If $x$ is a solution to the equation $dot y = f(y) quad f in C^infty (mathbb{R}^d)$, and $x(0) = x(T)$ for...
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In the book of The Elements of Mechanics by Gallovotti, at page 16, it is asked that
If $x$ is a solution to the equation $dot y = f(y) quad f in
C^infty (mathbb{R}^d)$, and $x(0) = x(T)$ for some T, prove that
$x(t) = x(t+T)$. Would this also be true if $f in C^1(mathbb{R}) ?$ [Hint: Use uniqueness.]
However, we do know that
$$x(t) = x(0) + int_{tau = 0}^{tau=t} f(x(tau)),$$
and if we looked at the difference
$$x(t+T) - x(t) = int_{tau=t}^{tau=t+T} f(x(tau)),$$
I cannot see any reason why this integral should be zero.
differential-equations classical-mechanics
asked Nov 22 at 3:43
onurcanbektas
3,3081936
add a comment |
up vote
2
down vote
favorite
In the book of The Elements of Mechanics by Gallovotti, at page 16, it is asked that
If $x$ is a solution to the equation $dot y = f(y) quad f in
C^infty (mathbb{R}^d)$, and $x(0) = x(T)$ for some T, prove that
$x(t) = x(t+T)$. Would this also be true if $f in C^1(mathbb{R}) ?$ [Hint: Use uniqueness.]
However, we do know that
$$x(t) = x(0) + int_{tau = 0}^{tau=t} f(x(tau)),$$
and if we looked at the difference
$$x(t+T) - x(t) = int_{tau=t}^{tau=t+T} f(x(tau)),$$
I cannot see any reason why this integral should be zero.
differential-equations classical-mechanics
asked Nov 22 at 3:43
onurcanbektas
3,3081936
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In the book of The Elements of Mechanics by Gallovotti, at page 16, it is asked that
If $x$ is a solution to the equation $dot y = f(y) quad f in
C^infty (mathbb{R}^d)$, and $x(0) = x(T)$ for some T, prove that
$x(t) = x(t+T)$. Would this also be true if $f in C^1(mathbb{R}) ?$ [Hint: Use uniqueness.]
However, we do know that
$$x(t) = x(0) + int_{tau = 0}^{tau=t} f(x(tau)),$$
and if we looked at the difference
$$x(t+T) - x(t) = int_{tau=t}^{tau=t+T} f(x(tau)),$$
I cannot see any reason why this integral should be zero.
differential-equations classical-mechanics
asked Nov 22 at 3:43
onurcanbektas
3,3081936
In the book of The Elements of Mechanics by Gallovotti, at page 16, it is asked that
If $x$ is a solution to the equation $dot y = f(y) quad f in
C^infty (mathbb{R}^d)$, and $x(0) = x(T)$ for some T, prove that
$x(t) = x(t+T)$. Would this also be true if $f in C^1(mathbb{R}) ?$ [Hint: Use uniqueness.]
However, we do know that
$$x(t) = x(0) + int_{tau = 0}^{tau=t} f(x(tau)),$$
and if we looked at the difference
$$x(t+T) - x(t) = int_{tau=t}^{tau=t+T} f(x(tau)),$$
I cannot see any reason why this integral should be zero.
differential-equations classical-mechanics
differential-equations classical-mechanics
asked Nov 22 at 3:43
onurcanbektas
3,3081936
asked Nov 22 at 3:43
onurcanbektas
3,3081936
asked Nov 22 at 3:43
onurcanbektas
3,3081936
asked Nov 22 at 3:43
onurcanbektas
3,3081936
asked Nov 22 at 3:43
onurcanbektas
3,3081936
3,3081936
add a comment |
add a comment |
2 Answers
2
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2
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accepted
Set
$z(t) = x(t + T); tag 1$
then
$z(0) = x(0 + T) = x(T) = x(0); tag 2$
also
$dot z(t) = dot x(t + T) = f(x(t + T)) = f(z(t)); tag{2.5}$
now since
$f(y) in C^infty(Bbb R^n) tag 3$
is differentiable, it follows that $f(y)$ is locally Lipschitz continuous at each point $y_0 in Bbb R^n$; and from this it further follows, via the Picard-Lindeloef theorem, that through any $y_0 in Bbb R^n$ there is precisely one solution to
$dot y = f(y), ; y(0) = y_0; tag 4$
therefore, in light of (2), which says that both $x(t)$ and $z(t)$ satisfy
$x(0) = z(0) = y_0 tag 5$
for some $y_0$, we conclude that
$x(t) = z(t) = x(t + T). tag 6$
Since $f(y) in C^1(Bbb R^n)$ is sufficient for $f(y)$ Lipschitz, the result holds in this case as well.
answered Nov 22 at 7:28
Robert Lewis
42.8k22862
add a comment |
up vote
4
down vote
Let $y_1(t) = x(t)$ and $y_2(t) = x(t+T)$. Then both $y_1$ and $y_2$ satisfy
begin{align}
y' = f(y)
end{align}
with the same initial condition. Hence by uniqueness we are done.
answered Nov 22 at 5:17
Jacky Chong
17.4k21128
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Set
$z(t) = x(t + T); tag 1$
then
$z(0) = x(0 + T) = x(T) = x(0); tag 2$
also
$dot z(t) = dot x(t + T) = f(x(t + T)) = f(z(t)); tag{2.5}$
now since
$f(y) in C^infty(Bbb R^n) tag 3$
is differentiable, it follows that $f(y)$ is locally Lipschitz continuous at each point $y_0 in Bbb R^n$; and from this it further follows, via the Picard-Lindeloef theorem, that through any $y_0 in Bbb R^n$ there is precisely one solution to
$dot y = f(y), ; y(0) = y_0; tag 4$
therefore, in light of (2), which says that both $x(t)$ and $z(t)$ satisfy
$x(0) = z(0) = y_0 tag 5$
for some $y_0$, we conclude that
$x(t) = z(t) = x(t + T). tag 6$
Since $f(y) in C^1(Bbb R^n)$ is sufficient for $f(y)$ Lipschitz, the result holds in this case as well.
answered Nov 22 at 7:28
Robert Lewis
42.8k22862
add a comment |
up vote
2
down vote
accepted
Set
$z(t) = x(t + T); tag 1$
then
$z(0) = x(0 + T) = x(T) = x(0); tag 2$
also
$dot z(t) = dot x(t + T) = f(x(t + T)) = f(z(t)); tag{2.5}$
now since
$f(y) in C^infty(Bbb R^n) tag 3$
is differentiable, it follows that $f(y)$ is locally Lipschitz continuous at each point $y_0 in Bbb R^n$; and from this it further follows, via the Picard-Lindeloef theorem, that through any $y_0 in Bbb R^n$ there is precisely one solution to
$dot y = f(y), ; y(0) = y_0; tag 4$
therefore, in light of (2), which says that both $x(t)$ and $z(t)$ satisfy
$x(0) = z(0) = y_0 tag 5$
for some $y_0$, we conclude that
$x(t) = z(t) = x(t + T). tag 6$
Since $f(y) in C^1(Bbb R^n)$ is sufficient for $f(y)$ Lipschitz, the result holds in this case as well.
answered Nov 22 at 7:28
Robert Lewis
42.8k22862
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Set
$z(t) = x(t + T); tag 1$
then
$z(0) = x(0 + T) = x(T) = x(0); tag 2$
also
$dot z(t) = dot x(t + T) = f(x(t + T)) = f(z(t)); tag{2.5}$
now since
$f(y) in C^infty(Bbb R^n) tag 3$
is differentiable, it follows that $f(y)$ is locally Lipschitz continuous at each point $y_0 in Bbb R^n$; and from this it further follows, via the Picard-Lindeloef theorem, that through any $y_0 in Bbb R^n$ there is precisely one solution to
$dot y = f(y), ; y(0) = y_0; tag 4$
therefore, in light of (2), which says that both $x(t)$ and $z(t)$ satisfy
$x(0) = z(0) = y_0 tag 5$
for some $y_0$, we conclude that
$x(t) = z(t) = x(t + T). tag 6$
Since $f(y) in C^1(Bbb R^n)$ is sufficient for $f(y)$ Lipschitz, the result holds in this case as well.
answered Nov 22 at 7:28
Robert Lewis
42.8k22862
Set
$z(t) = x(t + T); tag 1$
then
$z(0) = x(0 + T) = x(T) = x(0); tag 2$
also
$dot z(t) = dot x(t + T) = f(x(t + T)) = f(z(t)); tag{2.5}$
now since
$f(y) in C^infty(Bbb R^n) tag 3$
is differentiable, it follows that $f(y)$ is locally Lipschitz continuous at each point $y_0 in Bbb R^n$; and from this it further follows, via the Picard-Lindeloef theorem, that through any $y_0 in Bbb R^n$ there is precisely one solution to
$dot y = f(y), ; y(0) = y_0; tag 4$
therefore, in light of (2), which says that both $x(t)$ and $z(t)$ satisfy
$x(0) = z(0) = y_0 tag 5$
for some $y_0$, we conclude that
$x(t) = z(t) = x(t + T). tag 6$
Since $f(y) in C^1(Bbb R^n)$ is sufficient for $f(y)$ Lipschitz, the result holds in this case as well.
answered Nov 22 at 7:28
Robert Lewis
42.8k22862
answered Nov 22 at 7:28
Robert Lewis
42.8k22862
answered Nov 22 at 7:28
Robert Lewis
42.8k22862
answered Nov 22 at 7:28
Robert Lewis
42.8k22862
42.8k22862
add a comment |
add a comment |
up vote
4
down vote
Let $y_1(t) = x(t)$ and $y_2(t) = x(t+T)$. Then both $y_1$ and $y_2$ satisfy
begin{align}
y' = f(y)
end{align}
with the same initial condition. Hence by uniqueness we are done.
answered Nov 22 at 5:17
Jacky Chong
17.4k21128
add a comment |
up vote
4
down vote
Let $y_1(t) = x(t)$ and $y_2(t) = x(t+T)$. Then both $y_1$ and $y_2$ satisfy
begin{align}
y' = f(y)
end{align}
with the same initial condition. Hence by uniqueness we are done.
answered Nov 22 at 5:17
Jacky Chong
17.4k21128
add a comment |
up vote
4
down vote
up vote
4
down vote
Let $y_1(t) = x(t)$ and $y_2(t) = x(t+T)$. Then both $y_1$ and $y_2$ satisfy
begin{align}
y' = f(y)
end{align}
with the same initial condition. Hence by uniqueness we are done.
answered Nov 22 at 5:17
Jacky Chong
17.4k21128
Let $y_1(t) = x(t)$ and $y_2(t) = x(t+T)$. Then both $y_1$ and $y_2$ satisfy
begin{align}
y' = f(y)
end{align}
with the same initial condition. Hence by uniqueness we are done.
answered Nov 22 at 5:17
Jacky Chong
17.4k21128
answered Nov 22 at 5:17
Jacky Chong
17.4k21128
answered Nov 22 at 5:17
Jacky Chong
17.4k21128
answered Nov 22 at 5:17
Jacky Chong
17.4k21128
17.4k21128
add a comment |
add a comment |
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