Has this I.V.P. unique solution?
Clash Royale CLAN TAG#URR8PPP
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I know
$$begincases yâÂÂ=|y|\ y(0)=0 endcases $$
Has the solution $yequiv 0$... does it have another?
The question is about the theorem of existence and uniqueness because $partial_y f(x,y)$ is not continuous
differential-equations
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up vote
9
down vote
favorite
I know
$$begincases yâÂÂ=|y|\ y(0)=0 endcases $$
Has the solution $yequiv 0$... does it have another?
The question is about the theorem of existence and uniqueness because $partial_y f(x,y)$ is not continuous
differential-equations
add a comment |Â
up vote
9
down vote
favorite
up vote
9
down vote
favorite
I know
$$begincases yâÂÂ=|y|\ y(0)=0 endcases $$
Has the solution $yequiv 0$... does it have another?
The question is about the theorem of existence and uniqueness because $partial_y f(x,y)$ is not continuous
differential-equations
I know
$$begincases yâÂÂ=|y|\ y(0)=0 endcases $$
Has the solution $yequiv 0$... does it have another?
The question is about the theorem of existence and uniqueness because $partial_y f(x,y)$ is not continuous
differential-equations
differential-equations
asked Aug 17 at 3:35
Rubén Tobar
1046
1046
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2 Answers
2
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oldest
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10
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accepted
With
$f(y) = vert y vert, tag 1$
we have
$vert f(y_1) - f(y_2) vert = vert vert y_1 vert - vert y_2 vert vert le vert y_1 - y_2 vert, tag 2$
which shows that $f(y) = vert y vert$ is Lipschitz continuous with Lipschitz constant $1$; therefore by the Picard-Lindelof theorem, the equation
$dot y = vert y vert tag 3$
has a unique solution in some interval about any initial point $y_0 = y(t_0)$. So the only solution with $y(0) = 0$ is identically zero, $y(t) equiv 0$.
This shouldn't really come as too much of a surprise since $vert 0 vert = 0$, so $0$ is an equilibrium of (3).
Finally, we don't need $f(y)$ differentiable, merely Lipschitz continuous, for existence and uniqueness to apply.
Note Added in Edit, Friday 17 August 2018 11:14 PM PST: To really flesh this answer out, we should probably also show that the unique solution $y(t) equiv 0$ on an interval about $0$ extends uniquely to $(-infty, infty)$, but I'm leaving this discussion for a later date. End of Note.
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up vote
1
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Direct proof: any solution $y$ will be nondecreasing as $y'(x) = |y(x)|ge 0$. Then $y(x)ge 0$ for $xge 0$ and $y'(x) = y(x)$ for $xge 0$. Integrating,
$$y(x) = y(x) - y(0) = int_0^x y = 0 + int_0^x y.$$
Now, applying the Gronwall lemma, $y(x)le 0$ for $xge 0$.
EDIT: even more direct (Gronwall-less) proof.
Take $x_0in(0,1)$. Integrating as before and using that $y$ is nondecreasing,
$$y(x_0) = int_0^x_0 yle x_0,y(x_0)implies y(x_0) = 0,$$
i.e., $yequiv 0$ in $[0,1)$, so in $[0,1]$ by continuity. The same argument works in $[1,2]$, $[2,3]$,...
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
accepted
With
$f(y) = vert y vert, tag 1$
we have
$vert f(y_1) - f(y_2) vert = vert vert y_1 vert - vert y_2 vert vert le vert y_1 - y_2 vert, tag 2$
which shows that $f(y) = vert y vert$ is Lipschitz continuous with Lipschitz constant $1$; therefore by the Picard-Lindelof theorem, the equation
$dot y = vert y vert tag 3$
has a unique solution in some interval about any initial point $y_0 = y(t_0)$. So the only solution with $y(0) = 0$ is identically zero, $y(t) equiv 0$.
This shouldn't really come as too much of a surprise since $vert 0 vert = 0$, so $0$ is an equilibrium of (3).
Finally, we don't need $f(y)$ differentiable, merely Lipschitz continuous, for existence and uniqueness to apply.
Note Added in Edit, Friday 17 August 2018 11:14 PM PST: To really flesh this answer out, we should probably also show that the unique solution $y(t) equiv 0$ on an interval about $0$ extends uniquely to $(-infty, infty)$, but I'm leaving this discussion for a later date. End of Note.
add a comment |Â
up vote
10
down vote
accepted
With
$f(y) = vert y vert, tag 1$
we have
$vert f(y_1) - f(y_2) vert = vert vert y_1 vert - vert y_2 vert vert le vert y_1 - y_2 vert, tag 2$
which shows that $f(y) = vert y vert$ is Lipschitz continuous with Lipschitz constant $1$; therefore by the Picard-Lindelof theorem, the equation
$dot y = vert y vert tag 3$
has a unique solution in some interval about any initial point $y_0 = y(t_0)$. So the only solution with $y(0) = 0$ is identically zero, $y(t) equiv 0$.
This shouldn't really come as too much of a surprise since $vert 0 vert = 0$, so $0$ is an equilibrium of (3).
Finally, we don't need $f(y)$ differentiable, merely Lipschitz continuous, for existence and uniqueness to apply.
Note Added in Edit, Friday 17 August 2018 11:14 PM PST: To really flesh this answer out, we should probably also show that the unique solution $y(t) equiv 0$ on an interval about $0$ extends uniquely to $(-infty, infty)$, but I'm leaving this discussion for a later date. End of Note.
add a comment |Â
up vote
10
down vote
accepted
up vote
10
down vote
accepted
With
$f(y) = vert y vert, tag 1$
we have
$vert f(y_1) - f(y_2) vert = vert vert y_1 vert - vert y_2 vert vert le vert y_1 - y_2 vert, tag 2$
which shows that $f(y) = vert y vert$ is Lipschitz continuous with Lipschitz constant $1$; therefore by the Picard-Lindelof theorem, the equation
$dot y = vert y vert tag 3$
has a unique solution in some interval about any initial point $y_0 = y(t_0)$. So the only solution with $y(0) = 0$ is identically zero, $y(t) equiv 0$.
This shouldn't really come as too much of a surprise since $vert 0 vert = 0$, so $0$ is an equilibrium of (3).
Finally, we don't need $f(y)$ differentiable, merely Lipschitz continuous, for existence and uniqueness to apply.
Note Added in Edit, Friday 17 August 2018 11:14 PM PST: To really flesh this answer out, we should probably also show that the unique solution $y(t) equiv 0$ on an interval about $0$ extends uniquely to $(-infty, infty)$, but I'm leaving this discussion for a later date. End of Note.
With
$f(y) = vert y vert, tag 1$
we have
$vert f(y_1) - f(y_2) vert = vert vert y_1 vert - vert y_2 vert vert le vert y_1 - y_2 vert, tag 2$
which shows that $f(y) = vert y vert$ is Lipschitz continuous with Lipschitz constant $1$; therefore by the Picard-Lindelof theorem, the equation
$dot y = vert y vert tag 3$
has a unique solution in some interval about any initial point $y_0 = y(t_0)$. So the only solution with $y(0) = 0$ is identically zero, $y(t) equiv 0$.
This shouldn't really come as too much of a surprise since $vert 0 vert = 0$, so $0$ is an equilibrium of (3).
Finally, we don't need $f(y)$ differentiable, merely Lipschitz continuous, for existence and uniqueness to apply.
Note Added in Edit, Friday 17 August 2018 11:14 PM PST: To really flesh this answer out, we should probably also show that the unique solution $y(t) equiv 0$ on an interval about $0$ extends uniquely to $(-infty, infty)$, but I'm leaving this discussion for a later date. End of Note.
edited Aug 18 at 6:16
answered Aug 17 at 3:46
Robert Lewis
38.8k22358
38.8k22358
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up vote
1
down vote
Direct proof: any solution $y$ will be nondecreasing as $y'(x) = |y(x)|ge 0$. Then $y(x)ge 0$ for $xge 0$ and $y'(x) = y(x)$ for $xge 0$. Integrating,
$$y(x) = y(x) - y(0) = int_0^x y = 0 + int_0^x y.$$
Now, applying the Gronwall lemma, $y(x)le 0$ for $xge 0$.
EDIT: even more direct (Gronwall-less) proof.
Take $x_0in(0,1)$. Integrating as before and using that $y$ is nondecreasing,
$$y(x_0) = int_0^x_0 yle x_0,y(x_0)implies y(x_0) = 0,$$
i.e., $yequiv 0$ in $[0,1)$, so in $[0,1]$ by continuity. The same argument works in $[1,2]$, $[2,3]$,...
add a comment |Â
up vote
1
down vote
Direct proof: any solution $y$ will be nondecreasing as $y'(x) = |y(x)|ge 0$. Then $y(x)ge 0$ for $xge 0$ and $y'(x) = y(x)$ for $xge 0$. Integrating,
$$y(x) = y(x) - y(0) = int_0^x y = 0 + int_0^x y.$$
Now, applying the Gronwall lemma, $y(x)le 0$ for $xge 0$.
EDIT: even more direct (Gronwall-less) proof.
Take $x_0in(0,1)$. Integrating as before and using that $y$ is nondecreasing,
$$y(x_0) = int_0^x_0 yle x_0,y(x_0)implies y(x_0) = 0,$$
i.e., $yequiv 0$ in $[0,1)$, so in $[0,1]$ by continuity. The same argument works in $[1,2]$, $[2,3]$,...
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Direct proof: any solution $y$ will be nondecreasing as $y'(x) = |y(x)|ge 0$. Then $y(x)ge 0$ for $xge 0$ and $y'(x) = y(x)$ for $xge 0$. Integrating,
$$y(x) = y(x) - y(0) = int_0^x y = 0 + int_0^x y.$$
Now, applying the Gronwall lemma, $y(x)le 0$ for $xge 0$.
EDIT: even more direct (Gronwall-less) proof.
Take $x_0in(0,1)$. Integrating as before and using that $y$ is nondecreasing,
$$y(x_0) = int_0^x_0 yle x_0,y(x_0)implies y(x_0) = 0,$$
i.e., $yequiv 0$ in $[0,1)$, so in $[0,1]$ by continuity. The same argument works in $[1,2]$, $[2,3]$,...
Direct proof: any solution $y$ will be nondecreasing as $y'(x) = |y(x)|ge 0$. Then $y(x)ge 0$ for $xge 0$ and $y'(x) = y(x)$ for $xge 0$. Integrating,
$$y(x) = y(x) - y(0) = int_0^x y = 0 + int_0^x y.$$
Now, applying the Gronwall lemma, $y(x)le 0$ for $xge 0$.
EDIT: even more direct (Gronwall-less) proof.
Take $x_0in(0,1)$. Integrating as before and using that $y$ is nondecreasing,
$$y(x_0) = int_0^x_0 yle x_0,y(x_0)implies y(x_0) = 0,$$
i.e., $yequiv 0$ in $[0,1)$, so in $[0,1]$ by continuity. The same argument works in $[1,2]$, $[2,3]$,...
edited Aug 21 at 16:49
answered Aug 17 at 14:04
MartÃn-Blas Pérez Pinilla
33.5k42570
33.5k42570
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