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 yesterday
Joe Man Analysis
23319
add a comment |
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 yesterday
Joe Man Analysis
23319
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
yesterday
@LutzL Are you saying that the biggest interval would be the whole real line?
– Joe Man Analysis
yesterday
Yes, exactly that. You also got a global Lipschitz constant $1$, which is in itself an argument for global existence.
– LutzL
yesterday
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 yesterday
Joe Man Analysis
23319
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 yesterday
Joe Man Analysis
23319
asked yesterday
Joe Man Analysis
23319
asked yesterday
Joe Man Analysis
23319
asked yesterday
Joe Man Analysis
23319
asked yesterday
Joe Man Analysis
23319
23319
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
yesterday
@LutzL Are you saying that the biggest interval would be the whole real line?
– Joe Man Analysis
yesterday
Yes, exactly that. You also got a global Lipschitz constant $1$, which is in itself an argument for global existence.
– LutzL
yesterday
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
yesterday
@LutzL Are you saying that the biggest interval would be the whole real line?
– Joe Man Analysis
yesterday
Yes, exactly that. You also got a global Lipschitz constant $1$, which is in itself an argument for global existence.
– LutzL
yesterday
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
yesterday
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
yesterday
@LutzL Are you saying that the biggest interval would be the whole real line?
– Joe Man Analysis
yesterday
@LutzL Are you saying that the biggest interval would be the whole real line?
– Joe Man Analysis
yesterday
Yes, exactly that. You also got a global Lipschitz constant $1$, which is in itself an argument for global existence.
– LutzL
yesterday
Yes, exactly that. You also got a global Lipschitz constant $1$, which is in itself an argument for global existence.
– LutzL
yesterday
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
yesterday
@LutzL Are you saying that the biggest interval would be the whole real line?
– Joe Man Analysis
yesterday
Yes, exactly that. You also got a global Lipschitz constant $1$, which is in itself an argument for global existence.
– LutzL
yesterday