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Notebook[{
Cell["Spectral-based solution methods", "Title"],

Cell[TextData[{
  "This notebook has been written in ",
  StyleBox["Mathematica ",
    FontSlant->"Italic"],
  "by \n\n",
  StyleBox[
  "Mark J. McCready\nProfessor and Chair of Chemical Engineering\nUniversity \
of Notre Dame\nNotre Dame, IN 46556\nUSA",
    FontSize->14],
  "\n\n",
  ButtonBox["mjm@nd.edu",
    ButtonStyle->None],
  "\n",
  ButtonBox["http://www.nd.edu/~mjm/",
    ButtonData:>{
      URL[ "http://www.nd.edu/~mjm/"], None},
    ButtonStyle->"Hyperlink"],
  "\n\n\nIt is copyrighted to the extent allowed by which ever laws pertain \
to the World Wide Web and the Internet.\n\nI would hope that as a \
professional courtesy, this notice remain visible to other users. \nThere is \
no charge for copying and dissemination.   \n\nVersion: 11/25/98"
}], "Text"],

Cell[CellGroupData[{

Cell["Overview of the spectral approach", "Section"],

Cell[TextData[{
  "The basic ideal of anything \"spectral\" is the systematic \
representation/decomposition of a vector, function or solution in terms of \
some basis vectors, functions which are independent (in this case \
orthogonal/biorthogonal).  The basis vectors or functions must span the space \
and thus form a complete set so that ",
  StyleBox["any",
    FontSlant->"Italic"],
  " vector, function or solution can be represented.\n\nIn symbolic form, we \
are interested in solving, ",
  Cell[BoxData[
      \(TraditionalForm\`L\ [y\ ] = \ f(x)\)]],
  "  by representing ",
  Cell[BoxData[
      \(TraditionalForm\`f(x)\)]],
  " in terms of our basis functions and then ultimately finding ",
  StyleBox["y ",
    FontSlant->"Italic"],
  "in terms of these same functions.  Thus we are not directly finding the \
inverse operator solution as we did in the Green's function approach, i.e., ",
  
  Cell[BoxData[
      FormBox[
        RowBox[{"y", " ", "=", " ", 
          RowBox[{"L", 
            FormBox[\(\(\^\(-1\)\) \(f(x)\)\),
              "TraditionalForm"]}]}], TraditionalForm]]],
  ".  We look for these expansions, \n\n",
  Cell[BoxData[
      \(TraditionalForm
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  "  \n\n",
  Cell[BoxData[
      \(TraditionalForm
      \`\(y(x)\  = \[Sum]\+\(j = 1\)\%n d\_j\ \(\(\[Phi]\_j\)(x)\)\ \ \)\)]],
  "\n\nPresumably there will be a straightforward way of finding the ",
  Cell[BoxData[
      \(TraditionalForm\`c\_j' s\)]],
  " since ",
  Cell[BoxData[
      \(TraditionalForm\`f(x)\)]],
  " is a known function.  (This will be to form an inner product.)  We will \
find that we can obtain the ",
  Cell[BoxData[
      \(TraditionalForm\`d\_j\)]],
  "'s by substituting all that we know in the equation and forming an inner \
product for each term.\n\nOnce we have defined the problem as the equation, ",
  
  Cell[BoxData[
      \(TraditionalForm\`L\ y\  = \ f(x)\)]],
  ", and its boundary conditions, ",
  Cell[BoxData[
      \(TraditionalForm\`B(y)\  = 0. \)]],
  "  We need a systematic way to produce an appropriate set of basis \
functions.  The best possible set will usually be obtained from the \
eigenvalue problem,  ",
  Cell[BoxData[
      \(TraditionalForm\`L\ y\  = \ \(-\[Lambda]\)\ y\)]],
  " with ",
  Cell[BoxData[
      \(TraditionalForm\`B(y)\  = 0. \)]],
  " For a self adjoint operator we will generate real \[Lambda]'s and \
orthogonal eigenfunctions.  If ",
  StyleBox["L",
    FontSlant->"Italic"],
  " is not self-adjoint, we can take advantage of biorthonality.\n\nWe will \
define the expansion of a term in the equation by expansion in terms of \
eigenfunctions as a ",
  StyleBox["Finite Fourier Transform",
    FontSlant->"Italic"],
  " which is a linear operator and will, of course, have an inverse.\n\nIf we \
cannot readily solve ",
  Cell[BoxData[
      \(TraditionalForm\`L[\ y\ ] = \ \(-\[Lambda]\)\ y\)]],
  ", we may instead use a known set of eigenfunctions from a different \
operator.  This is effectively a ",
  StyleBox["numerical",
    FontSlant->"Italic"],
  " solution technique which we will discuss later. "
}], "Text"]
}, Closed]],

Cell[CellGroupData[{

Cell["\<\
Expansion of an arbitrary function in terms of a set of \
eigenfunctions.\
\>", "Section"],

Cell[CellGroupData[{

Cell["\<\
Here are two functions that we would like to fit with our \
Eigenfunctions.\
\>", "Subsubsection"],

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          \(\[Sum]\+\(j = 1\)\%n c\_j\ \(\[Phi]\_j\) \((x)\)\ \ \n\n\t\t\t\t
                now  form\ the\ inner\ product\ with\ the\ function\ 
                \(\[Phi]\_k\) \((x)\)\ \n\t\t\t\n\t\t\t\n\ 
                \(\[Integral]\_0\%1 \[Rho] \((x)\)\ \(\[Phi]\_k\) \((x)\)\ f 
                    \((x)\)\ \[DifferentialD]x\)\  = 
            \(\[Integral]\_0\%1 \[Rho] \((x)\) 
                  \(\(\ \ \(\[Phi]\_k\) \((x)\) 
                        \[Sum]\)\+\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
                      j = 1\)\%\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n\)\) 
                  \(c\_j\) \(\[Phi]\_j\) \((x)\) \[DifferentialD]x\ \ \n\nOn\ 
                  the\ right  side\ we\ can\ \ switch\ the\ summation\ and\ 
                  integration 
                  \((assuming\ that\ everything\ will\ converge)\)\n\t\t\t\n\ 
                  \(\[Integral]\_0\%1 \[Rho] \((x)\)\ \(\[Phi]\_k\) \((x)\)\ 
                      f \((x)\)\ \[DifferentialD]x\)\  = \ 
              \[Sum]\+\(j = 1\)\%n
                  \[Integral]\_0\%1 \[Rho] \((x)\)\ \(\[Phi]\_k\) 
                    \(\(\((x)\)\ \)\+\(\  \)\) \(c\_j\) \(\[Phi]\_j\) \((x)\) 
                    \[DifferentialD]x\ \ \n\nnext\ just  move\ the\ \(c\_j\) 
                    outside\ of\ the\ integration\)\), \n\n
        \t\t\t\t\t\t\t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
        \( = \ \ \[Sum]\+\(j = 1\)\%n\( c\_j\) 
              \(\[Integral]\_0\%1 \[Rho] \((x)\)\ \(\[Phi]\_k\) 
                  \(\(\((x)\)\ \)\+\(\  \)\) \(\[Phi]\_j\) \((x)\) 
                  \[DifferentialD]x\n\nFor\ properly\ normalized  
                  orthogonalized\ functions\ the\ integral\ just\ 
                  equals  \[Delta]\_kj\ which\ means\ that\ the\ right\ side\ 
                  is\ just\)\), \ c\_\(\(k\t . \)\n\t\)\)}], TextForm], 
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      RowBox[{
        FormBox[
          \(\[Integral]\_0\%1 \[Rho] \((x)\)\ \(\[Phi]\_k\) \((x)\)\ f 
                \((x)\) \[DifferentialD]x\  = \ 
            \(\(c\_k\)\_\(\t\n\t\) . \)\),
          "TextForm"], "\n", "\t\t", "\n"}], TextForm]}], "Text",
  CellLabel->"In[17]:="]
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Cell["First expand ff[x]", "Subsubsection"],

Cell[TextData[
"For the eigenfunctions we have chosen, \[Rho](x) = 1.  This gives then for \
ff[x], the first one and then a lot of others, "], "Text"],

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    \(\[Integral]\_0\%1 ff[x]\ \ \[Phi][x, 1] \[DifferentialD]x\)], "Input",
  CellLabel->"In[21]:=",
  AspectRatioFixed->True],

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    \(TraditionalForm\`\(6\ \@2\)\/\[Pi]\^3\)], "Output",
  CellLabel->"Out[21]="],

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Cell[BoxData[
    \(cks = 
      Table[Integrate[ff[x]\ \ \[Phi][x, i], {x, 0, 1}], {i, 1, 10}]\)], 
  "Input",
  CellLabel->"In[22]:=",
  AspectRatioFixed->True],

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    \(TraditionalForm
    \`{\(6\ \@2\)\/\[Pi]\^3, \(-\(3\/\(2\ \@2\ \[Pi]\^3\)\)\), 
      \(2\ \@2\)\/\(9\ \[Pi]\^3\), \(-\(3\/\(16\ \@2\ \[Pi]\^3\)\)\), 
      \(6\ \@2\)\/\(125\ \[Pi]\^3\), \(-\(1\/\(18\ \@2\ \[Pi]\^3\)\)\), 
      \(6\ \@2\)\/\(343\ \[Pi]\^3\), \(-\(3\/\(128\ \@2\ \[Pi]\^3\)\)\), 
      \(2\ \@2\)\/\(243\ \[Pi]\^3\), \(-\(3\/\(250\ \@2\ \[Pi]\^3\)\)\)}\)], 
  "Output",
  CellLabel->"Out[22]="]
}, Open  ]],

Cell["If you like numbers we have", "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(cksN = 
      Table[NIntegrate[ff[x]\ \ \[Phi][x, i], {x, 0, 1}], {i, 1, 10}]\)], 
  "Input",
  CellLabel->"In[23]:=",
  AspectRatioFixed->True],

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    FormBox[
      RowBox[{
      \(General::"spell1"\), \( : \ \), 
        "\<\"Possible spelling error: new symbol name \
\\\"\\!\\(TraditionalForm\\`cksN\\)\\\" is similar to existing symbol \\\"\\!\
\\(TraditionalForm\\`cks\\)\\\".\"\>"}], TraditionalForm]], "Message"],

Cell[BoxData[
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    \`{0.273663344416641507`, \(-0.0342079180520801884`\), 
      0.0101356794228385371`, \(-0.00427598975651000756`\), 
      0.00218930675533312024`, \(-0.00126695992785479116`\), 
      0.000797852316083490897`, \(-0.000534498719563740642`\), 
      0.000375395534179196666`, \(-0.000273663344416676101`\)}\)], "Output",
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Cell["Make some partial sums for the function", "Text"],

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    \(onex = 
      \[Sum]\+\(i = 1\)\%1 \[Phi][x, i]\ 
          cks\[LeftDoubleBracket]i\[RightDoubleBracket]\)], "Input",
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  AspectRatioFixed->True],

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Cell[BoxData[
    \(twox = 
      \[Sum]\+\(i = 1\)\%2 \[Phi][x, i]\ 
          cks\[LeftDoubleBracket]i\[RightDoubleBracket]\)], "Input",
  CellLabel->"In[25]:=",
  AspectRatioFixed->True],

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    \(TraditionalForm
    \`\(12\ \(sin(\[Pi]\ x)\)\)\/\[Pi]\^3 - 
      \(3\ \(sin(2\ \[Pi]\ x)\)\)\/\(2\ \[Pi]\^3\)\)], "Output",
  CellLabel->"Out[25]="]
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Cell[CellGroupData[{

Cell[BoxData[
    \(threex = 
      \[Sum]\+\(i = 1\)\%3 \[Phi][x, i]\ 
          cks\[LeftDoubleBracket]i\[RightDoubleBracket]\)], "Input",
  CellLabel->"In[26]:=",
  AspectRatioFixed->True],

Cell[BoxData[
    \(TraditionalForm
    \`\(12\ \(sin(\[Pi]\ x)\)\)\/\[Pi]\^3 - 
      \(3\ \(sin(2\ \[Pi]\ x)\)\)\/\(2\ \[Pi]\^3\) + 
      \(4\ \(sin(3\ \[Pi]\ x)\)\)\/\(9\ \[Pi]\^3\)\)], "Output",
  CellLabel->"Out[26]="]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(sixx = 
      \[Sum]\+\(i = 1\)\%6 \[Phi][x, i]\ 
          cks\[LeftDoubleBracket]i\[RightDoubleBracket]\)], "Input",
  CellLabel->"In[27]:=",
  AspectRatioFixed->True],

Cell[BoxData[
    \(TraditionalForm
    \`\(12\ \(sin(\[Pi]\ x)\)\)\/\[Pi]\^3 - 
      \(3\ \(sin(2\ \[Pi]\ x)\)\)\/\(2\ \[Pi]\^3\) + 
      \(4\ \(sin(3\ \[Pi]\ x)\)\)\/\(9\ \[Pi]\^3\) - 
      \(3\ \(sin(4\ \[Pi]\ x)\)\)\/\(16\ \[Pi]\^3\) + 
      \(12\ \(sin(5\ \[Pi]\ x)\)\)\/\(125\ \[Pi]\^3\) - 
      \(sin(6\ \[Pi]\ x)\)\/\(18\ \[Pi]\^3\)\)], "Output",
  CellLabel->"Out[27]="]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
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      \[Sum]\+\(i = 1\)\%10 \[Phi][x, i] 
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  CellLabel->"In[28]:=",
  AspectRatioFixed->True],

Cell[BoxData[
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      \(3\ \(sin(2\ \[Pi]\ x)\)\)\/\(2\ \[Pi]\^3\) + 
      \(4\ \(sin(3\ \[Pi]\ x)\)\)\/\(9\ \[Pi]\^3\) - 
      \(3\ \(sin(4\ \[Pi]\ x)\)\)\/\(16\ \[Pi]\^3\) + 
      \(12\ \(sin(5\ \[Pi]\ x)\)\)\/\(125\ \[Pi]\^3\) - 
      \(sin(6\ \[Pi]\ x)\)\/\(18\ \[Pi]\^3\) + 
      \(12\ \(sin(7\ \[Pi]\ x)\)\)\/\(343\ \[Pi]\^3\) - 
      \(3\ \(sin(8\ \[Pi]\ x)\)\)\/\(128\ \[Pi]\^3\) + 
      \(4\ \(sin(9\ \[Pi]\ x)\)\)\/\(243\ \[Pi]\^3\) - 
      \(3\ \(sin(10\ \[Pi]\ x)\)\)\/\(250\ \[Pi]\^3\)\)], "Output",
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We see that because gg is symmetric, the odd functions do not \
contribute.\
\>", "Text"],

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Cell["Now do hh", "Subsubsection"],

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Cell["\<\
We see that because hh is symmetric, the odd functions do not \
contribute.\
\>", "Text"],

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      2\ \(sin(21\ \[Pi]\ x)\) - 2\ \(sin(23\ \[Pi]\ x)\) + 
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Mfstroke
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Mfstroke
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Mistroke
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Mistroke
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Mistroke
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Mistroke
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Mistroke
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Mfstroke
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s
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s
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s
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s
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s
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Mistroke
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Mistroke
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Mistroke
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Cell["Again qwe can check the numerical values. ", "Text"],

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This one does not converge very well because the boundary \
conditions cannot be met.  We get the \"ringing\" that is shown in figure 3.7 \
of V&M.  

Think of the consequences of this observation for our problem below.  \
\>", 
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Cell[TextData[{
  "To recap, our function is known (at least symbolically) and we choose a \
set of known eigenfunctions from a convenient operator.  The expansion is \
easily accomplished by forming the inner product which directly gives values \
of the expansion coefficients for, say,\n",
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Cell["Finite Fourier transform", "Section"],

Cell[TextData[{
  "Now we are ready to solve an inhomogenous ODE with a spectral expansion in \
terms of the eigenfunctions for ",
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      \(TraditionalForm\`y(x)\)]],
  " and",
  Cell[BoxData[
      \(TraditionalForm\`\(\ f(x)\)\)]],
  ".\n\nA simple example equation (3.19.1) in V&M is\n",
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      \(TraditionalForm
      \`\(\(\(d\^2\) y\)\/dx\^2 - \ U\^2\ y\  = \ \(-\ \(f(x)\)\)\ \)\)]],
  ",\tx\[Element](0,L)\n\nwith ",
  Cell[BoxData[
      \(TraditionalForm\`y(0) = \ \(y \((L)\)\  = \ 0\)\)]],
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eigenvalue problem,\n\n",
  Cell[BoxData[
      \(TraditionalForm
      \`\(\(\(d\^2\) \[Phi]\)\/dx\^2 = \(-\ \[Lambda]\)\ \[Phi]\ \)\)]],
  " , with, ",
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  ",  is then ",
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  " although as V&M show, it could also be\n",
  Cell[BoxData[
      \(TraditionalForm
      \`\(\(d\^2\) \[Bullet]\)\/dx\^2 - \ U\^2\ \[Bullet]\)]],
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  Cell[BoxData[
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    FontWeight->"Bold"],
  "\n \nHere we go starting with:\n\n",
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  Cell[BoxData[
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        \(-\ \(\[Integral]\_0\%L\( \[Phi]\_n\) \(f(x)\)\ 
              \[DifferentialD]x\)\)\)]],
  "\t\t(eq1)\n\nWe will define an operator called the finite Fourier \
Transform, ",
  StyleBox["F",
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Cell["Here are some plots of our answer", "Subsubsection"],

Cell["\<\
Here is the solution to the example problem in V&M, p. 275.  I need \
to limit this to a finite number of terms, say 8.\
\>", "Text"],

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                      \[DifferentialD]t)\)\ 
                \(sin(\(5\ \[Pi]\ x\)\/L)\)\)\/\(U\^2 + 
                \(25\ \[Pi]\^2\)\/L\^2\) + 
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                      \[DifferentialD]t)\)\ 
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                      \[DifferentialD]t)\)\ 
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Cell["Choose a length of 1 and order 1 external cooling,", "Text"],

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                    \[DifferentialD]t)\)\ \(sin(2\ \[Pi]\ x)\)\)\/\(4 + 
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                    \[DifferentialD]t)\)\ \(sin(3\ \[Pi]\ x)\)\)\/\(4 + 
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Cell["We choose heating of the form ff[t] from above", "Text"],

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We see that it is hotest in the middle where the heat generation is \
largest.  But this is not an insightful answer because the ends are fixed at \
T=0 so it will always be hottest in the middle.  For example consider uniform \
heating\
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Here we compare the solution with f[t] = 1-t(1-t) for 4 terms and 8 \
terms.  There is not much difference perhaps because the end regions, where \
the fit to f(t) is not easy to do, does not contribute too much.  You may \
wish to check this further. \
\>", "Text"],

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Here is the solution to the example problem in V&M, p. 275.  I need \
to limit this to a finite number of terms, say 8.\
\>", "Text"],

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We see that it is hotest in the middle where the heat generation is \
largest.  But this is not an insightful answer because the ends are fixed at \
T=0 so it will always be hottest in the middle.  For example consider uniform \
heating\
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Let's see if this last one changes much if the number of terms is \
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Cell["\<\
Here we compare the solution with f[t] = 1-t(1-t) for 4 terms and 8 \
terms.  There is not much difference perhaps because the end regions, where \
the fit to f(t) is not easy to do, does not contribute too much.  You may \
wish to check this further. \
\>", "Text"],

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We might start solving this problem with a homogenous boundary \
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