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April 18th, 2017, 01:13 PM   #1
ZMD
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Non-linear Equations

Show that the (non-linear) 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)?
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April 18th, 2017, 05:42 PM   #2
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2xdx+2ydy=2udu+dv
2xdx+4ydy=6udu+8vdv

at (1/2,0,1/2,0)

dx=du
dx=3du

no solution
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April 18th, 2017, 06:12 PM   #3
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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.
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April 19th, 2017, 02:35 AM   #4
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Quote:
Originally Posted by ZMD View Post
Show that the (non-linear) 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)?
(1/2,0,1/2,0) is not a solution of above equations. If p0 is,

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.
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April 19th, 2017, 09:21 AM   #5
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Re-edit
The second equation has all +, no -
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April 20th, 2017, 06:30 AM   #6
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Quote:
Originally Posted by ZMD View Post
Re-edit
The second equation has all +, no -
(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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