hy) -b) u(y) - \int (ah \cdot sin(hy) u(y) [/tex]

    It doesn't look like there's an analytic solution.

    You should try solve it by using numerical methods for solving integral equations.

    I can't help more than this, sorry.
     
  4. Jul 6, 2011 #3
    thx for the input really appreciated
     
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