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












2












$begingroup$


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



$frac{d^2x_1}{dt^2}=k_{11}x_1+...k_{1n}x_n$



.



.



.



$frac{d^2x_n}{dt^2}=k_{n1}x_1+...k_{nn}x_n$



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



    $frac{d^2x_1}{dt^2}=k_{11}x_1+...k_{1n}x_n$



    .



    .



    .



    $frac{d^2x_n}{dt^2}=k_{n1}x_1+...k_{nn}x_n$



    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



      $frac{d^2x_1}{dt^2}=k_{11}x_1+...k_{1n}x_n$



      .



      .



      .



      $frac{d^2x_n}{dt^2}=k_{n1}x_1+...k_{nn}x_n$



      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



      $frac{d^2x_1}{dt^2}=k_{11}x_1+...k_{1n}x_n$



      .



      .



      .



      $frac{d^2x_n}{dt^2}=k_{n1}x_1+...k_{nn}x_n$



      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






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      asked 4 hours ago









      user446153user446153

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      1075






















          1 Answer
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          $begingroup$

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



          $$ frac{d^2}{dt^2} left[begin{array}{c} x_1 \ vdots \ x_n end{array}right] = left[begin{array}{ccc} k_{11} & cdots & k_{1n} \ vdots & ddots & vdots \ k_{1n} & cdots & k_{nn} end{array}right] left[ begin{array}{c} x_1 \ vdots \ x_nend{array}right]$$



          $$ frac{d^2}{dt^2} vec{x} = K vec{x}$$



          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,



          $$
          frac{d^2}{dt^2} vec{x} = V Lambda V^{-1} vec{x}
          $$



          $$
          V^{-1}frac{d^2}{dt^2} vec{x} = Lambda V^{-1} vec{x}
          $$



          $$
          frac{d^2}{dt^2} V^{-1}vec{x} = Lambda V^{-1} vec{x}
          $$



          Let $vec{y} = V^{-1} vec{x}$, then we have $
          frac{d^2}{dt^2} vec{y} = Lambda vec{y}
          $
          . This corresponds to the following system of equations.



          $$
          frac{d^2 y_1}{dt^2} = lambda_1 y_1
          $$

          $$
          frac{d^2 y_2}{dt^2} = lambda_2 y_2
          $$

          $$
          vdots
          $$

          $$
          frac{d^2 y_n}{dt^2} = lambda_n y_n
          $$



          Clearly the solutions are of the form,



          $$y_j(t) = C_1 e^{sqrt{lambda_j} t} + C_2 e^{-sqrt{lambda_j} t},$$



          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









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            active

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            3












            $begingroup$

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



            $$ frac{d^2}{dt^2} left[begin{array}{c} x_1 \ vdots \ x_n end{array}right] = left[begin{array}{ccc} k_{11} & cdots & k_{1n} \ vdots & ddots & vdots \ k_{1n} & cdots & k_{nn} end{array}right] left[ begin{array}{c} x_1 \ vdots \ x_nend{array}right]$$



            $$ frac{d^2}{dt^2} vec{x} = K vec{x}$$



            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,



            $$
            frac{d^2}{dt^2} vec{x} = V Lambda V^{-1} vec{x}
            $$



            $$
            V^{-1}frac{d^2}{dt^2} vec{x} = Lambda V^{-1} vec{x}
            $$



            $$
            frac{d^2}{dt^2} V^{-1}vec{x} = Lambda V^{-1} vec{x}
            $$



            Let $vec{y} = V^{-1} vec{x}$, then we have $
            frac{d^2}{dt^2} vec{y} = Lambda vec{y}
            $
            . This corresponds to the following system of equations.



            $$
            frac{d^2 y_1}{dt^2} = lambda_1 y_1
            $$

            $$
            frac{d^2 y_2}{dt^2} = lambda_2 y_2
            $$

            $$
            vdots
            $$

            $$
            frac{d^2 y_n}{dt^2} = lambda_n y_n
            $$



            Clearly the solutions are of the form,



            $$y_j(t) = C_1 e^{sqrt{lambda_j} t} + C_2 e^{-sqrt{lambda_j} t},$$



            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,



              $$ frac{d^2}{dt^2} left[begin{array}{c} x_1 \ vdots \ x_n end{array}right] = left[begin{array}{ccc} k_{11} & cdots & k_{1n} \ vdots & ddots & vdots \ k_{1n} & cdots & k_{nn} end{array}right] left[ begin{array}{c} x_1 \ vdots \ x_nend{array}right]$$



              $$ frac{d^2}{dt^2} vec{x} = K vec{x}$$



              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,



              $$
              frac{d^2}{dt^2} vec{x} = V Lambda V^{-1} vec{x}
              $$



              $$
              V^{-1}frac{d^2}{dt^2} vec{x} = Lambda V^{-1} vec{x}
              $$



              $$
              frac{d^2}{dt^2} V^{-1}vec{x} = Lambda V^{-1} vec{x}
              $$



              Let $vec{y} = V^{-1} vec{x}$, then we have $
              frac{d^2}{dt^2} vec{y} = Lambda vec{y}
              $
              . This corresponds to the following system of equations.



              $$
              frac{d^2 y_1}{dt^2} = lambda_1 y_1
              $$

              $$
              frac{d^2 y_2}{dt^2} = lambda_2 y_2
              $$

              $$
              vdots
              $$

              $$
              frac{d^2 y_n}{dt^2} = lambda_n y_n
              $$



              Clearly the solutions are of the form,



              $$y_j(t) = C_1 e^{sqrt{lambda_j} t} + C_2 e^{-sqrt{lambda_j} t},$$



              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,



                $$ frac{d^2}{dt^2} left[begin{array}{c} x_1 \ vdots \ x_n end{array}right] = left[begin{array}{ccc} k_{11} & cdots & k_{1n} \ vdots & ddots & vdots \ k_{1n} & cdots & k_{nn} end{array}right] left[ begin{array}{c} x_1 \ vdots \ x_nend{array}right]$$



                $$ frac{d^2}{dt^2} vec{x} = K vec{x}$$



                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,



                $$
                frac{d^2}{dt^2} vec{x} = V Lambda V^{-1} vec{x}
                $$



                $$
                V^{-1}frac{d^2}{dt^2} vec{x} = Lambda V^{-1} vec{x}
                $$



                $$
                frac{d^2}{dt^2} V^{-1}vec{x} = Lambda V^{-1} vec{x}
                $$



                Let $vec{y} = V^{-1} vec{x}$, then we have $
                frac{d^2}{dt^2} vec{y} = Lambda vec{y}
                $
                . This corresponds to the following system of equations.



                $$
                frac{d^2 y_1}{dt^2} = lambda_1 y_1
                $$

                $$
                frac{d^2 y_2}{dt^2} = lambda_2 y_2
                $$

                $$
                vdots
                $$

                $$
                frac{d^2 y_n}{dt^2} = lambda_n y_n
                $$



                Clearly the solutions are of the form,



                $$y_j(t) = C_1 e^{sqrt{lambda_j} t} + C_2 e^{-sqrt{lambda_j} t},$$



                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,



                $$ frac{d^2}{dt^2} left[begin{array}{c} x_1 \ vdots \ x_n end{array}right] = left[begin{array}{ccc} k_{11} & cdots & k_{1n} \ vdots & ddots & vdots \ k_{1n} & cdots & k_{nn} end{array}right] left[ begin{array}{c} x_1 \ vdots \ x_nend{array}right]$$



                $$ frac{d^2}{dt^2} vec{x} = K vec{x}$$



                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,



                $$
                frac{d^2}{dt^2} vec{x} = V Lambda V^{-1} vec{x}
                $$



                $$
                V^{-1}frac{d^2}{dt^2} vec{x} = Lambda V^{-1} vec{x}
                $$



                $$
                frac{d^2}{dt^2} V^{-1}vec{x} = Lambda V^{-1} vec{x}
                $$



                Let $vec{y} = V^{-1} vec{x}$, then we have $
                frac{d^2}{dt^2} vec{y} = Lambda vec{y}
                $
                . This corresponds to the following system of equations.



                $$
                frac{d^2 y_1}{dt^2} = lambda_1 y_1
                $$

                $$
                frac{d^2 y_2}{dt^2} = lambda_2 y_2
                $$

                $$
                vdots
                $$

                $$
                frac{d^2 y_n}{dt^2} = lambda_n y_n
                $$



                Clearly the solutions are of the form,



                $$y_j(t) = C_1 e^{sqrt{lambda_j} t} + C_2 e^{-sqrt{lambda_j} t},$$



                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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