Picard's method of succesive approximations












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$begingroup$



Construct first $3$ succesive approximations $x_0,x_1,x_2$ for the following Cauchy problems:




$$x'=-x+t^2$$
$$x(0)=2$$



I have no idea how to start this... any ideas?






ordinary-differential-equations approximation







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    0












    $begingroup$



    Construct first $3$ succesive approximations $x_0,x_1,x_2$ for the following Cauchy problems:




    $$x'=-x+t^2$$
    $$x(0)=2$$



    I have no idea how to start this... any ideas?






    ordinary-differential-equations approximation







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$



      Construct first $3$ succesive approximations $x_0,x_1,x_2$ for the following Cauchy problems:




      $$x'=-x+t^2$$
      $$x(0)=2$$



      I have no idea how to start this... any ideas?






      ordinary-differential-equations approximation







      share|cite|improve this question









      $endgroup$





      Construct first $3$ succesive approximations $x_0,x_1,x_2$ for the following Cauchy problems:




      $$x'=-x+t^2$$
      $$x(0)=2$$



      I have no idea how to start this... any ideas?






      ordinary-differential-equations approximation




      ordinary-differential-equations approximation






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 14 at 11:14









      C. CristiC. Cristi

      1,655318




      1,655318






















          2 Answers
          2






          active

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          1












          $begingroup$

          You compute successively the functions
          $$
          x_0(t)=x_0, ~~ x_{n+1}(t)=x_0+int_0^tf(s,x_n(s)),ds
          $$
          using your function $f(t,x)=-x+t^2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So my $x_0(t) = 2$?
            $endgroup$
            – C. Cristi
            Jan 14 at 11:25










          • $begingroup$
            Also when does the bounds of the integral changes? When $t_0$ is not 0?
            $endgroup$
            – C. Cristi
            Jan 14 at 11:25










          • $begingroup$
            Yes. And if $x(t_0)=x_0$, then the lower integral boundary is $t_0$ instead of $0$.
            $endgroup$
            – LutzL
            Jan 14 at 11:40



















          1












          $begingroup$

          Rewrite your equation as



          $$
          x(t) = x(0) + int_0^t[-x(u) + u^2]{rm d}u tag{1}
          $$



          The ideas is to approximate this with the expression



          $$
          x_{n + 1}(t) = x(0) + int_0^t[-x_n(u) + u^2]{rm d}u tag{2}
          $$



          Start with




          $x_0 = x(0) = 2$




          Replace that in (2) and you get



          $$
          x_1(t) = 2 + int_0^t [-x_0(u) + u^2]{rm d}u = 2 - 2t + frac{t^2}{3} tag{3}
          $$




          $x_1 = 2 - 2t + t^2/3$




          Evaluate this in Eq. (2) again and get



          $$
          x_2(t) = 2 + int_0^t[-x_1(u) + u^2] = 2 - 2t + 2t^2 + frac{t^3}{3} - t^4
          $$



          I will leave the other two for you to complete






          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












            $begingroup$

            You compute successively the functions
            $$
            x_0(t)=x_0, ~~ x_{n+1}(t)=x_0+int_0^tf(s,x_n(s)),ds
            $$
            using your function $f(t,x)=-x+t^2$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              So my $x_0(t) = 2$?
              $endgroup$
              – C. Cristi
              Jan 14 at 11:25










            • $begingroup$
              Also when does the bounds of the integral changes? When $t_0$ is not 0?
              $endgroup$
              – C. Cristi
              Jan 14 at 11:25










            • $begingroup$
              Yes. And if $x(t_0)=x_0$, then the lower integral boundary is $t_0$ instead of $0$.
              $endgroup$
              – LutzL
              Jan 14 at 11:40
















            1












            $begingroup$

            You compute successively the functions
            $$
            x_0(t)=x_0, ~~ x_{n+1}(t)=x_0+int_0^tf(s,x_n(s)),ds
            $$
            using your function $f(t,x)=-x+t^2$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              So my $x_0(t) = 2$?
              $endgroup$
              – C. Cristi
              Jan 14 at 11:25










            • $begingroup$
              Also when does the bounds of the integral changes? When $t_0$ is not 0?
              $endgroup$
              – C. Cristi
              Jan 14 at 11:25










            • $begingroup$
              Yes. And if $x(t_0)=x_0$, then the lower integral boundary is $t_0$ instead of $0$.
              $endgroup$
              – LutzL
              Jan 14 at 11:40














            1












            1








            1





            $begingroup$

            You compute successively the functions
            $$
            x_0(t)=x_0, ~~ x_{n+1}(t)=x_0+int_0^tf(s,x_n(s)),ds
            $$
            using your function $f(t,x)=-x+t^2$.






            share|cite|improve this answer









            $endgroup$



            You compute successively the functions
            $$
            x_0(t)=x_0, ~~ x_{n+1}(t)=x_0+int_0^tf(s,x_n(s)),ds
            $$
            using your function $f(t,x)=-x+t^2$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 14 at 11:23









            LutzLLutzL

            60.8k42157




            60.8k42157












            • $begingroup$
              So my $x_0(t) = 2$?
              $endgroup$
              – C. Cristi
              Jan 14 at 11:25










            • $begingroup$
              Also when does the bounds of the integral changes? When $t_0$ is not 0?
              $endgroup$
              – C. Cristi
              Jan 14 at 11:25










            • $begingroup$
              Yes. And if $x(t_0)=x_0$, then the lower integral boundary is $t_0$ instead of $0$.
              $endgroup$
              – LutzL
              Jan 14 at 11:40


















            • $begingroup$
              So my $x_0(t) = 2$?
              $endgroup$
              – C. Cristi
              Jan 14 at 11:25










            • $begingroup$
              Also when does the bounds of the integral changes? When $t_0$ is not 0?
              $endgroup$
              – C. Cristi
              Jan 14 at 11:25










            • $begingroup$
              Yes. And if $x(t_0)=x_0$, then the lower integral boundary is $t_0$ instead of $0$.
              $endgroup$
              – LutzL
              Jan 14 at 11:40
















            $begingroup$
            So my $x_0(t) = 2$?
            $endgroup$
            – C. Cristi
            Jan 14 at 11:25




            $begingroup$
            So my $x_0(t) = 2$?
            $endgroup$
            – C. Cristi
            Jan 14 at 11:25












            $begingroup$
            Also when does the bounds of the integral changes? When $t_0$ is not 0?
            $endgroup$
            – C. Cristi
            Jan 14 at 11:25




            $begingroup$
            Also when does the bounds of the integral changes? When $t_0$ is not 0?
            $endgroup$
            – C. Cristi
            Jan 14 at 11:25












            $begingroup$
            Yes. And if $x(t_0)=x_0$, then the lower integral boundary is $t_0$ instead of $0$.
            $endgroup$
            – LutzL
            Jan 14 at 11:40




            $begingroup$
            Yes. And if $x(t_0)=x_0$, then the lower integral boundary is $t_0$ instead of $0$.
            $endgroup$
            – LutzL
            Jan 14 at 11:40











            1












            $begingroup$

            Rewrite your equation as



            $$
            x(t) = x(0) + int_0^t[-x(u) + u^2]{rm d}u tag{1}
            $$



            The ideas is to approximate this with the expression



            $$
            x_{n + 1}(t) = x(0) + int_0^t[-x_n(u) + u^2]{rm d}u tag{2}
            $$



            Start with




            $x_0 = x(0) = 2$




            Replace that in (2) and you get



            $$
            x_1(t) = 2 + int_0^t [-x_0(u) + u^2]{rm d}u = 2 - 2t + frac{t^2}{3} tag{3}
            $$




            $x_1 = 2 - 2t + t^2/3$




            Evaluate this in Eq. (2) again and get



            $$
            x_2(t) = 2 + int_0^t[-x_1(u) + u^2] = 2 - 2t + 2t^2 + frac{t^3}{3} - t^4
            $$



            I will leave the other two for you to complete






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Rewrite your equation as



              $$
              x(t) = x(0) + int_0^t[-x(u) + u^2]{rm d}u tag{1}
              $$



              The ideas is to approximate this with the expression



              $$
              x_{n + 1}(t) = x(0) + int_0^t[-x_n(u) + u^2]{rm d}u tag{2}
              $$



              Start with




              $x_0 = x(0) = 2$




              Replace that in (2) and you get



              $$
              x_1(t) = 2 + int_0^t [-x_0(u) + u^2]{rm d}u = 2 - 2t + frac{t^2}{3} tag{3}
              $$




              $x_1 = 2 - 2t + t^2/3$




              Evaluate this in Eq. (2) again and get



              $$
              x_2(t) = 2 + int_0^t[-x_1(u) + u^2] = 2 - 2t + 2t^2 + frac{t^3}{3} - t^4
              $$



              I will leave the other two for you to complete






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Rewrite your equation as



                $$
                x(t) = x(0) + int_0^t[-x(u) + u^2]{rm d}u tag{1}
                $$



                The ideas is to approximate this with the expression



                $$
                x_{n + 1}(t) = x(0) + int_0^t[-x_n(u) + u^2]{rm d}u tag{2}
                $$



                Start with




                $x_0 = x(0) = 2$




                Replace that in (2) and you get



                $$
                x_1(t) = 2 + int_0^t [-x_0(u) + u^2]{rm d}u = 2 - 2t + frac{t^2}{3} tag{3}
                $$




                $x_1 = 2 - 2t + t^2/3$




                Evaluate this in Eq. (2) again and get



                $$
                x_2(t) = 2 + int_0^t[-x_1(u) + u^2] = 2 - 2t + 2t^2 + frac{t^3}{3} - t^4
                $$



                I will leave the other two for you to complete






                share|cite|improve this answer









                $endgroup$



                Rewrite your equation as



                $$
                x(t) = x(0) + int_0^t[-x(u) + u^2]{rm d}u tag{1}
                $$



                The ideas is to approximate this with the expression



                $$
                x_{n + 1}(t) = x(0) + int_0^t[-x_n(u) + u^2]{rm d}u tag{2}
                $$



                Start with




                $x_0 = x(0) = 2$




                Replace that in (2) and you get



                $$
                x_1(t) = 2 + int_0^t [-x_0(u) + u^2]{rm d}u = 2 - 2t + frac{t^2}{3} tag{3}
                $$




                $x_1 = 2 - 2t + t^2/3$




                Evaluate this in Eq. (2) again and get



                $$
                x_2(t) = 2 + int_0^t[-x_1(u) + u^2] = 2 - 2t + 2t^2 + frac{t^3}{3} - t^4
                $$



                I will leave the other two for you to complete







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 14 at 11:25









                caveraccaverac

                14.8k31130




                14.8k31130






























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