(************** Content-type: application/mathematica ************** CreatedBy='Mathematica 5.2' 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). NOTE: If you modify the data for this notebook not in a Mathematica- compatible application, you must delete the line below containing the word CacheID, otherwise Mathematica-compatible applications may try to use invalid cache data. 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[ 59990, 2311]*) (*NotebookOutlinePosition[ 61057, 2346]*) (* CellTagsIndexPosition[ 61013, 2342]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["Ordinary Differential Equations", "Title", Evaluatable->False, TextAlignment->Center, AspectRatioFixed->True], Cell[TextData[{ "Sean Mauch\nsean@caltech.edu\n", ButtonBox["http://www.its.caltech.edu/~sean", ButtonData:>{ URL[ "http://www.its.caltech.edu/~sean"], None}, ButtonStyle->"Hyperlink"], "\n", "This work is distributed under the GNU FDL. See ", ButtonBox["license.nb ", ButtonData:>{"license.nb", None}, ButtonStyle->"Hyperlink"], "for details." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Section"], ". First Order Differential Equations" }], "Section"], Cell[TextData[{ "Use the ", StyleBox["DSolve[]", "Input"], " function to solve differential equations." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(?DSolve\)\)], "Input"], Cell[TextData[{ "The first argument is a list of differential equations. The second \ argument is a list of dependent variables, (the functions to solve for); the \ final argument is the independent variable. Recall that in ", StyleBox["Mathematica", FontSlant->"Italic"], " equations have a ", Cell[BoxData[ \(TraditionalForm\` == \)]], " instead of the single ", Cell[BoxData[ \(TraditionalForm\` = \)]], " used for assignment. Below we obtain the general solution for a first \ order differential equation." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(Remove[y]; \nDSolve[\(y'\)[x] + \(1\/x\) y[x] == x\^2, y[x], x]\)], "Input"], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " uses ", Cell[BoxData[ \(TraditionalForm\`C[1], \ C[2], \ \[Ellipsis]\)]], " for the constants of integration. To impose an initial condition, add \ the constraint to the list of equations." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(DSolve[{\(y'\)[x] + \(1\/x\) y[x] == x\^2, y[1] == 1}, y[x], x]\)], "Input"], Cell["\<\ We can set a function equal to the solution to work with the \ result.\ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(f[x_] = y[x] /. First[%]\)], "Input"], Cell[BoxData[ \(Plot[f[x], {x, 0, 5}]; \nClear[f]; \)], "Input"], Cell[TextData[{ "Note that the general solution of the differential equation cannot satisfy \ the initial condition ", Cell[BoxData[ \(TraditionalForm\`y(0) = 1\)]], ". If you ask ", StyleBox["Mathematica", FontSlant->"Italic"], " to solve this ill-posed problem it will give you an error message and \ return the empty set of solutions." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(DSolve[{\(y'\)[x] + \(1\/x\) y[x] == x\^2, y[0] == 1}, y[x], x]\)], "Input"], Cell["\<\ Another problem can occur if a problem has a solution that is not \ unique.\ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(DSolve[\(y'\)[x] - \(1\/x\) y[x] == 0, y[x], x]\)], "Input"], Cell[TextData[{ "Note that there are an infinite number of solutions of the above \ differential equation that satisfy the initial condition ", Cell[BoxData[ \(TraditionalForm\`y(0) = 0\)]], ". Since ", StyleBox["Mathematica", FontSlant->"Italic"], " cannot find a unique solution, it returns the input." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(DSolve[{\(y'\)[x] - \(1\/x\) y[x] == 0, y[0] == 0}, y[x], x]\)], "Input"], Cell["\<\ To find the family of solutions, we could first calculate the \ general solution of the differential equation\ \>", "Text"], Cell[BoxData[ \(DSolve[{\(y'\)[x] - \(1\/x\) y[x] == 0}, y[x], x]\)], "Input"], Cell[BoxData[ \(f[x_] = y[x] /. First[%]\)], "Input"], Cell["\<\ and then note that this one parameter family of solutions satisfies \ the initial condition.\ \>", "Text"], Cell[BoxData[ \(f[0]\)], "Input"], Cell[CellGroupData[{ Cell[TextData[{ "Exercise ", CounterBox["Subsubsection"] }], "Subsubsection"], Cell["Solve the differential equation,", "Text"], Cell[BoxData[ \(TraditionalForm\`y' - \(1\/x\) y = \(\(x\^\[Alpha]\)\(.\)\)\)], "DisplayFormula", TextAlignment->Center] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Exercise ", CounterBox["Subsubsection"] }], "Subsubsection"], Cell["Show that the solution of the Bernoulli equation:", "Text"], Cell[BoxData[ \(TraditionalForm\`y' = \(2\/x\) y + y\^2\)], "DisplayFormula", TextAlignment->Center], Cell["is", "Text"], Cell[BoxData[ \(TraditionalForm\`y = \(3\ x\^2\)\/\(c - x\^3\)\)], "DisplayFormula", TextAlignment->Center] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Exercise ", CounterBox["Subsubsection"] }], "Subsubsection"], Cell["Show that the solution of the general Bernoulli equation:", "Text"], Cell[BoxData[ \(TraditionalForm\`y' = \(p(x)\)\ y + \(q(x)\) y\^\[Alpha], \ \ \ \ \ \ \ \ \[Alpha] \[NotEqual] 1, \)], "DisplayFormula", TextAlignment->Center], Cell["is", "Text"], Cell[BoxData[ \(TraditionalForm \`y(x) = \(\(( \(\[ExponentialE]\^\(\((1 - \[Alpha])\) \(\[Integral]\(p(x)\) \[DifferentialD]x\)\)\)( c - \((1 - \[Alpha])\) \(\[Integral]\[ExponentialE]\^\(\((\[Alpha] - 1)\) \(\[Integral]\(p(x)\) \[DifferentialD]x\)\)\ \(q(x)\) \[DifferentialD]x\))) \)\^\(1\/\(1 - \[Alpha]\)\) . \)\)], "DisplayFormula", TextAlignment->Center] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Section"], ". Higher Order Differential Equations" }], "Section"], Cell["Suppose we wish to solve,", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ FormBox[ RowBox[{\(\(y'\)' - 2 y' + 2 y == 0\), ",", " ", FormBox[\(y(0) = 0\), "TraditionalForm"], ",", " ", FormBox[\(y' \((0)\) = \(\(1\)\(.\)\)\), "TraditionalForm"]}], TraditionalForm]], "DisplayFormula", TextAlignment->Center, TextJustification->0], Cell["\<\ We could first obtain the general solution of the differential \ equation\ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(Remove[y]; \)\)], "Input"], Cell[BoxData[ \(DSolve[\(\(y'\)'\)[x] - 2 \( y'\)[x] + 2 y[x] == 0, y[x], x]\)], "Input"], Cell[BoxData[ \(f[x_] = y[x] /. First[%]\)], "Input"], Cell[TextData[{ "and then solve for ", Cell[BoxData[ \(TraditionalForm\`C[1]\)]], " and ", Cell[BoxData[ \(TraditionalForm\`C[2]\)]], " by applying the initial conditions." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(Solve[{f[0] == 0, \(f'\)[0] == 1}]\)], "Input"], Cell[BoxData[ \(f[x_] = f[x] /. %[\([1]\)]\)], "Input"], Cell[TextData[{ "Alternatively, we can include the initial conditions in the list of \ equations we give to ", Cell[BoxData[ FormBox[ StyleBox[\(DSolve[]\), "Input"], TraditionalForm]]], "." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(DSolve[{\(\(y'\)'\)[x] - 2 \( y'\)[x] + 2 y[x] == 0, y[0] == 0, \(y'\)[0] == 1}, y[x], x]\)], "Input"], Cell[BoxData[ \(f[x_] = y[x] /. First[%]\)], "Input"], Cell[BoxData[ \(Plot[f[x], {x, 0, 5}, PlotRange -> All]; \nClear[f]\)], "Input"], Cell[CellGroupData[{ Cell[TextData[{ "Exercise ", CounterBox["Subsubsection"] }], "Subsubsection"], Cell["Find the fundamental set of solutions for:", "Text"], Cell[BoxData[ \(TraditionalForm\`\(y'\)' + 2 a\ y' + \(a\^2\) y = 0\)], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "at the point ", Cell[BoxData[ \(TraditionalForm\`x = 0\)]], "." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Exercise ", CounterBox["Subsubsection"] }], "Subsubsection"], Cell["Find the general solutions of ", "Text"], Cell[BoxData[ \(TraditionalForm\`\(\(\(y'\)' + \(1\/x\) y' + \((1 - n\^2\/x\^2)\) y = 0\)\(,\)\)\)], "DisplayFormula", TextAlignment->Center], Cell[BoxData[ \(TraditionalForm\`\(\(\(y'\)' + \(1\/x\) y' + \((\[Lambda]\^2 - 1\/x\^2)\) y = 0\)\(,\)\)\)], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "for ", Cell[BoxData[ \(TraditionalForm\`x > 0\)]], ", where ", Cell[BoxData[ \(TraditionalForm\`n\)]], " is an integer." }], "Text"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Section"], ". Numerical Solutions" }], "Section"], Cell[CellGroupData[{ Cell["Single Differential Equation", "Subsection"], Cell["Consider the problem", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ FormBox[ RowBox[{\(\(y'\)' + \(1\/\(cos\ x\)\) y' + \[ExponentialE]\^x\ y = 0\), ",", " ", FormBox[\(y(0) = 1\), "TraditionalForm"], ",", " ", FormBox[\(y' \((0)\) = \(\(1\)\(.\)\)\), "TraditionalForm"]}], TraditionalForm]], "DisplayFormula", TextAlignment->Center, TextJustification->0], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " cannot solve this differential equation analytically so it returns the \ input. (For most differential equations, the solution cannot be expressed in \ terms of elementary functions.) " }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(Remove[y]\)], "Input"], Cell[BoxData[ \(DSolve[{\(\(y'\)'\)[x] + \(1\/Cos[x]\) \(y'\)[x] + Exp[x]\ y[x] == 0, y[0] == 1, \(y'\)[0] == 1}, y[x], x]\)], "Input"], Cell[TextData[{ "You can use the ", Cell[BoxData[ FormBox[ StyleBox[\(NDSolve[]\), "Input"], TraditionalForm]]], " function to obtain a numerical approximation of the solution. Instead of \ just specifying the independent variable as the last argument as you do with \ ", Cell[BoxData[ FormBox[ StyleBox[\(DSolve[]\), "Input"], TraditionalForm]]], ", you must specify the independent variable and a range on which to \ numerically solve the equation." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(?NDSolve\)\)], "Input"], Cell[BoxData[ \(NDSolve[{\(\(y'\)'\)[x] + \(1\/Cos[x]\) \(y'\)[x] + Exp[x]\ y[x] == 0, y[0] == 1, \(y'\)[0] == 1}, y[x], {x, 0, 4}]\)], "Input"], Cell[TextData[{ Cell[BoxData[ FormBox[ StyleBox[\(NDSolve[]\), "Input"], TraditionalForm]]], " returns an ", Cell[BoxData[ FormBox[ StyleBox[\(InterpolatingFunction[]\), "Input"], TraditionalForm]]], ". This is comprised of a table of values of the function and perhaps its \ derivatives and a domain on which these values are known. We can use ", Cell[BoxData[ FormBox[ StyleBox[\(InterpolatingFunction[]\), "Input"], TraditionalForm]]], "'s in much the same way that we use other functions. Below we set ", StyleBox["f[x]", "Input"], " to the solution." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(f[x_] = y[x] /. First[%]; \)\)], "Input"], Cell[BoxData[ \(\(?InterpolatingFunction\)\)], "Input"], Cell[TextData[{ "If we evaluate the ", Cell[BoxData[ FormBox[ StyleBox[\(InterpolatingFunction[]\), "Input"], TraditionalForm]]], " for a point that lies in its domain, then it will calculate a value by \ interpolating the known values of the function." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(f[1]\)], "Input"], Cell[TextData[{ "If we try to evaluate the ", Cell[BoxData[ FormBox[ StyleBox[\(InterpolatingFunction[]\), "Input"], TraditionalForm]]], " for a point outside its domain then it will use extrapolation to compute \ a value of the function. Be warned that extrapolation is typically wildly \ inaccurate." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(f[\(-1\)]\)], "Input"], Cell["\<\ Here is a plot of the numerical approximation of the solution.\ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(Plot[f[x], {x, 0, 4}]; \)\)], "Input"], Cell[TextData[{ "Like other functions, we can take derivatives of an ", Cell[BoxData[ FormBox[ StyleBox[\(InterpolatingFunction[]\), "Input"], TraditionalForm]]], "." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(Plot[\(f'\)[x], {x, 0, 4}]; \nClear[f]\)], "Input"], Cell[CellGroupData[{ Cell[TextData[{ "Exercise ", CounterBox["Subsubsection"] }], "Subsubsection"], Cell["Numerically solve the Van der Pol equation:", "Text"], Cell[BoxData[ \(TraditionalForm\`\(y'\)' - \[Mu]\ \((1 - y\^2)\) y' + y = 0, \ \ \ \ \[Mu] > 0. \)], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "The equation describes a non-linear oscillator. The ", Cell[BoxData[ \(TraditionalForm\`\(-\[Mu]\)\ \((1 - y\^2)\) y'\)]], " is a forcing/damping term. When ", Cell[BoxData[ \(TraditionalForm\`y < 1\)]], " it acts to increase the amplitude; when ", Cell[BoxData[ \(TraditionalForm\`y > 1\)]], " it is a damping that acts to decrease the amplitude. Verify that \ regardless of the initial conditions the amplitude of oscillation tends to 2. \ (Any initial condition except ", Cell[BoxData[ \(TraditionalForm\`y(0) = \(y' \((0)\) = 0\)\)]], ", that is. Take ", Cell[BoxData[ \(TraditionalForm\`\[Mu] = 1\)]], ". You cannot numerically solve a differential equation that involves \ parameters that don't have a numerical value.) Plot the solution for ", Cell[BoxData[ \(TraditionalForm\`\[Mu] = 0.1\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\[Mu] = 10\)]], " to see the effect of a large forcing/damping force." }], "Text", TextJustification->1] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["Systems of Differential Equations", "Subsection"], Cell["\<\ A nonlinear system of differential equations that describe a \ Lorentz attractor is\ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ FormBox[GridBox[{ {\(x' = 10\ y - 10\ x, \)}, {\(y' = 28\ x - y - x\ z, \)}, {\(z' = x\ y - \(8\/3\) \(z . \)\)} }, ColumnAlignments->{"="}], TraditionalForm]], "DisplayFormula", TextAlignment->Center, TextJustification->0], Cell[TextData[{ "Below we solve this set of equations numerically. The position at time ", Cell[BoxData[ \(TraditionalForm\`t = 0\)]], " is ", Cell[BoxData[ \(TraditionalForm\`\((\(-10\), \(-10\), 20)\)\)]], "." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[{ \(\(\[Sigma] = 10;\)\), "\n", \(\(r = 28;\)\), "\n", \(\(b = 8\/3;\)\)}], "Input"], Cell[BoxData[ \(NDSolve[\n\t{\(x'\)[t] == \[Sigma] \((y[t] - x[t])\), \n\t\t\(y'\)[t] == r\ x[t] - y[t] - x[t]\ z[t], \n\t\t\(z'\)[t] == x[t]\ y[t] - b\ z[t], \n\t\tx[0] == \(-10\), \n\t\ty[ 0] == \(-10\), \n\t\tz[0] == 20}, {x[t], y[t], z[t]}, {t, 0, 8}]\)], "Input"], Cell[TextData[{ "Note that ", StyleBox["NDSolve[]", FontWeight->"Bold"], " returned a list of interpolation functions. Below we define ", Cell[BoxData[ \(TraditionalForm\`\((xs, ys, zs)\)\)]], " to be this solution." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \({xs[t_], ys[t_], zs[t_]} = {x[t], y[t], z[t]} /. %[\([1]\)]\)], "Input"], Cell["\<\ Below is a plot of the solution. It looks pretty bad. This is \ because by default, the solution is being sampled at a low resolution for \ plotting purposes.\ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(ParametricPlot3D[{xs[t], ys[t], zs[t]}, {t, 0, 8}, PlotRange -> All];\)\)], "Input"], Cell[TextData[{ "To get a better picture, we just increase the number of ", StyleBox["PlotPoints", FontWeight->"Bold"], "." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(ParametricPlot3D[{xs[t], ys[t], zs[t]}, {t, 0, 8}, PlotRange -> All, PlotPoints -> 500];\)\)], "Input"], Cell[TextData[{ "Suppose we want to solve up to ", Cell[BoxData[ \(TraditionalForm\`t = 10\)]], " instead of ", Cell[BoxData[ \(TraditionalForm\`t = 8\)]], "." }], "Text"], Cell[BoxData[ \(NDSolve[\n\t{\(x'\)[t] == \[Sigma] \((y[t] - x[t])\), \n\t\t\(y'\)[t] == r\ x[t] - y[t] - x[t]\ z[t], \n\t\t\(z'\)[t] == x[t]\ y[t] - b\ z[t], \n\t\tx[0] == \(-10\), \n\t\ty[ 0] == \(-10\), \n\t\tz[0] == 20}, {x[t], y[t], z[t]}, {t, 0, 10}]\)], "Input"], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " tells us that it reached the maximum number of allowed steps before it \ got to ", Cell[BoxData[ \(TraditionalForm\`t = 10\)]], ". Usually this safe guard is a good thing. It will keep ", StyleBox["Mathematica", FontSlant->"Italic"], " from taking an infinite amount of time in computing the solution when it \ encounters a singularity. However there is no singularity here, the solution \ is just quite complicated. We solve this problem by increasing the number of \ allowed steps." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(NDSolve[\n\t{\(x'\)[t] == \[Sigma] \((y[t] - x[t])\), \n\t\t\(y'\)[t] == r\ x[t] - y[t] - x[t]\ z[t], \n\t\t\(z'\)[t] == x[t]\ y[t] - b\ z[t], \n\t\tx[0] == \(-10\), \n\t\ty[ 0] == \(-10\), \n\t\tz[0] == 20}, {x[t], y[t], z[t]}, {t, 0, 10}, MaxSteps -> \[Infinity]]\)], "Input"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Section"], ". Series Solutions" }], "Section"], Cell[CellGroupData[{ Cell["Computing the First Few Terms", "Subsection"], Cell["Consider the problem", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ FormBox[ RowBox[{\(y'' + \(1\/\(cos\ x\)\) y' + \[ExponentialE]\^x\ y = 0\), ",", " ", FormBox[\(y(0) = 1\), "TraditionalForm"], ",", " ", FormBox[\(y' \((0)\) = \(\(1\)\(.\)\)\), "TraditionalForm"]}], TraditionalForm]], "DisplayFormula", TextAlignment->Center, TextJustification->0], Cell["\<\ The solution cannot be expressed in terms of elementary \ functions.\ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(Remove[y]\)], "Input"], Cell[BoxData[ \(DSolve[\(\(y'\)'\)[x] + \(1\/Cos[x]\) \(y'\)[x] + Exp[x]\ y[x] == 0, y[x], x]\)], "Input"], Cell[TextData[{ "We want the series solution of this problem about the point ", Cell[BoxData[ \(TraditionalForm\`x = 0\)]], ". Since the coefficient functions are analytic there, it will be a Taylor \ series. The nearest singularity of the coefficient functions is at ", Cell[BoxData[ \(TraditionalForm\`x = \[Pi]\/2\)]], ", so the radius of convergence of the series will be at least ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], ".", "\n", "We can generate the series expansion of an unknown function." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(Series[y[x], {x, 0, 5}]\)], "Input"], Cell["\<\ Expanding the differential equation in a power series gives us\ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(Series[ \(\(y'\)'\)[x] + \(1\/Cos[x]\) \(y'\)[x] + Exp[x]\ y[x], {x, 0, 5}]\)], "Input"], Cell[BoxData[ \(\(eqnSeries = %;\)\)], "Input"], Cell[TextData[{ "Now we equate each coefficient in the expansion to zero. We will find ", Cell[BoxData[ \(TraditionalForm\`\(y\^\((n)\)\)(0)\)]], " in terms of ", Cell[BoxData[ \(TraditionalForm\`y(0)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`y' \((0)\)\)]], ". We use the ", Cell[BoxData[ FormBox[ StyleBox[\(SeriesCoefficient[]\), "Input"], TraditionalForm]]], " function to get the coefficients of terms in power series." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(?SeriesCoefficient\)\)], "Input"], Cell["The coefficient of the constant term in the series is", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(SeriesCoefficient[eqnSeries, 0]\)], "Input"], Cell[TextData[{ "We equate this to zero and solve for ", Cell[BoxData[ \(TraditionalForm\`\(y'\)' \((0)\)\)]], "." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(Solve[% == 0, \(\(y'\)'\)[0]]\)], "Input"], Cell[BoxData[ \(\(\(y'\)'\)[0] = \(\(y'\)'\)[0] /. %[\([1, 1]\)]\)], "Input"], Cell[TextData[{ "Now we do the same for the coefficients of the ", Cell[BoxData[ \(TraditionalForm\`x\)]], " and ", Cell[BoxData[ \(TraditionalForm\`x\^2\)]], " terms to find ", Cell[BoxData[ \(TraditionalForm\`\(\(y'\)'\)' \((0)\)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\(\(\(y'\)'\)'\)' \((0)\)\)]], "." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(Solve[ SeriesCoefficient[eqnSeries, 1] == 0, \(\(\(y'\)'\)'\)[0]]\)], "Input"], Cell[BoxData[ \(\(\(\(y'\)'\)'\)[0] = \(\(\(y'\)'\)'\)[0] /. %[\([1, 1]\)]\)], "Input"], Cell[BoxData[ \(Solve[ SeriesCoefficient[eqnSeries, 2] == 0, \(\(\(\(y'\)'\)'\)'\)[ 0]]\)], "Input"], Cell[BoxData[ \(\(\(\(\(y'\)'\)'\)'\)[ 0] = \(\(\(\(y'\)'\)'\)'\)[0] /. %[\([1, 1]\)]\)], "Input"], Cell[TextData[{ "We have determined ", Cell[BoxData[ \(TraditionalForm\`\(y'\)' \((0)\)\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\(\(y'\)'\)' \((0)\)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\(\(\(y'\)'\)'\)' \((0)\)\)]], " in terms of ", Cell[BoxData[ \(TraditionalForm\`y(0)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`y' \((0)\)\)]], ". The first few terms in the series expansion of ", Cell[BoxData[ \(TraditionalForm\`y\)]], " are" }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(Series[y[x], {x, 0, 4}]\)], "Input"], Cell[TextData[{ "We convert this series to a polynomial with the ", Cell[BoxData[ FormBox[ StyleBox[\(Normal[]\), "Input"], TraditionalForm]]], " function." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(Normal[%]\)], "Input"], Cell[TextData[{ "Now we apply the initial conditions to obtain the first few terms in the \ series expansion of the solution about the point ", Cell[BoxData[ \(TraditionalForm\`x = 0\)]], "." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(% /. {y[0] -> 1, \(y'\)[0] -> 1}\)], "Input"], Cell[BoxData[ \(Clear[y, eqnSeries]\)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["Computing More Terms", "Subsection"], Cell["\<\ We can automate the method of the previous section to compute the \ terms in the series expansion more quickly.\ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(Remove[u, y]\)], "Input"], Cell["First we specify the initial conditions.", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(y[0] = 1; \(y'\)[0] = 1; \)], "Input"], Cell[TextData[{ "We compute the series expansion of the differential equation up to the ", Cell[BoxData[ \(TraditionalForm\`x\^10\)]], " term." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(eqnSeries = Series[\(\(y'\)'\)[x] + \(1\/Cos[x]\) \(y'\)[x] + Exp[x]\ y[x], {x, 0, 10}];\)\)], "Input"], Cell["Now we make a table of the coefficients.", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(coefficients = Table[SeriesCoefficient[eqnSeries, n], {n, 0, 10}];\)\)], "Input"], Cell["Now we equate each of the coefficients to zero.", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(Solve[coefficients == 0]; \)\)], "Input"], Cell[TextData[{ "Finally we apply rules for ", Cell[BoxData[ \(TraditionalForm\`\(y\^\((n)\)\)(0)\)]], " to the series expansion of ", Cell[BoxData[ \(TraditionalForm\`y\)]], " and change the series to a polynomial." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(Series[y[x], {x, 0, 12}] /. %[\([1]\)] // Normal\)], "Input"], Cell[BoxData[ \(\(seriesSolution[x_] = %;\)\)], "Input"], Cell["\<\ We find a numerical approximation of the solution for comparison.\ \ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(NDSolve[{\(\(u'\)'\)[x] + \(1\/Cos[x]\) \(u'\)[x] + Exp[x]\ u[x] == 0, u[0] == 1, \(u'\)[0] == 1}, u[x], {x, 0, 2}]\)], "Input"], Cell[BoxData[ \(\(numericalSolution[x_] = u[x] /. %[\([1, 1]\)];\)\)], "Input"], Cell[TextData[{ "Here is a plot of the first thirteen terms in the series expansion of the \ solution and the numerical approximation of the solution. Note that the \ series expansion converges to the solution for ", Cell[BoxData[ \(TraditionalForm\`\( | x | \( < \[Pi]\/2\)\)\)]], "." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(Plot[{seriesSolution[x], numericalSolution[x]}, {x, 0, 2}]; \)\)], "Input"], Cell[BoxData[ \(Clear[y, eqnSeries, coefficients, seriesSolution, numericalSolution]\)], "Input"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Section"], ". Transformations" }], "Section"], Cell[CellGroupData[{ Cell["Bernoulli Equations", "Subsection"], Cell["Bernoulli equations have the form", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(TraditionalForm\`y' = \(p(x)\)\ y + \(q( x)\)\ \(\(y\^\[Alpha]\)\(.\)\)\)], "DisplayFormula", TextAlignment->Center, TextJustification->0], Cell[TextData[{ "They can be transformed to a first order linear equation with the change \ of variable ", Cell[BoxData[ \(TraditionalForm\`u = y\^\(1 - \[Alpha]\)\)]], ".", "\n", "Consider the Bernoulli equation" }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(TraditionalForm\`y' = \(2\/x\) y + \(\(y\^2\)\(.\)\)\)], "DisplayFormula", TextAlignment->Center, TextJustification->0], Cell["\<\ We can apply the change of variable by applying a rule to the \ equation.\ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(Remove[u, y]\)], "Input"], Cell[BoxData[ \(\(y'\)[x] - \(2\/x\) y[x] + y[x]\^2 /. {y[x] -> 1\/u[x], \(y'\)[x] -> \[PartialD]\_x\( 1\/u[x]\)}\)], "Input"], Cell[TextData[{ "We multiply by ", Cell[BoxData[ \(TraditionalForm\`u[x]\^2\)]], " to put this in a nice form." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(%*u[x]\^2 // Simplify\)], "Input"], Cell["Now we solve the linear differential equation.", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(DSolve[% == 0, u[x], x]\)], "Input"], Cell[TextData[{ "Finally we transform back to ", Cell[BoxData[ \(TraditionalForm\`y(x) = 1\/\(u(x)\)\)]], " to obtain the solution." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(1\/u[x] /. First[%] // Simplify\)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["Riccati Equations", "Subsection"], Cell["Differential equations of the form", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(TraditionalForm\`y' = \(a(x)\) y\^2 + \(b(x)\) y + c(x)\)], "DisplayFormula", TextAlignment->Center, TextJustification->0], Cell[TextData[{ "are called Riccati equations. Riccati equations can be transformed to \ second order linear differential equations with the change of variables ", Cell[BoxData[ \(TraditionalForm\`y = \(-\(u'\/\(a\ u\)\)\)\)]], ".", "\n", "Consider the Riccati equation" }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(TraditionalForm\`y' = y\^2 + \(1\/x\) y + 1\/\(\(x\^2\)\(.\)\)\)], "DisplayFormula", TextAlignment->Center, TextJustification->0], Cell[TextData[{ "We make the change of variables ", Cell[BoxData[ \(TraditionalForm\`y = \(-\(u'\/u\)\)\)]], "." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(Remove[u, y]; \)\)], "Input"], Cell[BoxData[ \(\(y'\)[x] - y[x]\^2 - \(1\/x\) y[x] - 1\/x\^2 /. {y[x] -> \(-\(\(u'\)[x]\/u[x]\)\), \(y'\)[x] -> D[\(-\(\(u'\)[x]\/u[x]\)\), x]}\)], "Input"], Cell[BoxData[ \(%*u[x] // Expand\)], "Input"], Cell["This is a second order Euler equation.", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(DSolve[% == 0, u[x], x]\)], "Input"], Cell[BoxData[ \(u[x_] = u[x] /. First[%]\)], "Input"], Cell[TextData[{ "We set ", Cell[BoxData[ \(TraditionalForm\`y = \(-\(u'\/u\)\)\)]], "." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(y[x_] = Simplify[\(-\(\(u'\)[x]\/u[x]\)\)]\)], "Input"], Cell[TextData[{ "Note that first order differential equation have only one constant of \ integration in the solution. We can put the answer in a better form by \ writing it in terms of the constant ", Cell[BoxData[ \(TraditionalForm\`c = C[2]/C[1]\)]], "." }], "Text", TextJustification->1], Cell[BoxData[ \(y[x_] = Simplify[y[x] /. {C[2] -> c\ C[1]}]\)], "Input"], Cell[BoxData[ \(Clear[y, u]\)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["Transformation to a Constant Coefficient Equation", "Subsection"], Cell["Consider the second order linear differential equation", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(TraditionalForm\`y'' + \(p(x)\) y' + \(q(x)\) y = 0. \)], "DisplayFormula", TextAlignment->Center, TextJustification->0], Cell["If the expression", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(TraditionalForm\`\(q' + 2 p\ q\)\/q\^\(3/2\)\)], "DisplayFormula", TextAlignment->Center, TextJustification->0], Cell["is a constant, then the change of variables", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ FormBox[ RowBox[{ \(\[Xi] = c \(\[Integral]\^x\(\@\( q(t)\)\) \[DifferentialD]t\)\), ",", " ", FormBox[\(u(\[Xi]) = y(x)\), "TraditionalForm"], ","}], TraditionalForm]], "DisplayFormula", TextAlignment->Center, TextJustification->0], Cell[TextData[{ "will transform the differential equation to a constant coefficient \ equation. ", Cell[BoxData[ \(TraditionalForm\`c\)]], " is an arbitrary constant.", "\n", "Consider the differential equation" }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(TraditionalForm\`\(y'\)' + 1\/x\ y' - 1\/x\^2\ y = 0. \)], "DisplayFormula", TextAlignment->Center, TextJustification->0], Cell[TextData[{ "Since ", Cell[BoxData[ \(TraditionalForm\`\(q' + 2 p\ q\)\/q\^\(3/2\)\)]], " is a constant, the equation can be transformed to a constant coefficient \ equation." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(p[x_] = 1/x; \nq[x_] = \(-1\)/x\^2; \n Simplify[\(\(q'\)[x] + 2 p[x] q[x]\)\/q[x]\^\(3/2\)]\)], "Input"], Cell[BoxData[ \(Clear[p, q]\)], "Input"], Cell["We make the change of independent variable", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ FormBox[ RowBox[{ \(\[Xi] = \(c \(\[Integral]\(\[ImaginaryI]\/x\) \[DifferentialD]x\) = log\ x\)\), ",", " ", RowBox[{ FormBox[\(u(\[Xi]) = y(x)\), "TraditionalForm"], "."}]}], TraditionalForm]], "DisplayFormula", TextAlignment->Center, TextJustification->0], Cell[BoxData[ \(Clear[x, \[Xi]]\)], "Input"], Cell[BoxData[ \(Remove[u, y]\)], "Input"], Cell[BoxData[ \(\(\(y'\)'\)[x] + \(1\/x\) \(y'\)[x] - \(1\/x\^2\) y[x] /. {y[x] -> u[Log[x]], \(y'\)[x] -> \[PartialD]\_x u[Log[x]], \(\(y'\)'\)[ x] -> \[PartialD]\_{x, 2}u[Log[x]]}\)], "Input"], Cell[TextData[{ "This gives a second order constant coefficient equation, once we have \ multiplied by ", Cell[BoxData[ \(TraditionalForm\`x\^2\)]], " to put it in the usual form." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(%*x\^2 // Expand\)], "Input"], Cell[TextData[{ "We substitute ", Cell[BoxData[ \(TraditionalForm\`x = \[ExponentialE]\^\[Xi]\)]], " to put the equation in terms of ", Cell[BoxData[ \(TraditionalForm\`\[Xi]\)]], "." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(% /. x -> Exp[\[Xi]]\)], "Input"], Cell[TextData[{ "We use ", Cell[BoxData[ FormBox[ StyleBox[\(PowerExpand[]\), "Input"], TraditionalForm]]], " to write ", Cell[BoxData[ \(TraditionalForm\`log(\[ExponentialE]\^x)\)]], " as ", Cell[BoxData[ \(TraditionalForm\`x\)]], "." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(PowerExpand[%]\)], "Input"], Cell[TextData[{ "Now we solve the constant coefficient equation for ", Cell[BoxData[ \(TraditionalForm\`u(\[Xi])\)]], "." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(DSolve[% == 0, u[\[Xi]], \[Xi]]\)], "Input"], Cell[BoxData[ \(u[\[Xi]_] = u[\[Xi]] /. First[%]\)], "Input"], Cell[TextData[{ "Finally we make the substitution ", Cell[BoxData[ \(TraditionalForm\`\[Xi] = log(x)\)]], " to obtain the solution ", Cell[BoxData[ \(TraditionalForm\`y(x)\)]], "." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(y[x_] = u[\[Xi]] /. \[Xi] -> Log[x]\)], "Input"], Cell[BoxData[ \(Clear[y, u]\)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["Normal Form", "Subsection"], Cell["The change of variables,", "Text"], Cell[BoxData[ \(TraditionalForm \`\(y(x) = \(exp(\(-\(1\/2\)\)\ \(\[Integral]\ \(p(x)\) \[DifferentialD]x\))\)\ \(u(x)\), \)\)], "DisplayFormula", TextAlignment->Center], Cell["transforms the differential equation,", "Text"], Cell[BoxData[ \(TraditionalForm\`\(\(\(y'\)' + \(p(x)\)\ y' + \(q(x)\)\ y = 0\)\(,\)\)\)], "DisplayFormula", TextAlignment->Center], Cell["into its normal form,", "Text"], Cell[BoxData[ \(TraditionalForm\`\(\(\(u'\)' + \(I(x)\)\ u = 0\)\(,\)\)\)], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "where the invariant of the equation, ", Cell[BoxData[ \(TraditionalForm\`I(x)\)]], ", is," }], "Text"], Cell[BoxData[ \(TraditionalForm\`I(x) = q - p\^2\/4 - \(p'\/2 . \)\)], "DisplayFormula",\ TextAlignment->Center], Cell[CellGroupData[{ Cell[TextData[{ "Exercise ", CounterBox["Subsubsection"] }], "Subsubsection"], Cell["Solve the differential equation,", "Text"], Cell[BoxData[ \(TraditionalForm\`\(\(\(y'\)' + \((2 + 4\/3\ x)\) y' + \(1\/9\) \((24 + 12 x + 4 x\^2)\) y = 0\)\(,\)\)\)], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "using the ", StyleBox["DSolve[]", FontWeight->"Bold"], " function. Write the solution as a real-valued function. Solve the \ differential equation by transforming the equation to normal form." }], "Text", TextJustification->1] }, Open ]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Section"], ". Vector Field of First Order Equations" }], "Section"], Cell["\<\ You can visualize the solutions of a first order differential \ equation by plotting its vector field. First load the PlotField \ package.\ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(<< Graphics`PlotField`\)], "Input"], Cell[TextData[{ "We will use the ", StyleBox["PlotVectorField[]", FontWeight->"Bold"], " function." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(?PlotVectorField\)\)], "Input"], Cell["\<\ The optional arguments (xu) and (yu) specify the spacing between \ vectors. Consider the problem\ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ FormBox[ RowBox[{\(y' = 2 y - y\^2\), ",", " ", FormBox[\(y(0) = \(\(y\_0\)\(.\)\)\), "TraditionalForm"]}], TraditionalForm]], "DisplayFormula", TextAlignment->Center, TextJustification->0], Cell[TextData[{ "The vector ", Cell[BoxData[ \(TraditionalForm\`\((x, 2 y - y\^2)\)\)]], " is tangent to a solution passing through the point ", Cell[BoxData[ \(TraditionalForm\`\((x, y)\)\)]], ". Below we plot this vector in the domain ", Cell[BoxData[ \(TraditionalForm\`\([0, 8]\)\[Times]\([\(-4\), 4]\)\)]], ". By default, ", StyleBox["PlotVectorField", FontWeight->"Bold"], " does not draw an axes or a frame around the graph. If you want a frame, \ you need to specify that in the options. You can make the graph easier to \ read by clicking on the graph and dragging a corner to enlarge it." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(PlotVectorField[{x, 2 y - y\^2}, {x, 0, 8}, {y, \(-4\), 4}, Frame -> True]; \)\)], "Input"], Cell[TextData[{ "Notice that the vectors are of different lengths. The lengths of the \ vectors is proportional to the length of ", Cell[BoxData[ \(TraditionalForm\`\((x, 2 y - y\^2)\)\)]], ". When drawing the vectors, ", StyleBox["Mathematica", FontSlant->"Italic"], " first uses a scale function to determine the length, by default this is \ the Cartesian length of the vector. Then ", StyleBox["Mathematica", FontSlant->"Italic"], " multiplies this length by a scale factor so that all the vectors fit in \ the graph.\nThe graph above is a little hard to read. For our purposes it \ would be more useful if the vectors all had the same length. We can \ accomplish this by changing the default scale function and scale factor. \ Below we specify that all vectors should be of length ", Cell[BoxData[ \(TraditionalForm\`1\/3\)]], " and there should be no rescaling of the length by a scale factor. In \ addition, we draw a vector at every half integer." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(vecField = PlotVectorField[{x, 2 y - y\^2}, {x, 0, 8, 1/2}, {y, \(-4\), 4, 1/2}, Frame -> True, ScaleFunction -> \((1\/3\ &)\), ScaleFactor -> None];\)\)], "Input"], Cell[TextData[{ "From the above vector field it appears that if the initial value, ", Cell[BoxData[ \(TraditionalForm\`y\_0\)]], " is in the ranges ", Cell[BoxData[ \(TraditionalForm\`\((0, 2)\)\)]], " or ", Cell[BoxData[ \(TraditionalForm\`\((2, 4)\)\)]], " then the solution will quickly approach ", Cell[BoxData[ \(TraditionalForm\`y = 2\)]], ". If the initial value is negative, then the solution stays negative and \ grows rapidly." }], "Text", TextAlignment->Left, TextJustification->1], Cell[TextData[{ "We can use ", StyleBox["DSolve[]", FontWeight->"Bold"], " to find the general solution of the differential equation." }], "Text"], Cell[BoxData[ \(DSolve[\(y'\)[x] == 2 y[x] - y[x]\^2, y[x], x]\)], "Input"], Cell[TextData[{ "Multiplying the numerator and denominator by ", Cell[BoxData[ \(TraditionalForm\`\[ExponentialE]\^\(\(-2\) x\)\)]], " and renaming the constant of integration, we see that the solution can be \ written" }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(TraditionalForm \`y = \(2\/\(1 - c\ \[ExponentialE]\^\(\(-2\) x\)\) . \)\)], "DisplayFormula", TextAlignment->Center, TextJustification->0], Cell["The solutions", "Text"], Cell[BoxData[ FormBox[ RowBox[{ \(y = 2\/\(1 - \(1\/3\) \[ExponentialE]\^\(\(-2\) x\)\)\), ",", " ", FormBox[\(y = 2\/\(1 + \[ExponentialE]\^\(\(-2\) x\)\)\), "TraditionalForm"], ",", " ", FormBox[\(y = 2\/\(1 - 3 \[ExponentialE]\^\(\(-2\) x\)\)\), "TraditionalForm"], ","}], TraditionalForm]], "DisplayFormula", TextAlignment->Center, TextJustification->0], Cell[TextData[{ "satisfy the initial conditions ", Cell[BoxData[ \(TraditionalForm\`y(0) = 3, 1, \(-1\)\)]], ", respectively. These solutions are plotted below. Note that the third \ solution is singular at ", Cell[BoxData[ \(TraditionalForm\`x = \(log\ 3\)\/2\)]], "." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(sol = Plot[{2\/\(1 - \(1\/3\) \[ExponentialE]\^\(\(-2\) x\)\), 2\/\(1 + \[ExponentialE]\^\(\(-2\) x\)\), 2\/\(1 - 3 \[ExponentialE]\^\(\(-2\) x\)\)}, {x, 0, 8}, PlotRange -> {\(-4\), 4}];\)\)], "Input"], Cell["\<\ By superimposing the vector field and the solutions we see how the \ solutions \"follow\" the vector field.\ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(Show[vecField, sol, PlotRange -> {\(-4.5\), 4.5}];\)\)], "Input"], Cell[BoxData[ \(Clear[vecField, sol]\)], "Input"], Cell[CellGroupData[{ Cell[TextData[{ "Exercise ", CounterBox["Subsubsection"] }], "Subsubsection"], Cell["\<\ The differential equation governing the motion of motion of a \ pendulum under the force of gravity is,\ \>", "Text", TextJustification->1], Cell[BoxData[ \(TraditionalForm\`\(\(\(\[Theta]'\)' + k\ \(sin(\[Theta])\) = 0\)\(,\)\)\)], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "where \[Theta] is the angle from the rest position. (For small \ oscillations one can approximate this equation with the familiar linear \ equation, ", Cell[BoxData[ \(TraditionalForm\`\(\[Theta]'\)' + k\ \[Theta] = 0\)]], ".) We can write this second order equation as a first order system:" }], "Text", TextJustification->1], Cell[BoxData[ FormBox[GridBox[{ {\(\[Theta]' = \[Phi], \)}, {\(\[Phi]' = \(-k\)\ \(sin(\[Theta]) . \)\)} }, ColumnAlignments->{Left}], TraditionalForm]], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "Plot the velocity field for ", Cell[BoxData[ \(TraditionalForm\`\((\[Theta], \[Phi])\)\)]], " in the phase space domain ", Cell[BoxData[ \(TraditionalForm\`\[Theta] \[Element] \((\(-\[Pi]\), \[Pi])\)\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\[Phi] \[Element] \((\(-\[Pi]\), \[Pi])\)\)]], ". That is, plot the vector field ", Cell[BoxData[ \(TraditionalForm \`\((\[Theta]', \[Phi]')\) = \((\[Phi], \(-k\)\ \(sin(\[Theta])\))\)\)]], ". (Take ", Cell[BoxData[ \(TraditionalForm\`k = 1\)]], ".) Numerically solve the differential equation ", Cell[BoxData[ \(TraditionalForm\`\(\[Theta]'\)' + k\ \(sin(\[Theta])\) = 0\)]], " with the initial conditions ", Cell[BoxData[ \(TraditionalForm\`\[Theta](0) = 0\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\[Theta]' \((0)\) = v\)]], " for several values of ", Cell[BoxData[ \(TraditionalForm\`v\)]], " in the range ", Cell[BoxData[ \(TraditionalForm\`\((0, 3)\)\)]], ". If the initial angular velocity is less than 2, the solution will be \ periodic in the phase space, the pendulum will swing back and forth. If the \ initial angular velocity is greater than 2, the solution is not periodic; the \ pendulum loops around. Show the velocity field together with several \ solutions in the phase space." }], "Text", TextJustification->1] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["Solutions", "Subtitle", TextAlignment->Center, TextJustification->0], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Section"], ". First Order Differential Equations" }], "Section"], Cell[CellGroupData[{ Cell[TextData[{ "Solution ", CounterBox["Subsubsection"] }], "Subsubsection"], Cell[BoxData[ \(TraditionalForm\`y' - \(1\/x\) y = \(\(x\^\[Alpha]\)\(.\)\)\)], "DisplayFormula", TextAlignment->Center], Cell[BoxData[ \(DSolve[\(y'\)[x] - \(1\/x\) y[x] == x\^\[Alpha], y[x], x]\)], "Input"], Cell[TextData[{ "For ", Cell[BoxData[ \(TraditionalForm\`\[Alpha] \[NotEqual] 0\)]], " the solution is" }], "Text"], Cell[BoxData[ \(TraditionalForm\`y = x\^\(\[Alpha] + 1\)\/\[Alpha] + c\ \(\(x\)\(.\)\)\)], "DisplayFormula", TextAlignment->Center], Cell[BoxData[ \(DSolve[\(y'\)[x] - \(1\/x\) y[x] == 1, y[x], x]\)], "Input"], Cell[TextData[{ "For ", Cell[BoxData[ \(TraditionalForm\`\[Alpha] = 0\)]], " the solution is" }], "Text"], Cell[BoxData[ \(TraditionalForm\`y = x\ log\ x + c\ \(\(x\)\(.\)\)\)], "DisplayFormula",\ TextAlignment->Center] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "Solution ", CounterBox["Subsubsection"] }], "Subsubsection"], Cell[BoxData[ \(DSolve[\(y'\)[x] == \(2\/x\) y[x] + y[x]\^2, y[x], x]\)], "Input"], Cell["\<\ We change the form of the constant of integration to obtain the \ desired answer.\ \>", "Text", TextJustification->1], Cell[BoxData[ \(\(y[x] /. First[%]\) /. C[1] -> 1/c // Simplify\)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "Solution ", CounterBox["Subsubsection"] }], "Subsubsection"], Cell[BoxData[ \(DSolve[\(y'\)[x] == p[x] y[x] + q[x] y[x]\^\[Alpha], y[x], x]\)], "Input"], Cell[BoxData[ \(y[x] /. First[%] // Simplify\)], "Input"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Section"], ". Higher Order Differential Equations" }], "Section"], Cell[CellGroupData[{ Cell[TextData[{ "Solution ", CounterBox["Subsubsection"] }], "Subsubsection"], Cell[BoxData[ \(TraditionalForm\`y'' + 2 a\ y' + \(a\^2\) y = 0\)], "DisplayFormula", TextAlignment->Center], Cell[BoxData[ \(DSolve[{\(\(y'\)'\)[x] + 2 a\ \(y'\)[x] + a\^2\ y[x] == 0, y[0] == 1, \(y'\)[0] == 0}, y[x], x]\)], "Input"], Cell[BoxData[ \(DSolve[{\(\(y'\)'\)[x] + 2 a\ \(y'\)[x] + a\^2\ y[x] == 0, y[0] == 0, \(y'\)[0] == 1}, y[x], x]\)], "Input"], Cell[TextData[{ "The fundamental set of solutions at the point ", Cell[BoxData[ \(TraditionalForm\`x = 0\)]], " is:" }], "Text"], Cell[BoxData[ \(TraditionalForm\`\(\({\((1 + a\ x)\) \[ExponentialE]\^\(\(-a\)\ x\), x\ \[ExponentialE]\^\(\(-a\)\ x\)}\)\(.\)\)\)], "DisplayFormula", TextAlignment->Center] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "Solution ", CounterBox["Subsubsection"] }], "Subsubsection"], Cell["First we consider,", "Text"], Cell[BoxData[ \(TraditionalForm\`\(y'\)' + \(1\/x\) y' + \((1 - n\^2\/x\^2)\) y = 0. \)], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "This is Bessel's equation. When ", Cell[BoxData[ \(TraditionalForm\`n\)]], " is not an integer, the general solution is ", Cell[BoxData[ \(TraditionalForm \`y = \(c\_1\) \(\(J\_n\)(x)\) + \(c\_2\) \(\(J\_\(-n\)\)(x)\)\)]], ". However, when ", Cell[BoxData[ \(TraditionalForm\`n\)]], " is an integer, the Bessel functions of first kind, ", Cell[BoxData[ \(TraditionalForm\`\(J\_n\)(x)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\(J\_\(-n\)\)(x)\)]], " are not linearly independent. The general solution is ", Cell[BoxData[ \(TraditionalForm \`y = \(c\_1\) \(\(J\_n\)(x)\) + \(c\_2\) \(\(Y\_n\)(x)\)\)]], "." }], "Text", TextJustification->1], Cell[BoxData[ \(DSolve[\(\(y'\)'\)[x] + \(1\/x\) \(y'\)[x] + \((1 - n\^2\/x\^2)\) y[x] == 0, y[x], x]\)], "Input"], Cell[TextData[{ "We can simplify the ", Cell[BoxData[ \(TraditionalForm\`\@x\^2\)]], " factor with ", StyleBox["PowerExpand[]", FontWeight->"Bold"], "." }], "Text", TextJustification->1], Cell[BoxData[ \(% // PowerExpand\)], "Input"], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " gives an answer that is not valid for the case we are considering. There \ is no way of telling it that ", Cell[BoxData[ \(TraditionalForm\`n\)]], " is an integer. If we substitute an integer for ", Cell[BoxData[ \(TraditionalForm\`n\)]], ", ", StyleBox["Mathematica", FontSlant->"Italic"], " will do the right thing, though." }], "Text", TextJustification->1], Cell[BoxData[ \(DSolve[\(\(y'\)'\)[x] + \(1\/x\) \(y'\)[x] + \((1 - 7\^2\/x\^2)\) y[x] == 0, y[x], x] // PowerExpand\)], "Input"], Cell["Now we consider the equation:", "Text"], Cell[BoxData[ \(TraditionalForm\`\(y'\)' + \(1\/x\) y' + \((\[Lambda]\^2 - 1\/x\^2)\) y = 0. \)], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "The general solution is ", Cell[BoxData[ \(TraditionalForm \`y = \(c\_1\) \(\(J\_1\)(\[Lambda]\ x)\) + \(c\_2\) \(\(Y\_1\)(\[Lambda]\ x)\)\)]], "." }], "Text"], Cell[BoxData[ \(DSolve[\(\(y'\)'\)[ x] + \(1\/x\) \(y'\)[x] + \((\[Lambda]\^2 - 1\/x\^2)\) y[x] == 0, y[x], x]\)], "Input"], Cell[BoxData[ \(% // TraditionalForm\)], "Input"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Section"], ". Numerical Solutions" }], "Section"], Cell[CellGroupData[{ Cell[TextData[{ "Solution ", CounterBox["Subsubsection"] }], "Subsubsection"], Cell["\<\ First we solve the equation when the initial amplitude is small and \ the initial velocity is zero.\ \>", "Text", TextJustification->1], Cell[BoxData[{ \(Clear[y]\), \(NDSolve[{\(\(y'\)'\)[x] - \((1 - y[x]\^2)\) \(y'\)[x] + y[x] == 0, y[0] == 0.01, \(y'\)[0] == 0}, y[x], {x, 0, 30}]\)}], "Input"], Cell[TextData[{ "We define ", Cell[BoxData[ \(TraditionalForm\`y[x]\)]], " to be the interpolating function and plot the solution." }], "Text", TextJustification->1], Cell[BoxData[ \(y[x_] = y[x] /. First[%]; \nPlot[y[x], {x, 0, 30}]; \)], "Input"], Cell["\<\ When the initial amplitude is zero and the initial velocity is \ small, the amplitude also tends to 2.\ \>", "Text", TextJustification->1], Cell[BoxData[{ \(Clear[y]\), \(y[x_] = y[x] /. First[ NDSolve[{\(\(y'\)'\)[x] - \((1 - y[x]\^2)\) \(y'\)[x] + y[x] == 0, y[0] == 0, \(y'\)[0] == 0.01}, y[x], {x, 0, 30}]]; \n Plot[y[x], {x, 0, 30}]; \)}], "Input"], Cell["Same story when the initial amplitude is large.", "Text", TextJustification->1], Cell[BoxData[{ \(Clear[y]\), \(y[x_] = y[x] /. First[ NDSolve[{\(\(y'\)'\)[x] - \((1 - y[x]\^2)\) \(y'\)[x] + y[x] == 0, y[0] == 5, \(y'\)[0] == 5}, y[x], {x, 0, 30}]]; \n Plot[y[x], {x, 0, 30}]; \)}], "Input"], Cell["\<\ When \[Mu] is small, the solution looks like an undamped harmonic \ oscillator.\ \>", "Text", TextJustification->1], Cell[BoxData[{ \(Clear[y]\), \(y[x_] = y[x] /. First[ NDSolve[{ \(\(y'\)'\)[x] - 0.1 \((1 - y[x]\^2)\) \(y'\)[x] + y[x] == 0, y[0] == 2, \(y'\)[0] == 0}, y[x], {x, 0, 30}]]; \n Plot[y[x], {x, 0, 30}]; \)}], "Input"], Cell[TextData[ "When \[Mu] is large, the frequency drops and the path of motion becomes \ asymmetric."], "Text", TextJustification->1], Cell[BoxData[{ \(Clear[y]\), \(y[x_] = y[x] /. First[ NDSolve[{ \(\(y'\)'\)[x] - 10 \((1 - y[x]\^2)\) \(y'\)[x] + y[x] == 0, y[0] == 2, \(y'\)[0] == 0}, y[x], {x, 0, 30}]]; \n Plot[y[x], {x, 0, 30}]; \)}], "Input"] }, Closed]] }, Closed]], Cell[TextData[{ CounterBox["Section"], ". Series Solutions" }], "Section"], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Section"], ". Transformations" }], "Section"], Cell[CellGroupData[{ Cell[TextData[{ "Solution ", CounterBox["Subsubsection"] }], "Subsubsection"], Cell[BoxData[{ \(Clear[y]\), \(DSolve[ \(\(y'\)'\)[x] + \((2 + \(4\/3\) x)\) \(y'\)[x] + \(1\/9\) \((24 + 12 x + 4 x\^2)\) y[x] == 0, y[x], x]\)}], "Input"], Cell[BoxData[ \(y[x] /. First[%] // Expand\)], "Input"], Cell["Writing the solution in terms of real and imaginary parts,", "Text"], Cell[BoxData[ \({E\^\(\((\(-1\) - I)\)\ x - x\^2\/3\)\ , E\^\(\((\(-1\) + I)\)\ x - x\^2\/3\)\ } // ComplexExpand\)], "Input"], Cell["we see that the general solution is,", "Text"], Cell[BoxData[ \(TraditionalForm \`y = c\_1\ \[ExponentialE]\^\(\(-x\) - x\^2/3\)\ \(cos(x)\) + c\_2\ \[ExponentialE]\^\(\(-x\) - x\^2/3\)\ \(sin(x) . \)\)], "DisplayFormula", TextAlignment->Center], Cell["\<\ Now we solve the differential equation by transforming it to normal \ form.\ \>", "Text", TextJustification->1], Cell[BoxData[ \(Exp[ \(-\(1\/2\)\) \(\[Integral]\((2 + \(4\/3\) x)\) \[DifferentialD]x\)] // Simplify \)], "Input"], Cell[TextData[{ "We make the change of variables ", Cell[BoxData[ \(TraditionalForm \`y \[Rule] \[ExponentialE]\^\(\(-\(x(x + 3)\)\)/3\)\ u\)]], "." }], "Text"], Cell[BoxData[ \(\(\(y'\)'\)[ x] + \((2 + \(4\/3\) x)\) \(y'\)[ x] + \(1\/9\) \((24 + 12 x + 4 x\^2)\) y[x] /. {y[x] -> Exp[\(-x\) \((x + 3)\)/3] u[x], \(y'\)[ x] -> \[PartialD]\_x\((Exp[\(-x\) \((x + 3)\)/3] u[x])\), \(\(y'\)'\)[ x] -> \[PartialD]\_{x, 2}\((Exp[\(-x\) \((x + 3)\)/3] u[x])\)} // Simplify\)], "Input"], Cell[BoxData[ \(%*Exp[x \((x + 3)\)/3]\)], "Input"], Cell["We solve the normal form of the equation.", "Text", TextJustification->1], Cell[BoxData[ \(DSolve[% == 0, u[x], x]\)], "Input"], Cell[TextData[{ "Finally, we write the solution in terms of ", Cell[BoxData[ \(TraditionalForm\`y\)]], "." }], "Text", TextJustification->1], Cell[BoxData[ \(TraditionalForm \`y = c\_1\ \[ExponentialE]\^\(\(-x\) - x\^2/3\)\ \(cos(x)\) + c\_2\ \[ExponentialE]\^\(\(-x\) - x\^2/3\)\ \(sin(x) . \)\)], "DisplayFormula", TextAlignment->Center] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Section"], ". Vector Field of First Order Equations" }], "Section"], Cell[CellGroupData[{ Cell[TextData[{ "Solution ", CounterBox["Subsubsection"] }], "Subsubsection"], Cell[TextData[{ "Here is the velocity field with the length of the vectors is proportional \ to the length of ", Cell[BoxData[ \(TraditionalForm\`\((\[Phi], \(-\(sin(\[Theta])\)\))\)\)]], "." }], "Text", TextJustification->1], Cell[BoxData[ \(\(PlotVectorField[{\[Phi], \(-Sin[\[Theta]]\)}, {\[Theta], \(-\[Pi]\), \[Pi]}, {\[Phi], \(-\[Pi]\), \[Pi]}]; \)\)], "Input"], Cell["\<\ Here we specify that all the vectors should have the same length.\ \ \>", "Text", TextJustification->1], Cell[BoxData[ \(\(plotvf = PlotVectorField[{\[Phi], \(-Sin[\[Theta]]\)}, {\[Theta], \(-\[Pi]\), \ \[Pi]}, {\[Phi], \(-\[Pi]\), \[Pi]}, ScaleFunction -> \((1 &)\)];\)\)], "Input"], Cell["\<\ We numerically solve the differential equation with an initial \ velocity of 1.5. Below is a plot of angle versus time.\ \>", "Text", TextJustification->1], Cell[BoxData[{ \(\(soln[t_] = \[Theta][t] /. First[NDSolve[{\(\(\[Theta]'\)'\)[t] + Sin[\[Theta][t]] == 0, \[Theta][0] == 0, \(\[Theta]'\)[0] == 1.5}, \[Theta][ t], {t, 0, 10}]];\)\), "\n", \(\(Plot[soln[t], {t, 0, 10}, PlotRange -> All];\)\)}], "Input"], Cell["Periodic solutions form closed curves in phase space.", "Text", TextJustification->1], Cell[BoxData[{ \(\(plot1 = ParametricPlot[{soln[t], \(soln'\)[t]}, {t, 0, 10}, PlotRange -> {{\(-\[Pi]\), \[Pi]}, {\(-\[Pi]\), \[Pi]}}];\)\), "\n\ ", \(\(Clear[soln];\)\)}], "Input"], Cell[TextData[{ "We numerically solve the differential equation for initial velocities ", Cell[BoxData[ \(TraditionalForm\`v = 1\/4, 1\/2, \[Ellipsis], 3\)]], "." }], "Text", TextJustification->1], Cell[BoxData[ \(Do[\n\t\(\(soln[n]\)[t_] = \[Theta][t] /. First[NDSolve[{\(\(\[Theta]'\)'\)[t] + Sin[\[Theta][t]] == 0, \[Theta][0] == 0, \(\[Theta]'\)[0] == n/4}, \[Theta][ t], {t, 0, 10}]];\), \n\t{n, 1, 12}]\)], "Input"], Cell[TextData[{ "In phase space we see the periodic and non-periodic solutions. When ", Cell[BoxData[ \(TraditionalForm\`v = 2\)]], " the pendulum approaches ", Cell[BoxData[ \(TraditionalForm\`\[Theta] = \[Pi]\)]], " as ", Cell[BoxData[ \(TraditionalForm\`t \[Rule] \[Infinity]\)]], ". " }], "Text", TextJustification->1], Cell[BoxData[ \(\(pp = ParametricPlot[ Evaluate[Table[{\(soln[n]\)[t], \(soln[n]'\)[t]}, {n, 1, 12}]], {t, 0, 10}, PlotRange -> {{\(-\[Pi]\), 3 \[Pi]}, {\(-\[Pi]\), \[Pi]}}];\)\)], "Input"], Cell["\<\ Below is a plot of the velocity field and the solution curves. The \ plotting domain is slightly extended so that the arrows on the boundary are \ not cut off.\ \>", "Text", TextJustification->1], Cell[BoxData[ \(\(Show[plotvf, pp, PlotRange -> {{\(-\[Pi]\) - 2 \[Pi]/15, \[Pi] + 2 \[Pi]/15}, {\(-\[Pi]\) - 2 \[Pi]/15, \[Pi] + 2 \[Pi]/15}}];\)\)], "Input"] }, Closed]] }, Closed]] }, Open ]] }, Open ]] }, FrontEndVersion->"5.2 for Microsoft Windows", ScreenRectangle->{{0, 1680}, {0, 963}}, WindowToolbars->"EditBar", CellGrouping->Automatic, WindowSize->{772, 780}, WindowMargins->{{Automatic, 0}, {Automatic, 0}}, PrintingPageRange->{Automatic, Automatic}, PrintingOptions->{"PaperSize"->{612, 792}, "PaperOrientation"->"Portrait", "Magnification"->1}, PrivateNotebookOptions->{"ColorPalette"->{RGBColor, 128}}, ShowCellLabel->True, ShowCellTags->False, RenderingOptions->{"ObjectDithering"->True, "RasterDithering"->False}, CharacterEncoding->"XAutomaticEncoding", Magnification->1.5 ] (******************************************************************* Cached data follows. If you edit this Notebook file directly, not using Mathematica, you must remove the line containing CacheID at the top of the file. The cache data will then be recreated when you save this file from within Mathematica. *******************************************************************) (*CellTagsOutline CellTagsIndex->{} *) (*CellTagsIndex CellTagsIndex->{} *) (*NotebookFileOutline Notebook[{ Cell[CellGroupData[{ Cell[1776, 53, 121, 3, 222, "Title", Evaluatable->False], Cell[1900, 58, 439, 14, 125, "Text", Evaluatable->False], Cell[CellGroupData[{ Cell[2364, 76, 99, 4, 109, "Section"], Cell[2466, 82, 168, 6, 46, "Text"], Cell[2637, 90, 44, 1, 39, "Input"], Cell[2684, 93, 602, 16, 121, "Text"], Cell[3289, 111, 99, 2, 94, "Input"], Cell[3391, 115, 331, 10, 71, "Text"], Cell[3725, 127, 99, 2, 68, "Input"], Cell[3827, 131, 141, 5, 46, "Text"], Cell[3971, 138, 57, 1, 39, "Input"], Cell[4031, 141, 68, 1, 62, "Input"], Cell[4102, 144, 418, 12, 96, "Text"], Cell[4523, 158, 99, 2, 68, "Input"], Cell[4625, 162, 146, 5, 46, "Text"], Cell[4774, 169, 80, 1, 68, "Input"], Cell[4857, 172, 384, 11, 96, "Text"], Cell[5244, 185, 93, 1, 68, "Input"], Cell[5340, 188, 133, 3, 71, "Text"], Cell[5476, 193, 82, 1, 68, "Input"], Cell[5561, 196, 57, 1, 39, "Input"], Cell[5621, 199, 116, 3, 46, "Text"], Cell[5740, 204, 37, 1, 39, "Input"], Cell[CellGroupData[{ Cell[5802, 209, 84, 4, 60, "Subsubsection"], Cell[5889, 215, 48, 0, 46, "Text"], Cell[5940, 217, 137, 3, 55, "DisplayFormula"] }, Open ]], Cell[CellGroupData[{ Cell[6114, 225, 84, 4, 60, "Subsubsection"], Cell[6201, 231, 65, 0, 46, "Text"], Cell[6269, 233, 106, 2, 55, "DisplayFormula"], Cell[6378, 237, 18, 0, 46, "Text"], Cell[6399, 239, 113, 2, 60, "DisplayFormula"] }, Open ]], Cell[CellGroupData[{ Cell[6549, 246, 84, 4, 60, "Subsubsection"], Cell[6636, 252, 73, 0, 46, "Text"], Cell[6712, 254, 183, 4, 35, "DisplayFormula"], Cell[6898, 260, 18, 0, 46, "Text"], Cell[6919, 262, 499, 10, 64, "DisplayFormula"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[7467, 278, 100, 4, 62, "Section"], Cell[7570, 284, 88, 2, 46, "Text"], Cell[7661, 288, 328, 8, 34, "DisplayFormula"], Cell[7992, 298, 144, 5, 46, "Text"], Cell[8139, 305, 48, 1, 39, "Input"], Cell[8190, 308, 97, 2, 39, "Input"], Cell[8290, 312, 57, 1, 39, "Input"], Cell[8350, 315, 253, 10, 46, "Text"], Cell[8606, 327, 67, 1, 39, "Input"], Cell[8676, 330, 59, 1, 39, "Input"], Cell[8738, 333, 279, 10, 71, "Text"], Cell[9020, 345, 132, 2, 62, "Input"], Cell[9155, 349, 57, 1, 39, "Input"], Cell[9215, 352, 84, 1, 62, "Input"], Cell[CellGroupData[{ Cell[9324, 357, 84, 4, 60, "Subsubsection"], Cell[9411, 363, 58, 0, 46, "Text"], Cell[9472, 365, 126, 3, 38, "DisplayFormula"], Cell[9601, 370, 104, 5, 46, "Text"] }, Open ]], Cell[CellGroupData[{ Cell[9742, 380, 84, 4, 60, "Subsubsection"], Cell[9829, 386, 46, 0, 46, "Text"], Cell[9878, 388, 154, 3, 61, "DisplayFormula"], Cell[10035, 393, 173, 4, 56, "DisplayFormula"], Cell[10211, 399, 171, 8, 46, "Text"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[10431, 413, 84, 4, 62, "Section"], Cell[CellGroupData[{ Cell[10540, 421, 50, 0, 70, "Subsection"], Cell[10593, 423, 83, 2, 46, "Text"], Cell[10679, 427, 368, 9, 56, "DisplayFormula"], Cell[11050, 438, 318, 8, 96, "Text"], Cell[11371, 448, 42, 1, 39, "Input"], Cell[11416, 451, 148, 2, 101, "Input"], Cell[11567, 455, 569, 17, 121, "Text"], Cell[12139, 474, 45, 1, 39, "Input"], Cell[12187, 477, 157, 2, 101, "Input"], Cell[12347, 481, 706, 21, 121, "Text"], Cell[13056, 504, 63, 1, 39, "Input"], Cell[13122, 507, 59, 1, 39, "Input"], Cell[13184, 510, 342, 10, 71, "Text"], Cell[13529, 522, 37, 1, 39, "Input"], Cell[13569, 525, 392, 11, 96, "Text"], Cell[13964, 538, 42, 1, 39, "Input"], Cell[14009, 541, 136, 5, 46, "Text"], Cell[14148, 548, 60, 1, 39, "Input"], Cell[14211, 551, 255, 9, 46, "Text"], Cell[14469, 562, 71, 1, 62, "Input"], Cell[CellGroupData[{ Cell[14565, 567, 84, 4, 60, "Subsubsection"], Cell[14652, 573, 59, 0, 46, "Text"], Cell[14714, 575, 154, 3, 41, "DisplayFormula"], Cell[14871, 580, 1045, 27, 196, "Text"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[15965, 613, 55, 0, 44, "Subsection"], Cell[16023, 615, 154, 5, 46, "Text"], Cell[16180, 622, 293, 8, 96, "DisplayFormula"], Cell[16476, 632, 293, 10, 71, "Text"], Cell[16772, 644, 111, 3, 117, "Input"], Cell[16886, 649, 315, 5, 177, "Input"], Cell[17204, 656, 296, 10, 71, "Text"], Cell[17503, 668, 92, 1, 39, "Input"], Cell[17598, 671, 231, 6, 71, "Text"], Cell[17832, 679, 115, 2, 62, "Input"], Cell[17950, 683, 192, 7, 46, "Text"], Cell[18145, 692, 134, 2, 62, "Input"], Cell[18282, 696, 193, 8, 46, "Text"], Cell[18478, 706, 316, 5, 177, "Input"], Cell[18797, 713, 637, 16, 146, "Text"], Cell[19437, 731, 339, 5, 200, "Input"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[19825, 742, 81, 4, 62, "Section"], Cell[CellGroupData[{ Cell[19931, 750, 51, 0, 70, "Subsection"], Cell[19985, 752, 83, 2, 46, "Text"], Cell[20071, 756, 364, 9, 56, "DisplayFormula"], Cell[20438, 767, 139, 5, 46, "Text"], Cell[20580, 774, 42, 1, 39, "Input"], Cell[20625, 777, 117, 2, 101, "Input"], Cell[20745, 781, 605, 16, 126, "Text"], Cell[21353, 799, 56, 1, 39, "Input"], Cell[21412, 802, 136, 5, 46, "Text"], Cell[21551, 809, 117, 3, 71, "Input"], Cell[21671, 814, 51, 1, 39, "Input"], Cell[21725, 817, 550, 18, 96, "Text"], Cell[22278, 837, 55, 1, 39, "Input"], Cell[22336, 840, 116, 2, 46, "Text"], Cell[22455, 844, 64, 1, 39, "Input"], Cell[22522, 847, 185, 7, 46, "Text"], Cell[22710, 856, 62, 1, 39, "Input"], Cell[22775, 859, 81, 1, 39, "Input"], Cell[22859, 862, 414, 16, 46, "Text"], Cell[23276, 880, 104, 2, 39, "Input"], Cell[23383, 884, 91, 1, 39, "Input"], Cell[23477, 887, 118, 3, 39, "Input"], Cell[23598, 892, 110, 2, 39, "Input"], Cell[23711, 896, 570, 22, 71, "Text"], Cell[24284, 920, 56, 1, 39, "Input"], Cell[24343, 923, 245, 9, 46, "Text"], Cell[24591, 934, 42, 1, 39, "Input"], Cell[24636, 937, 262, 8, 71, "Text"], Cell[24901, 947, 65, 1, 39, "Input"], Cell[24969, 950, 52, 1, 39, "Input"] }, Closed]], Cell[CellGroupData[{ Cell[25058, 956, 42, 0, 44, "Subsection"], Cell[25103, 958, 182, 5, 71, "Text"], Cell[25288, 965, 45, 1, 39, "Input"], Cell[25336, 968, 103, 2, 46, "Text"], Cell[25442, 972, 58, 1, 39, "Input"], Cell[25503, 975, 214, 7, 46, "Text"], Cell[25720, 984, 147, 3, 101, "Input"], Cell[25870, 989, 103, 2, 46, "Text"], Cell[25976, 993, 112, 2, 62, "Input"], Cell[26091, 997, 110, 2, 46, "Text"], Cell[26204, 1001, 63, 1, 39, "Input"], Cell[26270, 1004, 297, 10, 71, "Text"], Cell[26570, 1016, 81, 1, 39, "Input"], Cell[26654, 1019, 60, 1, 39, "Input"], Cell[26717, 1022, 138, 5, 46, "Text"], Cell[26858, 1029, 157, 2, 101, "Input"], Cell[27018, 1033, 83, 1, 39, "Input"], Cell[27104, 1036, 357, 9, 101, "Text"], Cell[27464, 1047, 100, 2, 39, "Input"], Cell[27567, 1051, 108, 2, 62, "Input"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[27724, 1059, 80, 4, 62, "Section"], Cell[CellGroupData[{ Cell[27829, 1067, 41, 0, 70, "Subsection"], Cell[27873, 1069, 96, 2, 46, "Text"], Cell[27972, 1073, 174, 4, 35, "DisplayFormula"], Cell[28149, 1079, 288, 10, 96, "Text"], Cell[28440, 1091, 154, 4, 55, "DisplayFormula"], Cell[28597, 1097, 144, 5, 46, "Text"], Cell[28744, 1104, 45, 1, 39, "Input"], Cell[28792, 1107, 150, 3, 71, "Input"], Cell[28945, 1112, 182, 7, 46, "Text"], Cell[29130, 1121, 54, 1, 44, "Input"], Cell[29187, 1124, 109, 2, 46, "Text"], Cell[29299, 1128, 56, 1, 39, "Input"], Cell[29358, 1131, 203, 7, 53, "Text"], Cell[29564, 1140, 64, 1, 71, "Input"] }, Closed]], Cell[CellGroupData[{ Cell[29665, 1146, 39, 0, 44, "Subsection"], Cell[29707, 1148, 97, 2, 46, "Text"], Cell[29807, 1152, 155, 4, 38, "DisplayFormula"], Cell[29965, 1158, 346, 10, 101, "Text"], Cell[30314, 1170, 160, 4, 56, "DisplayFormula"], Cell[30477, 1176, 183, 7, 51, "Text"], Cell[30663, 1185, 51, 1, 39, "Input"], Cell[30717, 1188, 181, 3, 129, "Input"], Cell[30901, 1193, 49, 1, 39, "Input"], Cell[30953, 1196, 101, 2, 46, "Text"], Cell[31057, 1200, 56, 1, 39, "Input"], Cell[31116, 1203, 57, 1, 39, "Input"], Cell[31176, 1206, 158, 7, 51, "Text"], Cell[31337, 1215, 75, 1, 71, "Input"], Cell[31415, 1218, 305, 8, 96, "Text"], Cell[31723, 1228, 76, 1, 39, "Input"], Cell[31802, 1231, 44, 1, 39, "Input"] }, Closed]], Cell[CellGroupData[{ Cell[31883, 1237, 71, 0, 44, "Subsection"], Cell[31957, 1239, 117, 2, 46, "Text"], Cell[32077, 1243, 151, 4, 34, "DisplayFormula"], Cell[32231, 1249, 80, 2, 46, "Text"], Cell[32314, 1253, 136, 3, 60, "DisplayFormula"], Cell[32453, 1258, 106, 2, 46, "Text"], Cell[32562, 1262, 301, 8, 56, "DisplayFormula"], Cell[32866, 1272, 287, 10, 96, "Text"], Cell[33156, 1284, 152, 4, 56, "DisplayFormula"], Cell[33311, 1290, 251, 8, 80, "Text"], Cell[33565, 1300, 129, 2, 129, "Input"], Cell[33697, 1304, 44, 1, 39, "Input"], Cell[33744, 1307, 105, 2, 46, "Text"], Cell[33852, 1311, 354, 10, 57, "DisplayFormula"], Cell[34209, 1323, 48, 1, 39, "Input"], Cell[34260, 1326, 45, 1, 39, "Input"], Cell[34308, 1329, 221, 3, 119, "Input"], Cell[34532, 1334, 251, 8, 71, "Text"], Cell[34786, 1344, 49, 1, 44, "Input"], Cell[34838, 1347, 264, 10, 46, "Text"], Cell[35105, 1359, 53, 1, 39, "Input"], Cell[35161, 1362, 344, 15, 46, "Text"], Cell[35508, 1379, 47, 1, 39, "Input"], Cell[35558, 1382, 192, 7, 46, "Text"], Cell[35753, 1391, 64, 1, 39, "Input"], Cell[35820, 1394, 65, 1, 39, "Input"], Cell[35888, 1397, 261, 10, 46, "Text"], Cell[36152, 1409, 68, 1, 39, "Input"], Cell[36223, 1412, 44, 1, 39, "Input"] }, Closed]], Cell[CellGroupData[{ Cell[36304, 1418, 33, 0, 44, "Subsection"], Cell[36340, 1420, 40, 0, 46, "Text"], Cell[36383, 1422, 195, 5, 57, "DisplayFormula"], Cell[36581, 1429, 53, 0, 46, "Text"], Cell[36637, 1431, 144, 3, 34, "DisplayFormula"], Cell[36784, 1436, 37, 0, 46, "Text"], Cell[36824, 1438, 129, 3, 34, "DisplayFormula"], Cell[36956, 1443, 131, 5, 46, "Text"], Cell[37090, 1450, 119, 3, 60, "DisplayFormula"], Cell[CellGroupData[{ Cell[37234, 1457, 84, 4, 60, "Subsubsection"], Cell[37321, 1463, 48, 0, 46, "Text"], Cell[37372, 1465, 189, 4, 56, "DisplayFormula"], Cell[37564, 1471, 257, 7, 71, "Text"] }, Open ]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[37882, 1485, 102, 4, 62, "Section"], Cell[37987, 1491, 210, 6, 71, "Text"], Cell[38200, 1499, 55, 1, 39, "Input"], Cell[38258, 1502, 168, 7, 46, "Text"], Cell[38429, 1511, 53, 1, 39, "Input"], Cell[38485, 1514, 168, 5, 71, "Text"], Cell[38656, 1521, 240, 6, 38, "DisplayFormula"], Cell[38899, 1529, 702, 18, 146, "Text"], Cell[39604, 1549, 121, 2, 75, "Input"], Cell[39728, 1553, 1059, 23, 226, "Text"], Cell[40790, 1578, 221, 4, 133, "Input"], Cell[41014, 1584, 539, 17, 96, "Text"], Cell[41556, 1603, 156, 5, 46, "Text"], Cell[41715, 1610, 80, 1, 47, "Input"], Cell[41798, 1613, 290, 8, 71, "Text"], Cell[42091, 1623, 170, 5, 56, "DisplayFormula"], Cell[42264, 1630, 29, 0, 46, "Text"], Cell[42296, 1632, 431, 10, 69, "DisplayFormula"], Cell[42730, 1644, 352, 11, 77, "Text"], Cell[43085, 1657, 271, 5, 116, "Input"], Cell[43359, 1664, 178, 5, 71, "Text"], Cell[43540, 1671, 87, 1, 39, "Input"], Cell[43630, 1674, 53, 1, 39, "Input"], Cell[CellGroupData[{ Cell[43708, 1679, 84, 4, 60, "Subsubsection"], Cell[43795, 1685, 151, 4, 71, "Text"], Cell[43949, 1691, 145, 3, 34, "DisplayFormula"], Cell[44097, 1696, 359, 8, 96, "Text"], Cell[44459, 1706, 238, 6, 58, "DisplayFormula"], Cell[44700, 1714, 1417, 39, 196, "Text"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[46166, 1759, 78, 2, 77, "Subtitle"], Cell[CellGroupData[{ Cell[46269, 1765, 99, 4, 109, "Section"], Cell[CellGroupData[{ Cell[46393, 1773, 84, 4, 60, "Subsubsection"], Cell[46480, 1779, 137, 3, 55, "DisplayFormula"], Cell[46620, 1784, 90, 1, 68, "Input"], Cell[46713, 1787, 127, 5, 46, "Text"], Cell[46843, 1794, 144, 3, 60, "DisplayFormula"], Cell[46990, 1799, 80, 1, 68, "Input"], Cell[47073, 1802, 117, 5, 46, "Text"], Cell[47193, 1809, 119, 3, 34, "DisplayFormula"] }, Closed]], Cell[CellGroupData[{ Cell[47349, 1817, 84, 4, 37, "Subsubsection"], Cell[47436, 1823, 86, 1, 68, "Input"], Cell[47525, 1826, 129, 4, 46, "Text"], Cell[47657, 1832, 80, 1, 39, "Input"] }, Closed]], Cell[CellGroupData[{ Cell[47774, 1838, 84, 4, 37, "Subsubsection"], Cell[47861, 1844, 101, 2, 40, "Input"], Cell[47965, 1848, 61, 1, 39, "Input"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[48075, 1855, 100, 4, 62, "Section"], Cell[CellGroupData[{ Cell[48200, 1863, 84, 4, 60, "Subsubsection"], Cell[48287, 1869, 115, 2, 38, "DisplayFormula"], Cell[48405, 1873, 138, 2, 75, "Input"], Cell[48546, 1877, 138, 2, 75, "Input"], Cell[48687, 1881, 140, 5, 46, "Text"], Cell[48830, 1888, 188, 3, 35, "DisplayFormula"] }, Closed]], Cell[CellGroupData[{ Cell[49055, 1896, 84, 4, 37, "Subsubsection"], Cell[49142, 1902, 34, 0, 46, "Text"], Cell[49179, 1904, 143, 3, 61, "DisplayFormula"], Cell[49325, 1909, 740, 23, 96, "Text"], Cell[50068, 1934, 127, 2, 77, "Input"], Cell[50198, 1938, 208, 9, 53, "Text"], Cell[50409, 1949, 49, 1, 39, "Input"], Cell[50461, 1952, 464, 15, 96, "Text"], Cell[50928, 1969, 144, 2, 104, "Input"], Cell[51075, 1973, 45, 0, 46, "Text"], Cell[51123, 1975, 155, 3, 56, "DisplayFormula"], Cell[51281, 1980, 205, 7, 46, "Text"], Cell[51489, 1989, 146, 3, 98, "Input"], Cell[51638, 1994, 53, 1, 39, "Input"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[51740, 2001, 84, 4, 62, "Section"], Cell[CellGroupData[{ Cell[51849, 2009, 84, 4, 60, "Subsubsection"], Cell[51936, 2015, 147, 4, 71, "Text"], Cell[52086, 2021, 178, 3, 97, "Input"], Cell[52267, 2026, 179, 6, 46, "Text"], Cell[52449, 2034, 85, 1, 62, "Input"], Cell[52537, 2037, 150, 4, 71, "Text"], Cell[52690, 2043, 258, 6, 166, "Input"], Cell[52951, 2051, 87, 1, 46, "Text"], Cell[53041, 2054, 255, 6, 166, "Input"], Cell[53299, 2062, 127, 4, 46, "Text"], Cell[53429, 2068, 274, 7, 166, "Input"], Cell[53706, 2077, 136, 3, 46, "Text"], Cell[53845, 2082, 273, 7, 166, "Input"] }, Closed]] }, Closed]], Cell[54145, 2093, 81, 4, 62, "Section"], Cell[CellGroupData[{ Cell[54251, 2101, 80, 4, 109, "Section"], Cell[CellGroupData[{ Cell[54356, 2109, 84, 4, 60, "Subsubsection"], Cell[54443, 2115, 184, 4, 124, "Input"], Cell[54630, 2121, 59, 1, 39, "Input"], Cell[54692, 2124, 74, 0, 46, "Text"], Cell[54769, 2126, 139, 2, 61, "Input"], Cell[54911, 2130, 52, 0, 46, "Text"], Cell[54966, 2132, 219, 5, 43, "DisplayFormula"], Cell[55188, 2139, 123, 4, 46, "Text"], Cell[55314, 2145, 144, 4, 68, "Input"], Cell[55461, 2151, 178, 6, 46, "Text"], Cell[55642, 2159, 416, 8, 142, "Input"], Cell[56061, 2169, 55, 1, 39, "Input"], Cell[56119, 2172, 81, 1, 46, "Text"], Cell[56203, 2175, 56, 1, 39, "Input"], Cell[56262, 2178, 154, 6, 46, "Text"], Cell[56419, 2186, 219, 5, 43, "DisplayFormula"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[56687, 2197, 102, 4, 45, "Section"], Cell[CellGroupData[{ Cell[56814, 2205, 84, 4, 60, "Subsubsection"], Cell[56901, 2211, 239, 7, 71, "Text"], Cell[57143, 2220, 153, 2, 39, "Input"], Cell[57299, 2224, 115, 4, 46, "Text"], Cell[57417, 2230, 202, 4, 62, "Input"], Cell[57622, 2236, 168, 4, 71, "Text"], Cell[57793, 2242, 309, 5, 131, "Input"], Cell[58105, 2249, 93, 1, 46, "Text"], Cell[58201, 2252, 211, 5, 85, "Input"], Cell[58415, 2259, 210, 6, 51, "Text"], Cell[58628, 2267, 279, 4, 154, "Input"], Cell[58910, 2273, 358, 12, 71, "Text"], Cell[59271, 2287, 243, 5, 85, "Input"], Cell[59517, 2294, 208, 5, 71, "Text"], Cell[59728, 2301, 210, 4, 85, "Input"] }, Closed]] }, Closed]] }, Open ]] }, Open ]] } ] *) (******************************************************************* End of Mathematica Notebook file. *******************************************************************)