How can I solve a system of BVPs with only boundary conditions given?

[amr07]

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

 

[HallsofIvy]

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

Of course, the third and fourth equations become
u'= -\fra{1}{r}sin(\theta)cos(\theta)u- \left[1- cos^2(\theta)\right]\frac{v}{r}
and
v'= \frac{1}{r}sin(\theta)cos(\theta)u+ \left[1+ sin^2(\theta)\right]\frac{u}{r}

 

[amr07]

thanks

 

[amr07]

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

Of course, the third and fourth equations become
u'= -\fra{1}{r}sin(\theta)cos(\theta)u- \left[1- cos^2(\theta)\right]\frac{v}{r}
and
v'= \frac{1}{r}sin(\theta)cos(\theta)u+ \left[1+ sin^2(\theta)\right]\frac{u}{r}

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