Solution to nonlinear ODE using Contraction Mapping Theorem.












2














I need to use the contraction mapping theorem to prove that a solution exists to the IVP $y' = sin(x)sin(y),; y(1)=2$, on the interval $;[1,1+h],; h>0$.



My attempt at the solution. Let $E = C^1([1,1+h])$ ($h$ remains to be chosen). Let $;d;$ be the usual function space metric. Then $(E,d)$ forms a complete metric space.

Define the map $T:Eto E$ by the rule $(Ty)(x)=2+intlimits_1^x sin(s)sin(y(s))ds$.

Let $y_1,y_2in E$.



WTS: $d(Ty_1,Ty_2)leq rd(y_1,y_2)$, with $0<r<1$.



By plugging in the expression for $;d(Ty_1,Ty_2),;$ I was able to perform a few algebraic manipulations to get to



$$d(Ty_1,Ty_2) leq Mintlimits_1^{1+h}|sin(y_1(s)-sin(y_2(s))|,$$
where $M$ came from the fact that $sin(x)$ is bounded. From here I got the expression:



$$d(Ty_1,Ty_2)leq Mhleft(sup_{xepsilon[1,1+h]}|sin(y_1(s)-sin(y_2(s))|right).$$



Here is where I got stuck. Usually, in the linear ODE and systems case, the absolute value is just $d(y_1,y_2)$, and then I can choose $h<frac{1}{M}$ and be done, but here I have the functions composed inside of the sine? How do I proceed from here?






real-analysis







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  • Please consider a better formatting for your question. Actually it is kinda hard to read.
    – nicomezi
    Nov 29 at 10:10






  • 1




    I rewrote the post to be more readable
    – Anonymous Snake
    Nov 29 at 10:23










  • Thank you for your edit.
    – nicomezi
    Nov 29 at 10:24
















2














I need to use the contraction mapping theorem to prove that a solution exists to the IVP $y' = sin(x)sin(y),; y(1)=2$, on the interval $;[1,1+h],; h>0$.



My attempt at the solution. Let $E = C^1([1,1+h])$ ($h$ remains to be chosen). Let $;d;$ be the usual function space metric. Then $(E,d)$ forms a complete metric space.

Define the map $T:Eto E$ by the rule $(Ty)(x)=2+intlimits_1^x sin(s)sin(y(s))ds$.

Let $y_1,y_2in E$.



WTS: $d(Ty_1,Ty_2)leq rd(y_1,y_2)$, with $0<r<1$.



By plugging in the expression for $;d(Ty_1,Ty_2),;$ I was able to perform a few algebraic manipulations to get to



$$d(Ty_1,Ty_2) leq Mintlimits_1^{1+h}|sin(y_1(s)-sin(y_2(s))|,$$
where $M$ came from the fact that $sin(x)$ is bounded. From here I got the expression:



$$d(Ty_1,Ty_2)leq Mhleft(sup_{xepsilon[1,1+h]}|sin(y_1(s)-sin(y_2(s))|right).$$



Here is where I got stuck. Usually, in the linear ODE and systems case, the absolute value is just $d(y_1,y_2)$, and then I can choose $h<frac{1}{M}$ and be done, but here I have the functions composed inside of the sine? How do I proceed from here?






real-analysis







share|cite|improve this question
























  • Please consider a better formatting for your question. Actually it is kinda hard to read.
    – nicomezi
    Nov 29 at 10:10






  • 1




    I rewrote the post to be more readable
    – Anonymous Snake
    Nov 29 at 10:23










  • Thank you for your edit.
    – nicomezi
    Nov 29 at 10:24














2












2








2


1





I need to use the contraction mapping theorem to prove that a solution exists to the IVP $y' = sin(x)sin(y),; y(1)=2$, on the interval $;[1,1+h],; h>0$.



My attempt at the solution. Let $E = C^1([1,1+h])$ ($h$ remains to be chosen). Let $;d;$ be the usual function space metric. Then $(E,d)$ forms a complete metric space.

Define the map $T:Eto E$ by the rule $(Ty)(x)=2+intlimits_1^x sin(s)sin(y(s))ds$.

Let $y_1,y_2in E$.



WTS: $d(Ty_1,Ty_2)leq rd(y_1,y_2)$, with $0<r<1$.



By plugging in the expression for $;d(Ty_1,Ty_2),;$ I was able to perform a few algebraic manipulations to get to



$$d(Ty_1,Ty_2) leq Mintlimits_1^{1+h}|sin(y_1(s)-sin(y_2(s))|,$$
where $M$ came from the fact that $sin(x)$ is bounded. From here I got the expression:



$$d(Ty_1,Ty_2)leq Mhleft(sup_{xepsilon[1,1+h]}|sin(y_1(s)-sin(y_2(s))|right).$$



Here is where I got stuck. Usually, in the linear ODE and systems case, the absolute value is just $d(y_1,y_2)$, and then I can choose $h<frac{1}{M}$ and be done, but here I have the functions composed inside of the sine? How do I proceed from here?






real-analysis







share|cite|improve this question















I need to use the contraction mapping theorem to prove that a solution exists to the IVP $y' = sin(x)sin(y),; y(1)=2$, on the interval $;[1,1+h],; h>0$.



My attempt at the solution. Let $E = C^1([1,1+h])$ ($h$ remains to be chosen). Let $;d;$ be the usual function space metric. Then $(E,d)$ forms a complete metric space.

Define the map $T:Eto E$ by the rule $(Ty)(x)=2+intlimits_1^x sin(s)sin(y(s))ds$.

Let $y_1,y_2in E$.



WTS: $d(Ty_1,Ty_2)leq rd(y_1,y_2)$, with $0<r<1$.



By plugging in the expression for $;d(Ty_1,Ty_2),;$ I was able to perform a few algebraic manipulations to get to



$$d(Ty_1,Ty_2) leq Mintlimits_1^{1+h}|sin(y_1(s)-sin(y_2(s))|,$$
where $M$ came from the fact that $sin(x)$ is bounded. From here I got the expression:



$$d(Ty_1,Ty_2)leq Mhleft(sup_{xepsilon[1,1+h]}|sin(y_1(s)-sin(y_2(s))|right).$$



Here is where I got stuck. Usually, in the linear ODE and systems case, the absolute value is just $d(y_1,y_2)$, and then I can choose $h<frac{1}{M}$ and be done, but here I have the functions composed inside of the sine? How do I proceed from here?






real-analysis




real-analysis






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edited Nov 29 at 17:29









user376343

2,7782822




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asked Nov 29 at 10:05









Anonymous Snake

162




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  • Please consider a better formatting for your question. Actually it is kinda hard to read.
    – nicomezi
    Nov 29 at 10:10






  • 1




    I rewrote the post to be more readable
    – Anonymous Snake
    Nov 29 at 10:23










  • Thank you for your edit.
    – nicomezi
    Nov 29 at 10:24


















  • Please consider a better formatting for your question. Actually it is kinda hard to read.
    – nicomezi
    Nov 29 at 10:10






  • 1




    I rewrote the post to be more readable
    – Anonymous Snake
    Nov 29 at 10:23










  • Thank you for your edit.
    – nicomezi
    Nov 29 at 10:24
















Please consider a better formatting for your question. Actually it is kinda hard to read.
– nicomezi
Nov 29 at 10:10




Please consider a better formatting for your question. Actually it is kinda hard to read.
– nicomezi
Nov 29 at 10:10




1




1




I rewrote the post to be more readable
– Anonymous Snake
Nov 29 at 10:23




I rewrote the post to be more readable
– Anonymous Snake
Nov 29 at 10:23












Thank you for your edit.
– nicomezi
Nov 29 at 10:24




Thank you for your edit.
– nicomezi
Nov 29 at 10:24










1 Answer
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For $a,b in mathbb R$ we have, by the mean value theorem:



$| sin(b)-sin(a)| le |b-a|$.



Thus $| sin(y_1(s))-sin(y_2(s))| le |y_1(s)-y_2(s)| le d(y_1,y_2).$






share|cite|improve this answer





















  • Thank you so much!
    – Anonymous Snake
    Nov 29 at 10:23











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For $a,b in mathbb R$ we have, by the mean value theorem:



$| sin(b)-sin(a)| le |b-a|$.



Thus $| sin(y_1(s))-sin(y_2(s))| le |y_1(s)-y_2(s)| le d(y_1,y_2).$






share|cite|improve this answer





















  • Thank you so much!
    – Anonymous Snake
    Nov 29 at 10:23
















1














For $a,b in mathbb R$ we have, by the mean value theorem:



$| sin(b)-sin(a)| le |b-a|$.



Thus $| sin(y_1(s))-sin(y_2(s))| le |y_1(s)-y_2(s)| le d(y_1,y_2).$






share|cite|improve this answer





















  • Thank you so much!
    – Anonymous Snake
    Nov 29 at 10:23














1












1








1






For $a,b in mathbb R$ we have, by the mean value theorem:



$| sin(b)-sin(a)| le |b-a|$.



Thus $| sin(y_1(s))-sin(y_2(s))| le |y_1(s)-y_2(s)| le d(y_1,y_2).$






share|cite|improve this answer












For $a,b in mathbb R$ we have, by the mean value theorem:



$| sin(b)-sin(a)| le |b-a|$.



Thus $| sin(y_1(s))-sin(y_2(s))| le |y_1(s)-y_2(s)| le d(y_1,y_2).$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 29 at 10:13









Fred

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  • Thank you so much!
    – Anonymous Snake
    Nov 29 at 10:23


















  • Thank you so much!
    – Anonymous Snake
    Nov 29 at 10:23
















Thank you so much!
– Anonymous Snake
Nov 29 at 10:23




Thank you so much!
– Anonymous Snake
Nov 29 at 10:23


















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