(* Content-type: application/vnd.wolfram.mathematica *) (*** Wolfram Notebook File ***) (* http://www.wolfram.com/nb *) (* CreatedBy='Mathematica 9.0' *) (*CacheID: 234*) (* Internal cache information: NotebookFileLineBreakTest NotebookFileLineBreakTest NotebookDataPosition[ 157, 7] NotebookDataLength[ 420606, 9779] NotebookOptionsPosition[ 402639, 9181] NotebookOutlinePosition[ 403774, 9220] CellTagsIndexPosition[ 403644, 9214] WindowFrame->Normal*) (* Beginning of Notebook Content *) Notebook[{ Cell[CellGroupData[{ Cell[TextData[{ "7-5-S-Laplace Transform Issues and Opportunities in ", StyleBox["Mathematica", FontSlant->"Italic"], " " }], "Section", CellDingbat->None, CellChangeTimes->{{3.63896138583011*^9, 3.6389613956101236`*^9}, { 3.642592284293995*^9, 3.642592292454006*^9}, 3.6441452927839127`*^9}, TextAlignment->Center], Cell["\<\ The Laplace Transform is a mathematical construct that has proven very useful \ in both solving and understanding differential equations. We introduce it and \ show its power here.\ \>", "Subsubsection", CellChangeTimes->{{3.638961315520012*^9, 3.63896136438008*^9}, { 3.638961410300144*^9, 3.638961412840148*^9}}], Cell[CellGroupData[{ Cell["\<\ Definition of Laplace Transform of a function f(t) if the integral converges:\ \>", "Subsubsection", CellChangeTimes->{{3.6389571797851124`*^9, 3.6389571977151375`*^9}}], Cell[BoxData[ StyleBox[ RowBox[{ RowBox[{"\[ScriptCapitalL]", StyleBox[ RowBox[{"(", RowBox[{"f", RowBox[{"(", "t", ")"}]}], ")"}], FontFamily->"Times New Roman"], StyleBox[" ", FontFamily->"Times New Roman"], StyleBox[ RowBox[{"(", "s", ")"}], FontFamily->"Times New Roman"]}], StyleBox["=", FontFamily->"Times New Roman"], StyleBox[" ", FontFamily->"Times New Roman"], StyleBox[ RowBox[{ RowBox[{"F", RowBox[{"(", "s", ")"}]}], " ", "=", " ", RowBox[{ SubsuperscriptBox["\[Integral]", "0", "\[Infinity]"], RowBox[{ SuperscriptBox["e", RowBox[{ RowBox[{"-", "s"}], " ", "t"}]], "f", RowBox[{"(", "t", ")"}], " ", "dt"}]}]}], FontFamily->"Times New Roman"]}], FontWeight->"Bold"]], "Text", CellChangeTimes->{{3.6389570020648637`*^9, 3.6389571204150295`*^9}, 3.638957188505125*^9, {3.6389573367453327`*^9, 3.638957337085333*^9}}, TextAlignment->Center], Cell["\<\ In order to evaluate such an integral we might be using integration by parts \ for f(t) of a different \"sort\" than exponential functions. However, we \ turn to Mathematica to do the integrations for us and use its own powerful \ computation, look-up, and pattern matching algorithms to do Laplace \ Transforms.\ \>", "Text", CellChangeTimes->{{3.6389578026940937`*^9, 3.6389579084342413`*^9}, { 3.6389614226601615`*^9, 3.6389614421501884`*^9}, 3.642592163913826*^9}] }, Open ]], Cell[CellGroupData[{ Cell["\<\ This \"strange\" integral transforms most functions (if the integral \ converges) from a t domain to an s domain where s is (in our case) a real \ positive number.\ \>", "Subsubsection"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Integrate", "[", RowBox[{ RowBox[{ RowBox[{"Exp", "[", RowBox[{ RowBox[{"-", " ", "s"}], " ", "t"}], "]"}], RowBox[{"f", "[", "t", "]"}]}], ",", RowBox[{"{", RowBox[{"t", ",", "0", ",", " ", "Infinity"}], "}"}]}], "]"}]], "Input"], Cell[BoxData[ RowBox[{"ConditionalExpression", "[", RowBox[{ FractionBox["1", "s"], ",", RowBox[{ RowBox[{"Re", "[", "s", "]"}], ">", "0"}]}], "]"}]], "Output", CellChangeTimes->{3.6389572189451675`*^9}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Integrate", "[", RowBox[{ RowBox[{ RowBox[{"Exp", "[", RowBox[{ RowBox[{"-", " ", "s"}], " ", "t"}], "]"}], RowBox[{"Cos", "[", RowBox[{"3", "t"}], "]"}]}], ",", RowBox[{"{", RowBox[{"t", ",", "0", ",", " ", "Infinity"}], "}"}]}], "]"}]], "Input"], Cell[BoxData[ RowBox[{"ConditionalExpression", "[", RowBox[{ FractionBox["s", RowBox[{"9", "+", SuperscriptBox["s", "2"]}]], ",", RowBox[{ RowBox[{"Re", "[", "s", "]"}], ">", "0"}]}], "]"}]], "Output", CellChangeTimes->{3.638957221305171*^9}] }, Open ]], Cell["\<\ Notice the restriction that Re[s]>0 in order for us to have convergence to \ the quantity, s/(9+s^2). Thus s/(9+s^2) IS the Laplace Transform of the \ function cos(3t). \ \>", "Text", CellChangeTimes->{{3.6389572318051853`*^9, 3.638957279765252*^9}, 3.638957945314293*^9}], Cell[TextData[{ "Tables of these exist and we could visit them, however since ", StyleBox["Mathematica", FontSlant->"Italic"], " can do these instantly without us looking them up we do our computations \ of Laplace Transforms in ", StyleBox["Mathematica", FontSlant->"Italic"], "." }], "Text", CellChangeTimes->{{3.6389572318051853`*^9, 3.638957279765252*^9}, { 3.638957945314293*^9, 3.6389579709543295`*^9}}] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Often the s domain is called the ", StyleBox["frequency domain", FontSlant->"Italic", FontVariations->{"Underline"->True}], ", and the t domain is called the ", StyleBox["time domain", FontSlant->"Italic", FontVariations->{"Underline"->True}], ", for if the following integral - the Laplace Transform - is to make sense \ with respect to units then the units of s must be 1/Time, for in this way s*t \ has no units and so Exp[-s*t] has no units. Thus the Laplace Transform \ integral is obtained by just integrating a function of t over an interval of \ time, [0, \[Infinity]) - resulting in a function of s, we call it \ \[ScriptCapitalL] (f(t))(s) = F(s)." }], "Subsubsection", CellChangeTimes->{{3.638957299935281*^9, 3.638957333045327*^9}, { 3.638958032234415*^9, 3.638958046554435*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"F", "[", "s_", "]"}], " ", "=", " ", RowBox[{"Integrate", "[", RowBox[{ RowBox[{ RowBox[{"Exp", "[", RowBox[{ RowBox[{"-", " ", "s"}], " ", "t"}], "]"}], RowBox[{"f", "[", "t", "]"}]}], ",", RowBox[{"{", RowBox[{"t", ",", "0", ",", " ", "Infinity"}], "}"}]}], "]"}]}]], "Input"], Cell[BoxData[ RowBox[{ SubsuperscriptBox["\[Integral]", "0", "\[Infinity]"], RowBox[{ RowBox[{ SuperscriptBox["\[ExponentialE]", RowBox[{ RowBox[{"-", "s"}], " ", "t"}]], " ", RowBox[{"f", "[", "t", "]"}]}], RowBox[{"\[DifferentialD]", "t"}]}]}]], "Output", CellChangeTimes->{{3.638957992074359*^9, 3.6389580092343826`*^9}}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " can compute these Laplace Transforms and we see how to do this. Here are \ the definitions brought up by query from ", StyleBox["Mathematica", FontSlant->"Italic"], ". If one needs more, such as an example, then go to Help Menu and type \ \"LaplaceTransform\" and you will have a window offering up a definition with \ an example worked out at the bottom.OR just type one of the following" }], "Subsection", CellChangeTimes->{{3.6389574226854525`*^9, 3.638957460435506*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"??", "LaplaceTransform"}]], "Input"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ StyleBox["\<\"\\!\\(\\*RowBox[{\\\"LaplaceTransform\\\", \\\"[\\\", \ RowBox[{StyleBox[\\\"expr\\\", \\\"TI\\\"], \\\",\\\", StyleBox[\\\"t\\\", \\\ \"TI\\\"], \\\",\\\", StyleBox[\\\"s\\\", \\\"TI\\\"]}], \\\"]\\\"}]\\) gives \ the Laplace transform of \\!\\(\\*StyleBox[\\\"expr\\\", \\\"TI\\\"]\\). \ \\n\\!\\(\\*RowBox[{\\\"LaplaceTransform\\\", \\\"[\\\", \ RowBox[{StyleBox[\\\"expr\\\", \\\"TI\\\"], \\\",\\\", RowBox[{\\\"{\\\", \ RowBox[{SubscriptBox[StyleBox[\\\"t\\\", \\\"TI\\\"], StyleBox[\\\"1\\\", \ \\\"TR\\\"]], \\\",\\\", SubscriptBox[StyleBox[\\\"t\\\", \\\"TI\\\"], \ StyleBox[\\\"2\\\", \\\"TR\\\"]], \\\",\\\", StyleBox[\\\"\[Ellipsis]\\\", \\\ \"TR\\\"]}], \\\"}\\\"}], \\\",\\\", RowBox[{\\\"{\\\", \ RowBox[{SubscriptBox[StyleBox[\\\"s\\\", \\\"TI\\\"], StyleBox[\\\"1\\\", \ \\\"TR\\\"]], \\\",\\\", SubscriptBox[StyleBox[\\\"s\\\", \\\"TI\\\"], \ StyleBox[\\\"2\\\", \\\"TR\\\"]], \\\",\\\", StyleBox[\\\"\[Ellipsis]\\\", \\\ \"TR\\\"]}], \\\"}\\\"}]}], \\\"]\\\"}]\\) gives the multidimensional Laplace \ transform of \\!\\(\\*StyleBox[\\\"expr\\\", \\\"TI\\\"]\\). \"\>", "MSG"], "\[NonBreakingSpace]", ButtonBox[ StyleBox["\[RightSkeleton]", "SR"], Active->True, BaseStyle->"Link", ButtonData->"paclet:ref/LaplaceTransform"]}]], "Print", "PrintUsage", CellChangeTimes->{3.6389575054555683`*^9}, CellTags->"Info3638943105-3626895"], Cell[BoxData[ InterpretationBox[GridBox[{ { RowBox[{ RowBox[{"Attributes", "[", "LaplaceTransform", "]"}], "=", RowBox[{"{", RowBox[{"Protected", ",", "ReadProtected"}], "}"}]}]}, {" "}, {GridBox[{ { RowBox[{ RowBox[{"Options", "[", "LaplaceTransform", "]"}], "=", RowBox[{"{", RowBox[{ RowBox[{"Assumptions", "\[RuleDelayed]", "$Assumptions"}], ",", RowBox[{"GenerateConditions", "\[Rule]", "False"}], ",", RowBox[{"PrincipalValue", "\[Rule]", "False"}], ",", RowBox[{"Analytic", "\[Rule]", "True"}]}], "}"}]}]} }, BaselinePosition->{Baseline, {1, 1}}, GridBoxAlignment->{ "Columns" -> {{Left}}, "ColumnsIndexed" -> {}, "Rows" -> {{Baseline}}, "RowsIndexed" -> {}}, GridBoxItemSize->{"Columns" -> {{ Scaled[0.999]}}, "ColumnsIndexed" -> {}, "Rows" -> {{1.}}, "RowsIndexed" -> {}}]} }, BaselinePosition->{Baseline, {1, 1}}, GridBoxAlignment->{ "Columns" -> {{Left}}, "ColumnsIndexed" -> {}, "Rows" -> {{Baseline}}, "RowsIndexed" -> {}}], Definition[LaplaceTransform], Editable->False]], "Print", CellChangeTimes->{3.6389575055055685`*^9}, CellTags->"Info3638943105-3626895"] }, Open ]] }, Open ]], Cell["\<\ We have a tougher trick in undoing the Laplace Transform and this is done \ using the InverseLaplaceTransform command.\ \>", "Text", CellChangeTimes->{{3.638957468005516*^9, 3.638957503145565*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"??", "InverseLaplaceTransform"}]], "Input"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ StyleBox["\<\"\\!\\(\\*RowBox[{\\\"InverseLaplaceTransform\\\", \\\"[\\\", \ RowBox[{StyleBox[\\\"expr\\\", \\\"TI\\\"], \\\",\\\", StyleBox[\\\"s\\\", \\\ \"TI\\\"], \\\",\\\", StyleBox[\\\"t\\\", \\\"TI\\\"]}], \\\"]\\\"}]\\) gives \ the inverse Laplace transform of \\!\\(\\*StyleBox[\\\"expr\\\", \ \\\"TI\\\"]\\). \\n\\!\\(\\*RowBox[{\\\"InverseLaplaceTransform\\\", \ \\\"[\\\", RowBox[{StyleBox[\\\"expr\\\", \\\"TI\\\"], \\\",\\\", \ RowBox[{\\\"{\\\", RowBox[{SubscriptBox[StyleBox[\\\"s\\\", \\\"TI\\\"], \ StyleBox[\\\"1\\\", \\\"TR\\\"]], \\\",\\\", SubscriptBox[StyleBox[\\\"s\\\", \ \\\"TI\\\"], StyleBox[\\\"2\\\", \\\"TR\\\"]], \\\",\\\", StyleBox[\\\"\ \[Ellipsis]\\\", \\\"TR\\\"]}], \\\"}\\\"}], \\\",\\\", RowBox[{\\\"{\\\", \ RowBox[{SubscriptBox[StyleBox[\\\"t\\\", \\\"TI\\\"], StyleBox[\\\"1\\\", \ \\\"TR\\\"]], \\\",\\\", SubscriptBox[StyleBox[\\\"t\\\", \\\"TI\\\"], \ StyleBox[\\\"2\\\", \\\"TR\\\"]], \\\",\\\", StyleBox[\\\"\[Ellipsis]\\\", \\\ \"TR\\\"]}], \\\"}\\\"}]}], \\\"]\\\"}]\\) gives the multidimensional inverse \ Laplace transform of \\!\\(\\*StyleBox[\\\"expr\\\", \\\"TI\\\"]\\). \"\>", "MSG"], "\[NonBreakingSpace]", ButtonBox[ StyleBox["\[RightSkeleton]", "SR"], Active->True, BaseStyle->"Link", ButtonData->"paclet:ref/InverseLaplaceTransform"]}]], "Print", "PrintUsage", CellChangeTimes->{3.6389575096555743`*^9}, CellTags->"Info3638943109-3626895"], Cell[BoxData[ InterpretationBox[GridBox[{ { RowBox[{ RowBox[{"Attributes", "[", "InverseLaplaceTransform", "]"}], "=", RowBox[{"{", RowBox[{"Protected", ",", "ReadProtected"}], "}"}]}]} }, BaselinePosition->{Baseline, {1, 1}}, GridBoxAlignment->{ "Columns" -> {{Left}}, "ColumnsIndexed" -> {}, "Rows" -> {{Baseline}}, "RowsIndexed" -> {}}], Definition[InverseLaplaceTransform], Editable->False]], "Print", CellChangeTimes->{3.6389575096855745`*^9}, CellTags->"Info3638943109-3626895"] }, Open ]] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["\<\ So here we go . . . taking Laplace Transforms of functions . . . . We Build \ Up a Table of Laplace Transforms and attempt to ascertain patterns.\ \>", "Subsection", CellChangeTimes->{{3.6389575298156023`*^9, 3.638957542895621*^9}, { 3.6389580636344585`*^9, 3.638958073654473*^9}, 3.638961540100326*^9}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"LaplaceTransform", "[", RowBox[{"1", ",", "t", ",", "s"}], "]"}]], "Input"], Cell[BoxData[ FractionBox["1", "s"]], "Output", CellChangeTimes->{3.638957546735626*^9}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"LaplaceTransform", "[", RowBox[{"t", ",", "t", ",", "s"}], "]"}]], "Input"], Cell[BoxData[ FractionBox["1", SuperscriptBox["s", "2"]]], "Output", CellChangeTimes->{3.638957546815626*^9}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"LaplaceTransform", "[", RowBox[{ RowBox[{"t", "^", "2"}], ",", "t", ",", "s"}], "]"}]], "Input"], Cell[BoxData[ FractionBox["2", SuperscriptBox["s", "3"]]], "Output", CellChangeTimes->{3.6389575468256264`*^9}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"LaplaceTransform", "[", RowBox[{ RowBox[{"t", "^", "3"}], ",", "t", ",", "s"}], "]"}]], "Input"], Cell[BoxData[ FractionBox["6", SuperscriptBox["s", "4"]]], "Output", CellChangeTimes->{3.6389575468456264`*^9}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"LaplaceTransform", "[", RowBox[{ RowBox[{"t", "^", "4"}], ",", "t", ",", "s"}], "]"}]], "Input"], Cell[BoxData[ FractionBox["24", SuperscriptBox["s", "5"]]], "Output", CellChangeTimes->{3.6389575468656263`*^9}] }, Open ]], Cell[TextData[{ "What do you believe we might get for the Laplace Transform of ", Cell[BoxData[ FormBox[ SuperscriptBox["t", "n"], TraditionalForm]], FormatType->"TraditionalForm"], "? Of course, we could stop and offer a formal induction proof of general \ formula, but we proceed to take the Laplace Transform of other classes of \ functions." }], "Text", CellChangeTimes->{{3.638958081544484*^9, 3.638958168814606*^9}, { 3.6389615455603333`*^9, 3.6389615525503435`*^9}}], Cell[CellGroupData[{ Cell[" Trig functions", "Subsubsection"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"LaplaceTransform", "[", RowBox[{ RowBox[{"Sin", "[", RowBox[{"3", "t"}], "]"}], ",", "t", ",", "s"}], "]"}]], "Input"], Cell[BoxData[ FractionBox["3", RowBox[{"9", "+", SuperscriptBox["s", "2"]}]]], "Output", CellChangeTimes->{3.6389575546956377`*^9}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"LaplaceTransform", "[", RowBox[{ RowBox[{"Sin", "[", RowBox[{"5", "t"}], "]"}], ",", "t", ",", "s"}], "]"}]], "Input", CellChangeTimes->{3.63895817869462*^9}], Cell[BoxData[ FractionBox["5", RowBox[{"25", "+", SuperscriptBox["s", "2"]}]]], "Output", CellChangeTimes->{3.638958179104621*^9}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"LaplaceTransform", "[", RowBox[{ RowBox[{"Cos", "[", RowBox[{"3", "t"}], "]"}], ",", "t", ",", "s"}], "]"}]], "Input"], Cell[BoxData[ FractionBox["s", RowBox[{"9", "+", SuperscriptBox["s", "2"]}]]], "Output", CellChangeTimes->{3.638957558685643*^9}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"LaplaceTransform", "[", RowBox[{ RowBox[{"Cos", "[", RowBox[{"5", "t"}], "]"}], ",", "t", ",", "s"}], "]"}]], "Input", CellChangeTimes->{3.6389581856946297`*^9}], Cell[BoxData[ FractionBox["s", RowBox[{"25", "+", SuperscriptBox["s", "2"]}]]], "Output", CellChangeTimes->{3.6389581860946302`*^9}] }, Open ]], Cell["See a pattern here?", "Text", CellChangeTimes->{{3.6389581897646356`*^9, 3.6389582006946507`*^9}}] }, Open ]], Cell[CellGroupData[{ Cell[". . . and exponential functions.", "Subsubsection"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"LaplaceTransform", "[", RowBox[{ RowBox[{"Exp", "[", RowBox[{"3", " ", "t"}], "]"}], ",", "t", ",", "s"}], "]"}]], "Input"], Cell[BoxData[ FractionBox["1", RowBox[{ RowBox[{"-", "3"}], "+", "s"}]]], "Output", CellChangeTimes->{3.638957561515647*^9}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"LaplaceTransform", "[", RowBox[{ RowBox[{"Exp", "[", RowBox[{"5", " ", "t"}], "]"}], ",", "t", ",", "s"}], "]"}]], "Input"], Cell[BoxData[ FractionBox["1", RowBox[{ RowBox[{"-", "5"}], "+", "s"}]]], "Output", CellChangeTimes->{3.6389575616256475`*^9}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"LaplaceTransform", "[", RowBox[{ RowBox[{"Exp", "[", RowBox[{ RowBox[{"-", "7"}], " ", "t"}], "]"}], ",", "t", ",", "s"}], "]"}]], "Input"], Cell[BoxData[ FractionBox["1", RowBox[{"7", "+", "s"}]]], "Output", CellChangeTimes->{3.638957561645647*^9}] }, Open ]], Cell[TextData[{ "Incidentally, the transform of functions like f(t) = ", Cell[BoxData[ FormBox[ SuperscriptBox["e", RowBox[{"a", " ", "t"}]], TraditionalForm]], FormatType->"TraditionalForm"], " and f(t) = k are easy by hand efforts! Try one." }], "Text", CellChangeTimes->{{3.6389582084446616`*^9, 3.6389582682547455`*^9}}] }, Open ]], Cell[CellGroupData[{ Cell["\<\ Continuing in this manner of discovery we could fill in a typical Laplace \ Transform Table. A sample, but short and simple form, table is the following:\ \>", "Subsubsection", CellChangeTimes->{{3.638957568835657*^9, 3.6389575785956707`*^9}, { 3.638957694867542*^9, 3.638957725153985*^9}, {3.6389582782947598`*^9, 3.63895828612477*^9}, 3.6389583226248217`*^9}], Cell[BoxData[ GraphicsBox[ TagBox[RasterBox[CompressedData[" 1:eJzsnQdUFOfXxmM3mpiYqPknJtE0k9h7jCUaG/aOvSAqiAgIiAgKgtIRAbsU FQSk2RBQEATEgtJUiiBIl44UaUrJ93w7xz3r7rIsC8uC3t858ZDZd955Z+be +9w75Z2fZFWWyXX85JNP1Lvjn2WbNf9VU9ustbwz/me61u5tsp3whzd+74z/ /v/v/wiCIAiCIAiCIAiCIAiCIAiCIAiCIAiCIAiCIAiCIAiCIAiCIAiCIAiC IAiCIAiCIAiCIAiCIAiCIAiitaipqcnIyHj0Pi9evKioqCgpKUlISIiPj3/1 6pWQvb1+/TopKenp06cFBQViHTbRNqmtrc3MzAwPD4dRwbSEacxle8nJyeXl 5UJurqioKC4uDlZaWloq5CpoifZYS3irJoiPAbiGg4PD8vextLRMTU2NiorS 1dXdu3dvWFiYkL0lJiaampoqKSkFBgaKddhE2wQphJOT04oVKxwdHcvKygQ3 Rsy/cOECl+0dPnwYciDk5u7du6ehoaGvr//kyRMhV3n8+LGent6ePXvu378v 5CoE8WHz9u3bvLy8yMhITU3NX9/Rt2/fzp07L1iwwMvLy9XVddmyZXBPPz8/ 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Let us do this. Our \ solution appears a bit on the \[OpenCurlyDoubleQuote]wild side\ \[CloseCurlyDoubleQuote] involving the constants a, b, and c, and the all too \ familiar radical", Cell[BoxData[ FormBox[ SqrtBox[ RowBox[{ SuperscriptBox["b", "2"], " ", "-", " ", RowBox[{"4", "ac"}]}]], TraditionalForm]], FormatType->"TraditionalForm"], " one obtains from the characteristic equation of the homogeneous portion of \ our original differential equation a y\[CloseCurlyQuote]\[CloseCurlyQuote](t) \ + b y\[CloseCurlyQuote](t) + c y(t) = sin(t) in using eigenvalue solution \ strategies." }], "Text", CellChangeTimes->{{3.638958601035211*^9, 3.6389586432052703`*^9}, { 3.638958681745324*^9, 3.638958831415534*^9}, {3.6389615976304064`*^9, 3.6389616189004364`*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"InverseLaplaceTransform", "[", RowBox[{ RowBox[{"YSol", "[", "s", "]"}], ",", "s", ",", "t"}], "]"}], "//", "Simplify"}]], "Input", CellChangeTimes->{{3.6389586447352724`*^9, 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FontWeight->"Bold", FontVariations->{"Underline"->True}], " of a differential equation such as \ a*y\[CloseCurlyQuote]\[CloseCurlyQuote](t) + b*y\[CloseCurlyQuote](t) + \ c*y(t) = sin(t) is Q(s) = 1/ (a*", Cell[BoxData[ FormBox[ SuperscriptBox["s", "2"], TraditionalForm]]], " + b*s + c) (1 over the denominator in the Laplace Transform of the \ solution) and notice the coefficients a, b, and c, of the LHS or of the \ system - not the driver sin(t). This characterizes the system response and \ the roots of the characteristic equation, a*", Cell[BoxData[ FormBox[ SuperscriptBox["s", "2"], TraditionalForm]]], " + b*s + c = 0, are the characteristic roots or eigenvalues of this system. \ Engineers get to \"know\" systems and the systems' behaviors by their \ Transfer functions and the eigenvalues; electrical engineers call the \ eigenvalues \"poles.\"\n\nIndeed, if we let y(0) = 0 and y'(0) = 0 then \ YSol[s] = L[s]*Q[s], and so Q[s] = Ysol[s]/L[s], where YSol[s] is the \ output's Laplace Transform and L[s] is the input's Laplace Transform. This \ effectively shows that the Laplace Transform of the Solution is \ \[OpenCurlyDoubleQuote]simply\[CloseCurlyDoubleQuote] a multiplication of \ Laplace Transform of the input or driver function, i.e. L[s], and the \ transfer function Q[s] = Q(s) = 1/ (a*", Cell[BoxData[ FormBox[ SuperscriptBox["s", "2"], TraditionalForm]]], " + b*s + c) . The latter contains all the \[OpenCurlyDoubleQuote]action\ \[CloseCurlyDoubleQuote] of the differential equation model terms a*y\ \[CloseCurlyQuote]\[CloseCurlyQuote](t) + b*y\[CloseCurlyQuote](t) + c y(t) \ while the former, L[s], contains the \[OpenCurlyDoubleQuote]action\ \[CloseCurlyDoubleQuote] of the driver function, in this case sin(t)." }], "Subsubsection", CellChangeTimes->{{3.6389588419255486`*^9, 3.638959193896041*^9}, { 3.6389593430162497`*^9, 3.6389593782462997`*^9}, {3.638961622310441*^9, 3.638961655700488*^9}}], Cell["\<\ So when we actually use reasonable initial conditions (y(0) = 0, no initial \ displacement, and y\[CloseCurlyQuote](0) = 0, no initial velocity - depending \ upon the driver to get some activity going) on the inverse Laplace Transform, \ YSol[s], of our differential equation \ a*y\[CloseCurlyQuote]\[CloseCurlyQuote](t) + b*y\[CloseCurlyQuote](t) + \ c*y(t) = sin(t)\ \>", "Text", CellChangeTimes->{{3.638959197346046*^9, 3.6389593409762473`*^9}, 3.6389616584804916`*^9}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"YSol", "[", "s", "]"}], "/.", RowBox[{"{", RowBox[{ RowBox[{ RowBox[{"y", "[", "0", "]"}], "\[Rule]", "0"}], ",", RowBox[{ RowBox[{ RowBox[{"y", "'"}], "[", "0", "]"}], "\[Rule]", "0"}]}], "}"}]}]], "Input"], Cell[BoxData[ FractionBox["1", RowBox[{ RowBox[{"(", RowBox[{"1", "+", SuperscriptBox["s", "2"]}], ")"}], " ", RowBox[{"(", RowBox[{"c", "+", RowBox[{"b", " ", "s"}], "+", RowBox[{"a", " ", SuperscriptBox["s", "2"]}]}], ")"}]}]]], "Output", CellChangeTimes->{3.6389591846560287`*^9}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "We try a specific example, one we know we can solve in several other ways - \ by hand, with ", StyleBox["Mathematica", FontSlant->"Italic"], " DSolve. We consider \n\n\t\ty''(t) + 6 y'(t) + 5 y(t) = sin(t), y(0) = \ 4, y'(0) = 0." }], "Subsubsection"], Cell["\<\ First we use DSolve and some formatting to grab the solution for comparison.\ \>", "Text", CellChangeTimes->{{3.638959476576437*^9, 3.6389595029564743`*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"ysol", "[", "t_", "]"}], " ", "=", " ", RowBox[{ RowBox[{ RowBox[{"y", "[", "t", "]"}], "/.", RowBox[{ RowBox[{"DSolve", "[", RowBox[{ RowBox[{"{", RowBox[{ RowBox[{ RowBox[{ RowBox[{"1", " ", RowBox[{ RowBox[{"y", "''"}], "[", "t", "]"}]}], " ", "+", " ", RowBox[{"6", " ", RowBox[{ RowBox[{"y", "'"}], "[", "t", "]"}]}], " ", "+", " ", RowBox[{"5", " ", RowBox[{"y", "[", "t", "]"}]}]}], "\[Equal]", RowBox[{"Sin", "[", "t", "]"}]}], ",", RowBox[{ RowBox[{"y", "[", "0", "]"}], "\[Equal]", "4"}], ",", RowBox[{ RowBox[{ RowBox[{"y", "'"}], "[", "0", "]"}], "\[Equal]", "0"}]}], "}"}], ",", RowBox[{"y", "[", "t", "]"}], ",", "t"}], "]"}], "[", RowBox[{"[", "1", "]"}], "]"}]}], "//", "Expand"}]}]], "Input", CellChangeTimes->{{3.638959403956335*^9, 3.638959472266431*^9}}], Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{"-", FractionBox["105", "104"]}], " ", SuperscriptBox["\[ExponentialE]", RowBox[{ RowBox[{"-", "5"}], " ", "t"}]]}], "+", FractionBox[ RowBox[{"41", " ", SuperscriptBox["\[ExponentialE]", RowBox[{"-", "t"}]]}], "8"], "-", FractionBox[ RowBox[{"3", " ", RowBox[{"Cos", "[", "t", "]"}]}], "26"], "+", FractionBox[ RowBox[{"Sin", "[", "t", "]"}], "13"]}]], "Output", CellChangeTimes->{{3.6389594620664167`*^9, 3.638959473646433*^9}}] }, Open ]], Cell["\<\ We consider this same differential equation and apply Laplace Transforms to \ all terms, i.e. both sides of the differential equation.\ \>", "Text", CellChangeTimes->{{3.638959518286495*^9, 3.6389595182964954`*^9}, { 3.638959556646549*^9, 3.6389595968866053`*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"eq", " ", "=", " ", RowBox[{ RowBox[{"LaplaceTransform", "[", RowBox[{ RowBox[{ RowBox[{ 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FractionBox["1", RowBox[{"1", "+", SuperscriptBox["s", "2"]}]]}]], "Output", CellChangeTimes->{3.638959508006481*^9}] }, Open ]], Cell["\<\ Now we solve for Y[s], the Laplace Transform of y[t] in order to get ready \ for the inverse Laplace Transform.\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"L", "[", "s_", "]"}], " ", "=", RowBox[{ RowBox[{"Y", "[", "s", "]"}], "/.", RowBox[{ RowBox[{"Solve", "[", RowBox[{"eq", ",", RowBox[{"Y", "[", "s", "]"}]}], "]"}], "[", RowBox[{"[", "1", "]"}], "]"}]}]}]], "Input"], Cell[BoxData[ FractionBox[ RowBox[{"25", "+", RowBox[{"4", " ", "s"}], "+", RowBox[{"24", " ", SuperscriptBox["s", "2"]}], "+", RowBox[{"4", " ", SuperscriptBox["s", "3"]}]}], RowBox[{ RowBox[{"(", RowBox[{"1", "+", SuperscriptBox["s", "2"]}], ")"}], " ", RowBox[{"(", RowBox[{"5", "+", RowBox[{"6", " ", "s"}], "+", SuperscriptBox["s", "2"]}], ")"}]}]]], "Output", CellChangeTimes->{3.6389596012866116`*^9}] }, Open ]], Cell["\<\ Now we apply the inverse transform to get the solution for y[t].\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"l", "[", "t_", "]"}], " ", "=", " ", RowBox[{ RowBox[{"InverseLaplaceTransform", "[", RowBox[{ RowBox[{"L", "[", "s", "]"}], ",", "s", ",", "t"}], "]"}], "//", "Expand"}]}]], "Input", CellChangeTimes->{{3.6389596113766255`*^9, 3.6389596129766283`*^9}}], Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{"-", FractionBox["105", "104"]}], " ", SuperscriptBox["\[ExponentialE]", RowBox[{ RowBox[{"-", "5"}], " ", "t"}]]}], "+", FractionBox[ RowBox[{"41", " ", SuperscriptBox["\[ExponentialE]", RowBox[{"-", "t"}]]}], "8"], "-", FractionBox[ RowBox[{"3", " ", RowBox[{"Cos", "[", "t", "]"}]}], "26"], "+", FractionBox[ RowBox[{"Sin", "[", "t", "]"}], "13"]}]], "Output", CellChangeTimes->{{3.638959604286616*^9, 3.6389596141466293`*^9}}] }, Open ]] }, Open ]], Cell[TextData[{ "Notice the two portions of this soloution, namely the Transient solution ", Cell[BoxData[ StyleBox[ RowBox[{ RowBox[{ RowBox[{"-", FractionBox["105", "104"]}], " ", SuperscriptBox["\[ExponentialE]", RowBox[{ RowBox[{"-", "5"}], " ", "t"}]]}], "+", FractionBox[ RowBox[{"41", " ", SuperscriptBox["\[ExponentialE]", RowBox[{"-", "t"}]]}], "8"]}], FontColor->RGBColor[0, 0, 1]]], CellChangeTimes->{{3.638959604286616*^9, 3.6389596141466293`*^9}}], "and the Teady State ", Cell[BoxData[ StyleBox[ RowBox[{ RowBox[{"-", FractionBox[ RowBox[{"3", " ", RowBox[{"Cos", "[", "t", "]"}]}], "26"]}], "+", FractionBox[ RowBox[{"Sin", "[", "t", "]"}], "13"]}], FontColor->RGBColor[0, 0, 1]]], CellChangeTimes->{{3.638959604286616*^9, 3.6389596141466293`*^9}}], ". These would be produced by various hand techniques or ", StyleBox["Mathematica", FontSlant->"Italic"], "\[CloseCurlyQuote]s DSolve approach which we show here." }], "Subsubsection", CellChangeTimes->{{3.6425924685642524`*^9, 3.642592566164389*^9}}], Cell[CellGroupData[{ Cell["\<\ How does that compare to the DSolve solution? Exactly the same as we see \ when, again, we use DSolve to obtain a solution.\ \>", "Subsubsection", CellChangeTimes->{{3.6389596297866516`*^9, 3.6389596673867044`*^9}, 3.638961668690506*^9}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"ysol", "[", "t_", "]"}], " ", "=", " ", RowBox[{ RowBox[{ RowBox[{"y", "[", "t", "]"}], "/.", RowBox[{ RowBox[{"DSolve", "[", RowBox[{ RowBox[{"{", RowBox[{ RowBox[{ RowBox[{ RowBox[{"1", " ", RowBox[{ RowBox[{"y", "''"}], "[", "t", "]"}]}], " ", "+", " ", RowBox[{"6", " ", RowBox[{ RowBox[{"y", "'"}], "[", "t", "]"}]}], " ", "+", " ", RowBox[{"5", " ", RowBox[{"y", "[", "t", "]"}]}]}], "\[Equal]", RowBox[{"Sin", "[", "t", "]"}]}], ",", RowBox[{ RowBox[{"y", "[", "0", "]"}], "\[Equal]", "4"}], ",", RowBox[{ RowBox[{ RowBox[{"y", "'"}], "[", "0", "]"}], "\[Equal]", "0"}]}], "}"}], ",", RowBox[{"y", "[", "t", "]"}], ",", "t"}], "]"}], "[", RowBox[{"[", "1", "]"}], "]"}]}], "//", "Expand"}]}]], "Input", CellChangeTimes->{{3.638959403956335*^9, 3.638959472266431*^9}}], Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{"-", FractionBox["105", "104"]}], " ", SuperscriptBox["\[ExponentialE]", RowBox[{ RowBox[{"-", "5"}], " ", "t"}]]}], "+", FractionBox[ RowBox[{"41", " ", SuperscriptBox["\[ExponentialE]", RowBox[{"-", "t"}]]}], "8"], "-", FractionBox[ RowBox[{"3", " ", RowBox[{"Cos", "[", "t", "]"}]}], "26"], "+", FractionBox[ RowBox[{"Sin", "[", "t", "]"}], "13"]}]], "Output", CellChangeTimes->{3.6389596698967075`*^9}] }, Open ]] }, Open ]], Cell["\<\ So what is the big deal? Who would someone choose Laplace Transforms to \ solve a problem if DSolve or by hand can do it directly? Here is the point (well one of the many points of advantage) of Laplace \ Transforms. They convert calculus to algebra, then we play in algebra land, \ and then we return to calculus land. More formally, the Laplace Transform \ transforms the problem from the time domain (t) to the frequency domain in \ (s). \ \>", "Subsubsection", CellChangeTimes->{{3.6389596831367264`*^9, 3.6389597323967953`*^9}, 3.6389616725905113`*^9}] }, Open ]], Cell[CellGroupData[{ Cell["\<\ We consider two additional \"Real Good\" reasons for Laplace Transforms\ \>", "Subsection", CellChangeTimes->{{3.638959736716801*^9, 3.638959738146803*^9}}], Cell[CellGroupData[{ Cell["\<\ Unit Step or Heaviside Function - what does this function do.\ \>", "Subsubsection"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"f1", "[", "x_", "]"}], " ", "=", " ", RowBox[{"UnitStep", "[", "x", "]"}]}]], "Input"], Cell[BoxData[ RowBox[{"UnitStep", "[", "x", "]"}]], "Output", CellChangeTimes->{3.6389178856880107`*^9}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Plot", "[", RowBox[{ RowBox[{"f1", "[", "x", "]"}], ",", RowBox[{"{", RowBox[{"x", ",", RowBox[{"-", "8"}], ",", "8"}], "}"}], ",", 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This function represents an \ instantaneous change. E.g., we change the salt concentration by \"dumping\" \ in more salt ALL AT ONCE! OR we \"shock\" a spring mass giving it an \ instantaneous force ALL AT ONCE!\ \>", "Text", CellChangeTimes->{{3.6389597780468593`*^9, 3.638959789386875*^9}}], Cell["\<\ The integral of an interval containing x = 0 for the DiracDelta[x] function \ is 1. What this says is the impulse offered by a driver of the form \ DiracDelta[x] imparts force of size 1 unit to our system, BUT all at once \ like the quick bang of a hammer, rather than a gradual application of force.\ \>", "Text", CellChangeTimes->{{3.638959799286889*^9, 3.6389599276580687`*^9}, 3.6389616804305224`*^9}], Cell["We check out some properties of the DiracDelta[x] function.", "Text", CellChangeTimes->{{3.6389599326980753`*^9, 3.638959952878104*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Integrate", "[", RowBox[{ RowBox[{"g", "[", "x", "]"}], ",", RowBox[{"{", RowBox[{"x", ",", RowBox[{"-", "1"}], ",", "1"}], "}"}]}], "]"}]], "Input"], Cell[BoxData["1"], "Output", CellChangeTimes->{3.6389179112080464`*^9}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Integrate", "[", RowBox[{ RowBox[{"g", "[", "x", "]"}], ",", RowBox[{"{", RowBox[{"x", ",", ".5", ",", "1"}], "}"}]}], "]"}]], "Input"], Cell[BoxData["0.`"], "Output", CellChangeTimes->{3.638917911218046*^9}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"g", "[", ".1", "]"}]], "Input"], Cell[BoxData["0"], "Output", CellChangeTimes->{3.6389179112280464`*^9}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"g", "[", "0", "]"}]], "Input"], Cell[BoxData[ RowBox[{"DiracDelta", "[", "0", "]"}]], "Output", CellChangeTimes->{3.6389179112280464`*^9}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"g", "[", "2", "]"}]], "Input"], Cell[BoxData["0"], "Output", CellChangeTimes->{3.6389179112380466`*^9}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["\<\ Transforms of UnitStep and Dirac Function - where does this fit into our \ Table of Laplace transforms?\ \>", "Subsubsection"], Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{"f", "[", "t_", "]"}], " ", "=", " ", RowBox[{"UnitStep", "[", "t", "]"}]}], ";"}]], "Input"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"LaplaceTransform", "[", RowBox[{ RowBox[{"f", "[", "t", "]"}], ",", "t", ",", "s"}], "]"}]], "Input"], Cell[BoxData[ FractionBox["1", "s"]], "Output", CellChangeTimes->{3.6389179226180625`*^9}] }, Open ]], Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{"g", "[", "t_", "]"}], " ", "=", " ", RowBox[{"DiracDelta", "[", "t", "]"}]}], ";"}]], "Input"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"LaplaceTransform", "[", RowBox[{ RowBox[{"g", "[", "t", "]"}], ",", "t", ",", "s"}], "]"}]], "Input"], Cell[BoxData["1"], "Output", CellChangeTimes->{3.638917922658063*^9}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["\<\ UnitStep function represents a \"step\" or finite immediate change in \ something. 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