global and unique solution for ODE












1















Let $y' = f(x) p(cos y) + g(x) q( sin y) $ and $y(0) =1$, with $f,g$ continuous and $p,q$ polynomials. Then there exists unique solution on all $mathbb{R}$ ?




Its easily solved if i assumed that $f,g$ are bounded, but the question asks about $f,g$ that are not necessarily bounded.



Can any one give a hint or a solution !






differential-equations







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    1















    Let $y' = f(x) p(cos y) + g(x) q( sin y) $ and $y(0) =1$, with $f,g$ continuous and $p,q$ polynomials. Then there exists unique solution on all $mathbb{R}$ ?




    Its easily solved if i assumed that $f,g$ are bounded, but the question asks about $f,g$ that are not necessarily bounded.



    Can any one give a hint or a solution !






    differential-equations







    share|cite|improve this question



























      1












      1








      1


      1






      Let $y' = f(x) p(cos y) + g(x) q( sin y) $ and $y(0) =1$, with $f,g$ continuous and $p,q$ polynomials. Then there exists unique solution on all $mathbb{R}$ ?




      Its easily solved if i assumed that $f,g$ are bounded, but the question asks about $f,g$ that are not necessarily bounded.



      Can any one give a hint or a solution !






      differential-equations







      share|cite|improve this question
















      Let $y' = f(x) p(cos y) + g(x) q( sin y) $ and $y(0) =1$, with $f,g$ continuous and $p,q$ polynomials. Then there exists unique solution on all $mathbb{R}$ ?




      Its easily solved if i assumed that $f,g$ are bounded, but the question asks about $f,g$ that are not necessarily bounded.



      Can any one give a hint or a solution !






      differential-equations




      differential-equations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 26 at 10:29









      LutzL

      55.8k42054




      55.8k42054










      asked Nov 26 at 9:39









      Ahmad

      2,5241625




      2,5241625






















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          $f$ and $g$ are bounded on every interval $[-R,R]$ for any $R=NinBbb N$. Thus you get unique solutions over these intervals, and by the uniqueness the solution for $R=N+1$ extends the solution for $R=N$. In the limit this gives a unique solution over all of $Bbb R$.






          share|cite|improve this answer





















          • Can you tell please, what theorem are you referring to?
            – Evgeny
            Nov 26 at 11:06










          • What do you mean, that a global Lipschitz constant on $(t,y)in [a,b]times Bbb R^n$ ensures the existence of a solution over $[a,b]$? This is a variant of the Picard-Lindelöf theorem.
            – LutzL
            Nov 26 at 11:21










          • If you remember the reference with a proof for this variant and can tell me, I would be very grateful!
            – Evgeny
            Nov 26 at 11:37










          • See math.stackexchange.com/a/2987844/115115 and linked posts. I know this variant from my lectures, I do not know which books one could cite. There is also a variant of this global proof that uses the Lipschitz constants of the iterates of the Picard iteration, which fall faster than simply the power of the Lipschitz constant of one iteration.
            – LutzL
            Nov 26 at 11:47












          • Thank you very much!
            – Evgeny
            Nov 26 at 12:01











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          $f$ and $g$ are bounded on every interval $[-R,R]$ for any $R=NinBbb N$. Thus you get unique solutions over these intervals, and by the uniqueness the solution for $R=N+1$ extends the solution for $R=N$. In the limit this gives a unique solution over all of $Bbb R$.






          share|cite|improve this answer





















          • Can you tell please, what theorem are you referring to?
            – Evgeny
            Nov 26 at 11:06










          • What do you mean, that a global Lipschitz constant on $(t,y)in [a,b]times Bbb R^n$ ensures the existence of a solution over $[a,b]$? This is a variant of the Picard-Lindelöf theorem.
            – LutzL
            Nov 26 at 11:21










          • If you remember the reference with a proof for this variant and can tell me, I would be very grateful!
            – Evgeny
            Nov 26 at 11:37










          • See math.stackexchange.com/a/2987844/115115 and linked posts. I know this variant from my lectures, I do not know which books one could cite. There is also a variant of this global proof that uses the Lipschitz constants of the iterates of the Picard iteration, which fall faster than simply the power of the Lipschitz constant of one iteration.
            – LutzL
            Nov 26 at 11:47












          • Thank you very much!
            – Evgeny
            Nov 26 at 12:01
















          1














          $f$ and $g$ are bounded on every interval $[-R,R]$ for any $R=NinBbb N$. Thus you get unique solutions over these intervals, and by the uniqueness the solution for $R=N+1$ extends the solution for $R=N$. In the limit this gives a unique solution over all of $Bbb R$.






          share|cite|improve this answer





















          • Can you tell please, what theorem are you referring to?
            – Evgeny
            Nov 26 at 11:06










          • What do you mean, that a global Lipschitz constant on $(t,y)in [a,b]times Bbb R^n$ ensures the existence of a solution over $[a,b]$? This is a variant of the Picard-Lindelöf theorem.
            – LutzL
            Nov 26 at 11:21










          • If you remember the reference with a proof for this variant and can tell me, I would be very grateful!
            – Evgeny
            Nov 26 at 11:37










          • See math.stackexchange.com/a/2987844/115115 and linked posts. I know this variant from my lectures, I do not know which books one could cite. There is also a variant of this global proof that uses the Lipschitz constants of the iterates of the Picard iteration, which fall faster than simply the power of the Lipschitz constant of one iteration.
            – LutzL
            Nov 26 at 11:47












          • Thank you very much!
            – Evgeny
            Nov 26 at 12:01














          1












          1








          1






          $f$ and $g$ are bounded on every interval $[-R,R]$ for any $R=NinBbb N$. Thus you get unique solutions over these intervals, and by the uniqueness the solution for $R=N+1$ extends the solution for $R=N$. In the limit this gives a unique solution over all of $Bbb R$.






          share|cite|improve this answer












          $f$ and $g$ are bounded on every interval $[-R,R]$ for any $R=NinBbb N$. Thus you get unique solutions over these intervals, and by the uniqueness the solution for $R=N+1$ extends the solution for $R=N$. In the limit this gives a unique solution over all of $Bbb R$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 26 at 10:31









          LutzL

          55.8k42054




          55.8k42054












          • Can you tell please, what theorem are you referring to?
            – Evgeny
            Nov 26 at 11:06










          • What do you mean, that a global Lipschitz constant on $(t,y)in [a,b]times Bbb R^n$ ensures the existence of a solution over $[a,b]$? This is a variant of the Picard-Lindelöf theorem.
            – LutzL
            Nov 26 at 11:21










          • If you remember the reference with a proof for this variant and can tell me, I would be very grateful!
            – Evgeny
            Nov 26 at 11:37










          • See math.stackexchange.com/a/2987844/115115 and linked posts. I know this variant from my lectures, I do not know which books one could cite. There is also a variant of this global proof that uses the Lipschitz constants of the iterates of the Picard iteration, which fall faster than simply the power of the Lipschitz constant of one iteration.
            – LutzL
            Nov 26 at 11:47












          • Thank you very much!
            – Evgeny
            Nov 26 at 12:01


















          • Can you tell please, what theorem are you referring to?
            – Evgeny
            Nov 26 at 11:06










          • What do you mean, that a global Lipschitz constant on $(t,y)in [a,b]times Bbb R^n$ ensures the existence of a solution over $[a,b]$? This is a variant of the Picard-Lindelöf theorem.
            – LutzL
            Nov 26 at 11:21










          • If you remember the reference with a proof for this variant and can tell me, I would be very grateful!
            – Evgeny
            Nov 26 at 11:37










          • See math.stackexchange.com/a/2987844/115115 and linked posts. I know this variant from my lectures, I do not know which books one could cite. There is also a variant of this global proof that uses the Lipschitz constants of the iterates of the Picard iteration, which fall faster than simply the power of the Lipschitz constant of one iteration.
            – LutzL
            Nov 26 at 11:47












          • Thank you very much!
            – Evgeny
            Nov 26 at 12:01
















          Can you tell please, what theorem are you referring to?
          – Evgeny
          Nov 26 at 11:06




          Can you tell please, what theorem are you referring to?
          – Evgeny
          Nov 26 at 11:06












          What do you mean, that a global Lipschitz constant on $(t,y)in [a,b]times Bbb R^n$ ensures the existence of a solution over $[a,b]$? This is a variant of the Picard-Lindelöf theorem.
          – LutzL
          Nov 26 at 11:21




          What do you mean, that a global Lipschitz constant on $(t,y)in [a,b]times Bbb R^n$ ensures the existence of a solution over $[a,b]$? This is a variant of the Picard-Lindelöf theorem.
          – LutzL
          Nov 26 at 11:21












          If you remember the reference with a proof for this variant and can tell me, I would be very grateful!
          – Evgeny
          Nov 26 at 11:37




          If you remember the reference with a proof for this variant and can tell me, I would be very grateful!
          – Evgeny
          Nov 26 at 11:37












          See math.stackexchange.com/a/2987844/115115 and linked posts. I know this variant from my lectures, I do not know which books one could cite. There is also a variant of this global proof that uses the Lipschitz constants of the iterates of the Picard iteration, which fall faster than simply the power of the Lipschitz constant of one iteration.
          – LutzL
          Nov 26 at 11:47






          See math.stackexchange.com/a/2987844/115115 and linked posts. I know this variant from my lectures, I do not know which books one could cite. There is also a variant of this global proof that uses the Lipschitz constants of the iterates of the Picard iteration, which fall faster than simply the power of the Lipschitz constant of one iteration.
          – LutzL
          Nov 26 at 11:47














          Thank you very much!
          – Evgeny
          Nov 26 at 12:01




          Thank you very much!
          – Evgeny
          Nov 26 at 12:01


















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