Mixing
A unique solution for $y'=cosleft(yright)$
$begingroup$
I've considered the non-linear problem
$$ displaystyle left(starright) begin{cases}
displaystyle y'(x)=cosleft(y(x)right)\
y(0)=alpha, alpha in mathbb{R}
end{cases}$$
I want to prove that it admits a unique solution on $mathbb{R}$. I've used the following lemma :
Grönwall Lemma : If $y$ satisfies for all $t in left[0,+inftyright[$ the inequality
$$
y'left(tright) leq betaleft(tright)yleft(tright)
$$
then for all $t in left[0,+inftyright[$
$$
yleft(tright) leq yleft(0right) text{exp}left(int_{0}^{t}betaleft(sright)text{d}sright)
$$
Then let $y_1$ and $y_2$ be two solutions of $(star)$ then we have
$$
(y_1-y_2)'=cosleft(y_1left(tright)right)-cosleft(y_2left(tright)right)=-2sinleft(frac{y_1+y_2}{2}right)sinleft(frac{y_1-y_2}{2}right)
$$
So letting $Y=(y_1-y_2)/2$ we have
$$
Y' leq sinleft(frac{y_1left(tright)+y_2left(tright)}{2}right)sinleft(-Yright)
$$
Using that $Yleft(0right)=0$, we have
$$
y_1(t)-y_2(t) leq 0
$$
I can then use the same argument with $Z=y_2-y_1$ to obtain $y_2(t)-y_1(t) leq 0$ to have
$$
y_1=y_2
$$
Can somebody tell me if it's true ( or correct me to fill the proof )? My problem is the majoration of $Y'$ because I dont know the sign of the sin ( can't be maxed by $1$ then ), and it depends on $y_1$ and $y_2$.
ordinary-differential-equations
asked Jan 27 at 22:33
AtmosAtmos
4,830420
$endgroup$
add a comment |
$begingroup$
I've considered the non-linear problem
$$ displaystyle left(starright) begin{cases}
displaystyle y'(x)=cosleft(y(x)right)\
y(0)=alpha, alpha in mathbb{R}
end{cases}$$
I want to prove that it admits a unique solution on $mathbb{R}$. I've used the following lemma :
Grönwall Lemma : If $y$ satisfies for all $t in left[0,+inftyright[$ the inequality
$$
y'left(tright) leq betaleft(tright)yleft(tright)
$$
then for all $t in left[0,+inftyright[$
$$
yleft(tright) leq yleft(0right) text{exp}left(int_{0}^{t}betaleft(sright)text{d}sright)
$$
Then let $y_1$ and $y_2$ be two solutions of $(star)$ then we have
$$
(y_1-y_2)'=cosleft(y_1left(tright)right)-cosleft(y_2left(tright)right)=-2sinleft(frac{y_1+y_2}{2}right)sinleft(frac{y_1-y_2}{2}right)
$$
So letting $Y=(y_1-y_2)/2$ we have
$$
Y' leq sinleft(frac{y_1left(tright)+y_2left(tright)}{2}right)sinleft(-Yright)
$$
Using that $Yleft(0right)=0$, we have
$$
y_1(t)-y_2(t) leq 0
$$
I can then use the same argument with $Z=y_2-y_1$ to obtain $y_2(t)-y_1(t) leq 0$ to have
$$
y_1=y_2
$$
Can somebody tell me if it's true ( or correct me to fill the proof )? My problem is the majoration of $Y'$ because I dont know the sign of the sin ( can't be maxed by $1$ then ), and it depends on $y_1$ and $y_2$.
ordinary-differential-equations
asked Jan 27 at 22:33
AtmosAtmos
4,830420
$endgroup$
$begingroup$
How do you know that there exists such a solution on $mathbb{R}$?
$endgroup$
– Eclipse Sun
Jan 27 at 22:39
$begingroup$
Because I've found it
$endgroup$
– Atmos
Jan 27 at 22:45
$begingroup$
If you have found a global solution, then the conclusion follows from Picard existence theorem, or known as Cauchy-Lipschitz theorem. However you don't need to know the solution because a stronger version of the theorem (mentioned in lonza's answer) applies as well.
$endgroup$
– Eclipse Sun
Jan 28 at 3:05
$begingroup$
I dont want to use this theorem but to complete the proof this way :/
$endgroup$
– Atmos
Jan 28 at 11:33
add a comment |
$begingroup$
I've considered the non-linear problem
$$ displaystyle left(starright) begin{cases}
displaystyle y'(x)=cosleft(y(x)right)\
y(0)=alpha, alpha in mathbb{R}
end{cases}$$
I want to prove that it admits a unique solution on $mathbb{R}$. I've used the following lemma :
Grönwall Lemma : If $y$ satisfies for all $t in left[0,+inftyright[$ the inequality
$$
y'left(tright) leq betaleft(tright)yleft(tright)
$$
then for all $t in left[0,+inftyright[$
$$
yleft(tright) leq yleft(0right) text{exp}left(int_{0}^{t}betaleft(sright)text{d}sright)
$$
Then let $y_1$ and $y_2$ be two solutions of $(star)$ then we have
$$
(y_1-y_2)'=cosleft(y_1left(tright)right)-cosleft(y_2left(tright)right)=-2sinleft(frac{y_1+y_2}{2}right)sinleft(frac{y_1-y_2}{2}right)
$$
So letting $Y=(y_1-y_2)/2$ we have
$$
Y' leq sinleft(frac{y_1left(tright)+y_2left(tright)}{2}right)sinleft(-Yright)
$$
Using that $Yleft(0right)=0$, we have
$$
y_1(t)-y_2(t) leq 0
$$
I can then use the same argument with $Z=y_2-y_1$ to obtain $y_2(t)-y_1(t) leq 0$ to have
$$
y_1=y_2
$$
Can somebody tell me if it's true ( or correct me to fill the proof )? My problem is the majoration of $Y'$ because I dont know the sign of the sin ( can't be maxed by $1$ then ), and it depends on $y_1$ and $y_2$.
ordinary-differential-equations
asked Jan 27 at 22:33
AtmosAtmos
4,830420
$endgroup$
I've considered the non-linear problem
$$ displaystyle left(starright) begin{cases}
displaystyle y'(x)=cosleft(y(x)right)\
y(0)=alpha, alpha in mathbb{R}
end{cases}$$
I want to prove that it admits a unique solution on $mathbb{R}$. I've used the following lemma :
Grönwall Lemma : If $y$ satisfies for all $t in left[0,+inftyright[$ the inequality
$$
y'left(tright) leq betaleft(tright)yleft(tright)
$$
then for all $t in left[0,+inftyright[$
$$
yleft(tright) leq yleft(0right) text{exp}left(int_{0}^{t}betaleft(sright)text{d}sright)
$$
Then let $y_1$ and $y_2$ be two solutions of $(star)$ then we have
$$
(y_1-y_2)'=cosleft(y_1left(tright)right)-cosleft(y_2left(tright)right)=-2sinleft(frac{y_1+y_2}{2}right)sinleft(frac{y_1-y_2}{2}right)
$$
So letting $Y=(y_1-y_2)/2$ we have
$$
Y' leq sinleft(frac{y_1left(tright)+y_2left(tright)}{2}right)sinleft(-Yright)
$$
Using that $Yleft(0right)=0$, we have
$$
y_1(t)-y_2(t) leq 0
$$
I can then use the same argument with $Z=y_2-y_1$ to obtain $y_2(t)-y_1(t) leq 0$ to have
$$
y_1=y_2
$$
Can somebody tell me if it's true ( or correct me to fill the proof )? My problem is the majoration of $Y'$ because I dont know the sign of the sin ( can't be maxed by $1$ then ), and it depends on $y_1$ and $y_2$.
ordinary-differential-equations
ordinary-differential-equations
asked Jan 27 at 22:33
AtmosAtmos
4,830420
asked Jan 27 at 22:33
AtmosAtmos
4,830420
asked Jan 27 at 22:33
AtmosAtmos
4,830420
asked Jan 27 at 22:33
AtmosAtmos
4,830420
asked Jan 27 at 22:33
AtmosAtmos
4,830420
4,830420
$begingroup$
How do you know that there exists such a solution on $mathbb{R}$?
$endgroup$
– Eclipse Sun
Jan 27 at 22:39
$begingroup$
Because I've found it
$endgroup$
– Atmos
Jan 27 at 22:45
$begingroup$
If you have found a global solution, then the conclusion follows from Picard existence theorem, or known as Cauchy-Lipschitz theorem. However you don't need to know the solution because a stronger version of the theorem (mentioned in lonza's answer) applies as well.
$endgroup$
– Eclipse Sun
Jan 28 at 3:05
$begingroup$
I dont want to use this theorem but to complete the proof this way :/
$endgroup$
– Atmos
Jan 28 at 11:33
add a comment |
$begingroup$
How do you know that there exists such a solution on $mathbb{R}$?
$endgroup$
– Eclipse Sun
Jan 27 at 22:39
$begingroup$
Because I've found it
$endgroup$
– Atmos
Jan 27 at 22:45
$begingroup$
If you have found a global solution, then the conclusion follows from Picard existence theorem, or known as Cauchy-Lipschitz theorem. However you don't need to know the solution because a stronger version of the theorem (mentioned in lonza's answer) applies as well.
$endgroup$
– Eclipse Sun
Jan 28 at 3:05
$begingroup$
I dont want to use this theorem but to complete the proof this way :/
$endgroup$
– Atmos
Jan 28 at 11:33
$begingroup$
How do you know that there exists such a solution on $mathbb{R}$?
$endgroup$
– Eclipse Sun
Jan 27 at 22:39
$begingroup$
How do you know that there exists such a solution on $mathbb{R}$?
$endgroup$
– Eclipse Sun
Jan 27 at 22:39
$begingroup$
Because I've found it
$endgroup$
– Atmos
Jan 27 at 22:45
$begingroup$
Because I've found it
$endgroup$
– Atmos
Jan 27 at 22:45
$begingroup$
If you have found a global solution, then the conclusion follows from Picard existence theorem, or known as Cauchy-Lipschitz theorem. However you don't need to know the solution because a stronger version of the theorem (mentioned in lonza's answer) applies as well.
$endgroup$
– Eclipse Sun
Jan 28 at 3:05
$begingroup$
If you have found a global solution, then the conclusion follows from Picard existence theorem, or known as Cauchy-Lipschitz theorem. However you don't need to know the solution because a stronger version of the theorem (mentioned in lonza's answer) applies as well.
$endgroup$
– Eclipse Sun
Jan 28 at 3:05
$begingroup$
I dont want to use this theorem but to complete the proof this way :/
$endgroup$
– Atmos
Jan 28 at 11:33
$begingroup$
I dont want to use this theorem but to complete the proof this way :/
$endgroup$
– Atmos
Jan 28 at 11:33
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since the function $ cos $ satisfies the Lipschitz condition $ mid cosleft(y_2right)-
cosleft(y_2right)mid le mid y_1 - y_2,mid$ for all $ y_1, y_2inmathbb R $, the existence and uniqueness of a solution follows from the global Cauchy-Lipschitz theorem.
answered Jan 28 at 0:45
lonza leggieralonza leggiera
1,20828
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Since the function $ cos $ satisfies the Lipschitz condition $ mid cosleft(y_2right)-
cosleft(y_2right)mid le mid y_1 - y_2,mid$ for all $ y_1, y_2inmathbb R $, the existence and uniqueness of a solution follows from the global Cauchy-Lipschitz theorem.
answered Jan 28 at 0:45
lonza leggieralonza leggiera
1,20828
$endgroup$
add a comment |
$begingroup$
Since the function $ cos $ satisfies the Lipschitz condition $ mid cosleft(y_2right)-
cosleft(y_2right)mid le mid y_1 - y_2,mid$ for all $ y_1, y_2inmathbb R $, the existence and uniqueness of a solution follows from the global Cauchy-Lipschitz theorem.
answered Jan 28 at 0:45
lonza leggieralonza leggiera
1,20828
$endgroup$
add a comment |
$begingroup$
Since the function $ cos $ satisfies the Lipschitz condition $ mid cosleft(y_2right)-
cosleft(y_2right)mid le mid y_1 - y_2,mid$ for all $ y_1, y_2inmathbb R $, the existence and uniqueness of a solution follows from the global Cauchy-Lipschitz theorem.
answered Jan 28 at 0:45
lonza leggieralonza leggiera
1,20828
$endgroup$
Since the function $ cos $ satisfies the Lipschitz condition $ mid cosleft(y_2right)-
cosleft(y_2right)mid le mid y_1 - y_2,mid$ for all $ y_1, y_2inmathbb R $, the existence and uniqueness of a solution follows from the global Cauchy-Lipschitz theorem.
answered Jan 28 at 0:45
lonza leggieralonza leggiera
1,20828
answered Jan 28 at 0:45
lonza leggieralonza leggiera
1,20828
answered Jan 28 at 0:45
lonza leggieralonza leggiera
1,20828
answered Jan 28 at 0:45
lonza leggieralonza leggiera
1,20828
1,20828
add a comment |
add a comment |
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$begingroup$
How do you know that there exists such a solution on $mathbb{R}$?
$endgroup$
– Eclipse Sun
Jan 27 at 22:39
$begingroup$
Because I've found it
$endgroup$
– Atmos
Jan 27 at 22:45
$begingroup$
If you have found a global solution, then the conclusion follows from Picard existence theorem, or known as Cauchy-Lipschitz theorem. However you don't need to know the solution because a stronger version of the theorem (mentioned in lonza's answer) applies as well.
$endgroup$
– Eclipse Sun
Jan 28 at 3:05
$begingroup$
I dont want to use this theorem but to complete the proof this way :/
$endgroup$
– Atmos
Jan 28 at 11:33