Picard theorem for $u' = sqrt{lvert u^2 -1 rvert}$ if we know $ u(pi / 2)= 0$











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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.










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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















up vote
0
down vote

favorite












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.










share|cite|improve this question
























  • 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













up vote
0
down vote

favorite









up vote
0
down vote

favorite











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.










share|cite|improve this question















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






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edited Nov 18 at 16:46

























asked Nov 18 at 16:24









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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


















  • 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















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