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We also presented \ applications of these packages to urban economic models[1]. However, applications are limited in the package for solving dynamic \ optimization problems, since there exist many problems for which the package \ cannot be applied. In this paper, we revise and expand it so that it can be \ widely used. By further expanding the package above, we can solve differential games. We \ first show a solution of simple differential games with Mathematica and view \ the optimal trajectories. We then present packages for differential games by \ upgrading the dynamic optimization package, and show some applications to \ economic problems.\ \>", "Text", CellMargins->{{58.5, 65.5}, {Inherited, Inherited}}] }, Open ]], Cell[CellGroupData[{ Cell["1.\tIntroduction", "Section"], Cell["\<\ Continuous-time dynamic optimization models play a vital role in analyzing \ time-dependent activities in an economic system. A system is represented by \ differential equations with respect to time, and its objective is given by \ integration over time. \ \>", "Text"], Cell[TextData[{ "Pontryagin's maximum principle are often employed to solve dynamic \ optimization problems. The principle states that the solutions must satisfy \ boundary-value differential equations, which in some cases can be solved by \ the ", StyleBox["Mathematica", FontSlant->"Italic"], " ", StyleBox["DSolve", FontFamily->"Courier New"], " command with both initial and terminal conditions. " }], "Text"], Cell["\<\ Unfortunately in many cases, however, analytical solutions cannot be \ obtained, thus we carry out calculations to obtain a numerical solution, \ which may be undesirable because of obscuring relationships between \ parameters and solutions. Sometimes, approximate solutions will suffice to \ investigate the relations. \ \>", "Text"], Cell["\<\ As described in our paper [1], we developed the commands which solves models \ approximately. We have upgraded some of the packages and show them in this \ paper. \ \>", "Text"], Cell["\<\ When several agents are involved in a model and they act independently or \ cooperatively, continuous-time dynamic optimization models of these agents \ are regarded as differential games. Each agent, or player, has its own \ objective to be maximized. We can use our packages to obtain approximate \ solutions of differential games and demonstrate their utilization in this \ paper.\ \>", "Text"], Cell["\<\ This paper is organized as follows: first, we present a simple economic \ differential game model; we then obtain its Pareto-optimal exact solution; we \ describe an open-loop Nash equilibria solution and get it approximately; we \ show the usage, applicability and advantages of our packages which solve \ models approximately. \ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["2.\tSimple economic model", "Section"], Cell[TextData[{ "Following references [2] and [3], we consider two communities that carry \ out a joint project, for example, serving public goods. They make each \ continuous contribution ", Cell[BoxData[ SubscriptBox[ StyleBox["x", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox["1", FontFamily->"Times New Roman", FontSize->7]]]], "(", StyleBox["t", FontSlant->"Italic"], ") and ", Cell[BoxData[ SubscriptBox[ StyleBox["x", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox["2", FontFamily->"Times New Roman", FontSize->7, FontWeight->"Plain", FontSlant->"Plain", FontColor->GrayLevel[0], FontVariations->{"Underline"->False, "StrikeThrough"->False}]]]], "(", StyleBox["t", FontSlant->"Italic"], ") over time ", StyleBox["t", FontSlant->"Italic"], ". 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"StrikeThrough"->False}]]], " is the same as that of ", Cell[BoxData[ StyleBox[ SubscriptBox[ StyleBox["x", FontSize->10, FontSlant->"Italic"], StyleBox["1", FontFamily->"Times New Roman", FontSize->7, FontWeight->"Plain", FontSlant->"Plain", FontColor->GrayLevel[0], FontVariations->{"Underline"->False, "StrikeThrough"->False}]], FontFamily->"Times New Roman", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontColor->GrayLevel[0], FontVariations->{"Underline"->False, "StrikeThrough"->False}]]], ", since we have assumed that \[Sigma] = ", Cell[BoxData[ \(1\/2\)]], "." }], "Text"], Cell[TextData[{ "We then try to obtain the exact solution when ", Cell[BoxData[ StyleBox[ SubscriptBox[ StyleBox["K", FontSize->10, FontSlant->"Italic"], StyleBox["0", FontFamily->"Times New Roman", FontSize->7, FontWeight->"Plain", FontSlant->"Plain", FontColor->GrayLevel[0], FontVariations->{"Underline"->False, "StrikeThrough"->False}]], FontFamily->"Times New Roman", FontSize->12, 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Hence, we delete ", Cell[BoxData[ StyleBox[ SubscriptBox[ StyleBox["K", FontFamily->"Courier New", FontSize->10, FontWeight->"Bold", FontSlant->"Plain"], StyleBox["0", FontFamily->"Times New Roman", FontSize->7, FontWeight->"Bold", FontSlant->"Plain", FontColor->GrayLevel[0], FontVariations->{"Underline"->False, "StrikeThrough"->False}]], FontFamily->"Times New Roman", FontSize->12, FontWeight->"Plain", FontSlant->"Italic", FontColor->GrayLevel[0], FontVariations->{"Underline"->False, "StrikeThrough"->False}]], FontFamily->"Courier New", FontWeight->"Bold"], StyleBox["->1", FontFamily->"Courier New", FontWeight->"Bold"], " and ", StyleBox["T->10", FontFamily->"Courier New", FontWeight->"Bold"], " above and repeat the similar calculation." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(deqT = { \(K'\)[t] == \((\(\(D[H, \[Phi]] /. xSol\) /. K -> K[t]\) /. \[Phi] -> \[Phi][t])\), \(\[Phi]'\)[t] == \((\(r\ \[Phi] - D[H, K] /. K -> K[t]\) /. \[Phi] -> \[Phi][t]) \), K[0] == K\_0, \[Phi][T] == 0} /. r -> 0.05\)], "Input", CellLabel->"In[33]:="], Cell[BoxData[ RowBox[{"{", RowBox[{ RowBox[{ RowBox[{ SuperscriptBox["K", "\[Prime]", MultilineFunction->None], "[", "t", "]"}], "==", \(\(-0.01`\)\ K[t] - \[Phi][t]\/25\)}], ",", RowBox[{ RowBox[{ SuperscriptBox["\[Phi]", "\[Prime]", MultilineFunction->None], "[", "t", "]"}], "==", \(4 - 2\ K[t] + 0.0600000000000000088`\ \[Phi][t]\)}], ",", \(K[0] == K\_0\), ",", \(\[Phi][T] == 0\)}], "}"}]], "Output", CellLabel->"Out[33]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(DSolve[deqT, {K[t], \[Phi][t]}, t]\)], "Input", CellLabel->"In[34]:="], Cell[BoxData[ RowBox[{"DSolve", "[", RowBox[{ RowBox[{"{", RowBox[{ RowBox[{ RowBox[{ SuperscriptBox["K", "\[Prime]", MultilineFunction->None], "[", "t", "]"}], "==", \(\(-0.01`\)\ K[t] - \[Phi][t]\/25\)}], ",", RowBox[{ RowBox[{ SuperscriptBox["\[Phi]", "\[Prime]", MultilineFunction->None], "[", "t", "]"}], "==", \(4 - 2\ K[t] + 0.0600000000000000088`\ \[Phi][t]\)}], ",", \(K[0] == K\_0\), ",", \(\[Phi][T] == 0\)}], "}"}], ",", \({K[t], \[Phi][t]}\), ",", "t"}], "]"}]], "Output", CellLabel->"Out[34]="] }, Open ]], Cell[TextData[{ StyleBox["DSolve", FontFamily->"Courier New"], " does not return a solution. Apparently we cannot obtain the exact \ solution of the boundary-value problem with the ", StyleBox["DSolve", FontFamily->"Courier New"], " command." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["3.2 an approximate solution", "Subsection"], Cell[TextData[{ "Our scheme here is that we will obtain a solution approximated by a \ polynomial of an arbitrary degree ", StyleBox["n", FontSlant->"Italic"], ". It is expected that a solution is closer to the exact one as ", StyleBox["n", FontSlant->"Italic"], " is larger, although it will take longer computation time. To solve \ differential equations with boundary values, we have developed packages for \ solving simultaneous nonlinear equations approximately and for solving \ differential equations approximately. " }], "Text"], Cell[CellGroupData[{ Cell["\<\ 3.2.1 approximate solution of simultaneous nonlinear equations\ \>", "Subsubsection"], Cell[TextData[{ "We cannot always obtain solutions of simultaneous nonlinear equations with \ the ", StyleBox["Mathematica ", FontSlant->"Italic"], "command ", StyleBox["Solve", FontFamily->"Courier New"], ", even if there exists the exact analytical solutions. Newton's method is \ often used to get numerical solutions. The method can be also used to obtain \ a symbolic solution approxiamted by a polynomial of parameters. " }], "Text"], Cell[TextData[{ "Consider simultaneous nonlinear equations ", StyleBox["f", FontFamily->"Times New Roman", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["(", FontFamily->"Times New Roman"], StyleBox["x", FontFamily->"Times New Roman", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[", ", FontFamily->"Times New Roman"], StyleBox["p", FontFamily->"Times New Roman", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[")=", FontFamily->"Times New Roman"], StyleBox["0", FontFamily->"Times New Roman", FontWeight->"Bold"], " where ", StyleBox["f", FontWeight->"Bold", FontSlant->"Italic"], " is a nonlinear vector function, ", StyleBox["x", FontWeight->"Bold", FontSlant->"Italic"], " is a variable vector, ", StyleBox["p", FontWeight->"Bold", FontSlant->"Italic"], " is a parameter vector. To solve ", StyleBox["f", FontWeight->"Bold", FontSlant->"Italic"], "(", StyleBox["x", FontWeight->"Bold", FontSlant->"Italic"], ", ", StyleBox["p", FontWeight->"Bold", FontSlant->"Italic"], ")=", StyleBox["0", FontWeight->"Bold"], " with respect to ", StyleBox["x", FontWeight->"Bold", FontSlant->"Italic"], ", by Newton's method we repeat the following calculation: " }], "Text"], Cell[TextData[{ Cell[BoxData[ SubscriptBox[ StyleBox["x", FontSlant->"Italic"], RowBox[{ StyleBox["n", FontSlant->"Italic"], "+", "1"}]]], FontFamily->"Times New Roman"], " = ", Cell[BoxData[ SubscriptBox[ StyleBox["x", FontSlant->"Italic"], StyleBox["n", FontSlant->"Italic"]]], FontFamily->"Times New Roman"], " ", StyleBox["-", FontFamily->"Courier New"], " ", Cell[BoxData[ FractionBox[ RowBox[{ StyleBox["f", FontFamily->"Times New Roman", FontSize->10, FontSlant->"Italic"], RowBox[{ StyleBox["(", FontFamily->"Times New Roman", FontSize->10, FontWeight->"Plain", FontSlant->"Plain", FontColor->GrayLevel[0], 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FontWeight->"Plain", FontSlant->"Plain", FontColor->GrayLevel[0], FontVariations->{"Underline"->False, "StrikeThrough"->False}], StyleBox["p", FontFamily->"Times New Roman", FontSize->10, FontWeight->"Plain", FontSlant->"Italic", FontColor->GrayLevel[0], FontVariations->{"Underline"->False, "StrikeThrough"->False}]}], StyleBox[")", FontFamily->"Times New Roman", FontSize->10, FontWeight->"Plain", FontSlant->"Plain", FontColor->GrayLevel[0], FontVariations->{"Underline"->False, "StrikeThrough"->False}]}]}], RowBox[{ StyleBox["\[PartialD]", FontFamily->"Times New Roman", FontSize->10, FontWeight->"Plain", FontSlant->"Plain", FontColor->GrayLevel[0], FontVariations->{"Underline"->False, "StrikeThrough"->False}], StyleBox["x", FontFamily->"Times New Roman", FontSize->10, FontWeight->"Plain", FontSlant->"Italic", FontColor->GrayLevel[0], FontVariations->{"Underline"->False, "StrikeThrough"->False}]}]], FontFamily->"Times New Roman", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontColor->GrayLevel[0], FontVariations->{"Underline"->False, "StrikeThrough"->False}]]]], "." }], "NumberedEquation", TextAlignment->Center, TextJustification->0], Cell[TextData[{ "To determine an initial ", Cell[BoxData[ SubscriptBox[ StyleBox["x", FontWeight->"Bold", FontSlant->"Italic"], "0"]], FontFamily->"Times New Roman"], ", with an approximate value ", Cell[BoxData[ SubscriptBox[ StyleBox["p", FontWeight->"Bold", FontSlant->"Italic"], "0"]], FontFamily->"Times New Roman"], " of ", StyleBox["p", FontWeight->"Bold", FontSlant->"Italic"], ", we use Newton's method again to solve ", StyleBox["f", FontWeight->"Bold", FontSlant->"Italic"], "(", Cell[BoxData[ StyleBox[ SubscriptBox[ StyleBox["x", FontFamily->"Times New Roman", FontSize->10, FontWeight->"Bold", FontSlant->"Italic", FontColor->GrayLevel[0], FontVariations->{"Underline"->False, "StrikeThrough"->False}], "0"], FontFamily->"Times New Roman", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontColor->GrayLevel[0], 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" }], "Text"], Cell["\<\ We have developed the approximateSolve package to carry out the above \ calculation. First, we read the package:\ \>", "Text"], Cell[BoxData[ \(<< approximate`Solve`\)], "Input", CellLabel->"In[1]:="], Cell[CellGroupData[{ Cell[BoxData[ \(\(?approximateSolve\)\)], "Input", CellLabel->"In[2]:="], Cell[BoxData[ \("approximateSolve[f,var,var0,prm,degree] solves systems of nonlinear \ equations approximately. The solution is approximated by degree-th degree \ polynomial of prm. f is a list of nonlinear equations, var is a list of \ variables, var0 is a list of initial values for var, and prm is a list of \ parameters with the corresponding approximate values. If the first element \ of the output is False, no solution is obtained."\)], "Print"] }, Open ]], Cell[TextData[{ "To solve ", Cell[BoxData[ SuperscriptBox[ StyleBox["x", FontFamily->"Times New Roman", FontSlant->"Italic"], "3"]]], StyleBox["-", FontFamily->"Courier New"], " ", Cell[BoxData[ SuperscriptBox[ StyleBox["p", FontFamily->"Times New Roman", FontSlant->"Italic"], "2"]]], "=0 around ", StyleBox["p", FontSlant->"Italic"], "=1 with the approximation by a polynomial of degree 5, execute the \ following command:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\(\ \ \ approximateSolve[{x^3 - p^2 == 0}, {x}, {1. }, {{p, 1. }}, 5]\)\)], "Input", CellLabel->"In[23]:="], Cell[BoxData[ \({x \[Rule] \(0.124828532235939038`\[InvisibleSpace]\) + 1.24828532235939881`\ p - 0.624142661179702074`\ p\^2 + 0.356652949245545203`\ p\^3 - 0.124828532235941169`\ p\^4 + 0.0192043895747602277`\ p\^5}\)], "Output", CellLabel->"Out[23]="] }, Open ]], Cell[TextData[{ "To see how close the approximate 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