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Cell[CellGroupData[{
Cell["Boundary Value problem in an open region", "Subsection"],

Cell[BoxData[
    \(Clear["\<`*\>"]\)], "Input"],

Cell[TextData[{
  StyleBox["The solution to Laplace's Equation in an open channel with \
boundary conditions \[CapitalPhi][0,y] = 0,  \[CapitalPhi][a,y] = 0, and \
\[CapitalPhi][x,0] = f[x], where ",
    FontFamily->"Arial"],
  Cell[BoxData[
      \(TraditionalForm\`f[x]\  = \(\(2\ x\)\/a\) V\)],
    FontFamily->"Arial"],
  StyleBox["  for 0 \[LessEqual] x \[LessEqual] a/2 and  ",
    FontFamily->"Arial"],
  Cell[BoxData[
      \(TraditionalForm\`f[x]\  = \(\(2 \((a - x)\)\)\/\[Placeholder]\) V\)],
    FontFamily->"Arial"],
  StyleBox[" for a/2 \[LessEqual] x \[LessEqual] a is\n\n\[CapitalPhi][x,y]  \
=  ",
    FontFamily->"Arial"],
  Cell[BoxData[
      \(TraditionalForm\`\(8\ V\)\/\[Pi]\^2\)],
    FontFamily->"Arial"],
  StyleBox["  ",
    FontFamily->"Arial"],
  Cell[BoxData[
      \(TraditionalForm\`\[Sum]\+\(m = 0\)\%\[Infinity]\ \ \
\((\(-1\))\)\^m\/\((2  m + 1)\)\^2\ Sin[\ \(\((2  m + 1)\)\ \[Pi]\ x\)\/a]\ \
Exp[\(-\(\(\((2  m + 1)\)\ \[Pi]\ y\)\/a\)\)]\)],
    FontFamily->"Arial"],
  StyleBox["\n\nLet us look at this for the case V = 1 and a = 1. We take 25 \
terms in the series",
    FontFamily->"Arial"]
}], "Text"],

Cell[BoxData[
    \(\[CapitalPhi] := 
      8/\[Pi]^2\ Sum[\((\(-1\))\)^
                m/\((2  m + 1)\)^2\ Sin[\((2  m + 
                    1)\)\ \[Pi]\ x]\ Exp[\(-\((2  m + 1)\)\) \[Pi]\ y], {m, 
            0, 25}]\)], "Input"],

Cell[BoxData[
    \(\(\(Plot3D[\[CapitalPhi], {x, 0, 1}, {y, 0, 1}, 
      ViewPoint \[Rule] {1.2, 1.2, 1}, PlotRange \[Rule] {0, 1}, 
      Boxed \[Rule] False, Ticks \[Rule] {Automatic, Automatic, {0,  .5, 1}}, 
      PlotLabel \[Rule] "\<\[CapitalPhi](x,y)\>", 
      AxesLabel \[Rule] {"\<x\>", "\<y\>", "\<\>"}]\)\(\n\)
    \)\)], "Input"],

Cell[BoxData[
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      RowBox[{\(Examine\ the\ charge\ on\ the\ surface\ y\), "=", 
        RowBox[{
          RowBox[{"0.", "   ", 
            FormBox[\(\[Sigma]\_x\),
              "TraditionalForm"]}], " ", "=", 
          RowBox[{
            FormBox[
              RowBox[{
                FormBox[\(\[Epsilon]\_\(\(0\)\(\ \)\)\),
                  "TraditionalForm"], \(E\_x\)}],
              "TraditionalForm"], "[", \(x, 0\), "]"}]}]}],
      FontFamily->"Arial"]], "Text"],

Cell[BoxData[
    \(Ex := \(-D[\[CapitalPhi], y]\)\)], "Input"],

Cell[BoxData[
    \(\[Sigma]x[x_] := Ex /. y \[Rule] 0\)], "Input"],

Cell[BoxData[
    \(Plot[\[Sigma]x[x], {x, 0, 1}, PlotRange \[Rule] {{0, 1}, {0, 5}}, 
      PlotStyle \[Rule] {Thickness[0.008], RGBColor[1, 0, 0]}]\)], "Input"],

Cell[TextData[{
  StyleBox["\nThe irreularities in ",
    FontFamily->"Arial"],
  Cell[BoxData[
      \(TraditionalForm\`\[Sigma]\_x\)],
    FontFamily->"Arial"],
  StyleBox[" near x=",
    FontFamily->"Arial"],
  Cell[BoxData[
      \(TraditionalForm\`1\/2\)],
    FontFamily->"Arial"],
  StyleBox[" are characteristic of Fourier Serirs near points of sudden \
change in value or slope. This behavior is referred to as the Stoke's  \
phenomenon.  Charge is seen to accumulate near sharp changes of potential.  \
The total charge on the surface is still integrable and has the folliring \
value. (This is the charge/length of the cylindrical region).",
    FontFamily->"Arial"]
}], "Text"],

Cell[BoxData[
    \(Qx = Integrate[\[Sigma]x[x], {x, 0, 1}] // N\)], "Input"],

Cell[BoxData[
    StyleBox[
      RowBox[{
        RowBox[{\(The\ capacitanve\ is\ Q/V . Therefore\ find\ C/L\), "=", 
          RowBox[{"1.485", 
            FormBox[\(\[Epsilon]\_\(\(0\)\(\ \)\(.\)\)\),
              "TraditionalForm"], "Now"}]}], 
        ",", \(look\ at\ the\ charge\ distribution\ on\ the\ x = 0\ face\)}],
      FontFamily->"Arial"]], "Text"],

Cell[BoxData[
    \(Ey := \(-D[\[CapitalPhi], x]\)\)], "Input"],

Cell[BoxData[
    \(\[Sigma]y[y_] := Ey /. x \[Rule] 0\)], "Input"],

Cell[BoxData[
    \(Plot[\[Sigma]y[y], {y, 0, 2}, 
      PlotStyle \[Rule] {Thickness[0.008], RGBColor[0, 1, 0]}]\)], "Input"],

Cell[BoxData[
    \(Qy = Integrate[\[Sigma]y[y], {y, 0, Infinity}] // N\)], "Input"],

Cell[TextData[{
  "\nThe charge on the face x=1 must be the same as the charge on x=0. \
Furthermore, the sum of the charges on all faces, Q = ",
  Cell[BoxData[
      FormBox[
        RowBox[{
          FormBox[\(Q\_x\),
            "TraditionalForm"], "+", \(2  Q\_y\)}], TraditionalForm]]],
  ", must vanish.    We find"
}], "Text",
  FontFamily->"Arial"],

Cell[BoxData[
    \(Q = Qx + 2  Qy\)], "Input"]
}, Open  ]],

Cell[CellGroupData[{

Cell["Green Function in the plane", "Subsection"],

Cell["\<\
Use the Green Function method to solve the 2D Laplace equation with boundary \
conditions 
\[CapitalPhi](0,y) = \[CapitalPhi](a,y) = \[CapitalPhi](x,b) =0  and \
\[CapitalPhi](x,0)  = V. (Set V=1,  a=1,  b=1/2)\
\>", "Text"],

Cell["Solution at y=0", "Text"],

Cell[BoxData[
    \(\(\(\ \)\(\[CapitalPhi]0 := \ \((4/\[Pi])\)\ Sum\ [\ 
          Sin[\((2  m + 1)\)\ \[Pi]\ x]/\((2\ m + 1)\), {m, 0, 
            25}]\)\)\)], "Input"],

Cell[BoxData[
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      PlotStyle \[Rule] {{Thickness[0.007], RGBColor[0, 0, 1]}, 
          Thickness[0.005]\ }]\)], "Input"],

Cell["Solution inside box", "Text"],

Cell[BoxData[
    \(\[CapitalPhi]\  := \ \((4/\[Pi])\)\ Sum\ [\ 
          Sin[\((2  m + 1)\)\ \[Pi]\ x]/\((\((2\ m + 
                      1)\)\ Sinh[\ \((2  m + 1)\)\ \[Pi]\ /
                        2])\)\ Sinh[\((2\ m + 1)\)\ \[Pi] \((1/2 - y)\)], {m, 
            0, 20}]\)], "Input"],

Cell[BoxData[
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      ViewPoint \[Rule] \ {1.2, 1.2, 1.2}, 
      Ticks \[Rule] {Automatic, Automatic, {0.5, 1}}]\)], "Input"],

Cell[BoxData[
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Cell[BoxData[
    \(\[Sigma]x\  = \ \(-D[\[CapitalPhi], y]\)\  /. \ 
        y \[Rule] \ 0\)], "Input"],

Cell[BoxData[
    \(Plot[\[Sigma]x, {x, 0, 1}]\)], "Input"],

Cell[BoxData[
    \(Q1\  = \ Integrate[\[Sigma]x, {x, 0, 1}] // N\)], "Input"],

Cell[BoxData[
    \(\[Sigma]y\  = \ \(-D[\[CapitalPhi], x]\)\  /. \ 
        x \[Rule] \ 0\)], "Input"],

Cell[BoxData[
    \(Q2\  = \ Integrate[\[Sigma]y, {y, 0, 1/2}] // N\)], "Input"],

Cell[BoxData[
    \(\[Sigma]xx\  = \ D[\[CapitalPhi], y] /. \ y \[Rule] 1/2\)], "Input"],

Cell[BoxData[
    \(Q3\  = \ Integrate[\[Sigma]xx, {x, 0, 1}] // N\)], "Input"],

Cell[BoxData[
    \(Q1 + \ 2\ Q2\  + \ Q3\)], "Input"]
}, Closed]]
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