NDSolve:PDE system, initial-boundary value problem:warning:NDSolve::mconly: For the method NDSolve`IDA, only...











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I tried to NDSolve the PDE system
$$partial_t w =xcdot wquadquadpartial_z x=w$$
for
$$(t,z)in[0,1]times[0,pi]$$
with boundary conditions
$$x(t,0)=w(t,0)=w(t,pi)=0$$
and initial conditions
$$w(0,z)=sin zquadquad x(0,z)=1-cos z$$
Here's my code:



s = NDSolve[{D[w[t, z], t] == w[t, z]*x[t, z], 
D[x[t, z], z] == w[t, z], w[0, z] == Sin[z], x[0, z] == 1 - Cos[z],
w[t, 0] == 0, w[t, π] == 0, x[t, 0] == 0}, {w , x}, {t, 0,
1}, {z, 0, π}]


Mathematica displays the following warning:




"NDSolve::mconly: For the method NDSolve`IDA, only machine real code
is available. Unable to continue with complex values or beyond
floating-point exceptions."




I would appreciate any help on how to overcome this error or solve numerically this kind of PDE system anyway.










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    up vote
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    down vote

    favorite












    I tried to NDSolve the PDE system
    $$partial_t w =xcdot wquadquadpartial_z x=w$$
    for
    $$(t,z)in[0,1]times[0,pi]$$
    with boundary conditions
    $$x(t,0)=w(t,0)=w(t,pi)=0$$
    and initial conditions
    $$w(0,z)=sin zquadquad x(0,z)=1-cos z$$
    Here's my code:



    s = NDSolve[{D[w[t, z], t] == w[t, z]*x[t, z], 
    D[x[t, z], z] == w[t, z], w[0, z] == Sin[z], x[0, z] == 1 - Cos[z],
    w[t, 0] == 0, w[t, π] == 0, x[t, 0] == 0}, {w , x}, {t, 0,
    1}, {z, 0, π}]


    Mathematica displays the following warning:




    "NDSolve::mconly: For the method NDSolve`IDA, only machine real code
    is available. Unable to continue with complex values or beyond
    floating-point exceptions."




    I would appreciate any help on how to overcome this error or solve numerically this kind of PDE system anyway.










    share|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I tried to NDSolve the PDE system
      $$partial_t w =xcdot wquadquadpartial_z x=w$$
      for
      $$(t,z)in[0,1]times[0,pi]$$
      with boundary conditions
      $$x(t,0)=w(t,0)=w(t,pi)=0$$
      and initial conditions
      $$w(0,z)=sin zquadquad x(0,z)=1-cos z$$
      Here's my code:



      s = NDSolve[{D[w[t, z], t] == w[t, z]*x[t, z], 
      D[x[t, z], z] == w[t, z], w[0, z] == Sin[z], x[0, z] == 1 - Cos[z],
      w[t, 0] == 0, w[t, π] == 0, x[t, 0] == 0}, {w , x}, {t, 0,
      1}, {z, 0, π}]


      Mathematica displays the following warning:




      "NDSolve::mconly: For the method NDSolve`IDA, only machine real code
      is available. Unable to continue with complex values or beyond
      floating-point exceptions."




      I would appreciate any help on how to overcome this error or solve numerically this kind of PDE system anyway.










      share|improve this question















      I tried to NDSolve the PDE system
      $$partial_t w =xcdot wquadquadpartial_z x=w$$
      for
      $$(t,z)in[0,1]times[0,pi]$$
      with boundary conditions
      $$x(t,0)=w(t,0)=w(t,pi)=0$$
      and initial conditions
      $$w(0,z)=sin zquadquad x(0,z)=1-cos z$$
      Here's my code:



      s = NDSolve[{D[w[t, z], t] == w[t, z]*x[t, z], 
      D[x[t, z], z] == w[t, z], w[0, z] == Sin[z], x[0, z] == 1 - Cos[z],
      w[t, 0] == 0, w[t, π] == 0, x[t, 0] == 0}, {w , x}, {t, 0,
      1}, {z, 0, π}]


      Mathematica displays the following warning:




      "NDSolve::mconly: For the method NDSolve`IDA, only machine real code
      is available. Unable to continue with complex values or beyond
      floating-point exceptions."




      I would appreciate any help on how to overcome this error or solve numerically this kind of PDE system anyway.







      differential-equations numerical-integration error boundary-conditions






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 27 at 15:45

























      asked Nov 27 at 13:29









      user61386

      396




      396






















          1 Answer
          1






          active

          oldest

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          up vote
          3
          down vote



          accepted










          This is a quasilinear hyperbolic system of equations. Not all initial data is valid, w=0 should be excluded from the initial data. An example of solving the problem



          s = NDSolve[{D[w[t, z], t] == w[t, z]*x[t, z], 
          D[x[t, z], z] == w[t, z], w[0, z] == 1, x[0, z] == 1 - Cos[z],
          w[t, 0] == 1, w[t, [Pi]] == 1, x[t, 0] == 0}, {w, x}, {t, 0,
          1}, {z, 0, [Pi]},
          Method -> {"MethodOfLines",
          "SpatialDiscretization" -> {"TensorProductGrid",
          "MinPoints" -> 80, "MaxPoints" -> 100,
          "DifferenceOrder" -> "Pseudospectral"}}];
          {ContourPlot[Evaluate[w[t, z] /. s], {t, 0, 1}, {z, 0, [Pi]},
          Contours -> 20, ColorFunction -> Hue, PlotLabel -> "w",
          FrameLabel -> {"t", "z"}, PlotLegends -> Automatic],
          ContourPlot[Evaluate[x[t, z] /. s], {t, 0, 1}, {z, 0, [Pi]},
          Contours -> 20, ColorFunction -> Hue, PlotLabel -> "x",
          FrameLabel -> {"t", "z"}, PlotLegends -> Automatic]}


          fig1






          share|improve this answer





















          • Thanks for great help! Also: how can I see that boundary condition $w(t,0)=0$ is invalid?
            – user61386
            Nov 27 at 15:05










          • @user61386 It is necessary to bring the system to a second order equation using $x=w_t/w$ and then $x_z=w=(w_t/w)_z$
            – Alex Trounev
            Nov 27 at 15:22










          • My only problem is that with the initial conditions used above, i.e. $w(0,z)=1$ and $x(0,z)=1-cos z$, the PDE system equation $partial_t x =w$ seems to fail for $t=0$.
            – user61386
            Nov 27 at 20:59












          • Sorry, where does this equation come from?
            – Alex Trounev
            Nov 27 at 21:12






          • 1




            Initial and boundary conditions must be consistent.
            – Alex Trounev
            Nov 28 at 10:28











          Your Answer





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          1 Answer
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          1 Answer
          1






          active

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          active

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          active

          oldest

          votes








          up vote
          3
          down vote



          accepted










          This is a quasilinear hyperbolic system of equations. Not all initial data is valid, w=0 should be excluded from the initial data. An example of solving the problem



          s = NDSolve[{D[w[t, z], t] == w[t, z]*x[t, z], 
          D[x[t, z], z] == w[t, z], w[0, z] == 1, x[0, z] == 1 - Cos[z],
          w[t, 0] == 1, w[t, [Pi]] == 1, x[t, 0] == 0}, {w, x}, {t, 0,
          1}, {z, 0, [Pi]},
          Method -> {"MethodOfLines",
          "SpatialDiscretization" -> {"TensorProductGrid",
          "MinPoints" -> 80, "MaxPoints" -> 100,
          "DifferenceOrder" -> "Pseudospectral"}}];
          {ContourPlot[Evaluate[w[t, z] /. s], {t, 0, 1}, {z, 0, [Pi]},
          Contours -> 20, ColorFunction -> Hue, PlotLabel -> "w",
          FrameLabel -> {"t", "z"}, PlotLegends -> Automatic],
          ContourPlot[Evaluate[x[t, z] /. s], {t, 0, 1}, {z, 0, [Pi]},
          Contours -> 20, ColorFunction -> Hue, PlotLabel -> "x",
          FrameLabel -> {"t", "z"}, PlotLegends -> Automatic]}


          fig1






          share|improve this answer





















          • Thanks for great help! Also: how can I see that boundary condition $w(t,0)=0$ is invalid?
            – user61386
            Nov 27 at 15:05










          • @user61386 It is necessary to bring the system to a second order equation using $x=w_t/w$ and then $x_z=w=(w_t/w)_z$
            – Alex Trounev
            Nov 27 at 15:22










          • My only problem is that with the initial conditions used above, i.e. $w(0,z)=1$ and $x(0,z)=1-cos z$, the PDE system equation $partial_t x =w$ seems to fail for $t=0$.
            – user61386
            Nov 27 at 20:59












          • Sorry, where does this equation come from?
            – Alex Trounev
            Nov 27 at 21:12






          • 1




            Initial and boundary conditions must be consistent.
            – Alex Trounev
            Nov 28 at 10:28















          up vote
          3
          down vote



          accepted










          This is a quasilinear hyperbolic system of equations. Not all initial data is valid, w=0 should be excluded from the initial data. An example of solving the problem



          s = NDSolve[{D[w[t, z], t] == w[t, z]*x[t, z], 
          D[x[t, z], z] == w[t, z], w[0, z] == 1, x[0, z] == 1 - Cos[z],
          w[t, 0] == 1, w[t, [Pi]] == 1, x[t, 0] == 0}, {w, x}, {t, 0,
          1}, {z, 0, [Pi]},
          Method -> {"MethodOfLines",
          "SpatialDiscretization" -> {"TensorProductGrid",
          "MinPoints" -> 80, "MaxPoints" -> 100,
          "DifferenceOrder" -> "Pseudospectral"}}];
          {ContourPlot[Evaluate[w[t, z] /. s], {t, 0, 1}, {z, 0, [Pi]},
          Contours -> 20, ColorFunction -> Hue, PlotLabel -> "w",
          FrameLabel -> {"t", "z"}, PlotLegends -> Automatic],
          ContourPlot[Evaluate[x[t, z] /. s], {t, 0, 1}, {z, 0, [Pi]},
          Contours -> 20, ColorFunction -> Hue, PlotLabel -> "x",
          FrameLabel -> {"t", "z"}, PlotLegends -> Automatic]}


          fig1






          share|improve this answer





















          • Thanks for great help! Also: how can I see that boundary condition $w(t,0)=0$ is invalid?
            – user61386
            Nov 27 at 15:05










          • @user61386 It is necessary to bring the system to a second order equation using $x=w_t/w$ and then $x_z=w=(w_t/w)_z$
            – Alex Trounev
            Nov 27 at 15:22










          • My only problem is that with the initial conditions used above, i.e. $w(0,z)=1$ and $x(0,z)=1-cos z$, the PDE system equation $partial_t x =w$ seems to fail for $t=0$.
            – user61386
            Nov 27 at 20:59












          • Sorry, where does this equation come from?
            – Alex Trounev
            Nov 27 at 21:12






          • 1




            Initial and boundary conditions must be consistent.
            – Alex Trounev
            Nov 28 at 10:28













          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          This is a quasilinear hyperbolic system of equations. Not all initial data is valid, w=0 should be excluded from the initial data. An example of solving the problem



          s = NDSolve[{D[w[t, z], t] == w[t, z]*x[t, z], 
          D[x[t, z], z] == w[t, z], w[0, z] == 1, x[0, z] == 1 - Cos[z],
          w[t, 0] == 1, w[t, [Pi]] == 1, x[t, 0] == 0}, {w, x}, {t, 0,
          1}, {z, 0, [Pi]},
          Method -> {"MethodOfLines",
          "SpatialDiscretization" -> {"TensorProductGrid",
          "MinPoints" -> 80, "MaxPoints" -> 100,
          "DifferenceOrder" -> "Pseudospectral"}}];
          {ContourPlot[Evaluate[w[t, z] /. s], {t, 0, 1}, {z, 0, [Pi]},
          Contours -> 20, ColorFunction -> Hue, PlotLabel -> "w",
          FrameLabel -> {"t", "z"}, PlotLegends -> Automatic],
          ContourPlot[Evaluate[x[t, z] /. s], {t, 0, 1}, {z, 0, [Pi]},
          Contours -> 20, ColorFunction -> Hue, PlotLabel -> "x",
          FrameLabel -> {"t", "z"}, PlotLegends -> Automatic]}


          fig1






          share|improve this answer












          This is a quasilinear hyperbolic system of equations. Not all initial data is valid, w=0 should be excluded from the initial data. An example of solving the problem



          s = NDSolve[{D[w[t, z], t] == w[t, z]*x[t, z], 
          D[x[t, z], z] == w[t, z], w[0, z] == 1, x[0, z] == 1 - Cos[z],
          w[t, 0] == 1, w[t, [Pi]] == 1, x[t, 0] == 0}, {w, x}, {t, 0,
          1}, {z, 0, [Pi]},
          Method -> {"MethodOfLines",
          "SpatialDiscretization" -> {"TensorProductGrid",
          "MinPoints" -> 80, "MaxPoints" -> 100,
          "DifferenceOrder" -> "Pseudospectral"}}];
          {ContourPlot[Evaluate[w[t, z] /. s], {t, 0, 1}, {z, 0, [Pi]},
          Contours -> 20, ColorFunction -> Hue, PlotLabel -> "w",
          FrameLabel -> {"t", "z"}, PlotLegends -> Automatic],
          ContourPlot[Evaluate[x[t, z] /. s], {t, 0, 1}, {z, 0, [Pi]},
          Contours -> 20, ColorFunction -> Hue, PlotLabel -> "x",
          FrameLabel -> {"t", "z"}, PlotLegends -> Automatic]}


          fig1







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 27 at 14:44









          Alex Trounev

          4,9201418




          4,9201418












          • Thanks for great help! Also: how can I see that boundary condition $w(t,0)=0$ is invalid?
            – user61386
            Nov 27 at 15:05










          • @user61386 It is necessary to bring the system to a second order equation using $x=w_t/w$ and then $x_z=w=(w_t/w)_z$
            – Alex Trounev
            Nov 27 at 15:22










          • My only problem is that with the initial conditions used above, i.e. $w(0,z)=1$ and $x(0,z)=1-cos z$, the PDE system equation $partial_t x =w$ seems to fail for $t=0$.
            – user61386
            Nov 27 at 20:59












          • Sorry, where does this equation come from?
            – Alex Trounev
            Nov 27 at 21:12






          • 1




            Initial and boundary conditions must be consistent.
            – Alex Trounev
            Nov 28 at 10:28


















          • Thanks for great help! Also: how can I see that boundary condition $w(t,0)=0$ is invalid?
            – user61386
            Nov 27 at 15:05










          • @user61386 It is necessary to bring the system to a second order equation using $x=w_t/w$ and then $x_z=w=(w_t/w)_z$
            – Alex Trounev
            Nov 27 at 15:22










          • My only problem is that with the initial conditions used above, i.e. $w(0,z)=1$ and $x(0,z)=1-cos z$, the PDE system equation $partial_t x =w$ seems to fail for $t=0$.
            – user61386
            Nov 27 at 20:59












          • Sorry, where does this equation come from?
            – Alex Trounev
            Nov 27 at 21:12






          • 1




            Initial and boundary conditions must be consistent.
            – Alex Trounev
            Nov 28 at 10:28
















          Thanks for great help! Also: how can I see that boundary condition $w(t,0)=0$ is invalid?
          – user61386
          Nov 27 at 15:05




          Thanks for great help! Also: how can I see that boundary condition $w(t,0)=0$ is invalid?
          – user61386
          Nov 27 at 15:05












          @user61386 It is necessary to bring the system to a second order equation using $x=w_t/w$ and then $x_z=w=(w_t/w)_z$
          – Alex Trounev
          Nov 27 at 15:22




          @user61386 It is necessary to bring the system to a second order equation using $x=w_t/w$ and then $x_z=w=(w_t/w)_z$
          – Alex Trounev
          Nov 27 at 15:22












          My only problem is that with the initial conditions used above, i.e. $w(0,z)=1$ and $x(0,z)=1-cos z$, the PDE system equation $partial_t x =w$ seems to fail for $t=0$.
          – user61386
          Nov 27 at 20:59






          My only problem is that with the initial conditions used above, i.e. $w(0,z)=1$ and $x(0,z)=1-cos z$, the PDE system equation $partial_t x =w$ seems to fail for $t=0$.
          – user61386
          Nov 27 at 20:59














          Sorry, where does this equation come from?
          – Alex Trounev
          Nov 27 at 21:12




          Sorry, where does this equation come from?
          – Alex Trounev
          Nov 27 at 21:12




          1




          1




          Initial and boundary conditions must be consistent.
          – Alex Trounev
          Nov 28 at 10:28




          Initial and boundary conditions must be consistent.
          – Alex Trounev
          Nov 28 at 10:28


















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