If $Vert f(x)-f(y) Vertleq kVert x-yVert,;;forall;x,yinBbbR^n,$ then $x(t,x_0)=x_0,forall ;tgeq 0,$
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We consider the following O.D.E
beginalign(1);;;begincasesx'(t)=f(x(t)) & tgeq 0,\x(0)=x_0in BbbR^n&endcasesendalign
where beginalignf: BbbR^nto BbbR^nendalign
Assuming that beginalignVert f(x)-f(y) Vertleq kVert x-yVert,;;forall;x,yinBbbR^n.endalign
Assuming that $f(x_0)=0.$ Please, do I prove that $x(t,x_0)=x_0,forall ;tgeq 0,$ where $x(cdot,,x_0)$ is the solution of $(1)$. If there are references, I also would appreciate!
calculus differential-equations
edited Aug 28 at 13:04
Adrian Keister
4,06541633
asked Aug 28 at 13:01
Mike
77415
 |Â
show 3 more comments
up vote
3
down vote
favorite
We consider the following O.D.E
beginalign(1);;;begincasesx'(t)=f(x(t)) & tgeq 0,\x(0)=x_0in BbbR^n&endcasesendalign
where beginalignf: BbbR^nto BbbR^nendalign
Assuming that beginalignVert f(x)-f(y) Vertleq kVert x-yVert,;;forall;x,yinBbbR^n.endalign
Assuming that $f(x_0)=0.$ Please, do I prove that $x(t,x_0)=x_0,forall ;tgeq 0,$ where $x(cdot,,x_0)$ is the solution of $(1)$. If there are references, I also would appreciate!
calculus differential-equations
edited Aug 28 at 13:04
Adrian Keister
4,06541633
asked Aug 28 at 13:01
Mike
77415
Any restrictions on $k?$
– Adrian Keister
Aug 28 at 13:08
@Adrian Keister: $k$ is assumed to be greater than $zero!$
– Mike
Aug 28 at 13:11
What I mean is, is $k$ arbitrary and not specified in advance? Or is it treated as a given?
– Adrian Keister
Aug 28 at 13:24
@Adrian Keister: Although, not given but I assume it to be $k<1.$
– Mike
Aug 28 at 13:26
Is the condition true for all $0<k<1?$
– Adrian Keister
Aug 28 at 13:27
 |Â
show 3 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
We consider the following O.D.E
beginalign(1);;;begincasesx'(t)=f(x(t)) & tgeq 0,\x(0)=x_0in BbbR^n&endcasesendalign
where beginalignf: BbbR^nto BbbR^nendalign
Assuming that beginalignVert f(x)-f(y) Vertleq kVert x-yVert,;;forall;x,yinBbbR^n.endalign
Assuming that $f(x_0)=0.$ Please, do I prove that $x(t,x_0)=x_0,forall ;tgeq 0,$ where $x(cdot,,x_0)$ is the solution of $(1)$. If there are references, I also would appreciate!
calculus differential-equations
edited Aug 28 at 13:04
Adrian Keister
4,06541633
asked Aug 28 at 13:01
Mike
77415
We consider the following O.D.E
beginalign(1);;;begincasesx'(t)=f(x(t)) & tgeq 0,\x(0)=x_0in BbbR^n&endcasesendalign
where beginalignf: BbbR^nto BbbR^nendalign
Assuming that beginalignVert f(x)-f(y) Vertleq kVert x-yVert,;;forall;x,yinBbbR^n.endalign
Assuming that $f(x_0)=0.$ Please, do I prove that $x(t,x_0)=x_0,forall ;tgeq 0,$ where $x(cdot,,x_0)$ is the solution of $(1)$. If there are references, I also would appreciate!
calculus differential-equations
edited Aug 28 at 13:04
Adrian Keister
4,06541633
asked Aug 28 at 13:01
Mike
77415
edited Aug 28 at 13:04
Adrian Keister
4,06541633
edited Aug 28 at 13:04
Adrian Keister
4,06541633
edited Aug 28 at 13:04
Adrian Keister
4,06541633
4,06541633
asked Aug 28 at 13:01
Mike
77415
asked Aug 28 at 13:01
Mike
77415
asked Aug 28 at 13:01
Mike
77415
77415
Any restrictions on $k?$
– Adrian Keister
Aug 28 at 13:08
@Adrian Keister: $k$ is assumed to be greater than $zero!$
– Mike
Aug 28 at 13:11
What I mean is, is $k$ arbitrary and not specified in advance? Or is it treated as a given?
– Adrian Keister
Aug 28 at 13:24
@Adrian Keister: Although, not given but I assume it to be $k<1.$
– Mike
Aug 28 at 13:26
Is the condition true for all $0<k<1?$
– Adrian Keister
Aug 28 at 13:27
 |Â
show 3 more comments
Any restrictions on $k?$
– Adrian Keister
Aug 28 at 13:08
@Adrian Keister: $k$ is assumed to be greater than $zero!$
– Mike
Aug 28 at 13:11
What I mean is, is $k$ arbitrary and not specified in advance? Or is it treated as a given?
– Adrian Keister
Aug 28 at 13:24
@Adrian Keister: Although, not given but I assume it to be $k<1.$
– Mike
Aug 28 at 13:26
Is the condition true for all $0<k<1?$
– Adrian Keister
Aug 28 at 13:27
Any restrictions on $k?$
– Adrian Keister
Aug 28 at 13:08
Any restrictions on $k?$
– Adrian Keister
Aug 28 at 13:08
@Adrian Keister: $k$ is assumed to be greater than $zero!$
– Mike
Aug 28 at 13:11
@Adrian Keister: $k$ is assumed to be greater than $zero!$
– Mike
Aug 28 at 13:11
What I mean is, is $k$ arbitrary and not specified in advance? Or is it treated as a given?
– Adrian Keister
Aug 28 at 13:24
What I mean is, is $k$ arbitrary and not specified in advance? Or is it treated as a given?
– Adrian Keister
Aug 28 at 13:24
@Adrian Keister: Although, not given but I assume it to be $k<1.$
– Mike
Aug 28 at 13:26
@Adrian Keister: Although, not given but I assume it to be $k<1.$
– Mike
Aug 28 at 13:26
Is the condition true for all $0<k<1?$
– Adrian Keister
Aug 28 at 13:27
Is the condition true for all $0<k<1?$
– Adrian Keister
Aug 28 at 13:27
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
$f$ being Lipschitz means, any solution to an initial value problem is unique (see e.g. https://en.wikipedia.org/wiki/Initial_value_problem#Existence_and_uniqueness_of_solutions)
Let us define a function $y(t):=x_0$, then $y'(t)=0$, since it is constant. We also have $y(0)=x_0$.
Furthermore $f(y(t))=f(x_0)=0$, hence $y'(t)=0=f(y(t))$. That means $y$ satisfies the differential equation and the initial conditions. Since solutions are unique, we are done, because then $x(t,x_0)=x_0$.
answered Aug 28 at 13:13
humanStampedist
2,036213
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
$f$ being Lipschitz means, any solution to an initial value problem is unique (see e.g. https://en.wikipedia.org/wiki/Initial_value_problem#Existence_and_uniqueness_of_solutions)
Let us define a function $y(t):=x_0$, then $y'(t)=0$, since it is constant. We also have $y(0)=x_0$.
Furthermore $f(y(t))=f(x_0)=0$, hence $y'(t)=0=f(y(t))$. That means $y$ satisfies the differential equation and the initial conditions. Since solutions are unique, we are done, because then $x(t,x_0)=x_0$.
answered Aug 28 at 13:13
humanStampedist
2,036213
add a comment |Â
up vote
4
down vote
accepted
$f$ being Lipschitz means, any solution to an initial value problem is unique (see e.g. https://en.wikipedia.org/wiki/Initial_value_problem#Existence_and_uniqueness_of_solutions)
Let us define a function $y(t):=x_0$, then $y'(t)=0$, since it is constant. We also have $y(0)=x_0$.
Furthermore $f(y(t))=f(x_0)=0$, hence $y'(t)=0=f(y(t))$. That means $y$ satisfies the differential equation and the initial conditions. Since solutions are unique, we are done, because then $x(t,x_0)=x_0$.
answered Aug 28 at 13:13
humanStampedist
2,036213
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
$f$ being Lipschitz means, any solution to an initial value problem is unique (see e.g. https://en.wikipedia.org/wiki/Initial_value_problem#Existence_and_uniqueness_of_solutions)
Let us define a function $y(t):=x_0$, then $y'(t)=0$, since it is constant. We also have $y(0)=x_0$.
Furthermore $f(y(t))=f(x_0)=0$, hence $y'(t)=0=f(y(t))$. That means $y$ satisfies the differential equation and the initial conditions. Since solutions are unique, we are done, because then $x(t,x_0)=x_0$.
answered Aug 28 at 13:13
humanStampedist
2,036213
$f$ being Lipschitz means, any solution to an initial value problem is unique (see e.g. https://en.wikipedia.org/wiki/Initial_value_problem#Existence_and_uniqueness_of_solutions)
Let us define a function $y(t):=x_0$, then $y'(t)=0$, since it is constant. We also have $y(0)=x_0$.
Furthermore $f(y(t))=f(x_0)=0$, hence $y'(t)=0=f(y(t))$. That means $y$ satisfies the differential equation and the initial conditions. Since solutions are unique, we are done, because then $x(t,x_0)=x_0$.
answered Aug 28 at 13:13
humanStampedist
2,036213
answered Aug 28 at 13:13
humanStampedist
2,036213
answered Aug 28 at 13:13
humanStampedist
2,036213
answered Aug 28 at 13:13
humanStampedist
2,036213
2,036213
add a comment |Â
add a comment |Â
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Any restrictions on $k?$
– Adrian Keister
Aug 28 at 13:08
@Adrian Keister: $k$ is assumed to be greater than $zero!$
– Mike
Aug 28 at 13:11
What I mean is, is $k$ arbitrary and not specified in advance? Or is it treated as a given?
– Adrian Keister
Aug 28 at 13:24
@Adrian Keister: Although, not given but I assume it to be $k<1.$
– Mike
Aug 28 at 13:26
Is the condition true for all $0<k<1?$
– Adrian Keister
Aug 28 at 13:27