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Proving that any solution to the differential equation of an oscillator can be written as a sum of sinusoids.



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)How can the general Green's function of a linear homogeneous differential equation be derived?Proving a system of n linear equations has only one solutionThe pair $x_1$ , $x_2$ are Linearly IndependentName/Solution of this Differential EquationEvery solution of some linear differential equation of order 2 is boundedSolution of Matrix differential equation $textbfX'(t)=textbfAtextbfX(t)$How to relate the solutions to a Fuchsian type differential equation to the solutions to the hypergeometric differential equation?Difference between real and complex solution in differential equationFinding the general solution to a system of differential equations using eigenvaluesTwo systems of linear equations equivalent










2












$begingroup$


Suppose you have a differential equation with n distinct functions of $t$ where



$fracd^2x_1dt^2=k_11x_1+...k_1nx_n$



.



.



.



$fracd^2x_ndt^2=k_n1x_1+...k_nnx_n$



I want to show that any set of solutions of this differential equation $(x_1,x_2,...,x_n) $



can be written as a linear combination of solutions of the form $(e^iw_1t,...,e^iw_1t), (e^iw_2t,...e^iw_2t), ...,(e^iw_mt,...e^iw_mt)$ where each $w_j$ is a real number.



i.e. I want to know why the motion of any oscillator can be written as a linear combination of its normal modes. I would also appreciate it if you could tell me if a proof of this fact has to do with eigenvalues and eigenvectors in general.










share|cite|improve this question









$endgroup$
















    2












    $begingroup$


    Suppose you have a differential equation with n distinct functions of $t$ where



    $fracd^2x_1dt^2=k_11x_1+...k_1nx_n$



    .



    .



    .



    $fracd^2x_ndt^2=k_n1x_1+...k_nnx_n$



    I want to show that any set of solutions of this differential equation $(x_1,x_2,...,x_n) $



    can be written as a linear combination of solutions of the form $(e^iw_1t,...,e^iw_1t), (e^iw_2t,...e^iw_2t), ...,(e^iw_mt,...e^iw_mt)$ where each $w_j$ is a real number.



    i.e. I want to know why the motion of any oscillator can be written as a linear combination of its normal modes. I would also appreciate it if you could tell me if a proof of this fact has to do with eigenvalues and eigenvectors in general.










    share|cite|improve this question









    $endgroup$














      2












      2








      2





      $begingroup$


      Suppose you have a differential equation with n distinct functions of $t$ where



      $fracd^2x_1dt^2=k_11x_1+...k_1nx_n$



      .



      .



      .



      $fracd^2x_ndt^2=k_n1x_1+...k_nnx_n$



      I want to show that any set of solutions of this differential equation $(x_1,x_2,...,x_n) $



      can be written as a linear combination of solutions of the form $(e^iw_1t,...,e^iw_1t), (e^iw_2t,...e^iw_2t), ...,(e^iw_mt,...e^iw_mt)$ where each $w_j$ is a real number.



      i.e. I want to know why the motion of any oscillator can be written as a linear combination of its normal modes. I would also appreciate it if you could tell me if a proof of this fact has to do with eigenvalues and eigenvectors in general.










      share|cite|improve this question









      $endgroup$




      Suppose you have a differential equation with n distinct functions of $t$ where



      $fracd^2x_1dt^2=k_11x_1+...k_1nx_n$



      .



      .



      .



      $fracd^2x_ndt^2=k_n1x_1+...k_nnx_n$



      I want to show that any set of solutions of this differential equation $(x_1,x_2,...,x_n) $



      can be written as a linear combination of solutions of the form $(e^iw_1t,...,e^iw_1t), (e^iw_2t,...e^iw_2t), ...,(e^iw_mt,...e^iw_mt)$ where each $w_j$ is a real number.



      i.e. I want to know why the motion of any oscillator can be written as a linear combination of its normal modes. I would also appreciate it if you could tell me if a proof of this fact has to do with eigenvalues and eigenvectors in general.







      linear-algebra ordinary-differential-equations physics






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 4 hours ago









      user446153user446153

      1075




      1075




















          1 Answer
          1






          active

          oldest

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          3












          $begingroup$

          The system of differential equations you wrote could be written as,



          $$ fracd^2dt^2 left[beginarrayc x_1 \ vdots \ x_n endarrayright] = left[beginarrayccc k_11 & cdots & k_1n \ vdots & ddots & vdots \ k_1n & cdots & k_nn endarrayright] left[ beginarrayc x_1 \ vdots \ x_nendarrayright]$$



          $$ fracd^2dt^2 vecx = K vecx$$



          The matrix $K=[k_ij]$ acts on the indices of the functions $x_1,dots,x_n$.



          We will suppose that $K$ is diagonalizable with eigenvalues $lambda_1, dots, lambda_n$.



          Let $Lambda$ be the diagonal form of $K$.



          Let $V$ be the matrix of eigenvectors of $K$. Note that $Lambda = V^-1 K V$.



          We can now write the system of differential equations as,



          $$
          fracd^2dt^2 vecx = V Lambda V^-1 vecx
          $$



          $$
          V^-1fracd^2dt^2 vecx = Lambda V^-1 vecx
          $$



          $$
          fracd^2dt^2 V^-1vecx = Lambda V^-1 vecx
          $$



          Let $vecy = V^-1 vecx$, then we have $
          fracd^2dt^2 vecy = Lambda vecy
          $
          . This corresponds to the following system of equations.



          $$
          fracd^2 y_1dt^2 = lambda_1 y_1
          $$

          $$
          fracd^2 y_2dt^2 = lambda_2 y_2
          $$

          $$
          vdots
          $$

          $$
          fracd^2 y_ndt^2 = lambda_n y_n
          $$



          Clearly the solutions are of the form,



          $$y_j(t) = C_1 e^sqrtlambda_j t + C_2 e^-sqrtlambda_j t,$$



          to obtain the $x_j$'s we just multiply by the $V$ matrix.



          $$ x_j(t) = sum_i V_ji y_i(t)$$



          Whether or not the solutions are oscillators depends on whether the eigenvalues are positive, negative, or complex. In physical applications it wouldn't be uncommon for $K$ to be a symmetric matrix with negative eigenvalues.






          share|cite|improve this answer









          $endgroup$













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            1 Answer
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            1 Answer
            1






            active

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            active

            oldest

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            active

            oldest

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            3












            $begingroup$

            The system of differential equations you wrote could be written as,



            $$ fracd^2dt^2 left[beginarrayc x_1 \ vdots \ x_n endarrayright] = left[beginarrayccc k_11 & cdots & k_1n \ vdots & ddots & vdots \ k_1n & cdots & k_nn endarrayright] left[ beginarrayc x_1 \ vdots \ x_nendarrayright]$$



            $$ fracd^2dt^2 vecx = K vecx$$



            The matrix $K=[k_ij]$ acts on the indices of the functions $x_1,dots,x_n$.



            We will suppose that $K$ is diagonalizable with eigenvalues $lambda_1, dots, lambda_n$.



            Let $Lambda$ be the diagonal form of $K$.



            Let $V$ be the matrix of eigenvectors of $K$. Note that $Lambda = V^-1 K V$.



            We can now write the system of differential equations as,



            $$
            fracd^2dt^2 vecx = V Lambda V^-1 vecx
            $$



            $$
            V^-1fracd^2dt^2 vecx = Lambda V^-1 vecx
            $$



            $$
            fracd^2dt^2 V^-1vecx = Lambda V^-1 vecx
            $$



            Let $vecy = V^-1 vecx$, then we have $
            fracd^2dt^2 vecy = Lambda vecy
            $
            . This corresponds to the following system of equations.



            $$
            fracd^2 y_1dt^2 = lambda_1 y_1
            $$

            $$
            fracd^2 y_2dt^2 = lambda_2 y_2
            $$

            $$
            vdots
            $$

            $$
            fracd^2 y_ndt^2 = lambda_n y_n
            $$



            Clearly the solutions are of the form,



            $$y_j(t) = C_1 e^sqrtlambda_j t + C_2 e^-sqrtlambda_j t,$$



            to obtain the $x_j$'s we just multiply by the $V$ matrix.



            $$ x_j(t) = sum_i V_ji y_i(t)$$



            Whether or not the solutions are oscillators depends on whether the eigenvalues are positive, negative, or complex. In physical applications it wouldn't be uncommon for $K$ to be a symmetric matrix with negative eigenvalues.






            share|cite|improve this answer









            $endgroup$

















              3












              $begingroup$

              The system of differential equations you wrote could be written as,



              $$ fracd^2dt^2 left[beginarrayc x_1 \ vdots \ x_n endarrayright] = left[beginarrayccc k_11 & cdots & k_1n \ vdots & ddots & vdots \ k_1n & cdots & k_nn endarrayright] left[ beginarrayc x_1 \ vdots \ x_nendarrayright]$$



              $$ fracd^2dt^2 vecx = K vecx$$



              The matrix $K=[k_ij]$ acts on the indices of the functions $x_1,dots,x_n$.



              We will suppose that $K$ is diagonalizable with eigenvalues $lambda_1, dots, lambda_n$.



              Let $Lambda$ be the diagonal form of $K$.



              Let $V$ be the matrix of eigenvectors of $K$. Note that $Lambda = V^-1 K V$.



              We can now write the system of differential equations as,



              $$
              fracd^2dt^2 vecx = V Lambda V^-1 vecx
              $$



              $$
              V^-1fracd^2dt^2 vecx = Lambda V^-1 vecx
              $$



              $$
              fracd^2dt^2 V^-1vecx = Lambda V^-1 vecx
              $$



              Let $vecy = V^-1 vecx$, then we have $
              fracd^2dt^2 vecy = Lambda vecy
              $
              . This corresponds to the following system of equations.



              $$
              fracd^2 y_1dt^2 = lambda_1 y_1
              $$

              $$
              fracd^2 y_2dt^2 = lambda_2 y_2
              $$

              $$
              vdots
              $$

              $$
              fracd^2 y_ndt^2 = lambda_n y_n
              $$



              Clearly the solutions are of the form,



              $$y_j(t) = C_1 e^sqrtlambda_j t + C_2 e^-sqrtlambda_j t,$$



              to obtain the $x_j$'s we just multiply by the $V$ matrix.



              $$ x_j(t) = sum_i V_ji y_i(t)$$



              Whether or not the solutions are oscillators depends on whether the eigenvalues are positive, negative, or complex. In physical applications it wouldn't be uncommon for $K$ to be a symmetric matrix with negative eigenvalues.






              share|cite|improve this answer









              $endgroup$















                3












                3








                3





                $begingroup$

                The system of differential equations you wrote could be written as,



                $$ fracd^2dt^2 left[beginarrayc x_1 \ vdots \ x_n endarrayright] = left[beginarrayccc k_11 & cdots & k_1n \ vdots & ddots & vdots \ k_1n & cdots & k_nn endarrayright] left[ beginarrayc x_1 \ vdots \ x_nendarrayright]$$



                $$ fracd^2dt^2 vecx = K vecx$$



                The matrix $K=[k_ij]$ acts on the indices of the functions $x_1,dots,x_n$.



                We will suppose that $K$ is diagonalizable with eigenvalues $lambda_1, dots, lambda_n$.



                Let $Lambda$ be the diagonal form of $K$.



                Let $V$ be the matrix of eigenvectors of $K$. Note that $Lambda = V^-1 K V$.



                We can now write the system of differential equations as,



                $$
                fracd^2dt^2 vecx = V Lambda V^-1 vecx
                $$



                $$
                V^-1fracd^2dt^2 vecx = Lambda V^-1 vecx
                $$



                $$
                fracd^2dt^2 V^-1vecx = Lambda V^-1 vecx
                $$



                Let $vecy = V^-1 vecx$, then we have $
                fracd^2dt^2 vecy = Lambda vecy
                $
                . This corresponds to the following system of equations.



                $$
                fracd^2 y_1dt^2 = lambda_1 y_1
                $$

                $$
                fracd^2 y_2dt^2 = lambda_2 y_2
                $$

                $$
                vdots
                $$

                $$
                fracd^2 y_ndt^2 = lambda_n y_n
                $$



                Clearly the solutions are of the form,



                $$y_j(t) = C_1 e^sqrtlambda_j t + C_2 e^-sqrtlambda_j t,$$



                to obtain the $x_j$'s we just multiply by the $V$ matrix.



                $$ x_j(t) = sum_i V_ji y_i(t)$$



                Whether or not the solutions are oscillators depends on whether the eigenvalues are positive, negative, or complex. In physical applications it wouldn't be uncommon for $K$ to be a symmetric matrix with negative eigenvalues.






                share|cite|improve this answer









                $endgroup$



                The system of differential equations you wrote could be written as,



                $$ fracd^2dt^2 left[beginarrayc x_1 \ vdots \ x_n endarrayright] = left[beginarrayccc k_11 & cdots & k_1n \ vdots & ddots & vdots \ k_1n & cdots & k_nn endarrayright] left[ beginarrayc x_1 \ vdots \ x_nendarrayright]$$



                $$ fracd^2dt^2 vecx = K vecx$$



                The matrix $K=[k_ij]$ acts on the indices of the functions $x_1,dots,x_n$.



                We will suppose that $K$ is diagonalizable with eigenvalues $lambda_1, dots, lambda_n$.



                Let $Lambda$ be the diagonal form of $K$.



                Let $V$ be the matrix of eigenvectors of $K$. Note that $Lambda = V^-1 K V$.



                We can now write the system of differential equations as,



                $$
                fracd^2dt^2 vecx = V Lambda V^-1 vecx
                $$



                $$
                V^-1fracd^2dt^2 vecx = Lambda V^-1 vecx
                $$



                $$
                fracd^2dt^2 V^-1vecx = Lambda V^-1 vecx
                $$



                Let $vecy = V^-1 vecx$, then we have $
                fracd^2dt^2 vecy = Lambda vecy
                $
                . This corresponds to the following system of equations.



                $$
                fracd^2 y_1dt^2 = lambda_1 y_1
                $$

                $$
                fracd^2 y_2dt^2 = lambda_2 y_2
                $$

                $$
                vdots
                $$

                $$
                fracd^2 y_ndt^2 = lambda_n y_n
                $$



                Clearly the solutions are of the form,



                $$y_j(t) = C_1 e^sqrtlambda_j t + C_2 e^-sqrtlambda_j t,$$



                to obtain the $x_j$'s we just multiply by the $V$ matrix.



                $$ x_j(t) = sum_i V_ji y_i(t)$$



                Whether or not the solutions are oscillators depends on whether the eigenvalues are positive, negative, or complex. In physical applications it wouldn't be uncommon for $K$ to be a symmetric matrix with negative eigenvalues.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 3 hours ago









                SpencerSpencer

                8,76812156




                8,76812156



























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