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Notebook[{

Cell[CellGroupData[{
Cell["Simplification of higher order splittings", "Title"],

Cell["The basic fourth order method has 9 steps requiring 6 FFT's", "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(\(SplitStep`Private`SplitStep[4, dt, \ v, \ L]\)[u]\)], "Input",
  CellLabel->"In[22]:="],

Cell[BoxData[
    \(\(SplitStep`Private`Linear[dt\/\(2\ \((2 - 2\^\(1/3\))\)\), L]\)[
      \(SplitStep`Private`Nonlinear[dt\/\(2 - 2\^\(1/3\)\), v]\)[
        \(SplitStep`Private`Linear[dt\/\(2\ \((2 - 2\^\(1/3\))\)\), L]\)[
          \(SplitStep`Private`Linear[
              \(-\(dt\/\(2\^\(2/3\)\ \((2 - 2\^\(1/3\))\)\)\)\), L]\)[
            \(SplitStep`Private`Nonlinear[
                \(-\(\(2\^\(1/3\)\ dt\)\/\(2 - 2\^\(1/3\)\)\)\), v]\)[
              \(SplitStep`Private`Linear[
                  \(-\(dt\/\(2\^\(2/3\)\ \((2 - 2\^\(1/3\))\)\)\)\), L]\)[
                \(SplitStep`Private`Linear[dt\/\(2\ \((2 - 2\^\(1/3\))\)\), 
                    L]\)[\(SplitStep`Private`Nonlinear[
                      dt\/\(2 - 2\^\(1/3\)\), v]\)[
                    \(SplitStep`Private`Linear[
                        dt\/\(2\ \((2 - 2\^\(1/3\))\)\), L]\)[u]]]]]]]]]\)], 
  "Output",
  CellLabel->"Out[22]="]
}, Open  ]],

Cell["Versus 7 requiring 4 FFT's with simplification", "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(SplitStep`Private`SimplifySplitStep[4, {dt, \ v, \ L}, \ Infinity]\)], 
  "Input",
  CellLabel->"In[23]:="],

Cell[BoxData[
    \(Function[{dt, v, L}, 
      \(SplitStep`Private`Linear[dt\/\(2\ \((2 - 2\^\(1/3\))\)\), L]\)[
          \(SplitStep`Private`Nonlinear[dt\/\(2 - 2\^\(1/3\)\), v]\)[
            \(SplitStep`Private`Linear[
                \(\((\(-1\) + 2\^\(1/3\))\)\ dt\)\/\(2\ \((\(-2\) + \
2\^\(1/3\))\)\), L]\)[
              \(SplitStep`Private`Nonlinear[
                  \(-\(\(2\^\(1/3\)\ dt\)\/\(2 - 2\^\(1/3\)\)\)\), v]\)[
                \(SplitStep`Private`Linear[
                    \(\((\(-1\) + 2\^\(1/3\))\)\ dt\)\/\(2\ \((\(-2\) + \
2\^\(1/3\))\)\), L]\)[
                  \(SplitStep`Private`Nonlinear[dt\/\(2 - 2\^\(1/3\)\), v]\)[
                    \(SplitStep`Private`Linear[
                        dt\/\(2\ \((2 - 2\^\(1/3\))\)\), L]\)[#1]]]]]]] &]
      \)], "Output",
  CellLabel->"Out[23]="]
}, Open  ]],

Cell["\<\
The difference becomes even greater for higher order. Since the expressions \
will be too large to really look at, a count is made.\
\>", "Text"],

Cell[CellGroupData[{

Cell[BoxData[{
    \(lc\  = \ \(nc\  = \ 0\); 
    Cases[\(SplitStep`Private`SplitStep[10, dt, \ v, \ L]\)[u], \ 
      \((x : \((SplitStep`Private`Linear\  | \ 
                SplitStep`Private`Nonlinear)\))\) \[RuleDelayed] \ 
        If[x\  === \ SplitStep`Private`Linear, \ \(lc++\), \ \(nc++\)], \ 
      {0, Infinity}, \ Heads \[Rule] True];\), "\[IndentingNewLine]", 
    \({lc, \ nc}\)}], "Input",
  CellLabel->"In[24]:="],

Cell[BoxData[
    \({162, 81}\)], "Output",
  CellLabel->"Out[25]="]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[{
    \(lc\  = \ \(nc\  = \ 0\); 
    Cases[SplitStep`Private`SimplifySplitStep[10, {dt, \ v, \ L}], \ 
      \((x : \((SplitStep`Private`Linear\  | \ 
                SplitStep`Private`Nonlinear)\))\) \[RuleDelayed] \ 
        If[x\  === \ SplitStep`Private`Linear, \ \(lc++\), \ \(nc++\)], \ 
      {0, Infinity}, \ Heads \[Rule] True];\), "\[IndentingNewLine]", 
    \({lc, \ nc}\)}], "Input",
  CellLabel->"In[26]:="],

Cell[BoxData[
    \({82, 81}\)], "Output",
  CellLabel->"Out[27]="]
}, Open  ]]
}, Open  ]],

Cell[CellGroupData[{

Cell["Application to another equation", "Title",
  CellTags->"ODExample"],

Cell["\<\
Here is a precomputed accurate solution of an ordinary differential equation.\
\
\>", "Text"],

Cell[BoxData[
    \(accsol\  = \ 
      SetPrecision[
        NDSolve[{\(x''\)[t]\  + \ x[t] \((1\  + \ Sin[x[t]])\)\  \[Equal] \ 
              0, x[0]\  \[Equal] \ 1, \ \(x'\)[0]\  \[Equal] \ 0}, x, t, 
          Method \[Rule] NDSolve`Extrapolation, \ 
          WorkingPrecision \[Rule] 40], 30]\)], "Input",
  Evaluatable->False],

Cell["\<\
Since this takes a while to compute, I have saved the value here:\
\>", "Text"],

Cell[BoxData[
    \(\(accsol\  = \ 
        {0.2409993005122375043176174260655668801334827`30, 
          \(-1.2389427677916917599937563887743513104124595`30\)};\)\)], 
  "Input",
  CellLabel->"In[28]:="],

Cell["\<\
This is a convenient equation to check the temporal convergence of the \
splitting on since it can be done without the expense of doing FFT's. All \
that is required is a local redefinition of Linear and Nonlinear.\
\>", "Text"],

Cell["\<\
Checking convergence will require repeating the same types of operations for \
different orders.  In cases like this, you often wind up saving time by \
writing a simple function:\
\>", "Text"],

Cell[BoxData[
    \(CheckConvergence[ord_Integer, \ range_List, prec___, \ opts___Rule]\  
      := \[IndentingNewLine]Block[
        {SplitStep`Private`Linear, \ SplitStep`Private`Nonlinear, dt, \ 
          v\ , \ L, \ split, \ err, \ errtab, \ r, \ tgen, \ 
          ttab}, \[IndentingNewLine]\(SplitStep`Private`Linear[dt_, p___]\)[
            {u_, v_}]\  = \ 
          MatrixExp[{{0, 1}, {\(-1\), 0}}\ dt] . 
            {u, v}; \[IndentingNewLine]\(SplitStep`Private`Nonlinear[dt_, 
              p___]\)[{u_, v_}]\  = \ 
          MatrixExp[{{0, 0}, {\(-Sin[u]\), 0}}\ dt]\  . 
            {u, v}; \[IndentingNewLine]tgen\  = \ 
          \(Timing[
              \(split\  = \ 
                  SplitStep`Private`SimplifySplitStep[ord, {dt, v, L}, 
                    prec];\)]\)[\([1]\)]; 
        Block[{$MaxPrecision, \ $MinPrecision}, \[IndentingNewLine]If[
            Length[{prec}]\  \[Equal] \ 1\  && \ 
              prec\  > \ 
                $MachinePrecision, \[IndentingNewLine]$MaxPrecision\  = \ 
              prec; \[IndentingNewLine]$MinPrecision\  = \ 
              prec]; \[IndentingNewLine]ttab\  = \ 
            \(Timing[
                \(errtab = 
                    Table[\[IndentingNewLine]nt\  = \ Ceiling[2^k]; 
                      h\  = \ 1/nt; \[IndentingNewLine]err\  = \ 
                        Nest[split[h, v, L], {1, 0}, nt]\  - \ 
                          accsol; \[IndentingNewLine]err\  = \ 
                        Sqrt[err . err]; \[IndentingNewLine]{h, 
                        err}, \[IndentingNewLine]Evaluate[
                        Join[{k}, \ range]]];\)]\)[
              \([1]\)];\[IndentingNewLine]]; \[IndentingNewLine]r\  = \ 
          Fit[Log[10, errtab], {1, x}, x]; \[IndentingNewLine]r\  = \ 
          \((r\  - \ \((r\  /. \ x \[Rule] 0)\))\)\  /. \ 
            x \[Rule] 1; \[IndentingNewLine]Print["\<Order = \>", ord, 
          "\< Rate = \>", r, "\<\nTime to generate method = \>", tgen, \ 
          "\<\ntime to run method = \>", ttab]; \[IndentingNewLine]ListPlot[
          Log[10, errtab], \ opts]]\)], "Input",
  CellLabel->"In[29]:="],

Cell[CellGroupData[{

Cell[BoxData[
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\[InvisibleSpace]2.062160312823319`\[InvisibleSpace]"\nTime to generate \
method = "\[InvisibleSpace]\(0.`\ Second\)\[InvisibleSpace]"\ntime to run \
method = "\[InvisibleSpace]\(0.22000000000002728`\ Second\)\),
      SequenceForm[ 
      "Order = ", 2, " Rate = ", 2.0621603128233188, 
        "\nTime to generate method = ", 
        Times[ 0.0, Second], "\ntime to run method = ", 
        Times[ .22000000000002728, Second]],
      Editable->False]], "Print",
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The function seems to work, so you can run it for a sequence of orders:\
\>", "Text"],

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  "We don't seem to be getting below about error of ",
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  "for any of the methods.  This is not surprising since machine precision on \
this machine is about 16 digits.  With more digits, perhaps the results \
should be better:"
}], "Text"],

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  "Why did I fix precision for calculations with bignums?  The basic reason \
is that the precision control mechanism is designed to estimate roundoff \
errors for basic arithmetic and computing functions in ",
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  ".  Because it is important that these estimates be correct, the estimates \
are done in a conservative way.  The error being measured here is of a \
different sort, typically called local truncation error -- i.e. the error in \
the approximation caused by truncating the series for the solution.  It is \
very often the case with iterative methods that roundoff error is small \
compared to the truncation error.  However, because the arbtrary precision \
control must treat each part of the computation as independent, over many \
iterations, you may lose all of your precision.  Here is an example with the \
fourth order method."
}], "Text"],

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    split\  = \ \ SplitStep`Private`SimplifySplitStep[2, {dt, v, L}, prec];
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    \(Table[
      nt\  = \ 2^k; \[IndentingNewLine]Precision[
        Nest[split[1/nt, Function[2\ #  Conjugate[#]], \ 1], \ 
          N[Range[32], prec], nt]], {k, 0, 4}]\)}], "Input",
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As you can see, even though the smaller time step should produce a better \
result, most of the precision is lost.\
\>", "Text"]
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