(************** 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[ 38189, 1366]*) (*NotebookOutlinePosition[ 39256, 1397]*) (* CellTagsIndexPosition[ 39212, 1393]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell[TextData[{ "Perturbation Solution of\n", Cell[BoxData[ \(TraditionalForm\`x\^2 + \[Epsilon]\ x\ - 1\ \ = 0\)]] }], "Title"], Cell[TextData[{ "Find an approximate solution of\n", Cell[BoxData[ \(TraditionalForm\`x\^2 + \[Epsilon]\ x\ - 1\ = 0\)]], "\nas\n", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(x(\[Epsilon]) = \(\(\[Sum]\+\(n = 0\)\%\[Infinity]\)\(\ \)\)\), "TraditionalForm"], \(x\_n\), \(\[Epsilon]\^n\)}], TraditionalForm]]], "\nThe equation obviously has an analytical solution, so the purpose\nhere \ is to demonstrate the perturbation method on an equation with \na known \ solution, so that we can see that the method works." }], "Text"], Cell[BoxData[ \(Off[General::spell1]\)], "Input"], Cell["First, define the equation:", "Text"], Cell[BoxData[ \(\(qe = x^2 + \[Epsilon]\ x - 1 \[Equal] 0;\)\)], "Input"], Cell["\<\ Find the exact solutions, which comes out as a list of 2 replacment \ rules:\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(qex = Solve[qe, x]\)], "Input"], Cell[BoxData[ \({{x \[Rule] 1\/2\ \((\(-\[Epsilon]\) - \@\(4 + \[Epsilon]\^2\))\)}, {x \[Rule] 1\/2\ \((\(-\[Epsilon]\) + \@\(4 + \[Epsilon]\^2\))\)}}\)], "Output"] }, Open ]], Cell["\<\ Show how to access the element of a list (in case you do not know):\ \ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(qex[\([2]\)]\)], "Input"], Cell[BoxData[ \({x \[Rule] 1\/2\ \((\(-\[Epsilon]\) + \@\(4 + \[Epsilon]\^2\))\)}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(qex[\([2, 1]\)]\)], "Input"], Cell[BoxData[ \(x \[Rule] 1\/2\ \((\(-\[Epsilon]\) + \@\(4 + \[Epsilon]\^2\))\)\)], "Output"] }, Open ]], Cell[TextData[{ "Check that these solutions really do satisfy ", StyleBox["qe", FontWeight->"Bold"], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(qe /. qex[\([1, 1]\)] // Simplify\)], "Input"], Cell[BoxData[ \(True\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(qe /. qex[\([2, 1]\)] // Simplify\)], "Input"], Cell[BoxData[ \(True\)], "Output"] }, Open ]], Cell[TextData[{ "Define functions ", Cell[BoxData[ \(TraditionalForm\`x(\[Epsilon])\)]], "for the two exact, analytical solutions" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(xe2[\[Epsilon]_] = x /. qex[\([1]\)]\)], "Input"], Cell[BoxData[ \(1\/2\ \((\(-\[Epsilon]\) - \@\(4 + \[Epsilon]\^2\))\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(xe1[\[Epsilon]_] = x /. qex[\([2]\)]\)], "Input"], Cell[BoxData[ \(1\/2\ \((\(-\[Epsilon]\) + \@\(4 + \[Epsilon]\^2\))\)\)], "Output"] }, Open ]], Cell[TextData[{ "Now prepare to find an approximate perturbation solution. Note\nthe O[\ \[Epsilon]] symbol is powerful in ", StyleBox["Mathematica, ", FontSlant->"Italic"], "try changing {n,0,2} to {n,0,10) and see what happens..." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(xs = Sum[\(x\_n\) \[Epsilon]\^n, {n, 0, 2}] + O[\[Epsilon]]^3\)], "Input"], Cell[BoxData[ InterpretationBox[ RowBox[{\(x\_0\), "+", \(x\_1\ \[Epsilon]\), "+", \(x\_2\ \[Epsilon]\^2\), "+", InterpretationBox[\(O[\[Epsilon]]\^3\), SeriesData[ \[Epsilon], 0, {}, 0, 3, 1]]}], SeriesData[ \[Epsilon], 0, { Subscript[ x, 0], Subscript[ x, 1], Subscript[ x, 2]}, 0, 3, 1]]], "Output"] }, Open ]], Cell[TextData[{ "Substitute the series solution for ", Cell[BoxData[ \(TraditionalForm\`x\)]], "into the equation ", StyleBox["qe", FontWeight->"Bold"], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(peq = qe /. x \[Rule] xs\)], "Input"], Cell[BoxData[ RowBox[{ InterpretationBox[ RowBox[{\((\(-1\) + x\_0\%2)\), "+", \(\((x\_0 + 2\ x\_0\ x\_1)\)\ \[Epsilon]\), "+", \(\((x\_1 + x\_1\%2 + 2\ x\_0\ x\_2)\)\ \[Epsilon]\^2\), "+", InterpretationBox[\(O[\[Epsilon]]\^3\), SeriesData[ \[Epsilon], 0, {}, 0, 3, 1]]}], SeriesData[ \[Epsilon], 0, { Plus[ -1, Power[ Subscript[ x, 0], 2]], Plus[ Subscript[ x, 0], Times[ 2, Subscript[ x, 0], Subscript[ x, 1]]], Plus[ Subscript[ x, 1], Power[ Subscript[ x, 1], 2], Times[ 2, Subscript[ x, 0], Subscript[ x, 2]]]}, 0, 3, 1]], "==", "0"}]], "Output"] }, Open ]], Cell[TextData[{ StyleBox["LogicalExpand", FontWeight->"Bold"], " is really cool, and works because of the presence of the O[\[Epsilon]] \ symbol :" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(eqns = LogicalExpand[peq]\)], "Input"], Cell[BoxData[ \(\(-1\) + x\_0\%2 == 0 && x\_0 + 2\ x\_0\ x\_1 == 0 && x\_1 + x\_1\%2 + 2\ x\_0\ x\_2 == 0\)], "Output"] }, Open ]], Cell["\<\ The \"logical and\" might just as well be a list of equations that \ need to be true. In fact the equation can be accessed as list, for \ example:\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(eqns[\([1]\)]\)], "Input"], Cell[BoxData[ \(\(-1\) + x\_0\%2 == 0\)], "Output"] }, Open ]], Cell[TextData[{ "Obviously, the above equation has solutionx ", Cell[BoxData[ \(TraditionalForm\`x\_0 = 1\)]], " and ", Cell[BoxData[ \(TraditionalForm\`x\_0 = \(-1\)\)]], ".\nTaking ", Cell[BoxData[ FormBox[ RowBox[{\(x\_0\), "=", RowBox[{ "1", "we", " ", "solve", " ", "for", " ", Cell[TextData[Cell[ BoxData[ \(TraditionalForm\`x\_1\)]]]]}]}], TraditionalForm]]], "and ", Cell[BoxData[ \(TraditionalForm\`x\_2\)]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(proot = Solve[{x\_0 \[Equal] 1, eqns[\([2]\)], eqns[\([3]\)]}, {x\_1, x\_2}]\)], "Input"], Cell[BoxData[ \({{x\_2 \[Rule] 1\/8, x\_1 \[Rule] \(-\(1\/2\)\)}}\)], "Output"] }, Open ]], Cell[TextData[{ "Similarly, with ", Cell[BoxData[ \(TraditionalForm\`x\_0 = \(-1\)\)]], " :" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(nroot = Solve[{x\_0 \[Equal] \(-1\), eqns[\([2]\)], eqns[\([3]\)]}, {x\_1, x\_2}]\)], "Input"], Cell[BoxData[ \({{x\_2 \[Rule] \(-\(1\/8\)\), x\_1 \[Rule] \(-\(1\/2\)\)}}\)], "Output"] }, Open ]], Cell[TextData[{ "So here are the two approximate solutions to ", StyleBox["qe:", FontWeight->"Bold"] }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(xa1[\[Epsilon]_] = \(Normal[xs] /. x\_0 \[Rule] 1\) /. proot[\([1]\)]\)], "Input"], Cell[BoxData[ \(1 - \[Epsilon]\/2 + \[Epsilon]\^2\/8\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(xa2[\[Epsilon]_] = \(Normal[xs] /. x\_0 \[Rule] \(-1\)\) /. nroot[\([1]\)]\)], "Input"], Cell[BoxData[ \(\(-1\) - \[Epsilon]\/2 - \[Epsilon]\^2\/8\)], "Output"] }, Open ]], Cell["Compare the exact (green) and approximate (red) solutions:", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Plot[{xa1[\[Epsilon]], xe1[\[Epsilon]]}, {\[Epsilon], \(-3. \), 3. }, PlotStyle \[Rule] {RGBColor[1, 0, 0], RGBColor[0, 1, 0]}]\)], "Input"], 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