Using Banach's Fixed Point Theorem, show that the following system has at least one solution: x = 0.000001 x 2 + 10 sin ⁡ y + 1 y = 0.000001 y 3 − 0.01 cos ⁡ x − 1Here is what I have tried:Consider the function defined by f ( x , y ) = ( 0.000001 x 2 + 10 sin ⁡ y + 1 , 0.000001 y 3 − 0.01 cos ⁡ x − 1 )and show that it has a fixed point. If it does, then that fixed point is clearly a solution to our system.I used the standard Euclidean distance between two points and attempted to show that d ( f ( x 1 , y 1 ) , f ( x 2 , y 2 ) ) ≤ d ( ( x 1 , y 1 ) , ( x 2 , y 2 ) )In doing so, I greatly struggle to manipulate the left-hand side so that it is clearly less than or equal to the right-hand side of the inequality above. I tremendously appreciate any advice, and if there is a smarter way to approach this, I appreciate any hints you are willing to offer.

Question

Using Banach&#039;s Fixed Point Theorem, show that the following system has at least one solution:  x  =  0.000001      x    2    +  10  sin  ⁡  y  +  1  y  =  0.000001      y    3    −  0.01  cos  ⁡  x  −  1Here is what I have tried:Consider the function defined by  f  (  x  ,  y  )  =  (  0.000001      x    2    +  10  sin  ⁡  y  +  1  ,  0.000001      y    3    −  0.01  cos  ⁡  x  −  1  )and show that it has a fixed point. If it does, then that fixed point is clearly a solution to our system.I used the standard Euclidean distance between two points and attempted to show that  d  (  f  (      x    1    ,      y    1    )  ,  f  (      x    2    ,      y    2    )  )  ≤  d  (  (      x    1    ,      y    1    )  ,  (      x    2    ,      y    2    )  )In doing so, I greatly struggle to manipulate the left-hand side so that it is clearly less than or equal to the right-hand side of the inequality above. I tremendously appreciate any advice, and if there is a smarter way to approach this, I appreciate any hints you are willing to offer.

Ethen Valentine · Accepted Answer

Idea: the mean value theorem shows that if   f is continuous on   [  a  ,  b  ], differentiable on   (  a  ,  b  ), and       f    ′    (  x  )  ≤  M on   (  a  ,  b  ), then   |  f  (  x  )  −  f  (  y  )  |  ≤  M  |  x  −  y  | on   (  a  ,  b  ). See if you can adapt that to a function of two variables.

Using Banach's Fixed Point Theorem, show that the following system has at least one solution: x

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