April 18th, 2017, 01:13 PM  #1 
Member Joined: Nov 2016 From: Kansas Posts: 48 Thanks: 0  Nonlinear Equations
Show that the (nonlinear) equation system: $x^{2}$ + $y^{2}$ $u^{2}$  v=0 $x^{2}$ + $2y^{2}$ $3u^{2}$  $4v^{2}$=1 can be solved in (u, v) in a neighbourhood of (1/2,0,1/2,0) How to show this and also calculate the derivative of u and v wrt (x,y)? 
April 18th, 2017, 05:42 PM  #2 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,053 Thanks: 85 
2xdx+2ydy=2udu+dv 2xdx+4ydy=6udu+8vdv at (1/2,0,1/2,0) dx=du dx=3du no solution 
April 18th, 2017, 06:12 PM  #3 
Senior Member Joined: Sep 2015 From: CA Posts: 1,238 Thanks: 637 
I don't really know how this sort of problem is done. But if it means anything the determinant of the Jacobian is zero at that point. 
April 19th, 2017, 02:35 AM  #4  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,053 Thanks: 85  Quote:
to find a local solution in a neighborhood of p0, take differential of above equations and solve for dx and dy in terms of du and dv at p0. 2xdx+2ydy=2udu+dv 2xdx+4ydy=6udu+8vdv dx= x$\displaystyle {_u}$du+x$\displaystyle {_v}$dv dy= y$\displaystyle {_u}$du+y$\displaystyle {_v}$dv and read off the partial derivatives. ZMD can do the algebra. EDIT (1/2,0,1/2,0) is a solution of: $x^{2}$ + $y^{2}$ $u^{2}$  v=0 $x^{2}$ + $2y^{2}$ +$3u^{2}$  $4v^{2}$=1 and proceed as above. Last edited by zylo; April 19th, 2017 at 03:11 AM.  
April 19th, 2017, 09:21 AM  #5 
Member Joined: Nov 2016 From: Kansas Posts: 48 Thanks: 0 
Reedit The second equation has all +, no  
April 20th, 2017, 06:30 AM  #6 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,053 Thanks: 85  (1/2,0,1/2,0) is a solution either way and procedure is the same.
Last edited by skipjack; April 20th, 2017 at 10:27 AM. 

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