Mixing
Lipschitz and Uniqueness of an IVP
Clash Royale CLAN TAG#URR8PPP
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Consider the IVP $$fracdxdt=1+x^2, x(0)=1.$$ I am trying to show that $f(x)=1+x^2$ is Lipschitz and hence the above IVP has a unique solution.
My attempt:
I considered $f(x)=1+x^2$. Now if $f$ is Lipschitz, then $exists LinmathbbR, textsuch that forall x,yinmathbbR$, $$|f(x)-f(y)|leq L|x-y|.$$
Now, beginalign
L&geq frac \
&=frac \
&=|x+y|
endalign
I'm a bit unsure of how to proceed. How can I use the initial value conditions to further my answer?
differential-equations proof-verification lipschitz-functions
asked 2 days ago
Bell
828314
add a comment |Â
up vote
4
down vote
favorite
Consider the IVP $$fracdxdt=1+x^2, x(0)=1.$$ I am trying to show that $f(x)=1+x^2$ is Lipschitz and hence the above IVP has a unique solution.
My attempt:
I considered $f(x)=1+x^2$. Now if $f$ is Lipschitz, then $exists LinmathbbR, textsuch that forall x,yinmathbbR$, $$|f(x)-f(y)|leq L|x-y|.$$
Now, beginalign
L&geq frac \
&=frac \
&=|x+y|
endalign
I'm a bit unsure of how to proceed. How can I use the initial value conditions to further my answer?
differential-equations proof-verification lipschitz-functions
asked 2 days ago
Bell
828314
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Consider the IVP $$fracdxdt=1+x^2, x(0)=1.$$ I am trying to show that $f(x)=1+x^2$ is Lipschitz and hence the above IVP has a unique solution.
My attempt:
I considered $f(x)=1+x^2$. Now if $f$ is Lipschitz, then $exists LinmathbbR, textsuch that forall x,yinmathbbR$, $$|f(x)-f(y)|leq L|x-y|.$$
Now, beginalign
L&geq frac \
&=frac \
&=|x+y|
endalign
I'm a bit unsure of how to proceed. How can I use the initial value conditions to further my answer?
differential-equations proof-verification lipschitz-functions
asked 2 days ago
Bell
828314
Consider the IVP $$fracdxdt=1+x^2, x(0)=1.$$ I am trying to show that $f(x)=1+x^2$ is Lipschitz and hence the above IVP has a unique solution.
My attempt:
I considered $f(x)=1+x^2$. Now if $f$ is Lipschitz, then $exists LinmathbbR, textsuch that forall x,yinmathbbR$, $$|f(x)-f(y)|leq L|x-y|.$$
Now, beginalign
L&geq frac \
&=frac \
&=|x+y|
endalign
I'm a bit unsure of how to proceed. How can I use the initial value conditions to further my answer?
differential-equations proof-verification lipschitz-functions
differential-equations proof-verification lipschitz-functions
asked 2 days ago
Bell
828314
asked 2 days ago
Bell
828314
asked 2 days ago
Bell
828314
asked 2 days ago
Bell
828314
asked 2 days ago
Bell
828314
828314
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
5
down vote
accepted
All you need for uniqueness is local Lipschitz property of $f(x)=1+x^2$: for any $x,y in [a,b]$, by the Lagrange Theorem,
$$|f(x)-f(y)|=|f'(t)||x-y|leq L|x-y|$$
where $L=max2t : tin [a,b]=2max(|a|,|b|)$ (or, according to you approach, we can also take $L=2max(|a|,|b|)geq |x|+ |y|geq |x+y|$).
By the way, in this case, uniqueness follows by solving the ODE after separating the variables:
$$int_1^x(t)fracdx1+x^2=int_0^tdtimplies arctan(x(t))-fracpi4=timplies x(t)=tanleft(t+fracpi4right).$$
answered 2 days ago
Robert Z
85.5k1055123
This helps a lot, thanks. For local Lipschitz, are there any restrictions on $a$ and $b$ in the interval $[a,b]$?
– Bell
2 days ago
For instance, if I considered the interval $[0,2]$, then by the theorem $$L=max_xin [0,2]|f'(x)|=max_xin [0,2]|2x|=2(2)=4.$$ Does this show $f$ is local Lipschitz and hence a unique solution exists to the IVP?
– Bell
2 days ago
@Bell No there are no restriction on $a$ and $b$. $f$ is local Lipschitz in $mathbbR$. This means that a unique solution exists for any initial value. Note that this ODE is autonomous en.wikipedia.org/wiki/Autonomous_system_(mathematics)
– Robert Z
2 days ago
So would my previous comment be correct? That is, is my example sufficient to show $f$ is local Lipschitz?
– Bell
2 days ago
@Bell Yes, it is correct.
– Robert Z
2 days ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
All you need for uniqueness is local Lipschitz property of $f(x)=1+x^2$: for any $x,y in [a,b]$, by the Lagrange Theorem,
$$|f(x)-f(y)|=|f'(t)||x-y|leq L|x-y|$$
where $L=max2t : tin [a,b]=2max(|a|,|b|)$ (or, according to you approach, we can also take $L=2max(|a|,|b|)geq |x|+ |y|geq |x+y|$).
By the way, in this case, uniqueness follows by solving the ODE after separating the variables:
$$int_1^x(t)fracdx1+x^2=int_0^tdtimplies arctan(x(t))-fracpi4=timplies x(t)=tanleft(t+fracpi4right).$$
answered 2 days ago
Robert Z
85.5k1055123
This helps a lot, thanks. For local Lipschitz, are there any restrictions on $a$ and $b$ in the interval $[a,b]$?
– Bell
2 days ago
For instance, if I considered the interval $[0,2]$, then by the theorem $$L=max_xin [0,2]|f'(x)|=max_xin [0,2]|2x|=2(2)=4.$$ Does this show $f$ is local Lipschitz and hence a unique solution exists to the IVP?
– Bell
2 days ago
@Bell No there are no restriction on $a$ and $b$. $f$ is local Lipschitz in $mathbbR$. This means that a unique solution exists for any initial value. Note that this ODE is autonomous en.wikipedia.org/wiki/Autonomous_system_(mathematics)
– Robert Z
2 days ago
So would my previous comment be correct? That is, is my example sufficient to show $f$ is local Lipschitz?
– Bell
2 days ago
@Bell Yes, it is correct.
– Robert Z
2 days ago
add a comment |Â
up vote
5
down vote
accepted
All you need for uniqueness is local Lipschitz property of $f(x)=1+x^2$: for any $x,y in [a,b]$, by the Lagrange Theorem,
$$|f(x)-f(y)|=|f'(t)||x-y|leq L|x-y|$$
where $L=max2t : tin [a,b]=2max(|a|,|b|)$ (or, according to you approach, we can also take $L=2max(|a|,|b|)geq |x|+ |y|geq |x+y|$).
By the way, in this case, uniqueness follows by solving the ODE after separating the variables:
$$int_1^x(t)fracdx1+x^2=int_0^tdtimplies arctan(x(t))-fracpi4=timplies x(t)=tanleft(t+fracpi4right).$$
answered 2 days ago
Robert Z
85.5k1055123
This helps a lot, thanks. For local Lipschitz, are there any restrictions on $a$ and $b$ in the interval $[a,b]$?
– Bell
2 days ago
For instance, if I considered the interval $[0,2]$, then by the theorem $$L=max_xin [0,2]|f'(x)|=max_xin [0,2]|2x|=2(2)=4.$$ Does this show $f$ is local Lipschitz and hence a unique solution exists to the IVP?
– Bell
2 days ago
@Bell No there are no restriction on $a$ and $b$. $f$ is local Lipschitz in $mathbbR$. This means that a unique solution exists for any initial value. Note that this ODE is autonomous en.wikipedia.org/wiki/Autonomous_system_(mathematics)
– Robert Z
2 days ago
So would my previous comment be correct? That is, is my example sufficient to show $f$ is local Lipschitz?
– Bell
2 days ago
@Bell Yes, it is correct.
– Robert Z
2 days ago
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
All you need for uniqueness is local Lipschitz property of $f(x)=1+x^2$: for any $x,y in [a,b]$, by the Lagrange Theorem,
$$|f(x)-f(y)|=|f'(t)||x-y|leq L|x-y|$$
where $L=max2t : tin [a,b]=2max(|a|,|b|)$ (or, according to you approach, we can also take $L=2max(|a|,|b|)geq |x|+ |y|geq |x+y|$).
By the way, in this case, uniqueness follows by solving the ODE after separating the variables:
$$int_1^x(t)fracdx1+x^2=int_0^tdtimplies arctan(x(t))-fracpi4=timplies x(t)=tanleft(t+fracpi4right).$$
answered 2 days ago
Robert Z
85.5k1055123
All you need for uniqueness is local Lipschitz property of $f(x)=1+x^2$: for any $x,y in [a,b]$, by the Lagrange Theorem,
$$|f(x)-f(y)|=|f'(t)||x-y|leq L|x-y|$$
where $L=max2t : tin [a,b]=2max(|a|,|b|)$ (or, according to you approach, we can also take $L=2max(|a|,|b|)geq |x|+ |y|geq |x+y|$).
By the way, in this case, uniqueness follows by solving the ODE after separating the variables:
$$int_1^x(t)fracdx1+x^2=int_0^tdtimplies arctan(x(t))-fracpi4=timplies x(t)=tanleft(t+fracpi4right).$$
answered 2 days ago
Robert Z
85.5k1055123
edited 2 days ago
answered 2 days ago
Robert Z
85.5k1055123
answered 2 days ago
Robert Z
85.5k1055123
answered 2 days ago
Robert Z
85.5k1055123
85.5k1055123
This helps a lot, thanks. For local Lipschitz, are there any restrictions on $a$ and $b$ in the interval $[a,b]$?
– Bell
2 days ago
For instance, if I considered the interval $[0,2]$, then by the theorem $$L=max_xin [0,2]|f'(x)|=max_xin [0,2]|2x|=2(2)=4.$$ Does this show $f$ is local Lipschitz and hence a unique solution exists to the IVP?
– Bell
2 days ago
@Bell No there are no restriction on $a$ and $b$. $f$ is local Lipschitz in $mathbbR$. This means that a unique solution exists for any initial value. Note that this ODE is autonomous en.wikipedia.org/wiki/Autonomous_system_(mathematics)
– Robert Z
2 days ago
So would my previous comment be correct? That is, is my example sufficient to show $f$ is local Lipschitz?
– Bell
2 days ago
@Bell Yes, it is correct.
– Robert Z
2 days ago
add a comment |Â
This helps a lot, thanks. For local Lipschitz, are there any restrictions on $a$ and $b$ in the interval $[a,b]$?
– Bell
2 days ago
For instance, if I considered the interval $[0,2]$, then by the theorem $$L=max_xin [0,2]|f'(x)|=max_xin [0,2]|2x|=2(2)=4.$$ Does this show $f$ is local Lipschitz and hence a unique solution exists to the IVP?
– Bell
2 days ago
@Bell No there are no restriction on $a$ and $b$. $f$ is local Lipschitz in $mathbbR$. This means that a unique solution exists for any initial value. Note that this ODE is autonomous en.wikipedia.org/wiki/Autonomous_system_(mathematics)
– Robert Z
2 days ago
So would my previous comment be correct? That is, is my example sufficient to show $f$ is local Lipschitz?
– Bell
2 days ago
@Bell Yes, it is correct.
– Robert Z
2 days ago
This helps a lot, thanks. For local Lipschitz, are there any restrictions on $a$ and $b$ in the interval $[a,b]$?
– Bell
2 days ago
This helps a lot, thanks. For local Lipschitz, are there any restrictions on $a$ and $b$ in the interval $[a,b]$?
– Bell
2 days ago
For instance, if I considered the interval $[0,2]$, then by the theorem $$L=max_xin [0,2]|f'(x)|=max_xin [0,2]|2x|=2(2)=4.$$ Does this show $f$ is local Lipschitz and hence a unique solution exists to the IVP?
– Bell
2 days ago
For instance, if I considered the interval $[0,2]$, then by the theorem $$L=max_xin [0,2]|f'(x)|=max_xin [0,2]|2x|=2(2)=4.$$ Does this show $f$ is local Lipschitz and hence a unique solution exists to the IVP?
– Bell
2 days ago
@Bell No there are no restriction on $a$ and $b$. $f$ is local Lipschitz in $mathbbR$. This means that a unique solution exists for any initial value. Note that this ODE is autonomous en.wikipedia.org/wiki/Autonomous_system_(mathematics)
– Robert Z
2 days ago
@Bell No there are no restriction on $a$ and $b$. $f$ is local Lipschitz in $mathbbR$. This means that a unique solution exists for any initial value. Note that this ODE is autonomous en.wikipedia.org/wiki/Autonomous_system_(mathematics)
– Robert Z
2 days ago
So would my previous comment be correct? That is, is my example sufficient to show $f$ is local Lipschitz?
– Bell
2 days ago
So would my previous comment be correct? That is, is my example sufficient to show $f$ is local Lipschitz?
– Bell
2 days ago
@Bell Yes, it is correct.
– Robert Z
2 days ago
@Bell Yes, it is correct.
– Robert Z
2 days ago
add a comment |Â
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