(*********************************************************************** Mathematica-Compatible Notebook This notebook can be used on any computer system with Mathematica 3.0, MathReader 3.0, or any compatible application. The data for the notebook starts with the line of 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). NOTE: If you modify the data for this notebook not in a Mathematica- compatible application, you must delete the line below containing the word CacheID, otherwise Mathematica-compatible applications may try to use invalid cache data. 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[ 122639, 4599]*) (*NotebookOutlinePosition[ 123783, 4639]*) (* CellTagsIndexPosition[ 123684, 4632]*) (*WindowFrame->Normal*) Notebook[{ Cell["\<\ CONSTRAINT SET TRACKING: OPTIMAL CONTROL PROBLEM \t\t\t\t by Paul J. Scott\ \>", "Title"], Cell["Introduction", "Section"], Cell["\<\ This notebook performs the numerical calculations proposed in a paper with \ same title. References below refer to this paper. This is a shooting method \ in which one varies the inital values of the costate until the constraint set \ is reached. \ \>", "Text"], Cell[CellGroupData[{ Cell["Part Zero", "Subsubtitle"], Cell["\<\ This function performs one Runge Kutta step for vector right hand side \ function df for initial vector y0 and step size dt. \ \>", "Text"], Cell[BoxData[ \(rkStep[\ \ y0_, \ dt_]\ := \n\t Module[{\ k1, \ k2, \ k3, \ k4\ }, \n\t\t k1\ = \ dt\ N[\ df[\ y0\ ]\ \ ]; \n\t\t k2\ = \ dt\ N[\ df[\ y0\ + \ k1/2]\ ]; \n\t\t k3\ = \ dt\ N[\ df[\ y0\ + \ k2/2\ ]\ ]; \n\t\t k4\ = \ dt\ N[\ df[\ y0\ + \ k3]\ ]; \n\t\t y0\ + \ \((k1\ + \ 2\ k2\ + \ 2\ k3\ + \ k4)\)/6\n\t]\)], "Input"], Cell["\<\ This function performs a series of fixed step size Runge Kutta steps to \ generate a trajectory.\ \>", "Text"], Cell[BoxData[ \(\(rk[\ y0_List, \ {t1_, \ dt_}]\ := \n\t NestList[\ rkStep[\ \ #, \ N[dt]]&, \ N[y0], \ Round[N[t1/dt]]\ ]\ \n \)\)], "Input"], Cell[TextData[{ "BOTH FUNCTIONS BASED ON A PACKAGE BY ROMAN E. MAEDER IN ", StyleBox["MathSource", FontSlant->"Italic"], "." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Part One", "Subsubtitle"], Cell["\<\ We set up equations using the same notation as in the paper. \ \>", "Text"], Cell["Declaring Variables", "Section"], Cell["Number of y state variables (called n).", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(n\ = \ 2\)], "Input"], Cell[BoxData[ \(2\)], "Output"] }, Open ]], Cell["Declaration of y state variables.", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(yy\ = \ Table[\ y\_i\ , \ {\ i\ , \ 1\ , \ n\ }\ ]\)], "Input"], Cell[BoxData[ \({y\_1, y\_2}\)], "Output"] }, Open ]], Cell[TextData[ "Declaration of y- costate variables (called \[Lambda])."], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Lambda]\[Lambda]\ = \ Table[\ \ \[Lambda]\_i\ \ , \ {\ i\ , 1\ , \ Length[\ yy\ ]\ }\ \ ]\)], "Input"], Cell[BoxData[ \({\[Lambda]\_1, \[Lambda]\_2}\)], "Output"] }, Open ]], Cell["Number of z state variables (called m).", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(m\ = \ 2\)], "Input"], Cell[BoxData[ \(2\)], "Output"] }, Open ]], Cell["Declaration of z-state variables.", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(zz\ = \ Table[\ \ z\_i\ \ , \ {\ i\ , 1\ , m\ }\ \ ]\)], "Input"], Cell[BoxData[ \({z\_1, z\_2}\)], "Output"] }, Open ]], Cell["Corresponding numerical variables.", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(nzz\ = \ Table[\ \ nz\_i\ \ , \ {\ i\ , 1\ , m\ }\ \ ]\)], "Input"], Cell[BoxData[ \({nz\_1, nz\_2}\)], "Output"] }, Open ]], Cell["Initial z-state variables at start of trajectory", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(initialz\ = \ Table[\ \ iz\_i\ \ , \ {\ i\ , 1\ , m\ }\ \ ]\)], "Input"], Cell[BoxData[ \({iz\_1, iz\_2}\)], "Output"] }, Open ]], Cell["\<\ Declaration of initial z-state variables. Used for initial approximation in \ solving (4.6).\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(ninitialz\ = \ Table[\ \ niz\_i\ \ , \ {\ i\ , 1\ , m\ }\ \ ]\)], "Input"], Cell[BoxData[ \(General::"spell1" \( : \ \) "Possible spelling error: new symbol name \"\!\(ninitialz\)\" is \ similar to existing symbol \"\!\(initialz\)\"."\)], "Message"], Cell[BoxData[ \({niz\_1, niz\_2}\)], "Output"] }, Open ]], Cell[TextData[{ "Transform variable ", Cell[BoxData[ \(TraditionalForm\`\(gg\&^\)\)]], ". See paper (4.6)." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(gg\&^\ = \ Table[\ \ \(g\_i\)\&^\ \ , \ {\ i\ , 1\ , m\ }\ \ ]\)], "Input"], Cell[BoxData[ \({\(\(g\_1\)\&^\), \(\(g\_2\)\&^\)}\)], "Output"] }, Open ]], Cell["Corresponding numerical variables.", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(ngg\&^\ = \ Table[\ \ \(ng\_i\)\&^\ \ , \ {\ i\ , 1\ , m\ }\ \ ]\)], "Input"], Cell[BoxData[ \({\(\(ng\_1\)\&^\), \(\(ng\_2\)\&^\)}\)], "Output"] }, Open ]], Cell[TextData[{ "Variable to hold rhs of DE for transformed variable ", Cell[BoxData[ \(TraditionalForm\`\(gg\&^\)\)]], ". See paper (4.8)." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(dgg\&^\ = \ Table[\ \ \(dg\_i\)\&^\ \ , \ {\ i\ , 1\ , m\ }\ \ ]\)], "Input"], Cell[BoxData[ \({\(\(dg\_1\)\&^\), \(\(dg\_2\)\&^\)}\)], "Output"] }, Open ]], Cell["Corresponding numerical variables.", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(ndgg\&^\ = \ Table[\ \ \(ndg\_i\)\&^\ \ , \ {\ i\ , 1\ , m\ }\ \ ]\)], "Input"], Cell[BoxData[ \(General::"spell" \( : \ \) "Possible spelling error: new symbol name \"\!\(ndgg\)\" is similar to \ existing symbols \!\({dgg, ngg}\)."\)], "Message"], Cell[BoxData[ \({\(\(ndg\_1\)\&^\), \(\(ndg\_2\)\&^\)}\)], "Output"] }, Open ]], Cell[TextData["Declaration of z-costate variables (called \[Xi])."], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Xi]\[Xi]\ = \ Table[\ \ \[Xi]\_i\ \ , \ {\ i\ , 1\ , Length[\ zz\ ]\ }\ \ ]\)], "Input"], Cell[BoxData[ \({\[Xi]\_1, \[Xi]\_2}\)], "Output"] }, Open ]], Cell["Declaration of error function e. See paper (1.4)", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(ee\ = \ Table[\ \ e\_i\ \ , \ {\ i\ , 1\ , \ Length[\ zz\ ]\ }\ \ ]\)], "Input"], Cell[BoxData[ \({e\_1, e\_2}\)], "Output"] }, Open ]], Cell["Corresponding numerical variable.", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(nee\ = \ Table[\ \ ne\_i\ \ , \ {\ i\ , 1\ , \ Length[\ zz\ ]\ }\ \ ]\)], "Input"], Cell[BoxData[ \({ne\_1, ne\_2}\)], "Output"] }, Open ]], Cell["\<\ Declaration for a matrix occuring frequently ( transpose of inverse of DzG). \ See (4.2), (4.3), (4.4), and (4.8).\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(TinDzG\ = \ Table[\ Tin[i, j]\ , \ {i, 1, Length[zz]}\n \t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ , \ {j, 1, Length[zz]}\ ]\)], "Input"], Cell[BoxData[ \({{Tin[1, 1], Tin[1, 2]}, {Tin[2, 1], Tin[2, 2]}}\)], "Output"] }, Open ]], Cell["\<\ Numerical form for matrix occuring frequently. See (4.2), (4.3), (4.4), and \ (4.8).\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(nTinDzG\ = \ Table[\ nTin[i, j]\ , \ {i, 1, Length[zz]}\n \t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ , \ {j, 1, Length[zz]}\ ]\)], "Input"], Cell[BoxData[ \(General::"spell1" \( : \ \) "Possible spelling error: new symbol name \"\!\(nTinDzG\)\" is similar \ to existing symbol \"\!\(TinDzG\)\"."\)], "Message"], Cell[BoxData[ \(General::"spell1" \( : \ \) "Possible spelling error: new symbol name \"\!\(nTin\)\" is similar to \ existing symbol \"\!\(Tin\)\"."\)], "Message"], Cell[BoxData[ \({{nTin[1, 1], nTin[1, 2]}, {nTin[2, 1], nTin[2, 2]}}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell["Defining the differential equation and constraint", "Section"], Cell[CellGroupData[{ Cell["Definition of rhs of ODE F. See paper (1.1)", "Subsection"], Cell["Each component of F is defined.", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(f\_1\ \ = \ \ \(-y\_1\)\ \ + \ \ z\_1\)], "Input"], Cell[BoxData[ \(\(-y\_1\) + z\_1\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(f\_2\ \ = \ \ \(-y\_2\)\ \ + \ \ z\_2\)], "Input"], Cell[BoxData[ \(\(-y\_2\) + z\_2\)], "Output"] }, Open ]], Cell["The components are listed in vector F.", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(F\ = \ Table[\ \ f\_i\ \ , \ {\ i\ , 1\ , Length[\ yy\ ]\ }\ \ ]\)], "Input"], Cell[BoxData[ \({\(-y\_1\) + z\_1, \(-y\_2\) + z\_2}\)], "Output"] }, Open ]] }, Open ]], Cell[TextData[{ "The Jacobian Matrix ", Cell[BoxData[ \(TraditionalForm\`\[PartialD]\_y\ F\)]] }], "Section"], Cell[CellGroupData[{ Cell[BoxData[ \(DyF\ \ = \ \ Transpose[\ \ Table[\ \[PartialD]\_\(y\_i\)\ F\ , \ {i, 1, Length[\ yy\ ]}\ ]\ \ ] \)], "Input"], Cell[BoxData[ \({{\(-1\), 0}, {0, \(-1\)}}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(MatrixForm[\ DyF\ ]\)], "Input"], Cell[BoxData[ TagBox[ RowBox[{"(", GridBox[{ {\(-1\), "0"}, {"0", \(-1\)} }], ")"}], (MatrixForm[ #]&)]], "Output"] }, Open ]], Cell[TextData[{ "The Jacobian Matrix ", Cell[BoxData[ \(TraditionalForm\`\[PartialD]\_z\ F\)]] }], "Section"], Cell[CellGroupData[{ Cell[BoxData[ \(DzF\ \ = \ \ Transpose[\ \ Table[\ \[PartialD]\_\(z\_i\)\ F\ , \ {i, 1, Length[\ zz\ ]}\ ]\ \ ] \)], "Input"], Cell[BoxData[ \({{1, 0}, {0, 1}}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(MatrixForm[\ DzF\ ]\)], "Input"], Cell[BoxData[ TagBox[ RowBox[{"(", GridBox[{ {"1", "0"}, {"0", "1"} }], ")"}], (MatrixForm[ #]&)]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell["Definition of constraint function G. See paper (1.2)", "Subsection"], Cell["Each component of g is defined.", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(sg\_1\ = \ \ y\_1^3\ \ + \ \ y\_2^2\ \ - \ \((z\_1^3\ \ + \ \ z\_2^2\ )\)\ \ \ + \ \ y\_1\ \ \((z\_2^2\ - \ 4\ )\)\)], "Input"], Cell[BoxData[ \(y\_1\%3 + y\_2\%2 - z\_1\%3 - z\_2\%2 + y\_1\ \((\(-4\) + z\_2\%2)\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\(sg\_2\ = \ \ \ \ y\_1\ \ - \ \ z\_1\ \ \ \)\)], "Input"], Cell[BoxData[ \(y\_1 - z\_1\)], "Output"] }, Open ]], Cell["The components are listed in a vector G.", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(G\ = \ Table[\ \ sg\_i\ \ , \ {\ i\ , 1\ , Length[\ zz\ ]\ }\ \ ]\)], "Input"], Cell[BoxData[ \({y\_1\%3 + y\_2\%2 - z\_1\%3 - z\_2\%2 + y\_1\ \((\(-4\) + z\_2\%2)\), y\_1 - z\_1}\)], "Output"] }, Open ]] }, Open ]] }, Open ]], Cell[TextData[{ "The Jacobian Matrix ", Cell[BoxData[ \(TraditionalForm\`\[PartialD]\_y\ G\)]] }], "Section"], Cell[CellGroupData[{ Cell[BoxData[ \(DyG\ \ = \ \ Transpose[\ \ Table[\ \[PartialD]\_\(y\_i\)\ G\ , \ {i, 1, Length[\ yy\ ]}\ ]\ \ ] \)], "Input"], Cell[BoxData[ \({{\(-4\) + 3\ y\_1\%2 + z\_2\%2, 2\ y\_2}, {1, 0}}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(MatrixForm[\ DyG\ ]\)], "Input"], Cell[BoxData[ TagBox[ RowBox[{"(", GridBox[{ {\(\(-4\) + 3\ y\_1\%2 + z\_2\%2\), \(2\ y\_2\)}, {"1", "0"} }], ")"}], (MatrixForm[ #]&)]], "Output"] }, Open ]], Cell[TextData[{ "The Jacobian Matrix ", Cell[BoxData[ \(TraditionalForm\`\[PartialD]\_z\ G\)]] }], "Section"], Cell[CellGroupData[{ Cell[BoxData[ \(DzG\ \ = \ \ Transpose[\ \ Table[\ \[PartialD]\_\(z\_i\)\ G\ , \ {i, 1, Length[\ zz\ ]}\ ]\ \ ] \)], "Input"], Cell[BoxData[ \({{\(-3\)\ z\_1\%2, \(-2\)\ z\_2 + 2\ y\_1\ z\_2}, {\(-1\), 0}}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(MatrixForm[\ DzG\ ]\)], "Input"], Cell[BoxData[ TagBox[ RowBox[{"(", GridBox[{ {\(\(-3\)\ z\_1\%2\), \(\(-2\)\ z\_2 + 2\ y\_1\ z\_2\)}, {\(-1\), "0"} }], ")"}], (MatrixForm[ #]&)]], "Output"] }, Open ]], Cell["Setting up equations to solve for z. See (4.6)", "Section"], Cell[CellGroupData[{ Cell[BoxData[ \(ls\ = \ {\ Thread[\ G\ - \ gg\&^\ ]\ }\)], "Input"], Cell[BoxData[ \({{\(-\(\(g\_1\)\&^\)\) + y\_1\%3 + y\_2\%2 - z\_1\%3 - z\_2\%2 + y\_1\ \((\(-4\) + z\_2\%2)\), \(-\(\(g\_2\)\&^\)\) + y\_1 - z\_1}} \)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(in\ = \ Thread[\ \ {\ zz\ , \ initialz\ }\ \ ]\)], "Input"], Cell[BoxData[ \({{z\_1, iz\_1}, {z\_2, iz\_2}}\)], "Output"] }, Open ]], Cell[BoxData[ \(Do[\ ls\ = \ Append[\ ls\ , \ in[\([i]\)]\ \ \ ]\ , \ {i, 1, Length[in]}\ ]\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(ls\)], "Input"], Cell[BoxData[ \({{\(-\(\(g\_1\)\&^\)\) + y\_1\%3 + y\_2\%2 - z\_1\%3 - z\_2\%2 + y\_1\ \((\(-4\) + z\_2\%2)\), \(-\(\(g\_2\)\&^\)\) + y\_1 - z\_1}, { z\_1, iz\_1}, {z\_2, iz\_2}}\)], "Output"] }, Open ]], Cell["Setting up the right hand side of state costate equations", "Section"], Cell[TextData[{ "\"stcst\" consists of the rhs of all differential equations. Its \ first component 1 is for integrating the time equation ", Cell[BoxData[ \(TraditionalForm\`\[PartialD]\_t\ t\)]], " = 1. The order of all the components is t,y, ", Cell[BoxData[ \(TraditionalForm\`\(g\&^\)\)]], ", \[Lambda], \[Xi]." }], "Text"], Cell["First we include the rhs of (1.1) and (1.3).", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(stcst\ = \ Flatten[\ {\ 1\ , \ F\ \ , \ dgg\&^\ }\ ]\)], "Input"], Cell[BoxData[ \({1, \(-y\_1\) + z\_1, \(-y\_2\) + z\_2, \(\(dg\_1\)\&^\), \(\(dg\_2\)\&^\)}\)], "Output"] }, Open ]], Cell["\<\ Next we compute the y-derivative of the error function, \"Dyee\" using (4.3) \ and (4.2).\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Dyee\ = \ TinDzG\ .\ Transpose[\ Table[\ \[PartialD]\_\(y\_i\)\((\ Transpose[DzG])\)\ .\ ee\ , \ {i, 1, Length[yy]}\ ]\ \ ]\)], "Input"], Cell[BoxData[ \({{2\ e\_1\ z\_2\ Tin[1, 2], 0}, {2\ e\_1\ z\_2\ Tin[2, 2], 0}}\)], "Output"] }, Open ]], Cell["\<\ Next we compute the y-derivative of the Hamiltonian using (3.6).\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(DyH\ = \ 2\ ee\ .\ Dyee\ \ + \ \ \[Lambda]\[Lambda]\ .\ DyF\)], "Input"], Cell[BoxData[ \({\(-\[Lambda]\_1\) + 2\ \((2\ e\_1\%2\ z\_2\ Tin[1, 2] + 2\ e\_1\ e\_2\ z\_2\ Tin[2, 2]) \), \(-\[Lambda]\_2\)}\)], "Output"] }, Open ]], Cell[TextData[ "Next we append the rhs of the \[Lambda]-costate DE. See paper (3.2). Note \ minus sign. "], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(stcst\ = \ Flatten[\ {\ stcst\ , \ \(-DyH\)\ }\ ]\)], "Input"], Cell[BoxData[ \({1, \(-y\_1\) + z\_1, \(-y\_2\) + z\_2, \(\(dg\_1\)\&^\), \(\(dg\_2\)\&^\), \[Lambda]\_1 - 2\ \((2\ e\_1\%2\ z\_2\ Tin[1, 2] + 2\ e\_1\ e\_2\ z\_2\ Tin[2, 2]) \), \[Lambda]\_2}\)], "Output"] }, Open ]], Cell["\<\ Next we compute the z-derivative of the error function, \"Dzee\" using (4.4) \ and (4.2).\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Dzee\ = \ TinDzG\ .\ Transpose[\ Table[\ \[PartialD]\_\(z\_i\)\((\ Transpose[DzG])\)\ .\ ee\ , \ {i, 1, Length[zz]}\ ]\ \ ]\)], "Input"], Cell[BoxData[ \(General::"spell1" \( : \ \) "Possible spelling error: new symbol name \"\!\(Dzee\)\" is similar to \ existing symbol \"\!\(Dyee\)\"."\)], "Message"], Cell[BoxData[ \({{\(-6\)\ e\_1\ z\_1\ Tin[1, 1], e\_1\ \((\(-2\) + 2\ y\_1)\)\ Tin[1, 2]}, { \(-6\)\ e\_1\ z\_1\ Tin[2, 1], e\_1\ \((\(-2\) + 2\ y\_1)\)\ Tin[2, 2]}}\)], "Output"] }, Open ]], Cell["\<\ Next we compute the z-derivative of the Hamiltonian using (3.7).\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(DzH\ = \ 2\ ee\ .\ Dzee\ \ + \ \ \[Lambda]\[Lambda]\ .\ DzF\)], "Input"], Cell[BoxData[ \({\[Lambda]\_1 + 2\ \((\(-6\)\ e\_1\%2\ z\_1\ Tin[1, 1] - 6\ e\_1\ e\_2\ z\_1\ Tin[2, 1])\), \[Lambda]\_2 + 2\ \((e\_1\%2\ \((\(-2\) + 2\ y\_1)\)\ Tin[1, 2] + e\_1\ e\_2\ \((\(-2\) + 2\ y\_1)\)\ Tin[2, 2])\)}\)], "Output"] }, Open ]], Cell[TextData[ "Next we append the rhs of the \[Xi]-costate DE. See paper (3.3). Note minus \ sign. "], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(stcst\ = \ Flatten[\ {\ stcst\ , \ \(-DzH\)\ }\ ]\)], "Input"], Cell[BoxData[ \({1, \(-y\_1\) + z\_1, \(-y\_2\) + z\_2, \(\(dg\_1\)\&^\), \(\(dg\_2\)\&^\), \[Lambda]\_1 - 2\ \((2\ e\_1\%2\ z\_2\ Tin[1, 2] + 2\ e\_1\ e\_2\ z\_2\ Tin[2, 2]) \), \[Lambda]\_2, \(-\[Lambda]\_1\) - 2\ \((\(-6\)\ e\_1\%2\ z\_1\ Tin[1, 1] - 6\ e\_1\ e\_2\ z\_1\ Tin[2, 1])\), \(-\[Lambda]\_2\) - 2\ \((e\_1\%2\ \((\(-2\) + 2\ y\_1)\)\ Tin[1, 2] + e\_1\ e\_2\ \((\(-2\) + 2\ y\_1)\)\ Tin[2, 2])\)}\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Part Two", "Subsubtitle"], Cell["Vector Input For Runge Kutta Step", "Section"], Cell[CellGroupData[{ Cell["Vector variables and substitutions.", "Subsubsection"], Cell[TextData[{ "\"st\" will be the vector input to rkStep. The order of the components of \ \"st\" is: t,y,", Cell[BoxData[ \(TraditionalForm\`\(g\&^\)\)]], ",\[Lambda], \[Xi]. " }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(st\ \ = \ \ Table[\ s[\ i\ ]\ , \ {i, 1\ , \ 1\ + \ n\ + \ m\ + \ n\ + \ m\ }\ ] \)], "Input"], Cell[BoxData[ \({s[1], s[2], s[3], s[4], s[5], s[6], s[7], s[8], s[9]}\)], "Output"] }, Open ]], Cell["\"st0\" will store initial values set up later.", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(st0\ \ = \ \ Table[\ \ int\ , \ {i, 1\ , \ 1\ + \ n\ + \ m\ + \ n\ + \ m\ }\ ] \)], "Input"], Cell[BoxData[ \({int, int, int, int, int, int, int, int, int}\)], "Output"] }, Open ]], Cell["\<\ In all the equations developed in Part One we must insert the correct \ components of \"st\". To do this we make the following lists of \ substitutions.\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(ysub\ = \ Table[\ yy[\([i]\)]\ -> \ st[\([\ 1\ + \ i\ ]\)]\ , \ {i, 1, Length[\ yy\ ]\ }\ ]\)], "Input"], Cell[BoxData[ \({y\_1 \[Rule] s[2], y\_2 \[Rule] s[3]}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(zsub\ = \ Table[\ zz[\([i]\)]\ -> \ nzz[\([\ i\ ]\)]\ , \ {i, 1, Length[\ zz\ ]\ }\ ]\)], "Input"], Cell[BoxData[ \(General::"spell1" \( : \ \) "Possible spelling error: new symbol name \"\!\(zsub\)\" is similar to \ existing symbol \"\!\(ysub\)\"."\)], "Message"], Cell[BoxData[ \({z\_1 \[Rule] nz\_1, z\_2 \[Rule] nz\_2}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(gsub\&^\ = \ Table[\ gg\&^[\([i]\)]\ -> \ st[\([\ 1\ + n\ + \ i\ ]\)]\ , \ {i, 1, Length[zz]}\ ]\)], "Input"], Cell[BoxData[ \(General::"spell" \( : \ \) "Possible spelling error: new symbol name \"\!\(gsub\)\" is similar to \ existing symbols \!\({ysub, zsub}\)."\)], "Message"], Cell[BoxData[ \({\(\(g\_1\)\&^\) \[Rule] s[4], \(\(g\_2\)\&^\) \[Rule] s[5]}\)], "Output"], Cell[CellGroupData[{ Cell[BoxData[ \(dgsub\&^\ = \ Table[\ dgg\&^[\([i]\)]\ -> \ ndgg\&^[\([\ i\ ]\)]\ , \ {i, 1, Length[zz]}\ ]\)], "Input"], Cell[BoxData[ \(General::"spell1" \( : \ \) "Possible spelling error: new symbol name \"\!\(dgsub\)\" is similar to \ existing symbol \"\!\(gsub\)\"."\)], "Message"], Cell[BoxData[ \({\(\(dg\_1\)\&^\) \[Rule] \(\(ndg\_1\)\&^\), \(\(dg\_2\)\&^\) \[Rule] \(\(ndg\_2\)\&^\)}\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\[Lambda]sub\ = \ Table[\ \[Lambda]\[Lambda][\([i]\)]\ -> \ st[\([\ 1\ \ + n\ + m\ + \ i\ ]\)]\ , \ \n \t\t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {i, 1, Length[\ yy\ ]} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ]\)], "Input"], Cell[BoxData[ \({\[Lambda]\_1 \[Rule] s[6], \[Lambda]\_2 \[Rule] s[7]}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\[Xi]sub\ = \ Table[\ \[Xi]\[Xi][\([i]\)]\ -> \ st[\([\ 1\ \ + \ n\ + m\ + n\ + \ i\ ]\)]\ , \n \t\t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {i, 1, Length[zz]} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ]\)], "Input"], Cell[BoxData[ \(General::"spell1" \( : \ \) "Possible spelling error: new symbol name \"\!\(\\[Xi]sub\)\" is \ similar to existing symbol \"\!\(\\[Lambda]sub\)\"."\)], "Message"], Cell[BoxData[ \({\[Xi]\_1 \[Rule] s[8], \[Xi]\_2 \[Rule] s[9]}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(esub\ = \ Table[\ ee[\([i]\)]\ -> \ nee[\([\ i\ ]\)]\ , \n \t\t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {i, 1, Length[zz]} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ]\)], "Input"], Cell[BoxData[ \(General::"spell" \( : \ \) "Possible spelling error: new symbol name \"\!\(esub\)\" is similar to \ existing symbols \!\({gsub, ysub, zsub}\)."\)], "Message"], Cell[BoxData[ \({e\_1 \[Rule] ne\_1, e\_2 \[Rule] ne\_2}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(Tsub\ = \ Table[\ TinDzG[\([i, j]\)]\ -> \ nTinDzG[\([i, j]\)]\ , \n \t\t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {i, 1, Length[zz]}, \ {j, 1, Length[zz]}\ ]\)], "Input"], Cell[BoxData[ \(General::"spell1" \( : \ \) "Possible spelling error: new symbol name \"\!\(Tsub\)\" is similar to \ existing symbol \"\!\(Stub\)\"."\)], "Message"], Cell[BoxData[ \({{Tin[1, 1] \[Rule] nTin[1, 1], Tin[1, 2] \[Rule] nTin[1, 2]}, { Tin[2, 1] \[Rule] nTin[2, 1], Tin[2, 2] \[Rule] nTin[2, 2]}}\)], "Output"] }, Open ]], Cell["The following, \"ssub\", is a master list of substitutions.", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(ssub1\ = \ Flatten[\ Union[\ ysub\ \ , \ \[Lambda]sub\ , \ \[Xi]sub\ \ , \ gsub\&^\ \ , \ dgsub\&^\ , \ esub\ \ , \ Tsub\ ]]\)], "Input"], Cell[BoxData[ \({Tin[1, 1] \[Rule] nTin[1, 1], Tin[1, 2] \[Rule] nTin[1, 2], Tin[2, 1] \[Rule] nTin[2, 1], Tin[2, 2] \[Rule] nTin[2, 2], \(\(dg\_1\)\&^\) \[Rule] \(\(ndg\_1\)\&^\), \(\(dg\_2\)\&^\) \[Rule] \(\(ndg\_2\)\&^\), \(\(g\_1\)\&^\) \[Rule] s[4], \(\(g\_2\)\&^\) \[Rule] s[5], e\_1 \[Rule] ne\_1, e\_2 \[Rule] ne\_2, y\_1 \[Rule] s[2], y\_2 \[Rule] s[3], \[Lambda]\_1 \[Rule] s[6], \[Lambda]\_2 \[Rule] s[7], \[Xi]\_1 \[Rule] s[8], \[Xi]\_2 \[Rule] s[9]} \)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(ssub\ = \ Union[\ ssub1\ , \ zsub\ \ ]\)], "Input"], Cell[BoxData[ \(General::"spell" \( : \ \) "Possible spelling error: new symbol name \"\!\(ssub\)\" is similar to \ existing symbols \!\({esub, gsub, ysub, zsub}\)."\)], "Message"], Cell[BoxData[ \({\(\(dg\_1\)\&^\) \[Rule] \(\(ndg\_1\)\&^\), \(\(dg\_2\)\&^\) \[Rule] \(\(ndg\_2\)\&^\), \(\(g\_1\)\&^\) \[Rule] s[4], \(\(g\_2\)\&^\) \[Rule] s[5], e\_1 \[Rule] ne\_1, e\_2 \[Rule] ne\_2, y\_1 \[Rule] s[2], y\_2 \[Rule] s[3], z\_1 \[Rule] nz\_1, z\_2 \[Rule] nz\_2, \[Lambda]\_1 \[Rule] s[6], \[Lambda]\_2 \[Rule] s[7], \[Xi]\_1 \[Rule] s[8], \[Xi]\_2 \[Rule] s[9], Tin[1, 1] \[Rule] nTin[1, 1], Tin[1, 2] \[Rule] nTin[1, 2], Tin[2, 1] \[Rule] nTin[2, 1], Tin[2, 2] \[Rule] nTin[2, 2]}\)], "Output"] }, Open ]], Cell["Objects needed in vector calculations", "Subsubsection"], Cell["\<\ In the following we insert the components of \"st\" into the objects set up \ in Part One. These objects have names starting with \"n\". These objects \ will be used in a function, called \"df\", for numerically computing the rhs \ of the DE's during the rkStep.\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(nls\ = \ ls\ //. \ ssub1\)], "Input"], Cell[BoxData[ \({{s[2]\^3 + s[3]\^2 - s[4] - z\_1\%3 - z\_2\%2 + s[2]\ \((\(-4\) + z\_2\%2)\), s[2] - s[5] - z\_1}, {z\_1, iz\_1}, { z\_2, iz\_2}}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(nDzG\ = \ DzG\ \ //. \ ssub\)], "Input"], Cell[BoxData[ \(General::"spell1" \( : \ \) "Possible spelling error: new symbol name \"\!\(nDzG\)\" is similar to \ existing symbol \"\!\(DzG\)\"."\)], "Message"], Cell[BoxData[ \({{\(-3\)\ nz\_1\%2, \(-2\)\ nz\_2 + 2\ s[2]\ nz\_2}, {\(-1\), 0}}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(n\[Xi]\[Xi]\ = \ \[Xi]\[Xi]\ //. \ ssub\)], "Input"], Cell[BoxData[ \({s[8], s[9]}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\(nstcst\ = \ stcst\ //. \ ssub\ \ \)\)], "Input"], Cell[BoxData[ \(General::"spell1" \( : \ \) "Possible spelling error: new symbol name \"\!\(nstcst\)\" is similar \ to existing symbol \"\!\(stcst\)\"."\)], "Message"], Cell[BoxData[ \({1, \(-s[2]\) + nz\_1, \(-s[3]\) + nz\_2, \(\(ndg\_1\)\&^\), \(\(ndg\_2\)\&^\), s[6] - 2\ \(( 2\ nTin[1, 2]\ ne\_1\%2\ nz\_2 + 2\ nTin[2, 2]\ ne\_1\ ne\_2\ nz\_2)\), s[7], \(-s[6]\) - 2\ \((\(-6\)\ nTin[1, 1]\ ne\_1\%2\ nz\_1 - 6\ nTin[2, 1]\ ne\_1\ ne\_2\ nz\_1)\), \(-s[7]\) - 2\ \((nTin[1, 2]\ \((\(-2\) + 2\ s[2])\)\ ne\_1\%2 + nTin[2, 2]\ \((\(-2\) + 2\ s[2])\)\ ne\_1\ ne\_2)\)}\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Part Three", "Subsubtitle"], Cell["\<\ The following module, called \"df\", numerically computes the right hand side \ of all DE's. It uses the numerical variables set up in Part Two. The module \ begins with computing the K-T multipliers and control.\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\(\ df[\ inst_\ ]\ \ := \ \(Module[\ {\ }\ , {\n (*\ The\ next\ command\ solves\ \((4.6)\)\ for\ \*"\"" \(z \*"\"\<\> "\)*) \n\t\t nzz\ = \ zz\ /. \ Apply[\ FindRoot\ , \ \((\((\ nls\ /. \ \n \t\t\t\t\t\t\t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Thread[\ st\ -> \ inst\ \ ]\ )\)\ \ /. \n \t\t\t\t\t\t\t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Thread[\ initialz\ -> \ ninitialz\ ]\ \ \ ) \)\ \ \ \ \ \ ]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ; \n\n \t\t\t\t\t\t\n\t (*\ The\ next\ command\ sets\ new\ initial\ values\n \t\t\t\t for\ the\ FindRoot \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ *) \n\t\t Do[\ niz\_i\ = \ nzz[\([i]\)]\ , \ {i, 1, Length[zz]}\ ] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ; \n \t\tDo[\ nz\_i\ = \ nzz[\([i]\)]\ , \ {i, 1, Length[zz]}\ ] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ; \n\t\t\t\n\t\t\t (*\ The\ next\ command\ finds\ the\ matrix\ occuring\ \n \t\t\t\t\ \ in\ many\ \(equations. \)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ *) \n\t\tnTinDzG\ = \ Transpose[\ \n\t\t\t\t\t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Inverse[\ nDzG\ /. \ Thread[\ st\ -> \ inst\ ]\ ]] \ \ \ \ \ \ \ \ \ \ \ \ \ ; \n\t\t\t\t\t Do[\ nTin[i, j]\ = \ nTinDzG[\([i, j]\)]\ , \n \t\t\t\t\t\t\ \ \ \ \ {i, 1, Length[zz]}, {j, 1, Length[zz]} \ \ ]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ; \n\t\t\t\t\t\n \t\t\t\t\t (*\ The\ next\ command\ finds\ the\ error\ function\n \t\t\t\t\t\t\ \ \*"\"" e \*"\"\< using (4.2)\> "*) \n \t\t\t\tnee\ = \ \(-\ nTinDzG\ . \((\ n\[Xi]\[Xi]\ /. \ Thread[\ st\ -> \ inst\ ])\)\)\ / \ 2\ \ \ \ \ \ ; \n\t\t\t\t\t Do[\ ne\_i\ = \ nee[\([i]\)]\ , \ {i, 1, Length[zz]}\ ] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ; \n \t\t\t\t\t\n\t\t\t\t\t (*\ The\ next\ command\ finds\ the\ rhs\ of\ the\ g\&^ - DE\n \t\t\t\t\t\t\ \ using\ \((4.8)\)\ and\ \(\((4.2)\). \)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ *) \n \t\t\t\tDo[\ \(ndg\_i\)\&^\ = \ nee[\([i]\)]\ , \ {i, 1, Length[zz]}\ ] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ; \n\t\t\t\n \t\t\t\t\t\t\ nstcst\ /. \ Thread[\ st\ -> \ inst\ ] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }]\)[\([1]\)]\n\)\)], "Input"], Cell[BoxData[ \(General::"spell1" \( : \ \) "Possible spelling error: new symbol name \"\!\(inst\)\" is similar to \ existing symbol \"\!\(int\)\"."\)], "Message"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Part Four", "Subsubtitle"], Cell[CellGroupData[{ Cell[TextData[{ "Set initial values for y-state and ", Cell[BoxData[ \(TraditionalForm\`\(g\&^\)\)]], "-state." }], "Section"], Cell[CellGroupData[{ Cell[BoxData[ \(initialy\ = \ {\ \ \(-1\)\ , \ \(-2\)\ }\)], "Input"], Cell[BoxData[ \(General::"spell1" \( : \ \) "Possible spelling error: new symbol name \"\!\(initialy\)\" is similar \ to existing symbol \"\!\(initialz\)\"."\)], "Message"], Cell[BoxData[ \({\(-1\), \(-2\)}\)], "Output"] }, Open ]], Cell["\<\ The following two commands are needs to give an initial approximation for \ FindRoot in the \"df\" module aaat the first attempts.\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(niz\_\(1\ \)\ = \ \(-1\)\)], "Input"], Cell[BoxData[ \(\(-1\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(niz\_2\ = \ 2\)], "Input"], Cell[BoxData[ \(2\)], "Output"] }, Open ]], Cell["(*^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^*)", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(initialg\&^\ = \ {\ \ 0\ , \ 0\ }\)], "Input"], Cell[BoxData[ \(General::"spell" \( : \ \) "Possible spelling error: new symbol name \"\!\(initialg\)\" is similar \ to existing symbols \!\({initialy, initialz}\)."\)], "Message"], Cell[BoxData[ \({0, 0}\)], "Output"] }, Open ]] }, Open ]], Cell["Set the initial time equal zero", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(st0[\([1]\)]\ = \ 0\)], "Input"], Cell[BoxData[ \(0\)], "Output"] }, Open ]], Cell["\<\ Put the initial values of y and z into the initial data vector \"st0\".\ \>", "Text"], Cell[BoxData[ \(Do[\ \ st0[\([1 + i]\)]\ = \ initialy[\([i]\)]\ , \ {i, 1, Length[\ yy\ ]}\ \ ]\)], "Input"], Cell[BoxData[ \(Do[\ st0[\([1 + Length[\ yy\ ] + i]\)]\ = \ initialg\&^[\([i]\)]\ , \n \t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {i, 1, Length[\ zz\ ]}\ \ ]\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(st0\)], "Input"], Cell[BoxData[ \({0, \(-1\), \(-2\), 0, 0, int, int, int, int}\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Part Five", "Subsubtitle"], Cell["\<\ Set initial values of costate. Run Runge-Kutta solver. View graphs of output.\ \ \>", "Text"], Cell[CellGroupData[{ Cell["Set initial costate values", "Section", CellTags->"costate"], Cell["\<\ Adjust initial costate values until constraint set is approached.\ \>", "Text"], Cell[BoxData[ \(initial\[Lambda]\ = \ {\ 0\ , \ 0\ \ }\)], "Input", CellTags->"il"], Cell[BoxData[ \({0, 0}\)], "Output"], Cell[CellGroupData[{ Cell[BoxData[ \(initial\[Xi]\ = \ {\ 0\ , \ 0\ }\)], "Input"], Cell[BoxData[ \(General::"spell1" \( : \ \) "Possible spelling error: new symbol name \"\!\(initial\\[Xi]\)\" is \ similar to existing symbol \"\!\(initial\\[Lambda]\)\"."\)], "Message"], Cell[BoxData[ \({0, 0}\)], "Output"], Cell[TextData[ "Put the initial values of \[Lambda] and \[Xi] into the initial data vector \ \"st0\"."], "Text"], Cell[BoxData[ \(Do[\ \ st0[\([1 + Length[\ yy\ ] + Length[\ zz\ ] + i]\)]\ = \ initial\[Lambda][\([i]\)]\ , \n \t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {i, 1, Length[\ yy\ ]}\ \ ]\)], "Input"], Cell[BoxData[ \(Do[\ st0[\([1 + Length[\ yy\ ] + Length[\ zz\ ] + Length[\ yy\ ] + i]\)]\ = \ \n\t\ \ \ \ \ \ \ initial\[Xi][\([i]\)]\ , \ \ {i, 1, Length[\ zz\ ]} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ]\)], "Input"], Cell[BoxData[ \(st0\)], "Input"], Cell[BoxData[ \({0, \(-1\), \(-2\), 0, 0, 0, 0, 0, 0}\)], "Output"] }, Open ]], Cell["\<\ \"tend\" is the end time for solving the trajectories. You might want to \ change this sometimes.\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(tend\ \ = \ \ 1.\)], "Input"], Cell[BoxData[ StyleBox["1.`", StyleBoxAutoDelete->True, PrintPrecision->1]], "Output"] }, Open ]] }, Open ]], Cell["\"dt\" is the step size used in the Runge-Kutta step.", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(ninitialz\)], "Input"], Cell[BoxData[ \({\(-1\), 2}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(dt\ = \ .05\)], "Input"], Cell[BoxData[ StyleBox["0.05`", StyleBoxAutoDelete->True, PrintPrecision->1]], "Output"] }, Open ]], Cell["The following is a set to hold the z-state trajectory", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(nDzG\ /. \ Thread[\ st\ -> \ st0\ ]\)], "Input"], Cell[BoxData[ \({{\(-3\)\ nz\_1\%2, \(-4\)\ nz\_2}, {\(-1\), 0}}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(nz\_2\)], "Input"], Cell[BoxData[ \(nz\_2\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(df[\ st0\ ]\)], "Input"], Cell[BoxData[ \({1, 0.`, 4.`, 0.`, 0.`, 0.`, 0, 0.`, 0.`}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(nzz\)], "Input"], Cell[BoxData[ \({\(-1.`\), 2.`}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(st0\)], "Input"], Cell[BoxData[ \({0, \(-1\), \(-2\), 0, 0, 0, 0, 0, 0}\)], "Output"] }, Open ]], Cell["Run Runge-Kutta Solver", "Section"], Cell["\<\ Solution is kept in table \"sol\". Since this is a big table I have ended \ the command with a \";\" so the table is not printed. Remove the \";\" to \ see the table.\ \>", "Text"], Cell["Reset these for the beginning of the trajectory", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(niz\_\(1\ \)\ = \ \(-1\)\)], "Input"], Cell[BoxData[ \(\(-1\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(niz\_2\ = \ 2\)], "Input"], Cell[BoxData[ \(2\)], "Output"] }, Open ]], Cell["(*^^^^^^^^^^^^^^^^^^^^^^^^^^^^*)", "Text"], Cell[BoxData[ \(\(\ \(sol\ = \ rk[\ st0, \ {\ tend, \ dt\ }]\ \ \ ; \)\)\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(\(sol\n\)\)], "Input"], Cell[BoxData[ \({{0, \(-1.`\), \(-2.`\), 0, 0, 0, 0, 0, 0}, {0.0499999999999999911`, \(-1.`\), \(-1.80727586819038918`\), 0.`, 0.`, 0.`, 0, 0.`, 0.`}, { 0.0999999999999999822`, \(-1.`\), \(-1.62824563664157757`\), 0.`, 0.`, 0.`, 0, 0.`, 0.`}, {0.15`, \(-1.`\), \(-1.46170639256335066`\), 0.`, 0.`, 0.`, 0, 0.`, 0.`}, {0.199999999999999982`, \(-1.`\), \(-1.30656050922263222`\), 0.`, 0.`, 0.`, 0, 0.`, 0.`}, { 0.249999999999999955`, \(-1.`\), \(-1.16180809472382251`\), 0.`, 0.`, 0.`, 0, 0.`, 0.`}, {0.3`, \(-1.`\), \(-1.02653984821490351`\), 0.`, 0.`, 0.`, 0, 0.`, 0.`}, {0.35`, \(-1.`\), \(-0.899930252213859915`\), 0.`, 0.`, 0.`, 0, 0.`, 0.`}, {0.399999999999999964`, \(-1.`\), \(-0.781231068075386136`\), 0.`, 0.`, 0.`, 0, 0.`, 0.`}, {0.45`, \(-1.`\), \(-0.669765086364312089`\), 0.`, 0.`, 0.`, 0, 0.`, 0.`}, { 0.499999999999999911`, \(-1.`\), \(-0.564920157069513351`\), 0.`, 0.`, 0.`, 0, 0.`, 0.`}, {0.549999999999999911`, \(-1.`\), \(-0.466143494480178244`\), 0.`, 0.`, 0.`, 0, 0.`, 0.`}, {0.6`, \(-1.`\), \(-0.372936278049554426`\), 0.`, 0.`, 0.`, 0, 0.`, 0.`}, { 0.65`, \(-1.`\), \(-0.284848569239385973`\), 0.`, 0.`, 0.`, 0, 0.`, 0.`}, {0.700000000000000088`, \(-1.`\), \(-0.201474561955042563`\), 0.`, 0.`, 0.`, 0, 0.`, 0.`}, {0.750000000000000088`, \(-1.`\), \(-0.122448178777756733`\), 0.`, 0.`, 0.`, 0, 0.`, 0.`}, { 0.800000000000000177`, \(-1.`\), \(-0.0474390182344587874`\), 0.`, 0.`, 0.`, 0, 0.`, 0.`}, {0.850000000000000177`, \(-1.`\), 0.0238513497753755476`, 0.`, 0.`, 0.`, 0, 0.`, 0.`}, { 0.900000000000000177`, \(-1.`\), 0.091692746412407331`, 0.`, 0.`, 0.`, 0, 0.`, 0.`}, {0.950000000000000355`, \(-1.`\), 0.156329417299347568`, 0.`, 0.`, 0.`, 0, 0.`, 0.`}, { 1.00000000000000022`, \(-1.`\), 0.217982773404089824`, 0.`, 0.`, 0.`, 0, 0.`, 0.`}}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell["Solving for the z-trajectory", "Section"], Cell["Reset the initial values of \"z\" at beginning of trajectory.", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(ninitialz\ = \ {\ \(-1\)\ , \ 2\ }\)], "Input"], Cell[BoxData[ \({\(-1\), 2}\)], "Output"] }, Open ]], Cell["(*^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^*)", "Text"], Cell["The following is a set that will hold the z-trajectory", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(zset\ = \ {\ }\)], "Input"], Cell[BoxData[ \({}\)], "Output"] }, Open ]], Cell[BoxData[ \(\(\ Do[ \ {\ \ nzz\ = \ zz\ /. Apply[\ FindRoot\ , \ \((ls\ /. \ \ \ Thread[\ \ \n\t\t\t\t\t\t\t\t\t\ \ Flatten[ \ {\ t\ , \ yy\ , \ gg\&^\ , \ \[Lambda]\[Lambda]\ , \ \[Xi]\[Xi]\ }\ ]\ -> \n \t\t\t\t\t\t\t\t\t\ \ \ sol[\([i]\)]\ ])\)\ /. \ Thread[\ initialz\ -> \ ninitialz\ ]\ ]\ \ \ ; \n\n \ \ \ \ \ \ \ \ \ \ \ \ \ \ ninitialz\ = \ nzz\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ; \n\n \ \ \ \ \ \ \ \ \ \ zset\ = \ Append[\ \ \ zset\ , \ \n\t\t\t\t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Flatten[\ {\ \(sol[\([i]\)]\)[\([1]\)]\ , \ nzz\ }\ \ ]\ \ ] \ \ \ \ \ \ \ ; \ \ \ \ }\ \ , \n \ \ \ \ \ \ \ \ \ \ \ \ {i, 1, Length[sol]} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ]\)\)], "Input"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(zset\)], "Input"], Cell[BoxData[ \({{0, \(-1.`\), 2.`}, {0.0499999999999999911`, \(-1.`\), 1.90607529543648563`}, {0.0999999999999999822`, \(-1.`\), 1.82362055445257791`}, {0.15`, \(-1.`\), 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"Section"], Cell[BoxData[ \(\(s\ = Flatten[\ sol[\([\ \ Range[1, Length[sol]\ ]\ , \ Range[1, 1]\ \ ]\)]] \ ; \)\)], "Input"], Cell[BoxData[ \(Do[\ gy[i]\ = Interpolation[\ \ Transpose[{\ s\ , \ Flatten[\ \n\t\t\t\t\t\t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ sol[\([\ \ Range[1, Length[sol]\ ]\ , \n \t\t\t\t\t\t\t\t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Range[1 + i, 1 + i]\ \ ] \)]]\ }]\ ]\ \ \ , \n \t\ \ \ \ \ {i, 1, Length[\ yy\ ]} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ]\)], "Input"], Cell[BoxData[ \(Do[\ ggg\&^[i]\ = Interpolation[\ \ Transpose[{\ s\ , \ Flatten[\ \n\t\t\t\t\t\t\ \ \ \ \ \ \ \ \ \ \ \ \ sol[\([\ \ Range[1, Length[sol]\ ]\ \ , \n \t\t\t\t\t\t\t\t\t\t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Range[1 + Length[yy] + i, 1 + \n\t\t\t\t\t\t\t\t\t\t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \t Length[yy] + i]\ \ 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