Using Picard-Lindelof to find a solution to $y'(t,y(t))=t+sin(y(t))$ where $y(2)=1.$
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Consider the initial value problem $y'(t,y(t))=t+sin(y(t))$ with initial condition $y(2)=1$. Find the largest interval $mathcal{I}subset mathbb{R}$ containing $t_0=2$ such that the problem has a unique solutions $y$ in $mathcal{I}$.
I've been trying to apply Picard-Lindelof theorem to this problem and I got that $M=underset{R}{sup}y'(t,y(t))=a+2+sin(b+1)$. Where $R={(t,y):|t-2|leq a, |y-1|leq b }$.
I know that the interval is $mathcal{I}=[2-varepsilon, 2 + varepsilon]$ where $varepsilon = min(a, b/M)$. I also know that in order for $mathcal{I}$ to be the largest possible, $a$ must equal $b/M$.
As $M$ depends on $a$, you cannot make $a$ to big without $b/M$ getting too small. This looks like a maximization problem here. I ended up getting a very nasty value for $varepsilon$. So I believe I am wrong.
Anyone have any hints?
real-analysis functional-analysis differential-equations approximation
asked Nov 14 at 5:47
Joe Man Analysis
25019
add a comment |
up vote
2
down vote
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Consider the initial value problem $y'(t,y(t))=t+sin(y(t))$ with initial condition $y(2)=1$. Find the largest interval $mathcal{I}subset mathbb{R}$ containing $t_0=2$ such that the problem has a unique solutions $y$ in $mathcal{I}$.
I've been trying to apply Picard-Lindelof theorem to this problem and I got that $M=underset{R}{sup}y'(t,y(t))=a+2+sin(b+1)$. Where $R={(t,y):|t-2|leq a, |y-1|leq b }$.
I know that the interval is $mathcal{I}=[2-varepsilon, 2 + varepsilon]$ where $varepsilon = min(a, b/M)$. I also know that in order for $mathcal{I}$ to be the largest possible, $a$ must equal $b/M$.
As $M$ depends on $a$, you cannot make $a$ to big without $b/M$ getting too small. This looks like a maximization problem here. I ended up getting a very nasty value for $varepsilon$. So I believe I am wrong.
Anyone have any hints?
real-analysis functional-analysis differential-equations approximation
asked Nov 14 at 5:47
Joe Man Analysis
25019
You got $y$, where it exists, satisfying $|y'|le 3+|t-2|$ so that $|y(t)-y(2)|le3|t-2|+frac12|t-2|^2$. Which means that there are no singularities at finite times.
– LutzL
Nov 14 at 8:42
@LutzL Are you saying that the biggest interval would be the whole real line?
– Joe Man Analysis
Nov 14 at 17:13
Yes, exactly that. You also got a global Lipschitz constant $1$, which is in itself an argument for global existence.
– LutzL
Nov 14 at 17:26
@LutzL So I attempted my own proof of it over here. math.stackexchange.com/questions/3004533/… Do you mind having a look and telling me if it is correct?
– Joe Man Analysis
Nov 19 at 6:51
1
Following the lines of math.stackexchange.com/a/2987844/115115 you get a modified norm $$|y|_w=sup_{tinBbb R} e^{-2|t-2|}|y(t)|,$$ and relative to this norm the Picard iteration is a contraction on $C(Bbb R)$, thus has a unique fixed point which is also a solution of the ODE by the Banach fixed-point theorem.
– LutzL
Nov 20 at 19:09
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Consider the initial value problem $y'(t,y(t))=t+sin(y(t))$ with initial condition $y(2)=1$. Find the largest interval $mathcal{I}subset mathbb{R}$ containing $t_0=2$ such that the problem has a unique solutions $y$ in $mathcal{I}$.
I've been trying to apply Picard-Lindelof theorem to this problem and I got that $M=underset{R}{sup}y'(t,y(t))=a+2+sin(b+1)$. Where $R={(t,y):|t-2|leq a, |y-1|leq b }$.
I know that the interval is $mathcal{I}=[2-varepsilon, 2 + varepsilon]$ where $varepsilon = min(a, b/M)$. I also know that in order for $mathcal{I}$ to be the largest possible, $a$ must equal $b/M$.
As $M$ depends on $a$, you cannot make $a$ to big without $b/M$ getting too small. This looks like a maximization problem here. I ended up getting a very nasty value for $varepsilon$. So I believe I am wrong.
Anyone have any hints?
real-analysis functional-analysis differential-equations approximation
asked Nov 14 at 5:47
Joe Man Analysis
25019
Consider the initial value problem $y'(t,y(t))=t+sin(y(t))$ with initial condition $y(2)=1$. Find the largest interval $mathcal{I}subset mathbb{R}$ containing $t_0=2$ such that the problem has a unique solutions $y$ in $mathcal{I}$.
I've been trying to apply Picard-Lindelof theorem to this problem and I got that $M=underset{R}{sup}y'(t,y(t))=a+2+sin(b+1)$. Where $R={(t,y):|t-2|leq a, |y-1|leq b }$.
I know that the interval is $mathcal{I}=[2-varepsilon, 2 + varepsilon]$ where $varepsilon = min(a, b/M)$. I also know that in order for $mathcal{I}$ to be the largest possible, $a$ must equal $b/M$.
As $M$ depends on $a$, you cannot make $a$ to big without $b/M$ getting too small. This looks like a maximization problem here. I ended up getting a very nasty value for $varepsilon$. So I believe I am wrong.
Anyone have any hints?
real-analysis functional-analysis differential-equations approximation
real-analysis functional-analysis differential-equations approximation
asked Nov 14 at 5:47
Joe Man Analysis
25019
asked Nov 14 at 5:47
Joe Man Analysis
25019
asked Nov 14 at 5:47
Joe Man Analysis
25019
asked Nov 14 at 5:47
Joe Man Analysis
25019
asked Nov 14 at 5:47
Joe Man Analysis
25019
25019
You got $y$, where it exists, satisfying $|y'|le 3+|t-2|$ so that $|y(t)-y(2)|le3|t-2|+frac12|t-2|^2$. Which means that there are no singularities at finite times.
– LutzL
Nov 14 at 8:42
@LutzL Are you saying that the biggest interval would be the whole real line?
– Joe Man Analysis
Nov 14 at 17:13
Yes, exactly that. You also got a global Lipschitz constant $1$, which is in itself an argument for global existence.
– LutzL
Nov 14 at 17:26
@LutzL So I attempted my own proof of it over here. math.stackexchange.com/questions/3004533/… Do you mind having a look and telling me if it is correct?
– Joe Man Analysis
Nov 19 at 6:51
1
Following the lines of math.stackexchange.com/a/2987844/115115 you get a modified norm $$|y|_w=sup_{tinBbb R} e^{-2|t-2|}|y(t)|,$$ and relative to this norm the Picard iteration is a contraction on $C(Bbb R)$, thus has a unique fixed point which is also a solution of the ODE by the Banach fixed-point theorem.
– LutzL
Nov 20 at 19:09
add a comment |
You got $y$, where it exists, satisfying $|y'|le 3+|t-2|$ so that $|y(t)-y(2)|le3|t-2|+frac12|t-2|^2$. Which means that there are no singularities at finite times.
– LutzL
Nov 14 at 8:42
@LutzL Are you saying that the biggest interval would be the whole real line?
– Joe Man Analysis
Nov 14 at 17:13
Yes, exactly that. You also got a global Lipschitz constant $1$, which is in itself an argument for global existence.
– LutzL
Nov 14 at 17:26
@LutzL So I attempted my own proof of it over here. math.stackexchange.com/questions/3004533/… Do you mind having a look and telling me if it is correct?
– Joe Man Analysis
Nov 19 at 6:51
1
Following the lines of math.stackexchange.com/a/2987844/115115 you get a modified norm $$|y|_w=sup_{tinBbb R} e^{-2|t-2|}|y(t)|,$$ and relative to this norm the Picard iteration is a contraction on $C(Bbb R)$, thus has a unique fixed point which is also a solution of the ODE by the Banach fixed-point theorem.
– LutzL
Nov 20 at 19:09
You got $y$, where it exists, satisfying $|y'|le 3+|t-2|$ so that $|y(t)-y(2)|le3|t-2|+frac12|t-2|^2$. Which means that there are no singularities at finite times.
– LutzL
Nov 14 at 8:42
You got $y$, where it exists, satisfying $|y'|le 3+|t-2|$ so that $|y(t)-y(2)|le3|t-2|+frac12|t-2|^2$. Which means that there are no singularities at finite times.
– LutzL
Nov 14 at 8:42
@LutzL Are you saying that the biggest interval would be the whole real line?
– Joe Man Analysis
Nov 14 at 17:13
@LutzL Are you saying that the biggest interval would be the whole real line?
– Joe Man Analysis
Nov 14 at 17:13
Yes, exactly that. You also got a global Lipschitz constant $1$, which is in itself an argument for global existence.
– LutzL
Nov 14 at 17:26
Yes, exactly that. You also got a global Lipschitz constant $1$, which is in itself an argument for global existence.
– LutzL
Nov 14 at 17:26
@LutzL So I attempted my own proof of it over here. math.stackexchange.com/questions/3004533/… Do you mind having a look and telling me if it is correct?
– Joe Man Analysis
Nov 19 at 6:51
@LutzL So I attempted my own proof of it over here. math.stackexchange.com/questions/3004533/… Do you mind having a look and telling me if it is correct?
– Joe Man Analysis
Nov 19 at 6:51
1
1
Following the lines of math.stackexchange.com/a/2987844/115115 you get a modified norm $$|y|_w=sup_{tinBbb R} e^{-2|t-2|}|y(t)|,$$ and relative to this norm the Picard iteration is a contraction on $C(Bbb R)$, thus has a unique fixed point which is also a solution of the ODE by the Banach fixed-point theorem.
– LutzL
Nov 20 at 19:09
Following the lines of math.stackexchange.com/a/2987844/115115 you get a modified norm $$|y|_w=sup_{tinBbb R} e^{-2|t-2|}|y(t)|,$$ and relative to this norm the Picard iteration is a contraction on $C(Bbb R)$, thus has a unique fixed point which is also a solution of the ODE by the Banach fixed-point theorem.
– LutzL
Nov 20 at 19:09
add a comment |
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You got $y$, where it exists, satisfying $|y'|le 3+|t-2|$ so that $|y(t)-y(2)|le3|t-2|+frac12|t-2|^2$. Which means that there are no singularities at finite times.
– LutzL
Nov 14 at 8:42
@LutzL Are you saying that the biggest interval would be the whole real line?
– Joe Man Analysis
Nov 14 at 17:13
Yes, exactly that. You also got a global Lipschitz constant $1$, which is in itself an argument for global existence.
– LutzL
Nov 14 at 17:26
@LutzL So I attempted my own proof of it over here. math.stackexchange.com/questions/3004533/… Do you mind having a look and telling me if it is correct?
– Joe Man Analysis
Nov 19 at 6:51
1
Following the lines of math.stackexchange.com/a/2987844/115115 you get a modified norm $$|y|_w=sup_{tinBbb R} e^{-2|t-2|}|y(t)|,$$ and relative to this norm the Picard iteration is a contraction on $C(Bbb R)$, thus has a unique fixed point which is also a solution of the ODE by the Banach fixed-point theorem.
– LutzL
Nov 20 at 19:09