Local implicit function theorem and Global implicit function theorem












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Suppose that $f=f(x,y)$ is $C^1$ in $x$ and $y$, where $x,yinmathbb{R}$ and $f(x_0,y_0)=0$. Moreover, $frac{partial f(x,y)}{partial y}>0$ for all $x,yinmathbb{R}$. Could we conclude that $y$ can be solved implicitly from $f(x,y)=0$ in terms of $x$ as $y=y(x)$ for all $xinmathbb{R}$? If the answer is no and an additional assumption is imposed, i.e. we can obtain $y'(x)>0$ for all $xinmathbb{R}$ by implicitly differentiating $f(x,y(x))=0$ with respect to $x$, then these assumptions are enough to get the same conclusion?






implicit-function-theorem







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  • $begingroup$
    maybe this is of interest for you: math.stackexchange.com/questions/678417/…
    $endgroup$
    – Thomas
    Jan 3 '15 at 9:46










  • $begingroup$
    Yes, that link helps much, thanks! I'm thinking what you said about using the mean value theorem to show that $varphi$ can be extended in $C^1$ to the boundary of any finite interval, hence to $mathbb{R}.$
    $endgroup$
    – LCH
    Jan 3 '15 at 12:50












  • $begingroup$
    Would you please give some hints?
    $endgroup$
    – LCH
    Jan 3 '15 at 13:46
















0












$begingroup$


Suppose that $f=f(x,y)$ is $C^1$ in $x$ and $y$, where $x,yinmathbb{R}$ and $f(x_0,y_0)=0$. Moreover, $frac{partial f(x,y)}{partial y}>0$ for all $x,yinmathbb{R}$. Could we conclude that $y$ can be solved implicitly from $f(x,y)=0$ in terms of $x$ as $y=y(x)$ for all $xinmathbb{R}$? If the answer is no and an additional assumption is imposed, i.e. we can obtain $y'(x)>0$ for all $xinmathbb{R}$ by implicitly differentiating $f(x,y(x))=0$ with respect to $x$, then these assumptions are enough to get the same conclusion?






implicit-function-theorem







share|cite|improve this question











$endgroup$












  • $begingroup$
    maybe this is of interest for you: math.stackexchange.com/questions/678417/…
    $endgroup$
    – Thomas
    Jan 3 '15 at 9:46










  • $begingroup$
    Yes, that link helps much, thanks! I'm thinking what you said about using the mean value theorem to show that $varphi$ can be extended in $C^1$ to the boundary of any finite interval, hence to $mathbb{R}.$
    $endgroup$
    – LCH
    Jan 3 '15 at 12:50












  • $begingroup$
    Would you please give some hints?
    $endgroup$
    – LCH
    Jan 3 '15 at 13:46














0












0








0





$begingroup$


Suppose that $f=f(x,y)$ is $C^1$ in $x$ and $y$, where $x,yinmathbb{R}$ and $f(x_0,y_0)=0$. Moreover, $frac{partial f(x,y)}{partial y}>0$ for all $x,yinmathbb{R}$. Could we conclude that $y$ can be solved implicitly from $f(x,y)=0$ in terms of $x$ as $y=y(x)$ for all $xinmathbb{R}$? If the answer is no and an additional assumption is imposed, i.e. we can obtain $y'(x)>0$ for all $xinmathbb{R}$ by implicitly differentiating $f(x,y(x))=0$ with respect to $x$, then these assumptions are enough to get the same conclusion?






implicit-function-theorem







share|cite|improve this question











$endgroup$




Suppose that $f=f(x,y)$ is $C^1$ in $x$ and $y$, where $x,yinmathbb{R}$ and $f(x_0,y_0)=0$. Moreover, $frac{partial f(x,y)}{partial y}>0$ for all $x,yinmathbb{R}$. Could we conclude that $y$ can be solved implicitly from $f(x,y)=0$ in terms of $x$ as $y=y(x)$ for all $xinmathbb{R}$? If the answer is no and an additional assumption is imposed, i.e. we can obtain $y'(x)>0$ for all $xinmathbb{R}$ by implicitly differentiating $f(x,y(x))=0$ with respect to $x$, then these assumptions are enough to get the same conclusion?






implicit-function-theorem




implicit-function-theorem






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share|cite|improve this question













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share|cite|improve this question








edited Jan 3 '15 at 9:32







LCH

















asked Jan 3 '15 at 9:26









LCHLCH

381111




381111












  • $begingroup$
    maybe this is of interest for you: math.stackexchange.com/questions/678417/…
    $endgroup$
    – Thomas
    Jan 3 '15 at 9:46










  • $begingroup$
    Yes, that link helps much, thanks! I'm thinking what you said about using the mean value theorem to show that $varphi$ can be extended in $C^1$ to the boundary of any finite interval, hence to $mathbb{R}.$
    $endgroup$
    – LCH
    Jan 3 '15 at 12:50












  • $begingroup$
    Would you please give some hints?
    $endgroup$
    – LCH
    Jan 3 '15 at 13:46


















  • $begingroup$
    maybe this is of interest for you: math.stackexchange.com/questions/678417/…
    $endgroup$
    – Thomas
    Jan 3 '15 at 9:46










  • $begingroup$
    Yes, that link helps much, thanks! I'm thinking what you said about using the mean value theorem to show that $varphi$ can be extended in $C^1$ to the boundary of any finite interval, hence to $mathbb{R}.$
    $endgroup$
    – LCH
    Jan 3 '15 at 12:50












  • $begingroup$
    Would you please give some hints?
    $endgroup$
    – LCH
    Jan 3 '15 at 13:46
















$begingroup$
maybe this is of interest for you: math.stackexchange.com/questions/678417/…
$endgroup$
– Thomas
Jan 3 '15 at 9:46




$begingroup$
maybe this is of interest for you: math.stackexchange.com/questions/678417/…
$endgroup$
– Thomas
Jan 3 '15 at 9:46












$begingroup$
Yes, that link helps much, thanks! I'm thinking what you said about using the mean value theorem to show that $varphi$ can be extended in $C^1$ to the boundary of any finite interval, hence to $mathbb{R}.$
$endgroup$
– LCH
Jan 3 '15 at 12:50






$begingroup$
Yes, that link helps much, thanks! I'm thinking what you said about using the mean value theorem to show that $varphi$ can be extended in $C^1$ to the boundary of any finite interval, hence to $mathbb{R}.$
$endgroup$
– LCH
Jan 3 '15 at 12:50














$begingroup$
Would you please give some hints?
$endgroup$
– LCH
Jan 3 '15 at 13:46




$begingroup$
Would you please give some hints?
$endgroup$
– LCH
Jan 3 '15 at 13:46










2 Answers
2






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$begingroup$

I'm not sure what you want to know beyond what is shown in this question and it's answer, but that question shows already where you have to expect the result to break down. As an example, consider



$$F(x,y) = - e^{-y} +x$$
Clearly, $frac{partial F}{partial y} = e^{-y}>0$ everywhere, but it is easy to see that $F$ is strictly negative for $xle0$ on the one hand side but admits a (obviously unique) solution of $F(x,y) = 0$ for each $x>0$. This shows that if the bounds on the derivates no longer hold, the solution may go off to infinity (Which is exactly what happens in this example).






share|cite|improve this answer











$endgroup$





















    -1












    $begingroup$

    For Global Implicit Function Theorem you need "somewhat" stronger assumption, namely, $frac{partial f(x,y)}{partial y}geq d>0,$ for all $(x,y)inmathbb{R}^2,$ $d$ is constant. See, for example, Lemma 1 and its proof in:



    Ge, S.S.; Wang, C., Adaptive NN control of uncertain nonlinear pure-feedback systems, Automatica 38, No.4, 671-682 (2002). ZBL0998.93025.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

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      active

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      0












      $begingroup$

      I'm not sure what you want to know beyond what is shown in this question and it's answer, but that question shows already where you have to expect the result to break down. As an example, consider



      $$F(x,y) = - e^{-y} +x$$
      Clearly, $frac{partial F}{partial y} = e^{-y}>0$ everywhere, but it is easy to see that $F$ is strictly negative for $xle0$ on the one hand side but admits a (obviously unique) solution of $F(x,y) = 0$ for each $x>0$. This shows that if the bounds on the derivates no longer hold, the solution may go off to infinity (Which is exactly what happens in this example).






      share|cite|improve this answer











      $endgroup$


















        0












        $begingroup$

        I'm not sure what you want to know beyond what is shown in this question and it's answer, but that question shows already where you have to expect the result to break down. As an example, consider



        $$F(x,y) = - e^{-y} +x$$
        Clearly, $frac{partial F}{partial y} = e^{-y}>0$ everywhere, but it is easy to see that $F$ is strictly negative for $xle0$ on the one hand side but admits a (obviously unique) solution of $F(x,y) = 0$ for each $x>0$. This shows that if the bounds on the derivates no longer hold, the solution may go off to infinity (Which is exactly what happens in this example).






        share|cite|improve this answer











        $endgroup$
















          0












          0








          0





          $begingroup$

          I'm not sure what you want to know beyond what is shown in this question and it's answer, but that question shows already where you have to expect the result to break down. As an example, consider



          $$F(x,y) = - e^{-y} +x$$
          Clearly, $frac{partial F}{partial y} = e^{-y}>0$ everywhere, but it is easy to see that $F$ is strictly negative for $xle0$ on the one hand side but admits a (obviously unique) solution of $F(x,y) = 0$ for each $x>0$. This shows that if the bounds on the derivates no longer hold, the solution may go off to infinity (Which is exactly what happens in this example).






          share|cite|improve this answer











          $endgroup$



          I'm not sure what you want to know beyond what is shown in this question and it's answer, but that question shows already where you have to expect the result to break down. As an example, consider



          $$F(x,y) = - e^{-y} +x$$
          Clearly, $frac{partial F}{partial y} = e^{-y}>0$ everywhere, but it is easy to see that $F$ is strictly negative for $xle0$ on the one hand side but admits a (obviously unique) solution of $F(x,y) = 0$ for each $x>0$. This shows that if the bounds on the derivates no longer hold, the solution may go off to infinity (Which is exactly what happens in this example).







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Apr 13 '17 at 12:21









          Community♦

          1




          1










          answered Jan 3 '15 at 15:12









          ThomasThomas

          16.9k21731




          16.9k21731























              -1












              $begingroup$

              For Global Implicit Function Theorem you need "somewhat" stronger assumption, namely, $frac{partial f(x,y)}{partial y}geq d>0,$ for all $(x,y)inmathbb{R}^2,$ $d$ is constant. See, for example, Lemma 1 and its proof in:



              Ge, S.S.; Wang, C., Adaptive NN control of uncertain nonlinear pure-feedback systems, Automatica 38, No.4, 671-682 (2002). ZBL0998.93025.






              share|cite|improve this answer









              $endgroup$


















                -1












                $begingroup$

                For Global Implicit Function Theorem you need "somewhat" stronger assumption, namely, $frac{partial f(x,y)}{partial y}geq d>0,$ for all $(x,y)inmathbb{R}^2,$ $d$ is constant. See, for example, Lemma 1 and its proof in:



                Ge, S.S.; Wang, C., Adaptive NN control of uncertain nonlinear pure-feedback systems, Automatica 38, No.4, 671-682 (2002). ZBL0998.93025.






                share|cite|improve this answer









                $endgroup$
















                  -1












                  -1








                  -1





                  $begingroup$

                  For Global Implicit Function Theorem you need "somewhat" stronger assumption, namely, $frac{partial f(x,y)}{partial y}geq d>0,$ for all $(x,y)inmathbb{R}^2,$ $d$ is constant. See, for example, Lemma 1 and its proof in:



                  Ge, S.S.; Wang, C., Adaptive NN control of uncertain nonlinear pure-feedback systems, Automatica 38, No.4, 671-682 (2002). ZBL0998.93025.






                  share|cite|improve this answer









                  $endgroup$



                  For Global Implicit Function Theorem you need "somewhat" stronger assumption, namely, $frac{partial f(x,y)}{partial y}geq d>0,$ for all $(x,y)inmathbb{R}^2,$ $d$ is constant. See, for example, Lemma 1 and its proof in:



                  Ge, S.S.; Wang, C., Adaptive NN control of uncertain nonlinear pure-feedback systems, Automatica 38, No.4, 671-682 (2002). ZBL0998.93025.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Apr 9 '18 at 6:10









                  RobertRobert

                  112




                  112






























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