Solving system of ODEs with extra parameter












3












$begingroup$


I would like to solve a $2times 2$ system of the form
$$frac{d}{dtheta}T=TA,quad T(0)=Id$$
where $theta$ is real and $A$ is of the form
$$A=begin{pmatrix} 0 & frac{e^{-i theta}}{lambda}\ frac{1}{36}e^{-itheta}left(9lambda + 2(lambda-1)^2 (6cos{theta} + cos{2theta} + 6)right) & 0end{pmatrix},$$
with $lambda$ a free parameter in the unit circle.



In particular I'm interested in obtaining numeric solutions at $theta=2pi$ depending on the extra parameter $lambda$. I'm fairly new using Mathematica, and this is what I have tried so far:



T[θ_] = {{T11[θ], T12[θ]}, {T21[θ], T22[θ]}};
A[θ_] = {
{0, E^(-I θ)/λ},
{1/36 E^(-I θ) (9 λ + 2 (-1 + λ)^2 (6 + 6 Cos[θ] + Cos[2 θ])), 0}
};
sys = {T'[θ] == T[θ].A[θ]};


The previous code sets the system that I want to solve and now I try to solve numerically. First I've tried



NSol = NDSolve[
{sys, T11[0] == 1, T12[0] == 0, T21[0] == 0, T22[0] == 1},
{T11[θ], T12[θ], T21[θ], T22[θ]},
{θ},
{θ, 0, 2 Pi}
];


which gives me the output



NDSolve::dupv: "Duplicate variable θ found in NDSolve[<<1>>]."


I have also tried



Nsol2 = ParametricNDSolve[
{sys, T11[0] == 1, T12[0] == 0, T21[0] == 0, T22[0] == 1},
{T11, T12, T21, T22},
{θ, 0, 2 Pi},
{λ}
];


which gives me as output $T_{11},dots,T_{22}$ as ParametricFunctions depending on each other and on $lambda$.



I don't know if this is the right approach and, if so, how to extract a numeric expression depending on $lambda$ from the last output - all that I've seen in the documentation are examples that are plotted for specific values of the parameter. Any help is much appreciated.



EDIT




Following the comments in one of the answers I'd like to explain further: the output that I would like to obtain is some sort of function depending of the parameter $lambda$ that I can manipulate afterwards. Say for instance, computing the series expansion of powers of $lambda$ of my solution. I don't know how to treat the ParametricFunction that I obtain to do such computations.











share|improve this question











$endgroup$

















    3












    $begingroup$


    I would like to solve a $2times 2$ system of the form
    $$frac{d}{dtheta}T=TA,quad T(0)=Id$$
    where $theta$ is real and $A$ is of the form
    $$A=begin{pmatrix} 0 & frac{e^{-i theta}}{lambda}\ frac{1}{36}e^{-itheta}left(9lambda + 2(lambda-1)^2 (6cos{theta} + cos{2theta} + 6)right) & 0end{pmatrix},$$
    with $lambda$ a free parameter in the unit circle.



    In particular I'm interested in obtaining numeric solutions at $theta=2pi$ depending on the extra parameter $lambda$. I'm fairly new using Mathematica, and this is what I have tried so far:



    T[θ_] = {{T11[θ], T12[θ]}, {T21[θ], T22[θ]}};
    A[θ_] = {
    {0, E^(-I θ)/λ},
    {1/36 E^(-I θ) (9 λ + 2 (-1 + λ)^2 (6 + 6 Cos[θ] + Cos[2 θ])), 0}
    };
    sys = {T'[θ] == T[θ].A[θ]};


    The previous code sets the system that I want to solve and now I try to solve numerically. First I've tried



    NSol = NDSolve[
    {sys, T11[0] == 1, T12[0] == 0, T21[0] == 0, T22[0] == 1},
    {T11[θ], T12[θ], T21[θ], T22[θ]},
    {θ},
    {θ, 0, 2 Pi}
    ];


    which gives me the output



    NDSolve::dupv: "Duplicate variable θ found in NDSolve[<<1>>]."


    I have also tried



    Nsol2 = ParametricNDSolve[
    {sys, T11[0] == 1, T12[0] == 0, T21[0] == 0, T22[0] == 1},
    {T11, T12, T21, T22},
    {θ, 0, 2 Pi},
    {λ}
    ];


    which gives me as output $T_{11},dots,T_{22}$ as ParametricFunctions depending on each other and on $lambda$.



    I don't know if this is the right approach and, if so, how to extract a numeric expression depending on $lambda$ from the last output - all that I've seen in the documentation are examples that are plotted for specific values of the parameter. Any help is much appreciated.



    EDIT




    Following the comments in one of the answers I'd like to explain further: the output that I would like to obtain is some sort of function depending of the parameter $lambda$ that I can manipulate afterwards. Say for instance, computing the series expansion of powers of $lambda$ of my solution. I don't know how to treat the ParametricFunction that I obtain to do such computations.











    share|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      I would like to solve a $2times 2$ system of the form
      $$frac{d}{dtheta}T=TA,quad T(0)=Id$$
      where $theta$ is real and $A$ is of the form
      $$A=begin{pmatrix} 0 & frac{e^{-i theta}}{lambda}\ frac{1}{36}e^{-itheta}left(9lambda + 2(lambda-1)^2 (6cos{theta} + cos{2theta} + 6)right) & 0end{pmatrix},$$
      with $lambda$ a free parameter in the unit circle.



      In particular I'm interested in obtaining numeric solutions at $theta=2pi$ depending on the extra parameter $lambda$. I'm fairly new using Mathematica, and this is what I have tried so far:



      T[θ_] = {{T11[θ], T12[θ]}, {T21[θ], T22[θ]}};
      A[θ_] = {
      {0, E^(-I θ)/λ},
      {1/36 E^(-I θ) (9 λ + 2 (-1 + λ)^2 (6 + 6 Cos[θ] + Cos[2 θ])), 0}
      };
      sys = {T'[θ] == T[θ].A[θ]};


      The previous code sets the system that I want to solve and now I try to solve numerically. First I've tried



      NSol = NDSolve[
      {sys, T11[0] == 1, T12[0] == 0, T21[0] == 0, T22[0] == 1},
      {T11[θ], T12[θ], T21[θ], T22[θ]},
      {θ},
      {θ, 0, 2 Pi}
      ];


      which gives me the output



      NDSolve::dupv: "Duplicate variable θ found in NDSolve[<<1>>]."


      I have also tried



      Nsol2 = ParametricNDSolve[
      {sys, T11[0] == 1, T12[0] == 0, T21[0] == 0, T22[0] == 1},
      {T11, T12, T21, T22},
      {θ, 0, 2 Pi},
      {λ}
      ];


      which gives me as output $T_{11},dots,T_{22}$ as ParametricFunctions depending on each other and on $lambda$.



      I don't know if this is the right approach and, if so, how to extract a numeric expression depending on $lambda$ from the last output - all that I've seen in the documentation are examples that are plotted for specific values of the parameter. Any help is much appreciated.



      EDIT




      Following the comments in one of the answers I'd like to explain further: the output that I would like to obtain is some sort of function depending of the parameter $lambda$ that I can manipulate afterwards. Say for instance, computing the series expansion of powers of $lambda$ of my solution. I don't know how to treat the ParametricFunction that I obtain to do such computations.











      share|improve this question











      $endgroup$




      I would like to solve a $2times 2$ system of the form
      $$frac{d}{dtheta}T=TA,quad T(0)=Id$$
      where $theta$ is real and $A$ is of the form
      $$A=begin{pmatrix} 0 & frac{e^{-i theta}}{lambda}\ frac{1}{36}e^{-itheta}left(9lambda + 2(lambda-1)^2 (6cos{theta} + cos{2theta} + 6)right) & 0end{pmatrix},$$
      with $lambda$ a free parameter in the unit circle.



      In particular I'm interested in obtaining numeric solutions at $theta=2pi$ depending on the extra parameter $lambda$. I'm fairly new using Mathematica, and this is what I have tried so far:



      T[θ_] = {{T11[θ], T12[θ]}, {T21[θ], T22[θ]}};
      A[θ_] = {
      {0, E^(-I θ)/λ},
      {1/36 E^(-I θ) (9 λ + 2 (-1 + λ)^2 (6 + 6 Cos[θ] + Cos[2 θ])), 0}
      };
      sys = {T'[θ] == T[θ].A[θ]};


      The previous code sets the system that I want to solve and now I try to solve numerically. First I've tried



      NSol = NDSolve[
      {sys, T11[0] == 1, T12[0] == 0, T21[0] == 0, T22[0] == 1},
      {T11[θ], T12[θ], T21[θ], T22[θ]},
      {θ},
      {θ, 0, 2 Pi}
      ];


      which gives me the output



      NDSolve::dupv: "Duplicate variable θ found in NDSolve[<<1>>]."


      I have also tried



      Nsol2 = ParametricNDSolve[
      {sys, T11[0] == 1, T12[0] == 0, T21[0] == 0, T22[0] == 1},
      {T11, T12, T21, T22},
      {θ, 0, 2 Pi},
      {λ}
      ];


      which gives me as output $T_{11},dots,T_{22}$ as ParametricFunctions depending on each other and on $lambda$.



      I don't know if this is the right approach and, if so, how to extract a numeric expression depending on $lambda$ from the last output - all that I've seen in the documentation are examples that are plotted for specific values of the parameter. Any help is much appreciated.



      EDIT




      Following the comments in one of the answers I'd like to explain further: the output that I would like to obtain is some sort of function depending of the parameter $lambda$ that I can manipulate afterwards. Say for instance, computing the series expansion of powers of $lambda$ of my solution. I don't know how to treat the ParametricFunction that I obtain to do such computations.








      differential-equations numerical-integration






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Mar 30 at 19:56









      Carl Woll

      72.8k396188




      72.8k396188










      asked Mar 30 at 13:31









      EduEdu

      1476




      1476






















          3 Answers
          3






          active

          oldest

          votes


















          2












          $begingroup$

          Here is how to obtain the series expansion of the matrix components. First, using a tweaked version (solving for t[2π] instead of t) of JM's formulation:



          A[θ_] = {
          {0, E^(-I θ)/λ},
          {1/36 E^(-I θ) (9 λ + 2 (-1 + λ)^2 (6 + 6 Cos[θ] + Cos[2 θ])), 0}
          };

          pf = ParametricNDSolveValue[
          {
          t'[θ] == t[θ].A[θ], t[0] == IdentityMatrix[2]
          },
          t[2π],
          {θ, 0, 2π},
          λ
          ];


          Then, pf will return the matrix value at $theta = 2 pi$. For example:



          pf[1]
          pf[Exp[I Pi/3]]



          {{1. + 3.62818*10^-9 I,
          6.82646*10^-8 - 5.73536*10^-9 I}, {1.70661*10^-8 - 1.43384*10^-9 I,
          1. + 3.62818*10^-9 I}}



          {{0.985595 + 1.17074 I, -0.425572 + 0.737112 I}, {-0.788363 - 1.36549 I,
          0.985595 - 1.17074 I}}




          Finding the series expansion is simple:



          DecimalForm[Series[pf[λ], {λ, 1, 5}], {4,4}] //TeXForm



          $left(
          begin{array}{cc}
          1.0000 & 6.2830 \
          1.5710 & 1.0000 \
          end{array}
          right)+left(
          begin{array}{cc}
          0.0000+0.0000 i & -6.2830+0.0000 i \
          1.5710+0.0000 i & 0.0000+0.0000 i \
          end{array}
          right) (lambda -1)+left(
          begin{array}{cc}
          0.0000-0.7869 i & 0.6691-0.0000 i \
          0.8799+0.0000 i & -0.0000+0.7869 i \
          end{array}
          right) (lambda -1)^2+left(
          begin{array}{cc}
          0.0000+0.7869 i & -1.3380+0.0000 i \
          0.0000+0.0000 i & 0.0000-0.7869 i \
          end{array}
          right) (lambda -1)^3+left(
          begin{array}{cc}
          -0.0152-0.4524 i & 1.8410-0.0000 i \
          -0.5708-0.0000 i & -0.0152+0.4524 i \
          end{array}
          right) (lambda -1)^4+left(
          begin{array}{cc}
          0.0305+0.1179 i & -2.1780+0.0000 i \
          0.5708-0.0000 i & 0.0305-0.1179 i \
          end{array}
          right) (lambda -1)^5+Oleft((lambda -1)^6right)$







          share|improve this answer









          $endgroup$





















            4












            $begingroup$

            As far as I can tell, you did everything right in your second approach. Classically, Mathematica returns lists of rules as results of solving functions, but in this case, I find this tradition rather confusing and prefer to use ParametricNDSolveValue; it returns a ParametricFunction object that, when applied to a numerical parameter, returns a list of 4 InterpolatingFunction for your 4 functions.



            T[θ_] = {{T11[θ], T12[θ]}, {T21[θ], T22[θ]}};
            A[θ_] = {{0, E^(-I θ)/λ}, {1/36 E^(-I θ) (9 λ + 2 (-1 + λ)^2 (6 + 6 Cos[θ] + Cos[2 θ])), 0}};
            sys = {T'[θ] == T[θ].A[θ]};

            Tsol = ParametricNDSolveValue[{sys, T11[0] == 1, T12[0] == 0,
            T21[0] == 0, T22[0] == 1},
            {T11, T12, T21, T22},
            {θ, 0, 2 Pi},
            {λ}
            ];


            In order to obtain the numerical values for all the solutions at θ = 2 Pi for a given parameter, say λ = 0.1, you may use Through:



            Through[Tsol[0.1][2. Pi]]



            {-0.545795 + 1.00532 I, -1.43497 - 7.95125*10^-7 I, -0.215035 -
            2.80298*10^-8 I, -0.545795 - 1.00532 I}




            In order to make that into a function, you may use



            f = λ [Function] Through[Tsol[λ][2. Pi]]





            share|improve this answer











            $endgroup$













            • $begingroup$
              Thanks for your answer. Only the last step differs from what I'd like to obtain. What I would like to see is a function depending on $lambda$, not the evaluation at a single point.
              $endgroup$
              – Edu
              Mar 30 at 14:17



















            3












            $begingroup$

            For this case, you don't even need to write out the components of your matrix function:



            pf = ParametricNDSolveValue[{t'[θ] == t[θ].{{0, Exp[-I θ]/λ},
            {Exp[-I θ] (9 λ + 2 (λ - 1)^2
            (6 Cos[θ] + Cos[2 θ] + 6))/36, 0}},
            t[0] == IdentityMatrix[2]}, t, {θ, 0, 2 π}, λ,
            Method -> "StiffnessSwitching"];

            sol = pf[(3 + 4 I)/5];

            ParametricPlot[ReIm[Tr[sol[θ]]], {θ, 0, 2 π}]


            some curve



            ParametricPlot[ReIm[Det[sol[θ]]], {θ, 0, 2 π}]


            some other curve



            You can even make a plot where the parameter is varying:



            Plot[Re[Tr[pf[Exp[I ϕ]][2 π]]], {ϕ, 0, 2 π}]


            yet another curve






            share|improve this answer











            $endgroup$













            • $begingroup$
              Thanks for your answer. Could I see the solution as a function of $lambda$ instead? I suppose is something possible to do...
              $endgroup$
              – Edu
              Mar 30 at 14:23










            • $begingroup$
              That is what the third plot is; I let $lambda=exp(ivarphi)$ (quote "with $lambda$ a free parameter in the unit circle") and plotted the real part of the trace of the matrix.
              $endgroup$
              – J. M. is away
              Mar 30 at 14:24












            • $begingroup$
              Sure, but I'd like to manipulate it further. Not just see its plot. See my point?
              $endgroup$
              – Edu
              Mar 30 at 14:26










            • $begingroup$
              You make no mention of what kind of manipulations you like to do in your question, so no, I do not see.
              $endgroup$
              – J. M. is away
              Mar 30 at 14:26












            • $begingroup$
              For example, can I get the solution as a function of $lambda$? Or as a series expansion of powers of $lambda$?
              $endgroup$
              – Edu
              Mar 30 at 14:41












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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Here is how to obtain the series expansion of the matrix components. First, using a tweaked version (solving for t[2π] instead of t) of JM's formulation:



            A[θ_] = {
            {0, E^(-I θ)/λ},
            {1/36 E^(-I θ) (9 λ + 2 (-1 + λ)^2 (6 + 6 Cos[θ] + Cos[2 θ])), 0}
            };

            pf = ParametricNDSolveValue[
            {
            t'[θ] == t[θ].A[θ], t[0] == IdentityMatrix[2]
            },
            t[2π],
            {θ, 0, 2π},
            λ
            ];


            Then, pf will return the matrix value at $theta = 2 pi$. For example:



            pf[1]
            pf[Exp[I Pi/3]]



            {{1. + 3.62818*10^-9 I,
            6.82646*10^-8 - 5.73536*10^-9 I}, {1.70661*10^-8 - 1.43384*10^-9 I,
            1. + 3.62818*10^-9 I}}



            {{0.985595 + 1.17074 I, -0.425572 + 0.737112 I}, {-0.788363 - 1.36549 I,
            0.985595 - 1.17074 I}}




            Finding the series expansion is simple:



            DecimalForm[Series[pf[λ], {λ, 1, 5}], {4,4}] //TeXForm



            $left(
            begin{array}{cc}
            1.0000 & 6.2830 \
            1.5710 & 1.0000 \
            end{array}
            right)+left(
            begin{array}{cc}
            0.0000+0.0000 i & -6.2830+0.0000 i \
            1.5710+0.0000 i & 0.0000+0.0000 i \
            end{array}
            right) (lambda -1)+left(
            begin{array}{cc}
            0.0000-0.7869 i & 0.6691-0.0000 i \
            0.8799+0.0000 i & -0.0000+0.7869 i \
            end{array}
            right) (lambda -1)^2+left(
            begin{array}{cc}
            0.0000+0.7869 i & -1.3380+0.0000 i \
            0.0000+0.0000 i & 0.0000-0.7869 i \
            end{array}
            right) (lambda -1)^3+left(
            begin{array}{cc}
            -0.0152-0.4524 i & 1.8410-0.0000 i \
            -0.5708-0.0000 i & -0.0152+0.4524 i \
            end{array}
            right) (lambda -1)^4+left(
            begin{array}{cc}
            0.0305+0.1179 i & -2.1780+0.0000 i \
            0.5708-0.0000 i & 0.0305-0.1179 i \
            end{array}
            right) (lambda -1)^5+Oleft((lambda -1)^6right)$







            share|improve this answer









            $endgroup$


















              2












              $begingroup$

              Here is how to obtain the series expansion of the matrix components. First, using a tweaked version (solving for t[2π] instead of t) of JM's formulation:



              A[θ_] = {
              {0, E^(-I θ)/λ},
              {1/36 E^(-I θ) (9 λ + 2 (-1 + λ)^2 (6 + 6 Cos[θ] + Cos[2 θ])), 0}
              };

              pf = ParametricNDSolveValue[
              {
              t'[θ] == t[θ].A[θ], t[0] == IdentityMatrix[2]
              },
              t[2π],
              {θ, 0, 2π},
              λ
              ];


              Then, pf will return the matrix value at $theta = 2 pi$. For example:



              pf[1]
              pf[Exp[I Pi/3]]



              {{1. + 3.62818*10^-9 I,
              6.82646*10^-8 - 5.73536*10^-9 I}, {1.70661*10^-8 - 1.43384*10^-9 I,
              1. + 3.62818*10^-9 I}}



              {{0.985595 + 1.17074 I, -0.425572 + 0.737112 I}, {-0.788363 - 1.36549 I,
              0.985595 - 1.17074 I}}




              Finding the series expansion is simple:



              DecimalForm[Series[pf[λ], {λ, 1, 5}], {4,4}] //TeXForm



              $left(
              begin{array}{cc}
              1.0000 & 6.2830 \
              1.5710 & 1.0000 \
              end{array}
              right)+left(
              begin{array}{cc}
              0.0000+0.0000 i & -6.2830+0.0000 i \
              1.5710+0.0000 i & 0.0000+0.0000 i \
              end{array}
              right) (lambda -1)+left(
              begin{array}{cc}
              0.0000-0.7869 i & 0.6691-0.0000 i \
              0.8799+0.0000 i & -0.0000+0.7869 i \
              end{array}
              right) (lambda -1)^2+left(
              begin{array}{cc}
              0.0000+0.7869 i & -1.3380+0.0000 i \
              0.0000+0.0000 i & 0.0000-0.7869 i \
              end{array}
              right) (lambda -1)^3+left(
              begin{array}{cc}
              -0.0152-0.4524 i & 1.8410-0.0000 i \
              -0.5708-0.0000 i & -0.0152+0.4524 i \
              end{array}
              right) (lambda -1)^4+left(
              begin{array}{cc}
              0.0305+0.1179 i & -2.1780+0.0000 i \
              0.5708-0.0000 i & 0.0305-0.1179 i \
              end{array}
              right) (lambda -1)^5+Oleft((lambda -1)^6right)$







              share|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Here is how to obtain the series expansion of the matrix components. First, using a tweaked version (solving for t[2π] instead of t) of JM's formulation:



                A[θ_] = {
                {0, E^(-I θ)/λ},
                {1/36 E^(-I θ) (9 λ + 2 (-1 + λ)^2 (6 + 6 Cos[θ] + Cos[2 θ])), 0}
                };

                pf = ParametricNDSolveValue[
                {
                t'[θ] == t[θ].A[θ], t[0] == IdentityMatrix[2]
                },
                t[2π],
                {θ, 0, 2π},
                λ
                ];


                Then, pf will return the matrix value at $theta = 2 pi$. For example:



                pf[1]
                pf[Exp[I Pi/3]]



                {{1. + 3.62818*10^-9 I,
                6.82646*10^-8 - 5.73536*10^-9 I}, {1.70661*10^-8 - 1.43384*10^-9 I,
                1. + 3.62818*10^-9 I}}



                {{0.985595 + 1.17074 I, -0.425572 + 0.737112 I}, {-0.788363 - 1.36549 I,
                0.985595 - 1.17074 I}}




                Finding the series expansion is simple:



                DecimalForm[Series[pf[λ], {λ, 1, 5}], {4,4}] //TeXForm



                $left(
                begin{array}{cc}
                1.0000 & 6.2830 \
                1.5710 & 1.0000 \
                end{array}
                right)+left(
                begin{array}{cc}
                0.0000+0.0000 i & -6.2830+0.0000 i \
                1.5710+0.0000 i & 0.0000+0.0000 i \
                end{array}
                right) (lambda -1)+left(
                begin{array}{cc}
                0.0000-0.7869 i & 0.6691-0.0000 i \
                0.8799+0.0000 i & -0.0000+0.7869 i \
                end{array}
                right) (lambda -1)^2+left(
                begin{array}{cc}
                0.0000+0.7869 i & -1.3380+0.0000 i \
                0.0000+0.0000 i & 0.0000-0.7869 i \
                end{array}
                right) (lambda -1)^3+left(
                begin{array}{cc}
                -0.0152-0.4524 i & 1.8410-0.0000 i \
                -0.5708-0.0000 i & -0.0152+0.4524 i \
                end{array}
                right) (lambda -1)^4+left(
                begin{array}{cc}
                0.0305+0.1179 i & -2.1780+0.0000 i \
                0.5708-0.0000 i & 0.0305-0.1179 i \
                end{array}
                right) (lambda -1)^5+Oleft((lambda -1)^6right)$







                share|improve this answer









                $endgroup$



                Here is how to obtain the series expansion of the matrix components. First, using a tweaked version (solving for t[2π] instead of t) of JM's formulation:



                A[θ_] = {
                {0, E^(-I θ)/λ},
                {1/36 E^(-I θ) (9 λ + 2 (-1 + λ)^2 (6 + 6 Cos[θ] + Cos[2 θ])), 0}
                };

                pf = ParametricNDSolveValue[
                {
                t'[θ] == t[θ].A[θ], t[0] == IdentityMatrix[2]
                },
                t[2π],
                {θ, 0, 2π},
                λ
                ];


                Then, pf will return the matrix value at $theta = 2 pi$. For example:



                pf[1]
                pf[Exp[I Pi/3]]



                {{1. + 3.62818*10^-9 I,
                6.82646*10^-8 - 5.73536*10^-9 I}, {1.70661*10^-8 - 1.43384*10^-9 I,
                1. + 3.62818*10^-9 I}}



                {{0.985595 + 1.17074 I, -0.425572 + 0.737112 I}, {-0.788363 - 1.36549 I,
                0.985595 - 1.17074 I}}




                Finding the series expansion is simple:



                DecimalForm[Series[pf[λ], {λ, 1, 5}], {4,4}] //TeXForm



                $left(
                begin{array}{cc}
                1.0000 & 6.2830 \
                1.5710 & 1.0000 \
                end{array}
                right)+left(
                begin{array}{cc}
                0.0000+0.0000 i & -6.2830+0.0000 i \
                1.5710+0.0000 i & 0.0000+0.0000 i \
                end{array}
                right) (lambda -1)+left(
                begin{array}{cc}
                0.0000-0.7869 i & 0.6691-0.0000 i \
                0.8799+0.0000 i & -0.0000+0.7869 i \
                end{array}
                right) (lambda -1)^2+left(
                begin{array}{cc}
                0.0000+0.7869 i & -1.3380+0.0000 i \
                0.0000+0.0000 i & 0.0000-0.7869 i \
                end{array}
                right) (lambda -1)^3+left(
                begin{array}{cc}
                -0.0152-0.4524 i & 1.8410-0.0000 i \
                -0.5708-0.0000 i & -0.0152+0.4524 i \
                end{array}
                right) (lambda -1)^4+left(
                begin{array}{cc}
                0.0305+0.1179 i & -2.1780+0.0000 i \
                0.5708-0.0000 i & 0.0305-0.1179 i \
                end{array}
                right) (lambda -1)^5+Oleft((lambda -1)^6right)$








                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Mar 30 at 18:58









                Carl WollCarl Woll

                72.8k396188




                72.8k396188























                    4












                    $begingroup$

                    As far as I can tell, you did everything right in your second approach. Classically, Mathematica returns lists of rules as results of solving functions, but in this case, I find this tradition rather confusing and prefer to use ParametricNDSolveValue; it returns a ParametricFunction object that, when applied to a numerical parameter, returns a list of 4 InterpolatingFunction for your 4 functions.



                    T[θ_] = {{T11[θ], T12[θ]}, {T21[θ], T22[θ]}};
                    A[θ_] = {{0, E^(-I θ)/λ}, {1/36 E^(-I θ) (9 λ + 2 (-1 + λ)^2 (6 + 6 Cos[θ] + Cos[2 θ])), 0}};
                    sys = {T'[θ] == T[θ].A[θ]};

                    Tsol = ParametricNDSolveValue[{sys, T11[0] == 1, T12[0] == 0,
                    T21[0] == 0, T22[0] == 1},
                    {T11, T12, T21, T22},
                    {θ, 0, 2 Pi},
                    {λ}
                    ];


                    In order to obtain the numerical values for all the solutions at θ = 2 Pi for a given parameter, say λ = 0.1, you may use Through:



                    Through[Tsol[0.1][2. Pi]]



                    {-0.545795 + 1.00532 I, -1.43497 - 7.95125*10^-7 I, -0.215035 -
                    2.80298*10^-8 I, -0.545795 - 1.00532 I}




                    In order to make that into a function, you may use



                    f = λ [Function] Through[Tsol[λ][2. Pi]]





                    share|improve this answer











                    $endgroup$













                    • $begingroup$
                      Thanks for your answer. Only the last step differs from what I'd like to obtain. What I would like to see is a function depending on $lambda$, not the evaluation at a single point.
                      $endgroup$
                      – Edu
                      Mar 30 at 14:17
















                    4












                    $begingroup$

                    As far as I can tell, you did everything right in your second approach. Classically, Mathematica returns lists of rules as results of solving functions, but in this case, I find this tradition rather confusing and prefer to use ParametricNDSolveValue; it returns a ParametricFunction object that, when applied to a numerical parameter, returns a list of 4 InterpolatingFunction for your 4 functions.



                    T[θ_] = {{T11[θ], T12[θ]}, {T21[θ], T22[θ]}};
                    A[θ_] = {{0, E^(-I θ)/λ}, {1/36 E^(-I θ) (9 λ + 2 (-1 + λ)^2 (6 + 6 Cos[θ] + Cos[2 θ])), 0}};
                    sys = {T'[θ] == T[θ].A[θ]};

                    Tsol = ParametricNDSolveValue[{sys, T11[0] == 1, T12[0] == 0,
                    T21[0] == 0, T22[0] == 1},
                    {T11, T12, T21, T22},
                    {θ, 0, 2 Pi},
                    {λ}
                    ];


                    In order to obtain the numerical values for all the solutions at θ = 2 Pi for a given parameter, say λ = 0.1, you may use Through:



                    Through[Tsol[0.1][2. Pi]]



                    {-0.545795 + 1.00532 I, -1.43497 - 7.95125*10^-7 I, -0.215035 -
                    2.80298*10^-8 I, -0.545795 - 1.00532 I}




                    In order to make that into a function, you may use



                    f = λ [Function] Through[Tsol[λ][2. Pi]]





                    share|improve this answer











                    $endgroup$













                    • $begingroup$
                      Thanks for your answer. Only the last step differs from what I'd like to obtain. What I would like to see is a function depending on $lambda$, not the evaluation at a single point.
                      $endgroup$
                      – Edu
                      Mar 30 at 14:17














                    4












                    4








                    4





                    $begingroup$

                    As far as I can tell, you did everything right in your second approach. Classically, Mathematica returns lists of rules as results of solving functions, but in this case, I find this tradition rather confusing and prefer to use ParametricNDSolveValue; it returns a ParametricFunction object that, when applied to a numerical parameter, returns a list of 4 InterpolatingFunction for your 4 functions.



                    T[θ_] = {{T11[θ], T12[θ]}, {T21[θ], T22[θ]}};
                    A[θ_] = {{0, E^(-I θ)/λ}, {1/36 E^(-I θ) (9 λ + 2 (-1 + λ)^2 (6 + 6 Cos[θ] + Cos[2 θ])), 0}};
                    sys = {T'[θ] == T[θ].A[θ]};

                    Tsol = ParametricNDSolveValue[{sys, T11[0] == 1, T12[0] == 0,
                    T21[0] == 0, T22[0] == 1},
                    {T11, T12, T21, T22},
                    {θ, 0, 2 Pi},
                    {λ}
                    ];


                    In order to obtain the numerical values for all the solutions at θ = 2 Pi for a given parameter, say λ = 0.1, you may use Through:



                    Through[Tsol[0.1][2. Pi]]



                    {-0.545795 + 1.00532 I, -1.43497 - 7.95125*10^-7 I, -0.215035 -
                    2.80298*10^-8 I, -0.545795 - 1.00532 I}




                    In order to make that into a function, you may use



                    f = λ [Function] Through[Tsol[λ][2. Pi]]





                    share|improve this answer











                    $endgroup$



                    As far as I can tell, you did everything right in your second approach. Classically, Mathematica returns lists of rules as results of solving functions, but in this case, I find this tradition rather confusing and prefer to use ParametricNDSolveValue; it returns a ParametricFunction object that, when applied to a numerical parameter, returns a list of 4 InterpolatingFunction for your 4 functions.



                    T[θ_] = {{T11[θ], T12[θ]}, {T21[θ], T22[θ]}};
                    A[θ_] = {{0, E^(-I θ)/λ}, {1/36 E^(-I θ) (9 λ + 2 (-1 + λ)^2 (6 + 6 Cos[θ] + Cos[2 θ])), 0}};
                    sys = {T'[θ] == T[θ].A[θ]};

                    Tsol = ParametricNDSolveValue[{sys, T11[0] == 1, T12[0] == 0,
                    T21[0] == 0, T22[0] == 1},
                    {T11, T12, T21, T22},
                    {θ, 0, 2 Pi},
                    {λ}
                    ];


                    In order to obtain the numerical values for all the solutions at θ = 2 Pi for a given parameter, say λ = 0.1, you may use Through:



                    Through[Tsol[0.1][2. Pi]]



                    {-0.545795 + 1.00532 I, -1.43497 - 7.95125*10^-7 I, -0.215035 -
                    2.80298*10^-8 I, -0.545795 - 1.00532 I}




                    In order to make that into a function, you may use



                    f = λ [Function] Through[Tsol[λ][2. Pi]]






                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited Mar 30 at 14:25

























                    answered Mar 30 at 13:47









                    Henrik SchumacherHenrik Schumacher

                    59.4k582165




                    59.4k582165












                    • $begingroup$
                      Thanks for your answer. Only the last step differs from what I'd like to obtain. What I would like to see is a function depending on $lambda$, not the evaluation at a single point.
                      $endgroup$
                      – Edu
                      Mar 30 at 14:17


















                    • $begingroup$
                      Thanks for your answer. Only the last step differs from what I'd like to obtain. What I would like to see is a function depending on $lambda$, not the evaluation at a single point.
                      $endgroup$
                      – Edu
                      Mar 30 at 14:17
















                    $begingroup$
                    Thanks for your answer. Only the last step differs from what I'd like to obtain. What I would like to see is a function depending on $lambda$, not the evaluation at a single point.
                    $endgroup$
                    – Edu
                    Mar 30 at 14:17




                    $begingroup$
                    Thanks for your answer. Only the last step differs from what I'd like to obtain. What I would like to see is a function depending on $lambda$, not the evaluation at a single point.
                    $endgroup$
                    – Edu
                    Mar 30 at 14:17











                    3












                    $begingroup$

                    For this case, you don't even need to write out the components of your matrix function:



                    pf = ParametricNDSolveValue[{t'[θ] == t[θ].{{0, Exp[-I θ]/λ},
                    {Exp[-I θ] (9 λ + 2 (λ - 1)^2
                    (6 Cos[θ] + Cos[2 θ] + 6))/36, 0}},
                    t[0] == IdentityMatrix[2]}, t, {θ, 0, 2 π}, λ,
                    Method -> "StiffnessSwitching"];

                    sol = pf[(3 + 4 I)/5];

                    ParametricPlot[ReIm[Tr[sol[θ]]], {θ, 0, 2 π}]


                    some curve



                    ParametricPlot[ReIm[Det[sol[θ]]], {θ, 0, 2 π}]


                    some other curve



                    You can even make a plot where the parameter is varying:



                    Plot[Re[Tr[pf[Exp[I ϕ]][2 π]]], {ϕ, 0, 2 π}]


                    yet another curve






                    share|improve this answer











                    $endgroup$













                    • $begingroup$
                      Thanks for your answer. Could I see the solution as a function of $lambda$ instead? I suppose is something possible to do...
                      $endgroup$
                      – Edu
                      Mar 30 at 14:23










                    • $begingroup$
                      That is what the third plot is; I let $lambda=exp(ivarphi)$ (quote "with $lambda$ a free parameter in the unit circle") and plotted the real part of the trace of the matrix.
                      $endgroup$
                      – J. M. is away
                      Mar 30 at 14:24












                    • $begingroup$
                      Sure, but I'd like to manipulate it further. Not just see its plot. See my point?
                      $endgroup$
                      – Edu
                      Mar 30 at 14:26










                    • $begingroup$
                      You make no mention of what kind of manipulations you like to do in your question, so no, I do not see.
                      $endgroup$
                      – J. M. is away
                      Mar 30 at 14:26












                    • $begingroup$
                      For example, can I get the solution as a function of $lambda$? Or as a series expansion of powers of $lambda$?
                      $endgroup$
                      – Edu
                      Mar 30 at 14:41
















                    3












                    $begingroup$

                    For this case, you don't even need to write out the components of your matrix function:



                    pf = ParametricNDSolveValue[{t'[θ] == t[θ].{{0, Exp[-I θ]/λ},
                    {Exp[-I θ] (9 λ + 2 (λ - 1)^2
                    (6 Cos[θ] + Cos[2 θ] + 6))/36, 0}},
                    t[0] == IdentityMatrix[2]}, t, {θ, 0, 2 π}, λ,
                    Method -> "StiffnessSwitching"];

                    sol = pf[(3 + 4 I)/5];

                    ParametricPlot[ReIm[Tr[sol[θ]]], {θ, 0, 2 π}]


                    some curve



                    ParametricPlot[ReIm[Det[sol[θ]]], {θ, 0, 2 π}]


                    some other curve



                    You can even make a plot where the parameter is varying:



                    Plot[Re[Tr[pf[Exp[I ϕ]][2 π]]], {ϕ, 0, 2 π}]


                    yet another curve






                    share|improve this answer











                    $endgroup$













                    • $begingroup$
                      Thanks for your answer. Could I see the solution as a function of $lambda$ instead? I suppose is something possible to do...
                      $endgroup$
                      – Edu
                      Mar 30 at 14:23










                    • $begingroup$
                      That is what the third plot is; I let $lambda=exp(ivarphi)$ (quote "with $lambda$ a free parameter in the unit circle") and plotted the real part of the trace of the matrix.
                      $endgroup$
                      – J. M. is away
                      Mar 30 at 14:24












                    • $begingroup$
                      Sure, but I'd like to manipulate it further. Not just see its plot. See my point?
                      $endgroup$
                      – Edu
                      Mar 30 at 14:26










                    • $begingroup$
                      You make no mention of what kind of manipulations you like to do in your question, so no, I do not see.
                      $endgroup$
                      – J. M. is away
                      Mar 30 at 14:26












                    • $begingroup$
                      For example, can I get the solution as a function of $lambda$? Or as a series expansion of powers of $lambda$?
                      $endgroup$
                      – Edu
                      Mar 30 at 14:41














                    3












                    3








                    3





                    $begingroup$

                    For this case, you don't even need to write out the components of your matrix function:



                    pf = ParametricNDSolveValue[{t'[θ] == t[θ].{{0, Exp[-I θ]/λ},
                    {Exp[-I θ] (9 λ + 2 (λ - 1)^2
                    (6 Cos[θ] + Cos[2 θ] + 6))/36, 0}},
                    t[0] == IdentityMatrix[2]}, t, {θ, 0, 2 π}, λ,
                    Method -> "StiffnessSwitching"];

                    sol = pf[(3 + 4 I)/5];

                    ParametricPlot[ReIm[Tr[sol[θ]]], {θ, 0, 2 π}]


                    some curve



                    ParametricPlot[ReIm[Det[sol[θ]]], {θ, 0, 2 π}]


                    some other curve



                    You can even make a plot where the parameter is varying:



                    Plot[Re[Tr[pf[Exp[I ϕ]][2 π]]], {ϕ, 0, 2 π}]


                    yet another curve






                    share|improve this answer











                    $endgroup$



                    For this case, you don't even need to write out the components of your matrix function:



                    pf = ParametricNDSolveValue[{t'[θ] == t[θ].{{0, Exp[-I θ]/λ},
                    {Exp[-I θ] (9 λ + 2 (λ - 1)^2
                    (6 Cos[θ] + Cos[2 θ] + 6))/36, 0}},
                    t[0] == IdentityMatrix[2]}, t, {θ, 0, 2 π}, λ,
                    Method -> "StiffnessSwitching"];

                    sol = pf[(3 + 4 I)/5];

                    ParametricPlot[ReIm[Tr[sol[θ]]], {θ, 0, 2 π}]


                    some curve



                    ParametricPlot[ReIm[Det[sol[θ]]], {θ, 0, 2 π}]


                    some other curve



                    You can even make a plot where the parameter is varying:



                    Plot[Re[Tr[pf[Exp[I ϕ]][2 π]]], {ϕ, 0, 2 π}]


                    yet another curve







                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    answered Mar 30 at 13:57


























                    community wiki





                    J. M. is away













                    • $begingroup$
                      Thanks for your answer. Could I see the solution as a function of $lambda$ instead? I suppose is something possible to do...
                      $endgroup$
                      – Edu
                      Mar 30 at 14:23










                    • $begingroup$
                      That is what the third plot is; I let $lambda=exp(ivarphi)$ (quote "with $lambda$ a free parameter in the unit circle") and plotted the real part of the trace of the matrix.
                      $endgroup$
                      – J. M. is away
                      Mar 30 at 14:24












                    • $begingroup$
                      Sure, but I'd like to manipulate it further. Not just see its plot. See my point?
                      $endgroup$
                      – Edu
                      Mar 30 at 14:26










                    • $begingroup$
                      You make no mention of what kind of manipulations you like to do in your question, so no, I do not see.
                      $endgroup$
                      – J. M. is away
                      Mar 30 at 14:26












                    • $begingroup$
                      For example, can I get the solution as a function of $lambda$? Or as a series expansion of powers of $lambda$?
                      $endgroup$
                      – Edu
                      Mar 30 at 14:41


















                    • $begingroup$
                      Thanks for your answer. Could I see the solution as a function of $lambda$ instead? I suppose is something possible to do...
                      $endgroup$
                      – Edu
                      Mar 30 at 14:23










                    • $begingroup$
                      That is what the third plot is; I let $lambda=exp(ivarphi)$ (quote "with $lambda$ a free parameter in the unit circle") and plotted the real part of the trace of the matrix.
                      $endgroup$
                      – J. M. is away
                      Mar 30 at 14:24












                    • $begingroup$
                      Sure, but I'd like to manipulate it further. Not just see its plot. See my point?
                      $endgroup$
                      – Edu
                      Mar 30 at 14:26










                    • $begingroup$
                      You make no mention of what kind of manipulations you like to do in your question, so no, I do not see.
                      $endgroup$
                      – J. M. is away
                      Mar 30 at 14:26












                    • $begingroup$
                      For example, can I get the solution as a function of $lambda$? Or as a series expansion of powers of $lambda$?
                      $endgroup$
                      – Edu
                      Mar 30 at 14:41
















                    $begingroup$
                    Thanks for your answer. Could I see the solution as a function of $lambda$ instead? I suppose is something possible to do...
                    $endgroup$
                    – Edu
                    Mar 30 at 14:23




                    $begingroup$
                    Thanks for your answer. Could I see the solution as a function of $lambda$ instead? I suppose is something possible to do...
                    $endgroup$
                    – Edu
                    Mar 30 at 14:23












                    $begingroup$
                    That is what the third plot is; I let $lambda=exp(ivarphi)$ (quote "with $lambda$ a free parameter in the unit circle") and plotted the real part of the trace of the matrix.
                    $endgroup$
                    – J. M. is away
                    Mar 30 at 14:24






                    $begingroup$
                    That is what the third plot is; I let $lambda=exp(ivarphi)$ (quote "with $lambda$ a free parameter in the unit circle") and plotted the real part of the trace of the matrix.
                    $endgroup$
                    – J. M. is away
                    Mar 30 at 14:24














                    $begingroup$
                    Sure, but I'd like to manipulate it further. Not just see its plot. See my point?
                    $endgroup$
                    – Edu
                    Mar 30 at 14:26




                    $begingroup$
                    Sure, but I'd like to manipulate it further. Not just see its plot. See my point?
                    $endgroup$
                    – Edu
                    Mar 30 at 14:26












                    $begingroup$
                    You make no mention of what kind of manipulations you like to do in your question, so no, I do not see.
                    $endgroup$
                    – J. M. is away
                    Mar 30 at 14:26






                    $begingroup$
                    You make no mention of what kind of manipulations you like to do in your question, so no, I do not see.
                    $endgroup$
                    – J. M. is away
                    Mar 30 at 14:26














                    $begingroup$
                    For example, can I get the solution as a function of $lambda$? Or as a series expansion of powers of $lambda$?
                    $endgroup$
                    – Edu
                    Mar 30 at 14:41




                    $begingroup$
                    For example, can I get the solution as a function of $lambda$? Or as a series expansion of powers of $lambda$?
                    $endgroup$
                    – Edu
                    Mar 30 at 14:41


















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Hall Of Fame””Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Bullet-For My Valentine booed at Metal Hammer Golden Gods Awards””Unholy Aliance””The End Of Slayer?””Slayer: We Could Thrash Out Two More Albums If We're Fast Enough...””'The Unholy Alliance: Chapter III' UK Dates Added”originalet”Megadeth And Slayer To Co-Headline 'Canadian Carnage' Trek”originalet”World Painted Blood””Release “World Painted Blood” by Slayer””Metallica Heading To Cinemas””Slayer, Megadeth To Join Forces For 'European Carnage' Tour - Dec. 18, 2010”originalet”Slayer's Hanneman Contracts Acute Infection; Band To Bring In Guest Guitarist””Cannibal Corpse's Pat O'Brien Will Step In As Slayer's Guest Guitarist”originalet”Slayer’s Jeff Hanneman Dead at 49””Dave Lombardo Says He Made Only $67,000 In 2011 While Touring With Slayer””Slayer: We Do Not Agree With Dave Lombardo's Substance Or Timeline Of Events””Slayer Welcomes Drummer Paul Bostaph Back To The Fold””Slayer Hope to Unveil Never-Before-Heard Jeff Hanneman Material on Next Album””Slayer Debut New Song 'Implode' During Surprise Golden Gods Appearance””Release group Repentless by Slayer””Repentless - Slayer - Credits””Slayer””Metal Storm Awards 2015””Slayer - to release comic book "Repentless #1"””Slayer To Release 'Repentless' 6.66" Vinyl Box Set””BREAKING NEWS: Slayer Announce Farewell Tour””Slayer Recruit Lamb of God, Anthrax, Behemoth + Testament for Final Tour””Slayer lägger ner efter 37 år””Slayer Announces Second North American Leg Of 'Final' Tour””Final World Tour””Slayer Announces Final European Tour With Lamb of God, Anthrax And Obituary””Slayer To Tour Europe With Lamb of God, Anthrax And Obituary””Slayer To Play 'Last French Show Ever' At Next Year's Hellfst””Slayer's Final World Tour Will Extend Into 2019””Death Angel's Rob Cavestany On Slayer's 'Farewell' Tour: 'Some Of Us Could See This Coming'””Testament Has No Plans To Retire Anytime Soon, Says Chuck Billy””Anthrax's Scott Ian On Slayer's 'Farewell' Tour Plans: 'I Was Surprised And I Wasn't Surprised'””Slayer””Slayer's Morbid Schlock””Review/Rock; For Slayer, the Mania Is the Message””Slayer - Biography””Slayer - Reign In Blood”originalet”Dave Lombardo””An exclusive oral history of Slayer”originalet”Exclusive! Interview With Slayer Guitarist Jeff Hanneman”originalet”Thinking Out Loud: Slayer's Kerry King on hair metal, Satan and being polite””Slayer Lyrics””Slayer - Biography””Most influential artists for extreme metal music””Slayer - Reign in Blood””Slayer guitarist Jeff Hanneman dies aged 49””Slatanic Slaughter: A Tribute to Slayer””Gateway to Hell: A Tribute to Slayer””Covered In Blood””Slayer: The Origins of Thrash in San Francisco, CA.””Why They Rule - #6 Slayer”originalet”Guitar World's 100 Greatest Heavy Metal Guitarists Of All Time”originalet”The fans have spoken: Slayer comes out on top in readers' polls”originalet”Tribute to Jeff Hanneman (1964-2013)””Lamb Of God Frontman: We Sound Like A Slayer Rip-Off””BEHEMOTH Frontman Pays Tribute To SLAYER's JEFF HANNEMAN””Slayer, Hatebreed Doing Double Duty On This Year's Ozzfest””System of a Down””Lacuna Coil’s Andrea Ferro Talks Influences, Skateboarding, Band Origins + More””Slayer - Reign in Blood””Into The Lungs of Hell””Slayer rules - en utställning om fans””Slayer and Their Fans Slashed Through a No-Holds-Barred Night at Gas Monkey””Home””Slayer””Gold & Platinum - The Big 4 Live from Sofia, Bulgaria””Exclusive! Interview With Slayer Guitarist Kerry King””2008-02-23: Wiltern, Los Angeles, CA, USA””Slayer's Kerry King To Perform With Megadeth Tonight! - Oct. 21, 2010”originalet”Dave Lombardo - Biography”Slayer Case DismissedArkiveradUltimate Classic Rock: Slayer guitarist Jeff Hanneman dead at 49.”Slayer: "We could never do any thing like Some Kind Of Monster..."””Cannibal Corpse'S Pat O'Brien Will Step In As Slayer'S Guest Guitarist | The Official Slayer Site”originalet”Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Kerrang! Awards 2006 Blog: Kerrang! Hall Of Fame””Kerrang! Awards 2013: Kerrang! Legend”originalet”Metallica, Slayer, Iron Maien Among Winners At Metal Hammer Awards””Metal Hammer Golden Gods Awards””Bullet For My Valentine Booed At Metal Hammer Golden Gods Awards””Metal Storm Awards 2006””Metal Storm Awards 2015””Slayer's Concert History””Slayer - Relationships””Slayer - Releases”Slayers officiella webbplatsSlayer på MusicBrainzOfficiell webbplatsSlayerSlayerr1373445760000 0001 1540 47353068615-5086262726cb13906545x(data)6033143kn20030215029