(************** Content-type: application/mathematica ************** CreatedBy='Mathematica 4.2' Mathematica-Compatible Notebook This notebook can be used with any Mathematica-compatible application, such as Mathematica, MathReader or Publicon. The data for the notebook starts with the line containing stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). 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For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. *******************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 37283, 1238]*) (*NotebookOutlinePosition[ 38242, 1268]*) (* CellTagsIndexPosition[ 38198, 1264]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell[TextData[{ "A Regular Perturbation Expansion\nfor \n", Cell[BoxData[ \(TraditionalForm\`y' = 1 + \((1 + \[Epsilon])\) y\^2\)]] }], "Title", TextAlignment->Center], Cell[TextData[{ "Solution of ", Cell[BoxData[ \(TraditionalForm\`y' \((x)\) = 1\ + \((1 + \[Epsilon])\) y\^2\)]], " with \ny(0)=1.\nWe seek a solution of form\n", Cell[BoxData[ \(TraditionalForm\`y( x) = \(y\_0\)(x) + \(\[Epsilon]y\_1\)( x) + \(\[Epsilon]\^2\) \(\(y\_2\)(x)\)\)]], "\nLogan 1.4 on page 50 (1st edition)." }], "Text"], Cell["Here is the proposed series solution:", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(ys = \ Sum[y\_n[t] \[Epsilon]\^n, {n, 0, 2}] + O[\[Epsilon]]^3\)], "Input"], Cell[BoxData[ InterpretationBox[ RowBox[{\(y\_0[t]\), "+", \(y\_1[t]\ \[Epsilon]\), "+", \(y\_2[t]\ \[Epsilon]\^2\), "+", InterpretationBox[\(O[\[Epsilon]]\^3\), SeriesData[ \[Epsilon], 0, {}, 0, 3, 1]]}], SeriesData[ \[Epsilon], 0, { Subscript[ y, 0][ t], Subscript[ y, 1][ t], Subscript[ y, 2][ t]}, 0, 3, 1]]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(D[ys, t]\)], "Input"], Cell[BoxData[ InterpretationBox[ RowBox[{ RowBox[{ SubsuperscriptBox["y", "0", "\[Prime]", MultilineFunction->None], "[", "t", "]"}], "+", RowBox[{ RowBox[{ SubsuperscriptBox["y", "1", "\[Prime]", MultilineFunction->None], "[", "t", "]"}], " ", "\[Epsilon]"}], "+", RowBox[{ RowBox[{ SubsuperscriptBox["y", "2", "\[Prime]", MultilineFunction->None], "[", "t", "]"}], " ", \(\[Epsilon]\^2\)}], "+", InterpretationBox[\(O[\[Epsilon]]\^3\), SeriesData[ \[Epsilon], 0, {}, 0, 3, 1]]}], SeriesData[ \[Epsilon], 0, { Derivative[ 1][ Subscript[ y, 0]][ t], Derivative[ 1][ Subscript[ y, 1]][ t], Derivative[ 1][ Subscript[ y, 2]][ t]}, 0, 3, 1]]], "Output"] }, Open ]], Cell["\<\ Here is the equation with the proposed perturbation solution:\ \>", \ "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(peq = D[ys, t] == 1 + \((1 + \[Epsilon])\) ys^2\)], "Input"], Cell[BoxData[ RowBox[{ InterpretationBox[ RowBox[{ RowBox[{ SubsuperscriptBox["y", "0", "\[Prime]", MultilineFunction->None], "[", "t", "]"}], "+", RowBox[{ RowBox[{ SubsuperscriptBox["y", "1", "\[Prime]", MultilineFunction->None], "[", "t", "]"}], " ", "\[Epsilon]"}], "+", RowBox[{ RowBox[{ SubsuperscriptBox["y", "2", "\[Prime]", MultilineFunction->None], "[", "t", "]"}], " ", \(\[Epsilon]\^2\)}], "+", InterpretationBox[\(O[\[Epsilon]]\^3\), SeriesData[ \[Epsilon], 0, {}, 0, 3, 1]]}], SeriesData[ \[Epsilon], 0, { Derivative[ 1][ Subscript[ y, 0]][ t], Derivative[ 1][ Subscript[ y, 1]][ t], Derivative[ 1][ Subscript[ y, 2]][ t]}, 0, 3, 1]], "==", InterpretationBox[ RowBox[{\((1 + y\_0[t]\^2)\), "+", \(\((y\_0[t]\^2 + 2\ y\_0[t]\ y\_1[t])\)\ \[Epsilon]\), "+", \(\((2\ y\_0[t]\ y\_1[t] + y\_1[t]\^2 + 2\ y\_0[t]\ y\_2[t])\)\ \[Epsilon]\^2\), "+", InterpretationBox[\(O[\[Epsilon]]\^3\), SeriesData[ \[Epsilon], 0, {}, 0, 3, 1]]}], SeriesData[ \[Epsilon], 0, { Plus[ 1, Power[ Subscript[ y, 0][ t], 2]], Plus[ Power[ Subscript[ y, 0][ t], 2], Times[ 2, Subscript[ y, 0][ t], Subscript[ y, 1][ t]]], Plus[ Times[ 2, Subscript[ y, 0][ t], Subscript[ y, 1][ t]], Power[ Subscript[ y, 1][ t], 2], Times[ 2, Subscript[ y, 0][ t], Subscript[ y, 2][ t]]]}, 0, 3, 1]]}]], "Output"] }, Open ]], Cell["\<\ With the O[\[Epsilon]] symbol, the LogicalExpand does a powerful \ step:\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(eqns = LogicalExpand[peq]\)], "Input"], Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{\(-1\), "-", \(y\_0[t]\^2\), "+", RowBox[{ SubsuperscriptBox["y", "0", "\[Prime]", MultilineFunction->None], "[", "t", "]"}]}], "==", "0"}], "&&", RowBox[{ RowBox[{\(-y\_0[t]\^2\), "-", \(2\ y\_0[t]\ y\_1[t]\), "+", RowBox[{ SubsuperscriptBox["y", "1", "\[Prime]", MultilineFunction->None], "[", "t", "]"}]}], "==", "0"}], "&&", RowBox[{ RowBox[{\(\(-2\)\ y\_0[t]\ y\_1[t]\), "-", \(y\_1[t]\^2\), "-", \(2\ y\_0[t]\ y\_2[t]\), "+", RowBox[{ SubsuperscriptBox["y", "2", "\[Prime]", MultilineFunction->None], "[", "t", "]"}]}], "==", "0"}]}]], "Output"] }, Open ]], Cell["Here is another way to look at it:", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(eqa = TableForm[ReplaceAll[eqns, And \[Rule] List]]\)], "Input"], Cell[BoxData[ InterpretationBox[GridBox[{ { RowBox[{ RowBox[{\(-1\), "-", \(y\_0[t]\^2\), "+", RowBox[{ SubsuperscriptBox["y", "0", "\[Prime]", MultilineFunction->None], "[", "t", "]"}]}], "==", "0"}]}, { RowBox[{ RowBox[{\(-y\_0[t]\^2\), "-", \(2\ y\_0[t]\ y\_1[t]\), "+", RowBox[{ SubsuperscriptBox["y", "1", "\[Prime]", MultilineFunction->None], "[", "t", "]"}]}], "==", "0"}]}, { RowBox[{ RowBox[{\(\(-2\)\ y\_0[t]\ y\_1[t]\), "-", \(y\_1[t]\^2\), "-", \(2\ y\_0[t]\ y\_2[t]\), "+", RowBox[{ SubsuperscriptBox["y", "2", "\[Prime]", MultilineFunction->None], "[", "t", "]"}]}], "==", "0"}]} }, RowSpacings->1, ColumnSpacings->3, RowAlignments->Baseline, ColumnAlignments->{Left}], TableForm[ { Equal[ Plus[ -1, Times[ -1, Power[ Subscript[ y, 0][ t], 2]], Derivative[ 1][ Subscript[ y, 0]][ t]], 0], Equal[ Plus[ Times[ -1, Power[ Subscript[ y, 0][ t], 2]], Times[ -2, Subscript[ y, 0][ t], Subscript[ y, 1][ t]], Derivative[ 1][ Subscript[ y, 1]][ t]], 0], Equal[ Plus[ Times[ -2, Subscript[ y, 0][ t], Subscript[ y, 1][ t]], Times[ -1, Power[ Subscript[ y, 1][ t], 2]], Times[ -2, Subscript[ y, 0][ t], Subscript[ y, 2][ t]], Derivative[ 1][ Subscript[ y, 2]][ t]], 0]}]]], "Output"] }, Open ]], Cell[TextData[{ "Here is the ODE for ", Cell[BoxData[ \(TraditionalForm\`y\_0\)]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(eqns[\([1]\)]\)], "Input"], Cell[BoxData[ RowBox[{ RowBox[{\(-1\), "-", \(y\_0[t]\^2\), "+", RowBox[{ SubsuperscriptBox["y", "0", "\[Prime]", MultilineFunction->None], "[", "t", "]"}]}], "==", "0"}]], "Output"] }, Open ]], Cell["...and it's solution:", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(sol0 = DSolve[{eqns[\([1]\)], y\_0[0] \[Equal] 1}, y\_0[t], t]\)], "Input"], Cell[BoxData[ \(Solve::"ifun" \(\(:\)\(\ \)\) "Inverse functions are being used by \!\(Solve\), so some solutions may \ not be found."\)], "Message"], Cell[BoxData[ \({{y\_0[t] \[Rule] Tan[\[Pi]\/4 + t]}}\)], "Output"] }, Open ]], Cell[TextData[{ "Here is the ODE for ", Cell[BoxData[ \(TraditionalForm\`y\_1\)]], ", making use of the known solution for ", Cell[BoxData[ \(TraditionalForm\`y\_0\)]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(eqn1 = eqns[\([2]\)] /. sol0[\([1]\)]\)], "Input"], Cell[BoxData[ RowBox[{ RowBox[{\(-Tan[\[Pi]\/4 + t]\^2\), "-", \(2\ Tan[\[Pi]\/4 + t]\ y\_1[t]\), "+", RowBox[{ SubsuperscriptBox["y", "1", "\[Prime]", MultilineFunction->None], "[", "t", "]"}]}], "==", "0"}]], "Output"] }, Open ]], Cell[TextData[{ "Here is the solution for ", Cell[BoxData[ \(TraditionalForm\`y\_1\)]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(sol1 = DSolve[{eqn1, y\_1[0] \[Equal] 0}, y\_1[t], t]\)], "Input"], Cell[BoxData[ \({{y\_1[ t] \[Rule] \(\(-1\) - 2\ t + Cos[2\ t]\)\/\(2\ \((\(-1\) + Sin[2\ \ t])\)\)}}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(sol0[\([1]\)]\)], "Input"], Cell[BoxData[ \({y\_0[t] \[Rule] Tan[\[Pi]\/4 + t]}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(eqns[\([3]\)]\)], "Input"], Cell[BoxData[ RowBox[{ RowBox[{\(\(-2\)\ y\_0[t]\ y\_1[t]\), "-", \(y\_1[t]\^2\), "-", \(2\ y\_0[t]\ y\_2[t]\), "+", RowBox[{ SubsuperscriptBox["y", "2", "\[Prime]", MultilineFunction->None], "[", "t", "]"}]}], "==", "0"}]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(eqn2 = \(eqns[\([3]\)] /. sol0[\([1]\)]\) /. sol1[\([1]\)]\)], "Input"], Cell[BoxData[ RowBox[{ RowBox[{\(-\(\((\(-1\) - 2\ t + Cos[2\ t])\)\^2\/\(4\ \((\(-1\) + Sin[2\ \ t])\)\^2\)\)\), "-", \(\(\((\(-1\) - 2\ t + Cos[2\ t])\)\ Tan[\[Pi]\/4 + t]\)\/\(\(-1\) + Sin[2\ t]\)\), "-", \(2\ Tan[\[Pi]\/4 + t]\ y\_2[t]\), "+", RowBox[{ SubsuperscriptBox["y", "2", "\[Prime]", MultilineFunction->None], "[", "t", "]"}]}], "==", "0"}]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(sol2 = DSolve[{eqn2, y\_2[0] == 0}, y\_2[t], t]\)], "Input"], Cell[BoxData[ \({{y\_2[t] \[Rule] 1\/8\ \((Cos[t] - Sin[t])\)\^\(\(-1\) - \(2\ Cos[2\ \ t]\)\/\(\((Cos[t] - Sin[t])\)\^3\ \((Cos[t] + Sin[t])\)\) + Sin[4\ \ t]\/\(\((Cos[t] - Sin[t])\)\^3\ \((Cos[t] + Sin[t])\)\)\)\ \((\(-3\)\ Cos[ t] - 2\ t\ Cos[t] + 4\ t\^2\ Cos[t] + 3\ Cos[t]\ Cos[2\ t] + 5\ Sin[t] + 10\ t\ Sin[t] + 4\ t\^2\ Sin[t] - 3\ Cos[2\ t]\ Sin[t])\)}}\)], "Output"] }, Open ]], Cell["\<\ Now convert these replacement rules into approximate solutions, of \ increasing O[\[Epsilon]], that we can plot:\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(ym0 = ys + O[\[Epsilon]]^1 /. sol0[\([1]\)]\)], "Input"], Cell[BoxData[ InterpretationBox[ RowBox[{\(Tan[\[Pi]\/4 + t]\), "+", InterpretationBox[\(O[\[Epsilon]]\^1\), SeriesData[ \[Epsilon], 0, {}, 0, 1, 1]]}], SeriesData[ \[Epsilon], 0, { Tan[ Plus[ Times[ Rational[ 1, 4], Pi], t]]}, 0, 1, 1]]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(ym1 = \(ys + O[\[Epsilon]]^2 /. sol0[\([1]\)]\) /. sol1[\([1]\)]\)], "Input"], Cell[BoxData[ InterpretationBox[ RowBox[{\(Tan[\[Pi]\/4 + t]\), "+", \(\(\((\(-1\) - 2\ t + Cos[2\ t])\)\ \[Epsilon]\)\/\(2\ \((\(-1\) + Sin[2\ t])\)\)\), "+", InterpretationBox[\(O[\[Epsilon]]\^2\), SeriesData[ \[Epsilon], 0, {}, 0, 2, 1]]}], SeriesData[ \[Epsilon], 0, { Tan[ Plus[ Times[ Rational[ 1, 4], Pi], t]], Times[ Rational[ 1, 2], Plus[ -1, Times[ -2, t], Cos[ Times[ 2, t]]], Power[ Plus[ -1, Sin[ Times[ 2, t]]], -1]]}, 0, 2, 1]]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(ym2 = \(\(ys + O[\[Epsilon]]^3 /. sol0[\([1]\)]\) /. sol1[\([1]\)]\) /. sol2[\([1]\)]\)], "Input"], Cell[BoxData[ InterpretationBox[ RowBox[{\(Tan[\[Pi]\/4 + t]\), "+", \(\(\((\(-1\) - 2\ t + Cos[2\ t])\)\ \[Epsilon]\)\/\(2\ \((\(-1\) + Sin[2\ t])\)\)\), "+", \(1\/8\ \((Cos[t] - Sin[t])\)\^\(\(-1\) - \(2\ Cos[2\ \ t]\)\/\(\((Cos[t] - Sin[t])\)\^3\ \((Cos[t] + Sin[t])\)\) + Sin[4\ \ t]\/\(\((Cos[t] - Sin[t])\)\^3\ \((Cos[t] + Sin[t])\)\)\)\ \((\(-3\)\ Cos[ t] - 2\ t\ Cos[t] + 4\ t\^2\ Cos[t] + 3\ Cos[t]\ Cos[2\ t] + 5\ Sin[t] + 10\ t\ Sin[t] + 4\ t\^2\ Sin[t] - 3\ Cos[2\ t]\ Sin[t])\)\ \[Epsilon]\^2\), "+", InterpretationBox[\(O[\[Epsilon]]\^3\), SeriesData[ \[Epsilon], 0, {}, 0, 3, 1]]}], SeriesData[ \[Epsilon], 0, { Tan[ Plus[ Times[ Rational[ 1, 4], Pi], t]], Times[ Rational[ 1, 2], Plus[ -1, Times[ -2, t], Cos[ Times[ 2, t]]], Power[ Plus[ -1, Sin[ Times[ 2, t]]], -1]], Times[ Rational[ 1, 8], Power[ Plus[ Cos[ t], Times[ -1, Sin[ t]]], Plus[ -1, Times[ -2, Cos[ Times[ 2, t]], Power[ Plus[ Cos[ t], Times[ -1, Sin[ t]]], -3], Power[ Plus[ Cos[ t], Sin[ t]], -1]], Times[ Power[ Plus[ Cos[ t], Times[ -1, Sin[ t]]], -3], Power[ Plus[ Cos[ t], Sin[ t]], -1], Sin[ Times[ 4, t]]]]], Plus[ Times[ -3, Cos[ t]], Times[ -2, t, Cos[ t]], Times[ 4, Power[ t, 2], Cos[ t]], Times[ 3, Cos[ t], Cos[ Times[ 2, t]]], Times[ 5, Sin[ t]], Times[ 10, t, Sin[ t]], Times[ 4, Power[ t, 2], Sin[ t]], Times[ -3, Cos[ Times[ 2, t]], Sin[ t]]]]}, 0, 3, 1]]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(ya0[t_] = Normal[ym0]\)], "Input"], Cell[BoxData[ \(Tan[\[Pi]\/4 + t]\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(ya1[t_, \[Epsilon]_] = Normal[ym1]\)], "Input"], Cell[BoxData[ \(\(\[Epsilon]\ \((\(-1\) - 2\ t + Cos[2\ t])\)\)\/\(2\ \((\(-1\) + Sin[2\ \ t])\)\) + Tan[\[Pi]\/4 + t]\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(ya2[t_, \[Epsilon]_] = Normal[ym2]\)], "Input"], Cell[BoxData[ \(1\/8\ \[Epsilon]\^2\ \((Cos[t] - Sin[t])\)\^\(\(-1\) - \(2\ Cos[2\ t]\)\ \/\(\((Cos[t] - Sin[t])\)\^3\ \((Cos[t] + Sin[t])\)\) + Sin[4\ \ t]\/\(\((Cos[t] - Sin[t])\)\^3\ \((Cos[t] + Sin[t])\)\)\)\ \((\(-3\)\ Cos[ t] - 2\ t\ Cos[t] + 4\ t\^2\ Cos[t] + 3\ Cos[t]\ Cos[2\ t] + 5\ Sin[t] + 10\ t\ Sin[t] + 4\ t\^2\ Sin[t] - 3\ Cos[2\ t]\ Sin[ t])\) + \(\[Epsilon]\ \((\(-1\) - 2\ t + Cos[2\ t])\)\)\/\(2\ \ \((\(-1\) + Sin[2\ t])\)\) + Tan[\[Pi]\/4 + t]\)], "Output"] }, Open ]], Cell["Test these functions:", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \({ya0[ .6], ya1[ .6, .1], ya2[ .6, .1]}\)], "Input"], Cell[BoxData[ \({5.331855223458727`, 6.6838398691016145`, 7.038699217661623`}\)], "Output"] }, Open ]], Cell["Lucky for us, we can find the exact solution:", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(ye = DSolve[{\(y'\)[t] \[Equal] 1 + \((1 + \[Epsilon])\) y[t]^2, y[0] \[Equal] 1}, y[t], t]\)], "Input"], Cell[BoxData[ \(Solve::"ifun" \(\(:\)\(\ \)\) "Inverse functions are being used by \!\(Solve\), so some solutions may \ not be found."\)], "Message"], Cell[BoxData[ \({{y[t] \[Rule] Tan[t\ \@\(1 + \[Epsilon]\) + ArcTan[\@\(1 + \[Epsilon]\)]]\/\@\(1 \ + \[Epsilon]\)}}\)], "Output"] }, Open ]], Cell["and here it is as a convenient function:", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(yexact[t_, \[Epsilon]_] = y[t] /. ye[\([1]\)]\)], "Input"], Cell[BoxData[ \(Tan[t\ \@\(1 + \[Epsilon]\) + ArcTan[\@\(1 + \[Epsilon]\)]]\/\@\(1 + \ \[Epsilon]\)\)], "Output"] }, Open ]], Cell["Let's test it:", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(ya0[ .6, .1]\)], "Input"], Cell[BoxData[ \(ya0[0.6`, 0.1`]\)], "Output"] }, Open ]], Cell[BoxData[ \(\(ep = .1;\)\)], "Input"], Cell[TextData[{ "Here we plot the exact solution (green), the ", Cell[BoxData[ \(TraditionalForm\`\(y\_0\)(x)\)]], "(red), ", Cell[BoxData[ \(TraditionalForm\`\(y\_0\)(x) + \[Epsilon]\)]], Cell[BoxData[ \(TraditionalForm\`\(y\_1\)(x)\)]], 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