Mixing
How to prove that a parametrized linear differential equation solution induces a lipschitz function with...
$begingroup$
Let's consider the following linear differential equation:
$X'_{lambda}(t) = A_{lambda}(t) X_{lambda}(t)$ where: $X_{lambda}(t) = begin{pmatrix} x_{lambda}(t) \ x'_{lambda}(t) end{pmatrix}$ and $A_{lambda}(t) = begin{bmatrix} 0 & 1 \ p(t) - lambda & 0 end{bmatrix}$ where $p$ is a continuous real function and $lambda in mathbb{R}$ and $X_{lambda}(0) = begin{pmatrix} 0 \ 1 end{pmatrix}$.
$newcommand{norm}[1]{lVert #1 rVert}$
$newcommand{abs}[1]{lvert #1 rvert}$
Let us assume $norm{cdot}$ is the usual norm over vectors and for matrices:
For all matrix $A$ of size $n times p$ : $norm{A} = maxlimits_{i in [[1, n]]} sumlimits_{j=1}^p abs{a_{ij}}$.
Let be $R > 0$.
Also, let us define $c = sup { norm{A_{lambda}(t)} mid (t, lambda) in [-R, R]^2 }$.
I am trying to show that: $norm{X_{lambda}(t) - X_{mu}(t)} leq Re^{2cR} abs{lambda - mu}$ for all $(lambda, mu, t) in [-R, R]^3$.
What I have already done:
- For all $t in [-R, R], norm{X_lambda}(t) leq e^{cabs{t}}$.
- For all $(s, t, lambda) in [-R, R]^3, norm{X_{lambda}(t) - X_{lambda}(s)} leq ce^{cR} abs{t - s} (*)$.
What I have tried:
- Creating a new differential equation which is verified by $X_{lambda}$ and $X_{mu}$, but I could not find one.
- Using $(*)$ and $X_{lambda}(0) = X_{mu}(0)$, but I cannot get $abs{lambda - mu}$ this way in my inequality…
real-analysis ordinary-differential-equations inequality
asked Apr 21 '18 at 13:52
RaitoRaito
686415
$endgroup$
add a comment |
$begingroup$
Let's consider the following linear differential equation:
$X'_{lambda}(t) = A_{lambda}(t) X_{lambda}(t)$ where: $X_{lambda}(t) = begin{pmatrix} x_{lambda}(t) \ x'_{lambda}(t) end{pmatrix}$ and $A_{lambda}(t) = begin{bmatrix} 0 & 1 \ p(t) - lambda & 0 end{bmatrix}$ where $p$ is a continuous real function and $lambda in mathbb{R}$ and $X_{lambda}(0) = begin{pmatrix} 0 \ 1 end{pmatrix}$.
$newcommand{norm}[1]{lVert #1 rVert}$
$newcommand{abs}[1]{lvert #1 rvert}$
Let us assume $norm{cdot}$ is the usual norm over vectors and for matrices:
For all matrix $A$ of size $n times p$ : $norm{A} = maxlimits_{i in [[1, n]]} sumlimits_{j=1}^p abs{a_{ij}}$.
Let be $R > 0$.
Also, let us define $c = sup { norm{A_{lambda}(t)} mid (t, lambda) in [-R, R]^2 }$.
I am trying to show that: $norm{X_{lambda}(t) - X_{mu}(t)} leq Re^{2cR} abs{lambda - mu}$ for all $(lambda, mu, t) in [-R, R]^3$.
What I have already done:
- For all $t in [-R, R], norm{X_lambda}(t) leq e^{cabs{t}}$.
- For all $(s, t, lambda) in [-R, R]^3, norm{X_{lambda}(t) - X_{lambda}(s)} leq ce^{cR} abs{t - s} (*)$.
What I have tried:
- Creating a new differential equation which is verified by $X_{lambda}$ and $X_{mu}$, but I could not find one.
- Using $(*)$ and $X_{lambda}(0) = X_{mu}(0)$, but I cannot get $abs{lambda - mu}$ this way in my inequality…
real-analysis ordinary-differential-equations inequality
asked Apr 21 '18 at 13:52
RaitoRaito
686415
$endgroup$
$begingroup$
A hint: try finding an integral inequality satisfied by $lVert X_lambda(t)-X_mu(t)rVert$ and applying Grönwall's inequality.
$endgroup$
– user539887
Apr 21 '18 at 17:06
$begingroup$
@user539887 Thank you, it did the job. Do you want me to post the details as the answer, or would you do it?
$endgroup$
– Raito
Apr 21 '18 at 17:55
$begingroup$
Yes, please do it
$endgroup$
– user539887
Apr 21 '18 at 19:57
add a comment |
$begingroup$
Let's consider the following linear differential equation:
$X'_{lambda}(t) = A_{lambda}(t) X_{lambda}(t)$ where: $X_{lambda}(t) = begin{pmatrix} x_{lambda}(t) \ x'_{lambda}(t) end{pmatrix}$ and $A_{lambda}(t) = begin{bmatrix} 0 & 1 \ p(t) - lambda & 0 end{bmatrix}$ where $p$ is a continuous real function and $lambda in mathbb{R}$ and $X_{lambda}(0) = begin{pmatrix} 0 \ 1 end{pmatrix}$.
$newcommand{norm}[1]{lVert #1 rVert}$
$newcommand{abs}[1]{lvert #1 rvert}$
Let us assume $norm{cdot}$ is the usual norm over vectors and for matrices:
For all matrix $A$ of size $n times p$ : $norm{A} = maxlimits_{i in [[1, n]]} sumlimits_{j=1}^p abs{a_{ij}}$.
Let be $R > 0$.
Also, let us define $c = sup { norm{A_{lambda}(t)} mid (t, lambda) in [-R, R]^2 }$.
I am trying to show that: $norm{X_{lambda}(t) - X_{mu}(t)} leq Re^{2cR} abs{lambda - mu}$ for all $(lambda, mu, t) in [-R, R]^3$.
What I have already done:
- For all $t in [-R, R], norm{X_lambda}(t) leq e^{cabs{t}}$.
- For all $(s, t, lambda) in [-R, R]^3, norm{X_{lambda}(t) - X_{lambda}(s)} leq ce^{cR} abs{t - s} (*)$.
What I have tried:
- Creating a new differential equation which is verified by $X_{lambda}$ and $X_{mu}$, but I could not find one.
- Using $(*)$ and $X_{lambda}(0) = X_{mu}(0)$, but I cannot get $abs{lambda - mu}$ this way in my inequality…
real-analysis ordinary-differential-equations inequality
asked Apr 21 '18 at 13:52
RaitoRaito
686415
$endgroup$
Let's consider the following linear differential equation:
$X'_{lambda}(t) = A_{lambda}(t) X_{lambda}(t)$ where: $X_{lambda}(t) = begin{pmatrix} x_{lambda}(t) \ x'_{lambda}(t) end{pmatrix}$ and $A_{lambda}(t) = begin{bmatrix} 0 & 1 \ p(t) - lambda & 0 end{bmatrix}$ where $p$ is a continuous real function and $lambda in mathbb{R}$ and $X_{lambda}(0) = begin{pmatrix} 0 \ 1 end{pmatrix}$.
$newcommand{norm}[1]{lVert #1 rVert}$
$newcommand{abs}[1]{lvert #1 rvert}$
Let us assume $norm{cdot}$ is the usual norm over vectors and for matrices:
For all matrix $A$ of size $n times p$ : $norm{A} = maxlimits_{i in [[1, n]]} sumlimits_{j=1}^p abs{a_{ij}}$.
Let be $R > 0$.
Also, let us define $c = sup { norm{A_{lambda}(t)} mid (t, lambda) in [-R, R]^2 }$.
I am trying to show that: $norm{X_{lambda}(t) - X_{mu}(t)} leq Re^{2cR} abs{lambda - mu}$ for all $(lambda, mu, t) in [-R, R]^3$.
What I have already done:
- For all $t in [-R, R], norm{X_lambda}(t) leq e^{cabs{t}}$.
- For all $(s, t, lambda) in [-R, R]^3, norm{X_{lambda}(t) - X_{lambda}(s)} leq ce^{cR} abs{t - s} (*)$.
What I have tried:
- Creating a new differential equation which is verified by $X_{lambda}$ and $X_{mu}$, but I could not find one.
- Using $(*)$ and $X_{lambda}(0) = X_{mu}(0)$, but I cannot get $abs{lambda - mu}$ this way in my inequality…
real-analysis ordinary-differential-equations inequality
real-analysis ordinary-differential-equations inequality
asked Apr 21 '18 at 13:52
RaitoRaito
686415
asked Apr 21 '18 at 13:52
RaitoRaito
686415
asked Apr 21 '18 at 13:52
RaitoRaito
686415
asked Apr 21 '18 at 13:52
RaitoRaito
686415
asked Apr 21 '18 at 13:52
RaitoRaito
686415
686415
$begingroup$
A hint: try finding an integral inequality satisfied by $lVert X_lambda(t)-X_mu(t)rVert$ and applying Grönwall's inequality.
$endgroup$
– user539887
Apr 21 '18 at 17:06
$begingroup$
@user539887 Thank you, it did the job. Do you want me to post the details as the answer, or would you do it?
$endgroup$
– Raito
Apr 21 '18 at 17:55
$begingroup$
Yes, please do it
$endgroup$
– user539887
Apr 21 '18 at 19:57
add a comment |
$begingroup$
A hint: try finding an integral inequality satisfied by $lVert X_lambda(t)-X_mu(t)rVert$ and applying Grönwall's inequality.
$endgroup$
– user539887
Apr 21 '18 at 17:06
$begingroup$
@user539887 Thank you, it did the job. Do you want me to post the details as the answer, or would you do it?
$endgroup$
– Raito
Apr 21 '18 at 17:55
$begingroup$
Yes, please do it
$endgroup$
– user539887
Apr 21 '18 at 19:57
$begingroup$
A hint: try finding an integral inequality satisfied by $lVert X_lambda(t)-X_mu(t)rVert$ and applying Grönwall's inequality.
$endgroup$
– user539887
Apr 21 '18 at 17:06
$begingroup$
A hint: try finding an integral inequality satisfied by $lVert X_lambda(t)-X_mu(t)rVert$ and applying Grönwall's inequality.
$endgroup$
– user539887
Apr 21 '18 at 17:06
$begingroup$
@user539887 Thank you, it did the job. Do you want me to post the details as the answer, or would you do it?
$endgroup$
– Raito
Apr 21 '18 at 17:55
$begingroup$
@user539887 Thank you, it did the job. Do you want me to post the details as the answer, or would you do it?
$endgroup$
– Raito
Apr 21 '18 at 17:55
$begingroup$
Yes, please do it
$endgroup$
– user539887
Apr 21 '18 at 19:57
$begingroup$
Yes, please do it
$endgroup$
– user539887
Apr 21 '18 at 19:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let us denote $M = begin{bmatrix} 0 & 0 \ 1 & 0 end{bmatrix}$.
Let's fix $(lambda, t, mu) in [-R, R]^3$.
We have, by the fundamental theorem of analysis, because $X_{lambda}$ and $X_{mu}$ are twice continuously differentiable.$newcommand{norm}[1]{lVert #1 rVert}newcommand{abs}[1]{lvert #1 rvert}$
$begin{align*}
X_{lambda}(t) - X_{mu}(t) & = X_{lambda}(t) - X_{lambda}(0) - (X_{mu}(t) - X_{mu}(0)) \
& = int_0^t X'_{lambda}(s) - X'_{mu}(s) textrm{d}s \
& = int_0^t A_{lambda}(s) X_{lambda}(s) - A_{mu}(s) X_{mu}(s) textrm{d}s \
& = int_0^t A_{lambda}(s)(X_{lambda}(s) - X_{mu}(s)) textrm{d}s + (lambda - mu)Mint_0^t X_{mu}(s) textrm{d}s
end{align*}$
By taking norms:
$begin{align*}
norm{X_{lambda}(t) - X_{mu}(t)} leq c int_0^t norm{X_{lambda}(s) - X_{mu}(s)} textrm{d}s + abs{lambda - mu}R e^{cR}
end{align*}$
By Gronwall's lemma:
$begin{equation*}
norm{X_{lambda}(t) - X_{mu}(t)} leq Re^{cR}e^{cR} abs{lambda - mu} = Re^{2cR} abs{lambda - mu}
end{equation*}$
edited Jan 31 at 21:45
Martin Sleziak
45k10122277
answered Apr 22 '18 at 7:37
RaitoRaito
686415
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let us denote $M = begin{bmatrix} 0 & 0 \ 1 & 0 end{bmatrix}$.
Let's fix $(lambda, t, mu) in [-R, R]^3$.
We have, by the fundamental theorem of analysis, because $X_{lambda}$ and $X_{mu}$ are twice continuously differentiable.$newcommand{norm}[1]{lVert #1 rVert}newcommand{abs}[1]{lvert #1 rvert}$
$begin{align*}
X_{lambda}(t) - X_{mu}(t) & = X_{lambda}(t) - X_{lambda}(0) - (X_{mu}(t) - X_{mu}(0)) \
& = int_0^t X'_{lambda}(s) - X'_{mu}(s) textrm{d}s \
& = int_0^t A_{lambda}(s) X_{lambda}(s) - A_{mu}(s) X_{mu}(s) textrm{d}s \
& = int_0^t A_{lambda}(s)(X_{lambda}(s) - X_{mu}(s)) textrm{d}s + (lambda - mu)Mint_0^t X_{mu}(s) textrm{d}s
end{align*}$
By taking norms:
$begin{align*}
norm{X_{lambda}(t) - X_{mu}(t)} leq c int_0^t norm{X_{lambda}(s) - X_{mu}(s)} textrm{d}s + abs{lambda - mu}R e^{cR}
end{align*}$
By Gronwall's lemma:
$begin{equation*}
norm{X_{lambda}(t) - X_{mu}(t)} leq Re^{cR}e^{cR} abs{lambda - mu} = Re^{2cR} abs{lambda - mu}
end{equation*}$
edited Jan 31 at 21:45
Martin Sleziak
45k10122277
answered Apr 22 '18 at 7:37
RaitoRaito
686415
$endgroup$
add a comment |
$begingroup$
Let us denote $M = begin{bmatrix} 0 & 0 \ 1 & 0 end{bmatrix}$.
Let's fix $(lambda, t, mu) in [-R, R]^3$.
We have, by the fundamental theorem of analysis, because $X_{lambda}$ and $X_{mu}$ are twice continuously differentiable.$newcommand{norm}[1]{lVert #1 rVert}newcommand{abs}[1]{lvert #1 rvert}$
$begin{align*}
X_{lambda}(t) - X_{mu}(t) & = X_{lambda}(t) - X_{lambda}(0) - (X_{mu}(t) - X_{mu}(0)) \
& = int_0^t X'_{lambda}(s) - X'_{mu}(s) textrm{d}s \
& = int_0^t A_{lambda}(s) X_{lambda}(s) - A_{mu}(s) X_{mu}(s) textrm{d}s \
& = int_0^t A_{lambda}(s)(X_{lambda}(s) - X_{mu}(s)) textrm{d}s + (lambda - mu)Mint_0^t X_{mu}(s) textrm{d}s
end{align*}$
By taking norms:
$begin{align*}
norm{X_{lambda}(t) - X_{mu}(t)} leq c int_0^t norm{X_{lambda}(s) - X_{mu}(s)} textrm{d}s + abs{lambda - mu}R e^{cR}
end{align*}$
By Gronwall's lemma:
$begin{equation*}
norm{X_{lambda}(t) - X_{mu}(t)} leq Re^{cR}e^{cR} abs{lambda - mu} = Re^{2cR} abs{lambda - mu}
end{equation*}$
edited Jan 31 at 21:45
Martin Sleziak
45k10122277
answered Apr 22 '18 at 7:37
RaitoRaito
686415
$endgroup$
add a comment |
$begingroup$
Let us denote $M = begin{bmatrix} 0 & 0 \ 1 & 0 end{bmatrix}$.
Let's fix $(lambda, t, mu) in [-R, R]^3$.
We have, by the fundamental theorem of analysis, because $X_{lambda}$ and $X_{mu}$ are twice continuously differentiable.$newcommand{norm}[1]{lVert #1 rVert}newcommand{abs}[1]{lvert #1 rvert}$
$begin{align*}
X_{lambda}(t) - X_{mu}(t) & = X_{lambda}(t) - X_{lambda}(0) - (X_{mu}(t) - X_{mu}(0)) \
& = int_0^t X'_{lambda}(s) - X'_{mu}(s) textrm{d}s \
& = int_0^t A_{lambda}(s) X_{lambda}(s) - A_{mu}(s) X_{mu}(s) textrm{d}s \
& = int_0^t A_{lambda}(s)(X_{lambda}(s) - X_{mu}(s)) textrm{d}s + (lambda - mu)Mint_0^t X_{mu}(s) textrm{d}s
end{align*}$
By taking norms:
$begin{align*}
norm{X_{lambda}(t) - X_{mu}(t)} leq c int_0^t norm{X_{lambda}(s) - X_{mu}(s)} textrm{d}s + abs{lambda - mu}R e^{cR}
end{align*}$
By Gronwall's lemma:
$begin{equation*}
norm{X_{lambda}(t) - X_{mu}(t)} leq Re^{cR}e^{cR} abs{lambda - mu} = Re^{2cR} abs{lambda - mu}
end{equation*}$
edited Jan 31 at 21:45
Martin Sleziak
45k10122277
answered Apr 22 '18 at 7:37
RaitoRaito
686415
$endgroup$
Let us denote $M = begin{bmatrix} 0 & 0 \ 1 & 0 end{bmatrix}$.
Let's fix $(lambda, t, mu) in [-R, R]^3$.
We have, by the fundamental theorem of analysis, because $X_{lambda}$ and $X_{mu}$ are twice continuously differentiable.$newcommand{norm}[1]{lVert #1 rVert}newcommand{abs}[1]{lvert #1 rvert}$
$begin{align*}
X_{lambda}(t) - X_{mu}(t) & = X_{lambda}(t) - X_{lambda}(0) - (X_{mu}(t) - X_{mu}(0)) \
& = int_0^t X'_{lambda}(s) - X'_{mu}(s) textrm{d}s \
& = int_0^t A_{lambda}(s) X_{lambda}(s) - A_{mu}(s) X_{mu}(s) textrm{d}s \
& = int_0^t A_{lambda}(s)(X_{lambda}(s) - X_{mu}(s)) textrm{d}s + (lambda - mu)Mint_0^t X_{mu}(s) textrm{d}s
end{align*}$
By taking norms:
$begin{align*}
norm{X_{lambda}(t) - X_{mu}(t)} leq c int_0^t norm{X_{lambda}(s) - X_{mu}(s)} textrm{d}s + abs{lambda - mu}R e^{cR}
end{align*}$
By Gronwall's lemma:
$begin{equation*}
norm{X_{lambda}(t) - X_{mu}(t)} leq Re^{cR}e^{cR} abs{lambda - mu} = Re^{2cR} abs{lambda - mu}
end{equation*}$
edited Jan 31 at 21:45
Martin Sleziak
45k10122277
answered Apr 22 '18 at 7:37
RaitoRaito
686415
edited Jan 31 at 21:45
Martin Sleziak
45k10122277
edited Jan 31 at 21:45
Martin Sleziak
45k10122277
edited Jan 31 at 21:45
Martin Sleziak
45k10122277
45k10122277
answered Apr 22 '18 at 7:37
RaitoRaito
686415
answered Apr 22 '18 at 7:37
RaitoRaito
686415
answered Apr 22 '18 at 7:37
RaitoRaito
686415
686415
add a comment |
add a comment |
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$begingroup$
A hint: try finding an integral inequality satisfied by $lVert X_lambda(t)-X_mu(t)rVert$ and applying Grönwall's inequality.
$endgroup$
– user539887
Apr 21 '18 at 17:06
$begingroup$
@user539887 Thank you, it did the job. Do you want me to post the details as the answer, or would you do it?
$endgroup$
– Raito
Apr 21 '18 at 17:55
$begingroup$
Yes, please do it
$endgroup$
– user539887
Apr 21 '18 at 19:57