why can't a non-constant solution of an autonomous DE intersect an equilibrium solution?
up vote
0
down vote
favorite
This question is related to Can a non-constant solution of DE intersect an equilibrium solution. and Why can't solutions to Autonomous ODE intersect?, where I didn't find sufficient details to resolve my question.
Specifically, the question is: for an autonomous DE, $dy/dx=f(y)$, with a critical point $c$, why can't a non-constant solution $y(x)$ intersect the equilibrium solution, $y=c$? (Assume $f$ and $f'$ are continuous functions of $y$.)
I am a beginner at differential equations, and am reading Zill and Wright's book on this topic. While I find this result plausible in light of the existence and uniqueness theorem of solutions (see below), I'm having difficulty coming up with a complete & rigorous proof. I state my attempt below, and would greatly appreciate it if someone'd confirm or refute it.
My attempted proof: Suppose $y(x)$ is a solution to the autonomous DE, $dy/dx=f(y)$, and $y(alpha)=c$ for some $alpha in mathbb R.$ I show that we must have $y(x)=c$ as follows.
Since $f$ and $f'$ are continuous functions (of $y$), by the uniqueness theorem below, $y(x)=c$ is the unique solution to $dy/dx=f(y), y(alpha)=c$ on interval $(alpha-h, alpha+h)$ for some $h>0.$ Let $beta=sup{t: y(x)=c, x in [alpha, t)}.$ Suppose $beta < infty$. Then $y(beta)ne c$, by the uniqueness theorem again. Hence $y(beta)=c+Delta, Deltane 0.$ But this means $y(x)$ is discontinuous at $beta$, contradicting the fact that $y(x)$ has to be continuous. So $beta$ has to be $infty$, and $y(x)=c$ on $[alpha, infty)$. The other half, $y(x)=c$ on $(-infty, alpha]$, is proved similarly.
Theorem Let $dy/dx=f(x,y), R={(x, y):ale xle b, c le y le d}$, and $(x_0, y_0)in R$. If $f$ and $partial f/partial y$ are continuous on $R$, then there exists a unique solution $y(x)$ on $(x_0-h, x_0+h)$ for some $h>0$ to the initial value problem, $dy/dx=f(x,y), y(x_0)=y_0.$
Is this proof correct? Is there a simpler proof? Thanks a lot!
differential-equations proof-verification
add a comment |
up vote
0
down vote
favorite
This question is related to Can a non-constant solution of DE intersect an equilibrium solution. and Why can't solutions to Autonomous ODE intersect?, where I didn't find sufficient details to resolve my question.
Specifically, the question is: for an autonomous DE, $dy/dx=f(y)$, with a critical point $c$, why can't a non-constant solution $y(x)$ intersect the equilibrium solution, $y=c$? (Assume $f$ and $f'$ are continuous functions of $y$.)
I am a beginner at differential equations, and am reading Zill and Wright's book on this topic. While I find this result plausible in light of the existence and uniqueness theorem of solutions (see below), I'm having difficulty coming up with a complete & rigorous proof. I state my attempt below, and would greatly appreciate it if someone'd confirm or refute it.
My attempted proof: Suppose $y(x)$ is a solution to the autonomous DE, $dy/dx=f(y)$, and $y(alpha)=c$ for some $alpha in mathbb R.$ I show that we must have $y(x)=c$ as follows.
Since $f$ and $f'$ are continuous functions (of $y$), by the uniqueness theorem below, $y(x)=c$ is the unique solution to $dy/dx=f(y), y(alpha)=c$ on interval $(alpha-h, alpha+h)$ for some $h>0.$ Let $beta=sup{t: y(x)=c, x in [alpha, t)}.$ Suppose $beta < infty$. Then $y(beta)ne c$, by the uniqueness theorem again. Hence $y(beta)=c+Delta, Deltane 0.$ But this means $y(x)$ is discontinuous at $beta$, contradicting the fact that $y(x)$ has to be continuous. So $beta$ has to be $infty$, and $y(x)=c$ on $[alpha, infty)$. The other half, $y(x)=c$ on $(-infty, alpha]$, is proved similarly.
Theorem Let $dy/dx=f(x,y), R={(x, y):ale xle b, c le y le d}$, and $(x_0, y_0)in R$. If $f$ and $partial f/partial y$ are continuous on $R$, then there exists a unique solution $y(x)$ on $(x_0-h, x_0+h)$ for some $h>0$ to the initial value problem, $dy/dx=f(x,y), y(x_0)=y_0.$
Is this proof correct? Is there a simpler proof? Thanks a lot!
differential-equations proof-verification
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question is related to Can a non-constant solution of DE intersect an equilibrium solution. and Why can't solutions to Autonomous ODE intersect?, where I didn't find sufficient details to resolve my question.
Specifically, the question is: for an autonomous DE, $dy/dx=f(y)$, with a critical point $c$, why can't a non-constant solution $y(x)$ intersect the equilibrium solution, $y=c$? (Assume $f$ and $f'$ are continuous functions of $y$.)
I am a beginner at differential equations, and am reading Zill and Wright's book on this topic. While I find this result plausible in light of the existence and uniqueness theorem of solutions (see below), I'm having difficulty coming up with a complete & rigorous proof. I state my attempt below, and would greatly appreciate it if someone'd confirm or refute it.
My attempted proof: Suppose $y(x)$ is a solution to the autonomous DE, $dy/dx=f(y)$, and $y(alpha)=c$ for some $alpha in mathbb R.$ I show that we must have $y(x)=c$ as follows.
Since $f$ and $f'$ are continuous functions (of $y$), by the uniqueness theorem below, $y(x)=c$ is the unique solution to $dy/dx=f(y), y(alpha)=c$ on interval $(alpha-h, alpha+h)$ for some $h>0.$ Let $beta=sup{t: y(x)=c, x in [alpha, t)}.$ Suppose $beta < infty$. Then $y(beta)ne c$, by the uniqueness theorem again. Hence $y(beta)=c+Delta, Deltane 0.$ But this means $y(x)$ is discontinuous at $beta$, contradicting the fact that $y(x)$ has to be continuous. So $beta$ has to be $infty$, and $y(x)=c$ on $[alpha, infty)$. The other half, $y(x)=c$ on $(-infty, alpha]$, is proved similarly.
Theorem Let $dy/dx=f(x,y), R={(x, y):ale xle b, c le y le d}$, and $(x_0, y_0)in R$. If $f$ and $partial f/partial y$ are continuous on $R$, then there exists a unique solution $y(x)$ on $(x_0-h, x_0+h)$ for some $h>0$ to the initial value problem, $dy/dx=f(x,y), y(x_0)=y_0.$
Is this proof correct? Is there a simpler proof? Thanks a lot!
differential-equations proof-verification
This question is related to Can a non-constant solution of DE intersect an equilibrium solution. and Why can't solutions to Autonomous ODE intersect?, where I didn't find sufficient details to resolve my question.
Specifically, the question is: for an autonomous DE, $dy/dx=f(y)$, with a critical point $c$, why can't a non-constant solution $y(x)$ intersect the equilibrium solution, $y=c$? (Assume $f$ and $f'$ are continuous functions of $y$.)
I am a beginner at differential equations, and am reading Zill and Wright's book on this topic. While I find this result plausible in light of the existence and uniqueness theorem of solutions (see below), I'm having difficulty coming up with a complete & rigorous proof. I state my attempt below, and would greatly appreciate it if someone'd confirm or refute it.
My attempted proof: Suppose $y(x)$ is a solution to the autonomous DE, $dy/dx=f(y)$, and $y(alpha)=c$ for some $alpha in mathbb R.$ I show that we must have $y(x)=c$ as follows.
Since $f$ and $f'$ are continuous functions (of $y$), by the uniqueness theorem below, $y(x)=c$ is the unique solution to $dy/dx=f(y), y(alpha)=c$ on interval $(alpha-h, alpha+h)$ for some $h>0.$ Let $beta=sup{t: y(x)=c, x in [alpha, t)}.$ Suppose $beta < infty$. Then $y(beta)ne c$, by the uniqueness theorem again. Hence $y(beta)=c+Delta, Deltane 0.$ But this means $y(x)$ is discontinuous at $beta$, contradicting the fact that $y(x)$ has to be continuous. So $beta$ has to be $infty$, and $y(x)=c$ on $[alpha, infty)$. The other half, $y(x)=c$ on $(-infty, alpha]$, is proved similarly.
Theorem Let $dy/dx=f(x,y), R={(x, y):ale xle b, c le y le d}$, and $(x_0, y_0)in R$. If $f$ and $partial f/partial y$ are continuous on $R$, then there exists a unique solution $y(x)$ on $(x_0-h, x_0+h)$ for some $h>0$ to the initial value problem, $dy/dx=f(x,y), y(x_0)=y_0.$
Is this proof correct? Is there a simpler proof? Thanks a lot!
differential-equations proof-verification
differential-equations proof-verification
edited Nov 21 at 8:18
asked Nov 21 at 8:04
syeh_106
1,105813
1,105813
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
It is fine to invoke the existence and uniqueness theorem, but it is not necessary to go into sup or $epsilon/delta$ discussions. It is sufficient to say "basta". At a hypothetical crossing point $(x_0,c)$ of the constant solution $xmapsto y_0(x)equiv c$ and some other solution $xmapsto y_1(x)$ that satisfies $y_1(x_0)=c$, but is not $equiv y_0(x)$ for $x_0-h< x<x_0+h$, we would have a violation of the uniqueness part of the theorem.
Thank you for the answer, but I'm a bit confused. Did you mean $y_1ne y_0$ on $(x_0-h, x_0+h)$? What if $y_1=y_0$ on $(x_0-h, x_0+h)$, but $y_1ne y_0$ elsewhere?
– syeh_106
Nov 21 at 13:20
1
In the question you asked whether some "other solution" could cross the graph of $y_0(cdot)$. I showed that this is not the case. "Global uniqueness" is some other matter, and needs a proof along the lines you have proposed.
– Christian Blatter
Nov 21 at 13:34
Thanks a lot! I appreciate the clarification.
– syeh_106
Nov 21 at 13:37
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
It is fine to invoke the existence and uniqueness theorem, but it is not necessary to go into sup or $epsilon/delta$ discussions. It is sufficient to say "basta". At a hypothetical crossing point $(x_0,c)$ of the constant solution $xmapsto y_0(x)equiv c$ and some other solution $xmapsto y_1(x)$ that satisfies $y_1(x_0)=c$, but is not $equiv y_0(x)$ for $x_0-h< x<x_0+h$, we would have a violation of the uniqueness part of the theorem.
Thank you for the answer, but I'm a bit confused. Did you mean $y_1ne y_0$ on $(x_0-h, x_0+h)$? What if $y_1=y_0$ on $(x_0-h, x_0+h)$, but $y_1ne y_0$ elsewhere?
– syeh_106
Nov 21 at 13:20
1
In the question you asked whether some "other solution" could cross the graph of $y_0(cdot)$. I showed that this is not the case. "Global uniqueness" is some other matter, and needs a proof along the lines you have proposed.
– Christian Blatter
Nov 21 at 13:34
Thanks a lot! I appreciate the clarification.
– syeh_106
Nov 21 at 13:37
add a comment |
up vote
1
down vote
It is fine to invoke the existence and uniqueness theorem, but it is not necessary to go into sup or $epsilon/delta$ discussions. It is sufficient to say "basta". At a hypothetical crossing point $(x_0,c)$ of the constant solution $xmapsto y_0(x)equiv c$ and some other solution $xmapsto y_1(x)$ that satisfies $y_1(x_0)=c$, but is not $equiv y_0(x)$ for $x_0-h< x<x_0+h$, we would have a violation of the uniqueness part of the theorem.
Thank you for the answer, but I'm a bit confused. Did you mean $y_1ne y_0$ on $(x_0-h, x_0+h)$? What if $y_1=y_0$ on $(x_0-h, x_0+h)$, but $y_1ne y_0$ elsewhere?
– syeh_106
Nov 21 at 13:20
1
In the question you asked whether some "other solution" could cross the graph of $y_0(cdot)$. I showed that this is not the case. "Global uniqueness" is some other matter, and needs a proof along the lines you have proposed.
– Christian Blatter
Nov 21 at 13:34
Thanks a lot! I appreciate the clarification.
– syeh_106
Nov 21 at 13:37
add a comment |
up vote
1
down vote
up vote
1
down vote
It is fine to invoke the existence and uniqueness theorem, but it is not necessary to go into sup or $epsilon/delta$ discussions. It is sufficient to say "basta". At a hypothetical crossing point $(x_0,c)$ of the constant solution $xmapsto y_0(x)equiv c$ and some other solution $xmapsto y_1(x)$ that satisfies $y_1(x_0)=c$, but is not $equiv y_0(x)$ for $x_0-h< x<x_0+h$, we would have a violation of the uniqueness part of the theorem.
It is fine to invoke the existence and uniqueness theorem, but it is not necessary to go into sup or $epsilon/delta$ discussions. It is sufficient to say "basta". At a hypothetical crossing point $(x_0,c)$ of the constant solution $xmapsto y_0(x)equiv c$ and some other solution $xmapsto y_1(x)$ that satisfies $y_1(x_0)=c$, but is not $equiv y_0(x)$ for $x_0-h< x<x_0+h$, we would have a violation of the uniqueness part of the theorem.
edited Nov 21 at 13:28
answered Nov 21 at 10:50
Christian Blatter
171k7111325
171k7111325
Thank you for the answer, but I'm a bit confused. Did you mean $y_1ne y_0$ on $(x_0-h, x_0+h)$? What if $y_1=y_0$ on $(x_0-h, x_0+h)$, but $y_1ne y_0$ elsewhere?
– syeh_106
Nov 21 at 13:20
1
In the question you asked whether some "other solution" could cross the graph of $y_0(cdot)$. I showed that this is not the case. "Global uniqueness" is some other matter, and needs a proof along the lines you have proposed.
– Christian Blatter
Nov 21 at 13:34
Thanks a lot! I appreciate the clarification.
– syeh_106
Nov 21 at 13:37
add a comment |
Thank you for the answer, but I'm a bit confused. Did you mean $y_1ne y_0$ on $(x_0-h, x_0+h)$? What if $y_1=y_0$ on $(x_0-h, x_0+h)$, but $y_1ne y_0$ elsewhere?
– syeh_106
Nov 21 at 13:20
1
In the question you asked whether some "other solution" could cross the graph of $y_0(cdot)$. I showed that this is not the case. "Global uniqueness" is some other matter, and needs a proof along the lines you have proposed.
– Christian Blatter
Nov 21 at 13:34
Thanks a lot! I appreciate the clarification.
– syeh_106
Nov 21 at 13:37
Thank you for the answer, but I'm a bit confused. Did you mean $y_1ne y_0$ on $(x_0-h, x_0+h)$? What if $y_1=y_0$ on $(x_0-h, x_0+h)$, but $y_1ne y_0$ elsewhere?
– syeh_106
Nov 21 at 13:20
Thank you for the answer, but I'm a bit confused. Did you mean $y_1ne y_0$ on $(x_0-h, x_0+h)$? What if $y_1=y_0$ on $(x_0-h, x_0+h)$, but $y_1ne y_0$ elsewhere?
– syeh_106
Nov 21 at 13:20
1
1
In the question you asked whether some "other solution" could cross the graph of $y_0(cdot)$. I showed that this is not the case. "Global uniqueness" is some other matter, and needs a proof along the lines you have proposed.
– Christian Blatter
Nov 21 at 13:34
In the question you asked whether some "other solution" could cross the graph of $y_0(cdot)$. I showed that this is not the case. "Global uniqueness" is some other matter, and needs a proof along the lines you have proposed.
– Christian Blatter
Nov 21 at 13:34
Thanks a lot! I appreciate the clarification.
– syeh_106
Nov 21 at 13:37
Thanks a lot! I appreciate the clarification.
– syeh_106
Nov 21 at 13:37
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3007411%2fwhy-cant-a-non-constant-solution-of-an-autonomous-de-intersect-an-equilibrium-s%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown