Picard theorem for $u' = sqrt{lvert u^2 -1 rvert}$ if we know $ u(pi / 2)= 0$
Problem: can we apply Picard theorem for $$u' = sqrt{lvert u^2 -1 rvert}$$ if $$
u(pi / 2)= 0$$
[$u$ is a function of a variable $x$ so $u = u(x)$]
My attempt:
Well, what I need to know is whether the function $f(x,u) = sqrt{lvert u^2 -1 rvert} $ is Lipshitz continuous when observing only the second variable: u, on some rectangle $(pi/2 - delta,pi/2 + delta)times(1- epsilon,1 + epsilon)$.
I'm however uncertaion whether this holds. I would say that if $epsilon geq 1$ the function is definitely not Lipshitz contiuous on the second variable, but I'm really not sure how to prove that it is or isn't if $epsilon > 1$, it seems to me it's supposed to be, but I can't find a way to prove it.
differential-equations lipschitz-functions
asked Nov 18 at 16:24
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716420
add a comment |
Problem: can we apply Picard theorem for $$u' = sqrt{lvert u^2 -1 rvert}$$ if $$
u(pi / 2)= 0$$
[$u$ is a function of a variable $x$ so $u = u(x)$]
My attempt:
Well, what I need to know is whether the function $f(x,u) = sqrt{lvert u^2 -1 rvert} $ is Lipshitz continuous when observing only the second variable: u, on some rectangle $(pi/2 - delta,pi/2 + delta)times(1- epsilon,1 + epsilon)$.
I'm however uncertaion whether this holds. I would say that if $epsilon geq 1$ the function is definitely not Lipshitz contiuous on the second variable, but I'm really not sure how to prove that it is or isn't if $epsilon > 1$, it seems to me it's supposed to be, but I can't find a way to prove it.
differential-equations lipschitz-functions
asked Nov 18 at 16:24
Collapse
716420
What about computing $partial f/partial u$ and checking if it is bounded?
– Mattos
Nov 18 at 16:50
@Mattos but I don't think you can compute it at any point (x,1) ? (because of the absolute value function..)
– Collapse
Nov 18 at 16:55
The Picard theorem is a local one, so you don't need to bother about $epsilonge1$. Here, the assumptions are not satisfied at $(x,-1)$ and at $(x,1)$ for any real $x$. Further, notice that $u(x)=sin{x}$ for $xin(-pi/2,pi/2)$. Do you see what happens at $x=pmpi/2$?
– user539887
Nov 19 at 10:15
add a comment |
Problem: can we apply Picard theorem for $$u' = sqrt{lvert u^2 -1 rvert}$$ if $$
u(pi / 2)= 0$$
[$u$ is a function of a variable $x$ so $u = u(x)$]
My attempt:
Well, what I need to know is whether the function $f(x,u) = sqrt{lvert u^2 -1 rvert} $ is Lipshitz continuous when observing only the second variable: u, on some rectangle $(pi/2 - delta,pi/2 + delta)times(1- epsilon,1 + epsilon)$.
I'm however uncertaion whether this holds. I would say that if $epsilon geq 1$ the function is definitely not Lipshitz contiuous on the second variable, but I'm really not sure how to prove that it is or isn't if $epsilon > 1$, it seems to me it's supposed to be, but I can't find a way to prove it.
differential-equations lipschitz-functions
asked Nov 18 at 16:24
Collapse
716420
Problem: can we apply Picard theorem for $$u' = sqrt{lvert u^2 -1 rvert}$$ if $$
u(pi / 2)= 0$$
[$u$ is a function of a variable $x$ so $u = u(x)$]
My attempt:
Well, what I need to know is whether the function $f(x,u) = sqrt{lvert u^2 -1 rvert} $ is Lipshitz continuous when observing only the second variable: u, on some rectangle $(pi/2 - delta,pi/2 + delta)times(1- epsilon,1 + epsilon)$.
I'm however uncertaion whether this holds. I would say that if $epsilon geq 1$ the function is definitely not Lipshitz contiuous on the second variable, but I'm really not sure how to prove that it is or isn't if $epsilon > 1$, it seems to me it's supposed to be, but I can't find a way to prove it.
differential-equations lipschitz-functions
differential-equations lipschitz-functions
asked Nov 18 at 16:24
Collapse
716420
asked Nov 18 at 16:24
Collapse
716420
edited Nov 18 at 16:46
asked Nov 18 at 16:24
Collapse
716420
asked Nov 18 at 16:24
Collapse
716420
asked Nov 18 at 16:24
Collapse
716420
716420
What about computing $partial f/partial u$ and checking if it is bounded?
– Mattos
Nov 18 at 16:50
@Mattos but I don't think you can compute it at any point (x,1) ? (because of the absolute value function..)
– Collapse
Nov 18 at 16:55
The Picard theorem is a local one, so you don't need to bother about $epsilonge1$. Here, the assumptions are not satisfied at $(x,-1)$ and at $(x,1)$ for any real $x$. Further, notice that $u(x)=sin{x}$ for $xin(-pi/2,pi/2)$. Do you see what happens at $x=pmpi/2$?
– user539887
Nov 19 at 10:15
add a comment |
What about computing $partial f/partial u$ and checking if it is bounded?
– Mattos
Nov 18 at 16:50
@Mattos but I don't think you can compute it at any point (x,1) ? (because of the absolute value function..)
– Collapse
Nov 18 at 16:55
The Picard theorem is a local one, so you don't need to bother about $epsilonge1$. Here, the assumptions are not satisfied at $(x,-1)$ and at $(x,1)$ for any real $x$. Further, notice that $u(x)=sin{x}$ for $xin(-pi/2,pi/2)$. Do you see what happens at $x=pmpi/2$?
– user539887
Nov 19 at 10:15
What about computing $partial f/partial u$ and checking if it is bounded?
– Mattos
Nov 18 at 16:50
What about computing $partial f/partial u$ and checking if it is bounded?
– Mattos
Nov 18 at 16:50
@Mattos but I don't think you can compute it at any point (x,1) ? (because of the absolute value function..)
– Collapse
Nov 18 at 16:55
@Mattos but I don't think you can compute it at any point (x,1) ? (because of the absolute value function..)
– Collapse
Nov 18 at 16:55
The Picard theorem is a local one, so you don't need to bother about $epsilonge1$. Here, the assumptions are not satisfied at $(x,-1)$ and at $(x,1)$ for any real $x$. Further, notice that $u(x)=sin{x}$ for $xin(-pi/2,pi/2)$. Do you see what happens at $x=pmpi/2$?
– user539887
Nov 19 at 10:15
The Picard theorem is a local one, so you don't need to bother about $epsilonge1$. Here, the assumptions are not satisfied at $(x,-1)$ and at $(x,1)$ for any real $x$. Further, notice that $u(x)=sin{x}$ for $xin(-pi/2,pi/2)$. Do you see what happens at $x=pmpi/2$?
– user539887
Nov 19 at 10:15
add a comment |
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What about computing $partial f/partial u$ and checking if it is bounded?
– Mattos
Nov 18 at 16:50
@Mattos but I don't think you can compute it at any point (x,1) ? (because of the absolute value function..)
– Collapse
Nov 18 at 16:55
The Picard theorem is a local one, so you don't need to bother about $epsilonge1$. Here, the assumptions are not satisfied at $(x,-1)$ and at $(x,1)$ for any real $x$. Further, notice that $u(x)=sin{x}$ for $xin(-pi/2,pi/2)$. Do you see what happens at $x=pmpi/2$?
– user539887
Nov 19 at 10:15