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