(************** Content-type: application/mathematica ************** CreatedBy='Mathematica 5.0' 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). 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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[ 125911, 4162]*) (*NotebookOutlinePosition[ 160535, 5311]*) (* CellTagsIndexPosition[ 160491, 5307]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["Solutions to Final Exam", "Title", TextAlignment->Center, TextJustification->0], Cell["Physics 604", "Subtitle", TextAlignment->Center, TextJustification->0], Cell["Dec. 15, 2003", "Subsubtitle", TextAlignment->Center, TextJustification->0], Cell["\<\ This CLOSED BOOK exam is to be completed in 2 hours. Choose THREE of the \ four problems \[LongDash] if you attempt all four, be sure to indicate \ clearly which three should be graded. You must explain your reasoning to \ receive full credit.\ \>", "Text"], Cell[CellGroupData[{ Cell["Initialization", "Subsubsection"], Cell[BoxData[{ \(\(ClearAll["\"];\)\), "\n", \(Off[General::spell, General::spell1]\)}], "Input", CellLabel->"In[58]:="], Cell[BoxData[{ \(\($DefaultFont = {"\", 12};\)\), "\n", \(\($TextStyle = {FontFamily \[Rule] "\", FontSize \[Rule] 12, FontSlant \[Rule] "\"};\)\)}], "Input", CellLabel->"In[60]:="], Cell[BoxData[{ \(\(Needs["\"];\)\), "\n", \(\(Needs["\"];\)\)}], "Input", CellLabel->"In[62]:="] }, Open ]], Cell[CellGroupData[{ Cell["Problem 1 (points per part: 4, 4, 2)", "ExerciseTitle"], Cell[TextData[{ "Sound waves in a pipe of slowly varying cross section ", Cell[BoxData[ \(TraditionalForm\`A[x]\)]], " satisfy the equation" }], "ExerciseText"], Cell[BoxData[ \(TraditionalForm\`\(A[ x]\/c\^2\) \[PartialD]\^2 \[Psi][x, t]\/\[PartialD]t\^2\ - \ \ \(\[PartialD]\/\[PartialD]x\) \((A[ x] \[PartialD]\[Psi][x, t]\/\[PartialD]x)\)\ \[Equal] \ 0\)], "DisplayFormula"], Cell[TextData[{ "Consider an ", StyleBox["exponential horn", FontSlant->"Italic"], " with ", Cell[BoxData[ \(TraditionalForm\`A[x] \[Proportional] \[ExponentialE]\^\(a\ x\)\)]], ". " }], "ExerciseText"], Cell["a) Show that there exist travelling waves of the form", "ExerciseText"], Cell[BoxData[ \(TraditionalForm\`\[Psi][x, t]\ = \ \(\[ExponentialE]\^\(\(-b\)\ x\)\) Cos[k\ x\ - \ \[Omega]\ t]\)], "DisplayFormula"], Cell[TextData[{ "where ", Cell[BoxData[ \(TraditionalForm\`b\)]], " is independent of ", Cell[BoxData[ \(TraditionalForm\`\[Omega]\)]], " and determine the dispersion relation ", Cell[BoxData[ \(TraditionalForm\`k[\[Omega]]\)]], ". " }], "ExerciseText"], Cell["\<\ b) Describe this solution. What happens for low frequencies? Develop and \ describe an alternative solution, if needed, for low frequencies.\ \>", "ExerciseText"], Cell[TextData[{ "c) Evaluate and sketch the phase and group velocities for propagating \ waves as functions of either ", Cell[BoxData[ \(TraditionalForm\`k\)]], " or ", Cell[BoxData[ \(TraditionalForm\`\[Omega]\)]], " (your choice). " }], "ExerciseText"], Cell[CellGroupData[{ Cell["a)", "Subsubsection"], Cell["\<\ We simply substitute the proposed solution into the wave equation to obtain\ \>", "Text"], Cell[BoxData[ \(TraditionalForm\`\(-\(\[Omega]\^2\/c\^2\)\) Exp[\((a - b)\) x] Cos[k\ x\ - \[Omega]\ t]\ + \ \(\[PartialD]\/\[PartialD]x\) \((\ Exp[\((a - b)\) x] \((b\ Cos[k\ x\ - \[Omega]\ t]\ + \ k\ Sin[k\ x\ - \[Omega]\ t])\)\ )\) \[Equal] \ 0\)], "DisplayFormula"], Cell["or", "Text"], Cell[BoxData[ \(TraditionalForm\`\(-\(\[Omega]\^2\/c\^2\)\) Cos[k\ x\ - \[Omega]\ t]\ + \ \((a - b)\) \((b\ Cos[k\ x\ - \[Omega]\ t] + k\ Sin[k\ x\ - \[Omega]\ t])\)\ + \ \((\(-k\)\ b\ Sin[ k\ x\ - \[Omega]\ t]\ + \ \(k\^2\) Cos[k\ x\ - \[Omega]\ t])\)\ \[Equal] \ 0\)], "DisplayFormula"], Cell["\<\ Collecting the coefficients of the sine and cosine functions provides the two \ equations\ \>", "Text"], Cell[BoxData[{ \(TextForm\`\(-\(\[Omega]\^2\/c\^2\)\) + \((a - b)\) b\ + \ k\^2\ \[Equal] \ 0\), "\[IndentingNewLine]", \(TextForm\`\((a - b)\) k\ - \ k\ b\ \[Equal] \ 0\)}], "DisplayFormula"], Cell["such that", "Text"], Cell[BoxData[{ \(TextForm\`b = a\/2\), "\[IndentingNewLine]", \(TextForm\`\(-\(\[Omega]\^2\/c\^2\)\) + a\^2\/4 + k\^2\ \[Equal] \ 0\ \ \[DoubleLongRightArrow]\ \ \ k\ = \ \@\(\[Omega]\^2\/c\^2 - \ a\^2\/4\)\)}], "DisplayFormula"], Cell[TextData[{ "This solution is an attenuated travelling wave where the attenuation \ length, ", Cell[BoxData[ \(TraditionalForm\`b\^\(-1\)\)]], ", is independent of frequency. Hence, the balance between high and \ moderately low frequencies is preserved as the sound propagates. " }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["b)", "Subsubsection"], Cell[TextData[{ "However, there is a low-frequency cut-off, ", Cell[BoxData[ \(TraditionalForm\`\[Omega]\_c = \(a\ c/2 = b\ c\)\)]], ", below which waves do not propagate because ", Cell[BoxData[ \(TraditionalForm\`k\)]], " becomes complex. For very low frequencies, we substitute a trial \ solution of the form" }], "Text"], Cell[BoxData[ \(TraditionalForm\`\[Psi][x, t]\ = \ \(\[ExponentialE]\^\(\(-b\)\ x\)\) Cos[\[Omega]\ t]\ \ \ \[DoubleLongRightArrow]\ \ \ \ \(-\(\[Omega]\^2\/c\^2\)\) + \ b(a - b)\ \[Equal] \ 0\)], "DisplayFormula"], Cell["to obtain", "Text"], Cell[BoxData[ \(TraditionalForm\`b\ = \ a\/2\ \[PlusMinus] \ \@\(a\^2\/4 - \[Omega]\^2\/c\^2\)\)], \ "DisplayFormula"], Cell[TextData[{ "for standing waves. Both damping coefficients are positive and must be \ retained, but after some distance the smaller one will dominate and \ determines the maximum penetration of low-frequencies vibrations into the \ horn. Notice that at ", Cell[BoxData[ \(TraditionalForm\`\[Omega] \[Rule] 0\)]], " there is a static solution with ", Cell[BoxData[ \(TraditionalForm\`b \[Rule] 0\)]], " that is not attenuated." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["c)", "Subsubsection"], Cell["The phase and group velocities are given by", "Text"], Cell[BoxData[{ \(TextForm\`v\_p = \(\[Omega]\/k\ = \ c \@\( 1 + a\^2\/\(4 k\^2\)\)\)\), "\[IndentingNewLine]", \(TextForm\`\[Omega]\ \[DifferentialD]\[Omega]\ \[Equal] \ \(c\^2\) k\ \[DifferentialD]k\ \ \[DoubleLongRightArrow]\ \ v\_g = \(\ \[DifferentialD]\[Omega]\/\[DifferentialD]k = c\^2\/v\_p\)\)}], "DisplayFormula"], Cell["such that", "Text"], Cell[BoxData[ \(TextForm\`\(v\_p\) v\_g = c\^2\)], "DisplayFormula"], Cell[TextData[{ "These velocities are sketched below as functions of ", Cell[BoxData[ \(TraditionalForm\`k/a\)]], ". 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-9.67027, 0.0410565, 0.0739016}, {{236.125, 266.563}, {56.4375, \ 39.5}} -> {-9.12981, -21.4997, 0.0478916, 0.15498}, {{236.125, 266.563}, \ {36.8125, 19.875}} -> {-9.12981, -23.9329, 0.0478916, 0.15498}}] }, Open ]] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Problem 2 (10 points)", "ExerciseTitle"], Cell["Use the Laplace transform to solve the initial-value problem", \ "ExerciseText"], Cell[BoxData[ \(TextForm\`\(y\^\[Prime]\)[ t]\ + \ \[Integral]\_0\%t y[\[Tau]] \[DifferentialD]\[Tau]\ \[Equal] \ \[ExponentialE]\^\ \(-t\)\ , \ y[0] \[Equal] 1\)], "DisplayFormula"], Cell[CellGroupData[{ Cell["solution", "Subsubsection"], Cell["\<\ The Laplace transform of this integro-differential equation reads\ \>", "Text"], Cell[BoxData[ \(TextForm\`s\ y\&~\ - 1\ + \ \(s\^\(-1\)\) y\&~\ \[Equal] \ \((s + 1)\)\^\(-1\)\ \ \[DoubleLongRightArrow]\ \ \ y\&~\ = \ \(s \((s + 2)\)\)\/\(\((s + 1)\) \((s\^2 + 1)\)\)\)], \ "DisplayFormula"], Cell["\<\ The solution transform is resolved into partial fractions according to\ \>", "Text"], Cell[BoxData[ \(TextForm\`A\/\(s + 1\) + B\/\(s + \[ImaginaryI]\) + C\/\(s - \[ImaginaryI]\) \[Equal] \(\(s \((s + 2)\)\)\/\(\((s + 1)\) \((s\^2 + 1)\)\)\)\ \ \[DoubleLongRightArrow]\ A \((s\^2 + 1)\) + B \((s + 1)\) \((s - \[ImaginaryI])\) + C \((s + 1)\) \((s + \[ImaginaryI])\) \[Equal] s\^2 + 2 s\)], "DisplayFormula"], Cell["which yields the system of equations", "Text"], Cell[BoxData[{ \(TextForm\`A + B + C \[Equal] 1\), "\[IndentingNewLine]", \(TextForm\`B \((1 - \[ImaginaryI])\) + C \((1 + \[ImaginaryI])\) \[Equal] 2\), "\[IndentingNewLine]", \(TextForm\`A - \[ImaginaryI]\ B + \[ImaginaryI]\ C\ \[Equal] 0\)}], "DisplayFormula"], Cell["The second gives", "Text"], Cell[BoxData[ \(TextForm\`B\ = \ \(2 - \((1 + \[ImaginaryI])\) C\)\/\(1 - \ \[ImaginaryI]\)\)], "DisplayFormula"], Cell["and substitution into the third gives", "Text"], Cell[BoxData[ \(TextForm\`\((1 - \[ImaginaryI])\) A\ - 2 \[ImaginaryI]\ + \ \[ImaginaryI]\ \((1 + \[ImaginaryI])\) C + \[ImaginaryI] \((1 - \[ImaginaryI])\) C \[Equal] 0\ \ \[DoubleLongRightArrow]\ \ \ A\ = \ \(\(2 \[ImaginaryI]\)\/\(1 \ - \[ImaginaryI]\)\) \((1 - C)\)\)], "DisplayFormula"], Cell["Finally, substitution into the first equation gives", "Text"], Cell[BoxData[ \(TextForm\`2 \[ImaginaryI] \((1 - C)\)\ + \ 2\ - \((1 - \[ImaginaryI])\) C + \((1 - \[ImaginaryI])\) C \[Equal] 0\ \ \[DoubleLongRightArrow]\ \ \ C\ = \ \(\(1 + 3 \ \[ImaginaryI]\)\/\(4 \[ImaginaryI]\) = \(\(\ \)\(3 - \ \[ImaginaryI]\)\)\/4\)\)], "DisplayFormula"], Cell["such that", "Text"], Cell[BoxData[{ \(TextForm\`A\ = \ \(\(\(2 \[ImaginaryI]\)\/\(1 - \[ImaginaryI]\)\) \((\ \(1 + \[ImaginaryI]\)\/4)\)\ = \ \(\(\[ImaginaryI]\/4\) \((1 + \ \[ImaginaryI])\)\^2\ = \ \(-\(1\/2\)\)\)\)\), "\[IndentingNewLine]", \(TextForm\`B\ = \ \(\(8 - \((1 + \[ImaginaryI])\) \((3 - \[ImaginaryI])\ \)\)\/\(4 \((1 - \[ImaginaryI])\)\)\ = \ \(\(8 - \((4 + 2 \ \[ImaginaryI])\)\)\/\(4 \((1 - \[ImaginaryI])\)\)\ = \ \(\(\((4 - 2 \ \[ImaginaryI])\) \((1 + \[ImaginaryI])\)\)\/8\ = \ \(3 + \ \[ImaginaryI]\)\/4\)\)\)\)}], "DisplayFormula"], Cell["Thus,", "Text"], Cell[BoxData[ \(TextForm\`y\&~\ = \ \(\(s \((s + 2)\)\)\/\(\((s + 1)\) \((s\^2 + \ 1)\)\)\ = \ \(-\(1\/2\)\) 1\/\(s + 1\)\ + \ \(\(3 + \[ImaginaryI]\)\/4\) 1\/\(s + \[ImaginaryI]\)\ + \ \(\(\(\ \)\(3 - \ \[ImaginaryI]\)\)\/4\) 1\/\(s - \[ImaginaryI]\)\)\)], "DisplayFormula"], Cell["can be inverted using", "Text"], Cell[BoxData[ \(TextForm\`\(1\/\(s + \[Alpha]\)\)\ \[LongRightArrow]\ \[ExponentialE]\^\ \(\(-\[Alpha]\)\ t\)\ \ \ \[DoubleLongRightArrow]\ \ y[ t]\ = \ \(-\(1\/2\)\) \[ExponentialE]\^\(-t\)\ + \ \(\(3 + \ \[ImaginaryI]\)\/4\) \[ExponentialE]\^\(\(-\[ImaginaryI]\)\ t\)\ + \ \(\(\(\ \ \)\(3 - \[ImaginaryI]\)\)\/4\) \[ExponentialE]\^\(\[ImaginaryI]\ t\)\)], \ "DisplayFormula"], Cell["Therefore, we obtain", "Text"], Cell[BoxData[ \(TextForm\`y[ t]\ = \ \(1\/2\) \((3 Cos[t] + Sin[t] - \[ExponentialE]\^\(-t\))\)\)], "DisplayFormula"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Problem 3 (10 points)", "ExerciseTitle"], Cell["Consider an integral of the form", "ExerciseText"], Cell[BoxData[ \(\[Integral]\_0\%\[Infinity] g[x] \[DifferentialD]x\)], "DisplayFormula",\ FontFamily->"Times New Roman"], Cell[TextData[{ "where ", Cell[BoxData[ \(TraditionalForm\`g[x]\)]], " is well-behaved on the positive real axis but is not symmetric with \ respect to the sign of ", Cell[BoxData[ \(TraditionalForm\`x\)]], ", such that the integration interval cannot be extended for use with a \ great semicircle. If the corresponding ", Cell[BoxData[ \(TraditionalForm\`g[z]\)]], " is a meromorphic function that decreases sufficiently rapidly for ", Cell[BoxData[ \(TraditionalForm\`z \[Rule] \[Infinity]\)]], " and is sufficiently small at the origin, one can often use ", Cell[BoxData[ \(TraditionalForm\`f[z] = g[z] Log[z]\)]], " with a branch cut on the positive real axis and a ", StyleBox["PacMan", FontSlant->"Italic"], " contour of the form shown below. Apply this method for the following \ integral." }], "ExerciseText"], Cell[BoxData[ FormBox[ RowBox[{Cell[TextData[StyleBox["", "Text"]]], FormBox[\(\[Integral]\_0\%\[Infinity]\[DifferentialD]x\/\(\((x + 1)\) \ \((x\^2 + 2 x + 2)\)\)\), "TraditionalForm"], " "}], TextForm]], "DisplayFormula"], Cell["\<\ For this integral one could resolve the integrand into partial fractions \ instead, but to obtain full credit you must employ the method described \ above, which can be useful for more complicated problems that are not \ suitable for partial-fraction decomposition. 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Cell[BoxData[ \(z = R\ \[ExponentialE]\^\(\[ImaginaryI]\ \[Theta]\)\ \ \ \ \[DoubleLongRightArrow]\ \ \[LeftBracketingBar]f[ z]\ \[DifferentialD]z\/\[DifferentialD]\[Theta]\ \[RightBracketingBar] \[TildeEqual] \ \[LeftBracketingBar]Log[R] + \ \[ImaginaryI]\ \[Theta]\[RightBracketingBar]\/R\^2\ \[Rule] \ 0\)], "DisplayFormula", FontFamily->"Times New Roman"], Cell["while near the origin", "Text"], Cell[BoxData[ \(z = \[CurlyEpsilon]\ \[ExponentialE]\^\(\[ImaginaryI]\ \[Theta]\)\ \ \ \ \[DoubleLongRightArrow]\ \[LeftBracketingBar]f[ z] \[DifferentialD]z\/\[DifferentialD]\[Theta]\ \[RightBracketingBar]\ \[TildeTilde] \ \[LeftBracketingBar]\[CurlyEpsilon] \ \((Log[\[CurlyEpsilon]] + \[ImaginaryI]\ \[Theta])\)\[RightBracketingBar]\ \ \[Rule] \ 0\)], "DisplayFormula", FontFamily->"Times New Roman"], Cell["Thus, we can neglect both circular portions and obtain", "Text"], Cell[BoxData[ \(TextForm\`\[Integral]\_0\%\[Infinity]\[DifferentialD]x\/\(\((x + 1)\) \ \((x\^2 + 2 x + 2)\)\)\ = \ \(-\(1\/\(2 \[Pi]\ \[ImaginaryI]\)\)\) \(\ \[ContourIntegral]\_C\( Log[ z]\/\(\((z + 1)\) \((z + 1 + \[ImaginaryI])\) \((z + 1 - \[ImaginaryI])\)\)\) \[DifferentialD]z\)\)], \ "DisplayFormula"], Cell[TextData[{ "The right-hand side can now be evaluated as the sum of the residues for ", Cell[BoxData[ \(TraditionalForm\`z = \(-1\), \(-1\) - \[ImaginaryI], \(-1\) + \ \[ImaginaryI]\)]], ", such that" }], "Text"], Cell[BoxData[{ \(TextForm\`\[Integral]\_0\%\[Infinity]\[DifferentialD]x\/\(\((x + 1)\) \ \((x\^2 + 2 x + 2)\)\)\ \ \[AlignmentMarker] = \ \(-\((Log[\(-1\)]\/\(\((\ \[ImaginaryI])\) \((\(-\[ImaginaryI]\))\)\)\ + \ Log[\(-1\) - \[ImaginaryI]]\/\(\((\(-\[ImaginaryI]\))\) \((\(-2\) \ \[ImaginaryI])\)\)\ + \ Log[\(-1\) + \[ImaginaryI]]\/\(\((\[ImaginaryI])\) \((2 \ \[ImaginaryI])\)\))\)\)\ \), "\[IndentingNewLine]", \(TextForm\`\[AlignmentMarker]\(\(=\)\(\ \)\(-\((\[ImaginaryI]\ \[Pi]\ - \ \(1\/2\) \((Log[\@2] + \(\(5 \[Pi]\)\/4\) \[ImaginaryI])\) - \(1\/2\) \ \((Log[\@2] + \(\(3 \[Pi]\)\/4\) \[ImaginaryI])\))\)\)\)\)}], \ "DisplayFormula", TextAlignment->AlignmentMarker], Cell["Therefore,", "Text"], Cell[BoxData[ \(TextForm\`\[Integral]\_0\%\[Infinity]\[DifferentialD]x\/\(\((x + 1)\) \ \((x\^2 + 2 x + 2)\)\)\ = \ Log[\@2]\)], "DisplayFormula", TextAlignment->AlignmentMarker] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Problem 4 (5 points per part)", "ExerciseTitle"], Cell[TextData[{ "The curved surface of a cylinder of radius ", Cell[BoxData[ \(TraditionalForm\`a\)]], " is grounded while the endcaps at ", Cell[BoxData[ \(TraditionalForm\`z = \(\[PlusMinus]L\)/2\)]], " are maintained at opposite potentials ", Cell[BoxData[ \(TraditionalForm\`\[Psi][\[Xi], \[Phi], \(\[PlusMinus]L\)/ 2] = \(\[PlusMinus]V[\[Xi], \[Phi]]\)\)]], "." }], "ExerciseText"], Cell[TextData[{ "a) Develop an expansion for the electrostatic potential ", Cell[BoxData[ \(TraditionalForm\`\[Psi][\[Xi], \[Phi], z]\)]], " within the cylinder and express the coefficients in terms of the \ appropriate integral over ", Cell[BoxData[ \(TraditionalForm\`V[\[Xi], \[Phi]]\)]], ".\nb) Determine the coefficients for the simple case ", Cell[BoxData[ \(TraditionalForm\`V[\[Xi], \[Phi]]\ = \ V\_0\)]], " where ", Cell[BoxData[ \(TraditionalForm\`V\_0\)]], " is constant." }], "ExerciseText"], Cell["Possibly useful information:", "ExerciseText"], Cell[BoxData[ \(TextForm\`\[Del]\^2\(\(=\)\(\ \)\(\(1\/\[Xi]\) \(\[PartialD]\/\ \[PartialD]\[Xi]\) \((\[Xi]\ \[PartialD]\/\[PartialD]\[Xi])\)\ + \ \ \(1\/\[Xi]\^2\) \[PartialD]\^2\/\[PartialD]\[Phi]\^2\ + \ \[PartialD]\^2\/\ \[PartialD]z\^2\)\)\)], "DisplayFormula"], Cell[BoxData[ \(TextForm\`\(\(\ \ \)\(\[DifferentialD]\^2 R\/\[DifferentialD]\[Xi]\^2 + 1\/\[Xi]\ \[DifferentialD]R\/\[DifferentialD]\[Xi]\ + \((k\^2 - \ m\^2\/\[Xi]\^2)\) R\ \[Equal] \ 0\ \ \[DoubleLongRightArrow]\ \ R\ = \ A\ J\_m[k\ \[Xi]]\ + \ B\ N\_m[k\ \[Xi]]\)\)\)], "DisplayFormula"], Cell[BoxData[ \(TextForm\`J\_m[\(k\_\(m, n\)\) a] = \(0\ \ \ \[DoubleLongRightArrow]\ \ \(\[Integral]\_0\%a\( \ J\_m[\(k\_\(m, n\)\) \[Xi]]\^2\) \[Xi] \[DifferentialD]\[Xi]\)\ = \ \ \(a\^2\/2\) J\_m\%\[Prime][\(k\_\(m, n\)\) a]\^2\)\)], "DisplayFormula"], Cell[BoxData[ \(TextForm\`m \[GreaterEqual] 1\ \ \[DoubleLongRightArrow]\ \ \(\[DifferentialD]\/\[DifferentialD]x\ \) \((\(x\^m\) J\_m[x])\)\ = \ \(x\^m\) J\_\(m - 1\)[x]\)], "DisplayFormula"], Cell[CellGroupData[{ Cell["a)", "Subsubsection"], Cell["The Laplacian takes the form", "Text"], Cell[BoxData[ \(TextForm\`\[Del]\^2\(\(=\)\(\ \)\(\(1\/\[Xi]\) \(\[PartialD]\/\ \[PartialD]\[Xi]\) \((\[Xi]\ \[PartialD]\/\[PartialD]\[Xi])\)\ + \ \ \(1\/\[Xi]\^2\) \[PartialD]\^2\/\[PartialD]\[Phi]\^2\ + \ \[PartialD]\^2\/\ \[PartialD]z\^2\)\)\)], "DisplayFormula"], Cell["in cylindrical coordinates. Laplace's equation", "Text"], Cell[BoxData[ \(TextForm\`\[Del]\^2 \[Psi]\ \[Equal] \ 0\ \ \ \ \ \ \ \ \ \ \[Psi][\[Xi], \[Phi], z]\ = \ R[\[Xi]] \[CapitalPhi][\[Phi]] Z[z]\)], "DisplayFormula"], Cell["with boundary conditions", "Text"], Cell[BoxData[{ \(TextForm\`R[0]\ finite, \ \ R[a] \[Equal] 0\), "\[IndentingNewLine]", \(TextForm\`\[CapitalPhi][\[Phi] + 2 \[Pi]]\ \[Equal] \ \[CapitalPhi][\[Phi]]\), \ "\[IndentingNewLine]", \(TextForm\`Z[\(-\(L\/2\)\)]\ \[Equal] \ \(-Z[ L\/2]\)\)}], "DisplayFormula"], Cell["then separates into", "Text"], Cell[BoxData[ \(TextForm\`\(1\/\[CapitalPhi]\) \[DifferentialD]\^2 \[CapitalPhi]\/\ \[DifferentialD]\[Phi]\^2\ \[Equal] \ \(-m\^2\ \ \ \ \ \ \[DoubleLongRightArrow]\ \ \ \ \[CapitalPhi]\)\ = \ \[ExponentialE]\^\(\ \[ImaginaryI]\ m\ \[Phi]\)\ , \ \ m = 0, \ \(\[PlusMinus]1\), \ \(\[PlusMinus]\ 2\), \ \[CenterEllipsis]\)], "DisplayFormula"], Cell[BoxData[ \(TextForm\`\(\(\(1\/Z\) \[DifferentialD]\^2 Z\/\[DifferentialD]z\^2\ \ \[Equal] \ k\^2\ \ \ \ \ \[DoubleLongRightArrow]\ \ \ \ Z\)\(\ \)\(=\)\(\ \ \)\(Sinh[k\ z]\)\(\ \)\)\)], "DisplayFormula"], Cell[BoxData[ \(TextForm\`\(\(\ \ \)\(\[DifferentialD]\^2 R\/\[DifferentialD]\[Xi]\^2 + 1\/\[Xi]\ \[DifferentialD]R\/\[DifferentialD]\[Xi]\ + \((k\^2 - \ m\^2\/\[Xi]\^2)\) R\ \[Equal] \ 0\ \ \[DoubleLongRightArrow]\ \ R\ = \ J\_m[k\ \[Xi]]\)\)\)], "DisplayFormula"], Cell[TextData[{ "where ", Cell[BoxData[ \(TraditionalForm\`J\_m\)]], " is the Bessel function of the first kind, the one that is finite at the \ origin. 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