Lipschitz constant for a second order nonlinear differential equation












1












$begingroup$


I'm trying to calculate the Lipschitz constant for a second order nonlinear differential equation:



$$y'' + y' + y^n = 0, ; y(0)=0, ; y'(0)= 0 text{ and } n>1$$



Should I solve for $y(x)$ and differentiate the solution to find a bound ($L$, Lipschitz constant) on $vert f'(x) vert$? Or does it mean that I have to differentiate $f(x)= y'' + y' + y^n$? I would like some tips to proceed as this is my first calculation of this kind. Thanks in advance.






ordinary-differential-equations lipschitz-functions







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I'm trying to calculate the Lipschitz constant for a second order nonlinear differential equation:



    $$y'' + y' + y^n = 0, ; y(0)=0, ; y'(0)= 0 text{ and } n>1$$



    Should I solve for $y(x)$ and differentiate the solution to find a bound ($L$, Lipschitz constant) on $vert f'(x) vert$? Or does it mean that I have to differentiate $f(x)= y'' + y' + y^n$? I would like some tips to proceed as this is my first calculation of this kind. Thanks in advance.






    ordinary-differential-equations lipschitz-functions







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I'm trying to calculate the Lipschitz constant for a second order nonlinear differential equation:



      $$y'' + y' + y^n = 0, ; y(0)=0, ; y'(0)= 0 text{ and } n>1$$



      Should I solve for $y(x)$ and differentiate the solution to find a bound ($L$, Lipschitz constant) on $vert f'(x) vert$? Or does it mean that I have to differentiate $f(x)= y'' + y' + y^n$? I would like some tips to proceed as this is my first calculation of this kind. Thanks in advance.






      ordinary-differential-equations lipschitz-functions







      share|cite|improve this question











      $endgroup$




      I'm trying to calculate the Lipschitz constant for a second order nonlinear differential equation:



      $$y'' + y' + y^n = 0, ; y(0)=0, ; y'(0)= 0 text{ and } n>1$$



      Should I solve for $y(x)$ and differentiate the solution to find a bound ($L$, Lipschitz constant) on $vert f'(x) vert$? Or does it mean that I have to differentiate $f(x)= y'' + y' + y^n$? I would like some tips to proceed as this is my first calculation of this kind. Thanks in advance.






      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








      edited Dec 27 '18 at 22:44







      Nora oBrist

















      asked Dec 26 '18 at 22:49









      Nora oBristNora oBrist

      83




      83






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          To apply the formalism you use the Lipschitz constant for, you first need to transform the ODE into a first order system,
          begin{align}
          y'&=v\
          v'&=-v-y^n.
          end{align}
          Next we know that non-linear polynomials are never globally Lipschitz, you only get a local Lipschitz constant on a set like $|y|<R$. Then
          begin{align}
          |F(y_2,v_2)-F(y_1,v_1)|_{max}&=max(|v_2-v_1|,|v_2-v_1+y_2^n-y_1^n|)\
          &le |v_2-v_1|+nR^{n-1}|y_2-y_1|\
          &le (1+nR^{n-1})|(y_2,v_2)-(y_1,v_1)|_{max}
          end{align}
          This is sufficient to prove uniqueness, but only local existence. No claim on the maximal domain follows from this bound.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you! I will still need some time to assimilate all the information but this is really helpfull. Happy hollydays LutzL!
            $endgroup$
            – Nora oBrist
            Dec 27 '18 at 22:34










          • $begingroup$
            After a while, I have been able to prove that $$|y_2^n -y_1^n| simeq nR^{n-1}|y_2 - y_1| $$ where R is the possible interval [-R,R] but I'm not sure how to get rid off $|v_2 - v_1|$ and combine it to get $|(y_2,v_2) - (y_1,v_1)|_{max}$. What book would you recommend in this matter? Again, thanks in advance.
            $endgroup$
            – Nora oBrist
            Dec 29 '18 at 18:10












          • $begingroup$
            I know that Lipschitz constant is $(1+nR^{n-1})$, however I'm not sure how would it affect if the second equation of the system was $v' = -2v - y^n$, for instance.
            $endgroup$
            – Nora oBrist
            Dec 29 '18 at 18:21










          • $begingroup$
            Then you get $(2+nR^{n-1})$ as Lipschitz constant. By playing with the norm, scaling one of the components, one can reduce the Lipschitz constant somewhat, distribute it over the equations.
            $endgroup$
            – LutzL
            Dec 29 '18 at 18:26










          • $begingroup$
            This is becoming a master class, but I have got one more question. To be a contraction mapping it is required L< 1, something that in this case we have not. Fixed point theorem is not allowed, however starting from a close enough solution it converges. From your experience, have you seen this happening sometimes? Also, when you told that it is allowed to have a L constant on a set $|y|< R$ but I get on $|x|< R$, is there something that I'm missing? Thank you, I appreciate your help because this is much like art to me.
            $endgroup$
            – Nora oBrist
            Jan 4 at 20:01











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          1 Answer
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          0












          $begingroup$

          To apply the formalism you use the Lipschitz constant for, you first need to transform the ODE into a first order system,
          begin{align}
          y'&=v\
          v'&=-v-y^n.
          end{align}
          Next we know that non-linear polynomials are never globally Lipschitz, you only get a local Lipschitz constant on a set like $|y|<R$. Then
          begin{align}
          |F(y_2,v_2)-F(y_1,v_1)|_{max}&=max(|v_2-v_1|,|v_2-v_1+y_2^n-y_1^n|)\
          &le |v_2-v_1|+nR^{n-1}|y_2-y_1|\
          &le (1+nR^{n-1})|(y_2,v_2)-(y_1,v_1)|_{max}
          end{align}
          This is sufficient to prove uniqueness, but only local existence. No claim on the maximal domain follows from this bound.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you! I will still need some time to assimilate all the information but this is really helpfull. Happy hollydays LutzL!
            $endgroup$
            – Nora oBrist
            Dec 27 '18 at 22:34










          • $begingroup$
            After a while, I have been able to prove that $$|y_2^n -y_1^n| simeq nR^{n-1}|y_2 - y_1| $$ where R is the possible interval [-R,R] but I'm not sure how to get rid off $|v_2 - v_1|$ and combine it to get $|(y_2,v_2) - (y_1,v_1)|_{max}$. What book would you recommend in this matter? Again, thanks in advance.
            $endgroup$
            – Nora oBrist
            Dec 29 '18 at 18:10












          • $begingroup$
            I know that Lipschitz constant is $(1+nR^{n-1})$, however I'm not sure how would it affect if the second equation of the system was $v' = -2v - y^n$, for instance.
            $endgroup$
            – Nora oBrist
            Dec 29 '18 at 18:21










          • $begingroup$
            Then you get $(2+nR^{n-1})$ as Lipschitz constant. By playing with the norm, scaling one of the components, one can reduce the Lipschitz constant somewhat, distribute it over the equations.
            $endgroup$
            – LutzL
            Dec 29 '18 at 18:26










          • $begingroup$
            This is becoming a master class, but I have got one more question. To be a contraction mapping it is required L< 1, something that in this case we have not. Fixed point theorem is not allowed, however starting from a close enough solution it converges. From your experience, have you seen this happening sometimes? Also, when you told that it is allowed to have a L constant on a set $|y|< R$ but I get on $|x|< R$, is there something that I'm missing? Thank you, I appreciate your help because this is much like art to me.
            $endgroup$
            – Nora oBrist
            Jan 4 at 20:01
















          0












          $begingroup$

          To apply the formalism you use the Lipschitz constant for, you first need to transform the ODE into a first order system,
          begin{align}
          y'&=v\
          v'&=-v-y^n.
          end{align}
          Next we know that non-linear polynomials are never globally Lipschitz, you only get a local Lipschitz constant on a set like $|y|<R$. Then
          begin{align}
          |F(y_2,v_2)-F(y_1,v_1)|_{max}&=max(|v_2-v_1|,|v_2-v_1+y_2^n-y_1^n|)\
          &le |v_2-v_1|+nR^{n-1}|y_2-y_1|\
          &le (1+nR^{n-1})|(y_2,v_2)-(y_1,v_1)|_{max}
          end{align}
          This is sufficient to prove uniqueness, but only local existence. No claim on the maximal domain follows from this bound.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you! I will still need some time to assimilate all the information but this is really helpfull. Happy hollydays LutzL!
            $endgroup$
            – Nora oBrist
            Dec 27 '18 at 22:34










          • $begingroup$
            After a while, I have been able to prove that $$|y_2^n -y_1^n| simeq nR^{n-1}|y_2 - y_1| $$ where R is the possible interval [-R,R] but I'm not sure how to get rid off $|v_2 - v_1|$ and combine it to get $|(y_2,v_2) - (y_1,v_1)|_{max}$. What book would you recommend in this matter? Again, thanks in advance.
            $endgroup$
            – Nora oBrist
            Dec 29 '18 at 18:10












          • $begingroup$
            I know that Lipschitz constant is $(1+nR^{n-1})$, however I'm not sure how would it affect if the second equation of the system was $v' = -2v - y^n$, for instance.
            $endgroup$
            – Nora oBrist
            Dec 29 '18 at 18:21










          • $begingroup$
            Then you get $(2+nR^{n-1})$ as Lipschitz constant. By playing with the norm, scaling one of the components, one can reduce the Lipschitz constant somewhat, distribute it over the equations.
            $endgroup$
            – LutzL
            Dec 29 '18 at 18:26










          • $begingroup$
            This is becoming a master class, but I have got one more question. To be a contraction mapping it is required L< 1, something that in this case we have not. Fixed point theorem is not allowed, however starting from a close enough solution it converges. From your experience, have you seen this happening sometimes? Also, when you told that it is allowed to have a L constant on a set $|y|< R$ but I get on $|x|< R$, is there something that I'm missing? Thank you, I appreciate your help because this is much like art to me.
            $endgroup$
            – Nora oBrist
            Jan 4 at 20:01














          0












          0








          0





          $begingroup$

          To apply the formalism you use the Lipschitz constant for, you first need to transform the ODE into a first order system,
          begin{align}
          y'&=v\
          v'&=-v-y^n.
          end{align}
          Next we know that non-linear polynomials are never globally Lipschitz, you only get a local Lipschitz constant on a set like $|y|<R$. Then
          begin{align}
          |F(y_2,v_2)-F(y_1,v_1)|_{max}&=max(|v_2-v_1|,|v_2-v_1+y_2^n-y_1^n|)\
          &le |v_2-v_1|+nR^{n-1}|y_2-y_1|\
          &le (1+nR^{n-1})|(y_2,v_2)-(y_1,v_1)|_{max}
          end{align}
          This is sufficient to prove uniqueness, but only local existence. No claim on the maximal domain follows from this bound.






          share|cite|improve this answer









          $endgroup$



          To apply the formalism you use the Lipschitz constant for, you first need to transform the ODE into a first order system,
          begin{align}
          y'&=v\
          v'&=-v-y^n.
          end{align}
          Next we know that non-linear polynomials are never globally Lipschitz, you only get a local Lipschitz constant on a set like $|y|<R$. Then
          begin{align}
          |F(y_2,v_2)-F(y_1,v_1)|_{max}&=max(|v_2-v_1|,|v_2-v_1+y_2^n-y_1^n|)\
          &le |v_2-v_1|+nR^{n-1}|y_2-y_1|\
          &le (1+nR^{n-1})|(y_2,v_2)-(y_1,v_1)|_{max}
          end{align}
          This is sufficient to prove uniqueness, but only local existence. No claim on the maximal domain follows from this bound.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 27 '18 at 11:06









          LutzLLutzL

          58.7k42055




          58.7k42055












          • $begingroup$
            Thank you! I will still need some time to assimilate all the information but this is really helpfull. Happy hollydays LutzL!
            $endgroup$
            – Nora oBrist
            Dec 27 '18 at 22:34










          • $begingroup$
            After a while, I have been able to prove that $$|y_2^n -y_1^n| simeq nR^{n-1}|y_2 - y_1| $$ where R is the possible interval [-R,R] but I'm not sure how to get rid off $|v_2 - v_1|$ and combine it to get $|(y_2,v_2) - (y_1,v_1)|_{max}$. What book would you recommend in this matter? Again, thanks in advance.
            $endgroup$
            – Nora oBrist
            Dec 29 '18 at 18:10












          • $begingroup$
            I know that Lipschitz constant is $(1+nR^{n-1})$, however I'm not sure how would it affect if the second equation of the system was $v' = -2v - y^n$, for instance.
            $endgroup$
            – Nora oBrist
            Dec 29 '18 at 18:21










          • $begingroup$
            Then you get $(2+nR^{n-1})$ as Lipschitz constant. By playing with the norm, scaling one of the components, one can reduce the Lipschitz constant somewhat, distribute it over the equations.
            $endgroup$
            – LutzL
            Dec 29 '18 at 18:26










          • $begingroup$
            This is becoming a master class, but I have got one more question. To be a contraction mapping it is required L< 1, something that in this case we have not. Fixed point theorem is not allowed, however starting from a close enough solution it converges. From your experience, have you seen this happening sometimes? Also, when you told that it is allowed to have a L constant on a set $|y|< R$ but I get on $|x|< R$, is there something that I'm missing? Thank you, I appreciate your help because this is much like art to me.
            $endgroup$
            – Nora oBrist
            Jan 4 at 20:01


















          • $begingroup$
            Thank you! I will still need some time to assimilate all the information but this is really helpfull. Happy hollydays LutzL!
            $endgroup$
            – Nora oBrist
            Dec 27 '18 at 22:34










          • $begingroup$
            After a while, I have been able to prove that $$|y_2^n -y_1^n| simeq nR^{n-1}|y_2 - y_1| $$ where R is the possible interval [-R,R] but I'm not sure how to get rid off $|v_2 - v_1|$ and combine it to get $|(y_2,v_2) - (y_1,v_1)|_{max}$. What book would you recommend in this matter? Again, thanks in advance.
            $endgroup$
            – Nora oBrist
            Dec 29 '18 at 18:10












          • $begingroup$
            I know that Lipschitz constant is $(1+nR^{n-1})$, however I'm not sure how would it affect if the second equation of the system was $v' = -2v - y^n$, for instance.
            $endgroup$
            – Nora oBrist
            Dec 29 '18 at 18:21










          • $begingroup$
            Then you get $(2+nR^{n-1})$ as Lipschitz constant. By playing with the norm, scaling one of the components, one can reduce the Lipschitz constant somewhat, distribute it over the equations.
            $endgroup$
            – LutzL
            Dec 29 '18 at 18:26










          • $begingroup$
            This is becoming a master class, but I have got one more question. To be a contraction mapping it is required L< 1, something that in this case we have not. Fixed point theorem is not allowed, however starting from a close enough solution it converges. From your experience, have you seen this happening sometimes? Also, when you told that it is allowed to have a L constant on a set $|y|< R$ but I get on $|x|< R$, is there something that I'm missing? Thank you, I appreciate your help because this is much like art to me.
            $endgroup$
            – Nora oBrist
            Jan 4 at 20:01
















          $begingroup$
          Thank you! I will still need some time to assimilate all the information but this is really helpfull. Happy hollydays LutzL!
          $endgroup$
          – Nora oBrist
          Dec 27 '18 at 22:34




          $begingroup$
          Thank you! I will still need some time to assimilate all the information but this is really helpfull. Happy hollydays LutzL!
          $endgroup$
          – Nora oBrist
          Dec 27 '18 at 22:34












          $begingroup$
          After a while, I have been able to prove that $$|y_2^n -y_1^n| simeq nR^{n-1}|y_2 - y_1| $$ where R is the possible interval [-R,R] but I'm not sure how to get rid off $|v_2 - v_1|$ and combine it to get $|(y_2,v_2) - (y_1,v_1)|_{max}$. What book would you recommend in this matter? Again, thanks in advance.
          $endgroup$
          – Nora oBrist
          Dec 29 '18 at 18:10






          $begingroup$
          After a while, I have been able to prove that $$|y_2^n -y_1^n| simeq nR^{n-1}|y_2 - y_1| $$ where R is the possible interval [-R,R] but I'm not sure how to get rid off $|v_2 - v_1|$ and combine it to get $|(y_2,v_2) - (y_1,v_1)|_{max}$. What book would you recommend in this matter? Again, thanks in advance.
          $endgroup$
          – Nora oBrist
          Dec 29 '18 at 18:10














          $begingroup$
          I know that Lipschitz constant is $(1+nR^{n-1})$, however I'm not sure how would it affect if the second equation of the system was $v' = -2v - y^n$, for instance.
          $endgroup$
          – Nora oBrist
          Dec 29 '18 at 18:21




          $begingroup$
          I know that Lipschitz constant is $(1+nR^{n-1})$, however I'm not sure how would it affect if the second equation of the system was $v' = -2v - y^n$, for instance.
          $endgroup$
          – Nora oBrist
          Dec 29 '18 at 18:21












          $begingroup$
          Then you get $(2+nR^{n-1})$ as Lipschitz constant. By playing with the norm, scaling one of the components, one can reduce the Lipschitz constant somewhat, distribute it over the equations.
          $endgroup$
          – LutzL
          Dec 29 '18 at 18:26




          $begingroup$
          Then you get $(2+nR^{n-1})$ as Lipschitz constant. By playing with the norm, scaling one of the components, one can reduce the Lipschitz constant somewhat, distribute it over the equations.
          $endgroup$
          – LutzL
          Dec 29 '18 at 18:26












          $begingroup$
          This is becoming a master class, but I have got one more question. To be a contraction mapping it is required L< 1, something that in this case we have not. Fixed point theorem is not allowed, however starting from a close enough solution it converges. From your experience, have you seen this happening sometimes? Also, when you told that it is allowed to have a L constant on a set $|y|< R$ but I get on $|x|< R$, is there something that I'm missing? Thank you, I appreciate your help because this is much like art to me.
          $endgroup$
          – Nora oBrist
          Jan 4 at 20:01




          $begingroup$
          This is becoming a master class, but I have got one more question. To be a contraction mapping it is required L< 1, something that in this case we have not. Fixed point theorem is not allowed, however starting from a close enough solution it converges. From your experience, have you seen this happening sometimes? Also, when you told that it is allowed to have a L constant on a set $|y|< R$ but I get on $|x|< R$, is there something that I'm missing? Thank you, I appreciate your help because this is much like art to me.
          $endgroup$
          – Nora oBrist
          Jan 4 at 20:01


















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          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
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