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why can't a non-constant solution of an autonomous DE intersect an equilibrium solution?
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
asked Nov 21 '18 at 8:04
syeh_106
1,122813
add a comment |
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
asked Nov 21 '18 at 8:04
syeh_106
1,122813
add a comment |
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
asked Nov 21 '18 at 8:04
syeh_106
1,122813
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
asked Nov 21 '18 at 8:04
syeh_106
1,122813
asked Nov 21 '18 at 8:04
syeh_106
1,122813
edited Nov 21 '18 at 8:18
asked Nov 21 '18 at 8:04
syeh_106
1,122813
asked Nov 21 '18 at 8:04
syeh_106
1,122813
asked Nov 21 '18 at 8:04
syeh_106
1,122813
1,122813
add a comment |
add a comment |
1 Answer
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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.
answered Nov 21 '18 at 10:50
Christian Blatter
172k7112326
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 '18 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 '18 at 13:34
Thanks a lot! I appreciate the clarification.
– syeh_106
Nov 21 '18 at 13:37
add a comment |
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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.
answered Nov 21 '18 at 10:50
Christian Blatter
172k7112326
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 '18 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 '18 at 13:34
Thanks a lot! I appreciate the clarification.
– syeh_106
Nov 21 '18 at 13:37
add a comment |
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.
answered Nov 21 '18 at 10:50
Christian Blatter
172k7112326
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 '18 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 '18 at 13:34
Thanks a lot! I appreciate the clarification.
– syeh_106
Nov 21 '18 at 13:37
add a comment |
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.
answered Nov 21 '18 at 10:50
Christian Blatter
172k7112326
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.
answered Nov 21 '18 at 10:50
Christian Blatter
172k7112326
edited Nov 21 '18 at 13:28
answered Nov 21 '18 at 10:50
Christian Blatter
172k7112326
answered Nov 21 '18 at 10:50
Christian Blatter
172k7112326
answered Nov 21 '18 at 10:50
Christian Blatter
172k7112326
172k7112326
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 '18 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 '18 at 13:34
Thanks a lot! I appreciate the clarification.
– syeh_106
Nov 21 '18 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 '18 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 '18 at 13:34
Thanks a lot! I appreciate the clarification.
– syeh_106
Nov 21 '18 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 '18 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 '18 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 '18 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 '18 at 13:34
Thanks a lot! I appreciate the clarification.
– syeh_106
Nov 21 '18 at 13:37
Thanks a lot! I appreciate the clarification.
– syeh_106
Nov 21 '18 at 13:37
add a comment |
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