Does $f(x(s)) = 0$ imply that $x(t) = x(s)$ for $tgeq s$ in an autonomous ODE?
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Given an autonomous ODE $x' = f(x)$ and $x(0) = x_0$, do we necessarily have that if $f(x(s)) = 0$ then $x(t) = x(s)$ for $tgeq s$? I understand that this similar to finding equilibrium solutions, and I understand intuitively that since $f(x(s)) = x'(s)= 0$ then there will be no change in $x(t)$ for all times after $t$. But I am having trouble formally showing this.
For example, what prevents scenarios where $x(t) = (t-1)^3$, in which $x'(1) = 0$ but $x(t)neq x(1)$ for $tgeq 1$? Although I know that in this example, $x(t)$ is not autonomous, I was wondering how we can show that autonomous ODEs don't run into problems where $x(t)$ have local maxima or minima or inflection points.
ordinary-differential-equations
asked Jan 12 at 20:30
J. PistachioJ. Pistachio
478212
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$begingroup$
Given an autonomous ODE $x' = f(x)$ and $x(0) = x_0$, do we necessarily have that if $f(x(s)) = 0$ then $x(t) = x(s)$ for $tgeq s$? I understand that this similar to finding equilibrium solutions, and I understand intuitively that since $f(x(s)) = x'(s)= 0$ then there will be no change in $x(t)$ for all times after $t$. But I am having trouble formally showing this.
For example, what prevents scenarios where $x(t) = (t-1)^3$, in which $x'(1) = 0$ but $x(t)neq x(1)$ for $tgeq 1$? Although I know that in this example, $x(t)$ is not autonomous, I was wondering how we can show that autonomous ODEs don't run into problems where $x(t)$ have local maxima or minima or inflection points.
ordinary-differential-equations
asked Jan 12 at 20:30
J. PistachioJ. Pistachio
478212
$endgroup$
add a comment |
$begingroup$
Given an autonomous ODE $x' = f(x)$ and $x(0) = x_0$, do we necessarily have that if $f(x(s)) = 0$ then $x(t) = x(s)$ for $tgeq s$? I understand that this similar to finding equilibrium solutions, and I understand intuitively that since $f(x(s)) = x'(s)= 0$ then there will be no change in $x(t)$ for all times after $t$. But I am having trouble formally showing this.
For example, what prevents scenarios where $x(t) = (t-1)^3$, in which $x'(1) = 0$ but $x(t)neq x(1)$ for $tgeq 1$? Although I know that in this example, $x(t)$ is not autonomous, I was wondering how we can show that autonomous ODEs don't run into problems where $x(t)$ have local maxima or minima or inflection points.
ordinary-differential-equations
asked Jan 12 at 20:30
J. PistachioJ. Pistachio
478212
$endgroup$
Given an autonomous ODE $x' = f(x)$ and $x(0) = x_0$, do we necessarily have that if $f(x(s)) = 0$ then $x(t) = x(s)$ for $tgeq s$? I understand that this similar to finding equilibrium solutions, and I understand intuitively that since $f(x(s)) = x'(s)= 0$ then there will be no change in $x(t)$ for all times after $t$. But I am having trouble formally showing this.
For example, what prevents scenarios where $x(t) = (t-1)^3$, in which $x'(1) = 0$ but $x(t)neq x(1)$ for $tgeq 1$? Although I know that in this example, $x(t)$ is not autonomous, I was wondering how we can show that autonomous ODEs don't run into problems where $x(t)$ have local maxima or minima or inflection points.
ordinary-differential-equations
ordinary-differential-equations
asked Jan 12 at 20:30
J. PistachioJ. Pistachio
478212
asked Jan 12 at 20:30
J. PistachioJ. Pistachio
478212
asked Jan 12 at 20:30
J. PistachioJ. Pistachio
478212
asked Jan 12 at 20:30
J. PistachioJ. Pistachio
478212
asked Jan 12 at 20:30
J. PistachioJ. Pistachio
478212
478212
add a comment |
add a comment |
1 Answer
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This only applies if $f$ is differentiable in that root, or at least Lipschitz in some interval around it.
See the usual counter example $x'=2sqrt{|x|}$ where $x(t)=0$ for $t<c$ and $x(t)=(t-c)^2$ for $tge c$ are all solutions for any $c>0$.
answered Jan 12 at 21:19
LutzLLutzL
59.6k42057
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1 Answer
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1 Answer
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$begingroup$
This only applies if $f$ is differentiable in that root, or at least Lipschitz in some interval around it.
See the usual counter example $x'=2sqrt{|x|}$ where $x(t)=0$ for $t<c$ and $x(t)=(t-c)^2$ for $tge c$ are all solutions for any $c>0$.
answered Jan 12 at 21:19
LutzLLutzL
59.6k42057
$endgroup$
add a comment |
$begingroup$
This only applies if $f$ is differentiable in that root, or at least Lipschitz in some interval around it.
See the usual counter example $x'=2sqrt{|x|}$ where $x(t)=0$ for $t<c$ and $x(t)=(t-c)^2$ for $tge c$ are all solutions for any $c>0$.
answered Jan 12 at 21:19
LutzLLutzL
59.6k42057
$endgroup$
add a comment |
$begingroup$
This only applies if $f$ is differentiable in that root, or at least Lipschitz in some interval around it.
See the usual counter example $x'=2sqrt{|x|}$ where $x(t)=0$ for $t<c$ and $x(t)=(t-c)^2$ for $tge c$ are all solutions for any $c>0$.
answered Jan 12 at 21:19
LutzLLutzL
59.6k42057
$endgroup$
This only applies if $f$ is differentiable in that root, or at least Lipschitz in some interval around it.
See the usual counter example $x'=2sqrt{|x|}$ where $x(t)=0$ for $t<c$ and $x(t)=(t-c)^2$ for $tge c$ are all solutions for any $c>0$.
answered Jan 12 at 21:19
LutzLLutzL
59.6k42057
answered Jan 12 at 21:19
LutzLLutzL
59.6k42057
answered Jan 12 at 21:19
LutzLLutzL
59.6k42057
answered Jan 12 at 21:19
LutzLLutzL
59.6k42057
59.6k42057
add a comment |
add a comment |
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