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












2














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?






differential-equations lipschitz-functions







share|cite|improve this question






















  • By Lipschitz in $y$ do you mean in $u$ in your notation? Or do you mean a bound of the form $K(|u_1 - u_2| + | v_1 -v_2|)$?
    – Arctic Char
    yesterday










  • Yes, exactly like the latter
    – Dylan Zammit
    yesterday






  • 1




    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.
    – LutzL
    yesterday


















2














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?






differential-equations lipschitz-functions







share|cite|improve this question






















  • By Lipschitz in $y$ do you mean in $u$ in your notation? Or do you mean a bound of the form $K(|u_1 - u_2| + | v_1 -v_2|)$?
    – Arctic Char
    yesterday










  • Yes, exactly like the latter
    – Dylan Zammit
    yesterday






  • 1




    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.
    – LutzL
    yesterday
















2












2








2







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?






differential-equations lipschitz-functions







share|cite|improve this question













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?






differential-equations lipschitz-functions




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 yesterday









Dylan Zammit

8801416




8801416












  • By Lipschitz in $y$ do you mean in $u$ in your notation? Or do you mean a bound of the form $K(|u_1 - u_2| + | v_1 -v_2|)$?
    – Arctic Char
    yesterday










  • Yes, exactly like the latter
    – Dylan Zammit
    yesterday






  • 1




    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.
    – LutzL
    yesterday




















  • By Lipschitz in $y$ do you mean in $u$ in your notation? Or do you mean a bound of the form $K(|u_1 - u_2| + | v_1 -v_2|)$?
    – Arctic Char
    yesterday










  • Yes, exactly like the latter
    – Dylan Zammit
    yesterday






  • 1




    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.
    – LutzL
    yesterday


















By Lipschitz in $y$ do you mean in $u$ in your notation? Or do you mean a bound of the form $K(|u_1 - u_2| + | v_1 -v_2|)$?
– Arctic Char
yesterday




By Lipschitz in $y$ do you mean in $u$ in your notation? Or do you mean a bound of the form $K(|u_1 - u_2| + | v_1 -v_2|)$?
– Arctic Char
yesterday












Yes, exactly like the latter
– Dylan Zammit
yesterday




Yes, exactly like the latter
– Dylan Zammit
yesterday




1




1




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.
– LutzL
yesterday






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.
– LutzL
yesterday












1 Answer
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oldest

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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 Answer
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    1 Answer
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    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


























      1














      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
























        1












        1








        1






        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












        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 yesterday









        LutzL

        56.3k42054




        56.3k42054






























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