Show monotonicity of solution of Delayed Differential Equation with respect to a parameter
$begingroup$
Short Description of the General Question
Suppose we have some Delayed Differential Equation (DDE) which depends on a parameter $a$, $x_a'(t)=f(a,x_a(t),x_a(t-s))$ for some fixed $a$ and $s$. I would like to prove that $x_a(t)$ is increasing in $a$. I.e. if we have $a<b$ and $x_a'(t) = f(a,x_a(t),x_a(t-s))$ and $x_b'(t)=f(b,x_b(t),x_b(t-s))$ then $x_a(t) leq x_b(t)$ for all $t$.
Specific Scenario
Consider the following Delayed Differential Equation:
begin{align*}
x_a(0) &= a\
x_a'(t) &= - a (1 - x_a(t)^2) & t leq 1\
x_a'(t) &= -a(x_a(t-1)^2 - x_a(t)^2) & t > 1.
end{align*}
I have found numerically that for all $a in (0,1)$ we have:
$$
a leq b Rightarrow x_a(t) leq x_b(t),
$$
but I am unable to prove this statement. I can solve the ODE in $[0,1]$ exactly, thus on this interval it is easily verified that $x_a(t) leq x_b(t)$. I then use that solution to find a solution on $[1,2]$ and so on. But as $t$ grows large we can't find an exact solution anymore.
real-analysis calculus ordinary-differential-equations delay-differential-equations
asked Dec 31 '18 at 13:07
HolyMonkHolyMonk
617416
$endgroup$
add a comment |
$begingroup$
Short Description of the General Question
Suppose we have some Delayed Differential Equation (DDE) which depends on a parameter $a$, $x_a'(t)=f(a,x_a(t),x_a(t-s))$ for some fixed $a$ and $s$. I would like to prove that $x_a(t)$ is increasing in $a$. I.e. if we have $a<b$ and $x_a'(t) = f(a,x_a(t),x_a(t-s))$ and $x_b'(t)=f(b,x_b(t),x_b(t-s))$ then $x_a(t) leq x_b(t)$ for all $t$.
Specific Scenario
Consider the following Delayed Differential Equation:
begin{align*}
x_a(0) &= a\
x_a'(t) &= - a (1 - x_a(t)^2) & t leq 1\
x_a'(t) &= -a(x_a(t-1)^2 - x_a(t)^2) & t > 1.
end{align*}
I have found numerically that for all $a in (0,1)$ we have:
$$
a leq b Rightarrow x_a(t) leq x_b(t),
$$
but I am unable to prove this statement. I can solve the ODE in $[0,1]$ exactly, thus on this interval it is easily verified that $x_a(t) leq x_b(t)$. I then use that solution to find a solution on $[1,2]$ and so on. But as $t$ grows large we can't find an exact solution anymore.
real-analysis calculus ordinary-differential-equations delay-differential-equations
asked Dec 31 '18 at 13:07
HolyMonkHolyMonk
617416
$endgroup$
$begingroup$
In general, x'$_a$ = f(a,b,c) = -a contradicts the desired conclusion. Also, some boundary conditions for the x's needs to be set.
$endgroup$
– William Elliot
Dec 31 '18 at 14:12
$begingroup$
Yes of course this statement isn't true in general, but I am looking for some method to try and show that it is true (a method which at least works on the provided example)
$endgroup$
– HolyMonk
Jan 2 at 0:16
add a comment |
$begingroup$
Short Description of the General Question
Suppose we have some Delayed Differential Equation (DDE) which depends on a parameter $a$, $x_a'(t)=f(a,x_a(t),x_a(t-s))$ for some fixed $a$ and $s$. I would like to prove that $x_a(t)$ is increasing in $a$. I.e. if we have $a<b$ and $x_a'(t) = f(a,x_a(t),x_a(t-s))$ and $x_b'(t)=f(b,x_b(t),x_b(t-s))$ then $x_a(t) leq x_b(t)$ for all $t$.
Specific Scenario
Consider the following Delayed Differential Equation:
begin{align*}
x_a(0) &= a\
x_a'(t) &= - a (1 - x_a(t)^2) & t leq 1\
x_a'(t) &= -a(x_a(t-1)^2 - x_a(t)^2) & t > 1.
end{align*}
I have found numerically that for all $a in (0,1)$ we have:
$$
a leq b Rightarrow x_a(t) leq x_b(t),
$$
but I am unable to prove this statement. I can solve the ODE in $[0,1]$ exactly, thus on this interval it is easily verified that $x_a(t) leq x_b(t)$. I then use that solution to find a solution on $[1,2]$ and so on. But as $t$ grows large we can't find an exact solution anymore.
real-analysis calculus ordinary-differential-equations delay-differential-equations
asked Dec 31 '18 at 13:07
HolyMonkHolyMonk
617416
$endgroup$
Short Description of the General Question
Suppose we have some Delayed Differential Equation (DDE) which depends on a parameter $a$, $x_a'(t)=f(a,x_a(t),x_a(t-s))$ for some fixed $a$ and $s$. I would like to prove that $x_a(t)$ is increasing in $a$. I.e. if we have $a<b$ and $x_a'(t) = f(a,x_a(t),x_a(t-s))$ and $x_b'(t)=f(b,x_b(t),x_b(t-s))$ then $x_a(t) leq x_b(t)$ for all $t$.
Specific Scenario
Consider the following Delayed Differential Equation:
begin{align*}
x_a(0) &= a\
x_a'(t) &= - a (1 - x_a(t)^2) & t leq 1\
x_a'(t) &= -a(x_a(t-1)^2 - x_a(t)^2) & t > 1.
end{align*}
I have found numerically that for all $a in (0,1)$ we have:
$$
a leq b Rightarrow x_a(t) leq x_b(t),
$$
but I am unable to prove this statement. I can solve the ODE in $[0,1]$ exactly, thus on this interval it is easily verified that $x_a(t) leq x_b(t)$. I then use that solution to find a solution on $[1,2]$ and so on. But as $t$ grows large we can't find an exact solution anymore.
real-analysis calculus ordinary-differential-equations delay-differential-equations
real-analysis calculus ordinary-differential-equations delay-differential-equations
asked Dec 31 '18 at 13:07
HolyMonkHolyMonk
617416
asked Dec 31 '18 at 13:07
HolyMonkHolyMonk
617416
asked Dec 31 '18 at 13:07
HolyMonkHolyMonk
617416
asked Dec 31 '18 at 13:07
HolyMonkHolyMonk
617416
asked Dec 31 '18 at 13:07
HolyMonkHolyMonk
617416
617416
$begingroup$
In general, x'$_a$ = f(a,b,c) = -a contradicts the desired conclusion. Also, some boundary conditions for the x's needs to be set.
$endgroup$
– William Elliot
Dec 31 '18 at 14:12
$begingroup$
Yes of course this statement isn't true in general, but I am looking for some method to try and show that it is true (a method which at least works on the provided example)
$endgroup$
– HolyMonk
Jan 2 at 0:16
add a comment |
$begingroup$
In general, x'$_a$ = f(a,b,c) = -a contradicts the desired conclusion. Also, some boundary conditions for the x's needs to be set.
$endgroup$
– William Elliot
Dec 31 '18 at 14:12
$begingroup$
Yes of course this statement isn't true in general, but I am looking for some method to try and show that it is true (a method which at least works on the provided example)
$endgroup$
– HolyMonk
Jan 2 at 0:16
$begingroup$
In general, x'$_a$ = f(a,b,c) = -a contradicts the desired conclusion. Also, some boundary conditions for the x's needs to be set.
$endgroup$
– William Elliot
Dec 31 '18 at 14:12
$begingroup$
In general, x'$_a$ = f(a,b,c) = -a contradicts the desired conclusion. Also, some boundary conditions for the x's needs to be set.
$endgroup$
– William Elliot
Dec 31 '18 at 14:12
$begingroup$
Yes of course this statement isn't true in general, but I am looking for some method to try and show that it is true (a method which at least works on the provided example)
$endgroup$
– HolyMonk
Jan 2 at 0:16
$begingroup$
Yes of course this statement isn't true in general, but I am looking for some method to try and show that it is true (a method which at least works on the provided example)
$endgroup$
– HolyMonk
Jan 2 at 0:16
add a comment |
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$begingroup$
In general, x'$_a$ = f(a,b,c) = -a contradicts the desired conclusion. Also, some boundary conditions for the x's needs to be set.
$endgroup$
– William Elliot
Dec 31 '18 at 14:12
$begingroup$
Yes of course this statement isn't true in general, but I am looking for some method to try and show that it is true (a method which at least works on the provided example)
$endgroup$
– HolyMonk
Jan 2 at 0:16