Riccati Comparison Principle












5












$begingroup$


This is a proposition in a book I'm reading whose proof I am unsure of how to fill in the details



Proposition



Suppose we have two smooth functions $rho_{1,2} : (0,b) to mathbb{R}$ such that
$$rho_1'+ rho_1^2 leq rho_2'+ rho_2^2 $$
then
$$rho_2 - rho_1 geq limsup_{t to 0},(, rho_2(t) - rho_1(t),)$$



Question



The author says that this follows from the easily verified fact that the function $(rho_2 - rho_1) e^F$ is increasing where $F$ is the antiderivative of $rho_2 + rho_1$ on $(0, b)$. This would be true if $rho_2 - rho_1$ is increasing but I don't think that this is the case. How does the lim sup inequality follow from the fact that $(rho_2 - rho_1) e^F$ is increasing?






real-analysis ordinary-differential-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Which book? $ $
    $endgroup$
    – Dap
    Jan 25 at 4:53










  • $begingroup$
    Riemannian Geometry by Peter Petersen. In the third edition, the Riccati Comparison Principle is on page 254.
    $endgroup$
    – Cookie
    Jan 29 at 3:08
















5












$begingroup$


This is a proposition in a book I'm reading whose proof I am unsure of how to fill in the details



Proposition



Suppose we have two smooth functions $rho_{1,2} : (0,b) to mathbb{R}$ such that
$$rho_1'+ rho_1^2 leq rho_2'+ rho_2^2 $$
then
$$rho_2 - rho_1 geq limsup_{t to 0},(, rho_2(t) - rho_1(t),)$$



Question



The author says that this follows from the easily verified fact that the function $(rho_2 - rho_1) e^F$ is increasing where $F$ is the antiderivative of $rho_2 + rho_1$ on $(0, b)$. This would be true if $rho_2 - rho_1$ is increasing but I don't think that this is the case. How does the lim sup inequality follow from the fact that $(rho_2 - rho_1) e^F$ is increasing?






real-analysis ordinary-differential-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Which book? $ $
    $endgroup$
    – Dap
    Jan 25 at 4:53










  • $begingroup$
    Riemannian Geometry by Peter Petersen. In the third edition, the Riccati Comparison Principle is on page 254.
    $endgroup$
    – Cookie
    Jan 29 at 3:08














5












5








5


1



$begingroup$


This is a proposition in a book I'm reading whose proof I am unsure of how to fill in the details



Proposition



Suppose we have two smooth functions $rho_{1,2} : (0,b) to mathbb{R}$ such that
$$rho_1'+ rho_1^2 leq rho_2'+ rho_2^2 $$
then
$$rho_2 - rho_1 geq limsup_{t to 0},(, rho_2(t) - rho_1(t),)$$



Question



The author says that this follows from the easily verified fact that the function $(rho_2 - rho_1) e^F$ is increasing where $F$ is the antiderivative of $rho_2 + rho_1$ on $(0, b)$. This would be true if $rho_2 - rho_1$ is increasing but I don't think that this is the case. How does the lim sup inequality follow from the fact that $(rho_2 - rho_1) e^F$ is increasing?






real-analysis ordinary-differential-equations







share|cite|improve this question











$endgroup$




This is a proposition in a book I'm reading whose proof I am unsure of how to fill in the details



Proposition



Suppose we have two smooth functions $rho_{1,2} : (0,b) to mathbb{R}$ such that
$$rho_1'+ rho_1^2 leq rho_2'+ rho_2^2 $$
then
$$rho_2 - rho_1 geq limsup_{t to 0},(, rho_2(t) - rho_1(t),)$$



Question



The author says that this follows from the easily verified fact that the function $(rho_2 - rho_1) e^F$ is increasing where $F$ is the antiderivative of $rho_2 + rho_1$ on $(0, b)$. This would be true if $rho_2 - rho_1$ is increasing but I don't think that this is the case. How does the lim sup inequality follow from the fact that $(rho_2 - rho_1) e^F$ is increasing?






real-analysis ordinary-differential-equations




real-analysis ordinary-differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 26 at 11:43









Daniele Tampieri

2,5222922




2,5222922










asked Jan 9 '17 at 3:26









SelfstudierSelfstudier

1016




1016












  • $begingroup$
    Which book? $ $
    $endgroup$
    – Dap
    Jan 25 at 4:53










  • $begingroup$
    Riemannian Geometry by Peter Petersen. In the third edition, the Riccati Comparison Principle is on page 254.
    $endgroup$
    – Cookie
    Jan 29 at 3:08


















  • $begingroup$
    Which book? $ $
    $endgroup$
    – Dap
    Jan 25 at 4:53










  • $begingroup$
    Riemannian Geometry by Peter Petersen. In the third edition, the Riccati Comparison Principle is on page 254.
    $endgroup$
    – Cookie
    Jan 29 at 3:08
















$begingroup$
Which book? $ $
$endgroup$
– Dap
Jan 25 at 4:53




$begingroup$
Which book? $ $
$endgroup$
– Dap
Jan 25 at 4:53












$begingroup$
Riemannian Geometry by Peter Petersen. In the third edition, the Riccati Comparison Principle is on page 254.
$endgroup$
– Cookie
Jan 29 at 3:08




$begingroup$
Riemannian Geometry by Peter Petersen. In the third edition, the Riccati Comparison Principle is on page 254.
$endgroup$
– Cookie
Jan 29 at 3:08










2 Answers
2






active

oldest

votes


















3












$begingroup$

Note that
$$begin{align}frac d{dt}left[(rho_2-rho_1)e^Fright]&=(rho'_2-rho'_1)e^F+(rho_2-rho_1)(rho_2+rho_1)e^F\
&=e^Fleft(rho'_2-rho'_1+rho_2^2-rho_1^2right)
end{align}$$
and since $e^F>0$, your inequality implies that the derivative of $(rho_2-rho_1)e^F$ is positive. So the function $(rho_2-rho_1)e^F$ is increasing.



On the other hand, a careless approach with some possible errors could be this:
$$begin{align}
rho_1'+ rho_1^2 leq rho_2'+ rho_2^2 &impliesrho_1'-rho_2' leq rho_2^2- rho_1^2\
&implies-dfrac{rho_1'-rho_2'}{rho_1-rho_2}gerho_1+rho_2\
&implies-ln|rho_1-rho_2|ge F\
&impliesfrac 1{|rho_1-rho_2|}le e^Fimplies(rho_2-rho_1)e^Fge 1
end{align}$$
or something like that....






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Right, but how does the fact that $rho_2 - rho_1 e^F$ is increasing imply that $rho_2 - rho_1 geq limsup _{t to 0}(rho_2(t) - rho_1(t))$
    $endgroup$
    – Selfstudier
    Jan 9 '17 at 13:29



















2





+50







$begingroup$

The book is wrong and it's easy to find counterexamples. For example if $ρ_1(x)=1/(x+2)$ and $ρ_2(x)=1/(x+1)$ then $ρ_i'+ρ_i^2=0$ and $ρ_2(0)-ρ_1(0)=1/2$ but $ρ_2(1)-ρ_1(1)=1/6.$



The author may have been thinking of something like the following argument which I will paraphrase from "New comparison theorems in Riemannian geometry" by Y. Han and coauthors, Theorem 5.1.
Suppose $ρ_1$ and $ρ_2$ are both $1/t+O(1).$
Define $ϕ_i(t)=texp(int_0^t (ρ_i(s)-1/s)ds).$ Then each $ϕ_i$ is a $C^1$ function on $[0,b)$ (i.e. the right-sided derivative exists at zero - in fact it's $1$), smooth and strictly positive on $(0,b),$ with $ϕ_i(0)=0.$ Differentiating gives $ϕ'_i(t)=ρ_i(t) ϕ_i(t)$ and hence $ϕ''_i(t)=(ρ'_i(t)+ρ_i(t)^2)ϕ_i(t).$ This leads to
$$(ϕ_2'ϕ_1-ϕ_1'ϕ_2)'=ϕ_2''ϕ_1-ϕ_1''ϕ_2=((ρ'_2+ρ_2^2)-(ρ'_1+ρ_1^2))ϕ_1ϕ_2geq 0.$$
Using the fact that $ϕ_2'(0)ϕ_1(0)-ϕ_1'(0)ϕ_2(0)=0$ we get $ϕ_2'ϕ_1geq ϕ_1'ϕ_2.$ Using our previous equation for $ϕ_i',$ this means $ρ_2(t)geq ρ_1(t).$






share|cite|improve this answer











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    2 Answers
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    3












    $begingroup$

    Note that
    $$begin{align}frac d{dt}left[(rho_2-rho_1)e^Fright]&=(rho'_2-rho'_1)e^F+(rho_2-rho_1)(rho_2+rho_1)e^F\
    &=e^Fleft(rho'_2-rho'_1+rho_2^2-rho_1^2right)
    end{align}$$
    and since $e^F>0$, your inequality implies that the derivative of $(rho_2-rho_1)e^F$ is positive. So the function $(rho_2-rho_1)e^F$ is increasing.



    On the other hand, a careless approach with some possible errors could be this:
    $$begin{align}
    rho_1'+ rho_1^2 leq rho_2'+ rho_2^2 &impliesrho_1'-rho_2' leq rho_2^2- rho_1^2\
    &implies-dfrac{rho_1'-rho_2'}{rho_1-rho_2}gerho_1+rho_2\
    &implies-ln|rho_1-rho_2|ge F\
    &impliesfrac 1{|rho_1-rho_2|}le e^Fimplies(rho_2-rho_1)e^Fge 1
    end{align}$$
    or something like that....






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      Right, but how does the fact that $rho_2 - rho_1 e^F$ is increasing imply that $rho_2 - rho_1 geq limsup _{t to 0}(rho_2(t) - rho_1(t))$
      $endgroup$
      – Selfstudier
      Jan 9 '17 at 13:29
















    3












    $begingroup$

    Note that
    $$begin{align}frac d{dt}left[(rho_2-rho_1)e^Fright]&=(rho'_2-rho'_1)e^F+(rho_2-rho_1)(rho_2+rho_1)e^F\
    &=e^Fleft(rho'_2-rho'_1+rho_2^2-rho_1^2right)
    end{align}$$
    and since $e^F>0$, your inequality implies that the derivative of $(rho_2-rho_1)e^F$ is positive. So the function $(rho_2-rho_1)e^F$ is increasing.



    On the other hand, a careless approach with some possible errors could be this:
    $$begin{align}
    rho_1'+ rho_1^2 leq rho_2'+ rho_2^2 &impliesrho_1'-rho_2' leq rho_2^2- rho_1^2\
    &implies-dfrac{rho_1'-rho_2'}{rho_1-rho_2}gerho_1+rho_2\
    &implies-ln|rho_1-rho_2|ge F\
    &impliesfrac 1{|rho_1-rho_2|}le e^Fimplies(rho_2-rho_1)e^Fge 1
    end{align}$$
    or something like that....






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      Right, but how does the fact that $rho_2 - rho_1 e^F$ is increasing imply that $rho_2 - rho_1 geq limsup _{t to 0}(rho_2(t) - rho_1(t))$
      $endgroup$
      – Selfstudier
      Jan 9 '17 at 13:29














    3












    3








    3





    $begingroup$

    Note that
    $$begin{align}frac d{dt}left[(rho_2-rho_1)e^Fright]&=(rho'_2-rho'_1)e^F+(rho_2-rho_1)(rho_2+rho_1)e^F\
    &=e^Fleft(rho'_2-rho'_1+rho_2^2-rho_1^2right)
    end{align}$$
    and since $e^F>0$, your inequality implies that the derivative of $(rho_2-rho_1)e^F$ is positive. So the function $(rho_2-rho_1)e^F$ is increasing.



    On the other hand, a careless approach with some possible errors could be this:
    $$begin{align}
    rho_1'+ rho_1^2 leq rho_2'+ rho_2^2 &impliesrho_1'-rho_2' leq rho_2^2- rho_1^2\
    &implies-dfrac{rho_1'-rho_2'}{rho_1-rho_2}gerho_1+rho_2\
    &implies-ln|rho_1-rho_2|ge F\
    &impliesfrac 1{|rho_1-rho_2|}le e^Fimplies(rho_2-rho_1)e^Fge 1
    end{align}$$
    or something like that....






    share|cite|improve this answer









    $endgroup$



    Note that
    $$begin{align}frac d{dt}left[(rho_2-rho_1)e^Fright]&=(rho'_2-rho'_1)e^F+(rho_2-rho_1)(rho_2+rho_1)e^F\
    &=e^Fleft(rho'_2-rho'_1+rho_2^2-rho_1^2right)
    end{align}$$
    and since $e^F>0$, your inequality implies that the derivative of $(rho_2-rho_1)e^F$ is positive. So the function $(rho_2-rho_1)e^F$ is increasing.



    On the other hand, a careless approach with some possible errors could be this:
    $$begin{align}
    rho_1'+ rho_1^2 leq rho_2'+ rho_2^2 &impliesrho_1'-rho_2' leq rho_2^2- rho_1^2\
    &implies-dfrac{rho_1'-rho_2'}{rho_1-rho_2}gerho_1+rho_2\
    &implies-ln|rho_1-rho_2|ge F\
    &impliesfrac 1{|rho_1-rho_2|}le e^Fimplies(rho_2-rho_1)e^Fge 1
    end{align}$$
    or something like that....







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 9 '17 at 8:23









    polfosolpolfosol

    5,93931945




    5,93931945








    • 2




      $begingroup$
      Right, but how does the fact that $rho_2 - rho_1 e^F$ is increasing imply that $rho_2 - rho_1 geq limsup _{t to 0}(rho_2(t) - rho_1(t))$
      $endgroup$
      – Selfstudier
      Jan 9 '17 at 13:29














    • 2




      $begingroup$
      Right, but how does the fact that $rho_2 - rho_1 e^F$ is increasing imply that $rho_2 - rho_1 geq limsup _{t to 0}(rho_2(t) - rho_1(t))$
      $endgroup$
      – Selfstudier
      Jan 9 '17 at 13:29








    2




    2




    $begingroup$
    Right, but how does the fact that $rho_2 - rho_1 e^F$ is increasing imply that $rho_2 - rho_1 geq limsup _{t to 0}(rho_2(t) - rho_1(t))$
    $endgroup$
    – Selfstudier
    Jan 9 '17 at 13:29




    $begingroup$
    Right, but how does the fact that $rho_2 - rho_1 e^F$ is increasing imply that $rho_2 - rho_1 geq limsup _{t to 0}(rho_2(t) - rho_1(t))$
    $endgroup$
    – Selfstudier
    Jan 9 '17 at 13:29











    2





    +50







    $begingroup$

    The book is wrong and it's easy to find counterexamples. For example if $ρ_1(x)=1/(x+2)$ and $ρ_2(x)=1/(x+1)$ then $ρ_i'+ρ_i^2=0$ and $ρ_2(0)-ρ_1(0)=1/2$ but $ρ_2(1)-ρ_1(1)=1/6.$



    The author may have been thinking of something like the following argument which I will paraphrase from "New comparison theorems in Riemannian geometry" by Y. Han and coauthors, Theorem 5.1.
    Suppose $ρ_1$ and $ρ_2$ are both $1/t+O(1).$
    Define $ϕ_i(t)=texp(int_0^t (ρ_i(s)-1/s)ds).$ Then each $ϕ_i$ is a $C^1$ function on $[0,b)$ (i.e. the right-sided derivative exists at zero - in fact it's $1$), smooth and strictly positive on $(0,b),$ with $ϕ_i(0)=0.$ Differentiating gives $ϕ'_i(t)=ρ_i(t) ϕ_i(t)$ and hence $ϕ''_i(t)=(ρ'_i(t)+ρ_i(t)^2)ϕ_i(t).$ This leads to
    $$(ϕ_2'ϕ_1-ϕ_1'ϕ_2)'=ϕ_2''ϕ_1-ϕ_1''ϕ_2=((ρ'_2+ρ_2^2)-(ρ'_1+ρ_1^2))ϕ_1ϕ_2geq 0.$$
    Using the fact that $ϕ_2'(0)ϕ_1(0)-ϕ_1'(0)ϕ_2(0)=0$ we get $ϕ_2'ϕ_1geq ϕ_1'ϕ_2.$ Using our previous equation for $ϕ_i',$ this means $ρ_2(t)geq ρ_1(t).$






    share|cite|improve this answer











    $endgroup$


















      2





      +50







      $begingroup$

      The book is wrong and it's easy to find counterexamples. For example if $ρ_1(x)=1/(x+2)$ and $ρ_2(x)=1/(x+1)$ then $ρ_i'+ρ_i^2=0$ and $ρ_2(0)-ρ_1(0)=1/2$ but $ρ_2(1)-ρ_1(1)=1/6.$



      The author may have been thinking of something like the following argument which I will paraphrase from "New comparison theorems in Riemannian geometry" by Y. Han and coauthors, Theorem 5.1.
      Suppose $ρ_1$ and $ρ_2$ are both $1/t+O(1).$
      Define $ϕ_i(t)=texp(int_0^t (ρ_i(s)-1/s)ds).$ Then each $ϕ_i$ is a $C^1$ function on $[0,b)$ (i.e. the right-sided derivative exists at zero - in fact it's $1$), smooth and strictly positive on $(0,b),$ with $ϕ_i(0)=0.$ Differentiating gives $ϕ'_i(t)=ρ_i(t) ϕ_i(t)$ and hence $ϕ''_i(t)=(ρ'_i(t)+ρ_i(t)^2)ϕ_i(t).$ This leads to
      $$(ϕ_2'ϕ_1-ϕ_1'ϕ_2)'=ϕ_2''ϕ_1-ϕ_1''ϕ_2=((ρ'_2+ρ_2^2)-(ρ'_1+ρ_1^2))ϕ_1ϕ_2geq 0.$$
      Using the fact that $ϕ_2'(0)ϕ_1(0)-ϕ_1'(0)ϕ_2(0)=0$ we get $ϕ_2'ϕ_1geq ϕ_1'ϕ_2.$ Using our previous equation for $ϕ_i',$ this means $ρ_2(t)geq ρ_1(t).$






      share|cite|improve this answer











      $endgroup$
















        2





        +50







        2





        +50



        2




        +50



        $begingroup$

        The book is wrong and it's easy to find counterexamples. For example if $ρ_1(x)=1/(x+2)$ and $ρ_2(x)=1/(x+1)$ then $ρ_i'+ρ_i^2=0$ and $ρ_2(0)-ρ_1(0)=1/2$ but $ρ_2(1)-ρ_1(1)=1/6.$



        The author may have been thinking of something like the following argument which I will paraphrase from "New comparison theorems in Riemannian geometry" by Y. Han and coauthors, Theorem 5.1.
        Suppose $ρ_1$ and $ρ_2$ are both $1/t+O(1).$
        Define $ϕ_i(t)=texp(int_0^t (ρ_i(s)-1/s)ds).$ Then each $ϕ_i$ is a $C^1$ function on $[0,b)$ (i.e. the right-sided derivative exists at zero - in fact it's $1$), smooth and strictly positive on $(0,b),$ with $ϕ_i(0)=0.$ Differentiating gives $ϕ'_i(t)=ρ_i(t) ϕ_i(t)$ and hence $ϕ''_i(t)=(ρ'_i(t)+ρ_i(t)^2)ϕ_i(t).$ This leads to
        $$(ϕ_2'ϕ_1-ϕ_1'ϕ_2)'=ϕ_2''ϕ_1-ϕ_1''ϕ_2=((ρ'_2+ρ_2^2)-(ρ'_1+ρ_1^2))ϕ_1ϕ_2geq 0.$$
        Using the fact that $ϕ_2'(0)ϕ_1(0)-ϕ_1'(0)ϕ_2(0)=0$ we get $ϕ_2'ϕ_1geq ϕ_1'ϕ_2.$ Using our previous equation for $ϕ_i',$ this means $ρ_2(t)geq ρ_1(t).$






        share|cite|improve this answer











        $endgroup$



        The book is wrong and it's easy to find counterexamples. For example if $ρ_1(x)=1/(x+2)$ and $ρ_2(x)=1/(x+1)$ then $ρ_i'+ρ_i^2=0$ and $ρ_2(0)-ρ_1(0)=1/2$ but $ρ_2(1)-ρ_1(1)=1/6.$



        The author may have been thinking of something like the following argument which I will paraphrase from "New comparison theorems in Riemannian geometry" by Y. Han and coauthors, Theorem 5.1.
        Suppose $ρ_1$ and $ρ_2$ are both $1/t+O(1).$
        Define $ϕ_i(t)=texp(int_0^t (ρ_i(s)-1/s)ds).$ Then each $ϕ_i$ is a $C^1$ function on $[0,b)$ (i.e. the right-sided derivative exists at zero - in fact it's $1$), smooth and strictly positive on $(0,b),$ with $ϕ_i(0)=0.$ Differentiating gives $ϕ'_i(t)=ρ_i(t) ϕ_i(t)$ and hence $ϕ''_i(t)=(ρ'_i(t)+ρ_i(t)^2)ϕ_i(t).$ This leads to
        $$(ϕ_2'ϕ_1-ϕ_1'ϕ_2)'=ϕ_2''ϕ_1-ϕ_1''ϕ_2=((ρ'_2+ρ_2^2)-(ρ'_1+ρ_1^2))ϕ_1ϕ_2geq 0.$$
        Using the fact that $ϕ_2'(0)ϕ_1(0)-ϕ_1'(0)ϕ_2(0)=0$ we get $ϕ_2'ϕ_1geq ϕ_1'ϕ_2.$ Using our previous equation for $ϕ_i',$ this means $ρ_2(t)geq ρ_1(t).$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 29 at 3:00









        Cookie

        8,786123884




        8,786123884










        answered Jan 25 at 4:50









        DapDap

        18.5k842




        18.5k842






























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            MongoDB - Not Authorized To Execute Command

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            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
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            How to fix TextFormField cause rebuild widget in Flutter

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            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
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            in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

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            1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
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