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Notebook[{

Cell[CellGroupData[{
Cell["\<\
Variational Solution to the Dirichlet BVP for Possion's Equation in \
Cylindrical Coordinates given in Chapter 1 of Jackson. \
\>", "Section"],

Cell[CellGroupData[{

Cell["Variational Approximation", "Subsection"],

Cell[TextData[{
  "We want to solve the Poisson equation ",
  Cell[BoxData[
      \(TraditionalForm\`\[EmptyDownTriangle]\^2\)]],
  "\[Phi] = -g[r] in cylindrical coordinates subject to the Dirichlet \
boundary condition \[Phi][1] = 0.  As a first step, we set up a variational \
function \[Phi][r,a,b,c] that has a polynomial dependence on r and vanishes \
at r=1.  Since the electric field must vanish at r=0 for a cylindrically \
symmetric charge distribution (by Gauss's law), we also must require ",
  Cell[BoxData[
      \(TraditionalForm\`d\[Phi]\/dr\)]],
  "=0 at r=0. To achieve this, we omit the linear term in the polynomial. \
Here is our ansatz for the potential."
}], "Text"],

Cell[BoxData[
    \(\[Phi][r_] = \ 
      a\ r^2\  + b\ r^3\  + \ c\ r^4\  - \((a + b + c)\)\)], "Input",
  CellLabel->"In[2]:="],

Cell["\<\
As a second step, we evaluate the gradient term that occurs in the action \
integral\
\>", "Text"],

Cell[BoxData[
    \(d\[Phi][x_]\  = \ \ D[\[Phi][x], x]\)], "Input",
  CellLabel->"In[3]:="],

Cell["Now, we form the first part of the \"action\" integral", "Text"],

Cell[BoxData[
    \(kin = \ Integrate[r\ d\[Phi][r]^2/2, {r, 0, 1}]\)], "Input",
  CellLabel->"In[5]:="],

Cell[TextData[{
  "Next, we form the second term in the action integral. In the equation \
below, we adopt the notation\n en = ",
  Cell[BoxData[
      \(TraditionalForm\`\[Integral]g[
            r]\ \((r\^n - 1)\)\ r \[DifferentialD]r\)]],
  "."
}], "Text"],

Cell[BoxData[
    \(pot\  = \ a\ e2\  + \ b\ e3\  + \ c\ e4\)], "Input",
  CellLabel->"In[6]:="],

Cell["Here is the complete action integral", "Text"],

Cell[BoxData[
    \(S\  = \ kin - pot\)], "Input",
  CellLabel->"In[7]:="],

Cell["Now find the derivatives of S with respect to a, b, and c.", "Text"],

Cell[BoxData[
    \(da\  = \ D[S, a]\)], "Input",
  CellLabel->"In[8]:="],

Cell[BoxData[
    \(db\  = \ D[S, b]\)], "Input",
  CellLabel->"In[9]:="],

Cell[BoxData[
    \(dc\  = \ D[S, c]\)], "Input",
  CellLabel->"In[10]:="],

Cell["\<\
Now, we solve the system of equations dS/da =0, dS/db=0, dS/dc=0.\
\>", "Text"],

Cell[BoxData[
    \(Solve[{da \[Equal] 0, db \[Equal] 0, dc \[Equal] 0}, {a, b, c}] // 
      First\)], "Input",
  CellLabel->"In[11]:="],

Cell[BoxData[
    \(a\  = \ a /. \ %\)], "Input",
  CellLabel->"In[12]:="],

Cell[BoxData[
    \(b\  = \ b /. \ %%\)], "Input",
  CellLabel->"In[13]:="],

Cell[BoxData[
    \(c\ \  = \ c\  /. \ %%%\)], "Input",
  CellLabel->"In[14]:="],

Cell["\<\
Now, lets examine the solution with a specific choice for the inhomogeneous \
term g[r]\
\>", "Text"],

Cell[BoxData[
    \(g[r_] = \ \(-5\) \((1 - r)\)\  + \ 10^4\ r^5\ \((1 - r)\)^5\)], "Input",\

  CellLabel->"In[15]:="],

Cell["With this choice, we can evaluate e2, e3, e4 explicitly", "Text"],

Cell[BoxData[
    \(e2 = \ Integrate[\((r^2 - 1)\) r\ g[r], {r, 0, 1}]\)], "Input",
  CellLabel->"In[16]:="],

Cell[BoxData[
    \(e3 = \ Integrate[\((r^3 - 1)\) r\ g[r], {r, 0, 1}]\)], "Input",
  CellLabel->"In[17]:="],

Cell[BoxData[
    \(e4 = \ Integrate[\((r^4 - 1)\) r\ g[r], {r, 0, 1}]\)], "Input",
  CellLabel->"In[18]:="],

Cell["\<\
What is the final result for \[Phi][r]?\
\>", "Text"],

Cell[BoxData[
    \(\[Phi][r]\)], "Input",
  CellLabel->"In[19]:="],

Cell["\<\
Let's take a look!  In the figure below, \[Phi][r] is in red and g[r]/20 is \
in cyan.\
\>", "Text"],

Cell[BoxData[
    \(Plot[{\[Phi][r], g[r]/20}, {r, 0, 1}, PlotRange -> {\(-0.3\), 0.7}, 
      PlotStyle \[Rule] \ {{Thickness[0.007], 
            RGBColor[1, 0, 0]}, {Thickness[0.01], 
            RGBColor[0, 1, 1]}}]\)], "Input",
  CellLabel->"In[20]:="]
}, Open  ]],

Cell[CellGroupData[{

Cell["Exact Answer", "Subsection"],

Cell[TextData[{
  "In this case, we can easily determine the exact answer \[Psi][r] by \
solving the radial dfferential equation\n ",
  Cell[BoxData[
      \(TraditionalForm\`1\/r\)]],
  Cell[BoxData[
      \(TraditionalForm\`d\/dr\)]],
  "r ",
  Cell[BoxData[
      \(TraditionalForm\`d\[Psi]\/dr\)]],
  "= -g[r] \n subject to boundary conditions ",
  Cell[BoxData[
      \(TraditionalForm\`d\[Psi]\/dr\)]],
  "[0] = 0 and \[Psi][1] = 0.  A first integral gives  \[Psi]' = ",
  Cell[BoxData[
      \(TraditionalForm\`d\[Psi]\/dr\)]]
}], "Text"],

Cell[BoxData[
    \(\(\[Psi]'\)[
        r_]\  = \ \(-Expand[Integrate[x\ g[x], {x, 0, r}]/r]\)\)], "Input",
  CellLabel->"In[21]:="],

Cell["A second integral gives \[Psi][r]", "Text"],

Cell[BoxData[
    \(\[Psi][r_]\  = \ Integrate[\(\[Psi]'\)[x], {x, 1, r}]\)], "Input",
  CellLabel->"In[22]:="],

Cell["Are the boundary conditions satisfied?", "Text"],

Cell[BoxData[
    \(\[Psi][1]\)], "Input",
  CellLabel->"In[23]:="],

Cell[BoxData[
    \(D[\[Psi][r], r]\  /. \ r \[Rule] \ 0\)], "Input",
  CellLabel->"In[24]:="],

Cell["\<\
Let's compare the exact solution (black) with the vatiational approximation \
(red)\
\>", "Text"],

Cell[BoxData[
    \(Plot[{\[Psi][r], \[Phi][r], g[r]/20}, {r, 0, 1}, 
      PlotRange -> {\(-0.3\), 0.7}, 
      PlotStyle \[Rule] \ {{Thickness[0.007], 
            RGBColor[0, 0, 0]}, {Thickness[0.007], 
            RGBColor[1, 0, 0]}, {Thickness[0.01], 
            RGBColor[0, 1, 1]}}]\)], "Input",
  CellLabel->"In[25]:="],

Cell["\<\
Let's now take a look at the difference between the \[Phi][r] and \[Psi][r]\
\>", "Text"],

Cell[BoxData[
    \(Plot[\((\[Phi][r] - \[Psi][r])\), {r, 0, 1}, 
      PlotRange \[Rule] \ {\(-0.015\), 0.015}, 
      PlotStyle \[Rule] {Thickness[0.01], Hue[0.94]}]\)], "Input"],

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      StyleBox[\(Clear["\<`*\>"]\),
        "MR"], " "}]], "Input",
  CellLabel->"In[26]:="]
}, Open  ]]
}, Open  ]]
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