Limit of a solution to a differential equation is a steady state.
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Suppose we have an initial value problem
$$dot{x}=f(x),$$
$$x(0)=x_0$$
where $fin C^1(E)$ for some open $E$. Moreover, suppose we have a solution $x(t)$ such that
$$underset{trightarrowinfty}{lim}x(t)=X$$
for $xin E$. The goal is to show that $f(X)=0$.
My initial idea is taking a limit of the differential equation, because continuity of $f$ gives $f(X)$ in the right-hand side. The problem I have with this approach is applying the limit to the left-hand side. I have also heard Mean Value Theorem can help with this but I don't see how. Anyone have guidance for this? Thanks in advance.
differential-equations limits analysis steady-state
asked Nov 19 at 2:36
user292256
424
add a comment |
up vote
0
down vote
favorite
Suppose we have an initial value problem
$$dot{x}=f(x),$$
$$x(0)=x_0$$
where $fin C^1(E)$ for some open $E$. Moreover, suppose we have a solution $x(t)$ such that
$$underset{trightarrowinfty}{lim}x(t)=X$$
for $xin E$. The goal is to show that $f(X)=0$.
My initial idea is taking a limit of the differential equation, because continuity of $f$ gives $f(X)$ in the right-hand side. The problem I have with this approach is applying the limit to the left-hand side. I have also heard Mean Value Theorem can help with this but I don't see how. Anyone have guidance for this? Thanks in advance.
differential-equations limits analysis steady-state
asked Nov 19 at 2:36
user292256
424
A more general question was asked and answered very recently, see Show that $p$ is a stationary solution.
– user539887
Nov 19 at 9:58
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose we have an initial value problem
$$dot{x}=f(x),$$
$$x(0)=x_0$$
where $fin C^1(E)$ for some open $E$. Moreover, suppose we have a solution $x(t)$ such that
$$underset{trightarrowinfty}{lim}x(t)=X$$
for $xin E$. The goal is to show that $f(X)=0$.
My initial idea is taking a limit of the differential equation, because continuity of $f$ gives $f(X)$ in the right-hand side. The problem I have with this approach is applying the limit to the left-hand side. I have also heard Mean Value Theorem can help with this but I don't see how. Anyone have guidance for this? Thanks in advance.
differential-equations limits analysis steady-state
asked Nov 19 at 2:36
user292256
424
Suppose we have an initial value problem
$$dot{x}=f(x),$$
$$x(0)=x_0$$
where $fin C^1(E)$ for some open $E$. Moreover, suppose we have a solution $x(t)$ such that
$$underset{trightarrowinfty}{lim}x(t)=X$$
for $xin E$. The goal is to show that $f(X)=0$.
My initial idea is taking a limit of the differential equation, because continuity of $f$ gives $f(X)$ in the right-hand side. The problem I have with this approach is applying the limit to the left-hand side. I have also heard Mean Value Theorem can help with this but I don't see how. Anyone have guidance for this? Thanks in advance.
differential-equations limits analysis steady-state
differential-equations limits analysis steady-state
asked Nov 19 at 2:36
user292256
424
asked Nov 19 at 2:36
user292256
424
asked Nov 19 at 2:36
user292256
424
asked Nov 19 at 2:36
user292256
424
asked Nov 19 at 2:36
user292256
424
424
A more general question was asked and answered very recently, see Show that $p$ is a stationary solution.
– user539887
Nov 19 at 9:58
add a comment |
A more general question was asked and answered very recently, see Show that $p$ is a stationary solution.
– user539887
Nov 19 at 9:58
A more general question was asked and answered very recently, see Show that $p$ is a stationary solution.
– user539887
Nov 19 at 9:58
A more general question was asked and answered very recently, see Show that $p$ is a stationary solution.
– user539887
Nov 19 at 9:58
add a comment |
1 Answer
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I believe you are correct about the mean value theorem, but I will, effectively, be using an integral form instead of the differential form.
The average rate of change is easier to work with than the derivative for precisely the limit reason you mentioned. Integrating the equation and dividing, we recover the average value
$$
frac{x(b)-x(a)}{b-a}=frac{1}{b-a}int_a^bf(x(t))dt.
$$
Extract $f(X)$:
$$
frac{x(b)-x(a)}{b-a}=frac{1}{b-a}int_a^b[f(x(t))-f(X)]dt+f(X).
$$
Now put $b=2a$ and send $atoinfty$. The left hand side vanishes, and the first term on the right hand side vanishes if we apply the integral mean value theorem to the continuous function $f$:
$$
frac{1}{a}int_a^{2a}[f(x(t))-f(X)]dt=f(x(s))-f(X),
$$
for some $sin[a,2a]$. This difference is small for all such $s$ if $a$ is large.
answered Nov 19 at 3:13
user254433
2,456712
1
Another way, which requires less calculation: Clearly the limit of $f(x(t))$ exists; this means the limit of $x'(t)$ exists. If this limit is anything but 0, for example $lim x'(t)=y>0$, then applying FTC for large $a$, we have $x(2a)=x(a)+int_a^{2a}x'(s)dsge x(a)+acdot (y-epsilon)$ for some small $epsilon$, which contradicts $x(2a)-x(a)to 0$.
– user254433
Nov 19 at 3:26
1
Thanks both! Very helpful!
– user292256
Nov 19 at 4:23
@user292256 Happy to help!
– user254433
Nov 19 at 5:20
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
accepted
I believe you are correct about the mean value theorem, but I will, effectively, be using an integral form instead of the differential form.
The average rate of change is easier to work with than the derivative for precisely the limit reason you mentioned. Integrating the equation and dividing, we recover the average value
$$
frac{x(b)-x(a)}{b-a}=frac{1}{b-a}int_a^bf(x(t))dt.
$$
Extract $f(X)$:
$$
frac{x(b)-x(a)}{b-a}=frac{1}{b-a}int_a^b[f(x(t))-f(X)]dt+f(X).
$$
Now put $b=2a$ and send $atoinfty$. The left hand side vanishes, and the first term on the right hand side vanishes if we apply the integral mean value theorem to the continuous function $f$:
$$
frac{1}{a}int_a^{2a}[f(x(t))-f(X)]dt=f(x(s))-f(X),
$$
for some $sin[a,2a]$. This difference is small for all such $s$ if $a$ is large.
answered Nov 19 at 3:13
user254433
2,456712
1
Another way, which requires less calculation: Clearly the limit of $f(x(t))$ exists; this means the limit of $x'(t)$ exists. If this limit is anything but 0, for example $lim x'(t)=y>0$, then applying FTC for large $a$, we have $x(2a)=x(a)+int_a^{2a}x'(s)dsge x(a)+acdot (y-epsilon)$ for some small $epsilon$, which contradicts $x(2a)-x(a)to 0$.
– user254433
Nov 19 at 3:26
1
Thanks both! Very helpful!
– user292256
Nov 19 at 4:23
@user292256 Happy to help!
– user254433
Nov 19 at 5:20
add a comment |
up vote
1
down vote
accepted
I believe you are correct about the mean value theorem, but I will, effectively, be using an integral form instead of the differential form.
The average rate of change is easier to work with than the derivative for precisely the limit reason you mentioned. Integrating the equation and dividing, we recover the average value
$$
frac{x(b)-x(a)}{b-a}=frac{1}{b-a}int_a^bf(x(t))dt.
$$
Extract $f(X)$:
$$
frac{x(b)-x(a)}{b-a}=frac{1}{b-a}int_a^b[f(x(t))-f(X)]dt+f(X).
$$
Now put $b=2a$ and send $atoinfty$. The left hand side vanishes, and the first term on the right hand side vanishes if we apply the integral mean value theorem to the continuous function $f$:
$$
frac{1}{a}int_a^{2a}[f(x(t))-f(X)]dt=f(x(s))-f(X),
$$
for some $sin[a,2a]$. This difference is small for all such $s$ if $a$ is large.
answered Nov 19 at 3:13
user254433
2,456712
1
Another way, which requires less calculation: Clearly the limit of $f(x(t))$ exists; this means the limit of $x'(t)$ exists. If this limit is anything but 0, for example $lim x'(t)=y>0$, then applying FTC for large $a$, we have $x(2a)=x(a)+int_a^{2a}x'(s)dsge x(a)+acdot (y-epsilon)$ for some small $epsilon$, which contradicts $x(2a)-x(a)to 0$.
– user254433
Nov 19 at 3:26
1
Thanks both! Very helpful!
– user292256
Nov 19 at 4:23
@user292256 Happy to help!
– user254433
Nov 19 at 5:20
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I believe you are correct about the mean value theorem, but I will, effectively, be using an integral form instead of the differential form.
The average rate of change is easier to work with than the derivative for precisely the limit reason you mentioned. Integrating the equation and dividing, we recover the average value
$$
frac{x(b)-x(a)}{b-a}=frac{1}{b-a}int_a^bf(x(t))dt.
$$
Extract $f(X)$:
$$
frac{x(b)-x(a)}{b-a}=frac{1}{b-a}int_a^b[f(x(t))-f(X)]dt+f(X).
$$
Now put $b=2a$ and send $atoinfty$. The left hand side vanishes, and the first term on the right hand side vanishes if we apply the integral mean value theorem to the continuous function $f$:
$$
frac{1}{a}int_a^{2a}[f(x(t))-f(X)]dt=f(x(s))-f(X),
$$
for some $sin[a,2a]$. This difference is small for all such $s$ if $a$ is large.
answered Nov 19 at 3:13
user254433
2,456712
I believe you are correct about the mean value theorem, but I will, effectively, be using an integral form instead of the differential form.
The average rate of change is easier to work with than the derivative for precisely the limit reason you mentioned. Integrating the equation and dividing, we recover the average value
$$
frac{x(b)-x(a)}{b-a}=frac{1}{b-a}int_a^bf(x(t))dt.
$$
Extract $f(X)$:
$$
frac{x(b)-x(a)}{b-a}=frac{1}{b-a}int_a^b[f(x(t))-f(X)]dt+f(X).
$$
Now put $b=2a$ and send $atoinfty$. The left hand side vanishes, and the first term on the right hand side vanishes if we apply the integral mean value theorem to the continuous function $f$:
$$
frac{1}{a}int_a^{2a}[f(x(t))-f(X)]dt=f(x(s))-f(X),
$$
for some $sin[a,2a]$. This difference is small for all such $s$ if $a$ is large.
answered Nov 19 at 3:13
user254433
2,456712
edited Nov 19 at 3:20
answered Nov 19 at 3:13
user254433
2,456712
answered Nov 19 at 3:13
user254433
2,456712
answered Nov 19 at 3:13
user254433
2,456712
2,456712
1
Another way, which requires less calculation: Clearly the limit of $f(x(t))$ exists; this means the limit of $x'(t)$ exists. If this limit is anything but 0, for example $lim x'(t)=y>0$, then applying FTC for large $a$, we have $x(2a)=x(a)+int_a^{2a}x'(s)dsge x(a)+acdot (y-epsilon)$ for some small $epsilon$, which contradicts $x(2a)-x(a)to 0$.
– user254433
Nov 19 at 3:26
1
Thanks both! Very helpful!
– user292256
Nov 19 at 4:23
@user292256 Happy to help!
– user254433
Nov 19 at 5:20
add a comment |
1
Another way, which requires less calculation: Clearly the limit of $f(x(t))$ exists; this means the limit of $x'(t)$ exists. If this limit is anything but 0, for example $lim x'(t)=y>0$, then applying FTC for large $a$, we have $x(2a)=x(a)+int_a^{2a}x'(s)dsge x(a)+acdot (y-epsilon)$ for some small $epsilon$, which contradicts $x(2a)-x(a)to 0$.
– user254433
Nov 19 at 3:26
1
Thanks both! Very helpful!
– user292256
Nov 19 at 4:23
@user292256 Happy to help!
– user254433
Nov 19 at 5:20
1
1
Another way, which requires less calculation: Clearly the limit of $f(x(t))$ exists; this means the limit of $x'(t)$ exists. If this limit is anything but 0, for example $lim x'(t)=y>0$, then applying FTC for large $a$, we have $x(2a)=x(a)+int_a^{2a}x'(s)dsge x(a)+acdot (y-epsilon)$ for some small $epsilon$, which contradicts $x(2a)-x(a)to 0$.
– user254433
Nov 19 at 3:26
Another way, which requires less calculation: Clearly the limit of $f(x(t))$ exists; this means the limit of $x'(t)$ exists. If this limit is anything but 0, for example $lim x'(t)=y>0$, then applying FTC for large $a$, we have $x(2a)=x(a)+int_a^{2a}x'(s)dsge x(a)+acdot (y-epsilon)$ for some small $epsilon$, which contradicts $x(2a)-x(a)to 0$.
– user254433
Nov 19 at 3:26
1
1
Thanks both! Very helpful!
– user292256
Nov 19 at 4:23
Thanks both! Very helpful!
– user292256
Nov 19 at 4:23
@user292256 Happy to help!
– user254433
Nov 19 at 5:20
@user292256 Happy to help!
– user254433
Nov 19 at 5:20
add a comment |
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A more general question was asked and answered very recently, see Show that $p$ is a stationary solution.
– user539887
Nov 19 at 9:58