estimates for parameter dependent differential equation
$begingroup$
Consider some function $f:mathbb{R}times X times Omega,$ $Omega subset X$ and the differential equation
begin{equation}
dot{x}(t)=f(t,x,p),
end{equation}
where $p in Omega$ is some parameter. Then, if $f$ is Lipschitz-continuous in $p$ we know that also $x$ is Lipschitz-continuous in $p$.
Now, what I try to see is some generalization. Given some sequence of $f_n$ such that $| f_n(t,x,cdot) |_{C^{0,1}(Omega)}leq c$ with $c$ independent of $n$ I would like to have $| x_n(t,cdot) |_{C^{0,1}(Omega)}leq c'$ for the corresponding solutions $x_n$. Unfortunately, I don't see how one could prove this. I assume, that the above statement is proved by showing
begin{equation}
|x(t,cdot)|_{C^{0,1}(Omega)} leq c | f(t,x,cdot) |_{C^{0,1}(Omega)}
end{equation}
for some further constant $c$, if this is the case my desired estimate would of course follow immediately. But I can't find a proof for the above statement, so I don't know if it really works like this. I appreciate any hint about how to prove any of the two statements. Thank you!
ordinary-differential-equations lipschitz-functions
asked Dec 23 '18 at 1:45
jason paperjason paper
13519
$endgroup$
add a comment |
$begingroup$
Consider some function $f:mathbb{R}times X times Omega,$ $Omega subset X$ and the differential equation
begin{equation}
dot{x}(t)=f(t,x,p),
end{equation}
where $p in Omega$ is some parameter. Then, if $f$ is Lipschitz-continuous in $p$ we know that also $x$ is Lipschitz-continuous in $p$.
Now, what I try to see is some generalization. Given some sequence of $f_n$ such that $| f_n(t,x,cdot) |_{C^{0,1}(Omega)}leq c$ with $c$ independent of $n$ I would like to have $| x_n(t,cdot) |_{C^{0,1}(Omega)}leq c'$ for the corresponding solutions $x_n$. Unfortunately, I don't see how one could prove this. I assume, that the above statement is proved by showing
begin{equation}
|x(t,cdot)|_{C^{0,1}(Omega)} leq c | f(t,x,cdot) |_{C^{0,1}(Omega)}
end{equation}
for some further constant $c$, if this is the case my desired estimate would of course follow immediately. But I can't find a proof for the above statement, so I don't know if it really works like this. I appreciate any hint about how to prove any of the two statements. Thank you!
ordinary-differential-equations lipschitz-functions
asked Dec 23 '18 at 1:45
jason paperjason paper
13519
$endgroup$
1
$begingroup$
Grönwall's inequality?
$endgroup$
– user539887
Dec 23 '18 at 8:36
$begingroup$
Thank you for the hint, this might work. I will check it!
$endgroup$
– jason paper
Dec 27 '18 at 2:11
add a comment |
$begingroup$
Consider some function $f:mathbb{R}times X times Omega,$ $Omega subset X$ and the differential equation
begin{equation}
dot{x}(t)=f(t,x,p),
end{equation}
where $p in Omega$ is some parameter. Then, if $f$ is Lipschitz-continuous in $p$ we know that also $x$ is Lipschitz-continuous in $p$.
Now, what I try to see is some generalization. Given some sequence of $f_n$ such that $| f_n(t,x,cdot) |_{C^{0,1}(Omega)}leq c$ with $c$ independent of $n$ I would like to have $| x_n(t,cdot) |_{C^{0,1}(Omega)}leq c'$ for the corresponding solutions $x_n$. Unfortunately, I don't see how one could prove this. I assume, that the above statement is proved by showing
begin{equation}
|x(t,cdot)|_{C^{0,1}(Omega)} leq c | f(t,x,cdot) |_{C^{0,1}(Omega)}
end{equation}
for some further constant $c$, if this is the case my desired estimate would of course follow immediately. But I can't find a proof for the above statement, so I don't know if it really works like this. I appreciate any hint about how to prove any of the two statements. Thank you!
ordinary-differential-equations lipschitz-functions
asked Dec 23 '18 at 1:45
jason paperjason paper
13519
$endgroup$
Consider some function $f:mathbb{R}times X times Omega,$ $Omega subset X$ and the differential equation
begin{equation}
dot{x}(t)=f(t,x,p),
end{equation}
where $p in Omega$ is some parameter. Then, if $f$ is Lipschitz-continuous in $p$ we know that also $x$ is Lipschitz-continuous in $p$.
Now, what I try to see is some generalization. Given some sequence of $f_n$ such that $| f_n(t,x,cdot) |_{C^{0,1}(Omega)}leq c$ with $c$ independent of $n$ I would like to have $| x_n(t,cdot) |_{C^{0,1}(Omega)}leq c'$ for the corresponding solutions $x_n$. Unfortunately, I don't see how one could prove this. I assume, that the above statement is proved by showing
begin{equation}
|x(t,cdot)|_{C^{0,1}(Omega)} leq c | f(t,x,cdot) |_{C^{0,1}(Omega)}
end{equation}
for some further constant $c$, if this is the case my desired estimate would of course follow immediately. But I can't find a proof for the above statement, so I don't know if it really works like this. I appreciate any hint about how to prove any of the two statements. Thank you!
ordinary-differential-equations lipschitz-functions
ordinary-differential-equations lipschitz-functions
asked Dec 23 '18 at 1:45
jason paperjason paper
13519
asked Dec 23 '18 at 1:45
jason paperjason paper
13519
asked Dec 23 '18 at 1:45
jason paperjason paper
13519
asked Dec 23 '18 at 1:45
jason paperjason paper
13519
asked Dec 23 '18 at 1:45
jason paperjason paper
13519
13519
1
$begingroup$
Grönwall's inequality?
$endgroup$
– user539887
Dec 23 '18 at 8:36
$begingroup$
Thank you for the hint, this might work. I will check it!
$endgroup$
– jason paper
Dec 27 '18 at 2:11
add a comment |
1
$begingroup$
Grönwall's inequality?
$endgroup$
– user539887
Dec 23 '18 at 8:36
$begingroup$
Thank you for the hint, this might work. I will check it!
$endgroup$
– jason paper
Dec 27 '18 at 2:11
1
1
$begingroup$
Grönwall's inequality?
$endgroup$
– user539887
Dec 23 '18 at 8:36
$begingroup$
Grönwall's inequality?
$endgroup$
– user539887
Dec 23 '18 at 8:36
$begingroup$
Thank you for the hint, this might work. I will check it!
$endgroup$
– jason paper
Dec 27 '18 at 2:11
$begingroup$
Thank you for the hint, this might work. I will check it!
$endgroup$
– jason paper
Dec 27 '18 at 2:11
add a comment |
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$begingroup$
Grönwall's inequality?
$endgroup$
– user539887
Dec 23 '18 at 8:36
$begingroup$
Thank you for the hint, this might work. I will check it!
$endgroup$
– jason paper
Dec 27 '18 at 2:11