How to prove the existence of these numbers?

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I have to prove that there are two numbers $x_1,x_2 in (-1,1)$ that are the unique solution of the system $3x=sin(x+y+13), 3y=cos(x+y)$.

Considering $x_1,x_2 in (frac-13,frac13)$ we can put the system in two 1-variable equations:

$x=frac13sin(fraccos(13)3sqrt1-9x^2+(1+sin(13))x+13)=:f(x)$

$y=frac13cos(fraccos(13)3sqrt1-9y^2+(1+sin(13))y)=:g(y)$

I don't know how to continue from this point because I'd have to prove that exist $L_1, L_2 in [0,1)$ such that $|f'(x)|leq L_1$ and $|g'(y)|leq L_2$ for all $x,yin (frac-13,frac13)$, then conclude that $f$ and $g$ have an unique fixed point, respectively, and those points are the solution of the main system.

My problem is working with the derivative functions to obtain the $L_1, L_2$, I think it's too hard bounding them.

systems-of-equations fixed-point-theorems

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asked Jul 30 at 23:46

Aarón David Arroyo Torres

201110

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up vote
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I have to prove that there are two numbers $x_1,x_2 in (-1,1)$ that are the unique solution of the system $3x=sin(x+y+13), 3y=cos(x+y)$.

Considering $x_1,x_2 in (frac-13,frac13)$ we can put the system in two 1-variable equations:

$x=frac13sin(fraccos(13)3sqrt1-9x^2+(1+sin(13))x+13)=:f(x)$

$y=frac13cos(fraccos(13)3sqrt1-9y^2+(1+sin(13))y)=:g(y)$

I don't know how to continue from this point because I'd have to prove that exist $L_1, L_2 in [0,1)$ such that $|f'(x)|leq L_1$ and $|g'(y)|leq L_2$ for all $x,yin (frac-13,frac13)$, then conclude that $f$ and $g$ have an unique fixed point, respectively, and those points are the solution of the main system.

My problem is working with the derivative functions to obtain the $L_1, L_2$, I think it's too hard bounding them.

systems-of-equations fixed-point-theorems

share|cite|improve this question

asked Jul 30 at 23:46

Aarón David Arroyo Torres

201110

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

I have to prove that there are two numbers $x_1,x_2 in (-1,1)$ that are the unique solution of the system $3x=sin(x+y+13), 3y=cos(x+y)$.

Considering $x_1,x_2 in (frac-13,frac13)$ we can put the system in two 1-variable equations:

$x=frac13sin(fraccos(13)3sqrt1-9x^2+(1+sin(13))x+13)=:f(x)$

$y=frac13cos(fraccos(13)3sqrt1-9y^2+(1+sin(13))y)=:g(y)$

I don't know how to continue from this point because I'd have to prove that exist $L_1, L_2 in [0,1)$ such that $|f'(x)|leq L_1$ and $|g'(y)|leq L_2$ for all $x,yin (frac-13,frac13)$, then conclude that $f$ and $g$ have an unique fixed point, respectively, and those points are the solution of the main system.

My problem is working with the derivative functions to obtain the $L_1, L_2$, I think it's too hard bounding them.

systems-of-equations fixed-point-theorems

share|cite|improve this question

asked Jul 30 at 23:46

Aarón David Arroyo Torres

201110

I have to prove that there are two numbers $x_1,x_2 in (-1,1)$ that are the unique solution of the system $3x=sin(x+y+13), 3y=cos(x+y)$.

Considering $x_1,x_2 in (frac-13,frac13)$ we can put the system in two 1-variable equations:

$x=frac13sin(fraccos(13)3sqrt1-9x^2+(1+sin(13))x+13)=:f(x)$

$y=frac13cos(fraccos(13)3sqrt1-9y^2+(1+sin(13))y)=:g(y)$

I don't know how to continue from this point because I'd have to prove that exist $L_1, L_2 in [0,1)$ such that $|f'(x)|leq L_1$ and $|g'(y)|leq L_2$ for all $x,yin (frac-13,frac13)$, then conclude that $f$ and $g$ have an unique fixed point, respectively, and those points are the solution of the main system.

My problem is working with the derivative functions to obtain the $L_1, L_2$, I think it's too hard bounding them.

systems-of-equations fixed-point-theorems

share|cite|improve this question

asked Jul 30 at 23:46

Aarón David Arroyo Torres

201110

share|cite|improve this question

share|cite|improve this question

asked Jul 30 at 23:46

Aarón David Arroyo Torres

201110

asked Jul 30 at 23:46

Aarón David Arroyo Torres

201110

asked Jul 30 at 23:46

Aarón David Arroyo Torres

201110

201110

add a comment | 

add a comment | 

1 Answer
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accepted

Let $f(x,y)=left(frac13sin(x+y+13),frac13cos(x+y)right)$ for $(x,y)in [-1,1]times[-1,1]=A$. Then $f$ maps $A$ to $A$, and for $x_1,x_2,y_1,y_2in[-1,1]$, beginalign*|f(x_1,y_1)-f(x_2,y_2)|^2&=frac19left(sin(x_1+y_1+13)-sin(x_2+y_2+13)right)^2\
&qquad+frac19left(cos(x_1+y_1)-cos(x_2+y_2)right)^2\
&leq frac19|x_1+y_1-x_2-y_2|^2+frac19|x_1+y_1-x_2-y_2|^2\
&leqfrac49left(|x_1-x_2|^2+|y_1-y_2|^2right),endalign* since $(a+b)^2leq 2a^2+2b^2$ for all $a,b$. Therefore, for any $P,Qin A$, $$|f(P)-f(Q)|leqfrac23|P-Q|,$$ hence $f$ is a contraction in $A$. Therefore $f$ has a unique fixed point in $A$.

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edited Jul 31 at 21:52

answered Jul 31 at 0:19

detnvvp

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1 Answer
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active

oldest

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1 Answer
1

active

oldest

votes

active

oldest

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active

oldest

votes

up vote
1
down vote



accepted

Let $f(x,y)=left(frac13sin(x+y+13),frac13cos(x+y)right)$ for $(x,y)in [-1,1]times[-1,1]=A$. Then $f$ maps $A$ to $A$, and for $x_1,x_2,y_1,y_2in[-1,1]$, beginalign*|f(x_1,y_1)-f(x_2,y_2)|^2&=frac19left(sin(x_1+y_1+13)-sin(x_2+y_2+13)right)^2\
&qquad+frac19left(cos(x_1+y_1)-cos(x_2+y_2)right)^2\
&leq frac19|x_1+y_1-x_2-y_2|^2+frac19|x_1+y_1-x_2-y_2|^2\
&leqfrac49left(|x_1-x_2|^2+|y_1-y_2|^2right),endalign* since $(a+b)^2leq 2a^2+2b^2$ for all $a,b$. Therefore, for any $P,Qin A$, $$|f(P)-f(Q)|leqfrac23|P-Q|,$$ hence $f$ is a contraction in $A$. Therefore $f$ has a unique fixed point in $A$.

share|cite|improve this answer

edited Jul 31 at 21:52

answered Jul 31 at 0:19

detnvvp

6,6291018

add a comment | 

up vote
1
down vote



accepted

Let $f(x,y)=left(frac13sin(x+y+13),frac13cos(x+y)right)$ for $(x,y)in [-1,1]times[-1,1]=A$. Then $f$ maps $A$ to $A$, and for $x_1,x_2,y_1,y_2in[-1,1]$, beginalign*|f(x_1,y_1)-f(x_2,y_2)|^2&=frac19left(sin(x_1+y_1+13)-sin(x_2+y_2+13)right)^2\
&qquad+frac19left(cos(x_1+y_1)-cos(x_2+y_2)right)^2\
&leq frac19|x_1+y_1-x_2-y_2|^2+frac19|x_1+y_1-x_2-y_2|^2\
&leqfrac49left(|x_1-x_2|^2+|y_1-y_2|^2right),endalign* since $(a+b)^2leq 2a^2+2b^2$ for all $a,b$. Therefore, for any $P,Qin A$, $$|f(P)-f(Q)|leqfrac23|P-Q|,$$ hence $f$ is a contraction in $A$. Therefore $f$ has a unique fixed point in $A$.

share|cite|improve this answer

edited Jul 31 at 21:52

answered Jul 31 at 0:19

detnvvp

6,6291018

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Let $f(x,y)=left(frac13sin(x+y+13),frac13cos(x+y)right)$ for $(x,y)in [-1,1]times[-1,1]=A$. Then $f$ maps $A$ to $A$, and for $x_1,x_2,y_1,y_2in[-1,1]$, beginalign*|f(x_1,y_1)-f(x_2,y_2)|^2&=frac19left(sin(x_1+y_1+13)-sin(x_2+y_2+13)right)^2\
&qquad+frac19left(cos(x_1+y_1)-cos(x_2+y_2)right)^2\
&leq frac19|x_1+y_1-x_2-y_2|^2+frac19|x_1+y_1-x_2-y_2|^2\
&leqfrac49left(|x_1-x_2|^2+|y_1-y_2|^2right),endalign* since $(a+b)^2leq 2a^2+2b^2$ for all $a,b$. Therefore, for any $P,Qin A$, $$|f(P)-f(Q)|leqfrac23|P-Q|,$$ hence $f$ is a contraction in $A$. Therefore $f$ has a unique fixed point in $A$.

share|cite|improve this answer

edited Jul 31 at 21:52

answered Jul 31 at 0:19

detnvvp

6,6291018

Let $f(x,y)=left(frac13sin(x+y+13),frac13cos(x+y)right)$ for $(x,y)in [-1,1]times[-1,1]=A$. Then $f$ maps $A$ to $A$, and for $x_1,x_2,y_1,y_2in[-1,1]$, beginalign*|f(x_1,y_1)-f(x_2,y_2)|^2&=frac19left(sin(x_1+y_1+13)-sin(x_2+y_2+13)right)^2\
&qquad+frac19left(cos(x_1+y_1)-cos(x_2+y_2)right)^2\
&leq frac19|x_1+y_1-x_2-y_2|^2+frac19|x_1+y_1-x_2-y_2|^2\
&leqfrac49left(|x_1-x_2|^2+|y_1-y_2|^2right),endalign* since $(a+b)^2leq 2a^2+2b^2$ for all $a,b$. Therefore, for any $P,Qin A$, $$|f(P)-f(Q)|leqfrac23|P-Q|,$$ hence $f$ is a contraction in $A$. Therefore $f$ has a unique fixed point in $A$.

share|cite|improve this answer

edited Jul 31 at 21:52

answered Jul 31 at 0:19

detnvvp

6,6291018

share|cite|improve this answer

share|cite|improve this answer

edited Jul 31 at 21:52

edited Jul 31 at 21:52

edited Jul 31 at 21:52

answered Jul 31 at 0:19

detnvvp

6,6291018

answered Jul 31 at 0:19

detnvvp

6,6291018

answered Jul 31 at 0:19

detnvvp

6,6291018

6,6291018

add a comment | 

add a comment | 

 

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