Showing that the system of a planar curve is Lipschitz in y












2












$begingroup$


This question below, which was given in an exam, goes as follows:




A planar curve $y(x)$ is such that its curvature,
$$k(x)=y''(x)/(1+y'(x)^2)^{3/2}$$ is equal to its height $y(x)$. Write
the equation as a pair of first order equations, and show that the
system is Lipschitz in $y$.




Now my working is as follows:
Let $$u=y, v=y'$$
$$implies u'=v, v'=(1+v^2)^{3/2}u$$
which answers the first part of the question. Now to show that it is Lipschitz in $y$:
$$|u_1'-u_2'|=|v_1-v_2|leq |u_1-u_2|+|v_1-v_2|$$



My problem is for the second part:
begin{align}
|v_1'-v_2'|=&|(1+v_1^2)^{3/2}u_1-(1+v_2^2)^{3/2}u_2|\
=&|(1+v_1^2)^{3/2}u_1-(1+v_1^2)^{3/2}u_2+(1+v_1^2)^{3/2}u_2-(1+v_2^2)^{3/2}u_2|\
leq &|1+v_1^2|^{3/2}|u_1-u_2|+|u_2|||(1+v_1^2)^{3/2}-(1+v_2^2)^{3/2}|
end{align}



Now I can always bound the first term, but for the second term, I have no idea how to simplify in such a way to get $K|v_1-v_2|$ for some constant $K$. Any hints? Am I going about this the wrong way?






ordinary-differential-equations lipschitz-functions







share|cite|improve this question









$endgroup$












  • $begingroup$
    By Lipschitz in $y$ do you mean in $u$ in your notation? Or do you mean a bound of the form $K(|u_1 - u_2| + | v_1 -v_2|)$?
    $endgroup$
    – Arctic Char
    Jan 3 at 14:40










  • $begingroup$
    Yes, exactly like the latter
    $endgroup$
    – Dylan Zammit
    Jan 3 at 15:49






  • 1




    $begingroup$
    That it is locally Lipschitz follows from the differentiability of the right sides, thus uniqueness of solutions follows. As you found, this is not globally Lipschitz. Integration after multiplication by $2y'$ gives $C=y^2+(1+y'^2)^{-1/2}$ as first integral.
    $endgroup$
    – LutzL
    Jan 3 at 16:43


















2












$begingroup$


This question below, which was given in an exam, goes as follows:




A planar curve $y(x)$ is such that its curvature,
$$k(x)=y''(x)/(1+y'(x)^2)^{3/2}$$ is equal to its height $y(x)$. Write
the equation as a pair of first order equations, and show that the
system is Lipschitz in $y$.




Now my working is as follows:
Let $$u=y, v=y'$$
$$implies u'=v, v'=(1+v^2)^{3/2}u$$
which answers the first part of the question. Now to show that it is Lipschitz in $y$:
$$|u_1'-u_2'|=|v_1-v_2|leq |u_1-u_2|+|v_1-v_2|$$



My problem is for the second part:
begin{align}
|v_1'-v_2'|=&|(1+v_1^2)^{3/2}u_1-(1+v_2^2)^{3/2}u_2|\
=&|(1+v_1^2)^{3/2}u_1-(1+v_1^2)^{3/2}u_2+(1+v_1^2)^{3/2}u_2-(1+v_2^2)^{3/2}u_2|\
leq &|1+v_1^2|^{3/2}|u_1-u_2|+|u_2|||(1+v_1^2)^{3/2}-(1+v_2^2)^{3/2}|
end{align}



Now I can always bound the first term, but for the second term, I have no idea how to simplify in such a way to get $K|v_1-v_2|$ for some constant $K$. Any hints? Am I going about this the wrong way?






ordinary-differential-equations lipschitz-functions







share|cite|improve this question









$endgroup$












  • $begingroup$
    By Lipschitz in $y$ do you mean in $u$ in your notation? Or do you mean a bound of the form $K(|u_1 - u_2| + | v_1 -v_2|)$?
    $endgroup$
    – Arctic Char
    Jan 3 at 14:40










  • $begingroup$
    Yes, exactly like the latter
    $endgroup$
    – Dylan Zammit
    Jan 3 at 15:49






  • 1




    $begingroup$
    That it is locally Lipschitz follows from the differentiability of the right sides, thus uniqueness of solutions follows. As you found, this is not globally Lipschitz. Integration after multiplication by $2y'$ gives $C=y^2+(1+y'^2)^{-1/2}$ as first integral.
    $endgroup$
    – LutzL
    Jan 3 at 16:43
















2












2








2





$begingroup$


This question below, which was given in an exam, goes as follows:




A planar curve $y(x)$ is such that its curvature,
$$k(x)=y''(x)/(1+y'(x)^2)^{3/2}$$ is equal to its height $y(x)$. Write
the equation as a pair of first order equations, and show that the
system is Lipschitz in $y$.




Now my working is as follows:
Let $$u=y, v=y'$$
$$implies u'=v, v'=(1+v^2)^{3/2}u$$
which answers the first part of the question. Now to show that it is Lipschitz in $y$:
$$|u_1'-u_2'|=|v_1-v_2|leq |u_1-u_2|+|v_1-v_2|$$



My problem is for the second part:
begin{align}
|v_1'-v_2'|=&|(1+v_1^2)^{3/2}u_1-(1+v_2^2)^{3/2}u_2|\
=&|(1+v_1^2)^{3/2}u_1-(1+v_1^2)^{3/2}u_2+(1+v_1^2)^{3/2}u_2-(1+v_2^2)^{3/2}u_2|\
leq &|1+v_1^2|^{3/2}|u_1-u_2|+|u_2|||(1+v_1^2)^{3/2}-(1+v_2^2)^{3/2}|
end{align}



Now I can always bound the first term, but for the second term, I have no idea how to simplify in such a way to get $K|v_1-v_2|$ for some constant $K$. Any hints? Am I going about this the wrong way?






ordinary-differential-equations lipschitz-functions







share|cite|improve this question









$endgroup$




This question below, which was given in an exam, goes as follows:




A planar curve $y(x)$ is such that its curvature,
$$k(x)=y''(x)/(1+y'(x)^2)^{3/2}$$ is equal to its height $y(x)$. Write
the equation as a pair of first order equations, and show that the
system is Lipschitz in $y$.




Now my working is as follows:
Let $$u=y, v=y'$$
$$implies u'=v, v'=(1+v^2)^{3/2}u$$
which answers the first part of the question. Now to show that it is Lipschitz in $y$:
$$|u_1'-u_2'|=|v_1-v_2|leq |u_1-u_2|+|v_1-v_2|$$



My problem is for the second part:
begin{align}
|v_1'-v_2'|=&|(1+v_1^2)^{3/2}u_1-(1+v_2^2)^{3/2}u_2|\
=&|(1+v_1^2)^{3/2}u_1-(1+v_1^2)^{3/2}u_2+(1+v_1^2)^{3/2}u_2-(1+v_2^2)^{3/2}u_2|\
leq &|1+v_1^2|^{3/2}|u_1-u_2|+|u_2|||(1+v_1^2)^{3/2}-(1+v_2^2)^{3/2}|
end{align}



Now I can always bound the first term, but for the second term, I have no idea how to simplify in such a way to get $K|v_1-v_2|$ for some constant $K$. Any hints? Am I going about this the wrong way?






ordinary-differential-equations lipschitz-functions




ordinary-differential-equations lipschitz-functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 3 at 13:18









Dylan ZammitDylan Zammit

9151416




9151416












  • $begingroup$
    By Lipschitz in $y$ do you mean in $u$ in your notation? Or do you mean a bound of the form $K(|u_1 - u_2| + | v_1 -v_2|)$?
    $endgroup$
    – Arctic Char
    Jan 3 at 14:40










  • $begingroup$
    Yes, exactly like the latter
    $endgroup$
    – Dylan Zammit
    Jan 3 at 15:49






  • 1




    $begingroup$
    That it is locally Lipschitz follows from the differentiability of the right sides, thus uniqueness of solutions follows. As you found, this is not globally Lipschitz. Integration after multiplication by $2y'$ gives $C=y^2+(1+y'^2)^{-1/2}$ as first integral.
    $endgroup$
    – LutzL
    Jan 3 at 16:43




















  • $begingroup$
    By Lipschitz in $y$ do you mean in $u$ in your notation? Or do you mean a bound of the form $K(|u_1 - u_2| + | v_1 -v_2|)$?
    $endgroup$
    – Arctic Char
    Jan 3 at 14:40










  • $begingroup$
    Yes, exactly like the latter
    $endgroup$
    – Dylan Zammit
    Jan 3 at 15:49






  • 1




    $begingroup$
    That it is locally Lipschitz follows from the differentiability of the right sides, thus uniqueness of solutions follows. As you found, this is not globally Lipschitz. Integration after multiplication by $2y'$ gives $C=y^2+(1+y'^2)^{-1/2}$ as first integral.
    $endgroup$
    – LutzL
    Jan 3 at 16:43


















$begingroup$
By Lipschitz in $y$ do you mean in $u$ in your notation? Or do you mean a bound of the form $K(|u_1 - u_2| + | v_1 -v_2|)$?
$endgroup$
– Arctic Char
Jan 3 at 14:40




$begingroup$
By Lipschitz in $y$ do you mean in $u$ in your notation? Or do you mean a bound of the form $K(|u_1 - u_2| + | v_1 -v_2|)$?
$endgroup$
– Arctic Char
Jan 3 at 14:40












$begingroup$
Yes, exactly like the latter
$endgroup$
– Dylan Zammit
Jan 3 at 15:49




$begingroup$
Yes, exactly like the latter
$endgroup$
– Dylan Zammit
Jan 3 at 15:49




1




1




$begingroup$
That it is locally Lipschitz follows from the differentiability of the right sides, thus uniqueness of solutions follows. As you found, this is not globally Lipschitz. Integration after multiplication by $2y'$ gives $C=y^2+(1+y'^2)^{-1/2}$ as first integral.
$endgroup$
– LutzL
Jan 3 at 16:43






$begingroup$
That it is locally Lipschitz follows from the differentiability of the right sides, thus uniqueness of solutions follows. As you found, this is not globally Lipschitz. Integration after multiplication by $2y'$ gives $C=y^2+(1+y'^2)^{-1/2}$ as first integral.
$endgroup$
– LutzL
Jan 3 at 16:43












1 Answer
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oldest

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$begingroup$

Apply either a mean value theorem or the binomial theorem as in
$$
(1+v_1^2)^{3/2}-(1+v_2^2)^{3/2}=frac{(1+v_1^2)^3-(1+v_2^2)^3}{(1+v_1^2)^{3/2}+(1+v_2^2)^{3/2}}
\
=frac{(v_1-v_2)(v_1+v_2)Bigl((1+v_1^2)^2+(1+v_1^2)(1+v_2^2)+(1+v_2^2)^2Bigr)}{(1+v_1^2)^{3/2}+(1+v_2^2)^{3/2}}
$$
where you get the simple difference as factors and can bound the other factors.






share|cite|improve this answer









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    1












    $begingroup$

    Apply either a mean value theorem or the binomial theorem as in
    $$
    (1+v_1^2)^{3/2}-(1+v_2^2)^{3/2}=frac{(1+v_1^2)^3-(1+v_2^2)^3}{(1+v_1^2)^{3/2}+(1+v_2^2)^{3/2}}
    \
    =frac{(v_1-v_2)(v_1+v_2)Bigl((1+v_1^2)^2+(1+v_1^2)(1+v_2^2)+(1+v_2^2)^2Bigr)}{(1+v_1^2)^{3/2}+(1+v_2^2)^{3/2}}
    $$
    where you get the simple difference as factors and can bound the other factors.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Apply either a mean value theorem or the binomial theorem as in
      $$
      (1+v_1^2)^{3/2}-(1+v_2^2)^{3/2}=frac{(1+v_1^2)^3-(1+v_2^2)^3}{(1+v_1^2)^{3/2}+(1+v_2^2)^{3/2}}
      \
      =frac{(v_1-v_2)(v_1+v_2)Bigl((1+v_1^2)^2+(1+v_1^2)(1+v_2^2)+(1+v_2^2)^2Bigr)}{(1+v_1^2)^{3/2}+(1+v_2^2)^{3/2}}
      $$
      where you get the simple difference as factors and can bound the other factors.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Apply either a mean value theorem or the binomial theorem as in
        $$
        (1+v_1^2)^{3/2}-(1+v_2^2)^{3/2}=frac{(1+v_1^2)^3-(1+v_2^2)^3}{(1+v_1^2)^{3/2}+(1+v_2^2)^{3/2}}
        \
        =frac{(v_1-v_2)(v_1+v_2)Bigl((1+v_1^2)^2+(1+v_1^2)(1+v_2^2)+(1+v_2^2)^2Bigr)}{(1+v_1^2)^{3/2}+(1+v_2^2)^{3/2}}
        $$
        where you get the simple difference as factors and can bound the other factors.






        share|cite|improve this answer









        $endgroup$



        Apply either a mean value theorem or the binomial theorem as in
        $$
        (1+v_1^2)^{3/2}-(1+v_2^2)^{3/2}=frac{(1+v_1^2)^3-(1+v_2^2)^3}{(1+v_1^2)^{3/2}+(1+v_2^2)^{3/2}}
        \
        =frac{(v_1-v_2)(v_1+v_2)Bigl((1+v_1^2)^2+(1+v_1^2)(1+v_2^2)+(1+v_2^2)^2Bigr)}{(1+v_1^2)^{3/2}+(1+v_2^2)^{3/2}}
        $$
        where you get the simple difference as factors and can bound the other factors.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 3 at 16:51









        LutzLLutzL

        59.4k42057




        59.4k42057






























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