Finding a principal fundamental matrix












1












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The question is as follows:




Let $$A=begin{pmatrix} sin(x) & 0 \ 0 & -1end{pmatrix} $$ This is periodic with period $T = 2pi$. Find a principal fundamental matrix $phi(x)$.




I am currently able to do problems similar to this where it is just constants in the matrix, but continue to struggle with this question. $($I have a feeling I need to use Floquet's theorem but I am not sure$)$. Any help would be much appreciated. Thanks.






real-analysis ordinary-differential-equations







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    1












    $begingroup$


    The question is as follows:




    Let $$A=begin{pmatrix} sin(x) & 0 \ 0 & -1end{pmatrix} $$ This is periodic with period $T = 2pi$. Find a principal fundamental matrix $phi(x)$.




    I am currently able to do problems similar to this where it is just constants in the matrix, but continue to struggle with this question. $($I have a feeling I need to use Floquet's theorem but I am not sure$)$. Any help would be much appreciated. Thanks.






    real-analysis ordinary-differential-equations







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      2



      $begingroup$


      The question is as follows:




      Let $$A=begin{pmatrix} sin(x) & 0 \ 0 & -1end{pmatrix} $$ This is periodic with period $T = 2pi$. Find a principal fundamental matrix $phi(x)$.




      I am currently able to do problems similar to this where it is just constants in the matrix, but continue to struggle with this question. $($I have a feeling I need to use Floquet's theorem but I am not sure$)$. Any help would be much appreciated. Thanks.






      real-analysis ordinary-differential-equations







      share|cite|improve this question











      $endgroup$




      The question is as follows:




      Let $$A=begin{pmatrix} sin(x) & 0 \ 0 & -1end{pmatrix} $$ This is periodic with period $T = 2pi$. Find a principal fundamental matrix $phi(x)$.




      I am currently able to do problems similar to this where it is just constants in the matrix, but continue to struggle with this question. $($I have a feeling I need to use Floquet's theorem but I am not sure$)$. Any help would be much appreciated. Thanks.






      real-analysis ordinary-differential-equations




      real-analysis ordinary-differential-equations






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      share|cite|improve this question













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      edited Jan 4 at 20:03









      mrtaurho

      5,68551540




      5,68551540










      asked Jan 4 at 19:55









      mallos98mallos98

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          $begingroup$

          I take it you're solving the differential equation system
          $$dfrac{dY}{dx} = pmatrix{sin(x) & 0cr 0 & -1cr} Y $$
          Hint: the system decouples into $$ eqalign{dfrac{dy_1}{dx} &= sin(x) y_1cr
          dfrac{dy_2}{dx} &= - y_2cr}$$






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Let's see now . . .



            If I recall correctly, a principal fundamental matrix corresponding to the coefficient matrix



            $A(x) = begin{bmatrix} sin x & 0 \ 0 & -1 end{bmatrix} tag 1$



            is the $2 times 2$ matrix function $phi(x, x_0)$ of $x$ satisfying



            $phi'(x, x_0) = dfrac{d phi(x, x_0)}{dx} = A(x) phi(x, x_0), ; phi(x_0, x_0) = I = begin{bmatrix} 1 & 0 \ 0 & 1 end{bmatrix}; tag 2$



            if we write



            $phi(x, x_0) = begin{bmatrix} phi_{11}(x, x_0) & phi_{12}(x, x_0) \ phi_{21}(x, x_0) & phi_{22}(x, x_0) end{bmatrix}, tag 3$



            then we see that the $phi_{ij}(x, x_0)$ satisfy



            $phi'_{1j}(x, x_0) = (sin x) phi_{1j}(x, x_0), tag 4$



            $phi'_{2j}(x, x_0) = - phi_{2j}(x, x_0), tag 5$



            where $j = 1, 2$; the solution to (5) is easily recognized to be



            $phi_{2j}(x, x_0) = phi_{2j}(x_0, x_0) e^{-(x - x_0)} = phi_{2j}(x_0, x_0)e^{x_0 - x}, tag 6$



            whereas that to (4) is only slightly less simple:



            $phi_{1j}(x, x_0) = phi_{1j}(x_0, x_0) e^{-(cos x - cos x_0)} = phi_{1j}(x_0, x_0) e^{cos x_0 - cos x}; tag 7$



            assembling (6) and (7) into $phi(x, x_0)$ whilst taking the initial conditions given in (2) into account yields



            $phi(x, x_0) = begin{bmatrix} e^{cos x_0 - cos x} & 0 \ 0 & e^{x_0 - x} end{bmatrix} tag 8$



            as the desired fundamental matrix.






            share|cite|improve this answer











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              2 Answers
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              2 Answers
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              2












              $begingroup$

              I take it you're solving the differential equation system
              $$dfrac{dY}{dx} = pmatrix{sin(x) & 0cr 0 & -1cr} Y $$
              Hint: the system decouples into $$ eqalign{dfrac{dy_1}{dx} &= sin(x) y_1cr
              dfrac{dy_2}{dx} &= - y_2cr}$$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                I take it you're solving the differential equation system
                $$dfrac{dY}{dx} = pmatrix{sin(x) & 0cr 0 & -1cr} Y $$
                Hint: the system decouples into $$ eqalign{dfrac{dy_1}{dx} &= sin(x) y_1cr
                dfrac{dy_2}{dx} &= - y_2cr}$$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  I take it you're solving the differential equation system
                  $$dfrac{dY}{dx} = pmatrix{sin(x) & 0cr 0 & -1cr} Y $$
                  Hint: the system decouples into $$ eqalign{dfrac{dy_1}{dx} &= sin(x) y_1cr
                  dfrac{dy_2}{dx} &= - y_2cr}$$






                  share|cite|improve this answer









                  $endgroup$



                  I take it you're solving the differential equation system
                  $$dfrac{dY}{dx} = pmatrix{sin(x) & 0cr 0 & -1cr} Y $$
                  Hint: the system decouples into $$ eqalign{dfrac{dy_1}{dx} &= sin(x) y_1cr
                  dfrac{dy_2}{dx} &= - y_2cr}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 4 at 19:59









                  Robert IsraelRobert Israel

                  326k23215469




                  326k23215469























                      1












                      $begingroup$

                      Let's see now . . .



                      If I recall correctly, a principal fundamental matrix corresponding to the coefficient matrix



                      $A(x) = begin{bmatrix} sin x & 0 \ 0 & -1 end{bmatrix} tag 1$



                      is the $2 times 2$ matrix function $phi(x, x_0)$ of $x$ satisfying



                      $phi'(x, x_0) = dfrac{d phi(x, x_0)}{dx} = A(x) phi(x, x_0), ; phi(x_0, x_0) = I = begin{bmatrix} 1 & 0 \ 0 & 1 end{bmatrix}; tag 2$



                      if we write



                      $phi(x, x_0) = begin{bmatrix} phi_{11}(x, x_0) & phi_{12}(x, x_0) \ phi_{21}(x, x_0) & phi_{22}(x, x_0) end{bmatrix}, tag 3$



                      then we see that the $phi_{ij}(x, x_0)$ satisfy



                      $phi'_{1j}(x, x_0) = (sin x) phi_{1j}(x, x_0), tag 4$



                      $phi'_{2j}(x, x_0) = - phi_{2j}(x, x_0), tag 5$



                      where $j = 1, 2$; the solution to (5) is easily recognized to be



                      $phi_{2j}(x, x_0) = phi_{2j}(x_0, x_0) e^{-(x - x_0)} = phi_{2j}(x_0, x_0)e^{x_0 - x}, tag 6$



                      whereas that to (4) is only slightly less simple:



                      $phi_{1j}(x, x_0) = phi_{1j}(x_0, x_0) e^{-(cos x - cos x_0)} = phi_{1j}(x_0, x_0) e^{cos x_0 - cos x}; tag 7$



                      assembling (6) and (7) into $phi(x, x_0)$ whilst taking the initial conditions given in (2) into account yields



                      $phi(x, x_0) = begin{bmatrix} e^{cos x_0 - cos x} & 0 \ 0 & e^{x_0 - x} end{bmatrix} tag 8$



                      as the desired fundamental matrix.






                      share|cite|improve this answer











                      $endgroup$


















                        1












                        $begingroup$

                        Let's see now . . .



                        If I recall correctly, a principal fundamental matrix corresponding to the coefficient matrix



                        $A(x) = begin{bmatrix} sin x & 0 \ 0 & -1 end{bmatrix} tag 1$



                        is the $2 times 2$ matrix function $phi(x, x_0)$ of $x$ satisfying



                        $phi'(x, x_0) = dfrac{d phi(x, x_0)}{dx} = A(x) phi(x, x_0), ; phi(x_0, x_0) = I = begin{bmatrix} 1 & 0 \ 0 & 1 end{bmatrix}; tag 2$



                        if we write



                        $phi(x, x_0) = begin{bmatrix} phi_{11}(x, x_0) & phi_{12}(x, x_0) \ phi_{21}(x, x_0) & phi_{22}(x, x_0) end{bmatrix}, tag 3$



                        then we see that the $phi_{ij}(x, x_0)$ satisfy



                        $phi'_{1j}(x, x_0) = (sin x) phi_{1j}(x, x_0), tag 4$



                        $phi'_{2j}(x, x_0) = - phi_{2j}(x, x_0), tag 5$



                        where $j = 1, 2$; the solution to (5) is easily recognized to be



                        $phi_{2j}(x, x_0) = phi_{2j}(x_0, x_0) e^{-(x - x_0)} = phi_{2j}(x_0, x_0)e^{x_0 - x}, tag 6$



                        whereas that to (4) is only slightly less simple:



                        $phi_{1j}(x, x_0) = phi_{1j}(x_0, x_0) e^{-(cos x - cos x_0)} = phi_{1j}(x_0, x_0) e^{cos x_0 - cos x}; tag 7$



                        assembling (6) and (7) into $phi(x, x_0)$ whilst taking the initial conditions given in (2) into account yields



                        $phi(x, x_0) = begin{bmatrix} e^{cos x_0 - cos x} & 0 \ 0 & e^{x_0 - x} end{bmatrix} tag 8$



                        as the desired fundamental matrix.






                        share|cite|improve this answer











                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Let's see now . . .



                          If I recall correctly, a principal fundamental matrix corresponding to the coefficient matrix



                          $A(x) = begin{bmatrix} sin x & 0 \ 0 & -1 end{bmatrix} tag 1$



                          is the $2 times 2$ matrix function $phi(x, x_0)$ of $x$ satisfying



                          $phi'(x, x_0) = dfrac{d phi(x, x_0)}{dx} = A(x) phi(x, x_0), ; phi(x_0, x_0) = I = begin{bmatrix} 1 & 0 \ 0 & 1 end{bmatrix}; tag 2$



                          if we write



                          $phi(x, x_0) = begin{bmatrix} phi_{11}(x, x_0) & phi_{12}(x, x_0) \ phi_{21}(x, x_0) & phi_{22}(x, x_0) end{bmatrix}, tag 3$



                          then we see that the $phi_{ij}(x, x_0)$ satisfy



                          $phi'_{1j}(x, x_0) = (sin x) phi_{1j}(x, x_0), tag 4$



                          $phi'_{2j}(x, x_0) = - phi_{2j}(x, x_0), tag 5$



                          where $j = 1, 2$; the solution to (5) is easily recognized to be



                          $phi_{2j}(x, x_0) = phi_{2j}(x_0, x_0) e^{-(x - x_0)} = phi_{2j}(x_0, x_0)e^{x_0 - x}, tag 6$



                          whereas that to (4) is only slightly less simple:



                          $phi_{1j}(x, x_0) = phi_{1j}(x_0, x_0) e^{-(cos x - cos x_0)} = phi_{1j}(x_0, x_0) e^{cos x_0 - cos x}; tag 7$



                          assembling (6) and (7) into $phi(x, x_0)$ whilst taking the initial conditions given in (2) into account yields



                          $phi(x, x_0) = begin{bmatrix} e^{cos x_0 - cos x} & 0 \ 0 & e^{x_0 - x} end{bmatrix} tag 8$



                          as the desired fundamental matrix.






                          share|cite|improve this answer











                          $endgroup$



                          Let's see now . . .



                          If I recall correctly, a principal fundamental matrix corresponding to the coefficient matrix



                          $A(x) = begin{bmatrix} sin x & 0 \ 0 & -1 end{bmatrix} tag 1$



                          is the $2 times 2$ matrix function $phi(x, x_0)$ of $x$ satisfying



                          $phi'(x, x_0) = dfrac{d phi(x, x_0)}{dx} = A(x) phi(x, x_0), ; phi(x_0, x_0) = I = begin{bmatrix} 1 & 0 \ 0 & 1 end{bmatrix}; tag 2$



                          if we write



                          $phi(x, x_0) = begin{bmatrix} phi_{11}(x, x_0) & phi_{12}(x, x_0) \ phi_{21}(x, x_0) & phi_{22}(x, x_0) end{bmatrix}, tag 3$



                          then we see that the $phi_{ij}(x, x_0)$ satisfy



                          $phi'_{1j}(x, x_0) = (sin x) phi_{1j}(x, x_0), tag 4$



                          $phi'_{2j}(x, x_0) = - phi_{2j}(x, x_0), tag 5$



                          where $j = 1, 2$; the solution to (5) is easily recognized to be



                          $phi_{2j}(x, x_0) = phi_{2j}(x_0, x_0) e^{-(x - x_0)} = phi_{2j}(x_0, x_0)e^{x_0 - x}, tag 6$



                          whereas that to (4) is only slightly less simple:



                          $phi_{1j}(x, x_0) = phi_{1j}(x_0, x_0) e^{-(cos x - cos x_0)} = phi_{1j}(x_0, x_0) e^{cos x_0 - cos x}; tag 7$



                          assembling (6) and (7) into $phi(x, x_0)$ whilst taking the initial conditions given in (2) into account yields



                          $phi(x, x_0) = begin{bmatrix} e^{cos x_0 - cos x} & 0 \ 0 & e^{x_0 - x} end{bmatrix} tag 8$



                          as the desired fundamental matrix.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Jan 4 at 23:51

























                          answered Jan 4 at 23:24









                          Robert LewisRobert Lewis

                          47.6k23067




                          47.6k23067






























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