Solving a second-order matrix differential equation












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I have the differential equation $frac{d^2 x}{dt^2}+Ax=0$ where $A$ is a matrix and $$frac{d^2 x}{dt^2}=left(frac{d^2 x_1}{dt^2},frac{d^2 x_2}{dt^2},ldots,frac{d^2 x_n}{dt^2}right)^Ttext{ for }x=(x_1,x_2,ldots,x_n).$$



I know the solution is $x=cos(sqrt{A}(t-t_0))x_0+(sqrt{A})^{-1}sin(sqrt{A}(t-t_0))dot{x}_0$, but I have no idea why this is.



How would I go about deriving this solution? The literature I have looked at simply states the solution with little or no explanation.






matrices ordinary-differential-equations







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    1












    $begingroup$


    I have the differential equation $frac{d^2 x}{dt^2}+Ax=0$ where $A$ is a matrix and $$frac{d^2 x}{dt^2}=left(frac{d^2 x_1}{dt^2},frac{d^2 x_2}{dt^2},ldots,frac{d^2 x_n}{dt^2}right)^Ttext{ for }x=(x_1,x_2,ldots,x_n).$$



    I know the solution is $x=cos(sqrt{A}(t-t_0))x_0+(sqrt{A})^{-1}sin(sqrt{A}(t-t_0))dot{x}_0$, but I have no idea why this is.



    How would I go about deriving this solution? The literature I have looked at simply states the solution with little or no explanation.






    matrices ordinary-differential-equations







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I have the differential equation $frac{d^2 x}{dt^2}+Ax=0$ where $A$ is a matrix and $$frac{d^2 x}{dt^2}=left(frac{d^2 x_1}{dt^2},frac{d^2 x_2}{dt^2},ldots,frac{d^2 x_n}{dt^2}right)^Ttext{ for }x=(x_1,x_2,ldots,x_n).$$



      I know the solution is $x=cos(sqrt{A}(t-t_0))x_0+(sqrt{A})^{-1}sin(sqrt{A}(t-t_0))dot{x}_0$, but I have no idea why this is.



      How would I go about deriving this solution? The literature I have looked at simply states the solution with little or no explanation.






      matrices ordinary-differential-equations







      share|cite|improve this question











      $endgroup$




      I have the differential equation $frac{d^2 x}{dt^2}+Ax=0$ where $A$ is a matrix and $$frac{d^2 x}{dt^2}=left(frac{d^2 x_1}{dt^2},frac{d^2 x_2}{dt^2},ldots,frac{d^2 x_n}{dt^2}right)^Ttext{ for }x=(x_1,x_2,ldots,x_n).$$



      I know the solution is $x=cos(sqrt{A}(t-t_0))x_0+(sqrt{A})^{-1}sin(sqrt{A}(t-t_0))dot{x}_0$, but I have no idea why this is.



      How would I go about deriving this solution? The literature I have looked at simply states the solution with little or no explanation.






      matrices ordinary-differential-equations




      matrices ordinary-differential-equations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Apr 26 '13 at 17:02









      Michael Hardy

      1




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      asked Apr 26 '13 at 15:41









      hello123hello123

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      211212






















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          $begingroup$

          Let $A$ be diagonalisable ($A=P^{-1}A_dP$ with $A_d$ diagonal). Then the equation $vec{x}''+Avec{x}=0$ can be rewritten as $Pvec{x}''+A_dPvec{x}=0$. Denote $vec{y}=Pvec{x}$, then
          $$vec{y}''+A_dvec{y}=0.$$
          But this is equivalent to $n$ independent scalar ODEs
          $$ y_k''+lambda_k y_k=0,$$
          where $lambda_k$ denotes the eigenvalue of $A$ standing in the $k$th place of the diagonal of $A_d$. Their solutions are
          $$y_k(t)=cossqrt{lambda_k}(t-t_0),y_k(t_0)+frac{1}{sqrt{lambda_k}}sinsqrt{lambda_k}(t-t_0),y'_k(t_0).$$
          Or, in matrix form
          $$ vec{y}(t)=cos sqrt{A_d}(t-t_0),vec{y}(t_0)+frac{1}{sqrt{A_d}}sin sqrt{A_d}(t-t_0),vec{y},'(t_0).
          $$
          Now using that $vec{x}=P^{-1}vec{y}$ and that $P^{-1}f(A_d)P=f(A)$ for any function $f$, we obtain the quoted result.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks. What about in the inhomogeneous case? I believe the answer is $x=cos(sqrt{A}(t-t_0))x_0+(sqrt{A})^{-1}sin(sqrt{A}(t-t_0))dot{x}_0 +(sqrt{A})^{-1}int_{t_0}^tsinleft(sqrt{A}(t-s)right)f(s)ds$, but this can't be derived in the same way since we would have a $Pf(t)$ on the RHS.
            $endgroup$
            – hello123
            Apr 27 '13 at 15:00








          • 1




            $begingroup$
            It will be fine for the very same reason. You will have the non-homogeneity $vec{F}=Pvec{f}$ for $vec{y}$, then solving for $vec{y}$, coming back to $vec{x}$ and using $P^{-1}mu(A_d)P=mu(A)$, you will get something like that.
            $endgroup$
            – Start wearing purple
            Apr 27 '13 at 15:18












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          1 Answer
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          1 Answer
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          active

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          active

          oldest

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          1












          $begingroup$

          Let $A$ be diagonalisable ($A=P^{-1}A_dP$ with $A_d$ diagonal). Then the equation $vec{x}''+Avec{x}=0$ can be rewritten as $Pvec{x}''+A_dPvec{x}=0$. Denote $vec{y}=Pvec{x}$, then
          $$vec{y}''+A_dvec{y}=0.$$
          But this is equivalent to $n$ independent scalar ODEs
          $$ y_k''+lambda_k y_k=0,$$
          where $lambda_k$ denotes the eigenvalue of $A$ standing in the $k$th place of the diagonal of $A_d$. Their solutions are
          $$y_k(t)=cossqrt{lambda_k}(t-t_0),y_k(t_0)+frac{1}{sqrt{lambda_k}}sinsqrt{lambda_k}(t-t_0),y'_k(t_0).$$
          Or, in matrix form
          $$ vec{y}(t)=cos sqrt{A_d}(t-t_0),vec{y}(t_0)+frac{1}{sqrt{A_d}}sin sqrt{A_d}(t-t_0),vec{y},'(t_0).
          $$
          Now using that $vec{x}=P^{-1}vec{y}$ and that $P^{-1}f(A_d)P=f(A)$ for any function $f$, we obtain the quoted result.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks. What about in the inhomogeneous case? I believe the answer is $x=cos(sqrt{A}(t-t_0))x_0+(sqrt{A})^{-1}sin(sqrt{A}(t-t_0))dot{x}_0 +(sqrt{A})^{-1}int_{t_0}^tsinleft(sqrt{A}(t-s)right)f(s)ds$, but this can't be derived in the same way since we would have a $Pf(t)$ on the RHS.
            $endgroup$
            – hello123
            Apr 27 '13 at 15:00








          • 1




            $begingroup$
            It will be fine for the very same reason. You will have the non-homogeneity $vec{F}=Pvec{f}$ for $vec{y}$, then solving for $vec{y}$, coming back to $vec{x}$ and using $P^{-1}mu(A_d)P=mu(A)$, you will get something like that.
            $endgroup$
            – Start wearing purple
            Apr 27 '13 at 15:18
















          1












          $begingroup$

          Let $A$ be diagonalisable ($A=P^{-1}A_dP$ with $A_d$ diagonal). Then the equation $vec{x}''+Avec{x}=0$ can be rewritten as $Pvec{x}''+A_dPvec{x}=0$. Denote $vec{y}=Pvec{x}$, then
          $$vec{y}''+A_dvec{y}=0.$$
          But this is equivalent to $n$ independent scalar ODEs
          $$ y_k''+lambda_k y_k=0,$$
          where $lambda_k$ denotes the eigenvalue of $A$ standing in the $k$th place of the diagonal of $A_d$. Their solutions are
          $$y_k(t)=cossqrt{lambda_k}(t-t_0),y_k(t_0)+frac{1}{sqrt{lambda_k}}sinsqrt{lambda_k}(t-t_0),y'_k(t_0).$$
          Or, in matrix form
          $$ vec{y}(t)=cos sqrt{A_d}(t-t_0),vec{y}(t_0)+frac{1}{sqrt{A_d}}sin sqrt{A_d}(t-t_0),vec{y},'(t_0).
          $$
          Now using that $vec{x}=P^{-1}vec{y}$ and that $P^{-1}f(A_d)P=f(A)$ for any function $f$, we obtain the quoted result.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks. What about in the inhomogeneous case? I believe the answer is $x=cos(sqrt{A}(t-t_0))x_0+(sqrt{A})^{-1}sin(sqrt{A}(t-t_0))dot{x}_0 +(sqrt{A})^{-1}int_{t_0}^tsinleft(sqrt{A}(t-s)right)f(s)ds$, but this can't be derived in the same way since we would have a $Pf(t)$ on the RHS.
            $endgroup$
            – hello123
            Apr 27 '13 at 15:00








          • 1




            $begingroup$
            It will be fine for the very same reason. You will have the non-homogeneity $vec{F}=Pvec{f}$ for $vec{y}$, then solving for $vec{y}$, coming back to $vec{x}$ and using $P^{-1}mu(A_d)P=mu(A)$, you will get something like that.
            $endgroup$
            – Start wearing purple
            Apr 27 '13 at 15:18














          1












          1








          1





          $begingroup$

          Let $A$ be diagonalisable ($A=P^{-1}A_dP$ with $A_d$ diagonal). Then the equation $vec{x}''+Avec{x}=0$ can be rewritten as $Pvec{x}''+A_dPvec{x}=0$. Denote $vec{y}=Pvec{x}$, then
          $$vec{y}''+A_dvec{y}=0.$$
          But this is equivalent to $n$ independent scalar ODEs
          $$ y_k''+lambda_k y_k=0,$$
          where $lambda_k$ denotes the eigenvalue of $A$ standing in the $k$th place of the diagonal of $A_d$. Their solutions are
          $$y_k(t)=cossqrt{lambda_k}(t-t_0),y_k(t_0)+frac{1}{sqrt{lambda_k}}sinsqrt{lambda_k}(t-t_0),y'_k(t_0).$$
          Or, in matrix form
          $$ vec{y}(t)=cos sqrt{A_d}(t-t_0),vec{y}(t_0)+frac{1}{sqrt{A_d}}sin sqrt{A_d}(t-t_0),vec{y},'(t_0).
          $$
          Now using that $vec{x}=P^{-1}vec{y}$ and that $P^{-1}f(A_d)P=f(A)$ for any function $f$, we obtain the quoted result.






          share|cite|improve this answer











          $endgroup$



          Let $A$ be diagonalisable ($A=P^{-1}A_dP$ with $A_d$ diagonal). Then the equation $vec{x}''+Avec{x}=0$ can be rewritten as $Pvec{x}''+A_dPvec{x}=0$. Denote $vec{y}=Pvec{x}$, then
          $$vec{y}''+A_dvec{y}=0.$$
          But this is equivalent to $n$ independent scalar ODEs
          $$ y_k''+lambda_k y_k=0,$$
          where $lambda_k$ denotes the eigenvalue of $A$ standing in the $k$th place of the diagonal of $A_d$. Their solutions are
          $$y_k(t)=cossqrt{lambda_k}(t-t_0),y_k(t_0)+frac{1}{sqrt{lambda_k}}sinsqrt{lambda_k}(t-t_0),y'_k(t_0).$$
          Or, in matrix form
          $$ vec{y}(t)=cos sqrt{A_d}(t-t_0),vec{y}(t_0)+frac{1}{sqrt{A_d}}sin sqrt{A_d}(t-t_0),vec{y},'(t_0).
          $$
          Now using that $vec{x}=P^{-1}vec{y}$ and that $P^{-1}f(A_d)P=f(A)$ for any function $f$, we obtain the quoted result.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 11 at 4:50









          Rócherz

          3,0263823




          3,0263823










          answered Apr 26 '13 at 16:05









          Start wearing purpleStart wearing purple

          47.4k12135192




          47.4k12135192












          • $begingroup$
            Thanks. What about in the inhomogeneous case? I believe the answer is $x=cos(sqrt{A}(t-t_0))x_0+(sqrt{A})^{-1}sin(sqrt{A}(t-t_0))dot{x}_0 +(sqrt{A})^{-1}int_{t_0}^tsinleft(sqrt{A}(t-s)right)f(s)ds$, but this can't be derived in the same way since we would have a $Pf(t)$ on the RHS.
            $endgroup$
            – hello123
            Apr 27 '13 at 15:00








          • 1




            $begingroup$
            It will be fine for the very same reason. You will have the non-homogeneity $vec{F}=Pvec{f}$ for $vec{y}$, then solving for $vec{y}$, coming back to $vec{x}$ and using $P^{-1}mu(A_d)P=mu(A)$, you will get something like that.
            $endgroup$
            – Start wearing purple
            Apr 27 '13 at 15:18


















          • $begingroup$
            Thanks. What about in the inhomogeneous case? I believe the answer is $x=cos(sqrt{A}(t-t_0))x_0+(sqrt{A})^{-1}sin(sqrt{A}(t-t_0))dot{x}_0 +(sqrt{A})^{-1}int_{t_0}^tsinleft(sqrt{A}(t-s)right)f(s)ds$, but this can't be derived in the same way since we would have a $Pf(t)$ on the RHS.
            $endgroup$
            – hello123
            Apr 27 '13 at 15:00








          • 1




            $begingroup$
            It will be fine for the very same reason. You will have the non-homogeneity $vec{F}=Pvec{f}$ for $vec{y}$, then solving for $vec{y}$, coming back to $vec{x}$ and using $P^{-1}mu(A_d)P=mu(A)$, you will get something like that.
            $endgroup$
            – Start wearing purple
            Apr 27 '13 at 15:18
















          $begingroup$
          Thanks. What about in the inhomogeneous case? I believe the answer is $x=cos(sqrt{A}(t-t_0))x_0+(sqrt{A})^{-1}sin(sqrt{A}(t-t_0))dot{x}_0 +(sqrt{A})^{-1}int_{t_0}^tsinleft(sqrt{A}(t-s)right)f(s)ds$, but this can't be derived in the same way since we would have a $Pf(t)$ on the RHS.
          $endgroup$
          – hello123
          Apr 27 '13 at 15:00






          $begingroup$
          Thanks. What about in the inhomogeneous case? I believe the answer is $x=cos(sqrt{A}(t-t_0))x_0+(sqrt{A})^{-1}sin(sqrt{A}(t-t_0))dot{x}_0 +(sqrt{A})^{-1}int_{t_0}^tsinleft(sqrt{A}(t-s)right)f(s)ds$, but this can't be derived in the same way since we would have a $Pf(t)$ on the RHS.
          $endgroup$
          – hello123
          Apr 27 '13 at 15:00






          1




          1




          $begingroup$
          It will be fine for the very same reason. You will have the non-homogeneity $vec{F}=Pvec{f}$ for $vec{y}$, then solving for $vec{y}$, coming back to $vec{x}$ and using $P^{-1}mu(A_d)P=mu(A)$, you will get something like that.
          $endgroup$
          – Start wearing purple
          Apr 27 '13 at 15:18




          $begingroup$
          It will be fine for the very same reason. You will have the non-homogeneity $vec{F}=Pvec{f}$ for $vec{y}$, then solving for $vec{y}$, coming back to $vec{x}$ and using $P^{-1}mu(A_d)P=mu(A)$, you will get something like that.
          $endgroup$
          – Start wearing purple
          Apr 27 '13 at 15:18


















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