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Howell, 1998", FontSize->18, FontColor->RGBColor[1, 0, 1]], StyleBox["\n", FontColor->RGBColor[1, 0, 0]], StyleBox["Complimentary software to accompany our textbook:", FontColor->RGBColor[0, 1, 0]] }], "Text", Editable->False, Evaluatable->False, TextAlignment->Center, AspectRatioFixed->True], Cell[TextData[{ StyleBox[ "COMPLEX ANALYSIS: for Mathematics and Engineering, \n3rd Edition, 1997, \ ISBN: 0-7637-0270-6", FontSize->18, FontColor->RGBColor[0, 0, 1]], StyleBox["\n", FontSize->18, FontColor->RGBColor[0, 1, 1]], StyleBox[ "Jones & Bartlett Publishers, Inc.\n40 Tall Pine Drive, Sudbury, MA 01776\n\ Tele. (800) 832-0034, FAX: (508) 443-8000\nE-mail: mkt@jbpub.com, \ http://www.jbpub.com/", FontSize->18, FontColor->RGBColor[0, 1, 0]], StyleBox["\n", FontSize->14], StyleBox[ "This free software is compliments of the authors.\nmathews@fullerton.edu, \ howell@westmont.edu", FontSize->14, FontColor->RGBColor[1, 0, 1]] }], "Text"] }, Closed]], Cell[TextData[{ StyleBox["CHAPTER 11 ", FontSize->18], StyleBox["FOURIER SERIES and the LAPLACE TRANSFORM\n", FontSize->18, FontColor->RGBColor[1, 0, 1]], StyleBox["Section 11.1 ", FontSize->18], ButtonBox["Fourier Series", ButtonData:>"Section 11.1", ButtonStyle->"Hyperlink"], StyleBox["\n", FontSize->18, FontColor->RGBColor[1, 0, 1]], StyleBox["Section 11.2 ", FontSize->18], ButtonBox["The Dirichlet Problem for the Unit Disk", ButtonData:>"Section 11.2", ButtonStyle->"Hyperlink"], StyleBox["\n", FontSize->18, FontColor->RGBColor[1, 0, 1]], StyleBox["Section 11.3 ", FontSize->18], ButtonBox["Vibrations in Mechanical Systems", ButtonData:>"Section 11.3", ButtonStyle->"Hyperlink"], StyleBox["\n", FontSize->18, FontColor->RGBColor[1, 0, 1]], StyleBox["Section 11.4 ", FontSize->18], ButtonBox["The Fourier Transform", ButtonData:>"Section 11.4", ButtonStyle->"Hyperlink"], StyleBox["\n", FontSize->18, FontColor->RGBColor[1, 0, 1]], StyleBox["Section 11.5 ", FontSize->18], ButtonBox["The Laplace Transform", ButtonData:>"Section 11.5", ButtonStyle->"Hyperlink"], StyleBox["\n", FontSize->18, FontColor->RGBColor[1, 0, 1]], StyleBox["Section 11.6 ", FontSize->18], ButtonBox["Laplace Transforms of Derivatives and Integrals", ButtonData:>"Section 11.6", ButtonStyle->"Hyperlink"], StyleBox["\n", FontSize->18, FontColor->RGBColor[1, 0, 1]], StyleBox["Section 11.7 ", FontSize->18], ButtonBox["Shifting Theorems and the Step Function", ButtonData:>"Section 11.7", ButtonStyle->"Hyperlink"], StyleBox["\n", FontSize->18, FontColor->RGBColor[1, 0, 1]], StyleBox["Section 11.8 ", FontSize->18], ButtonBox["Multiplication and Division by t", ButtonData:>"Section 11.8", ButtonStyle->"Hyperlink"], StyleBox["\n", FontSize->18, FontColor->RGBColor[1, 0, 1]], StyleBox["Section 11.9 ", FontSize->18], ButtonBox["Inverting the Laplace Transform", ButtonData:>"Section 11.9", ButtonStyle->"Hyperlink"], StyleBox["\n", FontSize->18, FontColor->RGBColor[1, 0, 1]], StyleBox["Section 11.10 ", FontSize->18], ButtonBox["Convolution\n", ButtonData:>"Section 11.10", ButtonStyle->"Hyperlink"], StyleBox["GoTo Chapter", FontSize->16, FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], StyleBox[" ", FontSize->18], ButtonBox["1", ButtonData:>{"C1.nb", "CHAPTER 1"}, ButtonStyle->"Hyperlink"], StyleBox[", ", FontSize->18], ButtonBox["2", ButtonData:>{"C2.nb", "CHAPTER 2"}, ButtonStyle->"Hyperlink"], StyleBox[", ", FontSize->18], ButtonBox["3", ButtonData:>{"C3.nb", "CHAPTER 3"}, ButtonStyle->"Hyperlink"], StyleBox[", ", FontSize->18], ButtonBox["4", ButtonData:>{"C4.nb", "CHAPTER 4"}, ButtonStyle->"Hyperlink"], StyleBox[", ", FontSize->18], ButtonBox["5", ButtonData:>{"C5.nb", "CHAPTER 5"}, ButtonStyle->"Hyperlink"], StyleBox[", ", FontSize->18], ButtonBox["6", ButtonData:>{"C6.nb", "CHAPTER 6"}, ButtonStyle->"Hyperlink"], StyleBox[", ", FontSize->18], ButtonBox["7", ButtonData:>{"C7.nb", "CHAPTER 7"}, ButtonStyle->"Hyperlink"], StyleBox[", ", FontSize->18], ButtonBox["8", ButtonData:>{"C8.nb", "CHAPTER 8"}, ButtonStyle->"Hyperlink"], StyleBox[", ", FontSize->18], ButtonBox["9", ButtonData:>{"C9.nb", "CHAPTER 9"}, ButtonStyle->"Hyperlink"], StyleBox[", ", FontSize->18], ButtonBox["10", ButtonData:>{"C10.nb", "CHAPTER 10"}, ButtonStyle->"Hyperlink"], StyleBox[", ", FontSize->18], ButtonBox["11", ButtonData:>{"C11.nb", "CHAPTER 11"}, ButtonStyle->"Hyperlink"], StyleBox[".\n", FontSize->18], StyleBox["GoTo ", FontSize->16, FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], ButtonBox["Contents", ButtonData:>{"Contents.nb", "CONTENTS"}, ButtonStyle->"Hyperlink"] }], "Text", CellTags->"CHAPTER"], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["CHAPTER 11 ", FontSize->18], StyleBox["FOURIER SERIES and the LAPLACE TRANSFORM", FontSize->18, FontColor->RGBColor[1, 0, 1]] }], "Text", Evaluatable->False, AspectRatioFixed->True, CellTags->"CHAPTER 11"], Cell[BoxData[ \(\(Clear[a, a1, a2, a3, A, A1, A2, b, b1, b2, B1, B2, B, c, d, EqA, EqB, f, f1, f2, f12, F, F1, F2, F12, F3, g, G, h, H, i, int, j, k, L, Ly, L1, L2, Lf, LDf, n, PQ, PR, r, res1, res2, s, solset, s1, s2, s3, S, S9, t, \[Tau], T, \[Theta], u, U, U1, U2, U7, v, v1, v2, v3, v4, val, w, w1, w2, y, y0, y1, Y, Yi, z1, z2]; \)\)], "Input", InitializationCell->True] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["Section 11.1 ", FontSize->18], StyleBox["Fourier Series", FontSize->18, FontColor->RGBColor[1, 0, 1]] }], "Text", Evaluatable->False, AspectRatioFixed->True, CellTags->"Section 11.1"], Cell[TextData[{ StyleBox["Theorem 11.1, Page 399", FontWeight->"Bold"], StyleBox[".", FontWeight->"Bold", FontSlant->"Italic"], " ", StyleBox["(Fourier Expansion)", FontColor->RGBColor[1, 0, 1]], " If ", Cell[BoxData[ \(U \((t)\)\)]], " has period ", Cell[BoxData[ \(2 \[Pi]\)]], " and ", Cell[BoxData[ RowBox[{ SuperscriptBox["U", "\[Prime]", MultilineFunction->None], \((t)\)}]]], " is piecewise continuous, the ", StyleBox["Fourier expansion", FontColor->RGBColor[1, 0, 1]], " is \n\t", Cell[BoxData[ RowBox[{\(U \((t)\)\), "=", RowBox[{ StyleBox[ FractionBox[ StyleBox[\(a\_0\), FontSize->14], StyleBox["2", FontSize->14]], FontSize->14], StyleBox[" ", FontSize->14], "+", " ", RowBox[{ StyleBox[ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], \(j = 1\), StyleBox["\[Infinity]", FontSize->12]], FontSize->12], RowBox[{"(", RowBox[{ RowBox[{ StyleBox[\(a\_j\), FontSize->14], " ", "cos", \((j\ t)\)}], " ", "+", " ", RowBox[{ SubscriptBox[ StyleBox["b", FontSize->14], "j"], " ", "sin", \((j\ t)\)}]}], ")"}]}]}]}]]], ".\n\t\n", StyleBox["Theorem 11.2, Page 400.", FontWeight->"Bold"], " If ", Cell[BoxData[ \(U \((t)\)\)]], " and ", Cell[BoxData[ \(V \((t)\)\)]], " have Fourier series representations, then their sum ", Cell[BoxData[ \(W \((t)\) = U \((t)\) + V \((t)\)\)]], " has a Fourier series representation, and the Fourier coefficients of W \ are obtained by adding the corresponding coefficients of U and V.\n\n", StyleBox["Theorem 11.3, Page 400", FontWeight->"Bold"], StyleBox[".", FontWeight->"Bold", FontSlant->"Italic"], " ", StyleBox["(Fourier Cosine Series)", FontColor->RGBColor[1, 0, 1]], " Assume that ", Cell[BoxData[ \(U \((t)\)\)]], " is an even function and has period ", Cell[BoxData[ \(2 \[Pi]\)]], ". If ", Cell[BoxData[ \(U \((t)\)\)]], " and ", Cell[BoxData[ RowBox[{ SuperscriptBox["U", "\[Prime]", MultilineFunction->None], \((t)\)}]]], " are piecewise continuous, the Fourier series for ", Cell[BoxData[ \(U \((t)\)\)]], " involves only the cosine terms, (i.e. ", Cell[BoxData[ RowBox[{ SubscriptBox[ StyleBox["b", FontSize->14], "j"], "=", \(0\ \ for\ all\ \ j\)}]]], "):\n\t", Cell[BoxData[ RowBox[{\(U \((t)\)\), "=", RowBox[{ FractionBox[ StyleBox[\(a\_0\), FontSize->14], StyleBox["2", FontSize->14]], " ", "+", " ", RowBox[{ StyleBox[ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], \(j = 1\), StyleBox["\[Infinity]", FontSize->12]], FontSize->12], RowBox[{ StyleBox[\(a\_j\), FontSize->14], " ", "cos", \((j\ t)\)}]}]}]}]]], ", where\n\t", Cell[BoxData[ RowBox[{\(a\_j\), "=", RowBox[{ FractionBox[ StyleBox["2", FontSize->12], StyleBox["\[Pi]", FontSize->12]], RowBox[{ SubsuperscriptBox[ StyleBox["\[Integral]", FontSize->18], StyleBox["0", FontSize->10], StyleBox["\[Pi]", FontSize->10]], \(U \((t)\) cos \((j\ t)\) \[DifferentialD]t\)}]}]}]]], ", for ", Cell[BoxData[ \(j = 0, 1, \[CenterEllipsis]\)]], ".\n\n", StyleBox["Theorem 11.4, Page 400", FontWeight->"Bold"], StyleBox[".", FontWeight->"Bold", FontSlant->"Italic"], " ", StyleBox["(Fourier Sine Series)", FontColor->RGBColor[1, 0, 1]], " Assume that ", Cell[BoxData[ \(U \((t)\)\)]], " is an odd function and has period ", Cell[BoxData[ \(2 \[Pi]\)]], ". If ", Cell[BoxData[ \(U \((t)\)\)]], " and ", Cell[BoxData[ RowBox[{ SuperscriptBox["U", "\[Prime]", MultilineFunction->None], \((t)\)}]]], " are piecewise continuous, the Fourier series for ", Cell[BoxData[ \(U \((t)\)\)]], " involves only the sine terms, (i.e. ", Cell[BoxData[ \(a\_j = 0\ \ for\ all\ \ j\)]], "):\n\t", Cell[BoxData[ RowBox[{\(U \((t)\)\), "=", RowBox[{ StyleBox[ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], \(j = 1\), StyleBox["\[Infinity]", FontSize->12]], FontSize->12], RowBox[{ SubscriptBox[ StyleBox["b", FontSize->14], "j"], " ", "sin", \((j\ t)\)}]}]}]]], ", where\n\t", Cell[BoxData[ RowBox[{\(b\_j\), "=", RowBox[{ FractionBox[ StyleBox["2", FontSize->12], StyleBox["\[Pi]", FontSize->12]], RowBox[{ SubsuperscriptBox[ StyleBox["\[Integral]", FontSize->18], StyleBox["0", FontSize->10], StyleBox["\[Pi]", FontSize->10]], \(U \((t)\) sin \((j\ t)\) \[DifferentialD]t\)}]}]}]]], ", for ", Cell[BoxData[ \(j = 1, 2, \[CenterEllipsis]\)]], ".\n\n", StyleBox["Theorem 11.5, Page 401", FontWeight->"Bold"], StyleBox[".", FontWeight->"Bold", FontSlant->"Italic"], " ", StyleBox["(Termwise Integration)", FontColor->RGBColor[1, 0, 1]], " If ", Cell[BoxData[ \(U \((t)\)\)]], " has the Fourier series representation ", Cell[BoxData[ RowBox[{\(U \((t)\)\), "=", RowBox[{ FractionBox[ StyleBox[\(a\_0\), FontSize->14], StyleBox["2", FontSize->14]], StyleBox[" ", FontSize->14], "+", " ", RowBox[{ StyleBox[ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], \(j = 1\), StyleBox["\[Infinity]", FontSize->12]], FontSize->12], RowBox[{"(", RowBox[{ RowBox[{ StyleBox[\(a\_j\), FontSize->14], " ", "cos", \((j\ t)\)}], " ", "+", " ", RowBox[{ SubscriptBox[ StyleBox["b", FontSize->14], "j"], " ", "sin", \((j\ t)\)}]}], ")"}]}]}]}]]], ". Then the integral of ", Cell[BoxData[ \(U \((t)\)\)]], " has a Fourier series representation which can be obtained by termwise \ integration of the Fourier series of ", Cell[BoxData[ \(U \((t)\)\)]], ", that is\n\t", Cell[BoxData[ RowBox[{ RowBox[{ SubsuperscriptBox[ StyleBox["\[Integral]", FontSize->18], StyleBox["0", FontSize->10], StyleBox["t", FontSize->10]], \(U \((\[Tau])\) \[DifferentialD]\[Tau]\)}], "=", RowBox[{ StyleBox[ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], \(j = 1\), StyleBox["\[Infinity]", FontSize->12]], FontSize->12], RowBox[{"(", RowBox[{ RowBox[{ StyleBox[ FractionBox[ RowBox[{ RowBox[{ StyleBox[\(a\_j\), FontSize->14], StyleBox[" ", FontSize->14], "+", " ", RowBox[{ StyleBox[ SuperscriptBox[ StyleBox[\((\(-1\))\), FontSize->12], StyleBox[\(j + 1\), FontSize->10]], FontSize->12], StyleBox[\(a\_0\), FontSize->14]}]}], " "}], StyleBox["j", FontSize->12]], FontSize->14], " ", "sin", \((j\ t)\)}], " ", "-", RowBox[{ FractionBox[ RowBox[{" ", SubscriptBox[ StyleBox["b", FontSize->14], "j"]}], StyleBox["j", FontSize->12]], " ", "cos", \((j\ t)\)}]}], ")"}]}]}]]], ",\nwhere we have used the expansion\n\t", Cell[BoxData[ RowBox[{ FractionBox[ RowBox[{ StyleBox[\(a\_0\), FontSize->14], StyleBox["t", FontSize->12]}], StyleBox["2", FontSize->14]], "=", RowBox[{ StyleBox[ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], \(j = 1\), StyleBox["\[Infinity]", FontSize->12]], FontSize->12], RowBox[{ StyleBox[ FractionBox[ RowBox[{ RowBox[{ StyleBox[ SuperscriptBox[ StyleBox[\((\(-1\))\), FontSize->12], StyleBox[\(j + 1\), FontSize->10]], FontSize->12], StyleBox[\(a\_0\), FontSize->14]}], " "}], StyleBox["j", FontSize->12]], FontSize->14], " ", "sin", \((j\ t)\)}]}]}]]], ".\n\n", StyleBox["Theorem 11.6, Page 401", FontWeight->"Bold"], StyleBox[".", FontWeight->"Bold", FontSlant->"Italic"], " ", StyleBox["(Termwise Differentiation)", FontColor->RGBColor[1, 0, 1]], " If ", Cell[BoxData[ \(U \((t)\)\)]], " has the Fourier series representation ", Cell[BoxData[ RowBox[{\(U \((t)\)\), "=", RowBox[{ StyleBox[ FractionBox[ StyleBox[\(a\_0\), FontSize->14], StyleBox["2", FontSize->14]], FontSize->14], StyleBox[" ", FontSize->14], "+", " ", RowBox[{ StyleBox[ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], \(j = 1\), StyleBox["\[Infinity]", FontSize->12]], FontSize->12], RowBox[{"(", RowBox[{ RowBox[{ StyleBox[\(a\_j\), FontSize->14], " ", "cos", \((j\ t)\)}], " ", "+", " ", RowBox[{ SubscriptBox[ StyleBox["b", FontSize->14], "j"], " ", "sin", \((j\ t)\)}]}], ")"}]}]}]}]]], ". Then the derivative of ", Cell[BoxData[ \(U \((t)\)\)]], " has a Fourier series representation which can be obtained by termwise \ differentiation of the Fourier series of ", Cell[BoxData[ \(U \((t)\)\)]], ", that is \n\t", Cell[BoxData[ RowBox[{ RowBox[{ SuperscriptBox["U", "\[Prime]", MultilineFunction->None], \((t)\)}], "=", " ", RowBox[{ StyleBox[ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], \(j = 1\), StyleBox["\[Infinity]", FontSize->12]], FontSize->12], RowBox[{"(", RowBox[{ RowBox[{"j", " ", SubscriptBox[ StyleBox["b", FontSize->14], "j"], " ", "cos", \((j\ t)\)}], "-", " ", RowBox[{"j", StyleBox[\(a\_j\), FontSize->14], " ", "sin", \((j\ t)\)}]}], ")"}]}]}]]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "Load ", StyleBox["Mathe", FontSlant->"Italic"], "matica's FourierTransform package.\nMake sure this is done only ONCE \ during a Mathematic", StyleBox["a", FontSlant->"Italic"], " session." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Get["\"]\)], "Input", AspectRatioFixed->True], Cell[TextData[StyleBox[ "First set up procedures to compute the coefficients and partial sum.", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[a, b, j, n, S, t, u]; \n a[j_]\ = \ \(1\/\[Pi]\) \(\[Integral]\_\(-\[Pi]\)\%\[Pi] u[t] Cos[j\ t] \[DifferentialD]t\); \nb[j_]\ = \ \(1\/\[Pi]\) \(\[Integral]\_\(-\[Pi]\)\%\[Pi] u[t] Sin[j\ t] \[DifferentialD]t\); \nS[n_, t_] = a[0]\/2 + \[Sum]\+\(j = 1\)\%n\((a[j] Cos[j\ t] + b[j] Sin[j\ t])\); \n Print[\*"\"\< \!\(a\_n\) = \>\"", a[j]]; \n Print[\*"\"\< \!\(b\_n\) = \>\"", b[j]]; \n Print[\*"\"\<\!\(S\_n\)[t] = \>\"", S[n, t]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 11.1, Page 399.", FontWeight->"Bold"], " Find the Fourier series expansion for u(t) = ", Cell[BoxData[ FractionBox[ StyleBox[\(t\ \), FontSize->14], StyleBox["2", FontSize->14]]]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[S9, t, u]; \nu[t_]\ = \ t\/2; \nPrint["\", u[t]]; \)], "Input", AspectRatioFixed->True], Cell["\<\ For illustration we can computer the Fourier series of degree n = \ 9.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(S9[t_]\ = \ S[9, t]; \nPrint[\*"\"\<\!\(S\_9\)[t] = \>\"", S9[t]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "We can also get this result using ", StyleBox["Mathematica", FontSlant->"Italic"], "'s built in procedure FourierTrigSeries.", StyleBox["", FontFamily->"Courier", FontWeight->"Bold"] }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(S9[t_]\ = \ FourierTrigSeries[u[t], {t, \(-Pi\), Pi}, 9]; \n Print[\*"\"\<\!\(S\_9\)[t] = \>\"", S9[t]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "A graph of u(t), ", Cell[BoxData[ StyleBox[\(S\_1\), FontSize->14]]], "(t), ", Cell[BoxData[ StyleBox[\(S\_2\), FontSize->14]]], "(t), ", Cell[BoxData[ StyleBox[\(S\_3\), FontSize->14]]], "(t) is given:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(S1[t_]\ = \ S[1, t]; \nS2[t_]\ = \ S[2, t]; \nS3[t_]\ = \ S[3, t]; \n Plot[{u[t], S1[t], S2[t], S3[t]}, {t, \(-Pi\), Pi}, \n PlotRange -> {{\(-\[Pi]\), \[Pi]}, {\(-1.5\), 1.5}}, PlotStyle \[Rule] {RGBColor[0, 0, 0], RGBColor[1, 0, 0], RGBColor[0, 1, 0], RGBColor[0, 0, 1]}, AxesLabel \[Rule] {"\", "\"}, Ticks -> {Range[\(-\[Pi]\), \[Pi], \[Pi]\/2], Range[\(-1.5\), 1.5, 0.5]}]; \nPrint["\", u[t]]; \n Print[\*"\"\<\!\(S\_1\)[t] = \>\"", S1[t]]; \n Print[\*"\"\<\!\(S\_2\)[t] = \>\"", S2[t]]; \n Print[\*"\"\<\!\(S\_3\)[t] = \>\"", S3[t]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 11.2, Page 401.", FontWeight->"Bold"], " The function ", Cell[BoxData[ \(U \((t)\) = \(\[VerticalSeparator]\( t \[VerticalSeparator] \)\)\)]], " has the Fourier series representation \n", Cell[BoxData[ RowBox[{\(U \((t)\)\), "=", RowBox[{ FractionBox[ StyleBox["\[Pi]", FontSize->12], StyleBox["2", FontSize->12]], "-", RowBox[{ FractionBox[ StyleBox["\[Pi]", FontSize->12], StyleBox["4", FontSize->12]], RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(j = 1\), FontSize->10], StyleBox["\[Infinity]", FontSize->10]], FractionBox[ StyleBox[\(cos \((\((2 j - 1)\) t)\)\), FontSize->12], StyleBox[\(\((2 j - 1)\)\^2\), FontSize->12]]}]}]}]}]]], ". \nSolution. The solution is given in the textbook" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["GoTo ", FontSize->16, FontColor->RGBColor[0, 1, 0]], ButtonBox["Chapter 11", ButtonData:>"CHAPTER", ButtonStyle->"Hyperlink"] }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "Section 11.2 ", StyleBox["The Dirichlet Problem for the Unit Disk", FontColor->RGBColor[1, 0, 1]] }], "Text", Evaluatable->False, AspectRatioFixed->True, FontSize->18, CellTags->"Section 11.2"], Cell[TextData[{ " The ", StyleBox["Dirichlet problem", FontColor->RGBColor[1, 0, 1]], " for the unit disk D: |z| < 1 is to find a real-valued function u(x,y) \ that is harmonic in the unit disk D and that takes on the boundary values:\n\ u(cos(", StyleBox["\[Theta]", FontFamily->"Symbol"], "), sin(", StyleBox["\[Theta]", FontFamily->"Symbol"], ")) = U(", StyleBox["\[Theta]", FontFamily->"Symbol"], ") for -", StyleBox["\[Pi]", FontFamily->"Symbol"], " < ", StyleBox["\[Theta]", FontFamily->"Symbol"], " \[LessEqual] ", StyleBox["\[Pi]", FontFamily->"Symbol"], " .\n \n", StyleBox["Theorem 11.7, Page 406.", FontWeight->"Bold"], " ", StyleBox["(The Dirichlet problem in D)", FontColor->RGBColor[1, 0, 1]], " Assume that ", Cell[BoxData[ \(U \((t)\)\)]], " has period ", Cell[BoxData[ \(2 \[Pi]\)]], ", ", Cell[BoxData[ RowBox[{ SuperscriptBox["U", "\[Prime]", MultilineFunction->None], \((t)\)}]]], " is piecewise continuous, and ", Cell[BoxData[ RowBox[{\(U \((t)\)\), "=", RowBox[{ StyleBox[ FractionBox[ StyleBox[\(a\_0\), FontSize->14], StyleBox["2", FontSize->12]], FontSize->14], StyleBox[" ", FontSize->14], "+", " ", RowBox[{ StyleBox[ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], \(j = 1\), StyleBox["\[Infinity]", FontSize->12]], FontSize->12], RowBox[{"(", RowBox[{ RowBox[{ StyleBox[\(a\_j\), FontSize->14], " ", "cos", \((j\ t)\)}], " ", "+", " ", RowBox[{ SubscriptBox[ StyleBox["b", FontSize->14], "j"], " ", "sin", \((j\ t)\)}]}], ")"}]}]}]}]]], ". \nThen the solution to the Dirichlet problem in D is\n ", Cell[BoxData[ RowBox[{\(u \((r\ cos\ \[Theta], r\ sin\ \[Theta])\)\), "=", RowBox[{ StyleBox[ FractionBox[ StyleBox[\(a\_0\), FontSize->14], StyleBox["2", FontSize->12]], FontSize->14], StyleBox[" ", FontSize->14], "+", " ", RowBox[{ StyleBox[ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], \(j = 1\), StyleBox["\[Infinity]", FontSize->12]], FontSize->12], RowBox[{"(", RowBox[{ RowBox[{ StyleBox[\(a\_j\), FontSize->14], " ", SuperscriptBox["r", StyleBox["j", FontSize->10]], " ", "cos", \((j\ \[Theta])\)}], " ", "+", " ", RowBox[{ SubscriptBox[ StyleBox["b", FontSize->14], "j"], " ", SuperscriptBox["r", StyleBox["j", FontSize->10]], " ", "sin", \((j\ \[Theta])\)}]}], ")"}]}]}]}]]], ",\nwhere x + \[ImaginaryI] y = r ", Cell[BoxData[ SuperscriptBox["\[ExponentialE]", StyleBox[\(\[ImaginaryI]\ \[Theta]\), FontSize->10]]]], " denotes a complex number in the closed disk ", Cell[BoxData[ \(\[VerticalSeparator] z \[VerticalSeparator] \( \[LessEqual] \ 1\)\)]], " .\n\n", StyleBox["Theorem 11.8, Page 407.", FontWeight->"Bold"], " ", StyleBox["(Poisson Integral Formula for the Unit Disk)", FontColor->RGBColor[1, 0, 1]], " Let ", Cell[BoxData[ \(u \((x, y)\)\)]], " be a function that is harmonic in a simply connected domain that \ contains the closed unit disk ", Cell[BoxData[ \(\(D : \) \[VerticalSeparator] \(z \[VerticalSeparator] \( < \ 1\)\)\)]], ". If ", Cell[BoxData[ \(u \((x, y)\)\)]], " takes on the boundary values\n\t", Cell[BoxData[ \(\(u \((r\ cos\ \[Theta], r\ sin\ \[Theta])\) = U \((\[Theta])\)\ \ \ for\ \ \ - \[Pi] < \[Theta] \[LessEqual] \[Pi]\ \)\)]], ",\nthen ", Cell[BoxData[ \(u \((x, y)\)\)]], " has the integral representation\n\t", Cell[BoxData[ RowBox[{ RowBox[{\(u \((r\ cos\ \[Theta], r\ sin\ \[Theta])\)\), "=", " ", RowBox[{ SubsuperscriptBox[ StyleBox["\[Integral]", FontSize->18], StyleBox[\(-\[Infinity]\), FontSize->10], StyleBox["\[Infinity]", FontSize->10]], RowBox[{ FractionBox[ StyleBox[\(\((1\ - \ r\^2)\) U \((t)\)\), FontSize->12], StyleBox[ \(1\ + \ r\^2\ - \ 2 r\ cos \((t\ - \ \[Theta])\)\), FontSize->12]], \(\[DifferentialD]t\)}]}]}], " "}]]], ",\nthat is valid for ", Cell[BoxData[ \(\[VerticalSeparator] z \[VerticalSeparator] \( < \ 1\)\)]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 11.3, Page 409.", FontWeight->"Bold"], " Find the function u(x,y) that is harmonic in the \nunit disk |z| < 1 \ and takes on the boundary values:\n u( cos ", StyleBox["\[Theta]", FontFamily->"Symbol"], " , sin ", StyleBox["\[Theta]", FontFamily->"Symbol"], " ) = U(", StyleBox["\[Theta]", FontFamily->"Symbol"], ") = ", Cell[BoxData[ FractionBox[ StyleBox["\[Theta]", FontSize->14], StyleBox["2", FontSize->14]]]], " for -", StyleBox["\[Pi]", FontFamily->"Symbol"], " < ", StyleBox["\[Theta]", FontFamily->"Symbol"], " < ", StyleBox["\[Pi]", FontFamily->"Symbol"], ".\nSolution: Using Example 11.1, Page 399. The Fourier series expansion \ is: \nU(", StyleBox["\[Theta]", FontFamily->"Symbol"], ") = ", Cell[BoxData[ RowBox[{ StyleBox[ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], \(j = 1\), StyleBox["\[Infinity]", FontSize->12]], FontSize->12], RowBox[{ FractionBox[\(\((\(-1\))\)\^\(n + 1\)\), StyleBox["n", FontSize->14]], " ", "sin", RowBox[{"(", RowBox[{"j", " ", StyleBox["\[Theta]", FontFamily->"Symbol"]}], ")"}]}]}]]], ". Using formula (3) for the solution of the \nDirichlet problem, to \ obtain the boundary value approximation ", Cell[BoxData[ StyleBox[\(U\_7\), FontSize->14]]], "(", StyleBox["\[Theta]", FontFamily->"Symbol"], "):" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[k, r, \[Theta], U7]; \n U7[r_, \[Theta]_]\ = \ \[Sum]\+\(k = 1\)\%7\((\(-1\))\)\^\(k + 1\)\/k\ r\^k\ Sin[k\ \[Theta]]; \nPrint[\*"\"\<\!\(U\_7\)[r,\[Theta]] = \>\"", U7[r, \[Theta]]]; \)], "Input"], Cell[BoxData[ \(Plot[U7[1, \[Theta]], {\[Theta], \(-Pi\), Pi}, PlotRange -> {{\(-\[Pi]\), \[Pi]}, {\(-1.7\), 1.7}}, PlotStyle \[Rule] {RGBColor[0, 0, 1]}, AxesLabel \[Rule] {"\<\[Theta]\>", "\"}, Ticks -> {Range[\(-\[Pi]\), \[Pi], \[Pi]\/2], Range[\(-1.5\), 1.5, 0.5]}]; \n Print[\*"\"\<\!\(U\_7\)[1,\[Theta]] = \>\"", U7[1, \[Theta]]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["GoTo ", FontSize->16, FontColor->RGBColor[0, 1, 0]], ButtonBox["Chapter 11", ButtonData:>"CHAPTER", ButtonStyle->"Hyperlink"] }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "Section 11.3 ", StyleBox["Vibrations in Mechanical Systems", FontColor->RGBColor[1, 0, 1]] }], "Text", Evaluatable->False, AspectRatioFixed->True, FontSize->18, CellTags->"Section 11.3"], Cell[TextData[{ " The forced motion of a mechanical system satisfies the nonhomogeneous \ linear\ndifferential equation m ", Cell[BoxData[ \(U\^\(''\)\)]], "(t) + c", Cell[BoxData[ \(\(U\^'\)\)]], "(t) + kU(t) = F(t). We investigate the solutions \nwhen F(t) is a \ Fourier series." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["Example 11.4, Page 416.", FontWeight->"Bold"], " Find the general solution to ", Cell[BoxData[ \(U\^\(''\)\)]], "(t) + 2", Cell[BoxData[ \(\(U\^'\)\)]], "(t) + U(t) = F(t).\nwhere F(t) is given by the Fourier series ", Cell[BoxData[ RowBox[{\(F \((t)\)\), "=", RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(n = 1\), FontSize->10], StyleBox["\[Infinity]", FontSize->10]], FractionBox[ StyleBox[\(cos \((\((2 n - 1)\) t)\)\), FontSize->12], StyleBox[ SuperscriptBox[ RowBox[{"(", RowBox[{ RowBox[{"2", StyleBox["n", FontSize->12]}], "-", "1"}], ")"}], "2"], FontSize->12]]}]}]]], ". \nSolution. The solution is given in the textbook." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["Exercise 1, Page 417.", FontWeight->"Bold"], " Find the general solution to ", Cell[BoxData[ \(U\^\(''\)\)]], "(t) + 3", Cell[BoxData[ \(\(U\^'\)\)]], "(t) + U(t) = F(t).\n", StyleBox["(a)", FontWeight->"Bold"], " Use F(t) = ", Cell[BoxData[ FractionBox[ StyleBox["t", FontSize->14], StyleBox["2", FontSize->14]]]], ". Recall that we know that: F(t) = ", Cell[BoxData[ RowBox[{ StyleBox[ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], \(j = 1\), StyleBox["\[Infinity]", FontSize->12]], FontSize->12], RowBox[{ FractionBox[\(\((\(-1\))\)\^\(n + 1\)\), StyleBox["n", FontSize->14]], " ", "sin", RowBox[{"(", RowBox[{"j", StyleBox[" ", FontSize->14], StyleBox["t", FontFamily->"Symbol", FontSize->14]}], ")"}]}]}]]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[a, b, A, B, EqA, EqB, f, F, L, n, t, T, S, U]; \n T[n_, t_]\ = \ A[n]\ Cos[n\ t]\ + \ B[n]\ Sin[n\ t]; \n F[0, t_]\ = \ 0; \n F[n_, t_]\ = \ \(\((\(-1\))\)\^\(n - 1\)\ Sin[n\ t]\)\/n; \n Print["\", T[n, t]]; \nPrint["\", F[n, t]]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Print[ Eq\ = \ \[PartialD]\_\(t, t\)T[n, t]\ + \ 3\ \[PartialD]\_t\ T[n, t]\ + \ T[n, t]\ == \ F[n, t]]; \n Print[EqA\ = \ ReplaceAll[Eq, {Cos[n\ t]\ \[Rule] \ 1, Sin[n\ t]\ \[Rule] \ 0}]]; \nPrint[EqB\ = \ ReplaceAll[Eq, {Sin[n\ t]\ \[Rule] \ 1, Cos[n\ t]\ \[Rule] \ 0}]]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(solset\ = \ Solve[{EqA, EqB}, \ {A[n], B[n]}]; \n a[n_]\ = \ solset\[LeftDoubleBracket]1, 1, 2\[RightDoubleBracket]; \n b[n_]\ = \ solset\[LeftDoubleBracket]1, 2, 2\[RightDoubleBracket]; \n a[0]\ \ = \ 2\ F[0, t]; \nPrint["\", a[0]]; \n Print["\", a[n]]; \nPrint["\", b[n]]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(U[n_, t_]\ = \ a[n]\ Cos[n\ t]\ + \ b[n]\ Sin[n\ t]; \n U[0, t]\ \ \ = \(\ a[0]\)\/2; \nPrint["\", U[0, t]]; \n Print["\", U[n, t]]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(L\ = \ \[PartialD]\_\(t, t\)U[n, t]\ + \ 3 \[PartialD]\_t U[n, t]\ + \ U[n, t]; \n Print[\*"\"\<\!\(\[PartialD]\_\(t, t\)\)U[n,t] + \ 3\!\(\[PartialD]\_t\)U[n,t] + U[n,t] = \>\"", F[n, t]]; \n Print[L, "\< = \>", F[n, t]]; \n Print[MapAll[Together, L], "\< = \>", F[n, t]]; \)], "Input"], Cell[BoxData[ \(S[t_]\ = \ \[Sum]\+\(n = 0\)\%5 U[n, t]; \n L\ = \ \(\(S'\)'\)[t]\ + \ 3\ \(S'\)[t]\ + \ S[t]; \n Print[\*"\"\<\!\(S\_5\)[t] = \!\(\[Sum]\+\(n = 0\)\%5\)U[n,t] = \>\"", S[t]]; \nPrint[\*"\"\<\!\(S\_5\)'[t] = \>\"", \(S'\)[t]]; \n Print[\*"\"\<\!\(S\_5\)''[t] = \>\"", \(\(S'\)'\)[t]]; \nPrint["\< \>"]; \nPrint[\*"\"\<\!\(S\_5\)''[t] + 3\!\(S\_5\)'[t] + \!\(S\_5\)[t] = \>\"", Expand[L]]; \nPrint["\< \>"]; \n f[t_]\ = \ \[Sum]\+\(n = 0\)\%5 F[n, t]; \n Print[\*"\"\<\!\(f\_5\)[t] = \!\(\[Sum]\+\(n = 0\)\%5\)f[n,t] = \>\"", f[t]]; \nPrint["\< \>"]; \n Print[\*"\"\<\!\(S\_5\)''[t] + 3\!\(S\_5\)'[t] + \!\(S\_5\)[t] == \ \!\(f\_5\)[t]\>\""]; \nPrint[Expand[L], "\< == \>", f[t]]; \n Print[Expand[L]\ == \ f[t]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Exercise 2, Page 418.", FontWeight->"Bold"], " Find the general solution to 2", Cell[BoxData[ \(\(\ U\^\(''\)\)\)]], "(t) + 2 ", Cell[BoxData[ \(\(U\^'\)\)]], "(t) + U(t) = F(t).\n", StyleBox["(a)", FontWeight->"Bold"], " Use F(t) = ", Cell[BoxData[ FractionBox[ StyleBox["t", FontSize->14], StyleBox["2", FontSize->14]]]], ". Recall that we know that: F(t) = ", Cell[BoxData[ RowBox[{ StyleBox[ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], \(j = 1\), StyleBox["\[Infinity]", FontSize->12]], FontSize->12], RowBox[{ FractionBox[\(\((\(-1\))\)\^\(n + 1\)\), StyleBox["n", FontSize->14]], " ", "sin", RowBox[{"(", RowBox[{"j", StyleBox[" ", FontSize->14], StyleBox["t", FontSize->14]}], ")"}]}]}]]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[a, b, A, B, EqA, EqB, f, F, L, n, t, T, S, U]; \n T[n_, t_]\ = \ A[n]\ Cos[n\ t]\ + \ B[n]\ Sin[n\ t]; \n F[0, t_]\ = \ 0; \n F[n_, t_]\ = \(\((\(-1\))\)\^\(n - 1\)\ Sin[n\ t]\)\/n; \n Print["\", T[n, t]]; \nPrint["\", F[n, t]]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Print[ Eq\ = \ 2\ \[PartialD]\_\(t, t\)T[n, t]\ + \ 2\ \[PartialD]\_t\ T[n, t]\ + \ T[n, t]\ == \ F[n, t]]; \n Print[EqA\ = \ ReplaceAll[Eq, {Cos[n\ t]\ \[Rule] \ 1, Sin[n\ t]\ \[Rule] \ 0}]]; \nPrint[EqB\ = \ ReplaceAll[Eq, {Sin[n\ t]\ \[Rule] \ 1, Cos[n\ t]\ \[Rule] \ 0}]]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(solset\ = \ Solve[{EqA, EqB}, \ {A[n], B[n]}]; \n a[n_]\ = \ solset\[LeftDoubleBracket]1, 1, 2\[RightDoubleBracket]; \n b[n_]\ = \ solset\[LeftDoubleBracket]1, 2, 2\[RightDoubleBracket]; \n a[0]\ \ = \ 2\ F[0, t]; \nPrint["\", a[0]]; \n Print["\", a[n]]; \nPrint["\", b[n]]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(U[n_, t_]\ = \ a[n]\ Cos[n\ t]\ + \ b[n]\ Sin[n\ t]; \n U[0, t]\ \ \ = \(\ a[0]\)\/2; \nPrint["\", U[0, t]]; \n Print["\", U[n, t]]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(L\ = \ 2 \[PartialD]\_\(t, t\)U[n, t]\ + \ 2 \[PartialD]\_t U[n, t]\ + \ U[n, t]; \n Print[\*"\"\<2\!\(\[PartialD]\_\(t, t\)\)U[n,t] + \ 2\!\(\[PartialD]\_t\)U[n,t] + U[n,t] = \>\"", F[n, t]]; \n Print[L, "\< = \>", F[n, t]]; \n Print[MapAll[Together, L], "\< = \>", F[n, t]]; \)], "Input"], Cell[BoxData[ \(S[t_]\ = \ \[Sum]\+\(n = 0\)\%5 U[n, t]; \n L\ = \ 2 \(\( S'\)'\)[t]\ + \ 2\ \(S'\)[t]\ + \ S[t]; \n Print[\*"\"\<\!\(S\_5\)[t] = \!\(\[Sum]\+\(n = 0\)\%5\)U[n,t] = \>\"", S[t]]; \nPrint[\*"\"\<\!\(S\_5\)'[t] = \>\"", \(S'\)[t]]; \n Print[\*"\"\<\!\(S\_5\)''[t] = \>\"", \(\(S'\)'\)[t]]; \nPrint["\< \>"]; \nPrint[\*"\"\<2\!\(S\_5\)''[t] + 2\!\(S\_5\)'[t] + \!\(S\_5\)[t] = \ \>\"", Expand[L]]; \nPrint["\< \>"]; \n f[t_]\ = \ \[Sum]\+\(n = 0\)\%5 F[n, t]; \n Print[\*"\"\<\!\(f\_5\)[t] = \!\(\[Sum]\+\(n = 0\)\%5\)f[n,t] = \>\"", f[t]]; \nPrint["\< \>"]; \n Print[\*"\"\<2\!\(S\_5\)''[t] + 2\!\(S\_5\)'[t] + \!\(S\_5\)[t] == \ \!\(f\_5\)[t]\>\""]; \nPrint[Expand[L], "\< == \>", f[t]]; \n Print[Expand[L]\ == \ f[t]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["GoTo ", FontSize->16, FontColor->RGBColor[0, 1, 0]], ButtonBox["Chapter 11", ButtonData:>"CHAPTER", ButtonStyle->"Hyperlink"] }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "Section 11.4 ", StyleBox["The Fourier Transform", FontColor->RGBColor[1, 0, 1]] }], "Text", Evaluatable->False, AspectRatioFixed->True, FontSize->18, CellTags->"Section 11.4"], Cell[TextData[{ StyleBox["Theorem 11.9, Page 420.", FontWeight->"Bold"], " ", StyleBox["(Fourier Transform)", FontColor->RGBColor[1, 0, 1]], " Let ", Cell[BoxData[ \(U \((t)\)\)]], " and ", Cell[BoxData[ RowBox[{ SuperscriptBox["U", "\[Prime]", MultilineFunction->None], \((t)\)}]]], " be piecewise continuous, and\n\t", Cell[BoxData[ RowBox[{ SubsuperscriptBox[ StyleBox["\[Integral]", FontSize->18], StyleBox[\(-\[Infinity]\), FontSize->12], StyleBox["\[Infinity]", FontSize->12]], RowBox[{"\[VerticalSeparator]", StyleBox[\(U \((t)\) \[VerticalSeparator] \[DifferentialD]t < M\), FontSize->14]}]}]]], ",\nfor some positive constant M. The ", StyleBox["Fourier transform", FontColor->RGBColor[1, 0, 1]], " ", Cell[BoxData[ \(F \((w)\)\)]], " of ", Cell[BoxData[ \(U \((t)\)\)]], " is defined as\n\t", Cell[BoxData[ RowBox[{\(F \((w)\)\), " ", "=", " ", RowBox[{ FractionBox[ StyleBox["1", FontSize->14], StyleBox[\(2 \[Pi]\), FontSize->14]], RowBox[{ SubsuperscriptBox[ StyleBox["\[Integral]", FontSize->18], StyleBox[\(-\[Infinity]\), FontSize->12], StyleBox["\[Infinity]", FontSize->12]], StyleBox[ \(U \((t)\) \(\[ExponentialE]\^\(\(-i\)\ w\ t\)\) \[DifferentialD]t\), FontSize->14]}]}]}]]], ".\nAt points of continuity, ", Cell[BoxData[ \(U \((t)\)\)]], " has the integral representation\n\t", Cell[BoxData[ RowBox[{\(U \((t)\)\), " ", "=", " ", RowBox[{ SubsuperscriptBox[ StyleBox["\[Integral]", FontSize->18], StyleBox[\(-\[Infinity]\), FontSize->12], StyleBox["\[Infinity]", FontSize->12]], StyleBox[ \(F \((w)\) \((t)\) \(\[ExponentialE]\^\(i\ w\ t\)\) \[DifferentialD]w\), FontSize->14]}]}]]], ",\nand at a point ", Cell[BoxData[ \(t = a\)]], " of discontinuity of ", Cell[BoxData[ \(U \((t)\)\)]], ", this integral converges to \n\t ", Cell[BoxData[ FractionBox[ StyleBox[\(U \((\(a\^-\))\) + U \((\(a\^+\))\)\), FontSize->12], StyleBox["2", FontSize->12]]]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "Load ", StyleBox["Mathe", FontSlant->"Italic"], "matica's FourierTransform package. Make sure this is done only ONCE \ during a Mathematic", StyleBox["a", FontSlant->"Italic"], " session." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Get["\"]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["Example 11.5, Page 421.", FontWeight->"Bold"], " Find the Fourier transform of U(t) = ", Cell[BoxData[ \(\[ExponentialE]\^\(-|\(t | \)\)\)], FontSize->14], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[F, F1, F2, F3, t, U, U1, U2, w, w1, w2]; \n U[t_]\ = \[ExponentialE]\^\(-Abs[t]\); \nU1[t_]\ = \[ExponentialE]\^t; \nU2[t_]\ = \[ExponentialE]\^\(-t\); \n w1[t_]\ = \[Integral]U1[t] \(\[ExponentialE]\^\(\(-\[ImaginaryI]\)\ w\ t\)\) \[DifferentialD]t; \n w2[t_]\ = \[Integral]U2[t] \(\[ExponentialE]\^\(\(-\[ImaginaryI]\)\ w\ t\)\) \[DifferentialD]t; \nPrint["\", U[t]]; \n Print[\*"\"\<\!\(U\_1\)[t] = \>\"", U1[t]]; \n Print[\*"\"\<\!\(U\_2\)[t] = \>\"", U2[t]]; \n Print[\*"\"\<\!\(w\_1\)[t] = \ \[Integral]\!\(U\_1\)[t]\!\(\[ExponentialE]\^\(\(-\[ImaginaryI]\)\\\ w\\\ t\)\ \)\[DifferentialD]t = \>\"", w1[t]]; \n Print[\*"\"\<\!\(w\_2\)[t] = \ \[Integral]\!\(U\_2\)[t]\!\(\[ExponentialE]\^\(\(-\[ImaginaryI]\)\\\ w\\\ \\\ \ t\)\)\[DifferentialD]t = \>\"", w2[t]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Because of the exponential order ", Cell[BoxData[ StyleBox[\(w\_1\), FontSize->14]]], "(-\[Infinity]) = 0 and ", Cell[BoxData[ StyleBox[\(w\_2\), FontSize->14]]], "(\[Infinity]) = 0 and the Fourier transform is given by the following \ computation:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(F[w_]\ = \ Together[\(w1[0] - w2[0]\)\/\(2 \[Pi]\)]; \n Print["\", U[t]]; \nPrint["\", F[w]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Or we can let ", StyleBox["Mathematica", FontSlant->"Italic"], " work with some very time consuming improper integrals." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(val1\ = \ Integrate[ U1[t] \[ExponentialE]\^\(\(-\[ImaginaryI]\)\ w\ t\), {t, \(-\[Infinity]\), 0}, Assumptions \[Rule] {Im[w] > \(-1\)}]; \n val2\ = \ Integrate[ U2[t] \[ExponentialE]\^\(\(-\[ImaginaryI]\)\ w\ t\), {t, 0, \[Infinity]}, Assumptions \[Rule] {Im[w] < 1}]; \n F1\ = \ \(val1\ + \ val2\)\/\(2 \[Pi]\); \n F2\ = \ Together[\(val1\ + \ val2\)\/\(2 \[Pi]\)]; \n F3\ = \ ExpandAll[Together[\(val1\ + \ val2\)\/\(2 \[Pi]\)]]; \n F[w_]\ = \ Factor[ExpandAll[Together[\(val1\ + \ val2\)\/\(2 \[Pi]\)]]]; \n Print["\", U[t]]; \n Print["\", F1, "\< = \>", F2, "\< = \>", F3]; \n Print["\", F[w]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Remark. The ", StyleBox["Mathematica", FontSlant->"Italic"], " FourierTransform package does not multiply the integral by ", Cell[BoxData[ FractionBox[ StyleBox["1", FontSize->14], StyleBox[\(2 \[Pi]\), FontSize->14]]]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Fm[w_]\ = \ FourierTransform[\[ExponentialE]\^\(-Abs[t]\), \ t, \ \[Omega]]; \n Print["\< U[t] = \>", \[ExponentialE]\^\(-Abs[t]\)]; \n Print["\", Fm[w]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 11.6, Page 421.", FontWeight->"Bold"], " Find the Fourier transform of U(t) = ", Cell[BoxData[ FractionBox[ StyleBox["1", FontSize->14], StyleBox[\(1\ + \ t\^2\), FontSize->14]]]], ".\nSolution: Using the result of Example 11.5 and the symmetry property \ to find the result." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[F, t, U, w]; \nU[t_]\ = 1\/\(1\ + \ t\^2\); \n F[w_]\ = \[ExponentialE]\^\(-Abs[t]\)\/\(2 \[Pi]\); \n Print["\", U[t]]; \nPrint["\", F[w]]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(F[w_]\ = \ FourierTransform[U[t], \ t, \ w]; \n Print["\", U[t]]; \nPrint["\", F[w]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Remark. The ", StyleBox["Mathematica", FontSlant->"Italic"], " FourierTransform package does not multiply the integral by ", Cell[BoxData[ FractionBox[ StyleBox["1", FontSize->14], StyleBox[\(2 \[Pi]\), FontSize->14]]]], ".\nHowever, the sum involving the UnitStep functions will reduce to the \ standard answer." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["GoTo ", FontSize->16, FontColor->RGBColor[0, 1, 0]], ButtonBox["Chapter 11", ButtonData:>"CHAPTER", ButtonStyle->"Hyperlink"] }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "Section 11.5 ", StyleBox["The Laplace Transform", FontColor->RGBColor[1, 0, 1]] }], "Text", Evaluatable->False, AspectRatioFixed->True, FontSize->18, CellTags->"Section 11.5"], Cell[TextData[{ StyleBox["Theorem 11.10, Page 424. ", FontWeight->"Bold"], StyleBox["(Existence of the Laplace Transform)", FontColor->RGBColor[1, 0, 1]], " If ", Cell[BoxData[ \(f \((t)\)\)]], " is of exponential order,\nthen its ", StyleBox["Laplace Transform", FontColor->RGBColor[1, 0, 1]], " ", Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{ StyleBox["\[ScriptCapitalL]", FontSize->16], \((f \((t)\))\)}], "=", \(F \((s)\)\)}], " "}]]], " is given by\n\t", Cell[BoxData[ RowBox[{\(F \((s)\)\), " ", "=", " ", RowBox[{ SubsuperscriptBox[ StyleBox["\[Integral]", FontSize->18], StyleBox[\(-\[Infinity]\), FontSize->12], StyleBox["\[Infinity]", FontSize->12]], StyleBox[ \(f \((t)\) \(\[ExponentialE]\^\(\(-s\)\ t\)\) \[DifferentialD]t\), FontSize->14]}]}]]], ", where ", Cell[BoxData[ \(s = \[Sigma] + \[ImaginaryI]\ \[Omega]\)]], ".\nThe defining integral for ", Cell[BoxData[ \(F \((s)\)\)]], " exists at points ", Cell[BoxData[ \(s = \[Sigma] + \[ImaginaryI]\ \[Tau]\)]], " in the right half plane ", Cell[BoxData[ \(\[Tau] > K\)]], ".\n\n", StyleBox["Theorem 11.11, Page 425. ", FontWeight->"Bold"], StyleBox["(Linearity of the Laplace Transform)", FontColor->RGBColor[1, 0, 1]], " Let ", Cell[BoxData[ \(f \((t)\)\)]], " and ", Cell[BoxData[ \(g \((t)\)\)]], " have Laplace transforms ", Cell[BoxData[ \(F \((s)\)\)]], " and ", Cell[BoxData[ \(G \((s)\)\)]], ", respectively. If a and b are constants, then ", Cell[BoxData[ RowBox[{ RowBox[{ StyleBox["\[ScriptCapitalL]", FontSize->16], \((a\ f \((t)\) + b\ g \((t)\))\)}], " ", "=", " ", \(a\ F \((s)\)\ + \ \ b\ G \((s)\)\)}]]], ".\n\n", StyleBox["Theorem 11.12, Page 425. ", FontWeight->"Bold"], StyleBox["(Uniqueness of the Laplace Transform)", FontColor->RGBColor[1, 0, 1]], " Let ", Cell[BoxData[ \(f \((t)\)\)]], " and ", Cell[BoxData[ \(g \((t)\)\)]], " have Laplace transforms ", Cell[BoxData[ \(F \((s)\)\)]], " and ", Cell[BoxData[ \(G \((s)\)\)]], ", respectively. If ", Cell[BoxData[ \(F \((s)\) \[Congruent] G \((s)\)\)]], " then ", Cell[BoxData[ \(f \((t)\) \[Congruent] g \((t)\)\)]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "Load ", StyleBox["Mathemati", FontSlant->"Italic"], "ca's LaplaceTransform package. Make sure this is done only ONCE during a \ Mathematic", StyleBox["a", FontSlant->"Italic"], " session." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Get["\"]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["Example 11.7, Page 426.", FontWeight->"Bold"], " Find the Laplace transform of the step function:\nf(t) = 1 for \ 0 \[LessEqual] t < c,\nf(t) = 0 for c < t." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[c, f, g, F, s, t]; \nf[t_]\ = \ 1; \n g[t_]\ = \[Integral]f[t] \(\[ExponentialE]\^\(\(-s\)\ t\)\) \[DifferentialD]t; \n F[s_]\ = \[Integral]\_0\%c f[t] \(\[ExponentialE]\^\(\(-s\)\ t\)\) \[DifferentialD]t; \nPrint["\", f[t]]; \n Print[\*"\"\\"", g[t]]; \nPrint["\< \>"]; \n Print["\", g[c] - g[0]]; \n Print[\*"\"\\"", F[s]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 11.8, Page 426.", FontWeight->"Bold"], " Find the Laplace transform of f(t) = ", Cell[BoxData[ StyleBox[\(\[ExponentialE]\^\(a\ t\)\), FontSize->14]]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[a, f, g, F, s, t]; \nf[t_]\ = \ \[ExponentialE]\^\(a\ t\); \n g[t_]\ = \ \[Integral]f[t] \(\[ExponentialE]\^\(\(-s\)\ t\)\) \[DifferentialD]t; \n F[s_]\ = \ Integrate[f[t] \[ExponentialE]\^\(\(-s\)\ t\), \ {t, 0, \[Infinity]}, Assumptions \[Rule] {Im[a] == 0, a < s}]; \n Print["\", f[t]]; \n Print[\*"\"\\"", g[t]]; \nPrint["\< \>"]; \n Print["\", g[\[Infinity]] - g[0]]; \n Print[\*"\"\\"", F[s]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[ "Assuming that f(t) is of exponential order, we have g(t)\[Rule]0 as \ t\[Rule]\[Infinity]. Then g(\[Infinity]) = 0, and the Laplace transform can \ be computed by the calculation g(\[Infinity]) - g(0) = - g(0)."], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Print["\", f[t]]; \n Print[\*"\"\\"", g[t]]; \nPrint["\", \(-g[0]\)]; \nPrint["\", Simplify[\(-g[0]\)]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "We can check this with ", StyleBox["Mathematica", FontSlant->"Italic"], "'s result using the LaplaceTransform package." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(F[s_]\ = \ LaplaceTransform[f[t], \ t, \ s]; \n Print["\", f[t]]; \nPrint["\", F[s]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 11.9, Page 426.", FontWeight->"Bold"], " Find the Laplace transform of of f(t) = sinh(a t). \nSolution: Using \ the result of Example 11.8. Since f(t) = sinh(at) = ", Cell[BoxData[ FractionBox[ StyleBox[ \(\[ExponentialE]\^\(a\ t\)\ \ - \ \[ExponentialE]\^\(\(-a\)\ t\)\ \), FontSize->14], StyleBox["2", FontSize->14]]]], ",\nwe use ", Cell[BoxData[ \(\[ScriptCapitalL]\)], FontSize->14], "(", Cell[BoxData[ \(\[ExponentialE]\^\(a\ t\)\)]], ") = ", Cell[BoxData[ FractionBox[ StyleBox["1", FontSize->14], StyleBox[\(s\ - \ a\), FontSize->14]]]], " and ", Cell[BoxData[ StyleBox["\[ScriptCapitalL]", FontSize->14]]], "(", Cell[BoxData[ \(\[ExponentialE]\^\(\(-a\)\ t\)\)]], ") = ", Cell[BoxData[ FractionBox[ StyleBox["1", FontSize->14], StyleBox[\(s\ + \ a\), FontSize->14]]]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[a, f, f1, f2, f12, F, F1, F2, F12, s, t]; \n f1[t_]\ = \ \[ExponentialE]\^\(a\ t\); \n f2[t_]\ = \ \[ExponentialE]\^\(\(-a\)\ t\); \n f12[t_]\ = \ \(\[ExponentialE]\^\(a\ t\) - \[ExponentialE]\^\(\(-a\)\ t\)\)\/2; \n f[t_]\ = \ Sinh[a\ t]; \nF1[s_]\ = \ LaplaceTransform[f1[t], \ t, \ s]; \nF2[s_]\ = \ LaplaceTransform[f2[t], \ t, \ s]; \n F12[s_]\ = \ LaplaceTransform[f[t], \ t, \ s]; \n F[s_]\ = \ LaplaceTransform[f[t], \ t, \ s]; \n Print[\*"\"\< \!\(f\_1\)[t] = \>\"", f1[t]]; \n Print[\*"\"\< \!\(f\_2\)[t] = \>\"", f2[t]]; \n Print[\*"\"\<\!\(f\_12\)[t] = \>\"", f12[t]]; \n Print[\*"\"\<\!\(f\_12\)[t] = \>\"", FullSimplify[f12[t]]]; \n Print["\< f[t] = \>", FullSimplify[f[t]]]; \nPrint["\< \>"]; \n Print[\*"\"\< \!\(F\_1\)[s] = \>\"", F1[s]]; \n Print[\*"\"\< \!\(F\_2\)[s] = \>\"", F2[s]]; \n Print[\*"\"\<\!\(F\_12\)[s] = \!\(\(F\_1[s] - F\_2[s]\)\/2\) = \>\"", \(F1[s] - F2[s]\)\/2]; \n Print[\*"\"\<\!\(F\_12\)[s] = \!\(\(F\_1[s] - F\_2[s]\)\/2\) = \>\"", Together[\(F1[s] - F2[s]\)\/2]]; \n Print["\< F[s] = L[Sinh[a t]] = \>", F[s]]; \)], "Input"], Cell[TextData[{ "\n", StyleBox["Example 11.10, Page 427.", FontWeight->"Bold"], " Find the Laplace transform of f(t) = t." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[f, g, F, s, t]; \nf[t_]\ = \ t; \n g[t_]\ = \ \[Integral]f[t] \(\[ExponentialE]\^\(\(-s\)\ t\)\) \[DifferentialD]t; \n F[s_]\ = \ Integrate[f[t] \[ExponentialE]\^\(\(-s\)\ t\), \ {t, 0, \[Infinity]}, Assumptions \[Rule] {Im[s] == 0, s > 0}]; \n Print["\", f[t]]; \n Print[\*"\"\\"", g[t]]; \nPrint["\< \>"]; \n Print["\", g[\[Infinity]] - g[0]]; \n Print[\*"\"\\"", F[s]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[ "Assuming that f(t) is of exponential order, we have g(t)\[Rule]0 as t\[Rule]\ \[Infinity]. Then g(\[Infinity]) = 0, and the Laplace transform can be \ computed by the calculation g(\[Infinity]) - g(0) = - g(0)."], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Print["\", f[t]]; \n Print[\*"\"\\"", g[t]]; \nPrint["\", \(-g[0]\)]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "We can check this with ", StyleBox["Mathematica", FontSlant->"Italic"], "'s result using the LaplaceTransform package." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(F[s_]\ = \ LaplaceTransform[t, \ t, \ s]; \nPrint["\", t]; \n Print["\", F[s]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 11.11, Page 427.", FontWeight->"Bold"], " Find the Laplace transform of f(t) = cos(b t)." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["Method 1.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]], " Using integration" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[b, f, g, F, s, t]; \nf[t_]\ = \ Cos[b\ t]; \n g[t_]\ = \ \[Integral]f[t] \(\[ExponentialE]\^\(\(-s\)\ t\)\) \[DifferentialD]t; \n F[s_]\ = \ Integrate[f[t] \[ExponentialE]\^\(\(-s\)\ t\), \ {t, 0, \[Infinity]}, Assumptions \[Rule] {Im[b] == 0, Im[s] == 0, s > 0}]; \n Print["\", f[t]]; \n Print[\*"\"\\"", g[t]]; \nPrint["\< \>"]; \n Print["\", g[\[Infinity]] - g[0]]; \n Print[\*"\"\\"", F[s]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[ "Assuming that f(t) is of exponential order, we have g(t) \[Rule] 0 as t \ \[Rule] \[Infinity]. Then g(\[Infinity]) = 0, \nand the Laplace transform \ can be computed by the calculation g(\[Infinity]) - g(0) = - g(0)."], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Print["\", f[t]]; \n Print[\*"\"\\"", g[t]]; \nPrint["\", \(-g[0]\)]; \nPrint["\", Simplify[\(-g[0]\)]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["Method 2", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]], StyleBox[".", FontColor->RGBColor[1, 0, 1]], " Since f(t) = cos(bt) = ", Cell[BoxData[ StyleBox[ FractionBox[ \(\[ExponentialE]\^\(i\ b\ t\)\ \ + \ \ \[ExponentialE]\^\(\(-i\)\ b\ t\)\), StyleBox["2", FontSize->14]], FontSize->14]]], ", \nwe use L", StyleBox["1", FontVariations->{"CompatibilityType"->"Subscript"}], "(", Cell[BoxData[ \(\[ExponentialE]\^\(i\ b\ t\)\)]], ") = ", Cell[BoxData[ FractionBox[ StyleBox["1", FontSize->14], StyleBox[\(s\ - \ ib\), FontSize->14]]]], " and L", StyleBox["1", FontVariations->{"CompatibilityType"->"Subscript"}], "(", Cell[BoxData[ \(\[ExponentialE]\^\(\(-i\)\ b\ t\)\)]], ") = ", Cell[BoxData[ FractionBox[ StyleBox["1", FontSize->14], StyleBox[\(s\ + \ ib\), FontSize->14]]]] }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[b, f, f1, f2, f12, F, F1, F2, F12, s, t]; \n f1[t_]\ = \ \[ExponentialE]\^\(\[ImaginaryI]\ b\ t\); \n f2[t_]\ = \ \[ExponentialE]\^\(\(-\[ImaginaryI]\)\ b\ t\); \n f12[t_]\ = \ \(\[ExponentialE]\^\(\[ImaginaryI]\ b\ t\) + \[ExponentialE]\^\(\(-\[ImaginaryI]\)\ b\ t\)\)\/2; \n f[t_]\ = \ Cos[b\ t]; \nF1[s_]\ = \ LaplaceTransform[f1[t], \ t, \ s]; \nF2[s_]\ = \ LaplaceTransform[f2[t], \ t, \ s]; \n F12[s_]\ = \ LaplaceTransform[f[t], \ t, \ s]; \n F[s_]\ = \ LaplaceTransform[f[t], \ t, \ s]; \n Print[\*"\"\< \!\(f\_1\)[t] = \>\"", f1[t]]; \n Print[\*"\"\< \!\(f\_2\)[t] = \>\"", f2[t]]; \n Print[\*"\"\<\!\(f\_12\)[t] = \>\"", f12[t]]; \n Print[\*"\"\<\!\(f\_12\)[t] = \>\"", FullSimplify[f12[t]]]; \n Print["\< f[t] = \>", FullSimplify[f[t]]]; \nPrint["\< \>"]; \n Print[\*"\"\< \!\(F\_1\)[s] = \>\"", F1[s]]; \n Print[\*"\"\< \!\(F\_2\)[s] = \>\"", F2[s]]; \n Print[\*"\"\<\!\(F\_12\)[s] = \!\(\(F\_1[s] + F\_2[s]\)\/2\) = \>\"", \(F1[s] + F2[s]\)\/2]; \n Print[\*"\"\<\!\(F\_12\)[s] = \!\(\(F\_1[s] + F\_2[s]\)\/2\) = \>\"", Together[\(F1[s] + F2[s]\)\/2]]; \n Print["\< F[s] = L[Cos[b t]] = \>", F[s]]; \)], "Input"], Cell[TextData[{ "\n", StyleBox["Example 11.12, Page 427.", FontWeight->"Bold"], " Find the inverse Laplace transform of \n F(s) = ", Cell[BoxData[ FractionBox[ StyleBox[\(3 s\ + \ 6\), FontSize->14], StyleBox[\(s\^2\ + \ 9\), FontSize->14]]]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[A, f, f1, f2, F, L1, L2, s, t]; \n F[s_]\ \ = \ \ \(3 s\ + \ 6\)\/\(s\^2\ + \ 9\); \n A\ = \ Expand[F[s]]; \nPrint["\", F[s]]; \n Print["\", A]; \)], "Input", AspectRatioFixed->True], Cell[TextData["The transform F(s) is a linear combination."], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(f1[t_]\ = Cos[3 t]; \nf2[t_]\ = Sin[3 t]; \n L1[s_]\ = \ s\/\(s\^2\ + \ 9\); \nL2[s_]\ = \ 3\/\(s\^2\ + \ 9\); \n Print[\*"\"\<\!\(f\_1\)[t] = \>\"", f1[t]]; \n Print[\*"\"\<\!\(f\_2\)[t] = \>\"", f2[t]]; \n Print[\*"\"\<\!\(L\_1\)[s] = \>\"", L1[s]]; \n Print[\*"\"\<\!\(L\_2\)[s] = \>\"", L2[s]]; \n Print["\< F[s] = \>", F[s]]; \n Print[\*"\"\\""]; \n Print[F[s]\ == 3\ L1[s]\ + \ 2\ L2[s], "\<, \>", Expand[F[s]]\ == \ 3\ L1[s]\ + \ 2\ L2[s]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "The inverse of ", Cell[BoxData[ StyleBox[\(L\_1\), FontSize->14]]], " is ", Cell[BoxData[ \(f\_1\)]], "(t) = cos(3t) and the inverse of ", Cell[BoxData[ StyleBox[\(L\_2\), FontSize->14]]], " is ", Cell[BoxData[ \(f\_2\)]], "(t) = sin(3t). \nHence f(t) = 3 ", Cell[BoxData[ \(f\_1\)]], "(t) + 2 ", Cell[BoxData[ \(f\_2\)]], "(t)" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(f[t_]\ = \ 3\ f1[t]\ + \ 2\ f2[t]; \n Print[\*"\"\\""]; \n Print["\", f[t]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "We can check this with ", StyleBox["Mathematica", FontSlant->"Italic"], "'s result using the LaplaceTransform package." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(f[t_]\ = \ InverseLaplaceTransform[F[s], \ s, \ t]; \n Print["\", F[s]]; \nPrint["\", f[t]]; \n Print[\*"\"\\"", 3 f1[t] + 2 f2[t]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["GoTo ", FontSize->16, FontColor->RGBColor[0, 1, 0]], ButtonBox["Chapter 11", ButtonData:>"CHAPTER", ButtonStyle->"Hyperlink"] }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "Section 11.6 ", StyleBox["Laplace Transforms of Derivatives and Integrals", FontColor->RGBColor[1, 0, 1]] }], "Text", Evaluatable->False, AspectRatioFixed->True, FontSize->18, CellTags->"Section 11.6"], Cell[TextData[{ StyleBox["Theorem 11.13, Page 430.", FontWeight->"Bold"], " ", StyleBox["(Differentiation of f(t))", FontColor->RGBColor[1, 0, 1]], " Let ", Cell[BoxData[ \(f \((t)\)\)]], " and ", Cell[BoxData[ RowBox[{ SuperscriptBox["f", "\[Prime]", MultilineFunction->None], \((t)\)}]]], " be continuous for ", Cell[BoxData[ \(0 \[LessEqual] t\)]], ", and of exponential order. Then ", Cell[BoxData[ RowBox[{ RowBox[{ StyleBox["\[ScriptCapitalL]", FontSize->16], RowBox[{"(", RowBox[{ SuperscriptBox["f", "\[Prime]", MultilineFunction->None], \((t)\)}], ")"}]}], "=", \(s\ F \((s)\) - f \((0)\)\)}]]], ", where ", Cell[BoxData[ RowBox[{ RowBox[{ StyleBox["\[ScriptCapitalL]", FontSize->16], \((f \((t)\))\)}], "=", \(F \((s)\)\)}]]], ".\n\n", StyleBox["Theorem 11.14, Page 431.", FontWeight->"Bold"], " ", StyleBox["(Integration of f(t))", FontColor->RGBColor[1, 0, 1]], " Let ", Cell[BoxData[ \(f \((t)\)\)]], " and ", Cell[BoxData[ RowBox[{ SuperscriptBox["f", "\[Prime]", MultilineFunction->None], \((t)\)}]]], " be continuous for ", Cell[BoxData[ \(0 \[LessEqual] t\)]], ", and of exponential order. Then ", Cell[BoxData[ RowBox[{ RowBox[{ StyleBox["\[ScriptCapitalL]", FontSize->16], RowBox[{"(", RowBox[{ StyleBox["\[Integral]", FontSize->18], \(f \((t)\) \[DifferentialD]t\)}], ")"}]}], "=", " ", FractionBox[ StyleBox[\(F \((s)\)\), FontSize->12], StyleBox["s", FontSize->12]]}]]], ", where ", Cell[BoxData[ RowBox[{ RowBox[{ StyleBox["\[ScriptCapitalL]", FontSize->16], \((f \((t)\))\)}], "=", \(F \((s)\)\)}]]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "Load the following ", StyleBox["Mathematica", FontSlant->"Italic"], " packages. Make sure this is done only ONCE in a ", StyleBox["Mathematica", FontSlant->"Italic"], " session." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Get["\"]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["Example 11.13, Page 430.", FontWeight->"Bold"], "\t Use Theorem 11.13 and find \[ScriptCapitalL](", Cell[BoxData[ StyleBox[\(cos\^2\), FontSize->14]]], "(t)) " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[f, F, Lf, LDf, s, t]; \nf[t_]\ = \ Cos[t]\^2; \n Print["\< f[t] = \>", f[t]]; \nPrint["\< f[0] = \>", f[0]]; \n Print["\", \(f'\)[t]]; \n Print["\", Simplify[\(f'\)[t]]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Use the fact that \[ScriptCapitalL](sin(2t)) = ", Cell[BoxData[ FractionBox[ StyleBox[\(-\ 2\), FontSize->14], StyleBox[\(s\^4\ + \ 4\), FontSize->14]]]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(LDf\ \ = \ \ \(-2\)\/\(s\^2\ + \ 4\); \n eqn\ \ = \ \ LDf\ == \ s\ Lf\ - \ f[0]; \nPrint[eqn]; \n solset\ = \ Solve[eqn, \ Lf]; \nPrint[solset]; \n Lf\ = \ solset\[LeftDoubleBracket]1, 1, 2\[RightDoubleBracket]; \n Print["\", LDf]; \n Print["\"]; \nPrint["\< L[f[t]] = \>", Lf]; \nPrint["\< L[f[t]] = \>", Simplify[Lf]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "We can check this with ", StyleBox["Mathematica", FontSlant->"Italic"], "'s result using the LaplaceTransform package." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(F[s_]\ = \ LaplaceTransform[f[t], \ t, \ s]; \n Print["\< f[t] = \>", f[t]]; \nPrint["\", F[s]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 11.14, Page 431.", FontWeight->"Bold"], "\t Use Theorem 11.14 and find \[ScriptCapitalL](f(t)).\n", StyleBox["(a)", FontWeight->"Bold"], " Find \[ScriptCapitalL](t", StyleBox["2", FontVariations->{"CompatibilityType"->"Superscript"}], ")." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[f, F, Lf, LDf, s, t]; \nf[t_]\ = \ t\^2; \n Print["\< f[t] = \>", f[t]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Use the fact that ", Cell[BoxData[ \(\(f\^'\)\)]], "(t) = 2t and \[ScriptCapitalL](2t) = ", Cell[BoxData[ FractionBox[ StyleBox["2", FontSize->14], StyleBox[\(s\^2\), FontSize->14]]]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(LDf\ \ = \ 2\/s\^2; \nLf\ = \ \ LDf\/s; \n Print["\< f'[t] = \>", \(f'\)[t]]; \n Print["\< L[f'[t]] = \>", LDf]; \n Print[\*"\"\\""]; \nPrint["\", Lf]; \n Print["\< L[f[t]] = \>", Lf]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "We can check this with ", StyleBox["Mathematica", FontSlant->"Italic"], "'s result using the LaplaceTransform package." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(F[s_]\ = \ LaplaceTransform[f[t], \ t, \ s]; \n Print["\< f[t] = \>", f[t]]; \nPrint["\", F[s]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["(b)", FontWeight->"Bold"], " Find \[ScriptCapitalL](", Cell[BoxData[ StyleBox[\(t\^3\), FontSize->14]]], ")." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[f, F, Lf, LDf, s, t]; \nf[t_]\ = \ t\^3; \n Print["\< f[t] = \>", f[t]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Use the fact that ", Cell[BoxData[ \(\(f\^'\)\)]], "(t) = 3", Cell[BoxData[ \(t\^2\)]], " and \[ScriptCapitalL](3", Cell[BoxData[ \(t\^2\)]], ") = ", Cell[BoxData[ FractionBox[ StyleBox["6", FontSize->14], StyleBox[\(s\^3\), FontSize->14]]]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(LDf\ \ = \ 6\/s\^3; \nLf\ = \ \ LDf\/s; \n Print["\< f'[t] = \>", \(f'\)[t]]; \n Print["\< L[f'[t]] = \>", LDf]; \n Print[\*"\"\\""]; \nPrint["\", Lf]; \n Print["\< L[f[t]] = \>", Lf]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "We can check this with ", StyleBox["Mathematica", FontSlant->"Italic"], "'s result using the LaplaceTransform package." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(F[s_]\ = \ LaplaceTransform[f[t], \ t, \ s]; \n Print["\< f[t] = \>", f[t]]; \nPrint["\", F[s]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 11.15, Page 432.", FontWeight->"Bold"], "\t Solve the initial value problem\n ", Cell[BoxData[ \(y\^\(''\)\)]], "(t) + y(t) = 0 with y(0) = 2 and ", Cell[BoxData[ \(\(y\^'\)\)]], "(0) = 3." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[eqn, f, F, s, solset, t, y, y0, y1, Y]; \ny0\ = \ 2; \n y1\ = \ 3; \nf[t_]\ = \ 0; \nF[s_]\ = \ 0; \n Print["\< y[0] = \>", y0]; \nPrint["\", y1]; \n Print["\< f[t] = \>", f[t]]; \nPrint["\< F[s] = \>", F[s]]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Clear[Y]; \n eqn\ \ = \ \ s\^2\ Y[s]\ - \ s\ y0\ - \ y1\ + \ Y[s]\ == \ F[s]; \n Print[eqn]; \nsolset\ = \ Solve[eqn, \ Y[s]]; \nPrint[solset]; \n Y[s_]\ = \ solset\[LeftDoubleBracket]1, 1, 2\[RightDoubleBracket]; \n Print["\", Y[s]]; \nPrint["\", Expand[Y[s]]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Using the known Laplace transforms \[ScriptCapitalL](sin t) = ", Cell[BoxData[ FractionBox[ StyleBox["1", FontSize->14], StyleBox[\(s\^2 + 1\), FontSize->14]]]], " and \[ScriptCapitalL](cos t) = ", Cell[BoxData[ FractionBox[ StyleBox["s", FontSize->14], StyleBox[\(s\^2 + 1\), FontSize->14]]]], "\nthe solution is the linear combination f(t) = 3 sin(t) + 2 \ cos(t)." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "We can check this with ", StyleBox["Mathematica", FontSlant->"Italic"], "'s result using the LaplaceTransform package." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(f[t_]\ = \ InverseLaplaceTransform[Y[s], \ s, \ t]; \n Print["\", Expand[Y[s]]]; \nPrint["\", f[t]]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Print["\", f[t]]; \nPrint["\", \(\(f'\)'\)[t]]; \n Print["\", f[t] + \(\(f'\)'\)[t]]; \n Print["\", f[0]]; \nPrint["\", \(f'\)[0]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "We can check this with ", StyleBox["Mathematica", FontSlant->"Italic"], "'s result using the DSolve package." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[y]; \neqn\ = \ \(\(y'\)'\)[t]\ + \ y[t]\ == \ 0; \nPrint[eqn]; \nsolset\ = \ DSolve[{eqn, \ y[0] == 2, \ \(y'\)[0] == 3}, \ y[t], \ t]; \nPrint[solset]; \n y[t_]\ = \ solset\[LeftDoubleBracket]1, 1, 2\[RightDoubleBracket]; \n Print["\", y[t]]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Clear[t, y]; \ny[t_]\ = \ 2\ Cos[t] + 3\ Sin[t]; \n Plot[y[t], {t, 0, \[Pi]}, \nPlotRange -> {{0, \[Pi]}, {\(-2\), 3.7}}, PlotStyle \[Rule] {RGBColor[0, 0, 1]}, AxesLabel \[Rule] {"\", "\"}, Ticks -> {Range[0, \[Pi], \[Pi]\/4], Range[\(-2\), 3.5, 1]}]; \n Print["\", y[t]]; \)], "Input"], Cell[TextData[{ "\n", StyleBox["Example 11.16, Page 432.", FontWeight->"Bold"], "\t Solve the initial value problem\n ", Cell[BoxData[ \(y\^\(''\)\)]], "(t) + ", Cell[BoxData[ \(\(y\^'\)\)]], "(t) - 2 y(t) = 0 with y(0) = 1 and ", Cell[BoxData[ \(\(y\^'\)\)]], "(0) = 4." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[eqn, f, F, s, solset, t, y, y0, y1, Y]; \ny0\ = \ 1; \n y1\ = \ 4; \nf[t_]\ = \ 0; \nF[s_]\ = \ 0; \n Print["\< y[0] = \>", y0]; \nPrint["\", y1]; \n Print["\< f[t] = \>", f[t]]; \nPrint["\< F[s] = \>", F[s]]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Clear[Y]; \n eqn\ \ = \ s\^2\ Y[s]\ - \ s\ y0\ - \ y1\ + \ s\ Y[s]\ - \ y0\ - \ 2\ Y[s]\ == \ F[s]; \nPrint[eqn]; \nsolset\ = \ Solve[eqn, \ Y[s]]; \n Print[solset]; \n Y[s_]\ = \ solset\[LeftDoubleBracket]1, 1, 2\[RightDoubleBracket]; \n Print["\", Y[s]]; \nPrint["\", Apart[Y[s]]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Using the known Laplace transforms \[ScriptCapitalL](", Cell[BoxData[ StyleBox[\(\[ExponentialE]\^t\), FontSize->14]]], ") = ", Cell[BoxData[ FractionBox[ StyleBox["1", FontSize->14], StyleBox[\(s - 1\), FontSize->14]]]], " and \[ScriptCapitalL](", Cell[BoxData[ StyleBox[\(\[ExponentialE]\^\(\(-2\) t\)\), FontSize->14]]], ") = ", Cell[BoxData[ FractionBox[ StyleBox["1", FontSize->14], StyleBox[\(s + 2\), FontSize->14]]]], "\nthe solution is the linear combination f(t) = 2 ", Cell[BoxData[ StyleBox[\(\[ExponentialE]\^t\), FontSize->14]]], " - ", Cell[BoxData[ StyleBox[\(\[ExponentialE]\^\(\(-2\) t\)\), FontSize->14]]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "We can check this with ", StyleBox["Mathematica", FontSlant->"Italic"], "'s result using the LaplaceTransform package." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(f[t_]\ = \ InverseLaplaceTransform[Y[s], \ s, \ t]; \n Print["\", Apart[Y[s]]]; \nPrint["\", f[t]]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Print["\", f[t]]; \nPrint["\", \(f'\)[t]]; \n Print["\", \(\(f'\)'\)[t]]; \n Print["\", \(\(f'\)'\)[t] + \(f'\)[t] - 2 f[t]]; \n Print["\", Expand[\(\(f'\)'\)[t] + \(f'\)[t] - 2 f[t]]]; \n Print["\", f[0]]; \nPrint["\", \(f'\)[0]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "We can check this with ", StyleBox["Mathematica", FontSlant->"Italic"], "'s result using the DSolve package." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[y]; \n eqn\ = \ \(\(y'\)'\)[t]\ + \ \(y'\)[t] - \ 2 y[t]\ == \ 0; \n Print[eqn]; \n solset\ = \ DSolve[{eqn, \ y[0] == 1, \ \(y'\)[0] == 4}, \ y[t], \ t]; \n Print[solset]; \n y[t_]\ = \ solset\[LeftDoubleBracket]1, 1, 2\[RightDoubleBracket]; \n Print["\", y[t]]; \nPrint["\", Expand[y[t]]]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Clear[t, y]; \n y[t_]\ = \ \(-\[ExponentialE]\^\(\(-2\)\ t\)\) + 2\ \[ExponentialE]\^t; \nPlot[y[t], {t, 0, 2}, \nPlotRange -> {{0, 2}, {0, 15}}, PlotStyle \[Rule] {RGBColor[0, 0, 1]}, AxesLabel \[Rule] {"\", "\"}, Ticks -> {Range[0, 2, 0.5], Range[0, 15, 2.5]}]; Print["\", y[t]]; \)], "Input"], Cell[TextData[{ StyleBox["GoTo ", FontSize->16, FontColor->RGBColor[0, 1, 0]], ButtonBox["Chapter 11", ButtonData:>"CHAPTER", ButtonStyle->"Hyperlink"] }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "Section 11.7 ", StyleBox["Shifting Theorems and the Step Function", FontColor->RGBColor[1, 0, 1]] }], "Text", Evaluatable->False, AspectRatioFixed->True, FontSize->18, CellTags->"Section 11.7"], Cell[TextData[{ StyleBox["Theorem 11.15, Page 434.", FontWeight->"Bold"], " ", StyleBox["(Shifting the Variable s)", FontColor->RGBColor[1, 0, 1]], " If ", Cell[BoxData[ \(F \((s)\)\)]], " is the Laplace transform of ", Cell[BoxData[ \(f \((t)\)\)]], ", then ", Cell[BoxData[ RowBox[{ RowBox[{ StyleBox["\[ScriptCapitalL]", FontSize->16], RowBox[{"(", RowBox[{ StyleBox[ SuperscriptBox["\[ExponentialE]", StyleBox["at", FontSize->12]], FontSize->14], "f", \((t)\)}], ")"}]}], "=", \(F \((s - a)\)\)}]]], ".\n\n", StyleBox["Theorem 11.16, Page 434.", FontWeight->"Bold"], " ", StyleBox["(Shifting the Variable t)", FontColor->RGBColor[1, 0, 1]], " If ", Cell[BoxData[ \(F \((s)\)\)]], " is the Laplace transform of ", Cell[BoxData[ \(f \((t)\)\)]], ", and ", Cell[BoxData[ \(0 \[LessEqual] a\)]], ", then ", Cell[BoxData[ RowBox[{ RowBox[{ StyleBox["\[ScriptCapitalL]", FontSize->16], RowBox[{"(", RowBox[{ SubscriptBox["U", StyleBox["a", FontSize->12]], \((t)\), "f", \((t - a)\)}], ")"}]}], "=", RowBox[{ StyleBox[ SuperscriptBox["\[ExponentialE]", StyleBox[\(-as\), FontSize->12]], FontSize->14], "F", \((s)\)}]}]]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "Load the following ", StyleBox["Mathematica", FontSlant->"Italic"], " packages. Make sure this is done only ONCE in a ", StyleBox["Mathematica", FontSlant->"Italic"], " session." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Get["\"]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["Example 11.17, Page 435.", FontWeight->"Bold"], "\t Use Theorem 11.15 and find ", Cell[BoxData[ RowBox[{ StyleBox["\[ScriptCapitalL]", FontSize->16], RowBox[{"(", RowBox[{\(t\^n\), StyleBox[ SuperscriptBox["\[ExponentialE]", StyleBox["at", FontSize->12]], FontSize->14]}], ")"}]}]]], ".\nSolution. Use the fact that ", Cell[BoxData[ RowBox[{ RowBox[{ StyleBox["\[ScriptCapitalL]", FontSize->16], \((t\^n)\)}], "=", FractionBox[ StyleBox[\(n!\), FontSize->12], StyleBox[\(s\^\(n + 1\)\), FontSize->12]]}]]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[a, F, G, n, s]; \nF[s_]\ = \ \(n!\)\/s\^\(n + 1\); \n Print[\*"\"\\"", F[s]]; \n Print[\*"\"\\"", F[s - a]]; \n Print[\*"\"\\"", FunctionExpand[F[s - a]]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "We can check this with ", StyleBox["Mathematica", FontSlant->"Italic"], "'s result using the LaplaceTransform package." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(G[s_]\ = \ LaplaceTransform[t\^n\ \[ExponentialE]\^\(a\ t\), \ t, \ s]; \n Print[\*"\"\\"", G[s]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 11.18, Page 435.", FontWeight->"Bold"], "\t Use Theorem 11.16 and find \[ScriptCapitalL](", Cell[BoxData[ StyleBox[ SubscriptBox["U", StyleBox["c", FontSize->10]], FontSize->14]]], "(t)).\nSolution: Let f(t) = 1 and F(s) = ", Cell[BoxData[ FractionBox[ StyleBox["1", FontSize->14], StyleBox["s", FontSize->14]]]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[c, f, F, g, G, n, s, t]; \nf[t_]\ = \ 1; \nF[s_]\ = \ 1\/s; \n Print["\< f[t] = \>", f[t]]; \nPrint["\< L[f[t]] = \>", F[s]]; \nPrint[\*"\"\\"", \(\(\[ExponentialE]\ \)\^\(\(-c\)\ s\)\) F[s]]; \)], "Input"], Cell[TextData[{ "We can check this with ", StyleBox["Mathematica", FontSlant->"Italic"], "'s result using the LaplaceTransform package.\nFor illustration purposes \ we use c = ", StyleBox["\[Pi]", FontFamily->"Symbol"], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(g[t_]\ = \ UnitStep[t - \[Pi]]; \n G[S_]\ = \ LaplaceTransform[UnitStep[t - \[Pi]], \ t, \ s]; \n Print["\< g[t] = \>", g[t]]; \n Print[\*"\"\\"", G[s]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 11.19, Page 435.", FontWeight->"Bold"], "\t Find \[ScriptCapitalL](f(t)) for f(t) given in Figure 11.24." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[f, F, s, t]; \n f[t_]\ = \ 1\ - \ UnitStep[t - 1]\ + \ UnitStep[t - 2]\ - \n\ \ \ \ \ \ \ \ UnitStep[t - 3]\ + \ UnitStep[t - 4]\ - \ UnitStep[t - 5]; \n Plot[f[t], \ {t, 0, 6}, Ticks \[Rule] {Range[0, 6, 1], Range[0, 1, 1]}, PlotStyle \[Rule] {RGBColor[0, 0, 1]}, AxesLabel \[Rule] {"\", "\"}, \ AspectRatio \[Rule] 1\/6]; \n Print["\", f[t]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Use ", StyleBox["Mathematica", FontSlant->"Italic"], "'s LaplaceTransform package." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(F[s_]\ = \ LaplaceTransform[f[t], \ t, \ s]; \n Print["\", f[t]]; \nPrint["\", F[s]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 11.20, Page 436.", FontWeight->"Bold"], "\t Solve the initial value problem\n ", Cell[BoxData[ \(y\^\(''\)\)]], "(t) + y(t) = ", Cell[BoxData[ StyleBox[\(U\_\[Pi]\), FontSize->14]]], "(t) with y(0) = 0 and ", Cell[BoxData[ \(y'\)]], "(0) = 0." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[eqn, f, F, s, solset, t, y, y0, y1, Y]; \ny0\ = \ 0; \n y1\ = \ 0; \nf[t_]\ = UnitStep[t - \[Pi]]; \n F[s_]\ = \ \[ExponentialE]\^\(\(-\[Pi]\)\ s\)\/s; \n Print["\< y[0] = \>", y0]; \nPrint["\", y1]; \n Print["\< f[t] = \>", f[t]]; \nPrint["\< F[s] = \>", F[s]]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Clear[Y]; \n eqn\ \ = \ s\^2\ Y[s]\ - \ s\ y0\ - \ y1\ + \ Y[s]\ == \ F[s]; \n Print[eqn]; \nsolset\ = \ Solve[eqn, \ Y[s]]; \nPrint[solset]; \n Y[s_]\ = \ solset\[LeftDoubleBracket]1, 1, 2\[RightDoubleBracket]; \n Print["\", Y[s]]; \n Y[s_]\ = \ \[ExponentialE]\^\(\(-\[Pi]\)\ s\)\ Apart[1\/\(s\ \((1\ + \ s\^2)\)\)]; \nPrint["\", Y[s]]; \n Print["\", Expand[Y[s]]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Use the facts that \[ScriptCapitalL](", Cell[BoxData[ StyleBox[\(U\_\[Pi]\), FontSize->14]]], "(t)) = ", Cell[BoxData[ FractionBox[ StyleBox[\(\[ExponentialE]\^\(-\[Pi]s\)\), FontSize->14], StyleBox["s", FontSize->14]]]], " and \[ScriptCapitalL](", Cell[BoxData[ StyleBox[\(U\_\[Pi]\), FontSize->14]]], "(t)cos(t - ", StyleBox["\[Pi]", FontFamily->"Symbol"], ")) = ", Cell[BoxData[ FractionBox[ RowBox[{" ", StyleBox[\(s\ \[ExponentialE]\^\(-\[Pi]s\)\), FontSize->14]}], StyleBox[\(1\ + \ s\^2\), FontSize->14]]]], " " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(y[t_]\ = \ UnitStep[t - \[Pi]]\ - \ UnitStep[t - \[Pi]] Cos[t - \[Pi]]; \n Print["\", y[t]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "We can check this with ", StyleBox["Mathematica", FontSlant->"Italic"], "'s result using the DSolve package." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(Clear[y]; \n eqn\ = \ \(\(y'\)'\)[t]\ + \ y[t]\ == \ UnitStep[t - \[Pi]]\), \(sol\ = \ DSolve[{eqn, \ y[0] == 0, \ \(y'\)[0] == 0}, \ y[t], \ t]\), \(y[t_]\ = \ Expand[sol[\([1, 1, 2]\)]\ ]; \nPrint["\", y[t]]; \)}], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Clear[t, y]; \n y[t_]\ = \ UnitStep[\(-\[Pi]\) + t] + Cos[t]\ UnitStep[\(-\[Pi]\) + t]; \nPlot[y[t], {t, 0, 2 \[Pi]}, PlotRange -> {{0, 2 \[Pi]}, {0, 2}}, PlotStyle \[Rule] {RGBColor[0, 0, 1]}, AxesLabel \[Rule] {"\", "\"}, Ticks -> {Range[0, 2 \[Pi], \[Pi]\/2], Range[0, 2, 0.5]}]; \n Print["\", y[t]]; \)], "Input"], Cell["\<\ This differs from the traditional mathematics solution by including \ a dirac function term.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["GoTo ", FontSize->16, FontColor->RGBColor[0, 1, 0]], ButtonBox["Chapter 11", ButtonData:>"CHAPTER", ButtonStyle->"Hyperlink"] }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "Section 11.8 ", StyleBox["Multiplication and Division by t", FontColor->RGBColor[1, 0, 1]] }], "Text", Evaluatable->False, AspectRatioFixed->True, FontSize->18, CellTags->"Section 11.8"], Cell[TextData[{ StyleBox["Theorem 11.17, Page 438.", FontWeight->"Bold"], " ", StyleBox["(Multiplication by t)", FontColor->RGBColor[1, 0, 1]], " If ", Cell[BoxData[ \(F \((s)\)\)]], " is the Laplace transform of ", Cell[BoxData[ \(f \((t)\)\)]], ", then ", Cell[BoxData[ RowBox[{ RowBox[{ StyleBox["\[ScriptCapitalL]", FontSize->16], \((t\ f \((t)\))\)}], "=", RowBox[{ RowBox[{"-", " ", SuperscriptBox["F", "\[Prime]", MultilineFunction->None]}], \((s)\)}]}]]], ".\n\n", StyleBox["Theorem 11.18, Page 438.", FontWeight->"Bold"], " ", StyleBox["(Division by t)", FontColor->RGBColor[1, 0, 1]], " Let both ", Cell[BoxData[ \(f \((t)\)\)]], " and ", Cell[BoxData[ FractionBox[ StyleBox[\(f \((t)\)\), FontSize->12], StyleBox["t", FontSize->12]]]], " have Laplace transforms and let ", Cell[BoxData[ \(F \((s)\)\)]], " denote the Laplace transform of ", Cell[BoxData[ \(f \((t)\)\)]], ". If ", Cell[BoxData[ RowBox[{ StyleBox[ SubscriptBox["lim", StyleBox[\(t \[Rule] \(0\^+\)\), FontSize->10]], FontSize->12], FractionBox[ StyleBox[\(f \((t)\)\), FontSize->12], StyleBox["t", FontSize->12]]}]]], " exists\nthen ", Cell[BoxData[ RowBox[{ RowBox[{ StyleBox["\[ScriptCapitalL]", FontSize->16], RowBox[{"(", FractionBox[ StyleBox[\(f \((t)\)\), FontSize->12], StyleBox["t", FontSize->12]], ")"}]}], "=", RowBox[{ SubsuperscriptBox[ StyleBox["\[Integral]", FontSize->18], StyleBox["s", FontSize->10], StyleBox["\[Infinity]", FontSize->10]], \(F \((\[Sigma])\) \[DifferentialD]\[Sigma]\)}]}]]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "Load the following ", StyleBox["Mathematica", FontSlant->"Italic"], " packages. Make sure this is done only ONCE in a ", StyleBox["Mathematica", FontSlant->"Italic"], " session." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Get["\"]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["Example 11.21, Page 439.", FontWeight->"Bold"], "\t Use Theorem 11.17 and find \[ScriptCapitalL](t cos(b t)).\n\n\ Solution: Use the fact that \[ScriptCapitalL](cos(bt)) = ", Cell[BoxData[ FractionBox[ StyleBox["s", FontSize->14], StyleBox[\(s\^2\ + \ b\^2\), FontSize->14]]]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[b, f, F, s, t]; \nf[t_]\ = \ Cos[b\ t]; \n F[s_]\ = \(\ s\)\/\(s\^2\ + \ b\^2\); \n Print["\< f[t] = \>", f[t]]; \nPrint["\< L[f[t]] = \>", F[s]]; \n Print["\< \>"]; \nPrint["\< t f[t] = \>", t\ f[t]]; \n Print["\"]; \nPrint["\", \(F'\)[s]]; \nPrint["\", Together[\(F'\)[s]]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "We can check this with ", StyleBox["Mathematica", FontSlant->"Italic"], "'s result using the LaplaceTransform package." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Ft[s_]\ = \ Together[LaplaceTransform[t\ f[t], \ t, \ s]\ ]; \n Print["\< t f[t] = \>", t\ f[t]]; \nPrint["\", Ft[s]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 11.22, Page 439.", FontWeight->"Bold"], "\t Use Theorem 11.18 and find \[ScriptCapitalL]( ", Cell[BoxData[ FractionBox[ StyleBox[\(sin \((t)\)\), FontSize->14], StyleBox["t", FontSize->14]]]], ").\nSolution: Use the fact that \[ScriptCapitalL](sin(t)) = ", Cell[BoxData[ FractionBox[ StyleBox["1", FontSize->14], StyleBox[\(s\^2\ + \ 1\), FontSize->14]]]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[f, F, s, t, Y]; \nf[t_]\ = \ Sin[t]; \n F[s_]\ = \ 1\/\(s\^2\ + \ 1\); \n Y[s_]\ = \ Integrate[F[r], {r, s, \[Infinity]}, Assumptions \[Rule] {Im[s] == 0, s > 0}]; \n Print["\< f[t] = \>", f[t]]; \nPrint["\< F[s]= \>", F[s]]; \n Print["\< \>"]; \nPrint[\*"\"\< \!\(f[t]\/t\) = \>\"", f[t]\/t]; \n Print[\*"\"\\""]; \nPrint[\*"\"\\"", Y[s]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "We can check this with ", StyleBox["Mathematica", FontSlant->"Italic"], "'s result using the LaplaceTransform package." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Y[s_]\ = \ LaplaceTransform[f[t]\/t, \ t, \ s]; \n Print[\*"\"\< \!\(f[t]\/t\) = \>\"", f[t]\/t]; \n Print[\*"\"\\"", Y[s]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 11.23, Page 439.", FontWeight->"Bold"], "\t Solve the initial value problem\n t ", Cell[BoxData[ \(y\^\(''\)\)]], "(t) - t ", Cell[BoxData[ \(\(y\^'\)\)]], "(t) - y(t) = 0 with y(0) = 0 and ", Cell[BoxData[ \(\(y\^'\)\)]], "(0) = 0." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[eqn, f, F, Ly, s, solset, t, y, y0, Y]; \ny0\ = \ 0; \n f[t_]\ = 0; \nF[s_]\ = 0; \nPrint["\< y[0] = \>", y0]; \n Print["\< f[t] = \>", f[t]]; \nPrint["\< F[s] = \>", F[s]]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Clear[Y]; \n eqn\ \ = \ \ \((\(-\ s\^2\) \(Y'\)[s]\ - \ 2\ s\ Y[s]\ + \ y0)\)\ - \((\(-\ s\)\ \(Y'\)[s]\ - \ Y[s])\)\ - \ Y[s]\ \ == \ F[s]; \n Print[eqn]; \nsolset\ = \ DSolve[eqn, \ Y[s], s]; \nPrint[solset]; \n Y[s_]\ = \ solset\[LeftDoubleBracket]1, 1, 2\[RightDoubleBracket]; \n y[t_]\ = \ InverseLaplaceTransform[Y[s], \ s, \ t]; \n Ly\ = \ t\ \(\(y'\)'\)[t]\ - \ t\ \(y'\)[t]\ - \ y[t]; \n Print["\", Y[s]]; \nPrint["\< \>"]; \n Print["\", y[t]]; \nPrint["\", \(y'\)[t]]; \n Print["\", \(\(y'\)'\)[t]]; \nPrint["\< \>"]; \n Print["\", Ly]; \n Print["\", Expand[Ly]]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Clear[t, y]; \ny[t_]\ = \ t\ \[ExponentialE]\^t; \n Plot[y[t], {t, 0, 1}, PlotRange -> {{0, 1}, {0, 3}}, PlotStyle \[Rule] {RGBColor[0, 0, 1]}, AxesLabel \[Rule] {"\", "\"}, Ticks -> {Range[0, 1, 0.25], Range[0, 3, 0.5]}]; \n Print["\", y[t]]; \)], "Input"], Cell[TextData[{ StyleBox["GoTo ", FontSize->16, FontColor->RGBColor[0, 1, 0]], ButtonBox["Chapter 11", ButtonData:>"CHAPTER", ButtonStyle->"Hyperlink"] }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "Section 11.9 ", StyleBox["Inverting the Laplace Transform", FontColor->RGBColor[1, 0, 1]] }], "Text", Evaluatable->False, AspectRatioFixed->True, FontSize->18, CellTags->"Section 11.9"], Cell[TextData[{ StyleBox["Theorem 11.19, Page 441.", FontWeight->"Bold"], " ", StyleBox["(Nonrepeated Linear Factors)", FontColor->RGBColor[1, 0, 1]], " Let ", Cell[BoxData[ \(P \((s)\)\)]], " be a poynomial of degree at most ", Cell[BoxData[ \(n - 1\)]], ". If ", Cell[BoxData[ \(Q \((s)\)\)]], " has degree n, and has distinct complex roots ", Cell[BoxData[ RowBox[{ StyleBox[\(a\_1\), FontSize->14], ",", StyleBox[\(a\_2\), FontSize->14], StyleBox[",", FontSize->14], "\[CenterEllipsis]", StyleBox[",", FontSize->14], StyleBox[\(a\_n\), FontSize->14]}]]], ", then\n\t", Cell[BoxData[ RowBox[{\(Y \((s)\)\), "=", RowBox[{ FractionBox[ StyleBox[\(P \((s)\)\), FontSize->12], StyleBox[\(Q \((s)\)\), FontSize->12]], "=", RowBox[{ FractionBox[ StyleBox[\(P \((s)\)\), FontSize->12], StyleBox[ \(\((s - a\_1)\) \((s - a\_2)\) \[CenterEllipsis] \((s - a\_n)\)\), FontSize->12]], "=", " ", RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(k = 1\), FontSize->10], StyleBox["n", FontSize->10]], FractionBox[ StyleBox[\(Res[Y, a\_k]\), FontSize->14], StyleBox[\(s\ - \ a\_k\), FontSize->14]]}]}]}]}]]], ".\n\n", StyleBox["Theorem 11.20, Page 441.", FontWeight->"Bold"], " ", StyleBox["(A Repeated Linear Factors)", FontColor->RGBColor[1, 0, 1]], " If ", Cell[BoxData[ \(P \((s)\)\)]], " and ", Cell[BoxData[ \(Q \((s)\)\)]], " are polynomials of degree ", Cell[BoxData[ \(\[Mu]\)]], " and ", Cell[BoxData[ \(\[Nu]\)]], ", respectively and ", Cell[BoxData[ \(\[Mu] < \[Nu] + n\)]], " and ", Cell[BoxData[ \(Q \((a)\) \[NotEqual] 0\)]], ", then\n\t", Cell[BoxData[ RowBox[{\(Y \((s)\)\), "=", RowBox[{ FractionBox[ StyleBox[\(P \((s)\)\), FontSize->12], StyleBox[ RowBox[{ SuperscriptBox[ RowBox[{"(", RowBox[{"s", " ", "-", StyleBox[" ", FontSize->12], "a"}], ")"}], "n"], "Q", \((s)\)}], FontSize->12]], "=", " ", RowBox[{ RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(k = 1\), FontSize->10], StyleBox["n", FontSize->10]], FractionBox[ SubscriptBox[ StyleBox["A", FontSize->14], "k"], StyleBox[ SuperscriptBox[ RowBox[{"(", RowBox[{"s", " ", "-", StyleBox[" ", FontSize->12], "a"}], ")"}], "k"], FontSize->12]]}], " ", "+", " ", \(R \((s)\)\)}]}]}]]], ",\nwhere ", Cell[BoxData[ \(R \((s)\)\)]], " is the sum of all partial fractions that do not involve factors of the \ form ", Cell[BoxData[ StyleBox[ SuperscriptBox[ RowBox[{"(", RowBox[{"s", " ", "-", StyleBox[" ", FontSize->12], "a"}], ")"}], "j"], FontSize->12]]], ". Furthermore, the coefficients ", Cell[BoxData[ SubscriptBox[ StyleBox["A", FontSize->14], "k"]]], " can be computed with the formula\n\t", Cell[BoxData[ RowBox[{ SubscriptBox[ StyleBox["A", FontSize->14], "k"], "=", RowBox[{ FractionBox[ StyleBox["1", FontSize->12], StyleBox[\(\((n\ - \ k)\)!\), FontSize->12]], RowBox[{ UnderscriptBox["lim", StyleBox[\(s \[Rule] a\), FontSize->10]], RowBox[{ SubscriptBox["\[PartialD]", StyleBox[\({s, n - k}\), FontSize->10]], FractionBox[ StyleBox[\(P \((s)\)\), FontSize->12], StyleBox[\(Q \((s)\)\), FontSize->12]]}]}]}]}]]], " for ", Cell[BoxData[ \(k = 1, 2, \[CenterEllipsis], n\)]], ". \n\n", StyleBox["Theorem 11.21, Page 442.", FontWeight->"Bold"], " ", StyleBox["(Irreducible Quadratic Factors)", FontColor->RGBColor[1, 0, 1]], " Let ", Cell[BoxData[ \(P \((s)\)\)]], " and ", Cell[BoxData[ \(P \((s)\)\)]], " be polynomials with real coefficients such that the degree of ", Cell[BoxData[ \(P \((s)\)\)]], " is at most 1 larger than the degree of ", Cell[BoxData[ \(Q \((s)\)\)]], ". If ", Cell[BoxData[ \(T \((s)\)\)]], " does not have a factor of the form ", Cell[BoxData[ \(\((s - a)\)\^2 + b\^2\)]], ", then\n\t", Cell[BoxData[ RowBox[{\(Y \((s)\)\), "=", RowBox[{ FractionBox[ StyleBox[\(P \((s)\)\), FontSize->12], StyleBox[\(Q \((s)\)\), FontSize->12]], "=", RowBox[{ FractionBox[ StyleBox[\(P \((s)\)\), FontSize->12], StyleBox[ RowBox[{ RowBox[{"(", RowBox[{ SuperscriptBox[ RowBox[{"(", RowBox[{"s", StyleBox[" ", FontSize->12], "-", " ", "a"}], ")"}], "2"], " ", "+", " ", \(b\^2\)}], " ", ")"}], "T", \((s)\)}], FontSize->12]], "=", RowBox[{ FractionBox[ StyleBox[\(2\ A \((s\ - \ a\ )\) - \ 2 B\ b\), FontSize->12], StyleBox[\(\((s\ - \ a\ )\)\^2\ \ + \ \ b\^2\), FontSize->12]], " ", "+", " ", \(R \((s)\)\)}]}]}]}]]], ",\nwhere ", Cell[BoxData[ RowBox[{\(A\ + \[ImaginaryI]\ B\), "=", FractionBox[ StyleBox[\(P \((a\ + \ \[ImaginaryI]\ b)\)\), FontSize->12], StyleBox[ RowBox[{ SuperscriptBox["Q", "\[Prime]", MultilineFunction->None], \((a\ + \ \[ImaginaryI]\ b)\)}], FontSize->12]]}]]], ".\n\n", StyleBox["Theorem 11.22, Page 445.", FontWeight->"Bold"], " ", StyleBox["(Inverse Laplace Transform)", FontColor->RGBColor[1, 0, 1]], " Let ", Cell[BoxData[ RowBox[{\(F \((s)\)\), "=", FractionBox[ StyleBox[\(P \((s)\)\), FontSize->12], StyleBox[\(Q \((s)\)\), FontSize->12]]}]]], ", where ", Cell[BoxData[ \(P \((s)\)\)]], " and ", Cell[BoxData[ \(Q \((s)\)\)]], " are polynomials of degree m and n, respectively, and ", Cell[BoxData[ \(n > m\)]], ". The inverse Laplace transform of ", Cell[BoxData[ \(F \((s)\)\)]], " is given by\n\t", Cell[BoxData[ RowBox[{\(f \((t)\)\), "=", RowBox[{ RowBox[{ StyleBox[ SuperscriptBox[ StyleBox["\[ScriptCapitalL]", FontSize->16], \(-1\)], FontSize->16], RowBox[{"(", RowBox[{"F", StyleBox[\((s)\), FontSize->12]}], StyleBox[")", FontSize->12]}]}], "=", " ", RowBox[{"\[Sum]", " ", RowBox[{ StyleBox["Res", FontSize->14], StyleBox["[", FontSize->14], RowBox[{ RowBox[{ StyleBox["F", FontSize->14], StyleBox[\((s)\), FontSize->14], SuperscriptBox["\[ExponentialE]", StyleBox[\(s\ t\), FontSize->10]]}], StyleBox[",", FontSize->14], SubscriptBox[ StyleBox["s", FontSize->14], "k"]}], StyleBox["]", FontSize->14]}]}]}]}]]], ",\nwhere the sum is taken over all the residues of the complex function ", Cell[BoxData[ RowBox[{ StyleBox["F", FontSize->14], StyleBox[\((s)\), FontSize->14], SuperscriptBox["\[ExponentialE]", StyleBox[\(s\ t\), FontSize->10]]}]]], ".\n\n", StyleBox["Theorem 11.23, Page 446.", FontWeight->"Bold"], " ", StyleBox["(Heaviside Expansion Theorem)", FontColor->RGBColor[1, 0, 1]], " Let ", Cell[BoxData[ \(P \((s)\)\)]], " and ", Cell[BoxData[ \(Q \((s)\)\)]], " be polynomials of degree m and n, respectively, where ", Cell[BoxData[ \(n > m\)]], ". If ", Cell[BoxData[ \(Q \((s)\)\)]], " has n distinct simple zeros at the points ", Cell[BoxData[ RowBox[{ SubscriptBox[ StyleBox["s", FontSize->14], "1"], ",", SubscriptBox[ StyleBox["s", FontSize->14], "2"], StyleBox[",", FontSize->14], "\[CenterEllipsis]", StyleBox[",", FontSize->14], SubscriptBox[ StyleBox["s", FontSize->14], "n"]}]]], ", then ", Cell[BoxData[ FractionBox[ StyleBox[\(P \((s)\)\), FontSize->12], StyleBox[\(Q \((s)\)\), FontSize->12]]]], " is the Laplace transform of the function ", Cell[BoxData[ \(f \((t)\)\)]], " given by\n\t", Cell[BoxData[ RowBox[{\(f \((t)\)\), "=", RowBox[{ RowBox[{ StyleBox[ SuperscriptBox[ StyleBox["\[ScriptCapitalL]", FontSize->16], \(-1\)], FontSize->16], RowBox[{"(", FractionBox[ StyleBox[\(P \((s)\)\), FontSize->12], StyleBox[\(Q \((s)\)\), FontSize->12]], StyleBox[")", FontSize->12]}]}], "=", " ", RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(k = 1\), FontSize->10], StyleBox["n", FontSize->10]], RowBox[{ FractionBox[ StyleBox[ RowBox[{"P", RowBox[{"(", SubscriptBox[ StyleBox["s", FontSize->14], "k"], ")"}]}], FontSize->12], StyleBox[ RowBox[{ SuperscriptBox["Q", "\[Prime]", MultilineFunction->None], RowBox[{"(", SubscriptBox[ StyleBox["s", FontSize->14], "k"], ")"}]}], FontSize->12]], SuperscriptBox["\[ExponentialE]", RowBox[{ SubscriptBox[ StyleBox["s", FontSize->12], "k"], StyleBox[" ", FontSize->10], StyleBox["t", FontSize->11]}]]}]}]}]}]]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "Load the following ", StyleBox["Mathematica", FontSlant->"Italic"], " packages. Make sure this is done only ONCE in a ", StyleBox["Mathematica", FontSlant->"Italic"], " session." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Get["\"]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["Example 11.24, Page 442.", FontWeight->"Bold"], " Find the inverse Laplace transform of \n Y(s) = ", Cell[BoxData[ FractionBox[ StyleBox[\(s\^3\ - \ 4 s\ + \ 1\), FontSize->14], StyleBox[\(s \((s - 1)\)\^3\), FontSize->14]]]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[a1, a2, a3, b1, F, PQ, Pr, s, Y]; \n Y[s_]\ = \ \(s\^3\ - \ 4 s\ + \ 1\)\/\(s \((s - 1)\)\^3\); \n Print["\", Y[s]]; \nPrint["\", Apart[Y[s]]]; \)], "Input", AspectRatioFixed->True], Cell["\<\ The partial fraction expansion can be obtained with residues. At s = 1, we compute as follows:\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(PQ\ = \ \(s\^3\ - \ 4 s\ + \ 1\)\/s; \n a3\ = \ Limit[PQ, s \[Rule] 1]; \n a2\ = \ Limit[\[PartialD]\_s\ PQ, s \[Rule] 1]; \n a1\ = 1\/\(2!\)\ Limit[\[PartialD]\_\(s, s\)PQ, s \[Rule] 1]; \n Print[\*"\"\<\!\(a\_3\) = \!\(lim\+\(s \[Rule] 1\)\) \>\"", PQ, "\< = \>", a3]; \n Print[\*"\"\<\!\(a\_2\) = \!\(lim\+\(s \[Rule] 1\)\) \>\"", Together[\[PartialD]\_s\ PQ], "\< = \>", a2]; \n Print[\*"\"\<\!\(a\_1\) = \!\(lim\+\(s \[Rule] 1\)\) \>\"", Together[\[PartialD]\_\(s, s\)PQ], "\< = \>", a1]; \)], "Input", AspectRatioFixed->True], Cell[TextData["At s = 0, we compute as follows:"], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(PR\ = \ \(s\^3\ - \ 4 s\ + \ 1\)\/\((s - 1)\)\^3; \n b1\ = \ Limit[PR, s \[Rule] 0]; \n Print[\*"\"\<\!\(b\_1\) = \!\(lim\+\(s \[Rule] 0\)\) \>\"", PR, "\< = \>", b1]; \)], "Input", AspectRatioFixed->True], Cell[TextData["The partial fraction expansion is:"], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(F[s_]\ = \(\ a3\)\/\((s\ - \ 1)\)\^3\ + \(\ a2\)\/\((s\ - \ 1)\)\^2 + \ a1\/\(s\ - \ 1\)\ + \(\ b1\)\/s; \nPrint["\", F[s]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Or we can use ", StyleBox["Mathematica", FontSlant->"Italic"], "'s command to find the partial fraction expansion." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\(Print["\", Apart[Y[s]]]; \)\)], "Input", AspectRatioFixed->True], Cell[TextData[ "Using the table of Laplace transforms, the solution is:"], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(For[i = 1, i \[LessEqual] 4, \(i++\), Yi\ = \ \(Apart[Y[s]]\)[\([i]\)]; \n yi\ = \ InverseLaplaceTransform[\(Apart[Y[s]]\)[\([i]\)], \ s, \ t]; \n Print[Yi, "\< \>", yi]]; \n y[t_]\ = \ \(-\ t\^2\)\ \[ExponentialE]\^t\ + \ t\ \[ExponentialE]\^t\ + \ 2\ \[ExponentialE]\^t\ - \ 1; \nPrint["\", y[t]]; \)], "Input"], Cell[TextData[{ "We can check this with ", StyleBox["Mathematica", FontSlant->"Italic"], "'s result using the InverseLaplaceTransform package." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(y[t_]\ = \ InverseLaplaceTransform[Y[s], \ s, \ t]; \n Print["\", Apart[Y[s]]]; \nPrint["\", y[t]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 11.25, Page 443.", FontWeight->"Bold"], " Find the inverse Laplace transform of \n F(s) = ", Cell[BoxData[ RowBox[{" ", FractionBox[ StyleBox[\(5 s\), FontSize->14], StyleBox[\(\((s\^2 + 4)\) \((s\^2 + 9)\)\), FontSize->14]]}]]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[a1, a2, A1, A2, b1, b2, B1, B2, F, F1, F2, i, res1, res2, s, Y, Yi, z1, z2]; \nY[s_]\ = \ \(5 s\)\/\(\((s\^2 + 4)\) \((s\^2 + 9)\)\); \nPrint["\", Y[s]]; \nPrint["\", Apart[Y[s]]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "The simple poles in the upper half plane occur at ", Cell[BoxData[ \(z\_1\)], FontSize->14], " = 2i and ", Cell[BoxData[ StyleBox[\(z\_2\), FontSize->14]]], " = 3i." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(z1\ = \ 2 \[ImaginaryI]; \nz2\ = \ 3 \[ImaginaryI]; \n res1\ = \ Residue[Y[s], {s, z1}]; \nres2\ = \ Residue[Y[s], {s, z2}]; \n Print["\", z1, "\<] = \>", res1]; \n Print["\", z2, "\<] = \>", res2]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(a1\ = \ Re[z1]; \nb1\ = \ Im[z1]; \nA1\ = \ Re[res1]; \n B1\ = \ Im[res1]; \na2\ = \ Re[z2]; \nb2\ = \ Im[z2]; \n A2\ = \ Re[res2]; \nB2\ = \ Im[res2]; \n F1[s_]\ = \((2\ A1 \((s\ - \ a1)\)\ - \ 2\ B1\ b1)\)\/\(( \((s\ - \ a1)\)\^2\ + \ b1\^2)\); \n F2[s_]\ = \((2\ A2 \((s\ - \ a2)\)\ - \ 2\ B2\ b2)\)\/\(( \((s\ - \ a2)\)\^2\ + \ b2\^2)\); \nPrint["\", F1[s]]; \nPrint["\", F2[s]]; \n Print["\< F[s] = \>", F1[s]\ + \ F2[s]]; \)], "Input", AspectRatioFixed->True], Cell["The partial fraction expansion for Y(s) is:", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\(Print["\", Apart[Y[s]]]; \)\)], "Input"], Cell[TextData["Using the table of Laplace transforms, the answer is:"], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(For[i = 1, i \[LessEqual] 2, \(i++\), Yi\ = \ \(Apart[Y[s]]\)[\([i]\)]; \n yi\ = \ InverseLaplaceTransform[\(Apart[Y[s]]\)[\([i]\)], \ s, \ t]; \n Print[Yi, "\< \>", yi]]; \n y[t_]\ = \ Cos[2 t]\ - \ Cos[3 t]; \nPrint["\", y[t]]; \)], "Input"], Cell[TextData[{ "We can check this with ", StyleBox["Mathematica", FontSlant->"Italic"], "'s result using the InverseLaplaceTransform package." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(y[t_]\ = \ InverseLaplaceTransform[Y[s], \ s, \ t]; \n Print["\", Apart[Y[s]]]; \nPrint["\", y[t]]; \n Print["\", Expand[y[t]]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 11.26, Page 444.", FontWeight->"Bold"], " Find the inverse Laplace transform of:\n Y(s) = ", Cell[BoxData[ FractionBox[ StyleBox[\(s\^3 + \ s\^2 - s + 1\), FontSize->14], StyleBox[\(s \(\((s + 1)\)\^2\) \((s\^2 + 1)\)\), FontSize->14]]]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[i, s, t, y, Y, Yi]; \n Y[s_]\ = \ \(s\^3 + 3 s\^2 - s + 1\)\/\(s \(\((s + 1)\)\^2\) \((s\^2 + 1)\)\); \n Print["\", Y[s]]; \nPrint["\", Apart[Y[s]]]; \)], "Input"], Cell[TextData["Using the table of Laplace transforms, the answer is:"], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(For[i = 1, i \[LessEqual] 4, \(i++\), Yi\ = \ \(Apart[Y[s]]\)[\([i]\)]; \n yi\ = \ InverseLaplaceTransform[\(Apart[Y[s]]\)[\([i]\)], \ s, \ t]; \n Print[Yi, "\< \>", yi]]; \n y[t_]\ = \ 1\ - \ 2\ t\ \[ExponentialE]\^\(-t\) - \ 2\ \[ExponentialE]\^\(-t\) + \ Sin[t]\ + \ Cos[t]; \nPrint["\", y[t]]; \)], "Input"], Cell[TextData[{ "\n", StyleBox["Example 11.27, Page 444.", FontWeight->"Bold"], "\t Solve the initial value problem\n ", Cell[BoxData[ \(\(y\^'\)\)]], "(t) = y(t) - x(t) with y(0) = 1,\n ", Cell[BoxData[ \(\(x\^'\)\)]], "(t) = 5y(t) - 3x(t) x(0) = 2." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[eqn1, eqn2, s, solset, t, x, x0, X, y, y0, Y]; \nx0\ = \ 2; \n y0\ = \ 1; \n eqn1\ \ = \ \ s\ Y[s]\ - \ y0\ \ == \ \ \ Y[s]\ - \ X[s]; \n eqn2\ \ = \ \ s\ X[s]\ - \ x0\ \ == \ \ 5 Y[s]\ - \ 3 X[s]; \n Print["\", x0]; \nPrint["\", y0]; \nPrint[eqn1]; \n Print[eqn1]; \nsolset\ = \ Solve[{eqn1, eqn2}, \ {X[s], Y[s]}]; \n Print[solset]; \n X[s_]\ = \ solset\[LeftDoubleBracket]1, 1, 2\[RightDoubleBracket]; \n Y[s_]\ = \ solset\[LeftDoubleBracket]1, 2, 2\[RightDoubleBracket]; \n y[t_]\ = \ InverseLaplaceTransform[Y[s], \ s, \ t]; \n x[t_]\ = \ InverseLaplaceTransform[X[s], \ s, \ t]; \n Print["\", X[s]]; \nPrint["\", Y[s]]; \n Print["\< \>"]; \nPrint["\", x[t]]; \n Print["\", y[t]]; \nPrint["\< \>"]; \n Print["\< y'[t] = \>", \(y'\)[t]]; \n Print["\< y[t] - x[t] = \>", Expand[y[t] - x[t]]\ ]; \nPrint["\< \>"]; \n Print["\< x'[t] = \>", \(x'\)[t]]; \n Print["\<5y[t] - 3x[t] = \>", Expand[5 y[t]\ - \ 3 x[t]]]; \)], "Input",\ AspectRatioFixed->True], Cell[TextData[{ "We can check this with ", StyleBox["Mathematica", FontSlant->"Italic"], "'s result using the DSolve package." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[eqn1, eqn2, solset, t, x, y]; \n eqn1\ = \ \(y'\)[t]\ == \ \ y[t]\ - \ x[t]; \n eqn2\ = \ \(x'\)[t]\ == \ 5 y[t]\ - \ 3 x[t]; \nPrint[eqn1]; \n Print[eqn1]; \n solset\ = \ DSolve[{eqn1, eqn2, \ x[0] == 2, \ y[0] == 1}, \ \n \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {y[t], \ x[t]}, \ t]; \n Print[solset]; \n y[t_]\ = \ FullSimplify[solset\[LeftDoubleBracket]1, 1, 2\[RightDoubleBracket]]; \n x[t_]\ = \ FullSimplify[solset\[LeftDoubleBracket]1, 2, 2\[RightDoubleBracket]]; \n Print["\< \>"]; \nPrint["\", x[t]]; \n Print["\", y[t]]; \nPrint["\< \>"]; \n Print["\< y'[t] = \>", \(y'\)[t]]; \n Print["\< y[t] - x[t] = \>", Expand[y[t] - x[t]]\ ]; \nPrint["\< \>"]; \n Print["\< x'[t] = \>", \(x'\)[t]]; \n Print["\<5y[t] - 3x[t] = \>", Expand[5 y[t]\ - \ 3 x[t]]]; \)], "Input",\ AspectRatioFixed->True], Cell[BoxData[ \(ParametricPlot[{x[t], y[t]}, {t, 0, 5}, PlotRange -> {{\(-0.1\), 2}, {\(-0.1\), 1}}, PlotStyle \[Rule] {RGBColor[0, 0, 1]}, AxesLabel \[Rule] {"\", "\"}, Ticks -> {Range[0, 2, 0.5], Range[0, 1, 0.25]}]; \n Print["\", x[t]]; \nPrint["\", y[t]]; \)], "Input"], Cell[TextData[{ "\n", StyleBox["Example 11.28, Page 446.", FontWeight->"Bold"], " \tFind the inverse Laplace transform of \n F(s) = ", Cell[BoxData[ StyleBox[ FractionBox[ StyleBox[\(4 s\ + \ 3\), FontSize->14], StyleBox[\(\ s\^3\ + \ 2 s\^2\ + \ s\ + \ 2\), FontSize->14]], FontSize->14]]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[F, P, Q, s, solset, s1, s2, s3]; \nP[s_]\ = \ 4 s\ + \ 3; \n Q[s_]\ = \ s\^3\ + \ 2 s\^2\ + \ s\ + \ 2; \nF[s_]\ = \ P[s]\/Q[s]; \nPrint["\", P[s]]; \nPrint["\", Q[s]]; \n Print["\", F[s]]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(solset\ = \ Solve[Q[s] == 0, s]; \n s1\ = \ solset\[LeftDoubleBracket]1, 1, 2\[RightDoubleBracket]; \n s2\ = \ solset\[LeftDoubleBracket]2, 1, 2\[RightDoubleBracket]; \n s3\ = \ solset\[LeftDoubleBracket]3, 1, 2\[RightDoubleBracket]; \n Print["\", Q[s]]; \nPrint["\", Factor[Q[s]]]; \n Print["\"]; \nPrint[solset]; \n Print[\*"\"\<\!\(s\_1\) = \>\"", s1]; \n Print[\*"\"\<\!\(s\_2\) = \>\"", s2]; \n Print[\*"\"\<\!\(s\_3\) = \>\"", s3]; \)], "Input"], Cell[TextData[{ "Q(s) has simple zeros at the points ", Cell[BoxData[ StyleBox[\(s\_1\), FontSize->14]]], " = - 2, ", Cell[BoxData[ StyleBox[\(s\_2\), FontSize->14]]], " = i, and ", Cell[BoxData[ StyleBox[\(s\_3\), FontSize->14]]], " = - i." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Print[\*"\"\<\!\(P[\(-2\)]\/\(Q'\)[\(-2\)]\) = \>\"", P[\(-2\)]\/\(Q'\)[\(-2\)]]; \n Print[\*"\"\<\!\(P[\[ImaginaryI]]\/\(Q'\)[\[ImaginaryI]]\) = \>\"", P[\[ImaginaryI]]\/\(Q'\)[\[ImaginaryI]]]; \n Print[\*"\"\<\!\(P[\(-\[ImaginaryI]\)]\/\(Q'\)[\(-\[ImaginaryI]\)]\) = \>\ \"", P[\(-\[ImaginaryI]\)]\/\(Q'\)[\(-\[ImaginaryI]\)]]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(f[t_]\ = P[\(-2\)]\/\(Q'\)[\(-2\)]\ \[ExponentialE]\^\(\(-2\) t\) + P[\[ImaginaryI]]\/\(Q'\)[\[ImaginaryI]]\ \[ExponentialE]\^\(\[ImaginaryI]\ t\) + P[\(-\[ImaginaryI]\)]\/\(Q'\)[\(-\[ImaginaryI]\)]\ \[ExponentialE]\^\(\(-\[ImaginaryI]\)\ t\); \n Print["\", f[t]]; \nPrint["\", ComplexExpand[f[t]]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "We can check this with ", StyleBox["Mathematica", FontSlant->"Italic"], "'s result using the InverseLaplaceTransform package." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(f[t_]\ = \ InverseLaplaceTransform[F[s], \ s, \ t]; \n Print["\", Apart[F[s]]]; \nPrint["\", f[t]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["GoTo ", FontSize->16, FontColor->RGBColor[0, 1, 0]], ButtonBox["Chapter 11", ButtonData:>"CHAPTER", ButtonStyle->"Hyperlink"] }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "Section 11.10 ", StyleBox["Convolution", FontColor->RGBColor[1, 0, 1]] }], "Text", Evaluatable->False, AspectRatioFixed->True, FontSize->18, CellTags->"Section 11.10"], Cell[TextData[{ StyleBox["Theorem 10.24, Page 448.", FontWeight->"Bold"], " ", StyleBox["(Convolution Theorem)", FontColor->RGBColor[1, 0, 1]], " Let", StyleBox[" ", FontSize->14], Cell[BoxData[ \(F \((s)\)\)]], " and", StyleBox[" ", FontSize->14], Cell[BoxData[ \(G \((s)\)\)]], " denote the Laplace transforms of", StyleBox[" ", FontSize->14], Cell[BoxData[ \(f \((t)\)\)]], " and", StyleBox[" ", FontSize->14], Cell[BoxData[ \(g \((t)\)\)]], ", respectively. Then the product ", StyleBox[" ", FontSize->14], Cell[BoxData[ RowBox[{ RowBox[{\(H \((s)\)\), StyleBox[" ", FontSize->14], StyleBox["=", FontSize->14], \(F \((s)\) G \((s)\)\)}], " "}]]], "is the Laplace transform of the convolution of", StyleBox[" ", FontSize->14], Cell[BoxData[ \(f \((t)\)\)]], " and ", StyleBox[" ", FontSize->14], Cell[BoxData[ \(g \((t)\)\)]], ", and is denoted by ", Cell[BoxData[ \(h \((t)\) = \((f*g)\) \((t)\)\)]], ", and has the integral representation \n", Cell[BoxData[{ RowBox[{ RowBox[{\(h \((t)\)\), "=", RowBox[{\(\((f*g)\) \((t)\)\), "=", " ", RowBox[{ SubsuperscriptBox[ StyleBox["\[Integral]", FontSize->18], StyleBox["0", FontSize->10], StyleBox["t", FontSize->10]], \(f \((\[Tau])\)\ g \((t\ - \ \[Tau])\) \[DifferentialD]\[Tau]\)}]}]}], ",", " ", "or"}], RowBox[{\(h \((t)\)\), "=", RowBox[{\(\((g*f)\) \((t)\)\), "=", " ", RowBox[{ SubsuperscriptBox[ StyleBox["\[Integral]", FontSize->18], StyleBox["0", FontSize->10], StyleBox["t", FontSize->10]], \(g \((\[Tau])\)\ f \((t\ - \ \[Tau])\) \(\[DifferentialD]\[Tau].\)\)}]}]}]}]] }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "Load the following ", StyleBox["Mathematica", FontSlant->"Italic"], " packages. Make sure this is done only ONCE in a ", StyleBox["Mathematica", FontSlant->"Italic"], " session." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Get["\"]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["Example 11.29, Page 449.", FontWeight->"Bold"], " Use convolution to find the inverse Laplace transform of \n H(s) \ =", Cell[BoxData[ RowBox[{" ", FractionBox[ StyleBox[\(2 s\), FontSize->14], StyleBox[\(\((s\^2\ + \ 1)\)\^2\), FontSize->14]]}]]], ".\nSolution: H(s) is the product of F(s) = ", Cell[BoxData[ FractionBox[ StyleBox["1", FontSize->14], StyleBox[\(s\^2\ + \ 1\), FontSize->14]]]], " and G(s) = ", Cell[BoxData[ FractionBox[ StyleBox[\(2 s\), FontSize->14], StyleBox[\(s\^2\ + \ 1\), FontSize->14]]]], ", \nwhich arethe Laplace transfoms of f(t) = sin(t) and g(t) = 2 \ cos(t), respectively." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[f, F, g, G, h, H, int, s, t, \[Tau], val]; \n F[s_]\ = \ 1\/\(s\^2\ + \ 1\); \n G[s_]\ = \ \(2 s\)\/\(s\^2\ + \ 1\); \n H[s_]\ = \ \(2 s\)\/\((s\^2\ + \ 1)\)\^2; \n Print["\< F[s] = \>", F[s]]; \nPrint["\< G[s] = \>", G[s]]; \n Print["\", F[s] G[s]]; \nPrint["\< H[s] = \>", H[s]]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(f[t_]\ = \ Sin[t]; \ng[t_]\ = \ 2\ Cos[t]; \n int\ = \ \[Integral]f[t - \[Tau]] g[\[Tau]] \[DifferentialD]\[Tau]; \n h[t_]\ = \ \[Integral]\_0\%t f[t - \[Tau]] g[\[Tau]] \[DifferentialD]\[Tau]; \n Print["\< f[t] = \>", f[t]]; \nPrint["\< g[t] = \>", g[t]]; \n Print["\", f[t - \[Tau]] g[\[Tau]]]; \n Print["\<\[Integral]f[t-\[Tau]]g[\[Tau]]\[DifferentialD]\[Tau] = \>", int]; \nPrint[ "\<\[Integral]f[t-\[Tau]]g[\[Tau]]\[DifferentialD]\[Tau] = \>", Expand[int]]; \n Print[\*"\"\< h[t] = \!\(\[Integral]\_0\%t\)f[t-\[Tau]]g[\[Tau]]\ \[DifferentialD]\[Tau]\>\""]; \nPrint["\< h[t] = \>", h[t]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "We can check this with ", StyleBox["Mathematica", FontSlant->"Italic"], "'s result using the InverseLaplaceTransform package." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(h[t_]\ = \ InverseLaplaceTransform[H[s], \ s, \ t]; \n Print["\", H[s]]; \nPrint["\", h[t]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 11.30, Page 450.", FontWeight->"Bold"], " Use the convolution theorem to solve the integral equation\nf(t) = 2 \ cos(t) - ", StyleBox["\[Integral] (t-\[Tau])f(\[Tau])\[DifferentialD] \[Tau]", FontSize->14], ", where the integral is taken over the interval [0,t].\nSolution: Take \ the Laplace transform of each of the terms and get:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[eqn, f, F, s, solset]; \n eqn\ = \ F[s]\ == \ \(2 s\)\/\(s\^2 + \ 1\)\ - \ \ 1\/s\^2\ F[s]; \n solset\ = \ Solve[eqn, \ F[s]]; \nPrint[eqn]; \nPrint[solset]; \n F[s_]\ = \ Apart[solset\[LeftDoubleBracket]1, 1, 2\[RightDoubleBracket]]; \n Print["\", F[s]]; \)], "Input", AspectRatioFixed->True], Cell[TextData["Using a table of Laplace transforms, the solution is:"], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(f[t_]\ = \ \(-t\)\ Sin[t]\ + \ 2\ Cos[t]; \n Print["\", f[t]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "We can check this with ", StyleBox["Mathematica", FontSlant->"Italic"], "'s result using the InverseLaplaceTransform package." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(f[t_]\ = \ InverseLaplaceTransform[F[s], \ s, \ t]; \n Print["\", F[s]]; \nPrint["\", f[t]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 11.31, Page 451.", FontWeight->"Bold"], " Find the Laplace transform of the Dirac delta function ", StyleBox["\[Delta]", FontFamily->"Symbol"], "(t).\nSolution: For any a > 0 we define ", Cell[BoxData[ StyleBox[\(\[Delta]\_a\), FontSize->14]]], "(t) on the interval [0,a] as follows:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(Clear[a, d, f, F, s, t]\), \(d[a_, t_]\ = \ 1\/a; \n F[a_, s_]\ \ = \ \ \[Integral]\_0\%a d[a, t] \(\[ExponentialE]\^\(\(-s\)\ t\)\) \[DifferentialD]t; \nPrint["\", d[a, t]]; \n Print[\*"\"\\""]; \nPrint["\", F[a, s]]; \n Print["\", Together[F[a, s]]]; \)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "Let a\[Rule]0 to obtain the Laplace transform of ", StyleBox["\[Delta]", FontFamily->"Symbol"], "(t)." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\(Print[\*"\"\\"", Limit[F[a, s], a \[Rule] 0]]; \)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "We can check this with ", StyleBox["Mathematica", FontSlant->"Italic"], "'s result using the LaplaceTransform package." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(f[t_]\ = \ DiracDelta[t]; \n F[s_]\ = \ LaplaceTransform[DiracDelta[t], \ t, \ s]; \n Print["\", f[t]]; \nPrint["\", F[s]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 11.32, Page 452.", FontWeight->"Bold"], " Solve the initial value problem\n ", Cell[BoxData[ \(y\^\(''\)\)]], "(t) + 4", Cell[BoxData[ \(\(y\^'\)\)]], "(t) + 13y(t) = 3 ", StyleBox["\[Delta]", FontFamily->"Symbol"], "(t) with y(0) = 0 and ", Cell[BoxData[ \(\(y\^'\)\)]], "(0) = 0.\nSolution: Taking transforms results in the equation:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[eqn, s, solset, t, y, Y]; \n eqn\ \ = \ \ \((s\^2\ + \ 4 s\ + \ 13)\)\ Y[s]\ \ == \ \ 3; \n solset\ = \ Solve[eqn, \ Y[s]]; \nPrint[eqn]; \nPrint[solset]; \n Y[s_]\ = \ solset\[LeftDoubleBracket]1, 1, 2\[RightDoubleBracket]; \n Print["\", Y[s]]; \)], "Input", AspectRatioFixed->True], Cell["Using a table of Laplace transforms, the solution is:", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(y[_t]\ = \ \(\[ExponentialE]\^\(\(-2\) t\)\) Sin[3 t]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "We can check this with ", StyleBox["Mathematica", FontSlant->"Italic"], "'s result using the InverseLaplaceTransform package." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(y[t_]\ = \ InverseLaplaceTransform[Y[s], \ s, \ t]; \n Print["\", Y[s]]; \nPrint["\", y[t]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 11.33, Page 453.", FontWeight->"Bold"], " Use convolution to solve the initial value problem\n\t", Cell[BoxData[ \(y\^\(''\)\)]], "(t) + y(t) = tan(t)\t with y(0) = 1 and ", Cell[BoxData[ \(\(y\^'\)\)]], "(0) = 2.\nSolution. First solve ", Cell[BoxData[ \(u\^\(''\)\)]], "(t) + u(t) = 0 with u(0) = 1 and ", Cell[BoxData[ \(\(u\^'\)\)]], "(0) = 2.\nSecond solve for ", Cell[BoxData[ \(v\^\(''\)\)]], "(t) + v(t) = tan(t) with v(0) = 0 and ", Cell[BoxData[ \(\(v\^'\)\)]], "(0) = 0.\nThen the desired solution is y(t) = u(t) + v(t). 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