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Notebook[{
Cell["Higher-Order Ordinary Differential Equations", "Title"],

Cell[CellGroupData[{

Cell["Background: Another example of for Forward Differencing", "Subtitle"],

Cell[TextData[{
  "Recall how we used forward finite differencing to simulate the behavior of \
a function that changed proportional to its current size in Lecture 19.  \
Consider now a function that changes \"",
  StyleBox["how fast it changes",
    FontSlant->"Italic"],
  "\" proportional to its current change and its current value--in other \
words its current acceleration is proportional to its velocity and to its \
size:"
}], "Text"],

Cell["Let's work out how to find a finite differencing function:", "Text"],

Cell["This is the current change or approximation to velocity", "Text"],

Cell[BoxData[
    \(CurrentChangePerDelta[F_, \ 
        i_, \ \[CapitalDelta]_]\  := \ \(F[i + 1]\  - \ F[i]\)\/\
\[CapitalDelta]\)], "Input",
  CellTags->{
  "mmtag:21:foward_differencing__higher_order_ODEs", 
    "mmtag:21:foward_differencing__example_function"}],

Cell["\<\
This is the current change in change or approximation to \
acceleration\
\>", "Text"],

Cell[BoxData[
    \(CurrentChangeinCurrentChangeperDelta[F_, \ 
        i_, \ \[CapitalDelta]_]\  := \ \
\[IndentingNewLine]Simplify[\(CurrentChangePerDelta[F, i + 1, \
\[CapitalDelta]]\  - \ CurrentChangePerDelta[F, i, \[CapitalDelta]]\)\/\
\[CapitalDelta]]\)], "Input",
  CellTags->{
  "mmtag:21:discrete_approximation_to_derivative", 
    "mmtag:21:derivatives__discrete_approximation"}],

Cell["\<\
Our model is that the acceleration is proportional to  size of the \
current function and its velocity, let these proportions be: -\[Alpha] and -\
\[Beta]\
\>", "Text"],

Cell[BoxData[
    \(DifferenceRelation\  = \ 
      CurrentChangeinCurrentChangeperDelta[F, 
          i, \[CapitalDelta]]\  == \ \(-\[Beta]\)\ \ CurrentChangePerDelta[F, 
              i, \[CapitalDelta]] - \ \[Alpha]\ F[i]\)], "Input",
  CellTags->{
  "mmtag:21:derivatives__discrete_approximation_second_derivative", 
    "mmtag:21:second_derivatives__discrete_approximation"}],

Cell["Solve this for the \"latest\" values", "Text"],

Cell[BoxData[
    \(ForDiffSol\  = \ 
      Solve[DifferenceRelation, F[i + 2]] // Flatten\)], "Input"],

Cell["Replace to find the form of the solution", "Text"],

Cell[BoxData[
    \(ForDiffSolV2\  = \ ForDiffSol /. i \[Rule] \ j - 2\)], "Input"],

Cell["\<\
Notice that this forward differnce needs TWO consectutive past \
values before it can calculate the current value, create a function to \
Appends the current value to a list by using the last two values:\
\>", "Text"],

Cell["\<\
The fact that TWO consecutive values are needed is not surprising. \
Because this iteration procedure is giving an approximation to the solution \
of a second-order ODE, we need to make a numerical approximation to a second \
derivative. THREE points\[LongDash]the two consecutive ones and the current \
one\[LongDash]are needed to do this. (If only two points were used, you could \
only compute the slope of the straight line connecting the points; if three \
points are used, a circle can be fit to the three points and the inverse of \
its radius gives the curvature.)\
\>", "Subsubsection"],

Cell[BoxData[
    \(GrowList[
        ValuesList_List, \ \[CapitalDelta]_\ , \ \[Alpha]_\ , \ \[Beta]_]\  := 
      Module[{Minus1 = ValuesList[\([\(-1\)]\)], \ 
          Minus2\  = \ ValuesList[\([\(-2\)]\)]}, \[IndentingNewLine]Append[
          ValuesList, \[IndentingNewLine]2*Minus1\  - \ 
            Minus2\  + \ \[CapitalDelta]\ *\((\[Beta]*\((Minus2\  - \ 
                        Minus1)\)\  - \[Alpha]*\[CapitalDelta]*
                    Minus2)\)]]\)], "Input"],

Cell[BoxData[
    \(result\  = \ GrowList[{1, 1},  .001, 1,  .1]\)], "Input"],

Cell[BoxData[
    \(result\  = \ GrowList[result,  .001, 1,  .1]\)], "Input"],

Cell["\<\
Generate a sequence of length 20 from initial values {1,1} for \
\[CapitalDelta]=0.001, \[Alpha]=1, \[Beta]=0.1\
\>", "Text"],

Cell[BoxData[
    \(Nest[GrowList[#,  .001, 1,  .1] &, {1, 1}, 20]\)], "Input"],

Cell["\<\
Visualize results (it takes about 45 seconds to visualize 20000 \
iterations)\
\>", "Text"],

Cell[BoxData[
    \(ListPlot[Nest[GrowList[#,  .001, 1,  .1] &, {1, 1}, 20000]]\)], "Input"],

Cell[TextData[{
  "Change parameters for Growth Function (this example shows that the \
numerical solution ",
  StyleBox["does not converge",
    FontSlant->"Italic"],
  " to the accurate solution):"
}], "Text"],

Cell[BoxData[
    \(ListPlot[Nest[GrowList[#, 0.01, 0.5, 0] &, {1, 1}, 20000]]\)], "Input"]
}, Closed]],

Cell[CellGroupData[{

Cell["\<\
General Solution to the Second-Order Linear Homogeneous ODEs with \
Constant Coefficients\
\>", "Subtitle"],

Cell["\<\
Analysis of basis solutions to  y'' + \[Beta]y' + \[Gamma]y = 0 in \
terms of constant coefficients \[Beta] and \[Gamma]\
\>", "Text",
  CellTags->"mmtag:21:second_order_odes__characterizing_solution_behavior"],

Cell[CellGroupData[{

Cell["First write a general expression for this homogeneous ODE:", \
"Subsubsection"],

Cell[BoxData[
    \(TheODE[function_, \ var_]\  := \ 
      D[function[var], {var, 2}]\  + \ \[Beta]\ D[function[var], 
            var]\  + \ \[Gamma]\ function[var]\)], "Input"],

Cell[BoxData[
    \(TheODE[y, x]\)], "Input"]
}, Open  ]],

Cell[CellGroupData[{

Cell["\<\
Guess a solution and substitute it into the left-hand side of the \
ODE:\
\>", "Subsubsection"],

Cell[BoxData[
    \(TheGuess[x_]\  := \ Exp[\[Lambda]\ x]\)], "Input"],

Cell[BoxData[
    \(TheODE[TheGuess, x]\)], "Input"]
}, Open  ]],

Cell[CellGroupData[{

Cell["\<\
This will be a solution when the resulting quadratic expression in \
\[Lambda] is equal to 0:\
\>", "Subsubsection"],

Cell[BoxData[
    \(\[Lambda]Solution\  = \ 
      Solve[TheODE[TheGuess, x] \[Equal] 0, \[Lambda]]\)], "Input"]
}, Open  ]],

Cell["The two roots \[Lambda]+ and \[Lambda]+ are:", "Subsubsection",
  CellTags->
  "mmtag:21:second_order_odes__general solution_for_constant_coefficients"],

Cell[BoxData[
    \({\[Lambda]Minus, \[Lambda]Plus} = \ \[Lambda] /. \[Lambda]Solution\)], \
"Input"],

Cell[BoxData[
    \(GeneralSolution[x_]\  := \ 
      C[LPlus]\ Exp[\[Lambda]Plus\ x]\  + \ \ C[
            LMinus]\ Exp[\[Lambda]Minus\ x]\)], "Input"],

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    \(TheODE[GeneralSolution, z]\)], "Input"],

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}, Closed]],

Cell[CellGroupData[{

Cell["\<\
Graphical Characterization of General Solution to the Second Order \
ODE with Constant Coefficients\
\>", "Subtitle"],

Cell[BoxData[
    \(Reduce[\[Lambda]Plus \[Element] \ 
          Reals\  && \ \ \[Lambda]Minus\  \[Element] \ 
          Reals, {\[Beta], \[Gamma]}, Reals]\)], "Input"],

Cell[TextData[{
  "Assuming that \[Beta] and \[Gamma] are real, the solutions can be \
characterized by the roots of the quadratic equation. That is, our solution \
is known to be ",
  Cell[BoxData[
      \(TraditionalForm\`e\^\[Lambda]t\)]],
  ":  the solution monotonically grows or shrinks in time if the roots are \
real-positive or real-negative. The condition that the roots are real is \
\[Gamma] <",
  Cell[BoxData[
      \(TraditionalForm\`\(\(\ \)\(\[Beta]\^2\)\)\/4\)]],
  "If the roots have an imaginary part, the solutions can oscillate. Whether \
the oscillation amplitudes grow or shrink depends on whether the real part of \
\[Lambda] is positive or negative."
}], "Text"],

Cell["\<\
We now do some calculations to make an annotated map on the \[Beta]\
\[Dash]\[Gamma] plane of solution scenarios for the \[Lambda] values:\
\>", \
"Subsubsection"],

Cell[BoxData[
    \(CplexReal\  = \ 
      Plot[\[Beta]^2/4, {\[Beta], \(-1\), 1}, 
        AxesLabel \[Rule] {"\<\[Beta]\>", "\<\[Gamma]\>"}]\)], "Input",
  CellTags->"mmtag:21:graphics__building_piece_by_piece"],

Cell[BoxData[
    \(CplexRealAnnote\  = 
      Show[CplexReal, 
        Graphics[
          Text["\<Complex\nCongugate\nRoots\>", {0.25, 0.25}, {\(-1\), 1}]], 
        Graphics[Text["\<Real Roots\>", {0.75, 0.05}]]]\)], "Input"],

Cell[TextData[{
  "If the roots complex, then the sign of the real part is determined only by \
the sign of \[Beta]; i.e., \[Lambda] = ",
  Cell[BoxData[
      \(TraditionalForm\`\(-\[Beta]\)\/2\)]],
  "\[PlusMinus] ",
  Cell[BoxData[
      FormBox[
        SqrtBox[
          RowBox[{
            FormBox[\(\((\[Beta]\/2)\)\^2\),
              "TraditionalForm"], "-", " ", "\[Gamma]"}]], TraditionalForm]]]
}], "Text"],

Cell[BoxData[
    \(CplexPosNeg\  = 
      ParametricPlot[{0, t}, {t, 0,  .25}, 
        PlotStyle \[Rule] {Thickness[0.015], Hue[0]}, 
        DisplayFunction \[Rule] Identity]\)], "Input"],

Cell[BoxData[
    \(CplexPosNegAnnote\  = \ 
      Show[CplexPosNeg, 
        Graphics[
          Text[StyleForm["\<Positive\nReal\nPart\>", 
              FontColor \[Rule] Hue[0]], {\(- .5\), 0.15}, {\(-1\), 1}]], 
        Graphics[
          Text[StyleForm["\<Negative\nReal\nPart\>", 
              FontColor \[Rule] Hue[0]], { .5, 0.15}, {1, 1}]], 
        DisplayFunction \[Rule] $DisplayFunction]\)], "Input"],

Cell[BoxData[
    \(CplexPlot\  = Show[CplexRealAnnote, CplexPosNegAnnote]\)], "Input"],

Cell["\<\
Find the conditions where the real roots are positive and/or \
negative\
\>", "Text"],

Cell[BoxData[
    \(Reduce[{\[Lambda]Plus\  > 0, \[Lambda]Minus > 0}]\)], "Input",
  CellTags->"mmtag:21:Reduce[]"],

Cell[BoxData[
    \(AnnotePosRealRoots\  = 
      Graphics[Text[
          StyleForm["\<Positive Roots\>", 
            FontColor \[Rule] Hue[ .6]], {\(-1.0\), 0.025}, {\(-1\), 
            0}]]\)], "Input"],

Cell[BoxData[
    \(Reduce[{\[Lambda]Plus\  > 0, \[Lambda]Minus < 0}]\)], "Input"],

Cell[BoxData[{
    \(MixedRealRoots\  = \ 
      Plot[0, {t, \(-1\), 1}, PlotStyle \[Rule] {Hue[0.6], Thickness[0.015]}, 
        DisplayFunction \[Rule] Identity]\), "\[IndentingNewLine]", 
    \(AnnoteMixedRealRoots\  = 
      Show[MixedRealRoots, 
        Graphics[
          Text[StyleForm["\<Mixed Real Roots\>", 
              FontColor \[Rule] Hue[ .6]], {0.2, \(-0.1\)}, {\(-1\), 0}]], 
        DisplayFunction \[Rule] $DisplayFunction]\)}], "Input"],

Cell[BoxData[
    \(Reduce[{\[Lambda]Plus\  < 0, \[Lambda]Minus < 0}]\)], "Input"],

Cell[BoxData[
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      Graphics[Text[
          StyleForm["\<Negative Roots\>", FontColor \[Rule] Hue[ .6]], {1.0, 
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Cell[BoxData[
    \(Show[CplexPlot, AnnotePosRealRoots, AnnoteMixedRealRoots, 
      AnnoteNegRealRoots]\)], "Input"]
}, Closed]],

Cell[CellGroupData[{

Cell["\<\
Determination of the Specific Solution for Particular Boundary \
Values\
\>", "Subtitle"],

Cell[TextData[{
  "Second order ODEs require that ",
  StyleBox["two",
    FontWeight->"Bold"],
  " conditions  be specified to generate a particular solution."
}], "Text",
  CellTags->"mmtag:21:second_order_odes__using_boundary_values"],

Cell["\<\
Example of y(0) = 0 and y(L)=1
The general solution will have the form:\
\>", "Subsubtitle"],

Cell[BoxData[
    \(GeneralSolution[x_]\  := \ 
      CPlus\ Exp[\[Lambda]Plus\ x]\  + \ 
        CMinus\ Exp[\[Lambda]Minus\ x]\)], "Input"],

Cell[TextData[{
  "Apply the two boundary conditions and solve the resulting equations for \
constants ",
  StyleBox["CPlus",
    FontWeight->"Bold"],
  " and ",
  StyleBox["CMinus",
    FontWeight->"Bold"],
  ":"
}], "Subsubsection"],

Cell[BoxData[
    \(SolutionOne\  = \ 
      Solve[{GeneralSolution[0] == 0, GeneralSolution[L] == 1}, {CPlus, 
          CMinus}]\)], "Input"],

Cell["Write the resulting solution:", "Subsubsection"],

Cell[BoxData[
    \(SpecificSolutionOne\  = \ 
      Simplify[GeneralSolution[x] /. SolutionOne]\)], "Input"],

Cell["\<\
Second example with different form of boundary condition:  y(0) = 1 \
and y'(0)=0\
\>", "Text"],

Cell[BoxData[
    \(DGen\  = \ D[GeneralSolution[x], x]\)], "Input"],

Cell["Now the constants CPlus and CMinus are found by solving:", \
"Subsubsection"],

Cell[BoxData[
    \(SolutionTwo\  = \ 
      Solve[{GeneralSolution[0] \[Equal] 
            1, \((DGen /. x \[Rule] 0)\) \[Equal] 0}, {CPlus, 
          CMinus}]\)], "Input"],

Cell["And the new solution is:", "Subsubsection"],

Cell[BoxData[
    \(SpecificSolutionTwo\  = \ 
      Simplify[GeneralSolution[x] /. SolutionTwo]\)], "Input"]
}, Closed]],

Cell[CellGroupData[{

Cell["\<\
Example of Higher-Order equation and Visualization of Solutions:
The Beam Equation\
\>", "Subtitle",
  CellTags->{
  "mmtag:21:beam_theory", "mmtag:21:fourth_order_odes__beam_theory"}],

Cell["Visualizing Beam Deflections:", "Text"],

Cell["\<\
Set up the ODE solution (we will specify a unit beam stiffness) EI \
= 1\
\>", "Text"],

Cell[BoxData[
    \(BeamEquation[y_\ , x_\ , \ w_, \ BC1_\ , \ BC2_]\  := \ 
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        Flatten\)], "Input",
  CellTags->"mmtag:21:beam_equation"],

Cell["Set up some typical boundary conditions:", "Text",
  CellTags->{
  "mmtag:21:beam_boundary_conditions", 
    "mmtag:21:boundary_conditions__for_beams"}],

Cell[BoxData[{
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    \(PointLoad[y_, \ x_, \ position_, \ 
        moment_]\  := \ {y[x] \[Equal] \ position, \ \(y''\)[x]\  \[Equal] \ 
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    \(FreeEnd[y_\ , \ 
        x_]\  := \ {\(y''\)[x] \[Equal] 0, \ \(y'''\)[x] \[Equal] 
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Cell[BoxData[
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Cell["Clamped Beam, loaded at the end", "Text"],

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Cell["Clamped Beam, Free end, unit applied load", "Text"],

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Point Loaded on both ends, with linear distributed load positive at \
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\>", "Text",
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      PointLoad[y, 1, 0, 0]]\)], "Input"],

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Cell[BoxData[
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