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













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      asked Jan 29 at 12:16









      Muhammad MubarakMuhammad Mubarak

      9810




      9810






















          2 Answers
          2






          active

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          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











            $endgroup$














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              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






























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