Existence of Solution for Initial value Problem $y'(t)=cos(ty), t>0$ and $y(0)=frac{1}{n}, ninmathbb{N}$...












1












$begingroup$


Consider the cauchy problem



(1)begin{cases}
y'(t)=frac{1}{1+ty}, & t>0 \
y(0)=1+frac{1}{n} & ninmathbb{N}
end{cases}



(2)begin{cases}
y'(t)=cos(ty), & t>0 \
y(0)=frac{1}{n} & ninmathbb{N}
end{cases}
(a) Prove that for every $$ there exists a solution $y_n(.)$ defined on the whole $mathbb{R}^+$.



(b) Prove that the sequence $(y_n)$ admits a subsequence uniformly converging on any compact subinterval of $mathbb{R}^+$ to the solution of the Cauchy problem $y'(t)=cos(ty), t>0$ and $y(0)=0$.



My attempt



(1a) Since $t>0, y>0$, $y'(t)>0$. This and definition of $y'$ implies $0<y_n'(t)<1, forall ninmathbb{N}$. Thus
begin{equation}
0<y_n(t)=y_n(0)+int_0^ty_n'(s)dsleq 2+t, t>0
end{equation}
which is the required solution.



(1b) Let $I=[0,1]$. Since $0leq y_n'(t)leq 1$ and $y_n'$ is Lipschitz.Thus, $(y_n)$ is Equicontinous and by Arzela Ascoli the result follows.



(2) My problem is here. Since $-1leq y'(t)leq 1, t>0$, graph of $y$ alternate and I do not know how to get boundedness for the family $(y_n)$






functional-analysis ordinary-differential-equations initial-value-problems equicontinuity arzela-ascoli







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Consider the cauchy problem



    (1)begin{cases}
    y'(t)=frac{1}{1+ty}, & t>0 \
    y(0)=1+frac{1}{n} & ninmathbb{N}
    end{cases}



    (2)begin{cases}
    y'(t)=cos(ty), & t>0 \
    y(0)=frac{1}{n} & ninmathbb{N}
    end{cases}
    (a) Prove that for every $$ there exists a solution $y_n(.)$ defined on the whole $mathbb{R}^+$.



    (b) Prove that the sequence $(y_n)$ admits a subsequence uniformly converging on any compact subinterval of $mathbb{R}^+$ to the solution of the Cauchy problem $y'(t)=cos(ty), t>0$ and $y(0)=0$.



    My attempt



    (1a) Since $t>0, y>0$, $y'(t)>0$. This and definition of $y'$ implies $0<y_n'(t)<1, forall ninmathbb{N}$. Thus
    begin{equation}
    0<y_n(t)=y_n(0)+int_0^ty_n'(s)dsleq 2+t, t>0
    end{equation}
    which is the required solution.



    (1b) Let $I=[0,1]$. Since $0leq y_n'(t)leq 1$ and $y_n'$ is Lipschitz.Thus, $(y_n)$ is Equicontinous and by Arzela Ascoli the result follows.



    (2) My problem is here. Since $-1leq y'(t)leq 1, t>0$, graph of $y$ alternate and I do not know how to get boundedness for the family $(y_n)$






    functional-analysis ordinary-differential-equations initial-value-problems equicontinuity arzela-ascoli







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Consider the cauchy problem



      (1)begin{cases}
      y'(t)=frac{1}{1+ty}, & t>0 \
      y(0)=1+frac{1}{n} & ninmathbb{N}
      end{cases}



      (2)begin{cases}
      y'(t)=cos(ty), & t>0 \
      y(0)=frac{1}{n} & ninmathbb{N}
      end{cases}
      (a) Prove that for every $$ there exists a solution $y_n(.)$ defined on the whole $mathbb{R}^+$.



      (b) Prove that the sequence $(y_n)$ admits a subsequence uniformly converging on any compact subinterval of $mathbb{R}^+$ to the solution of the Cauchy problem $y'(t)=cos(ty), t>0$ and $y(0)=0$.



      My attempt



      (1a) Since $t>0, y>0$, $y'(t)>0$. This and definition of $y'$ implies $0<y_n'(t)<1, forall ninmathbb{N}$. Thus
      begin{equation}
      0<y_n(t)=y_n(0)+int_0^ty_n'(s)dsleq 2+t, t>0
      end{equation}
      which is the required solution.



      (1b) Let $I=[0,1]$. Since $0leq y_n'(t)leq 1$ and $y_n'$ is Lipschitz.Thus, $(y_n)$ is Equicontinous and by Arzela Ascoli the result follows.



      (2) My problem is here. Since $-1leq y'(t)leq 1, t>0$, graph of $y$ alternate and I do not know how to get boundedness for the family $(y_n)$






      functional-analysis ordinary-differential-equations initial-value-problems equicontinuity arzela-ascoli







      share|cite|improve this question









      $endgroup$




      Consider the cauchy problem



      (1)begin{cases}
      y'(t)=frac{1}{1+ty}, & t>0 \
      y(0)=1+frac{1}{n} & ninmathbb{N}
      end{cases}



      (2)begin{cases}
      y'(t)=cos(ty), & t>0 \
      y(0)=frac{1}{n} & ninmathbb{N}
      end{cases}
      (a) Prove that for every $$ there exists a solution $y_n(.)$ defined on the whole $mathbb{R}^+$.



      (b) Prove that the sequence $(y_n)$ admits a subsequence uniformly converging on any compact subinterval of $mathbb{R}^+$ to the solution of the Cauchy problem $y'(t)=cos(ty), t>0$ and $y(0)=0$.



      My attempt



      (1a) Since $t>0, y>0$, $y'(t)>0$. This and definition of $y'$ implies $0<y_n'(t)<1, forall ninmathbb{N}$. Thus
      begin{equation}
      0<y_n(t)=y_n(0)+int_0^ty_n'(s)dsleq 2+t, t>0
      end{equation}
      which is the required solution.



      (1b) Let $I=[0,1]$. Since $0leq y_n'(t)leq 1$ and $y_n'$ is Lipschitz.Thus, $(y_n)$ is Equicontinous and by Arzela Ascoli the result follows.



      (2) My problem is here. Since $-1leq y'(t)leq 1, t>0$, graph of $y$ alternate and I do not know how to get boundedness for the family $(y_n)$






      functional-analysis ordinary-differential-equations initial-value-problems equicontinuity arzela-ascoli




      functional-analysis ordinary-differential-equations initial-value-problems equicontinuity arzela-ascoli






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 29 at 12:16









      Muhammad MubarakMuhammad Mubarak

      9810




      9810






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          The task is easily solvable using the Grönwall theorem, as the right sides have easy Lipschitz constants $L(t)=t$.



          For two solutions $y,z$ you get for the difference the upper bounds
          $$
          |dot z(t)-dot y(t)|le t|z(t)-y(t)|
          implies
          |z(t)-y(t)|le e^{t^2/2},|z(0)-y(0)|.
          $$
          This implies uniform convergence on any finite intervals, as for $tin[0,T]$ one has
          $$
          |y_n(t)-y_infty(t)|le frac1n, e^{T^2/2}
          $$






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            (a) Since $-1leq y'(t)leq 1, t>0$, we have that $-1leq y_n'(t)leq 1, t>0, ninmathbb{N}$. Thus
            begin{equation} y_n(t)=frac{1}{n}+int_0^t y_n'(s)dsend{equation}
            which implies
            begin{equation}-tleq y_n(t)leq frac{1}{n}+tleq 1+t.end{equation}
            Thus $(y_n)_n$ is the required solution.



            (b) Let $I=[0,1]$. This implies $-1leq y_n(t)leq 2$ equivalently $|y_n(t)|leq 2$ (uniformly bounded).



            Since $y_n'$ is continuous for each $ninmathbb{N}$, $(y_n)_n$ is a Lipschitz family. So they are Equicontinuous. By Arzela Ascoli, there exist $(y_{n_k})_k$ converging to $y$ where $y(0)=lim frac{1}{n_k}=0$. Hence, the result.






            share|cite|improve this answer











            $endgroup$














              Your Answer





              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("mathjaxEditing", function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              });
              });
              }, "mathjax-editing");

              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "69"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });














              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3092102%2fexistence-of-solution-for-initial-value-problem-yt-costy-t0-and-y0%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              The task is easily solvable using the Grönwall theorem, as the right sides have easy Lipschitz constants $L(t)=t$.



              For two solutions $y,z$ you get for the difference the upper bounds
              $$
              |dot z(t)-dot y(t)|le t|z(t)-y(t)|
              implies
              |z(t)-y(t)|le e^{t^2/2},|z(0)-y(0)|.
              $$
              This implies uniform convergence on any finite intervals, as for $tin[0,T]$ one has
              $$
              |y_n(t)-y_infty(t)|le frac1n, e^{T^2/2}
              $$






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                The task is easily solvable using the Grönwall theorem, as the right sides have easy Lipschitz constants $L(t)=t$.



                For two solutions $y,z$ you get for the difference the upper bounds
                $$
                |dot z(t)-dot y(t)|le t|z(t)-y(t)|
                implies
                |z(t)-y(t)|le e^{t^2/2},|z(0)-y(0)|.
                $$
                This implies uniform convergence on any finite intervals, as for $tin[0,T]$ one has
                $$
                |y_n(t)-y_infty(t)|le frac1n, e^{T^2/2}
                $$






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  The task is easily solvable using the Grönwall theorem, as the right sides have easy Lipschitz constants $L(t)=t$.



                  For two solutions $y,z$ you get for the difference the upper bounds
                  $$
                  |dot z(t)-dot y(t)|le t|z(t)-y(t)|
                  implies
                  |z(t)-y(t)|le e^{t^2/2},|z(0)-y(0)|.
                  $$
                  This implies uniform convergence on any finite intervals, as for $tin[0,T]$ one has
                  $$
                  |y_n(t)-y_infty(t)|le frac1n, e^{T^2/2}
                  $$






                  share|cite|improve this answer











                  $endgroup$



                  The task is easily solvable using the Grönwall theorem, as the right sides have easy Lipschitz constants $L(t)=t$.



                  For two solutions $y,z$ you get for the difference the upper bounds
                  $$
                  |dot z(t)-dot y(t)|le t|z(t)-y(t)|
                  implies
                  |z(t)-y(t)|le e^{t^2/2},|z(0)-y(0)|.
                  $$
                  This implies uniform convergence on any finite intervals, as for $tin[0,T]$ one has
                  $$
                  |y_n(t)-y_infty(t)|le frac1n, e^{T^2/2}
                  $$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 29 at 13:46

























                  answered Jan 29 at 13:09









                  LutzLLutzL

                  60.1k42057




                  60.1k42057























                      0












                      $begingroup$

                      (a) Since $-1leq y'(t)leq 1, t>0$, we have that $-1leq y_n'(t)leq 1, t>0, ninmathbb{N}$. Thus
                      begin{equation} y_n(t)=frac{1}{n}+int_0^t y_n'(s)dsend{equation}
                      which implies
                      begin{equation}-tleq y_n(t)leq frac{1}{n}+tleq 1+t.end{equation}
                      Thus $(y_n)_n$ is the required solution.



                      (b) Let $I=[0,1]$. This implies $-1leq y_n(t)leq 2$ equivalently $|y_n(t)|leq 2$ (uniformly bounded).



                      Since $y_n'$ is continuous for each $ninmathbb{N}$, $(y_n)_n$ is a Lipschitz family. So they are Equicontinuous. By Arzela Ascoli, there exist $(y_{n_k})_k$ converging to $y$ where $y(0)=lim frac{1}{n_k}=0$. Hence, the result.






                      share|cite|improve this answer











                      $endgroup$


















                        0












                        $begingroup$

                        (a) Since $-1leq y'(t)leq 1, t>0$, we have that $-1leq y_n'(t)leq 1, t>0, ninmathbb{N}$. Thus
                        begin{equation} y_n(t)=frac{1}{n}+int_0^t y_n'(s)dsend{equation}
                        which implies
                        begin{equation}-tleq y_n(t)leq frac{1}{n}+tleq 1+t.end{equation}
                        Thus $(y_n)_n$ is the required solution.



                        (b) Let $I=[0,1]$. This implies $-1leq y_n(t)leq 2$ equivalently $|y_n(t)|leq 2$ (uniformly bounded).



                        Since $y_n'$ is continuous for each $ninmathbb{N}$, $(y_n)_n$ is a Lipschitz family. So they are Equicontinuous. By Arzela Ascoli, there exist $(y_{n_k})_k$ converging to $y$ where $y(0)=lim frac{1}{n_k}=0$. Hence, the result.






                        share|cite|improve this answer











                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          (a) Since $-1leq y'(t)leq 1, t>0$, we have that $-1leq y_n'(t)leq 1, t>0, ninmathbb{N}$. Thus
                          begin{equation} y_n(t)=frac{1}{n}+int_0^t y_n'(s)dsend{equation}
                          which implies
                          begin{equation}-tleq y_n(t)leq frac{1}{n}+tleq 1+t.end{equation}
                          Thus $(y_n)_n$ is the required solution.



                          (b) Let $I=[0,1]$. This implies $-1leq y_n(t)leq 2$ equivalently $|y_n(t)|leq 2$ (uniformly bounded).



                          Since $y_n'$ is continuous for each $ninmathbb{N}$, $(y_n)_n$ is a Lipschitz family. So they are Equicontinuous. By Arzela Ascoli, there exist $(y_{n_k})_k$ converging to $y$ where $y(0)=lim frac{1}{n_k}=0$. Hence, the result.






                          share|cite|improve this answer











                          $endgroup$



                          (a) Since $-1leq y'(t)leq 1, t>0$, we have that $-1leq y_n'(t)leq 1, t>0, ninmathbb{N}$. Thus
                          begin{equation} y_n(t)=frac{1}{n}+int_0^t y_n'(s)dsend{equation}
                          which implies
                          begin{equation}-tleq y_n(t)leq frac{1}{n}+tleq 1+t.end{equation}
                          Thus $(y_n)_n$ is the required solution.



                          (b) Let $I=[0,1]$. This implies $-1leq y_n(t)leq 2$ equivalently $|y_n(t)|leq 2$ (uniformly bounded).



                          Since $y_n'$ is continuous for each $ninmathbb{N}$, $(y_n)_n$ is a Lipschitz family. So they are Equicontinuous. By Arzela Ascoli, there exist $(y_{n_k})_k$ converging to $y$ where $y(0)=lim frac{1}{n_k}=0$. Hence, the result.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Jan 30 at 10:37

























                          answered Jan 29 at 13:19









                          Muhammad MubarakMuhammad Mubarak

                          9810




                          9810






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid …



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3092102%2fexistence-of-solution-for-initial-value-problem-yt-costy-t0-and-y0%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              Popular posts from this blog

                              Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

                              Image
                              28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
                              Read more

                              Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

                              Image
                              up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
                              Read more

                              A Topological Invariant for $pi_3(U(n))$

                              Image
                              3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
                              Read more