Proving there exists a unique solution close to a point in a non linear system of equations.
2
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Consider the system of non linear equations $$begin{cases}x^2y^3+x^3y^2+x^5y+1=a \ xy^2-2x^2y^4+3x^3y=bend{cases}$$
How can I prove that for a $(a,b)$ close to $(4,2)$ there is a unique solution $x=f(a,b)$, $y=g(a,b)$ close to $(1,1)$?
Note that when $a=4,b=2$ the system has the solution $x=1,y=1$
multivariable-calculus
asked Dec 15 '18 at 23:17
codingnightcodingnight
677
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add a comment |
2
$begingroup$
Consider the system of non linear equations $$begin{cases}x^2y^3+x^3y^2+x^5y+1=a \ xy^2-2x^2y^4+3x^3y=bend{cases}$$
How can I prove that for a $(a,b)$ close to $(4,2)$ there is a unique solution $x=f(a,b)$, $y=g(a,b)$ close to $(1,1)$?
Note that when $a=4,b=2$ the system has the solution $x=1,y=1$
multivariable-calculus
asked Dec 15 '18 at 23:17
codingnightcodingnight
677
$endgroup$
1
$begingroup$
Inverse function theorem (en.wikipedia.org/wiki/Inverse_function_theorem).
$endgroup$
– Jeff
Dec 15 '18 at 23:22
1
$begingroup$
So if I calculate the Jacobian of the vector-valued function of $mathbb{R^2}$ defined by $F(x,y)=begin{pmatrix} x^2y^3+x^3y^2+x^5y+1-a \ xy^2-2x^2y^4+3x^3y-b end{pmatrix}$ And this Jacobian is non-zero in $x=1,y=1$ that would mean that there is a unique solution $x=f(a,b)$, $y=g(a,b)$ close to $(1,1)$?
$endgroup$
– codingnight
Dec 15 '18 at 23:38
$begingroup$
Yes, not only that, but the mappings $f$ and $g$ are infinitely differentiable!
$endgroup$
– Jeff
Dec 17 '18 at 16:56
add a comment |
2
2
2
1
$begingroup$
Consider the system of non linear equations $$begin{cases}x^2y^3+x^3y^2+x^5y+1=a \ xy^2-2x^2y^4+3x^3y=bend{cases}$$
How can I prove that for a $(a,b)$ close to $(4,2)$ there is a unique solution $x=f(a,b)$, $y=g(a,b)$ close to $(1,1)$?
Note that when $a=4,b=2$ the system has the solution $x=1,y=1$
multivariable-calculus
asked Dec 15 '18 at 23:17
codingnightcodingnight
677
$endgroup$
Consider the system of non linear equations $$begin{cases}x^2y^3+x^3y^2+x^5y+1=a \ xy^2-2x^2y^4+3x^3y=bend{cases}$$
How can I prove that for a $(a,b)$ close to $(4,2)$ there is a unique solution $x=f(a,b)$, $y=g(a,b)$ close to $(1,1)$?
Note that when $a=4,b=2$ the system has the solution $x=1,y=1$
multivariable-calculus
multivariable-calculus
asked Dec 15 '18 at 23:17
codingnightcodingnight
677
asked Dec 15 '18 at 23:17
codingnightcodingnight
677
asked Dec 15 '18 at 23:17
codingnightcodingnight
677
asked Dec 15 '18 at 23:17
codingnightcodingnight
677
asked Dec 15 '18 at 23:17
codingnightcodingnight
677
677
1
$begingroup$
Inverse function theorem (en.wikipedia.org/wiki/Inverse_function_theorem).
$endgroup$
– Jeff
Dec 15 '18 at 23:22
1
$begingroup$
So if I calculate the Jacobian of the vector-valued function of $mathbb{R^2}$ defined by $F(x,y)=begin{pmatrix} x^2y^3+x^3y^2+x^5y+1-a \ xy^2-2x^2y^4+3x^3y-b end{pmatrix}$ And this Jacobian is non-zero in $x=1,y=1$ that would mean that there is a unique solution $x=f(a,b)$, $y=g(a,b)$ close to $(1,1)$?
$endgroup$
– codingnight
Dec 15 '18 at 23:38
$begingroup$
Yes, not only that, but the mappings $f$ and $g$ are infinitely differentiable!
$endgroup$
– Jeff
Dec 17 '18 at 16:56
add a comment |
1
$begingroup$
Inverse function theorem (en.wikipedia.org/wiki/Inverse_function_theorem).
$endgroup$
– Jeff
Dec 15 '18 at 23:22
1
$begingroup$
So if I calculate the Jacobian of the vector-valued function of $mathbb{R^2}$ defined by $F(x,y)=begin{pmatrix} x^2y^3+x^3y^2+x^5y+1-a \ xy^2-2x^2y^4+3x^3y-b end{pmatrix}$ And this Jacobian is non-zero in $x=1,y=1$ that would mean that there is a unique solution $x=f(a,b)$, $y=g(a,b)$ close to $(1,1)$?
$endgroup$
– codingnight
Dec 15 '18 at 23:38
$begingroup$
Yes, not only that, but the mappings $f$ and $g$ are infinitely differentiable!
$endgroup$
– Jeff
Dec 17 '18 at 16:56
1
1
$begingroup$
Inverse function theorem (en.wikipedia.org/wiki/Inverse_function_theorem).
$endgroup$
– Jeff
Dec 15 '18 at 23:22
$begingroup$
Inverse function theorem (en.wikipedia.org/wiki/Inverse_function_theorem).
$endgroup$
– Jeff
Dec 15 '18 at 23:22
1
1
$begingroup$
So if I calculate the Jacobian of the vector-valued function of $mathbb{R^2}$ defined by $F(x,y)=begin{pmatrix} x^2y^3+x^3y^2+x^5y+1-a \ xy^2-2x^2y^4+3x^3y-b end{pmatrix}$ And this Jacobian is non-zero in $x=1,y=1$ that would mean that there is a unique solution $x=f(a,b)$, $y=g(a,b)$ close to $(1,1)$?
$endgroup$
– codingnight
Dec 15 '18 at 23:38
$begingroup$
So if I calculate the Jacobian of the vector-valued function of $mathbb{R^2}$ defined by $F(x,y)=begin{pmatrix} x^2y^3+x^3y^2+x^5y+1-a \ xy^2-2x^2y^4+3x^3y-b end{pmatrix}$ And this Jacobian is non-zero in $x=1,y=1$ that would mean that there is a unique solution $x=f(a,b)$, $y=g(a,b)$ close to $(1,1)$?
$endgroup$
– codingnight
Dec 15 '18 at 23:38
$begingroup$
Yes, not only that, but the mappings $f$ and $g$ are infinitely differentiable!
$endgroup$
– Jeff
Dec 17 '18 at 16:56
$begingroup$
Yes, not only that, but the mappings $f$ and $g$ are infinitely differentiable!
$endgroup$
– Jeff
Dec 17 '18 at 16:56
add a comment |
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1
$begingroup$
Inverse function theorem (en.wikipedia.org/wiki/Inverse_function_theorem).
$endgroup$
– Jeff
Dec 15 '18 at 23:22
1
$begingroup$
So if I calculate the Jacobian of the vector-valued function of $mathbb{R^2}$ defined by $F(x,y)=begin{pmatrix} x^2y^3+x^3y^2+x^5y+1-a \ xy^2-2x^2y^4+3x^3y-b end{pmatrix}$ And this Jacobian is non-zero in $x=1,y=1$ that would mean that there is a unique solution $x=f(a,b)$, $y=g(a,b)$ close to $(1,1)$?
$endgroup$
– codingnight
Dec 15 '18 at 23:38
$begingroup$
Yes, not only that, but the mappings $f$ and $g$ are infinitely differentiable!
$endgroup$
– Jeff
Dec 17 '18 at 16:56