Limit of a solution to a differential equation is a steady state.












0














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







share|cite|improve this question






















  • A more general question was asked and answered very recently, see Show that $p$ is a stationary solution.
    – user539887
    Nov 19 '18 at 9:58


















0














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







share|cite|improve this question






















  • A more general question was asked and answered very recently, see Show that $p$ is a stationary solution.
    – user539887
    Nov 19 '18 at 9:58
















0












0








0







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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











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asked Nov 19 '18 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 '18 at 9:58




















  • A more general question was asked and answered very recently, see Show that $p$ is a stationary solution.
    – user539887
    Nov 19 '18 at 9:58


















A more general question was asked and answered very recently, see Show that $p$ is a stationary solution.
– user539887
Nov 19 '18 at 9:58






A more general question was asked and answered very recently, see Show that $p$ is a stationary solution.
– user539887
Nov 19 '18 at 9:58












1 Answer
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oldest

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1














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.






share|cite|improve this answer



















  • 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 '18 at 3:26








  • 1




    Thanks both! Very helpful!
    – user292256
    Nov 19 '18 at 4:23










  • @user292256 Happy to help!
    – user254433
    Nov 19 '18 at 5:20











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1 Answer
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active

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1














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.






share|cite|improve this answer



















  • 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 '18 at 3:26








  • 1




    Thanks both! Very helpful!
    – user292256
    Nov 19 '18 at 4:23










  • @user292256 Happy to help!
    – user254433
    Nov 19 '18 at 5:20
















1














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.






share|cite|improve this answer



















  • 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 '18 at 3:26








  • 1




    Thanks both! Very helpful!
    – user292256
    Nov 19 '18 at 4:23










  • @user292256 Happy to help!
    – user254433
    Nov 19 '18 at 5:20














1












1








1






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.






share|cite|improve this answer














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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 19 '18 at 3:20

























answered Nov 19 '18 at 3:13









user254433

2,531712




2,531712








  • 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 '18 at 3:26








  • 1




    Thanks both! Very helpful!
    – user292256
    Nov 19 '18 at 4:23










  • @user292256 Happy to help!
    – user254433
    Nov 19 '18 at 5:20














  • 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 '18 at 3:26








  • 1




    Thanks both! Very helpful!
    – user292256
    Nov 19 '18 at 4:23










  • @user292256 Happy to help!
    – user254433
    Nov 19 '18 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 '18 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 '18 at 3:26






1




1




Thanks both! Very helpful!
– user292256
Nov 19 '18 at 4:23




Thanks both! Very helpful!
– user292256
Nov 19 '18 at 4:23












@user292256 Happy to help!
– user254433
Nov 19 '18 at 5:20




@user292256 Happy to help!
– user254433
Nov 19 '18 at 5:20


















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