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Boundary value problem

  1. Dec 2, 2009 #1
    Dear all,

    I have system(4 equations) of BVPs. Could anybody recommend me, how to solve this system(whatever numericaly or analytical):

    x'=-y/sqrt(x^2+y^2) + u
    y'=x/sqrt(x^2+y^2) + v
    u' = -xy/(x^2+y^2)^3/2 u - [1/sqrt(x^2+y^2) - x^2 /(x^2+y^2)^3/2] v
    v' = xy/(x^2+y^2)^3/2 v - [-1/sqrt(x^2+y^2) + y^2 /(x^2+y^2)^3/2] u

    and we have only boundary condations for x and y....x(0)=y(0)=-1, x(pi/2)=y(pi/2)=1

    thanks
     
    Last edited: Dec 3, 2009
  2. jcsd
  3. Dec 3, 2009 #2

    HallsofIvy

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    With all those "[itex]x^2+ y^2[/itex]" in there, the first thing I would try is to convert x and y to polar coordinates: [itex]x= r cos(\theta)[/itex] so that [itex]x^2+ y^2= r^2[/itex], [itex]x'= r' cos(\theta)- r sin(\theta)\theta'[/itex], and [itex]y= r sin(\theta)[/itex] so that [itex]y'= r' sin(\theta)+ r cos(\theta)\theta'[/itex].
    Your first equation becomes [itex]r' cos(\theta)- r sin(\theta)\theta'= - sin(\theta)+ u[/itex] and your second equation [itex]r' sin(\theta)+ r cos(\theta)\theta'= cos(\theta)+ v[/itex]. If you multiply the first equation by [itex]cos(\theta)[/itex] and the second by [itex]sin(\theta)[/itex] and add you get [itex]r'= u cos(\theta)+ v\sin(\theta)[/itex]. If you multiply the first equation by [itex]sin(\theta)[/itex], the second by [itex]cos(\theta)[/itex] and subtract first from second you get [itex]r \theta'= 1+ v cos(\theta)- u sin(\theta)[/itex].

    Of course, the third and fourth equations become
    [tex]u'= -\fra{1}{r}sin(\theta)cos(\theta)u- \left[1- cos^2(\theta)\right]\frac{v}{r}[/tex]
    and
    [tex]v'= \frac{1}{r}sin(\theta)cos(\theta)u+ \left[1+ sin^2(\theta)\right]\frac{u}{r}[/tex]
     
  4. Dec 3, 2009 #3
    thanks
     
  5. Dec 3, 2009 #4

    Dear HallsofIvy,

    many thanks for your helping. the second term in the fourth equation(for v') should be [1-sin^2]. so we have no also four differential equations. it seems easily but how i can get the solution?

    thanks again
     
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