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Each paragraph in bold is a \ ", StyleBox["Mathematica", FontSlant->"Italic"], " command. Click anywhere on the command, and press Ctrl+Enter to execute \ that command. If you do this for all of the ", StyleBox["Mathematica", FontSlant->"Italic"], " commands in order, ", StyleBox["Mathematica", FontSlant->"Italic"], " will provide the proofs of the main results of the paper.\n\nExterior \ matrices convert problems involving the determinant of a product of matrices \ to just a product of matrices. Thus, a problem of the form\n\n", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"\[VerticalSeparator]", RowBox[{ RowBox[{"(", GridBox[{ {"*", "*", "*", "*"}, {"*", "*", "*", "*"} }], ")"}], RowBox[{"(", GridBox[{ {"*", "*", "*", "*"}, {"*", "*", "*", "*"}, {"*", "*", "*", "*"}, {"*", "*", "*", "*"} }], ")"}], RowBox[{"(", GridBox[{ {"*", "*", "*", "*"}, {"*", "*", "*", "*"}, {"*", "*", "*", "*"}, {"*", "*", "*", "*"} }], ")"}], RowBox[{"\[CenterEllipsis]", "(", GridBox[{ {"*", "*", "*", "*"}, {"*", "*", "*", "*"}, {"*", "*", "*", "*"}, {"*", "*", "*", "*"} }], ")"}], RowBox[{"(", GridBox[{ {"*", "*"}, {"*", "*"}, {"*", "*"}, {"*", "*"} }], ")"}]}], "\[VerticalSeparator]"}], " ", "=", " ", "0"}], TraditionalForm]]], "\n\nis converted to a problem of the form\n\n", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{"(", GridBox[{ {"*", "*", "*", "*", "*", "*"} }], ")"}], RowBox[{"(", GridBox[{ {"*", "*", "*", "*", "*", "*"}, {"*", "*", "*", "*", "*", "*"}, {"*", "*", "*", "*", "*", "*"}, {"*", "*", "*", "*", "*", "*"}, {"*", "*", "*", "*", "*", "*"}, {"*", "*", "*", "*", "*", "*"} }], ")"}], RowBox[{"(", GridBox[{ {"*", "*", "*", "*", "*", "*"}, {"*", "*", "*", "*", "*", "*"}, {"*", "*", "*", "*", "*", "*"}, {"*", "*", "*", "*", "*", "*"}, {"*", "*", "*", "*", "*", "*"}, {"*", "*", "*", "*", "*", "*"} }], ")"}], "\[CenterEllipsis]", RowBox[{"(", GridBox[{ {"*", "*", "*", "*", "*", "*"}, {"*", "*", "*", "*", "*", "*"}, {"*", "*", "*", "*", "*", "*"}, {"*", "*", "*", "*", "*", "*"}, {"*", "*", "*", "*", "*", "*"}, {"*", "*", "*", "*", "*", "*"} }], ")"}], RowBox[{"(", GridBox[{ {"*"}, {"*"}, {"*"}, {"*"}, {"*"}, {"*"} }], ")"}]}], "=", "0."}], TraditionalForm]]], "\n\nWe eliminate the determinant, but we have to use slightly larger \ matrices. \n\nThe exterior matrix of a 4 by 4 matrix is a 6 by 6 matrix \ consisting of all of the determinants of 2 by 2 submatrices. Here is the \ definition:" }], "Text"], Cell[BoxData[ \(Exterior[ M_]\ := \ {{\(M[\([1]\)]\)[\([1]\)]\ \(M[\([2]\)]\)[\([2]\)]\ - \ \ \(M[\([1]\)]\)[\([2]\)]\ \(M[\([2]\)]\)[\([1]\)], \ \[IndentingNewLine]\(M[\([1]\)]\)[\([1]\)]\ \(M[\([2]\)]\)[\([3]\)]\ - \ \ \(M[\([1]\)]\)[\([3]\)]\ \(M[\([2]\)]\)[\([1]\)], \ \[IndentingNewLine]\(M[\([1]\)]\)[\([1]\)]\ \(M[\([2]\)]\)[\([4]\)]\ - \ \ \(M[\([1]\)]\)[\([4]\)]\ \(M[\([2]\)]\)[\([1]\)], \ \[IndentingNewLine]\(M[\([1]\)]\)[\([3]\)]\ \(M[\([2]\)]\)[\([4]\)]\ - \ \ \(M[\([1]\)]\)[\([4]\)]\ \(M[\([2]\)]\)[\([3]\)], \ \[IndentingNewLine]\(M[\([1]\)]\)[\([4]\)]\ \(M[\([2]\)]\)[\([2]\)]\ - \ \ \(M[\([1]\)]\)[\([2]\)]\ \(M[\([2]\)]\)[\([4]\)], \ \[IndentingNewLine]\(M[\([1]\)]\)[\([2]\)]\ \(M[\([2]\)]\)[\([3]\)]\ - \ \ \(M[\([1]\)]\)[\([3]\)]\ \(M[\([2]\)]\)[\([2]\)]}, \ \[IndentingNewLine]{\(M[\([1]\)]\)[\([1]\)]\ \(M[\([3]\)]\)[\([2]\)]\ - \ \ \(M[\([1]\)]\)[\([2]\)]\ \(M[\([3]\)]\)[\([1]\)], \ \[IndentingNewLine]\(M[\([1]\)]\)[\([1]\)]\ \(M[\([3]\)]\)[\([3]\)]\ - \ \ \(M[\([1]\)]\)[\([3]\)]\ \(M[\([3]\)]\)[\([1]\)], \ \[IndentingNewLine]\(M[\([1]\)]\)[\([1]\)]\ \(M[\([3]\)]\)[\([4]\)]\ - \ \ \(M[\([1]\)]\)[\([4]\)]\ \(M[\([3]\)]\)[\([1]\)], \ \[IndentingNewLine]\(M[\([1]\)]\)[\([3]\)]\ \(M[\([3]\)]\)[\([4]\)]\ - \ \ \(M[\([1]\)]\)[\([4]\)]\ \(M[\([3]\)]\)[\([3]\)], \ \[IndentingNewLine]\(M[\([1]\)]\)[\([4]\)]\ \(M[\([3]\)]\)[\([2]\)]\ - \ \ \(M[\([1]\)]\)[\([2]\)]\ \(M[\([3]\)]\)[\([4]\)], \ \[IndentingNewLine]\(M[\([1]\)]\)[\([2]\)]\ \(M[\([3]\)]\)[\([3]\)]\ - \ \ \(M[\([1]\)]\)[\([3]\)]\ \(M[\([3]\)]\)[\([2]\)]}, \ \[IndentingNewLine]{\(M[\([1]\)]\)[\([1]\)]\ \(M[\([4]\)]\)[\([2]\)]\ - \ \ \(M[\([1]\)]\)[\([2]\)]\ \(M[\([4]\)]\)[\([1]\)], \ \[IndentingNewLine]\(M[\([1]\)]\)[\([1]\)]\ \(M[\([4]\)]\)[\([3]\)]\ - \ \ \(M[\([1]\)]\)[\([3]\)]\ \(M[\([4]\)]\)[\([1]\)], \ \[IndentingNewLine]\(M[\([1]\)]\)[\([1]\)]\ \(M[\([4]\)]\)[\([4]\)]\ - \ \ \(M[\([1]\)]\)[\([4]\)]\ \(M[\([4]\)]\)[\([1]\)], \ \[IndentingNewLine]\(M[\([1]\)]\)[\([3]\)]\ \(M[\([4]\)]\)[\([4]\)]\ - \ \ \(M[\([1]\)]\)[\([4]\)]\ \(M[\([4]\)]\)[\([3]\)], \ \[IndentingNewLine]\(M[\([1]\)]\)[\([4]\)]\ \(M[\([4]\)]\)[\([2]\)]\ - \ \ \(M[\([1]\)]\)[\([2]\)]\ \(M[\([4]\)]\)[\([4]\)], \ \[IndentingNewLine]\(M[\([1]\)]\)[\([2]\)]\ \(M[\([4]\)]\)[\([3]\)]\ - \ \ \(M[\([1]\)]\)[\([3]\)]\ \(M[\([4]\)]\)[\([2]\)]}, \ \[IndentingNewLine]{\(M[\([3]\)]\)[\([1]\)]\ \(M[\([4]\)]\)[\([2]\)]\ - \ \ \(M[\([3]\)]\)[\([2]\)]\ \(M[\([4]\)]\)[\([1]\)], \ \[IndentingNewLine]\(M[\([3]\)]\)[\([1]\)]\ \(M[\([4]\)]\)[\([3]\)]\ - \ \ \(M[\([3]\)]\)[\([3]\)]\ \(M[\([4]\)]\)[\([1]\)], \ \[IndentingNewLine]\(M[\([3]\)]\)[\([1]\)]\ \(M[\([4]\)]\)[\([4]\)]\ - \ \ \(M[\([3]\)]\)[\([4]\)]\ \(M[\([4]\)]\)[\([1]\)], \ \[IndentingNewLine]\(M[\([3]\)]\)[\([3]\)]\ \(M[\([4]\)]\)[\([4]\)]\ - \ \ \(M[\([3]\)]\)[\([4]\)]\ \(M[\([4]\)]\)[\([3]\)], \ \[IndentingNewLine]\(M[\([3]\)]\)[\([4]\)]\ \(M[\([4]\)]\)[\([2]\)]\ - \ \ \(M[\([3]\)]\)[\([2]\)]\ \(M[\([4]\)]\)[\([4]\)], \ \[IndentingNewLine]\(M[\([3]\)]\)[\([2]\)]\ \(M[\([4]\)]\)[\([3]\)]\ - \ \ \(M[\([3]\)]\)[\([3]\)]\ \(M[\([4]\)]\)[\([2]\)]}, \ \[IndentingNewLine]{\(M[\([4]\)]\)[\([1]\)]\ \(M[\([2]\)]\)[\([2]\)]\ - \ \ \(M[\([4]\)]\)[\([2]\)]\ \(M[\([2]\)]\)[\([1]\)], \ \[IndentingNewLine]\(M[\([4]\)]\)[\([1]\)]\ \(M[\([2]\)]\)[\([3]\)]\ - \ \ \(M[\([4]\)]\)[\([3]\)]\ \(M[\([2]\)]\)[\([1]\)], \ \[IndentingNewLine]\(M[\([4]\)]\)[\([1]\)]\ \(M[\([2]\)]\)[\([4]\)]\ - \ \ \(M[\([4]\)]\)[\([4]\)]\ \(M[\([2]\)]\)[\([1]\)], \ \[IndentingNewLine]\(M[\([4]\)]\)[\([3]\)]\ \(M[\([2]\)]\)[\([4]\)]\ - \ \ \(M[\([4]\)]\)[\([4]\)]\ \(M[\([2]\)]\)[\([3]\)], \ \[IndentingNewLine]\(M[\([4]\)]\)[\([4]\)]\ \(M[\([2]\)]\)[\([2]\)]\ - \ \ \(M[\([4]\)]\)[\([2]\)]\ \(M[\([2]\)]\)[\([4]\)], \ \[IndentingNewLine]\(M[\([4]\)]\)[\([2]\)]\ \(M[\([2]\)]\)[\([3]\)]\ - \ \ \(M[\([4]\)]\)[\([3]\)]\ \(M[\([2]\)]\)[\([2]\)]}, \ \[IndentingNewLine]{\(M[\([2]\)]\)[\([1]\)]\ \(M[\([3]\)]\)[\([2]\)]\ - \ \ \(M[\([2]\)]\)[\([2]\)]\ \(M[\([3]\)]\)[\([1]\)], \ \[IndentingNewLine]\(M[\([2]\)]\)[\([1]\)]\ \(M[\([3]\)]\)[\([3]\)]\ - \ \ \(M[\([2]\)]\)[\([3]\)]\ \(M[\([3]\)]\)[\([1]\)], \ \[IndentingNewLine]\(M[\([2]\)]\)[\([1]\)]\ \(M[\([3]\)]\)[\([4]\)]\ - \ \ \(M[\([2]\)]\)[\([4]\)]\ \(M[\([3]\)]\)[\([1]\)], \ \[IndentingNewLine]\(M[\([2]\)]\)[\([3]\)]\ \(M[\([3]\)]\)[\([4]\)]\ - \ \ \(M[\([2]\)]\)[\([4]\)]\ \(M[\([3]\)]\)[\([3]\)], \ \[IndentingNewLine]\(M[\([2]\)]\)[\([4]\)]\ \(M[\([3]\)]\)[\([2]\)]\ - \ \ \(M[\([2]\)]\)[\([2]\)]\ \(M[\([3]\)]\)[\([4]\)], \ \[IndentingNewLine]\(M[\([2]\)]\)[\([2]\)]\ \(M[\([3]\)]\)[\([3]\)]\ - \ \ \(M[\([2]\)]\)[\([3]\)]\ \(M[\([3]\)]\)[\([2]\)]}}\)], "Input"], Cell["A typical matrix would be given as", "Text"], Cell[BoxData[ \(M\ = \ {{m11, m12, m13, m14}, {m21, m22, m23, m24}, {m31, m32, m33, m34}, {m41, m42, m43, m44}}\)], "Input"], Cell[BoxData[ \(MatrixForm[M]\)], "Input"], Cell[TextData[{ "Here is the exterior matrix of ", Cell[BoxData[ \(TraditionalForm\`M\)]], ", denoted by ", Cell[BoxData[ \(TraditionalForm\`ext(M)\)]], "." }], "Text"], Cell[BoxData[ \(P\ = \ Exterior[M]\)], "Input"], Cell[BoxData[ \(MatrixForm[P]\)], "Input"], Cell[TextData[{ "The key property of exterior matrices is that ", Cell[BoxData[ \(TraditionalForm\`ext( M\[CenterDot]A)\ = \ \(\(\(ext(M)\)\[CenterDot]\(ext( A)\)\)\(\ \)\)\)]], ". If we define another matrix ", Cell[BoxData[ \(TraditionalForm\`A\)]], "," }], "Text"], Cell[BoxData[ \(A\ = \ {{a11, a12, a13, a14}, {a21, a22, a23, a24}, {a31, a32, a33, a34}, {a41, a42, a43, a44}}\)], "Input"], Cell[TextData[{ "we can have ", StyleBox["Mathematica", FontSlant->"Italic"], " verify this fact." }], "Text"], Cell[BoxData[ \(Simplify[Exterior[M . A]\ - \ Exterior[M] . Exterior[A]]\)], "Input"], Cell["\<\ This proves Lemma 1 in the paper. We also can define the exterior matrix of a 4 by 2 or a 2 by 4 matrix. \ \>", "Text"], Cell[BoxData[ \(ExteriorF[ F_]\ := \ {{F[\([1, 1]\)]\ F[\([2, 2]\)]\ - \ F[\([1, 2]\)]\ F[\([2, 1]\)], \[IndentingNewLine]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ F[\([1, 1]\)]\ F[\([2, 3]\)]\ - \ F[\([1, 3]\)]\ F[\([2, 1]\)], \[IndentingNewLine]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ F[\([1, 1]\)]\ F[\([2, 4]\)]\ - \ F[\([1, 4]\)]\ F[\([2, 1]\)], \[IndentingNewLine]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ F[\([1, 3]\)]\ F[\([2, 4]\)]\ - \ F[\([1, 4]\)]\ F[\([2, 3]\)], \[IndentingNewLine]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ F[\([1, 4]\)]\ F[\([2, 2]\)]\ - \ F[\([1, 2]\)]\ F[\([2, 4]\)], \[IndentingNewLine]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ F[\([1, 2]\)] F[\([2, 3]\)]\ - \ F[\([1, 3]\)]\ F[\([2, 2]\)]}}\)], "Input"], Cell[BoxData[ \(ExteriorB[ B_]\ := \ {{B[\([1, 1]\)]\ B[\([2, 2]\)]\ - \ B[\([1, 2]\)] B[\([2, 1]\)]}, \[IndentingNewLine]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {B[\([1, 1]\)]\ B[\([3, 2]\)]\ - \ B[\([1, 2]\)] B[\([3, 1]\)]}, \[IndentingNewLine]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {B[\([1, 1]\)]\ B[\([4, 2]\)]\ - \ B[\([1, 2]\)] B[\([4, 1]\)]}, \[IndentingNewLine]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {B[\([3, 1]\)]\ B[\([4, 2]\)]\ - \ B[\([3, 2]\)] B[\([4, 1]\)]}, \[IndentingNewLine]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {B[\([4, 1]\)]\ B[\([2, 2]\)]\ - \ B[\([4, 2]\)] B[\([2, 1]\)]}, \[IndentingNewLine]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {B[\([2, 1]\)]\ B[\([3, 2]\)]\ - \ B[\([2, 2]\)] B[\([3, 1]\)]}}\)], "Input"], Cell["Then typical for typical matrices B and F,", "Text"], Cell[BoxData[ \(B\ = \ {{b11, b12}, {b21, b22}, {b31, b32}, {b41, b42}}\)], "Input"], Cell[BoxData[ \(MatrixForm[B]\)], "Input"], Cell[BoxData[ \(F\ = \ {{f11, f12, f13, f14}, {f21, f22, f23, f24}}\)], "Input"], Cell[BoxData[ \(MatrixForm[F]\)], "Input"], Cell["The exterior matrices for these are column and row matrices.", "Text"], Cell[BoxData[ \(ExteriorB[B]\)], "Input"], Cell[BoxData[ \(MatrixForm[%]\)], "Input"], Cell[BoxData[ \(ExteriorF[F]\)], "Input"], Cell[BoxData[ \(MatrixForm[%]\)], "Input"], Cell[TextData[{ "Now, we can compute the determinant of a product ", Cell[BoxData[ \(TraditionalForm\`det(F\[CenterDot]M\[CenterDot]B)\)]], " by computing the product of exterior matrices ", Cell[BoxData[ \(TraditionalForm\`\(ext(F)\)\[CenterDot]\(ext(M)\)\[CenterDot]\(ext( B)\)\)]], "." }], "Text"], Cell[BoxData[ \(Simplify[{{Det[F . M . B]}} - \((ExteriorF[F] . Exterior[M] . ExteriorB[B])\)\ ]\)], "Input"], Cell[TextData[{ "Note that we had to make a 1 by 1 matrix out of the determinant to compare \ the two. This result, when combined with lemma 1, shows us that \n\n \ ", Cell[BoxData[ \(TraditionalForm\`det( F\[CenterDot]M\_\(n - 1\)\[CenterDot] M\_\(n - 2\) \[CenterEllipsis]\ M\_2\[CenterDot]M\_1\[CenterDot] B)\ = \ P\_n\[CenterDot]P\_\(n - 1\)\[CenterDot] P\_\(n - 2\) \[CenterEllipsis]\ P\_2\[CenterDot]P\_1\[CenterDot] P\_0\)]], "\n \n Where ", Cell[BoxData[ \(TraditionalForm\`\(\(P\_i\)\(\ \)\(=\)\(\ \)\(ext( M\_i)\)\(\ \)\)\)]], " for ", Cell[BoxData[ \(TraditionalForm\`1 \[LessEqual] i \[LessEqual] n - 1\)]], ", ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ FormBox[\(P\_n\), "TraditionalForm"], "TraditionalForm"], " ", "=", " ", \(ext(F)\)}], TraditionalForm]]], ", and ", Cell[BoxData[ \(TraditionalForm\`P\_0\ = \ ext(B)\)]], ". This is propositon 1 in the paper.\n \n But there is another \ important property of the exterior matrices. If we define the matrix" }], "Text"], Cell[BoxData[ \(J\ = \ {{0, 0, 0, 1, 0, 0}, {0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, 0, 1}, {1, 0, 0, 0, 0, 0}, {0, 1, 0, 0, 0, 0}, {0, 0, 1, 0, 0, 0}}\)], "Input"], Cell[BoxData[ \(MatrixForm[J]\)], "Input"], Cell[TextData[{ "then if ", Cell[BoxData[ \(TraditionalForm\`P\ = \ ext(M)\)]], ", then ", Cell[BoxData[ \(TraditionalForm\`P\^T\[CenterDot]J\[CenterDot]P\[CenterDot] J\ = \ \(det(M)\)\[CenterDot]I\)]] }], "Text"], Cell[BoxData[ \(Simplify[ Transpose[P] . J . P . J\ - \ Det[M]*IdentityMatrix[6]]\)], "Input"], Cell[TextData[{ "This proves proposition 2 in the paper. In particular, this says that ", Cell[BoxData[ \(TraditionalForm\`A\ = \ P/\@\(det(M)\)\)]], " will have an orthogonal-like property: \n \ ", Cell[BoxData[ \(TraditionalForm\`A\^T\[CenterDot]J\[CenterDot]A\ = \ J\)]], ". \nThis is not quite the same as a symplectic matrix, for the ", Cell[BoxData[ \(TraditionalForm\`J\)]], " used in symplectic matrices have the lower half of the elements negative.\ \n\nWe now can discuss how exterior matrices can be used to solve fourth \ order sequentially coupled PDE's. Suppose we have a system of partial \ differential equations for\n ", Cell[BoxData[ \(TraditionalForm\`\(\(u\_1\)(x, t)\)\ \ \ 0\ \[LessEqual] \ x\ \[LessEqual] \ L\_1, \ \ t\ \[GreaterEqual] \ 0\)]], "\n ", Cell[BoxData[ \(TraditionalForm\`\(\(u\_2\)(x, t)\)\ \ \ 0\ \[LessEqual] \ x\ \[LessEqual] \ L\_2, \ \ t\ \[GreaterEqual] \ 0\)]], "\n \ \[VerticalEllipsis]\n ", Cell[BoxData[ \(TraditionalForm\`\(\(u\_m\)(x, t)\)\ \ \ 0\ \[LessEqual] \ x\ \[LessEqual] \ L\_m, \ \ t\ \[GreaterEqual] \ 0\)]], "\n \nfor which the boundary conditions \ of one function is tied to the boundary conditions of the next\n\n \ ", Cell[BoxData[ FormBox[ RowBox[{" ", RowBox[{ FormBox[ RowBox[{"(", GridBox[{ {\(\(u\_i\)(0, t)\)}, {\(\(u\_i\%\[Prime]\)(0, t)\)}, {\(\(u\_i\%\[Prime]\[Prime]\)(0, t)\)}, {\(\(u\_i\%\[Prime]\[Prime]\[Prime]\)(0, t)\)} }], ")"}], "TraditionalForm"], "=", " ", RowBox[{ FormBox[\(C\_i\), "TraditionalForm"], FormBox[ RowBox[{"(", GridBox[{ {\(\(u\_\(i - 1\)\)(L\_\(i - 1\), t)\)}, {\(\(u\_\(i - 1\)\%\[Prime]\)(L\_\(i - 1\), t)\)}, {\(\(u\_\(i - 1\)\%\[Prime]\[Prime]\)(L\_\(i - 1\), t)\)}, {\(\(u\_\(i - 1\)\%\[Prime]\[Prime]\[Prime]\)( L\_\(i - 1\), t)\)} }], ")"}], "TraditionalForm"]}]}]}], TraditionalForm]]], "\n \nwith ", Cell[BoxData[ \(TraditionalForm\`C\_i\)]], " being a 4 by 4 matrix, which could represent, among other things, a \ damper, a bend, or a point mass. There would also be two boundary conditions \ at each end. It should be pointed out that each of the ", Cell[BoxData[ \(TraditionalForm\`\(u\_i\)(x, t)\)]], " can satisfy a different PDE. For example, we could have a \ Euler-Bernoulli beam suspended by two inclined cables. The only restriction \ is that each of the PDE's be a fourth order equation in the ", Cell[BoxData[ \(TraditionalForm\`x\)]], " variable.\n\nThe classical method for solving such a system is to express \ each\n ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ RowBox[{\(\(u\_i\)(x, t)\), " ", "=", " ", UnderscriptBox[ StyleBox["\[CapitalSigma]", FontSize->24], "\[Lambda]"]}], "TraditionalForm"], " ", FormBox[\(\(\(y\_\(i, \[Lambda]\)\)(x)\)\ e\^\(\[Lambda]\ t\)\), "TraditionalForm"]}], TraditionalForm]]], "\nwhere the sum is taken over all of the eigenfrequencies ", Cell[BoxData[ \(TraditionalForm\`\[Lambda]\)]], " of the\nsystem. Each ", Cell[BoxData[ \(TraditionalForm\`\(y\_\(i, \[Lambda]\)\)(x)\)]], " will solve an ODE\n\n ", Cell[BoxData[ \(TraditionalForm\`y\_i\%iv\ + \ \(p( x, \[Lambda])\)\ y\_i\%\[Prime]\[Prime]\[Prime]\ + \ \(q( x, \[Lambda])\)\ y\_i\%\[Prime]\[Prime]\ + \ \(r( x, \[Lambda])\)\ y\_i\%\[Prime]\ + \ \(s( x, \[Lambda])\)\ y\_i\ = \ 0. \)]], "\n \nformed by plugging the sum into the corresponding PDE. \ The main problem is finding the eigenfrequencies ", Cell[BoxData[ \(TraditionalForm\`\[Lambda]\)]], ". The most common way to do this is with the transfer matrix method, in \ which we find the general solution to each of the ODE's,\n ", Cell[BoxData[ \(TraditionalForm\`y\_\(i, \[Lambda]\)\ = \ \(c\_\(i, 1\)\) \(\(y\_\(i, 1\)\)( x, \[Lambda])\)\ + \ \(c\_\(i, 2\)\) \(\(y\_\(i, 2\)\)( x, \[Lambda])\)\ + \ \(c\_\(i, 3\)\) \(\(y\_\(i, 3\)\)( x, \[Lambda])\)\ + \ \(c\_\(i, 4\)\) \(\(y\_\(i, 4\)\)( x, \[Lambda])\)\)]], "\n \nsuch that the solutions satisfy the initial conditions\n\ \n", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{ FormBox[\(\(\(y\_\(i, 1\)\)(0, \[Lambda])\ = \ 1\)\(,\)\(\ \ \)\), "TraditionalForm"], FormBox[\(\(y\_\(i, 1\)\%\[Prime]\)(0, \[Lambda])\), "TraditionalForm"]}], " ", "=", " ", "0"}], ",", " ", \(\(y\_\(i, 1\)\%\[Prime]\[Prime]\)(0, \[Lambda])\ = \ 0\), ",", " ", \(\(y\_\(i, 1\)\%\[Prime]\[Prime]\[Prime]\)( 0, \[Lambda])\ = \ 0\), ","}], TraditionalForm]]], "\n", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{ FormBox[\(\(\(y\_\(i, 2\)\)(0, \[Lambda])\ = \ 0\)\(,\)\(\ \ \)\), "TraditionalForm"], FormBox[\(\(y\_\(i, 2\)\%\[Prime]\)(0, \[Lambda])\), "TraditionalForm"]}], " ", "=", " ", "1"}], ",", " ", \(\(y\_\(i, 2\)\%\[Prime]\[Prime]\)(0, \[Lambda])\ = \ 0\), ",", " ", \(\(y\_\(i, 2\)\%\[Prime]\[Prime]\[Prime]\)( 0, \[Lambda])\ = \ 0\), ","}], TraditionalForm]]], "\n", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{ FormBox[\(\(\(y\_\(i, 3\)\)(0, \[Lambda])\ = \ 0\)\(,\)\(\ \ \)\), "TraditionalForm"], FormBox[\(\(y\_\(i, 3\)\%\[Prime]\)(0, \[Lambda])\), "TraditionalForm"]}], " ", "=", " ", "0"}], ",", " ", \(\(y\_\(i, 3\)\%\[Prime]\[Prime]\)(0, \[Lambda])\ = \ 1\), ",", " ", \(\(y\_\(i, 3\)\%\[Prime]\[Prime]\[Prime]\)( 0, \[Lambda])\ = \ 0\), ","}], TraditionalForm]]], "\n", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{ FormBox[\(\(\(y\_\(i, 4\)\)(0, \[Lambda])\ = \ 0\)\(,\)\(\ \ \)\), "TraditionalForm"], FormBox[\(\(y\_\(i, 4\)\%\[Prime]\)(0, \[Lambda])\), "TraditionalForm"]}], " ", "=", " ", "0"}], ",", " ", \(\(y\_\(i, 4\)\%\[Prime]\[Prime]\)(0, \[Lambda])\ = \ 0\), ",", " ", \(\(y\_\(i, 4\)\%\[Prime]\[Prime]\[Prime]\)( 0, \[Lambda])\ = \ 1. \)}], TraditionalForm]]], "\n\nWe then form the matrix\n\n ", Cell[BoxData[ FormBox[ RowBox[{\(W\_i\), " ", "=", " ", RowBox[{"(", GridBox[{ {\(\(y\_\(i, 1\)\)(L\_i, \[Lambda])\), \(\(y\_\(i, 2\)\)( L\_i, \[Lambda])\), \(\(y\_\(i, 3\)\)( L\_i, \[Lambda])\), \(\(y\_\(i, 4\)\)( L\_i, \[Lambda])\)}, {\(\(y\_\(i, 1\)\%\[Prime]\)( L\_i, \[Lambda])\), \(\(y\_\(i, 2\)\%\[Prime]\)( L\_i, \[Lambda])\), \(\(y\_\(i, 3\)\%\[Prime]\)( L\_i, \[Lambda])\), \(\(y\_\(i, 4\)\%\[Prime]\)( L\_i, \[Lambda])\)}, {\(\(y\_\(i, 1\)\%\[Prime]\[Prime]\)( L\_i, \[Lambda])\), \(\(y\_\(i, 2\)\%\[Prime]\[Prime]\)( L\_i, \[Lambda])\), \(\(y\_\(i, 3\)\%\[Prime]\[Prime]\)( L\_i, \[Lambda])\), \(\(y\_\(i, 4\)\%\[Prime]\[Prime]\)( L\_i, \[Lambda])\)}, {\(\(y\_\(i, 1\)\%\[Prime]\[Prime]\[Prime]\)( L\_i, \[Lambda])\), \(\(\(y\_\(i, 2\)\%\[Prime]\[Prime]\[Prime]\)( L\_i, \[Lambda])\)\(\ \)\), \(\(y\_\(i, 3\)\%\[Prime]\[Prime]\[Prime]\)( L\_i, \[Lambda])\), \(\(y\_\(i, 4\)\%\[Prime]\[Prime]\[Prime]\)(L\_i, \[Lambda])\)} }], ")"}]}], TraditionalForm]]], "\nso that \n ", Cell[BoxData[ FormBox[ RowBox[{" ", RowBox[{ FormBox[ RowBox[{"(", GridBox[{ {\(\(y\_\(i, \[Lambda]\)\)(L\_i)\)}, {\(\(y\_\(i, \[Lambda]\)\%\[Prime]\)(L\_i)\)}, {\(\(y\_\(i, \[Lambda]\)\%\[Prime]\[Prime]\)(L\_i)\)}, {\(\(y\_\(i, \[Lambda]\)\%\[Prime]\[Prime]\[Prime]\)( L\_i)\)} }], ")"}], "TraditionalForm"], "=", " ", RowBox[{ FormBox[\(W\_i\), "TraditionalForm"], FormBox[ RowBox[{"(", GridBox[{ {\(\(y\_\(i, \[Lambda]\)\)(0)\)}, {\(\(y\_\(i, \[Lambda]\)\%\[Prime]\)(0)\)}, {\(\(y\_\(i, \[Lambda]\)\%\[Prime]\[Prime]\)(0)\)}, {\(\(y\_\(i, \[Lambda]\)\%\[Prime]\[Prime]\[Prime]\)( 0)\)} }], ")"}], "TraditionalForm"]}]}]}], TraditionalForm]]], ".\n Thus, we see that\n ", Cell[BoxData[ FormBox[ RowBox[{" ", RowBox[{ FormBox[ RowBox[{"(", GridBox[{ {\(\(y\_\(m, \[Lambda]\)\)(L\_m)\)}, {\(\(y\_\(m, \[Lambda]\)\%\[Prime]\)(L\_m)\)}, {\(\(y\_\(m, \[Lambda]\)\%\[Prime]\[Prime]\)(L\_m)\)}, {\(\(y\_\(m, \[Lambda]\)\%\[Prime]\[Prime]\[Prime]\)( L\_m)\)} }], ")"}], "TraditionalForm"], "=", " ", RowBox[{ FormBox[\(W\_m\[CenterDot]C\_m\[CenterDot] W\_\(m - 1\)\[CenterDot] C\_\(m - 1\) \[CenterEllipsis]\ C\_2\[CenterDot]W\_1\), "TraditionalForm"], FormBox[ RowBox[{"(", GridBox[{ {\(\(y\_\(1, \[Lambda]\)\)(0)\)}, {\(\(y\_\(1, \[Lambda]\)\%\[Prime]\)(0)\)}, {\(\(y\_\(1, \[Lambda]\)\%\[Prime]\[Prime]\)(0)\)}, {\(\(y\_\(1, \[Lambda]\)\%\[Prime]\[Prime]\[Prime]\)( 0)\)} }], ")"}], "TraditionalForm"]}]}]}], TraditionalForm]]], ".\nNow,we can form a 4 by 2 matrix ", Cell[BoxData[ \(TraditionalForm\`B\)]], " from the two beginning boundary conditions,and a 2 by 4 matrix ", Cell[BoxData[ \(TraditionalForm\`F\)]], "\nusing the boundary conditions at the final end, so that the \ eigenfrequencies are the solutions to the equation\n\n \ ", Cell[BoxData[ \(TraditionalForm\`det( F\[CenterDot]W\_m\[CenterDot]C\_m\[CenterDot]W\_\(m - 1\)\ \[CenterDot]C\_\(m - 1\) \[CenterEllipsis]\ \ C\_2\[CenterDot]W\_1\[CenterDot]B)\ = \ 0\)]], "\n \nFor example, if both ends are \ clamped,we have ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\(y\_\(1, \[Lambda]\)\)(0)\), "TraditionalForm"], " ", "=", " ", RowBox[{ FormBox[\(\(y\_\(1, \[Lambda]\)\%\[Prime]\)(0)\), "TraditionalForm"], " ", "=", " ", "0"}]}], TraditionalForm]]], ", and ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\(y\_\(m, \[Lambda]\)\)(L\_m)\), "TraditionalForm"], " ", "=", " ", RowBox[{ FormBox[\(\(y\_\(m, \[Lambda]\)\%\[Prime]\)(L\_m)\), "TraditionalForm"], " ", "=", " ", "0"}]}], TraditionalForm]]], ". This can be expressed by\n ", Cell[BoxData[ FormBox[ RowBox[{\(B\_clamped\), " ", "=", " ", RowBox[{"(", GridBox[{ {"0", "0"}, {"0", "0"}, {"1", "0"}, {"0", "1"} }], ")"}]}], TraditionalForm]]], " and ", Cell[BoxData[ FormBox[ RowBox[{\(F\_clamped\), " ", "=", " ", RowBox[{"(", GridBox[{ {"1", "0", "0", "0"}, {"0", "1", "0", "0"} }], ")"}]}], TraditionalForm]]], ".\nIt is now clear how the exterior matrices can play a role in finding \ the eigenfrequencies \[Lambda]. By renumbering the matrices with ", Cell[BoxData[ \(TraditionalForm\`n\ = \ 2 m\)]], ", ", Cell[BoxData[ \(TraditionalForm\`W\_i = M\_\(2 i - 1\)\)]], ", and ", Cell[BoxData[ \(TraditionalForm\`C\_i = M\_\(2 i - 2\)\)]], ", we can express the equation\n ", Cell[BoxData[ \(TraditionalForm\`det( F\[CenterDot]W\_m\[CenterDot]C\_m\[CenterDot]W\_\(m - 1\)\ \[CenterDot]C\_\(m - 1\) \[CenterEllipsis]\ \ C\_2\[CenterDot]W\_1\[CenterDot]B)\ = \ 0\)]], "\nas \n ", Cell[BoxData[ \(TraditionalForm\`det( F\[CenterDot]M\_\(n - 1\)\[CenterDot]M\_\(n - 2\)\[CenterDot]M\_\(n \ - 3\)\[CenterDot]M\_\(n - 4\) \[CenterEllipsis]\ M\_2\[CenterDot] M\_1\[CenterDot]B)\ = \ 0\)]], ".\nTo avoid confusion, we will index the ", Cell[BoxData[ \(TraditionalForm\`M\)]], " matrices with ", Cell[BoxData[ \(TraditionalForm\`j\)]], " instead of ", Cell[BoxData[ \(TraditionalForm\`i\)]], ". We can now use proposition 1, and solve the equation\n \ ", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(P\_n\[CenterDot]P\_\(n - 1\)\[CenterDot] P\_\(n - 2\) \[CenterEllipsis]\ P\_2\[CenterDot]P\_1\[CenterDot] P\_0\)\)\)]], " = 0\n\nfor \[Lambda]. Here, ", Cell[BoxData[ \(TraditionalForm\`\(\(P\_j\)\(\ \)\(=\)\(\ \)\(ext( M\_j)\)\(\ \)\)\)]], " for ", Cell[BoxData[ \(TraditionalForm\`1 \[LessEqual] j \[LessEqual] n - 1\)]], ", ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ FormBox[\(P\_n\), "TraditionalForm"], "TraditionalForm"], " ", "=", " ", \(ext(F)\)}], TraditionalForm]]], ", and ", Cell[BoxData[ \(TraditionalForm\`P\_0\ = \ ext(B)\)]], ". When ", Cell[BoxData[ \(TraditionalForm\`j\)]], " is odd, ", Cell[BoxData[ \(TraditionalForm\`M\_j\)]], " will represent\nthe solutions to the forth order equation at ", Cell[BoxData[ \(TraditionalForm\`L\_j\)]] }], "Text"], Cell[BoxData[ \(Mj\ = \ {{y1[Lj], \ y2[Lj], \ y3[Lj], \ y4[Lj]}, {\(y1'\)[Lj], \ \(y2'\)[Lj], \ \(y3'\)[Lj], \ \(y4'\)[ Lj]}, {\(y1''\)[Lj], \ \(y2''\)[Lj], \ \(y3''\)[Lj], \ \(y4''\)[ Lj]}, {\(y1'''\)[Lj], \ \(y2'''\)[Lj], \ \(y3'''\)[ Lj], \ \(y4'''\)[Lj]}}\)], "Input"], Cell[BoxData[ \(MatrixForm[Mj]\)], "Input"], Cell[TextData[{ "Then ", Cell[BoxData[ \(TraditionalForm\`P\_j\)]], " is given by" }], "Text"], Cell[BoxData[ \(Pj\ = \ Exterior[Mj]\)], "Input"], Cell[BoxData[ \(MatrixForm[Pj]\)], "Input"], Cell[TextData[{ "Not only will this matrix satisfy the properties that we have already \ observed, but there are other interesting properties of this matrix. If we \ take the derivative of the first row of this matrix with respect to ", Cell[BoxData[ \(TraditionalForm\`L\_j\)]] }], "Text"], Cell[BoxData[ \(D[Pj[\([1]\)], Lj]\)], "Input"], Cell[TextData[{ "we get the second row of the matrix. That is, ", Cell[BoxData[ \(TraditionalForm\`\(\(\(\[PartialD]\)\(\ \)\)\/\[PartialD]L\_j\) P\_\(j, 1 v\)\ = \ P\_\(j, 2 v\)\)]], ", where ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, u\ v\)\)]], " represents the (u,v) element of the matrix ", Cell[BoxData[ \(TraditionalForm\`P\_j\)]], ". We can test this by finding" }], "Text"], Cell[BoxData[ \(D[Pj[\([1]\)], Lj]\ - \ Pj[\([2]\)]\)], "Input"], Cell[TextData[{ "Likewise, we find that \n ", Cell[BoxData[ \(TraditionalForm\`\(\(\(\[PartialD]\)\(\ \)\)\/\[PartialD]L\_j\) P\_\(j, 6 v\)\ = \ \(-P\_\(j, 5 v\)\)\)]], ", \n ", Cell[BoxData[ \(TraditionalForm\`\(\(\(\[PartialD]\)\(\ \)\)\/\[PartialD]L\_j\) P\_\(j, 2 v\)\ = \ P\_\(j, 3 v\)\ + \ P\_\(j, 6 v\)\)]], "\n ", Cell[BoxData[ \(TraditionalForm\`\(\(\(\[PartialD]\)\(\ \)\)\/\[PartialD]L\_j\) P\_\(j, 3 v\)\ = \ \(-P\_\(j, 5 v\)\)\ - \ \(p(L\_j)\) P\_\(j, 3 v\)\ - \ \ \(q(L\_j)\) P\_\(j, 2 v\)\ - \ \ \(r(L\_j)\) P\_\(j, 1 v\)\)]], "\n ", Cell[BoxData[ \(TraditionalForm\`\(\(\(\[PartialD]\)\(\ \)\)\/\[PartialD]L\_j\) P\_\(j, 4 v\)\ = \ \ \(-\ \(p(L\_j)\)\) P\_\(j, 4 v\)\ + \ \ \(r(L\_j)\) P\_\(j, 6 v\)\ + \ \ \(s(L\_j)\) P\_\(j, 2 v\)\)]], "\n ", Cell[BoxData[ \(TraditionalForm\`\(\(\(\[PartialD]\)\(\ \)\)\/\[PartialD]L\_j\) P\_\(j, 5 v\)\ = \ \(-P\_\(j, 4 v\)\)\ - \ \(p(L\_j)\) P\_\(j, 5 v\)\ + \ \ \(q(L\_j)\) P\_\(j, 6 v\)\ - \ \ \(s(L\_j)\) P\_\(j, 1 v\)\)]] }], "Text"], Cell[BoxData[ \(D[Pj[\([6]\)], Lj]\ + \ Pj[\([5]\)]\)], "Input"], Cell[BoxData[ \(D[Pj[\([2]\)], Lj]\ - \ Pj[\([3]\)] - Pj[\([6]\)]\)], "Input"], Cell[BoxData[ \(D[Pj[\([3]\)], Lj]\ + \ Pj[\([5]\)]\ + \ p[Lj]\ Pj[\([3]\)]\ + \ q[Lj]\ Pj[\([2]\)]\ + \ r[Lj]\ Pj[\([1]\)]\)], "Input"], Cell["\<\ Since this involves fourth derivatives, we have to use the differential \ equation to simplify.\ \>", "Text"], Cell[BoxData[ RowBox[{"%", " ", "/.", " ", RowBox[{"{", RowBox[{ RowBox[{ RowBox[{ SuperscriptBox["y1", TagBox[\((4)\), Derivative], MultilineFunction->None], "[", "Lj", "]"}], " ", "\[Rule]", " ", \(\(-p[Lj]\)\ \(y1'''\)[Lj]\ - \ q[Lj]\ \(y1''\)[Lj]\ - \ r[Lj]\ \(y1'\)[Lj]\ - \ s[Lj]\ y1[Lj]\)}], ",", "\[IndentingNewLine]", RowBox[{ RowBox[{ SuperscriptBox["y2", TagBox[\((4)\), Derivative], MultilineFunction->None], "[", "Lj", "]"}], " ", "\[Rule]", " ", \(\(-p[Lj]\)\ \(y2'''\)[Lj]\ - \ q[Lj]\ \(y2''\)[Lj]\ - \ r[Lj]\ \(y2'\)[Lj]\ - \ s[Lj]\ y2[Lj]\)}], ",", "\[IndentingNewLine]", RowBox[{ RowBox[{ SuperscriptBox["y3", TagBox[\((4)\), Derivative], MultilineFunction->None], "[", "Lj", "]"}], " ", "\[Rule]", " ", \(\(-p[Lj]\)\ \(y3'''\)[Lj]\ - \ q[Lj]\ \(y3''\)[Lj]\ - \ r[Lj]\ \(y3'\)[Lj]\ - \ s[Lj]\ y3[Lj]\)}], ",", "\[IndentingNewLine]", RowBox[{ RowBox[{ SuperscriptBox["y4", TagBox[\((4)\), Derivative], MultilineFunction->None], "[", "Lj", "]"}], " ", "\[Rule]", " ", \(\(-p[Lj]\)\ \(y4'''\)[Lj]\ - \ q[Lj]\ \(y4''\)[Lj]\ - \ r[Lj]\ \(y4'\)[Lj]\ - \ s[Lj]\ y4[Lj]\)}]}], "}"}]}]], "Input"], Cell[BoxData[ \(Simplify[%]\)], "Input"], Cell[BoxData[ \(D[Pj[\([4]\)], Lj]\ + \ p[Lj]\ Pj[\([4]\)]\ - \ r[Lj]\ Pj[\([6]\)]\ - \ s[Lj]\ Pj[\([2]\)]\)], "Input"], Cell[BoxData[ RowBox[{"%", " ", "/.", " ", RowBox[{"{", RowBox[{ RowBox[{ RowBox[{ SuperscriptBox["y1", TagBox[\((4)\), Derivative], MultilineFunction->None], "[", "Lj", "]"}], " ", "\[Rule]", " ", \(\(-p[Lj]\)\ \(y1'''\)[Lj]\ - \ q[Lj]\ \(y1''\)[Lj]\ - \ r[Lj]\ \(y1'\)[Lj]\ - \ s[Lj]\ y1[Lj]\)}], ",", "\[IndentingNewLine]", RowBox[{ RowBox[{ SuperscriptBox["y2", TagBox[\((4)\), Derivative], MultilineFunction->None], "[", "Lj", "]"}], " ", "\[Rule]", " ", \(\(-p[Lj]\)\ \(y2'''\)[Lj]\ - \ q[Lj]\ \(y2''\)[Lj]\ - \ r[Lj]\ \(y2'\)[Lj]\ - \ s[Lj]\ y2[Lj]\)}], ",", "\[IndentingNewLine]", RowBox[{ RowBox[{ SuperscriptBox["y3", TagBox[\((4)\), Derivative], MultilineFunction->None], "[", "Lj", "]"}], " ", "\[Rule]", " ", \(\(-p[Lj]\)\ \(y3'''\)[Lj]\ - \ q[Lj]\ \(y3''\)[Lj]\ - \ r[Lj]\ \(y3'\)[Lj]\ - \ s[Lj]\ y3[Lj]\)}], ",", "\[IndentingNewLine]", RowBox[{ RowBox[{ SuperscriptBox["y4", TagBox[\((4)\), Derivative], MultilineFunction->None], "[", "Lj", "]"}], " ", "\[Rule]", " ", \(\(-p[Lj]\)\ \(y4'''\)[Lj]\ - \ q[Lj]\ \(y4''\)[Lj]\ - \ r[Lj]\ \(y4'\)[Lj]\ - \ s[Lj]\ y4[Lj]\)}]}], "}"}]}]], "Input"], Cell[BoxData[ \(Simplify[%]\)], "Input"], Cell[BoxData[ \(D[Pj[\([5]\)], Lj]\ + \ Pj[\([4]\)]\ + \ p[Lj]\ Pj[\([5]\)]\ - \ q[Lj]\ Pj[\([6]\)]\ + \ s[Lj]\ Pj[\([1]\)]\)], "Input"], Cell[BoxData[ RowBox[{"%", " ", "/.", " ", RowBox[{"{", RowBox[{ RowBox[{ RowBox[{ SuperscriptBox["y1", TagBox[\((4)\), Derivative], MultilineFunction->None], "[", "Lj", "]"}], " ", "\[Rule]", " ", \(\(-p[Lj]\)\ \(y1'''\)[Lj]\ - \ q[Lj]\ \(y1''\)[Lj]\ - \ r[Lj]\ \(y1'\)[Lj]\ - \ s[Lj]\ y1[Lj]\)}], ",", "\[IndentingNewLine]", RowBox[{ RowBox[{ SuperscriptBox["y2", TagBox[\((4)\), Derivative], MultilineFunction->None], "[", "Lj", "]"}], " ", "\[Rule]", " ", \(\(-p[Lj]\)\ \(y2'''\)[Lj]\ - \ q[Lj]\ \(y2''\)[Lj]\ - \ r[Lj]\ \(y2'\)[Lj]\ - \ s[Lj]\ y2[Lj]\)}], ",", "\[IndentingNewLine]", RowBox[{ RowBox[{ SuperscriptBox["y3", TagBox[\((4)\), Derivative], MultilineFunction->None], "[", "Lj", "]"}], " ", "\[Rule]", " ", \(\(-p[Lj]\)\ \(y3'''\)[Lj]\ - \ q[Lj]\ \(y3''\)[Lj]\ - \ r[Lj]\ \(y3'\)[Lj]\ - \ s[Lj]\ y3[Lj]\)}], ",", "\[IndentingNewLine]", RowBox[{ RowBox[{ SuperscriptBox["y4", TagBox[\((4)\), Derivative], MultilineFunction->None], "[", "Lj", "]"}], " ", "\[Rule]", " ", \(\(-p[Lj]\)\ \(y4'''\)[Lj]\ - \ q[Lj]\ \(y4''\)[Lj]\ - \ r[Lj]\ \(y4'\)[Lj]\ - \ s[Lj]\ y4[Lj]\)}]}], "}"}]}]], "Input"], Cell[BoxData[ \(Simplify[%]\)], "Input"], Cell[TextData[{ "This verifies the equations (5.3) in the paper. Furthermore, we can \ combine the first 4 of these equations to find ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, 3 v\)\)]], " in terms of ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, 1 v\)\)]], ":\n ", Cell[BoxData[ \(TraditionalForm\`2 \(\(\(\[PartialD]\)\(\ \)\)\/\[PartialD]L\_j\) P\_\(j, 3 v\) + \ \ \(p(L\_j)\) P\_\(j, 3 v\) = \ \(\(\(\[PartialD]\^3\)\(\ \ \)\)\/\[PartialD]L\_j\^3\) P\_\(j, 1 v\)\ - \ \ \(q( L\_j)\) \(\(\(\[PartialD]\)\(\ \)\)\/\[PartialD]L\_j\) P\_\(j, 1 v\)\ - \ \ \(r(L\_j)\) P\_\(j, 1 v\)\)]], "." }], "Text"], Cell[BoxData[ \(2\ D[Pj[\([3]\)], Lj]\ \ + \ p[Lj]\ Pj[\([3]\)]\ - \ D[Pj[\([1]\)], {Lj, 3}]\ + q[Lj]\ D[Pj[\([1]\)], Lj]\ + \ r[Lj]\ Pj[\([1]\)]\)], "Input"], Cell[BoxData[ RowBox[{"%", " ", "/.", " ", RowBox[{"{", RowBox[{ RowBox[{ RowBox[{ SuperscriptBox["y1", TagBox[\((4)\), Derivative], MultilineFunction->None], "[", "Lj", "]"}], " ", "\[Rule]", " ", \(\(-p[Lj]\)\ \(y1'''\)[Lj]\ - \ q[Lj]\ \(y1''\)[Lj]\ - \ r[Lj]\ \(y1'\)[Lj]\ - \ s[Lj]\ y1[Lj]\)}], ",", "\[IndentingNewLine]", RowBox[{ RowBox[{ SuperscriptBox["y2", TagBox[\((4)\), Derivative], MultilineFunction->None], "[", "Lj", "]"}], " ", "\[Rule]", " ", \(\(-p[Lj]\)\ \(y2'''\)[Lj]\ - \ q[Lj]\ \(y2''\)[Lj]\ - \ r[Lj]\ \(y2'\)[Lj]\ - \ s[Lj]\ y2[Lj]\)}], ",", "\[IndentingNewLine]", RowBox[{ RowBox[{ SuperscriptBox["y3", TagBox[\((4)\), Derivative], MultilineFunction->None], "[", "Lj", "]"}], " ", "\[Rule]", " ", \(\(-p[Lj]\)\ \(y3'''\)[Lj]\ - \ q[Lj]\ \(y3''\)[Lj]\ - \ r[Lj]\ \(y3'\)[Lj]\ - \ s[Lj]\ y3[Lj]\)}], ",", "\[IndentingNewLine]", RowBox[{ RowBox[{ SuperscriptBox["y4", TagBox[\((4)\), Derivative], MultilineFunction->None], "[", "Lj", "]"}], " ", "\[Rule]", " ", \(\(-p[Lj]\)\ \(y4'''\)[Lj]\ - \ q[Lj]\ \(y4''\)[Lj]\ - \ r[Lj]\ \(y4'\)[Lj]\ - \ s[Lj]\ y4[Lj]\)}]}], "}"}]}]], "Input"], Cell[BoxData[ \(Simplify[%]\)], "Input"], Cell[TextData[{ "This verifies equation (5.4) from the paper. This also shows that if we \ know the first row of ", Cell[BoxData[ \(TraditionalForm\`P\_j\)]], ", we can reconstruct the entire matrix ", Cell[BoxData[ \(TraditionalForm\`P\_j\)]], ". \n\nThe goal is to be able to reconstruct ", Cell[BoxData[ \(TraditionalForm\`P\_j\)]], " from a single function. However, in order to do this, we need to \ introduce another variable \[Xi]. Rather than determining the solutions at \ ", Cell[BoxData[ \(TraditionalForm\`x = P\_j\)]], " in terms of the solutions at ", Cell[BoxData[ \(TraditionalForm\`x = 0\)]], ", we will determine the solutions at ", Cell[BoxData[ \(TraditionalForm\`x = P\_j\)]], " in terms of the solutions at ", Cell[BoxData[ \(TraditionalForm\`x = \[Xi]\)]], ". That is, the functions ", Cell[BoxData[ \(TraditionalForm\`y\_\(i, 1\)\)]], ", ", Cell[BoxData[ \(TraditionalForm\`y\_\(i, 2\)\)]], ",. ", Cell[BoxData[ \(TraditionalForm\`y\_\(i, 3\)\)]], ", and ", Cell[BoxData[ \(TraditionalForm\`y\_\(i, 4\)\)]], " now must satisfy\n\n ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{ FormBox[\(\(\(y\_\(i, 1\)\)(\[Xi], \[Lambda])\ = \ 1\)\(,\)\(\ \ \)\), "TraditionalForm"], FormBox[\(\(y\_\(i, 1\)\%\[Prime]\)(\[Xi], \[Lambda])\), "TraditionalForm"]}], " ", "=", " ", "0"}], ",", " ", \(\(y\_\(i, 1\)\%\[Prime]\[Prime]\)(\[Xi], \[Lambda])\ = \ 0\), ",", " ", \(\(y\_\(i, 1\)\%\[Prime]\[Prime]\[Prime]\)(\[Xi], \[Lambda])\ = \ 0\), ","}], TraditionalForm]]], "\n ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{ FormBox[\(\(\(y\_\(i, 2\)\)(\[Xi], \[Lambda])\ = \ 0\)\(,\)\(\ \ \)\), "TraditionalForm"], FormBox[\(\(y\_\(i, 2\)\%\[Prime]\)(\[Xi], \[Lambda])\), "TraditionalForm"]}], " ", "=", " ", "1"}], ",", " ", \(\(y\_\(i, 2\)\%\[Prime]\[Prime]\)(\[Xi], \[Lambda])\ = \ 0\), ",", " ", \(\(y\_\(i, 2\)\%\[Prime]\[Prime]\[Prime]\)(\[Xi], \[Lambda])\ = \ 0\), ","}], TraditionalForm]]], "\n ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{ FormBox[\(\(\(y\_\(i, 3\)\)(\[Xi], \[Lambda])\ = \ 0\)\(,\)\(\ \ \)\), "TraditionalForm"], FormBox[\(\(y\_\(i, 3\)\%\[Prime]\)(\[Xi], \[Lambda])\), "TraditionalForm"]}], " ", "=", " ", "0"}], ",", " ", \(\(y\_\(i, 3\)\%\[Prime]\[Prime]\)(\[Xi], \[Lambda])\ = \ 1\), ",", " ", \(\(y\_\(i, 3\)\%\[Prime]\[Prime]\[Prime]\)(\[Xi], \[Lambda])\ = \ 0\), ","}], TraditionalForm]]], "\n ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{ FormBox[\(\(\(y\_\(i, 4\)\)(\[Xi], \[Lambda])\ = \ 0\)\(,\)\(\ \ \)\), "TraditionalForm"], FormBox[\(\(y\_\(i, 4\)\%\[Prime]\)(\[Xi], \[Lambda])\), "TraditionalForm"]}], " ", "=", " ", "0"}], ",", " ", \(\(y\_\(i, 4\)\%\[Prime]\[Prime]\)(\[Xi], \[Lambda])\ = \ 0\), ",", " ", \(\(y\_\(i, 4\)\%\[Prime]\[Prime]\[Prime]\)(\[Xi], \[Lambda])\ = \ 1. \)}], TraditionalForm]]], "\n\nThen, when we create ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ FormBox[\(W\_i\), "TraditionalForm"], "TraditionalForm"], "(", RowBox[{ FormBox[\(L\_i\), "TraditionalForm"], ",", "\[Xi]"}], ")"}], TraditionalForm]]], " from these new functions, the new transfer matrix will satisfy\n\n \ ", Cell[BoxData[ FormBox[ RowBox[{" ", RowBox[{ FormBox[ RowBox[{"(", GridBox[{ {\(\(y\_\(i, \[Lambda]\)\)(L\_i)\)}, {\(\(y\_\(i, \[Lambda]\)\%\[Prime]\)(L\_i)\)}, {\(\(y\_\(i, \[Lambda]\)\%\[Prime]\[Prime]\)(L\_i)\)}, {\(\(y\_\(i, \[Lambda]\)\%\[Prime]\[Prime]\[Prime]\)( L\_i)\)} }], ")"}], "TraditionalForm"], "=", " ", RowBox[{ FormBox[\(W\_i\), "TraditionalForm"], FormBox[ RowBox[{"(", GridBox[{ {\(\(y\_\(i, \[Lambda]\)\)(\[Xi])\)}, {\(\(y\_\(i, \[Lambda]\)\%\[Prime]\)(\[Xi])\)}, {\(\(y\_\(i, \[Lambda]\)\%\[Prime]\[Prime]\)(\[Xi])\)}, {\(\(y\_\(i, \[Lambda]\)\%\[Prime]\[Prime]\[Prime]\)(\ \[Xi])\)} }], ")"}], "TraditionalForm"]}]}]}], TraditionalForm]]], ".\nFrom this, it is easy to see that exchanging ", Cell[BoxData[ \(TraditionalForm\`L\_i\)]], " and \[Xi], or ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ FormBox[\(W\_i\), "TraditionalForm"], "TraditionalForm"], "(", RowBox[{"\[Xi]", ",", FormBox[\(L\_i\), "TraditionalForm"]}], ")"}], TraditionalForm]]], ", produces the inverse of ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ FormBox[\(W\_i\), "TraditionalForm"], "TraditionalForm"], "(", RowBox[{ FormBox[\(L\_i\), "TraditionalForm"], ",", "\[Xi]"}], ")"}], TraditionalForm]]], ". This means that if we create the exterior matrix from ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ FormBox[\(W\_i\), "TraditionalForm"], "TraditionalForm"], "(", RowBox[{ FormBox[\(L\_i\), "TraditionalForm"], ",", "\[Xi]"}], ")"}], TraditionalForm]]], " = ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ FormBox[\(M\_j\), "TraditionalForm"], "TraditionalForm"], "(", RowBox[{ FormBox[\(L\_i\), "TraditionalForm"], ",", "\[Xi]"}], ")"}], TraditionalForm]]], ", lemma 1 shows we will get a matrix ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ FormBox[\(P\_j\), "TraditionalForm"], "TraditionalForm"], "(", RowBox[{ FormBox[\(L\_i\), "TraditionalForm"], ",", "\[Xi]"}], ")"}], TraditionalForm]]], " for which \n \n \ ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ FormBox[\(P\_j\), "TraditionalForm"], "TraditionalForm"], "(", RowBox[{"\[Xi]", ",", FormBox[\(L\_j\), "TraditionalForm"]}], ")"}], TraditionalForm]]], " = ", Cell[BoxData[ FormBox[ SuperscriptBox[ RowBox[{"(", RowBox[{ FormBox[ FormBox[\(P\_j\), "TraditionalForm"], "TraditionalForm"], "(", RowBox[{ FormBox[\(L\_j\), "TraditionalForm"], ",", "\[Xi]"}], ")"}], ")"}], \(-1\)], TraditionalForm]]], ". \n \ \nBut proposition 2 shows us another way to find the inverse. \n \ ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ RowBox[{ SuperscriptBox[ RowBox[{"(", RowBox[{ FormBox[ FormBox[\(P\_j\), "TraditionalForm"], "TraditionalForm"], "(", RowBox[{ FormBox[\(L\_j\), "TraditionalForm"], ",", "\[Xi]"}], ")"}], ")"}], \(-1\)], "=", " ", RowBox[{"J", "\[CenterDot]", SuperscriptBox[ RowBox[{"(", RowBox[{ FormBox[ FormBox[\(P\_j\), "TraditionalForm"], "TraditionalForm"], "(", RowBox[{ FormBox[\(L\_j\), "TraditionalForm"], ",", "\[Xi]"}], ")"}], ")"}], "T"], "\[CenterDot]", \(\(J\)\(/\)\)}]}], "TraditionalForm"], " ", RowBox[{"det", "(", RowBox[{\(M\_j\), "(", RowBox[{ FormBox[\(L\_j\), "TraditionalForm"], ",", "\[Xi]"}], ")"}], ")"}]}], TraditionalForm]]], ".\nBut Abel's formula allows us to find ", Cell[BoxData[ FormBox[ RowBox[{"det", "(", RowBox[{\(M\_j\), "(", RowBox[{ FormBox[\(L\_j\), "TraditionalForm"], ",", "\[Xi]"}], ")"}], ")"}], TraditionalForm]]], " for odd ", Cell[BoxData[ \(TraditionalForm\`j\)]], ". Since ", Cell[BoxData[ \(TraditionalForm\`\(y\_\(i, 1\)\)(x)\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\(y\_\(i, 2\)\)(x)\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\(y\_\(i, 3\)\)(x)\)]], ", and ", Cell[BoxData[ \(TraditionalForm\`\(y\_\(i, 4\)\)(x)\)]], " all satisfy\n\n ", Cell[BoxData[ \(TraditionalForm\`y\_i\%iv\ + \ \(p( x, \[Lambda])\)\ y\_i\%\[Prime]\[Prime]\[Prime]\ + \ \(q( x, \[Lambda])\)\ y\_i\%\[Prime]\[Prime]\ + \ \(r( x, \[Lambda])\)\ y\_i\%\[Prime]\ + \ \(s( x, \[Lambda])\)\ y\_i\ = \ 0\)]], ", then\n\n ", Cell[BoxData[ FormBox[ RowBox[{" ", RowBox[{ FormBox[ RowBox[{"\[VerticalSeparator]", GridBox[{ {\(\(y\_\(i, 1\)\)(L\_i, \[Lambda])\), \(\(y\_\(i, 2\)\)( L\_i, \[Lambda])\), \(\(y\_\(i, 3\)\)( L\_i, \[Lambda])\), \(\(y\_\(i, 4\)\)( L\_i, \[Lambda])\)}, {\(\(y\_\(i, 1\)\%\[Prime]\)( L\_i, \[Lambda])\), \(\(y\_\(i, 2\)\%\[Prime]\)( L\_i, \[Lambda])\), \(\(y\_\(i, 3\)\%\[Prime]\)( L\_i, \[Lambda])\), \(\(y\_\(i, 4\)\%\[Prime]\)( L\_i, \[Lambda])\)}, {\(\(y\_\(i, 1\)\%\[Prime]\[Prime]\)( L\_i, \[Lambda])\), \(\(y\_\(i, 2\)\%\[Prime]\[Prime]\)( L\_i, \[Lambda])\), \(\(y\_\(i, 3\)\%\[Prime]\[Prime]\)( L\_i, \[Lambda])\), \(\(y\_\(i, 4\)\%\[Prime]\[Prime]\)(L\_i, \[Lambda])\)}, {\(\(y\_\(i, 1\)\%\[Prime]\[Prime]\[Prime]\)( L\_i, \[Lambda])\), \(\(\(y\_\(i, 2\)\%\[Prime]\[Prime]\[Prime]\)( L\_i, \[Lambda])\)\(\ \)\), \(\(y\_\(i, 3\)\%\[Prime]\[Prime]\[Prime]\)( L\_i, \[Lambda])\), \(\(y\_\(i, 4\)\%\[Prime]\[Prime]\[Prime]\)( L\_i, \[Lambda])\)} }], "\[VerticalSeparator]"}], "TraditionalForm"], " ", "=", RowBox[{"K", " ", RowBox[{ FormBox[\(e\^\(\[Integral]\_\(\(\ \)\(\[Xi]\)\)\%\(L\_i\)\ \ \(-\(p(x)\)\) \[DifferentialD]x\)\), "TraditionalForm"], " ", ".", " "}]}]}]}], TraditionalForm]]], " \n \nWhen ", Cell[BoxData[ \(TraditionalForm\`L\_i = \[Xi]\)]], ", the matrix ", Cell[BoxData[ \(TraditionalForm\`M\_j\)]], " becomes the identity matrix, and so we see that the constant ", Cell[BoxData[ \(TraditionalForm\`K\)]], " = 1. We define\n\n ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\[CapitalDelta]\_j\), "TraditionalForm"], FormBox[\(\((L\_j, \[Xi])\)\ = \ \(-\(\[Integral]\_\(\(\ \)\(\[Xi]\ \)\)\%\(\(\ \)\(L\_j\)\)\(p(x)\) \[DifferentialD]x\)\)\), "TraditionalForm"]}], TraditionalForm]]], ".\n Then we have that \n \ ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\[CapitalDelta]\_j\), "TraditionalForm"], FormBox[\(\((L\_j, \[Xi])\)\(\[CenterDot]\)\), "TraditionalForm"]}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{ FormBox[ FormBox[\(P\_j\), "TraditionalForm"], "TraditionalForm"], "(", RowBox[{"\[Xi]", ",", FormBox[\(L\_j\), "TraditionalForm"]}], ")"}], TraditionalForm]]], " = ", Cell[BoxData[ FormBox[ RowBox[{"J", "\[CenterDot]", SuperscriptBox[ RowBox[{"(", RowBox[{ FormBox[ FormBox[\(P\_j\), "TraditionalForm"], "TraditionalForm"], "(", RowBox[{ FormBox[\(L\_j\), "TraditionalForm"], ",", "\[Xi]"}], ")"}], ")"}], "T"], "\[CenterDot]", "J"}], TraditionalForm]]], ".\nThis property of the exterior matrices (equation (5.1) in the paper) is \ the key to reconstructing the entire matrix ", Cell[BoxData[ \(TraditionalForm\`P\_j\)]], " from a single function. Because ", Cell[BoxData[ FormBox[ SqrtBox[ RowBox[{ FormBox[\(\[CapitalDelta]\_j\), "TraditionalForm"], FormBox[\((L\_j, \[Xi])\), "TraditionalForm"]}]], TraditionalForm]]], " is often needed, this notebook will use ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\[Delta]\_j\), "TraditionalForm"], FormBox[\((L\_j, \[Xi])\), "TraditionalForm"]}], TraditionalForm]]], " for the square root of ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\[CapitalDelta]\_j\), "TraditionalForm"], FormBox[\((L\_j, \[Xi])\), "TraditionalForm"]}], TraditionalForm]]], "." }], "Text"], Cell[BoxData[ \(\[Delta][Lj_, \[Xi]_]\ := \ E^\(-Integrate[p[s]/2, {s, \[Xi], Lj}]\)\)], "Input"], Cell[BoxData[ \(\[Delta][Lj, \[Xi]]\)], "Input"], Cell["But another key is determining whether the quantity", "Text"], Cell[BoxData[ \(Q[x_]\ := \ \(-\ p[x]^3\)\ + \ 4\ p[x]\ q[x] - 8\ r[x] - 6 p[x]\ \(p'\)[x] + \ 8\ \(q'\)[x]\ - \ 4\ \(p''\)[x]\)], "Input"], Cell[TextData[{ "is zero or not. For if ", Cell[BoxData[ \(TraditionalForm\`Q(x)\ = \ 0\)]], ", then \n ", Cell[BoxData[ \(TraditionalForm\`2 \(\(\(\[PartialD]\)\(\ \)\)\/\[PartialD]L\_j\) P\_\(j, 3 v\) + \ \ \(p(L\_j)\) P\_\(j, 3 v\) = \ \(\(\(\[PartialD]\^3\)\(\ \ \)\)\/\[PartialD]L\_j\^3\) P\_\(j, 1 v\)\ - \ \ \(q( L\_j)\) \(\(\(\[PartialD]\)\(\ \)\)\/\[PartialD]L\_j\) P\_\(j, 1 v\)\ - \ \ \(r(L\_j)\) P\_\(j, 1 v\)\)]], "\ncan be solved exactly via integration by parts. In fact, the solution \ is that \n ", Cell[BoxData[ FormBox[ RowBox[{\(P\_\(j, 3 v\)\), "=", " ", RowBox[{\(\(1\/2\) \(\(\(\[PartialD]\^2\)\(\ \ \)\)\/\[PartialD]L\_j\^2\) P\_\(j, 1 v\)\), " ", "-", " ", \(1\/4\ \(p( L\_j)\) \(\(\(\[PartialD]\)\(\ \)\)\/\[PartialD]L\_j\) P\_\(j, 1 v\)\), " ", "+", " ", \(\((\(1\/4\) \(\(p\^\[Prime]\)( L\_j)\)\ + \(1\/8\) \(p(L\_j)\)\^2 - \(1\/2\) \(q( L\_j)\))\) P\_\(j, 1 v\)\), "+", " ", RowBox[{\(\(C\_v\)(\[Xi])\), " ", SqrtBox[ RowBox[{ FormBox[\(\[CapitalDelta]\_j\), "TraditionalForm"], FormBox[\((L\_j, \[Xi])\), "TraditionalForm"]}]]}]}]}], TraditionalForm]]], ".\nwhere the function ", Cell[BoxData[ \(TraditionalForm\`\(C\_v\)(\[Xi])\)]], " depends only on \[Xi]. To see this, let" }], "Text"], Cell[BoxData[ \(P3v\ = \ \(P1v''\)[Lj]/2\ - \ p[Lj]\ \(P1v'\)[Lj]/ 4\ + \ \((\(p'\)[Lj]/4\ + \ p[Lj]^2/8\ - \ q[Lj]/2)\) P1v[Lj]\ + \ C\ \[Delta][Lj, \[Xi]]\)], "Input"], Cell[TextData[{ "Then ", Cell[BoxData[ FormBox[ RowBox[{" ", RowBox[{ FormBox[\(2 \(\(\(\[PartialD]\)\(\ \)\)\/\[PartialD]L\_j\) P\_\(j, 3 v\) + \ \ \(p(L\_j)\) P\_\(j, 3 v\)\), "TraditionalForm"], FormBox[\(\(-\(\(\(\[PartialD]\^3\)\(\ \)\)\/\[PartialD]L\_j\^3\)\ \) P\_\(j, 1 v\)\ + \ \ \(q( L\_j)\) \(\(\(\[PartialD]\)\(\ \)\)\/\[PartialD]L\_j\) P\_\(j, 1 v\)\ + \ \ \(r(L\_j)\) P\_\(j, 1 v\)\), "TraditionalForm"]}]}], TraditionalForm]]], " becomes" }], "Text"], Cell[BoxData[ \(2\ D[P3v, Lj]\ + \ p[Lj]\ P3v\ - \ \(P1v'''\)[Lj]\ + \ q[Lj]\ \(P1v'\)[Lj]\ + \ r[Lj]\ P1v[Lj]\)], "Input"], Cell[BoxData[ \(Factor[%]\)], "Input"], Cell[TextData[{ "And we see that if ", Cell[BoxData[ \(TraditionalForm\`Q(x)\ = \ 0\)]], ", that this would become 0." }], "Text"], Cell[BoxData[ \(Expand[%\ + \ P1v[Lj]*Q[Lj]/8]\)], "Input"], Cell[TextData[{ "This verifies equation (5.5) in the paper. In particular, if we consider \ the case ", Cell[BoxData[ \(TraditionalForm\`v = 4\)]], ", then using the equations in (5.3) we have \n\n", Cell[BoxData[ FormBox[ RowBox[{\(P\_\(j, 34\)\), "=", " ", RowBox[{\(\(1\/2\) \((P\_\(j, 34\)\ + P\_\(j, 64\))\)\), "-", " ", \(1\/4\ \(p(L\_j)\) P\_\(j, 24\)\), " ", "+", " ", \(\((\(1\/4\) \(\(p\^\[Prime]\)( L\_j)\)\ + \(1\/8\) \(p(L\_j)\)\^2 - \(1\/2\) \(q( L\_j)\))\) P\_\(j, 14\)\), "+", " ", RowBox[{\(\(C\_4\)(\[Xi])\), " ", SqrtBox[ RowBox[{ FormBox[\(\[CapitalDelta]\_j\), "TraditionalForm"], FormBox[\((L\_j, \[Xi])\), "TraditionalForm"]}]]}]}]}], TraditionalForm]]], ".\n\nBut we know that when ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(L\_j\), "TraditionalForm"], "=", " ", "\[Xi]"}], TraditionalForm]]], ", ", Cell[BoxData[ \(TraditionalForm\`P\_j\)]], " becomes the identity matrix. Thus we have\n\n \ ", Cell[BoxData[ FormBox[ RowBox[{\(\(P\_\(j, u\ v\)\)\( \[VerticalSeparator] \_\(L\_j = \[Xi]\)\)\), " ", "=", " ", TagBox[ StyleBox[ RowBox[{"{", StyleBox[GridBox[{ {\(0\ \ \ if\ \ \ u \[NotEqual] v\)}, {\(1\ \ \ if\ \ \ u = v\)} }], ShowAutoStyles->True]}], ShowAutoStyles->False], (#&)]}], TraditionalForm]]], "\n \nand since ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\[CapitalDelta]\_j\), "TraditionalForm"], FormBox[\(\((\[Xi], \[Xi])\) = 1\), "TraditionalForm"]}], TraditionalForm]]], ", we find that ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\(C\_4\)(\[Xi])\), "TraditionalForm"], " ", "=", "0"}], TraditionalForm]]], ". \n\nNow that we have ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, 34\)\)]], " expressed in terms of ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, 14\)\)]], ", we can find a differential equation for which ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, 14\)\)]], " solves." }], "Text"], Cell[BoxData[ \(P34[Lj_]\ := \ \(P14''\)[Lj]/2\ - \ p[Lj]\ \(P14'\)[Lj]/ 4\ + \ \((\(p'\)[Lj]/4\ + \ p[Lj]^2/8\ - \ q[Lj]/2)\) P14[Lj]\)], "Input"], Cell[BoxData[ \(P44[Lj_]\ := \ \(-\ \(P54'\)[Lj]\)\ - \ p[Lj]\ P54[Lj]\ + \ q[Lj]\ P64[Lj]\ - \ s[Lj]\ P14[Lj]\)], "Input"], Cell[BoxData[ \(P54[Lj_]\ := \ \(-\ \(P64'\)[Lj]\)\)], "Input"], Cell[BoxData[ \(P64[Lj_]\ := \ \(P24'\)[Lj]\ - \ P34[Lj]\)], "Input"], Cell[BoxData[ \(P24[Lj_]\ := \ \(P14'\)[Lj]\)], "Input"], Cell[TextData[{ "Now any expression involving ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, 24\)\)]], ", ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, 34\)\)]], ", ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, 44\)\)]], ", ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, 54\)\)]], ", or ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, 64\)\)]], " will be converted to an expression involving only ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, 14\)\)]], ". In particular, we have the last equation from the set of equations in \ (5.3), which will give us a differential equation for which ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, 14\)\)]], " solves." }], "Text"], Cell[BoxData[ \(\(P44'\)[Lj]\ + \ p[Lj]\ P44[Lj]\ - \ r[Lj]\ P64[Lj]\ - \ s[Lj]\ P24[Lj]\)], "Input"], Cell[BoxData[ \(Expand[%]\)], "Input"], Cell["Multiply by the common denominator, 8,", "Text"], Cell[BoxData[ \(Expand[8*%]\)], "Input"], Cell[TextData[{ "and after rearranging the terms, we get equation (5.13) from the paper.\n\n\ However, instead of an equation for ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, 14\)\)]], ", which is not symmetric in ", Cell[BoxData[ \(TraditionalForm\`L\_j\)]], " and \[Xi], we want a differential equation for\n", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"f", "(", RowBox[{ FormBox[\(L\_j\), "TraditionalForm"], ",", "\[Xi]"}], ")"}], " ", "=", " ", FormBox[ RowBox[{ RowBox[{\(P\_\(j, 14\)\), "(", RowBox[{ FormBox[\(L\_j\), "TraditionalForm"], ",", "\[Xi]"}], ")"}], "/", SqrtBox[ RowBox[{ FormBox[\(\[CapitalDelta]\_j\), "TraditionalForm"], FormBox[\((L\_j, \[Xi])\), "TraditionalForm"]}]]}], "TraditionalForm"]}], TraditionalForm]]], ". We can find this equation easily by setting" }], "Text"], Cell[BoxData[ \(P14[Lj_]\ := \ f[Lj]\ \[Delta][Lj, \[Xi]]\)], "Input"], Cell[BoxData[ \(%%\)], "Input"], Cell[TextData[{ "We of course need to divide by ", Cell[BoxData[ FormBox[ RowBox[{" ", FormBox[\(e\^\(-\(\[Integral]\_\(\(\ \)\(\[Xi]\)\)\%\(L\_j\)\ \(p( s)\)/2 \[DifferentialD]s\)\)\), "TraditionalForm"]}], TraditionalForm]]], " and expand" }], "Text"], Cell[BoxData[ \(Expand[%/\[ExponentialE]\^\(-\(\[Integral]\_\[Xi]\%Lj\( p[s]\/2\) \ \[DifferentialD]s\)\)]\)], "Input"], Cell[TextData[{ "This can be simplified by subtracting a multiple of ", Cell[BoxData[ FormBox[ RowBox[{"Q", "(", FormBox[\(L\_j\), "TraditionalForm"], ")"}], TraditionalForm]]], ", which of course we are assuming to be 0." }], "Text"], Cell[BoxData[ \(%\ - \ Q[Lj] \((\(f''\)[Lj]/2\ - \ p[Lj]\ \(f'\)[Lj]/4 - \ p[Lj]^2 f[Lj]/8\ + \ q[Lj]\ f[Lj]/2\ - \ \(p'\)[Lj] f[Lj])\)\)], "Input"], Cell[BoxData[ \(Expand[%]\)], "Input"], Cell["Now we multiply by 4 to eliminate that one fraction.", "Text"], Cell[BoxData[ \(Expand[4*%]\)], "Input"], Cell[TextData[{ "After rearranging the terms, this gives us equation (5.7) from the paper. \ But we also need to see the initial conditions that are satisfied by ", Cell[BoxData[ FormBox[ RowBox[{"f", "(", RowBox[{ FormBox[\(L\_j\), "TraditionalForm"], ",", "\[Xi]"}], ")"}], TraditionalForm]]], ". To do this, we will define ", Cell[BoxData[ FormBox[ RowBox[{"f", "(", RowBox[{ FormBox[\(L\_j\), "TraditionalForm"], ",", "\[Xi]"}], ")"}], TraditionalForm]]], " back in terms of ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, 14\)\)]], "." }], "Text"], Cell[BoxData[ \(fj[Lj_]\ := \ Pj14[Lj]\ /\[Delta][Lj, \[Xi]]\)], "Input"], Cell[TextData[{ "Now we define how to take derivatives of ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, u\ 4\)\)]], ", where ", Cell[BoxData[ \(TraditionalForm\`u\)]], " is between 1 and 6." }], "Text"], Cell[BoxData[ \(\(Pj14'\)[Lj]\ := \ Pj24[Lj]\)], "Input"], Cell[BoxData[ \(\(Pj24'\)[Lj]\ := \ Pj34[Lj]\ + \ Pj64[Lj]\)], "Input"], Cell[BoxData[ \(\(Pj34'\)[Lj]\ := \ \(-\ Pj54[Lj]\) - p[Lj]\ Pj34[Lj]\ - \ q[Lj]\ Pj24[Lj]\ - \ r[Lj]\ Pj14[Lj]\)], "Input"], Cell[BoxData[ \(\(Pj44'\)[Lj]\ := \ \(-\ p[Lj]\)\ Pj44[Lj]\ + \ r[Lj]\ Pj64[Lj]\ + \ s[Lj]\ Pj24[Lj]\)], "Input"], Cell[BoxData[ \(\(Pj54'\)[Lj]\ := \ \(-\ Pj44[Lj]\)\ - \ p[Lj]\ Pj54[Lj]\ + \ q[Lj]\ Pj64[Lj]\ - \ s[Lj] Pj14[Lj]\)], "Input"], Cell[BoxData[ \(\(Pj64'\)[Lj]\ := \ \(-\ Pj54[Lj]\)\)], "Input"], Cell[TextData[{ "Now each time we take a derivative, it expresses the answer in terms of \ ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, u\ 4\)\)]], " without derivatives. Plugging in ", Cell[BoxData[ \(TraditionalForm\`L\_j = \[Xi]\)]], " amounts to setting ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, 44\)\)]], " to 1, and the other ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, u\ 4\)\)]], " to 0. " }], "Text"], Cell[BoxData[ \(D[fj[Lj], Lj]\)], "Input"], Cell[BoxData[ \(%\ /. \ {Pj14[Lj] \[Rule] 0, \ Pj24[Lj] \[Rule] 0, Pj34[Lj] \[Rule] 0, Pj44[Lj] \[Rule] 1, Pj54[Lj] \[Rule] 0, Pj64[Lj] \[Rule] 0}\)], "Input"], Cell[BoxData[ \(D[fj[Lj], {Lj, 2}]\)], "Input"], Cell[BoxData[ \(%\ /. \ {Pj14[Lj] \[Rule] 0, \ Pj24[Lj] \[Rule] 0, Pj34[Lj] \[Rule] 0, Pj44[Lj] \[Rule] 1, Pj54[Lj] \[Rule] 0, Pj64[Lj] \[Rule] 0}\)], "Input"], Cell[BoxData[ \(D[fj[Lj], {Lj, 3}]\)], "Input"], Cell[BoxData[ \(%\ /. \ {Pj14[Lj] \[Rule] 0, \ Pj24[Lj] \[Rule] 0, Pj34[Lj] \[Rule] 0, Pj44[Lj] \[Rule] 1, Pj54[Lj] \[Rule] 0, Pj64[Lj] \[Rule] 0}\)], "Input"], Cell[BoxData[ \(D[fj[Lj], {Lj, 4}]\)], "Input"], Cell[BoxData[ \(%\ /. \ {Pj14[Lj] \[Rule] 0, \ Pj24[Lj] \[Rule] 0, Pj34[Lj] \[Rule] 0, Pj44[Lj] \[Rule] 1, Pj54[Lj] \[Rule] 0, Pj64[Lj] \[Rule] 0}\)], "Input"], Cell[BoxData[ \(%\ /. \ Lj\ \[Rule] \ \[Xi]\)], "Input"], Cell[TextData[{ "Thus, ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(f(\[Xi], \[Xi])\ = \ \(\(\(\(\(\(\[PartialD]\)\(\ \)\)\/\ \[PartialD]L\_j\) f\)\( \[VerticalSeparator] \_\(\_\(L\_j = \[Xi]\)\)\)\)\ \ = \(\(\(\(\(\(\[PartialD]\^2\)\(\ \)\)\/\[PartialD]L\_j\^2\) f\)\( \[VerticalSeparator] \_\(\_\(L\_j = \[Xi]\)\)\)\)\ \(\ \)\(=\)\)\)\), "TraditionalForm"], FormBox[\(\(\ \)\(\(\(\(\(\(\[PartialD]\^3\)\(\ \ \)\)\/\[PartialD]L\_j\^3\) f\)\( \[VerticalSeparator] \_\(\_\(L\_j = \[Xi]\)\)\)\)\(\ \ \)\(=\)\)\), "TraditionalForm"], " ", "0"}], TraditionalForm]]], ", ", Cell[BoxData[ \(TraditionalForm\`\(\(\(\(\(\[PartialD]\^4\)\(\ \ \)\)\/\[PartialD]L\_j\^4\) f\)\( \[VerticalSeparator] \_\(\_\(L\_j = \[Xi]\)\)\)\)\ = \ 2. \)]], "\n\nThis gives us equation (5.8) in the paper. We now are ready to \ construct the full matrix ", Cell[BoxData[ \(TraditionalForm\`P\_j\)]], " in terms of the function ", Cell[BoxData[ FormBox[ RowBox[{"f", "(", RowBox[{ FormBox[\(L\_j\), "TraditionalForm"], ",", "\[Xi]"}], ")"}], TraditionalForm]]], ". First we define a blank matrix" }], "Text"], Cell[BoxData[ \(P\ = \ {{0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0}}\)], "Input"], Cell[BoxData[ \(\[Delta][Lj_, \[Xi]_]\ := \ E^\(-Integrate[p[s]/2, {s, \[Xi], Lj}]\)\)], "Input"], Cell[TextData[{ "We know that ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(P\_\(j, 14\)\), "TraditionalForm"], " ", "=", RowBox[{ FormBox[ RowBox[{"f", "(", RowBox[{ FormBox[\(L\_j\), "TraditionalForm"], ",", "\[Xi]"}], ")"}], "TraditionalForm"], FormBox[ SqrtBox[ RowBox[{ FormBox[\(\[CapitalDelta]\_j\), "TraditionalForm"], FormBox[\((L\_j, \[Xi])\), "TraditionalForm"]}]], "TraditionalForm"]}]}], TraditionalForm]]], ", and we found a formula for ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, 34\)\)]], " in terms of ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, 14\)\)]], "." }], "Text"], Cell[BoxData[ \(P[\([1, 4]\)]\ = \ f[Lj, \[Xi]]\ \[Delta][Lj, \[Xi]]\)], "Input"], Cell[BoxData[ \(P[\([3, 4]\)]\ = \ Simplify[\ D[P[\([1, 4]\)], {Lj, 2}]/2\ - \ p[Lj]\ D[P[\([1, 4]\)], Lj]/ 4\ + \ \((\(p'\)[Lj]/4\ + \ p[Lj]^2/8\ - \ q[Lj]/2)\) P[\([1, 4]\)]\ ]\)], "Input"], Cell["\<\ and we can use the equations in (5.3) to fill in the rest of the 4th \ column.\ \>", "Text"], Cell[BoxData[ \(P[\([2, 4]\)]\ = \ Simplify[\ D[P[\([1, 4]\)], Lj]\ ]\)], "Input"], Cell[BoxData[ \(P[\([6, 4]\)]\ = \ Simplify[\ D[P[\([2, 4]\)], Lj]\ - \ P[\([3, 4]\)]\ ]\)], "Input"], Cell[BoxData[ \(P[\([5, 4]\)]\ = \ Simplify[\ \(-\ D[P[\([6, 4]\)], Lj]\)\ ]\)], "Input"], Cell[BoxData[ \(P[\([4, 4]\)] = \ Simplify[\ \(-\ D[P[\([5, 4]\)], Lj]\) - \ p[Lj]\ P[\([5, 4]\)]\ + \ q[Lj]\ P[\([6, 4]\)]\ - \ s[Lj]\ P[\([1, 4]\)]\ ]\)], "Input"], Cell[TextData[{ "But we can find the first row by using the equation ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\[CapitalDelta]\_j\), "TraditionalForm"], FormBox[\(\((L\_j, \[Xi])\)\(\[CenterDot]\)\), "TraditionalForm"]}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{ FormBox[ FormBox[\(P\_j\), "TraditionalForm"], "TraditionalForm"], "(", RowBox[{"\[Xi]", ",", FormBox[\(L\_j\), "TraditionalForm"]}], ")"}], TraditionalForm]]], " = ", Cell[BoxData[ FormBox[ RowBox[{"J", "\[CenterDot]", SuperscriptBox[ RowBox[{"(", RowBox[{ FormBox[ FormBox[\(P\_j\), "TraditionalForm"], "TraditionalForm"], "(", RowBox[{ FormBox[\(L\_j\), "TraditionalForm"], ",", "\[Xi]"}], ")"}], ")"}], "T"], "\[CenterDot]", "J"}], TraditionalForm]]], ". First we need a function that switches the roles of ", Cell[BoxData[ \(TraditionalForm\`L\_j\)]], " and \[Xi]: " }], "Text"], Cell[BoxData[ RowBox[{\(Exchange[qq_]\), " ", ":=", " ", RowBox[{"Module", "[", RowBox[{\({temp}\), ",", "\[IndentingNewLine]", RowBox[{\(temp\ = \ qq\ /. \ Lj\ \[Rule] \ xxx\), ";", "\[IndentingNewLine]", \(temp\ = \ temp\ /. \ \[Xi]\ \[Rule] \ Lj\), ";", "\[IndentingNewLine]", \(temp\ = \ temp\ /. \ xxx\ \[Rule] \ \[Xi]\), ";", "\[IndentingNewLine]", \(temp\ = \ temp\ /. \ f[\[Xi], Lj]\ \[Rule] \ f[Lj, \[Xi]]\), ";", "\[IndentingNewLine]", RowBox[{"temp", " ", "=", " ", RowBox[{"temp", " ", "/.", " ", RowBox[{ RowBox[{ SuperscriptBox["f", TagBox[\((ii_, jj_)\), Derivative], MultilineFunction->None], "[", \(\[Xi], Lj\), "]"}], " ", "\[Rule]", " ", RowBox[{ SuperscriptBox["f", TagBox[\((jj, ii)\), Derivative], MultilineFunction->None], "[", \(Lj, \[Xi]\), "]"}]}]}]}], ";", "\[IndentingNewLine]", \(temp\ = \ temp\ /. \ \ \[ExponentialE]\^\(-\(\[Integral]\_Lj\%\[Xi]\( \ p[s]\/2\) \[DifferentialD]s\)\)\ \[Rule] \ \[ExponentialE]\^\(\[Integral]\_\ \[Xi]\%Lj\( p[s]\/2\) \[DifferentialD]s\)\), ";", "\[IndentingNewLine]", \(Return[temp]\)}]}], "]"}]}]], "Input"], Cell[TextData[{ "This routine allows us to find ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ FormBox[\(P\_\(j, u\ v\)\), "TraditionalForm"], "TraditionalForm"], "(", RowBox[{"\[Xi]", ",", FormBox[\(L\_j\), "TraditionalForm"]}], ")"}], TraditionalForm]]], " in terms of ", Cell[BoxData[ FormBox[ FormBox[ RowBox[{"f", "(", RowBox[{ FormBox[\(L\_j\), "TraditionalForm"], ",", "\[Xi]"}], ")"}], "TraditionalForm"], TraditionalForm]]], ". For example, ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ FormBox[\(P\_\(j, 24\)\), "TraditionalForm"], "TraditionalForm"], "(", RowBox[{"\[Xi]", ",", FormBox[\(L\_j\), "TraditionalForm"]}], ")"}], TraditionalForm]]], " would be" }], "Text"], Cell[BoxData[ \(Exchange[P[\([2, 4]\)]]\)], "Input"], Cell[TextData[{ "which now has derivatives with respect to \[Xi]. Since ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ FormBox[ FormBox[\(P\_\(j, 15\)\), "TraditionalForm"], "TraditionalForm"], "(", FormBox[\(L\_j, \[Xi]\), "TraditionalForm"], ")"}], " ", "=", " "}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\[CapitalDelta]\_j\), "TraditionalForm"], FormBox[\(\((L\_j, \[Xi])\)\(\[CenterDot]\)\), "TraditionalForm"]}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{ FormBox[ FormBox[\(P\_\(j, 24\)\), "TraditionalForm"], "TraditionalForm"], "(", RowBox[{"\[Xi]", ",", FormBox[\(L\_j\), "TraditionalForm"]}], ")"}], TraditionalForm]]], ", we have" }], "Text"], Cell[BoxData[ \(P[\([1, 5]\)]\ = \ Exchange[P[\([2, 4]\)]]\ \[Delta][Lj, \[Xi]]^2\)], "Input"], Cell["Likewise, we have", "Text"], Cell[BoxData[ \(P[\([1, 1]\)]\ = \ Exchange[P[\([4, 4]\)]]\ \[Delta][Lj, \[Xi]]^2\)], "Input"], Cell[BoxData[ \(P[\([1, 2]\)]\ = \ Exchange[P[\([5, 4]\)]] \[Delta][Lj, \[Xi]]^2\)], "Input"], Cell[BoxData[ \(P[\([1, 3]\)]\ = Exchange[P[\([6, 4]\)]] \[Delta][Lj, \[Xi]]^2\)], "Input"], Cell[BoxData[ \(P[\([1, 6]\)]\ = \ Exchange[P[\([3, 4]\)]] \[Delta][Lj, \[Xi]]^2\)], "Input"], Cell[TextData[{ "Finding the element ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ FormBox[ FormBox[\(P\_\(j, \ 36\)\), "TraditionalForm"], "TraditionalForm"], "(", FormBox[\(L\_j, \[Xi]\), "TraditionalForm"], ")"}], " "}], TraditionalForm]]], "is a bit trickier. If we eliminate the derivatives in the equation\n\n ", Cell[BoxData[ FormBox[ RowBox[{\(P\_\(j, 3 v\)\), "=", " ", RowBox[{\(\(1\/2\) \(\(\(\[PartialD]\^2\)\(\ \ \)\)\/\[PartialD]L\_j\^2\) P\_\(j, 1 v\)\), " ", "-", " ", \(1\/4\ \(p( L\_j)\) \(\(\(\[PartialD]\)\(\ \)\)\/\[PartialD]L\_j\) P\_\(j, 1 v\)\), " ", "+", " ", \(\((\(1\/4\) \(\(p\^\[Prime]\)( L\_j)\)\ + \(1\/8\) \(p(L\_j)\)\^2 - \(1\/2\) \(q( L\_j)\))\) P\_\(j, 1 v\)\), "+", " ", RowBox[{\(\(C\_v\)(\[Xi])\), " ", SqrtBox[ RowBox[{ FormBox[\(\[CapitalDelta]\_j\), "TraditionalForm"], FormBox[\((L\_j, \[Xi])\), "TraditionalForm"]}]]}]}]}], TraditionalForm]]], "\n \n we get, for ", Cell[BoxData[ \(TraditionalForm\`v\ = \ 6\)]], ", \n \n ", Cell[BoxData[ FormBox[ RowBox[{\(P\_\(j, 36\)\), "=", " ", RowBox[{\(\(1\/2\) \((P\_\(j, 36\)\ + P\_\(j, 66\))\)\), "-", " ", \(1\/4\ \(p(L\_j)\) P\_\(j, 26\)\), " ", "+", " ", \(\((\(1\/4\) \(\(p\^\[Prime]\)( L\_j)\)\ + \(1\/8\) \(p(L\_j)\)\^2 - \(1\/2\) \(q( L\_j)\))\) P\_\(j, 16\)\), "+", " ", RowBox[{\(\(C\_6\)(\[Xi])\), " ", SqrtBox[ RowBox[{ FormBox[\(\[CapitalDelta]\_j\), "TraditionalForm"], FormBox[\((L\_j, \[Xi])\), "TraditionalForm"]}]]}]}]}], TraditionalForm]]], ".\n \n Now setting ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(L\_j\), "TraditionalForm"], "=", "\[Xi]"}], TraditionalForm]]], " tells us that ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\(C\_6\)(\[Xi])\), "TraditionalForm"], "=", \(\(-1\)/2\)}], TraditionalForm]]], ". Thus, " }], "Text"], Cell[BoxData[ \(P[\([3, 6]\)]\ = \ Simplify[\ D[P[\([1, 6]\)], {Lj, 2}]/2\ - \ p[Lj]\ D[P[\([1, 6]\)], Lj]/ 4\ + \ \((\(p'\)[Lj]/4\ + \ p[Lj]^2/8\ - \ q[Lj]/2)\) P[\([1, 6]\)]\ \ - \ \[Delta][Lj, \[Xi]]/2]\)], "Input"], Cell["We now can finish the 6th column.", "Text"], Cell[BoxData[ \(P[\([2, 6]\)]\ = \ Simplify[\ D[P[\([1, 6]\)], Lj]\ ]\)], "Input"], Cell[BoxData[ \(P[\([6, 6]\)]\ = \ Simplify[\ D[P[\([2, 6]\)], Lj]\ - \ P[\([3, 6]\)]\ ]\)], "Input"], Cell[BoxData[ \(P[\([5, 6]\)]\ = \ Simplify[\ \(-\ D[P[\([6, 6]\)], Lj]\)\ ]\)], "Input"], Cell[BoxData[ \(P[\([4, 6]\)] = \ Simplify[\ \(-\ D[P[\([5, 6]\)], Lj]\) - \ p[Lj]\ P[\([5, 6]\)]\ + \ q[Lj]\ P[\([6, 6]\)]\ - \ s[Lj]\ P[\([1, 6]\)]\ ]\)], "Input"], Cell["We can now use the Exchange function to fill in the 3rd row.", "Text"], Cell[BoxData[ \(P[\([3, 1]\)]\ = \ Exchange[P[\([4, 6]\)]]\ \[Delta][Lj, \[Xi]]^2\)], "Input"], Cell[BoxData[ \(P[\([3, 2]\)]\ = \ Exchange[P[\([5, 6]\)]]\ \[Delta][Lj, \[Xi]]^2\)], "Input"], Cell[BoxData[ \(P[\([3, 3]\)]\ = \ Exchange[P[\([6, 6]\)]]\ \[Delta][Lj, \[Xi]]^2\)], "Input"], Cell[BoxData[ \(P[\([3, 5]\)]\ = \ Exchange[P[\([2, 6]\)]]\ \[Delta][Lj, \[Xi]]^2\)], "Input"], Cell["\<\ We now can use the equations in (5.3) to complete the matrix.\ \>", "Text"], Cell[BoxData[ \(P[\([2, 1]\)]\ = \ Simplify[\ D[P[\([1, 1]\)], Lj]\ ]\)], "Input"], Cell[BoxData[ \(P[\([2, 2]\)]\ = \ Simplify[\ D[P[\([1, 2]\)], Lj]\ ]\)], "Input"], Cell[BoxData[ \(P[\([2, 3]\)]\ = \ Simplify[\ D[P[\([1, 3]\)], Lj]\ ]\)], "Input"], Cell[BoxData[ \(P[\([2, 5]\)]\ = \ Simplify[\ D[P[\([1, 5]\)], Lj]\ ]\)], "Input"], Cell[BoxData[ \(P[\([6, 1]\)]\ = \ Simplify[\ D[P[\([2, 1]\)], Lj]\ - \ P[\([3, 1]\)]\ ]\)], "Input"], Cell[BoxData[ \(P[\([6, 2]\)]\ = \ Simplify[\ D[P[\([2, 2]\)], Lj]\ - \ P[\([3, 2]\)]\ ]\)], "Input"], Cell[BoxData[ \(P[\([6, 3]\)]\ = \ Simplify[\ D[P[\([2, 3]\)], Lj]\ - \ P[\([3, 3]\)]\ ]\)], "Input"], Cell[BoxData[ \(P[\([6, 5]\)]\ = \ Simplify[\ D[P[\([2, 5]\)], Lj]\ - \ P[\([3, 5]\)]\ ]\)], "Input"], Cell[BoxData[ \(P[\([5, 1]\)]\ = \ Simplify[\ \(-\ D[P[\([6, 1]\)], Lj]\)\ ]\)], "Input"], Cell[BoxData[ \(P[\([5, 2]\)]\ = \ Simplify[\ \(-\ D[P[\([6, 2]\)], Lj]\)\ ]\)], "Input"], Cell[BoxData[ \(P[\([5, 3]\)]\ = \ Simplify[\ \(-\ D[P[\([6, 3]\)], Lj]\)\ ]\)], "Input"], Cell[BoxData[ \(P[\([5, 5]\)]\ = \ Simplify[\ \(-\ D[P[\([6, 5]\)], Lj]\)\ ]\)], "Input"], Cell[BoxData[ \(P[\([4, 1]\)] = \ Simplify[\ \(-\ D[P[\([5, 1]\)], Lj]\) - \ p[Lj]\ P[\([5, 1]\)]\ + \ q[Lj]\ P[\([6, 1]\)]\ - \ s[Lj]\ P[\([1, 1]\)]\ ]\)], "Input"], Cell[BoxData[ \(P[\([4, 2]\)] = \ Simplify[\ \(-\ D[P[\([5, 2]\)], Lj]\) - \ p[Lj]\ P[\([5, 2]\)]\ + \ q[Lj]\ P[\([6, 2]\)]\ - \ s[Lj]\ P[\([1, 2]\)]\ ]\)], "Input"], Cell[BoxData[ \(P[\([4, 3]\)] = \ Simplify[\ \(-\ D[P[\([5, 3]\)], Lj]\) - \ p[Lj]\ P[\([5, 3]\)]\ + \ q[Lj]\ P[\([6, 3]\)]\ - \ s[Lj]\ P[\([1, 3]\)]\ ]\)], "Input"], Cell[BoxData[ \(P[\([4, 5]\)] = \ Simplify[\ \(-\ D[P[\([5, 5]\)], Lj]\) - \ p[Lj]\ P[\([5, 5]\)]\ + \ q[Lj]\ P[\([6, 5]\)]\ - \ s[Lj]\ P[\([1, 5]\)]\ ]\)], "Input"], Cell[TextData[{ "We have finished defining the matrix ", Cell[BoxData[ \(TraditionalForm\`P\_j\)]], " in terms of the single function ", Cell[BoxData[ FormBox[ RowBox[{"f", "(", RowBox[{ FormBox[\(L\_j\), "TraditionalForm"], ",", "\[Xi]"}], ")"}], TraditionalForm]]], ". But there is a more efficient way of doing this! If we define the \ operator matrix\n \n ", Cell[BoxData[ FormBox[ RowBox[{" ", RowBox[{ FormBox[\(\(D\_x\)\(\ \)\(=\)\), "TraditionalForm"], FormBox[ RowBox[{"(", GridBox[{ {"1"}, {\(\(\(\[PartialD]\)\(\ \)\)\/\[PartialD]x - \ \(p(x)\)\/2\)}, {\(\(1\/2\) \(\(\[PartialD]\^2\)\(\ \)\)\/\[PartialD]x\^2 \ - \(3\/4\) \(p( x)\) \(\(\[PartialD]\)\(\ \)\)\/\[PartialD]x + \ \ \(3 \( p(x)\)\^2 - \ 4\ \(q(x)\)\)\/8\)}, { RowBox[{"(", GridBox[{ {\(\(1\/2\) \(\(\[PartialD]\^4\)\(\ \)\)\/\ \[PartialD]x\^4 - \(1\/4\) \(p( x)\) \(\(\[PartialD]\^3\)\(\ \)\)\/\ \[PartialD]x\^3 + \(\(4 \( q(x)\) - 5 p' \((x)\) - \(p(x)\)\^2\)\/4\) \(\(\ \[PartialD]\^2\)\(\ \)\)\/\[PartialD]x\^2 + \(\(\(p(x)\)\^3 - 4 \( p(x)\) \(q(x)\) - 6 \( p(x)\) p' \((x)\) + 16 q' \((x)\) - 20 p'' \((x)\)\)\/16\) \(\(\ \[PartialD]\)\(\ \)\)\/\[PartialD]x\)}, {\(+\(\(\(p(x)\)\^4 - 8 \(\( p(x)\)\^2\) \(q(x)\) + 16 \( q(x)\)\^2 - 32 \( s(x)\) + 6 \(\( p(x)\)\^2\) p' \((x)\) - 24 \( q(x)\) p' \((x)\) - 8 \( p(x)\) p'' \((x)\) + 16 q'' \((x)\) - 16 p''' \((x)\)\)\/32\)\)} }], ")"}]}, {\(\(-\(1\/2\)\) \(\(\[PartialD]\^3\)\(\ \ \)\)\/\[PartialD]x\^3 + \(1\/2\) \(p( x)\) \(\(\[PartialD]\^2\)\(\ \ \)\)\/\[PartialD]x\^2 + \(\(3 p' \((x)\) - 2 \( q( x)\)\)\/4\) \(\(\[PartialD]\)\(\ \)\)\/\ \[PartialD]x + \(4 \( p(x)\) \(q(x)\) - \(p(x)\)\^3 - 8 q' \((x)\) + 8 p'' \ \((x)\)\)\/16\)}, {\(\(1\/2\) \(\(\[PartialD]\^2\)\(\ \)\)\/\[PartialD]x\^2 \ - \(1\/4\) \(p( x)\) \(\(\[PartialD]\)\(\ \)\)\/\[PartialD]x + \ \(4 \( q(x)\) - 4 p' \((x)\) - \(p(x)\)\^2\)\/8\)} }], ")"}], "TraditionalForm"]}]}], TraditionalForm]]], "\n \nwhere the partial derivatives are applied \ to whatever follows. Thus, ", Cell[BoxData[ \(TraditionalForm\`D\_x\)]], " can be applied to a single function, producing a column matrix, or can be \ applied to a row matrix, producing a matrix with 6 rows. Also, the variable \ ", Cell[BoxData[ \(TraditionalForm\`x\)]], " can be replaced by any variable, so that ", Cell[BoxData[ \(TraditionalForm\`D\_\[Xi]\)]], " would have all partial derivatives taken with respect to \[Xi]. Here is \ this operator matrix defined in ", StyleBox["Mathematica.", FontSlant->"Italic"] }], "Text"], Cell[BoxData[ \(D1[x_, y_]\ := \[IndentingNewLine]{\(#\ &\)\ @\ y, \[IndentingNewLine]\(D[#, x]\ - \ p[x]/2\ *\ #\ &\)\ @\ y, \[IndentingNewLine]\(D[#, {x, 2}]/2\ - \ 3\ p[x]\ D[#, x]/4\ + \((3\ p[x]^2 - \ 4\ q[x])\)/ 8*#\ &\)\ @\ y, \[IndentingNewLine]\(D[#, {x, 4}]/2\ - \ p[x]\ D[#, {x, 3}]/ 4\ + \ \((4\ q[x]\ - 5 \( p'\)[x] - p[x]^2)\)\ D[#, {x, 2}]/ 4\ + \ \((p[x]^3\ - \ 4\ p[x]\ q[x]\ - \ 6\ p[x]\ \(p'\)[x]\ + \ 16\ \(q'\)[x]\ - \ 20 \( p''\)[x])\) D[#, x]/ 16\ + \((p[x]^4\ - \ 8\ p[x]^2\ q[x] + \ 16\ q[x]^2\ - \ 32\ s[x]\ + \ 6\ p[x]^2 \( p'\)[x] - 24 q[x] \(p'\)[x] - 8 p[x] \(p''\)[x] + \ 16\ \(q''\)[x] - \ 16\ \(p'''\)[x])\)*#/32\ &\)\ @\ y, \[IndentingNewLine]\(\(-D[#, {x, 3}]\)/2\ + \ p[x]\ D[#, {x, 2}]/ 2\ + \ \((3\ \(p'\)[x]\ - \ 2\ q[x])\)\ D[#, x]/ 4\ + \ \((4\ p[x]\ q[x]\ - \ p[x]^3\ - \ 8\ \(q'\)[x]\ + \ 8\ \(p''\)[x])\)*#/16\ &\)\ @\ y, \[IndentingNewLine]\(D[#, {x, 2}]/2\ - \ p[x]\ D[#, x]/ 4\ + \ \((4\ q[x]\ - \ 4\ \(p'\)[x] - \ p[x]^2)\)*#/ 8\ &\)@y}\)], "Input"], Cell[TextData[{ "We also need to define the row vector ", Cell[BoxData[ FormBox[ RowBox[{\(G(x)\), " ", "=", " ", RowBox[{"(", GridBox[{ {"0", "0", \(-1\), \(\(4 \( q(x)\) - \(p(x)\)\^2 - 2 p' \((x)\)\)\/4\), \(\(p(x)\)\/2\), "1"} }], ")"}]}], TraditionalForm]]], " ." }], "Text"], Cell[BoxData[ \(G[x_]\ := \ {{0, 0, \(-1\), \ \((4 q[x]\ - \ p[x]^2\ - \ 2\ \(p'\)[x])\)/4, \ p[x]/2, 1}}\)], "Input"], Cell[TextData[{ "Then the matrix ", Cell[BoxData[ \(TraditionalForm\`P\_j\)]], " can be determined by the simple formula\n\n ", Cell[BoxData[ FormBox[ RowBox[{" ", RowBox[{ FormBox[\(P\_j\), "TraditionalForm"], " ", "=", " ", FormBox[ RowBox[{ FormBox[ SqrtBox[ RowBox[{ FormBox[\(\[CapitalDelta]\_j\), "TraditionalForm"], FormBox[\((L\_j, \[Xi])\), "TraditionalForm"]}]], "TraditionalForm"], "\[CenterDot]", RowBox[{"(", RowBox[{ RowBox[{\(D\_\(L\_j\)\), "(", SuperscriptBox[ RowBox[{"(", RowBox[{"J", "\[CenterDot]", RowBox[{\(D\_\[Xi]\), "(", " ", FormBox[ RowBox[{"f", "(", RowBox[{ FormBox[\(L\_j\), "TraditionalForm"], ",", "\[Xi]"}], ")"}], "TraditionalForm"], ")"}]}], ")"}], "T"], ")"}], " ", "-", " ", RowBox[{\(\(G(L\_j)\)\^T\[CenterDot]\(G(\[Xi])\)\ \[CenterDot]J/2\), Cell[""]}]}], ")"}]}], "TraditionalForm"]}]}], TraditionalForm]]], ".\n \n To see this, we can have ", StyleBox["Mathematica", FontSlant->"Italic"], " construct the right hand side of this equation." }], "Text"], Cell[BoxData[ \(D1[\[Xi], {f[Lj, \[Xi]]}]\)], "Input"], Cell[BoxData[ \(J . %\)], "Input"], Cell[BoxData[ \(Transpose[%]\)], "Input"], Cell["\<\ We need to convert this row matrix into a vector before proceeding.\ \>", "Text"], Cell[BoxData[ \(%[\([1]\)]\)], "Input"], Cell[BoxData[ \(D1[Lj, %]\)], "Input"], Cell[BoxData[ \(%\ - \(\(Transpose[G[Lj]] . G[\[Xi]] . J/2\)\(\ \)\)\)], "Input"], Cell[BoxData[ \(\[Delta][Lj, \[Xi]]\ *\ %\)], "Input"], Cell[TextData[{ "Finally, we can see if this is the same as the ", Cell[BoxData[ \(TraditionalForm\`P\_j\)]], " found the hard way is the same." }], "Text"], Cell[BoxData[ \(Simplify[P\ - \ %]\)], "Input"], Cell["\<\ And indeed, they are. So this proves theorem 1 from the paper.\ \>", "Text"], Cell[TextData[{ "If ", Cell[BoxData[ \(TraditionalForm\`Q(x) \[NotEqual] 0\)]], ", we have to work harder to construct the matrix ", Cell[BoxData[ \(TraditionalForm\`P\_j\)]], ", since we need to be able to solve the equation\n\n ", Cell[BoxData[ \(TraditionalForm\`2 \(\(\(\[PartialD]\)\(\ \)\)\/\[PartialD]L\_j\) P\_\(j, 3 v\) + \ \ \(p(L\_j)\) P\_\(j, 3 v\) = \ \(\(\(\[PartialD]\^3\)\(\ \ \)\)\/\[PartialD]L\_j\^3\) P\_\(j, 1 v\)\ - \ \ \(q( L\_j)\) \(\(\(\[PartialD]\)\(\ \)\)\/\[PartialD]L\_j\) P\_\(j, 1 v\)\ - \ \ \(r(L\_j)\) P\_\(j, 1 v\)\)]], ".\n \n The general solution is found using ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\[CapitalDelta]\_j\), "TraditionalForm"], FormBox[\(\((L\_j, \[Xi])\)\^\(\(-1\)/2\)\), "TraditionalForm"]}], TraditionalForm]]], " as the integrating factor. \n \n ", Cell[BoxData[ FormBox[ RowBox[{\(P\_\(j, 3 v\)\), "=", " ", RowBox[{\(\(1\/2\) \(\(\(\[PartialD]\^2\)\(\ \ \)\)\/\[PartialD]L\_j\^2\) P\_\(j, 1 v\)\), " ", "-", " ", \(1\/4\ \(p( L\_j)\) \(\(\(\[PartialD]\)\(\ \)\)\/\[PartialD]L\_j\) P\_\(j, 1 v\)\), " ", "+", " ", \(\((\(1\/4\) \(\(p\^\[Prime]\)( L\_j)\)\ + \(1\/8\) \(p(L\_j)\)\^2 - \(1\/2\) \(q( L\_j)\))\) P\_\(j, 1 v\)\), "+", RowBox[{\(1\/16\), SqrtBox[ RowBox[{ FormBox[\(\[CapitalDelta]\_j\), "TraditionalForm"], FormBox[\((L\_j, \[Xi])\), "TraditionalForm"]}]], RowBox[{\(\[Integral]\_\[Xi]\%\(L\_j\)\), RowBox[{ FormBox[\(\[CapitalDelta]\_j\), "TraditionalForm"], FormBox[\(\(\((s, \[Xi])\)\^\(\(-1\)/2\)\) \(Q(s)\)\), "TraditionalForm"], \(\(P\_\(j, 1 v\)\)( s, \[Xi])\), \(\[DifferentialD]s\)}]}]}], " ", "+", " ", RowBox[{\(\(C\_v\)(\[Xi])\), " ", SqrtBox[ RowBox[{ FormBox[\(\[CapitalDelta]\_j\), "TraditionalForm"], FormBox[\((L\_j, \[Xi])\), "TraditionalForm"]}]]}]}]}], TraditionalForm]]], ".\n \n The integral will mess up our progress unless we choose ", Cell[BoxData[ FormBox[ RowBox[{"f", "(", RowBox[{ FormBox[\(L\_j\), "TraditionalForm"], ",", "\[Xi]"}], ")"}], TraditionalForm]]], " to be a function of ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, 14\)\)]], " such that the integral can be computed. The most natural choice would be\ \n \n ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ RowBox[{"f", "(", RowBox[{ FormBox[\(L\_j\), "TraditionalForm"], ",", "\[Xi]"}], ")"}], "TraditionalForm"], " ", "=", " ", RowBox[{ FormBox[ RowBox[{\(\[Integral]\_0\%\[Xi]\), RowBox[{\(\[Integral]\_0\%\(L\_j\)\), RowBox[{ FormBox[\(\[CapitalDelta]\_j\), "TraditionalForm"], FormBox[\(\(\((s, t)\)\^\(\(-1\)/2\)\) \(\(P\_\(j, 14\)\)( s, t)\) \(Q(s)\)\), "TraditionalForm"], \(Q( t)\), \(\[DifferentialD]s\), \ \(\[DifferentialD]t\)}]}]}], "TraditionalForm"], " ", "+", " ", RowBox[{ FormBox[\(C\_1\), "TraditionalForm"], "(", FormBox[\(L\_j\), "TraditionalForm"], ")"}], " ", "+", " ", FormBox[\(\(C\_2\)(\[Xi])\), "TraditionalForm"], "+", " ", FormBox[\(C\_3\), "TraditionalForm"]}]}], TraditionalForm]]], ",\n \n where ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(C\_1\), "TraditionalForm"], "(", FormBox[\(L\_j\), "TraditionalForm"], ")"}], TraditionalForm]]], ", ", Cell[BoxData[ \(TraditionalForm\`\(C\_2\)(\[Xi])\)]], ", and ", Cell[BoxData[ \(TraditionalForm\`C\_3\)]], " are too be determined later. This allows ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, 14\)\)]], " to be expressible in terms of ", Cell[BoxData[ FormBox[ RowBox[{"f", "(", RowBox[{ FormBox[\(L\_j\), "TraditionalForm"], ",", "\[Xi]"}], ")"}], TraditionalForm]]], ", but also ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, 34\)\)]], ", ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, 16\)\)]], ", and ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, 36\)\)]], ". We find that \n \n ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(P\_\(j, 14\)\), "TraditionalForm"], " ", "=", " ", RowBox[{ FormBox[ RowBox[{ FractionBox[ SqrtBox[ RowBox[{ FormBox[\(\[CapitalDelta]\_j\), "TraditionalForm"], FormBox[\((L\_j, \[Xi])\), "TraditionalForm"]}]], \(\(Q( L\_j)\) \(Q(\[Xi])\)\)], \(\(\(\[PartialD]\^2\)\(\ \)\)\ \/\(\[PartialD]L\_j \[PartialD]\[Xi]\)\)}], "TraditionalForm"], FormBox[ RowBox[{"f", "(", RowBox[{ FormBox[\(L\_j\), "TraditionalForm"], ",", "\[Xi]"}], ")"}], "TraditionalForm"]}]}], TraditionalForm]]], ",\nbut we can choose ", Cell[BoxData[ \(TraditionalForm\`\(C\_2\)(\[Xi])\)]], " so that \n ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ FormBox[\(P\_\(j, 34\) = \ 1\/2\), "TraditionalForm"], FormBox[\(\(\(\(\[PartialD]\^2\)\(\ \)\)\/\[PartialD]L\_j\^2\) P\_\(j, 14\)\), "TraditionalForm"]}], "-", \(\(1\/4\) \(p(L\_j)\) \(\[PartialD]\/\[PartialD]L\_j\) P\_\(j, 14\)\), "+", \(\(1\/4\) p' \((L\_j)\) P\_\(j, 14\)\), " ", "+", \(\(1\/8\) \(\(p(L\_j)\)\^2\) P\_\(j, 14\)\), "-", \(\(1\/2\) \(q(L\_j)\) P\_\(j, 14\)\), "+", RowBox[{ FractionBox[ SqrtBox[ RowBox[{ FormBox[\(\[CapitalDelta]\_j\), "TraditionalForm"], FormBox[\((L\_j, \[Xi])\), "TraditionalForm"]}]], RowBox[{ "16", \(Q(\[Xi])\), Cell[ ""]}]], \(\(\(\[PartialD]\)\(\ \)\)\/\[PartialD]\[Xi]\), RowBox[{"f", "(", RowBox[{ FormBox[\(L\_j\), "TraditionalForm"], ",", "\[Xi]"}], ")"}]}]}], TraditionalForm]]], ".\n Likewise, we can choose ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(C\_1\), "TraditionalForm"], "(", FormBox[\(L\_j\), "TraditionalForm"], ")"}], TraditionalForm]]], " such that \n ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ FormBox[\(P\_\(j, 16\) = \ 1\/2\), "TraditionalForm"], FormBox[\(\(\(\(\[PartialD]\^2\)\(\ \)\)\/\[PartialD]\[Xi]\^2\) P\_\(j, 14\)\), "TraditionalForm"]}], "-", \(\(5\/4\) \(p(\[Xi])\) \(\[PartialD]\/\[PartialD]\[Xi]\) P\_\(j, 14\)\), "-", \(\(1\/4\) p' \((\[Xi])\) P\_\(j, 14\)\), " ", "+", \(\(7\/8\) \(\(p(\[Xi])\)\^2\) P\_\(j, 14\)\), "-", \(\(1\/2\) \(q(\[Xi])\) P\_\(j, 14\)\), "+", RowBox[{ FractionBox[ SqrtBox[ RowBox[{ FormBox[\(\[CapitalDelta]\_j\), "TraditionalForm"], FormBox[\((L\_j, \[Xi])\), "TraditionalForm"]}]], RowBox[{ "16", \(Q(L\_j)\), Cell[ ""]}]], \(\(\(\[PartialD]\)\(\ \)\)\/\[PartialD]L\_j\), RowBox[{"f", "(", RowBox[{ FormBox[\(L\_j\), "TraditionalForm"], ",", "\[Xi]"}], ")"}]}]}], TraditionalForm]]], ",\nso we have now defined ", Cell[BoxData[ FormBox[ RowBox[{"f", "(", RowBox[{ FormBox[\(L\_j\), "TraditionalForm"], ",", "\[Xi]"}], ")"}], TraditionalForm]]], " up to an arbitrary constant ", Cell[BoxData[ \(TraditionalForm\`C\_3\)]], ". Putting these into a new matrix, we have" }], "Text"], Cell[BoxData[ \(P\ = \ {{0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0}}\)], "Input"], Cell[BoxData[ \(\[Delta][Lj_, \[Xi]_]\ := \ E^\(-Integrate[p[s]/2, {s, \[Xi], Lj}]\)\)], "Input"], Cell[BoxData[ \(P[\([1, 4]\)]\ = \ Simplify[\ D[D[f[Lj, \[Xi]], Lj], \[Xi]]\ \[Delta][Lj, \[Xi]] R[Lj]\ R[\[Xi]]]\)], "Input"], Cell[BoxData[ \(P[\([3, 4]\)]\ = \ Simplify[D[P[\([1, 4]\)], {Lj, 2}]/2\ - \ p[Lj]\ D[P[\([1, 4]\)], Lj]/ 4 + \ \((\(p'\)[Lj]/4\ + \ p[Lj]^2/8 - q[Lj]/2)\) P[\([1, 4]\)]\ + \ D[f[Lj, \[Xi]], \[Xi]]\ \[Delta][Lj, \[Xi]] R[\[Xi]]/16]\)], "Input"], Cell[BoxData[ \(Factor[%]\)], "Input"], Cell[BoxData[ \(P[\([1, 6]\)]\ = \ Simplify[D[P[\([1, 4]\)], {\[Xi], 2}]/2\ - \ 5\ p[\[Xi]]\ D[P[\([1, 4]\)], \[Xi]]/ 4 + \ \((\(-\(p'\)[\[Xi]]\)/4\ + \ 7\ p[\[Xi]]^2/8 - q[\[Xi]]/2)\) P[\([1, 4]\)]\ + \ D[f[Lj, \[Xi]], Lj]\ \[Delta][Lj, \[Xi]] R[Lj]/16\ ]\)], "Input"], Cell[BoxData[ \(Factor[%]\)], "Input"], Cell[TextData[{ "Here, we introduced the function ", Cell[BoxData[ \(TraditionalForm\`R(x)\ = \ 1/\(Q(x)\)\)]], ". In order to find ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, 36\)\)]], ", we have to integrate either \n\n ", Cell[BoxData[ FormBox[ RowBox[{"\[Integral]", RowBox[{ FormBox[\(\[CapitalDelta]\_j\), "TraditionalForm"], SuperscriptBox[ FormBox[\((L\_j, \[Xi])\), "TraditionalForm"], \(\(-1\)/2\)], \(Q( L\_j)\), \(P\_\(j, 16\)\), \(\[DifferentialD]L\_j\)}]}], TraditionalForm]]], " or ", Cell[BoxData[ FormBox[ RowBox[{"\[Integral]", RowBox[{ FormBox[\(\[CapitalDelta]\_j\), "TraditionalForm"], SuperscriptBox[ FormBox[\((L\_j, \[Xi])\), "TraditionalForm"], \(\(-1\)/2\)], \(Q(\[Xi])\), \(P\_\(j, 34\)\), \(\[DifferentialD]\[Xi]\)}]}], TraditionalForm]]], ".\nThe first method gives us" }], "Text"], Cell[BoxData[ RowBox[{\(P[\([3, 6]\)]\), " ", "=", " ", RowBox[{"Simplify", "[", RowBox[{\(D[P[\([1, 6]\)], {Lj, 2}]/2\), " ", "-", " ", \(p[Lj]\ D[P[\([1, 6]\)], Lj]/4\), "+", " ", \(\((\(p'\)[Lj]/4\ + \ p[Lj]^2/8 - q[Lj]/2)\) P[\([1, 6]\)]\), " ", "+", " ", "\[IndentingNewLine]", RowBox[{\(\[Delta][Lj, \[Xi]]/16\), "*", RowBox[{ RowBox[{"(", RowBox[{ RowBox[{ SuperscriptBox["f", TagBox[\((0, 0)\), Derivative], MultilineFunction->None], "[", \(Lj, \[Xi]\), "]"}], "+", RowBox[{"2", " ", RowBox[{"(", RowBox[{ RowBox[{ "3", " ", \(p[\[Xi]]\^2\), " ", \(R[\[Xi]]\), " ", RowBox[{ SuperscriptBox["f", TagBox[\((0, 1)\), Derivative], MultilineFunction->None], "[", \(Lj, \[Xi]\), "]"}]}], "-", RowBox[{ "4", " ", \(q[\[Xi]]\), " ", \(R[\[Xi]]\), " ", RowBox[{ SuperscriptBox["f", TagBox[\((0, 1)\), Derivative], MultilineFunction->None], "[", \(Lj, \[Xi]\), "]"}]}], "+", RowBox[{"4", " ", RowBox[{ SuperscriptBox["R", "\[Prime]\[Prime]", MultilineFunction->None], "[", "\[Xi]", "]"}], " ", RowBox[{ SuperscriptBox["f", TagBox[\((0, 1)\), Derivative], MultilineFunction->None], "[", \(Lj, \[Xi]\), "]"}]}], "+", RowBox[{"8", " ", RowBox[{ SuperscriptBox["R", "\[Prime]", MultilineFunction->None], "[", "\[Xi]", "]"}], " ", RowBox[{ SuperscriptBox["f", TagBox[\((0, 2)\), Derivative], MultilineFunction->None], "[", \(Lj, \[Xi]\), "]"}]}], "-", RowBox[{"6", " ", \(p[\[Xi]]\), " ", RowBox[{"(", RowBox[{ RowBox[{ RowBox[{ SuperscriptBox["R", "\[Prime]", MultilineFunction->None], "[", "\[Xi]", "]"}], " ", RowBox[{ SuperscriptBox["f", TagBox[\((0, 1)\), Derivative], MultilineFunction->None], "[", \(Lj, \[Xi]\), "]"}]}], "+", RowBox[{\(R[\[Xi]]\), " ", RowBox[{ SuperscriptBox["f", TagBox[\((0, 2)\), Derivative], MultilineFunction->None], "[", \(Lj, \[Xi]\), "]"}]}]}], ")"}]}], "+", RowBox[{"4", " ", \(R[\[Xi]]\), " ", RowBox[{ SuperscriptBox["f", TagBox[\((0, 3)\), Derivative], MultilineFunction->None], "[", \(Lj, \[Xi]\), "]"}]}]}], ")"}]}]}], ")"}], "/", "16"}]}]}], "]"}]}]], "Input"], Cell[TextData[{ "Here,the integral ", Cell[BoxData[ FormBox[ RowBox[{"\[Integral]", RowBox[{ FormBox[\(\[CapitalDelta]\_j\), "TraditionalForm"], SuperscriptBox[ FormBox[\((L\_j, \[Xi])\), "TraditionalForm"], \(\(-1\)/2\)], \(Q( L\_j)\), \(P\_\(j, 16\)\), \(\[DifferentialD]L\_j\)}]}], TraditionalForm]]], " was done by copying and pasting the above result for ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, 16\)\)]], ", and manually decrease each derivative in the ", Cell[BoxData[ \(TraditionalForm\`L\_j\)]], " direction by 1. The term with the arbitrary constant was left out, but if \ we did this using ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, 34\)\)]], " instead, " }], "Text"], Cell[BoxData[ RowBox[{"Simplify", "[", RowBox[{\(D[P[\([3, 4]\)], {\[Xi], 2}]/2\), " ", "-", " ", \(5\ p[\[Xi]]\ D[P[\([3, 4]\)], \[Xi]]/4\), "+", " ", \(\((\(-\(p'\)[\[Xi]]\)/4\ + \ 7\ p[\[Xi]]^2/8 - q[\[Xi]]/2)\) P[\([3, 4]\)]\), " ", "+", " ", "\[IndentingNewLine]", " ", RowBox[{\(\[Delta][Lj, \[Xi]]/16\), "*", RowBox[{ RowBox[{"(", RowBox[{ RowBox[{ SuperscriptBox["f", TagBox[\((0, 0)\), Derivative], MultilineFunction->None], "[", \(Lj, \[Xi]\), "]"}], "+", RowBox[{"2", " ", RowBox[{"(", RowBox[{ RowBox[{"3", " ", \(p[Lj]\^2\), " ", \(R[Lj]\), " ", RowBox[{ SuperscriptBox["f", TagBox[\((1, 0)\), Derivative], MultilineFunction->None], "[", \(Lj, \[Xi]\), "]"}]}], "-", RowBox[{"4", " ", \(q[Lj]\), " ", \(R[Lj]\), " ", RowBox[{ SuperscriptBox["f", TagBox[\((1, 0)\), Derivative], MultilineFunction->None], "[", \(Lj, \[Xi]\), "]"}]}], "+", RowBox[{"4", " ", RowBox[{ SuperscriptBox["R", "\[Prime]\[Prime]", MultilineFunction->None], "[", "Lj", "]"}], " ", RowBox[{ SuperscriptBox["f", TagBox[\((1, 0)\), Derivative], MultilineFunction->None], "[", \(Lj, \[Xi]\), "]"}]}], "+", RowBox[{"8", " ", RowBox[{ SuperscriptBox["R", "\[Prime]", MultilineFunction->None], "[", "Lj", "]"}], " ", RowBox[{ SuperscriptBox["f", TagBox[\((2, 0)\), Derivative], MultilineFunction->None], "[", \(Lj, \[Xi]\), "]"}]}], "-", RowBox[{"6", " ", \(p[Lj]\), " ", RowBox[{"(", RowBox[{ RowBox[{ RowBox[{ SuperscriptBox["R", "\[Prime]", MultilineFunction->None], "[", "Lj", "]"}], " ", RowBox[{ SuperscriptBox["f", TagBox[\((1, 0)\), Derivative], MultilineFunction->None], "[", \(Lj, \[Xi]\), "]"}]}], "+", RowBox[{\(R[Lj]\), " ", RowBox[{ SuperscriptBox["f", TagBox[\((2, 0)\), Derivative], MultilineFunction->None], "[", \(Lj, \[Xi]\), "]"}]}]}], ")"}]}], "+", RowBox[{"4", " ", \(R[Lj]\), " ", RowBox[{ SuperscriptBox["f", TagBox[\((3, 0)\), Derivative], MultilineFunction->None], "[", \(Lj, \[Xi]\), "]"}]}]}], ")"}]}]}], ")"}], "/", "16"}]}]}], " ", "]"}]], "Input"], Cell["we get exactly the same thing.", "Text"], Cell[BoxData[ \(%\ - \ %%\)], "Input"], Cell[TextData[{ "Since doing it one way could introduce an extra term of ", Cell[BoxData[ FormBox[ RowBox[{\(\(C\_6\)(\[Xi])\), SqrtBox[ RowBox[{ FormBox[\(\[CapitalDelta]\_j\), "TraditionalForm"], FormBox[\((L\_j, \[Xi])\), "TraditionalForm"]}]]}], TraditionalForm]]], ", where ", Cell[BoxData[ \(TraditionalForm\`C\_6\)]], " is a function of only \[Xi], and doing it the other way could introduce \ an extra term of ", Cell[BoxData[ FormBox[ RowBox[{\(\(C\_6\)(L\_j)\), SqrtBox[ RowBox[{ FormBox[\(\[CapitalDelta]\_j\), "TraditionalForm"], FormBox[\((L\_j, \[Xi])\), "TraditionalForm"]}]]}], TraditionalForm]]], ". Thus, the extra term must be a constant times ", Cell[BoxData[ FormBox[ SqrtBox[ RowBox[{ FormBox[\(\[CapitalDelta]\_j\), "TraditionalForm"], FormBox[\((L\_j, \[Xi])\), "TraditionalForm"]}]], TraditionalForm]]], ", and we can even absorb this term in the first term of the above \ expression. (Recall that ", Cell[BoxData[ FormBox[ RowBox[{"f", "(", RowBox[{ FormBox[\(L\_j\), "TraditionalForm"], ",", "\[Xi]"}], ")"}], TraditionalForm]]], " was defined before up to a multiplicative constant.) Thus, we can define \ the above expression to be ", Cell[BoxData[ \(TraditionalForm\`P\_\(j, 36\)\)]], ", and we have finally finished defining ", Cell[BoxData[ FormBox[ RowBox[{"f", "(", RowBox[{ FormBox[\(L\_j\), "TraditionalForm"], ",", "\[Xi]"}], ")"}], TraditionalForm]]], ". We can now define the rest of the matrix ", Cell[BoxData[ \(TraditionalForm\`P\_j\)]], "." }], "Text"], Cell[BoxData[ \(P[\([2, 4]\)]\ = \ Simplify[\ D[P[\([1, 4]\)], Lj]\ ]\)], "Input"], Cell[BoxData[ \(P[\([6, 4]\)]\ = \ Simplify[\ D[P[\([2, 4]\)], Lj]\ - \ P[\([3, 4]\)]\ ]\)], "Input"], Cell[BoxData[ \(P[\([5, 4]\)]\ = \ Simplify[\ \(-\ D[P[\([6, 4]\)], Lj]\)\ ]\)], "Input"], Cell[BoxData[ \(P[\([4, 4]\)] = \ Simplify[\ \(-\ D[P[\([5, 4]\)], Lj]\) - \ p[Lj]\ P[\([5, 4]\)]\ + \ q[Lj]\ P[\([6, 4]\)]\ - \ s[Lj]\ P[\([1, 4]\)]\ ]\)], "Input"], Cell[BoxData[ \(P[\([1, 1]\)]\ = \ Exchange[P[\([4, 4]\)]]\ \[Delta][Lj, \[Xi]]^2\)], "Input"], Cell[BoxData[ \(P[\([1, 2]\)]\ = \ Exchange[P[\([5, 4]\)]] \[Delta][Lj, \[Xi]]^2\)], "Input"], Cell[BoxData[ \(P[\([1, 3]\)]\ = Exchange[P[\([6, 4]\)]] \[Delta][Lj, \[Xi]]^2\)], "Input"], Cell[BoxData[ \(P[\([1, 5]\)]\ = \ Exchange[P[\([2, 4]\)]]\ \[Delta][Lj, \[Xi]]^2\)], "Input"], Cell[BoxData[ \(P[\([1, 6]\)]\ = \ Exchange[P[\([3, 4]\)]] \[Delta][Lj, \[Xi]]^2\)], "Input"], Cell[BoxData[ \(P[\([2, 6]\)]\ = \ Simplify[\ D[P[\([1, 6]\)], Lj]\ ]\)], "Input"], Cell[BoxData[ \(P[\([6, 6]\)]\ = \ Simplify[\ D[P[\([2, 6]\)], Lj]\ - \ P[\([3, 6]\)]\ ]\)], "Input"], Cell[BoxData[ \(P[\([5, 6]\)]\ = \ Simplify[\ \(-\ D[P[\([6, 6]\)], Lj]\)\ ]\)], "Input"], Cell[BoxData[ \(P[\([4, 6]\)] = \ Simplify[\ \(-\ D[P[\([5, 6]\)], Lj]\) - \ p[Lj]\ P[\([5, 6]\)]\ + \ q[Lj]\ P[\([6, 6]\)]\ - \ s[Lj]\ P[\([1, 6]\)]\ ]\)], "Input"], Cell[BoxData[ \(P[\([3, 1]\)]\ = \ Exchange[P[\([4, 6]\)]]\ \[Delta][Lj, \[Xi]]^2\)], "Input"], Cell[BoxData[ \(P[\([3, 2]\)]\ = \ Exchange[P[\([5, 6]\)]]\ \[Delta][Lj, \[Xi]]^2\)], "Input"], Cell[BoxData[ \(P[\([3, 3]\)]\ = \ Exchange[P[\([6, 6]\)]]\ \[Delta][Lj, \[Xi]]^2\)], "Input"], Cell[BoxData[ \(P[\([3, 5]\)]\ = \ Exchange[P[\([2, 6]\)]]\ \[Delta][Lj, \[Xi]]^2\)], "Input"], Cell[BoxData[ \(P[\([2, 1]\)]\ = \ Simplify[\ D[P[\([1, 1]\)], Lj]\ ]\)], "Input"], Cell[BoxData[ \(P[\([2, 2]\)]\ = \ Simplify[\ D[P[\([1, 2]\)], Lj]\ ]\)], "Input"], Cell[BoxData[ \(P[\([2, 3]\)]\ = \ Simplify[\ D[P[\([1, 3]\)], Lj]\ ]\)], "Input"], Cell[BoxData[ \(P[\([2, 5]\)]\ = \ Simplify[\ D[P[\([1, 5]\)], Lj]\ ]\)], "Input"], Cell[BoxData[ \(P[\([6, 1]\)]\ = \ Simplify[\ D[P[\([2, 1]\)], Lj]\ - \ P[\([3, 1]\)]\ ]\)], "Input"], Cell[BoxData[ \(P[\([6, 2]\)]\ = \ Simplify[\ D[P[\([2, 2]\)], Lj]\ - \ P[\([3, 2]\)]\ ]\)], "Input"], Cell[BoxData[ \(P[\([6, 3]\)]\ = \ Simplify[\ D[P[\([2, 3]\)], Lj]\ - \ P[\([3, 3]\)]\ ]\)], "Input"], Cell[BoxData[ \(P[\([6, 5]\)]\ = \ Simplify[\ D[P[\([2, 5]\)], Lj]\ - \ P[\([3, 5]\)]\ ]\)], "Input"], Cell[BoxData[ \(P[\([5, 1]\)]\ = \ Simplify[\ \(-\ D[P[\([6, 1]\)], Lj]\)\ ]\)], "Input"], Cell[BoxData[ \(P[\([5, 2]\)]\ = \ Simplify[\ \(-\ D[P[\([6, 2]\)], Lj]\)\ ]\)], "Input"], Cell[BoxData[ \(P[\([5, 3]\)]\ = \ Simplify[\ \(-\ D[P[\([6, 3]\)], Lj]\)\ ]\)], "Input"], Cell[BoxData[ \(P[\([5, 5]\)]\ = \ Simplify[\ \(-\ D[P[\([6, 5]\)], Lj]\)\ ]\)], "Input"], Cell[BoxData[ \(P[\([4, 1]\)] = \ Simplify[\ \(-\ D[P[\([5, 1]\)], Lj]\) - \ p[Lj]\ P[\([5, 1]\)]\ + \ q[Lj]\ P[\([6, 1]\)]\ - \ s[Lj]\ P[\([1, 1]\)]\ ]\)], "Input"], Cell[BoxData[ \(P[\([4, 2]\)] = \ Simplify[\ \(-\ D[P[\([5, 2]\)], Lj]\) - \ p[Lj]\ P[\([5, 2]\)]\ + \ q[Lj]\ P[\([6, 2]\)]\ - \ s[Lj]\ P[\([1, 2]\)]\ ]\)], "Input"], Cell[BoxData[ \(P[\([4, 3]\)] = \ Simplify[\ \(-\ D[P[\([5, 3]\)], Lj]\) - \ p[Lj]\ P[\([5, 3]\)]\ + \ q[Lj]\ P[\([6, 3]\)]\ - \ s[Lj]\ P[\([1, 3]\)]\ ]\)], "Input"], Cell[BoxData[ \(P[\([4, 5]\)] = \ Simplify[\ \(-\ D[P[\([5, 5]\)], Lj]\) - \ p[Lj]\ P[\([5, 5]\)]\ + \ q[Lj]\ P[\([6, 5]\)]\ - \ s[Lj]\ P[\([1, 5]\)]\ ]\)], "Input"], Cell[TextData[{ "However, if we define the matrix operator ", Cell[BoxData[ \(TraditionalForm\`D\_x\)]], " by \n\n ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(D\_x\), "TraditionalForm"], " ", "=", " ", RowBox[{"(", GridBox[{ {\(\(R(x)\)\ \[PartialD]\/\[PartialD]x\)}, {\(\(R( x)\) \[PartialD]\^2\/\[PartialD]x\^2 + \(\(\(-\(p( x)\)\) \(R(x)\) + 2 R\ ' \((x)\)\)\/2\) \ \[PartialD]\/\[PartialD]x\)}, {\(\(\(R( x)\)\/2\) \[PartialD]\^3\/\[PartialD]x\^3 + \ \(\(\(-3\) \(p(x)\) \(R(x)\) + 4 R\ ' \((x)\)\)\/4\) \ \[PartialD]\^2\/\[PartialD]x\^2 + \(\(3 \(\( p(x)\)\^2\) \(R(x)\) - 4 \( q(x)\) \(R(x)\) - 6 \( p(x)\) R\ ' \((x)\) + 4 \( R\ '\)\ ' \((x)\)\)\/8\) \[PartialD]\/\ \[PartialD]x + 1\/16\)}, { RowBox[{"(", GridBox[{ {\(\(\(R( x)\)\/2\) \[PartialD]\^5\/\[PartialD]x\^5 \ + \(\(\(-\(p(x)\)\) \(R(x)\) + 8 R\ ' \((x)\)\)\/4\) \[PartialD]\^4\/\ \[PartialD]x\^4 + \(\(\(-\(p(x)\)\^2\) \(R(x)\) + 4 \( q(x)\) \(R(x)\) - 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1\/16\)} }], ")"}]}], TraditionalForm]]], "," }], "Text"], Cell[BoxData[ \(D2[x_, y_]\ := \[IndentingNewLine]{\(R[x]\ D[#, x]\ &\)\ @\ y, \[IndentingNewLine]\(R[x] D[#, {x, 2}]\ + \ \ \((2\ \(R'\)[x] - \ p[x] R[x])\) D[#, x]/2\ &\)\ @\ y, \[IndentingNewLine]\(R[x]\ D[#, {x, 3}]/ 2\ + \ \((4\ \(R'\)[x]\ - \ 3\ p[x]\ R[x])\)\ D[#, {x, 2}]/ 4\ + \ \((3\ p[x]^2\ R[x]\ - \ 4\ q[x]\ R[x]\ - \ 6\ p[x]\ \(R'\)[x]\ + \ 4\ \(R''\)[x])\) D[#, x]/8\ + \ #/16\ &\)\ @\ y, \[In