global and unique solution for ODE
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Let $y' = f(x) p(cos y) + g(x) q( sin y) $ and $y(0) =1$, with $f,g$ continuous and $p,q$ polynomials. Then there exists unique solution on all $mathbb{R}$ ?
Its easily solved if i assumed that $f,g$ are bounded, but the question asks about $f,g$ that are not necessarily bounded.
Can any one give a hint or a solution !
differential-equations
edited Nov 26 at 10:29
LutzL
54.6k41953
asked Nov 26 at 9:39
Ahmad
2,4921625
add a comment |
up vote
1
down vote
favorite
Let $y' = f(x) p(cos y) + g(x) q( sin y) $ and $y(0) =1$, with $f,g$ continuous and $p,q$ polynomials. Then there exists unique solution on all $mathbb{R}$ ?
Its easily solved if i assumed that $f,g$ are bounded, but the question asks about $f,g$ that are not necessarily bounded.
Can any one give a hint or a solution !
differential-equations
edited Nov 26 at 10:29
LutzL
54.6k41953
asked Nov 26 at 9:39
Ahmad
2,4921625
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $y' = f(x) p(cos y) + g(x) q( sin y) $ and $y(0) =1$, with $f,g$ continuous and $p,q$ polynomials. Then there exists unique solution on all $mathbb{R}$ ?
Its easily solved if i assumed that $f,g$ are bounded, but the question asks about $f,g$ that are not necessarily bounded.
Can any one give a hint or a solution !
differential-equations
edited Nov 26 at 10:29
LutzL
54.6k41953
asked Nov 26 at 9:39
Ahmad
2,4921625
Let $y' = f(x) p(cos y) + g(x) q( sin y) $ and $y(0) =1$, with $f,g$ continuous and $p,q$ polynomials. Then there exists unique solution on all $mathbb{R}$ ?
Its easily solved if i assumed that $f,g$ are bounded, but the question asks about $f,g$ that are not necessarily bounded.
Can any one give a hint or a solution !
differential-equations
differential-equations
edited Nov 26 at 10:29
LutzL
54.6k41953
asked Nov 26 at 9:39
Ahmad
2,4921625
edited Nov 26 at 10:29
LutzL
54.6k41953
asked Nov 26 at 9:39
Ahmad
2,4921625
edited Nov 26 at 10:29
LutzL
54.6k41953
edited Nov 26 at 10:29
LutzL
54.6k41953
edited Nov 26 at 10:29
LutzL
54.6k41953
54.6k41953
asked Nov 26 at 9:39
Ahmad
2,4921625
asked Nov 26 at 9:39
Ahmad
2,4921625
asked Nov 26 at 9:39
Ahmad
2,4921625
2,4921625
add a comment |
add a comment |
1 Answer
1
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oldest
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1
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accepted
$f$ and $g$ are bounded on every interval $[-R,R]$ for any $R=NinBbb N$. Thus you get unique solutions over these intervals, and by the uniqueness the solution for $R=N+1$ extends the solution for $R=N$. In the limit this gives a unique solution over all of $Bbb R$.
answered Nov 26 at 10:31
LutzL
54.6k41953
Can you tell please, what theorem are you referring to?
– Evgeny
Nov 26 at 11:06
What do you mean, that a global Lipschitz constant on $(t,y)in [a,b]times Bbb R^n$ ensures the existence of a solution over $[a,b]$? This is a variant of the Picard-Lindelöf theorem.
– LutzL
Nov 26 at 11:21
If you remember the reference with a proof for this variant and can tell me, I would be very grateful!
– Evgeny
Nov 26 at 11:37
See math.stackexchange.com/a/2987844/115115 and linked posts. I know this variant from my lectures, I do not know which books one could cite. There is also a variant of this global proof that uses the Lipschitz constants of the iterates of the Picard iteration, which fall faster than simply the power of the Lipschitz constant of one iteration.
– LutzL
Nov 26 at 11:47
Thank you very much!
– Evgeny
Nov 26 at 12:01
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$f$ and $g$ are bounded on every interval $[-R,R]$ for any $R=NinBbb N$. Thus you get unique solutions over these intervals, and by the uniqueness the solution for $R=N+1$ extends the solution for $R=N$. In the limit this gives a unique solution over all of $Bbb R$.
answered Nov 26 at 10:31
LutzL
54.6k41953
Can you tell please, what theorem are you referring to?
– Evgeny
Nov 26 at 11:06
What do you mean, that a global Lipschitz constant on $(t,y)in [a,b]times Bbb R^n$ ensures the existence of a solution over $[a,b]$? This is a variant of the Picard-Lindelöf theorem.
– LutzL
Nov 26 at 11:21
If you remember the reference with a proof for this variant and can tell me, I would be very grateful!
– Evgeny
Nov 26 at 11:37
See math.stackexchange.com/a/2987844/115115 and linked posts. I know this variant from my lectures, I do not know which books one could cite. There is also a variant of this global proof that uses the Lipschitz constants of the iterates of the Picard iteration, which fall faster than simply the power of the Lipschitz constant of one iteration.
– LutzL
Nov 26 at 11:47
Thank you very much!
– Evgeny
Nov 26 at 12:01
add a comment |
up vote
1
down vote
accepted
$f$ and $g$ are bounded on every interval $[-R,R]$ for any $R=NinBbb N$. Thus you get unique solutions over these intervals, and by the uniqueness the solution for $R=N+1$ extends the solution for $R=N$. In the limit this gives a unique solution over all of $Bbb R$.
answered Nov 26 at 10:31
LutzL
54.6k41953
Can you tell please, what theorem are you referring to?
– Evgeny
Nov 26 at 11:06
What do you mean, that a global Lipschitz constant on $(t,y)in [a,b]times Bbb R^n$ ensures the existence of a solution over $[a,b]$? This is a variant of the Picard-Lindelöf theorem.
– LutzL
Nov 26 at 11:21
If you remember the reference with a proof for this variant and can tell me, I would be very grateful!
– Evgeny
Nov 26 at 11:37
See math.stackexchange.com/a/2987844/115115 and linked posts. I know this variant from my lectures, I do not know which books one could cite. There is also a variant of this global proof that uses the Lipschitz constants of the iterates of the Picard iteration, which fall faster than simply the power of the Lipschitz constant of one iteration.
– LutzL
Nov 26 at 11:47
Thank you very much!
– Evgeny
Nov 26 at 12:01
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$f$ and $g$ are bounded on every interval $[-R,R]$ for any $R=NinBbb N$. Thus you get unique solutions over these intervals, and by the uniqueness the solution for $R=N+1$ extends the solution for $R=N$. In the limit this gives a unique solution over all of $Bbb R$.
answered Nov 26 at 10:31
LutzL
54.6k41953
$f$ and $g$ are bounded on every interval $[-R,R]$ for any $R=NinBbb N$. Thus you get unique solutions over these intervals, and by the uniqueness the solution for $R=N+1$ extends the solution for $R=N$. In the limit this gives a unique solution over all of $Bbb R$.
answered Nov 26 at 10:31
LutzL
54.6k41953
answered Nov 26 at 10:31
LutzL
54.6k41953
answered Nov 26 at 10:31
LutzL
54.6k41953
answered Nov 26 at 10:31
LutzL
54.6k41953
54.6k41953
Can you tell please, what theorem are you referring to?
– Evgeny
Nov 26 at 11:06
What do you mean, that a global Lipschitz constant on $(t,y)in [a,b]times Bbb R^n$ ensures the existence of a solution over $[a,b]$? This is a variant of the Picard-Lindelöf theorem.
– LutzL
Nov 26 at 11:21
If you remember the reference with a proof for this variant and can tell me, I would be very grateful!
– Evgeny
Nov 26 at 11:37
See math.stackexchange.com/a/2987844/115115 and linked posts. I know this variant from my lectures, I do not know which books one could cite. There is also a variant of this global proof that uses the Lipschitz constants of the iterates of the Picard iteration, which fall faster than simply the power of the Lipschitz constant of one iteration.
– LutzL
Nov 26 at 11:47
Thank you very much!
– Evgeny
Nov 26 at 12:01
add a comment |
Can you tell please, what theorem are you referring to?
– Evgeny
Nov 26 at 11:06
What do you mean, that a global Lipschitz constant on $(t,y)in [a,b]times Bbb R^n$ ensures the existence of a solution over $[a,b]$? This is a variant of the Picard-Lindelöf theorem.
– LutzL
Nov 26 at 11:21
If you remember the reference with a proof for this variant and can tell me, I would be very grateful!
– Evgeny
Nov 26 at 11:37
See math.stackexchange.com/a/2987844/115115 and linked posts. I know this variant from my lectures, I do not know which books one could cite. There is also a variant of this global proof that uses the Lipschitz constants of the iterates of the Picard iteration, which fall faster than simply the power of the Lipschitz constant of one iteration.
– LutzL
Nov 26 at 11:47
Thank you very much!
– Evgeny
Nov 26 at 12:01
Can you tell please, what theorem are you referring to?
– Evgeny
Nov 26 at 11:06
Can you tell please, what theorem are you referring to?
– Evgeny
Nov 26 at 11:06
What do you mean, that a global Lipschitz constant on $(t,y)in [a,b]times Bbb R^n$ ensures the existence of a solution over $[a,b]$? This is a variant of the Picard-Lindelöf theorem.
– LutzL
Nov 26 at 11:21
What do you mean, that a global Lipschitz constant on $(t,y)in [a,b]times Bbb R^n$ ensures the existence of a solution over $[a,b]$? This is a variant of the Picard-Lindelöf theorem.
– LutzL
Nov 26 at 11:21
If you remember the reference with a proof for this variant and can tell me, I would be very grateful!
– Evgeny
Nov 26 at 11:37
If you remember the reference with a proof for this variant and can tell me, I would be very grateful!
– Evgeny
Nov 26 at 11:37
See math.stackexchange.com/a/2987844/115115 and linked posts. I know this variant from my lectures, I do not know which books one could cite. There is also a variant of this global proof that uses the Lipschitz constants of the iterates of the Picard iteration, which fall faster than simply the power of the Lipschitz constant of one iteration.
– LutzL
Nov 26 at 11:47
See math.stackexchange.com/a/2987844/115115 and linked posts. I know this variant from my lectures, I do not know which books one could cite. There is also a variant of this global proof that uses the Lipschitz constants of the iterates of the Picard iteration, which fall faster than simply the power of the Lipschitz constant of one iteration.
– LutzL
Nov 26 at 11:47
Thank you very much!
– Evgeny
Nov 26 at 12:01
Thank you very much!
– Evgeny
Nov 26 at 12:01
add a comment |
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