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We make the assumption that the \ mass m is much larger than the mass of the spring so that we can neglect \ the mass of the spring. If there is no motion then the system is in ", StyleBox["static equilibrium", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], ". If the mass is pulled down further and released, then it will undergo \ an oscillatory motion.\n\n\tSuppose that there is no friction to slow down \ the motion of the mass, then we say that the system is ", StyleBox["undamped", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], ". We will determine the ", ButtonBox["Simple Harmonic Motion", ButtonData:>{ URL[ "http://mathworld.wolfram.com/SimpleHarmonicMotion.html"], None}, ButtonStyle->"Hyperlink"], " of this mechanical system by considering the forces acting on the mass \ during the motion. This will lead to a differential equation relating the \ displacement as a function of time. For details see page 412 in the \ textbook. The undamped mechanical system is governed by the linear \ differential equation\n\n\t", Cell[BoxData[ \(m\ U'' \((t)\)\ + \ k\ U \((t)\)\ = \ 0\)]], ", \n\nand the general solution to this D. E. is known to be ", Cell[BoxData[ \(U \((t)\)\ = \ A\ cos \((\[Omega]\ t)\) + B\ sin \((\[Omega]\ t)\), \ \ where\ \ \[Omega] = \@\(k\/m\)\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\tIf we consider frictional forces that slow down the motion of the mass, \ then we say that the system is ", StyleBox["damped", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], ". This is visualized by connecting a dashpot (or shock adsorber) to the \ mass. For small velocities it is assumed that the frictional force is \ proportional to velocity. The damping constant c must be positive. For \ details see page 414 in the textbook. The ", ButtonBox["Damped Mechanical Motion", ButtonData:>{ URL[ "http://mathworld.wolfram.com/DampedSimpleHarmonicMotion.html"], None}, ButtonStyle->"Hyperlink"], " is governed by the linear differential equation\n\n\t", Cell[BoxData[ \(m\ U'' \((t)\)\ + \ c\ U' \((t)\)\ + \ k\ U \((t)\)\ = \ 0\)]], ". " }], "Text"], Cell[TextData[{ "\tThe two cases mentioned above are called ", StyleBox["free vibrations", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " because all the forces that affect the motion of the system are internal \ to the system. We extend our analysis to cover the case in which an external \ force F(t) acts on the mass. Such a force might occur from vibrations of \ the support to which the top of the spring is attached, or from the effect of \ a magnetic field on a mass made of iron. For details see page 414 in the \ textbook. The forced motion of the mechanical system satisfies the \ nonhomogeneous linear differential equation \n\n\t", Cell[BoxData[ \(m\ U'' \((t)\)\ + \ c\ U' \((t)\)\ + \ k\ U \((t)\)\ = \ F \((t)\)\)]], ". \n\nThe function F(t) is called the ", StyleBox["input", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " or ", StyleBox["driving force", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " and the solution U(t) is called the ", StyleBox["output", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " or ", StyleBox["response", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], ". Of particular interest are periodic inputs F(t) that can be \ represented by Fourier Series. " }], "Text"], Cell[TextData[{ "\tFor damped mechanical systems that are driven by a periodic input F(t), \ the general solution involves a transient part that vanishes as ", Cell[BoxData[ \(t\ \[Rule] \ \(+\[Infinity]\)\)]], ", and a steady state part that is periodic. The transient part of the \ solution ", Cell[BoxData[ \(\(U\_h\) \((t)\)\)]], " is found by solving the homogeneous differential equation\n\n\t", Cell[BoxData[ \(m\ \(U\_h\%\(\(\ \)\(''\)\)\) \((t)\)\ + \ c\ \(U\_h\%\(\(\ \)\('\)\)\) \((t)\)\ + \ k\ \(U\_h\) \((t)\)\ = \ 0\)]], ". \n\nThis D. E. leads to the characteristic equation ", Cell[BoxData[ \(m\ \[Lambda]\^2\ + \ c\ \[Lambda]\ + \ k\ = \ 0\)]], ", which has roots ", Cell[BoxData[ \(\[Lambda]\ = \ \(\(-\ c\)\ \ \[PlusMinus] \ \ \@\(c\^2\ \ - \ \ 4 \ m\ k\)\)\/\(2 m\)\)]], ". The constants m, c, and k are all positive, and there are three cases \ to consider.\n\n\tIf ", Cell[BoxData[ \(c\^2\ \ - \ \ 4 m\ k\ > 0\)]], ", then the roots are real and distinct, and since ", Cell[BoxData[ \(\@\(c\^2\ \ - \ \ 4 m\ k\) < c\)]], ", it follows that the roots ", Cell[BoxData[ \(\[Lambda]\_1\ \ and\ \ \[Lambda]\_2\)]], " are negative real numbers. Thus for this case we have \n\n\t", Cell[BoxData[ \(\(U\_h\) \((t)\)\ = \ \(A\_1\) \[ExponentialE]\^\(\(\[Lambda]\_1\) t\ \)\ + \ \(A\_2\) \[ExponentialE]\^\(\(\[Lambda]\_2\) t\)\)]], ", and ", Cell[BoxData[ \(lim\+\(t\ \[Rule] \ \(+\[Infinity]\)\)\ \(U\_h\) \((t)\) = 0\)]], ".\n\nIf ", Cell[BoxData[ \(c\^2\ \ - \ \ 4 m\ k\ = 0\)]], ", then the roots are real and equal, ", Cell[BoxData[ \(\[Lambda]\_1\ = \ \(\[Lambda]\_2 = \[Lambda]\)\)]], ", where ", Cell[BoxData[ \(\[Lambda]\)]], " is a negative real number, and for this case we have \n\n\t", Cell[BoxData[ \(\(U\_h\) \((t)\)\ = \ \(A\_1\) \[ExponentialE]\^\[Lambda]t\ + \ \(A\ \_2\) t\ \[ExponentialE]\^\[Lambda]t\)]], ", and ", Cell[BoxData[ \(lim\+\(t\ \[Rule] \ \(+\[Infinity]\)\)\ \(U\_h\) \((t)\) = 0\)]], ".\n\nIf ", Cell[BoxData[ \(c\^2\ \ - \ \ 4 m\ k\ < 0\)]], ", then the roots are complex conjugates, ", Cell[BoxData[ \(\[Lambda] = \(-\ \[Alpha]\)\ \[PlusMinus] \ \[ImaginaryI]\ \ \[Beta]\)]], ", where ", Cell[BoxData[ \(\[Alpha]\ \ and\ \ \[Beta]\)]], " are positive real number, and it follows that\n\n\t", Cell[BoxData[ \(\(U\_h\) \((t)\)\ = \ \(A\_1\) \(\[ExponentialE]\^\(\(-\ \[Alpha]\)\ \ t\)\) cos \((\[Beta]\ t)\)\ + \ \(A\_2\) \[ExponentialE]\^\(\(-\ \[Alpha]\)\ \ t\)\ sin \((\[Beta]\ t)\)\)]], ", and ", Cell[BoxData[ \(lim\+\(t\ \[Rule] \ \(+\[Infinity]\)\)\ \(U\_h\) \((t)\) = 0\)]], ".\n\nIn all three cases, we see that the homogeneous solution ", Cell[BoxData[ \(\(U\_h\) \((t)\)\)]], " decays to 0 as ", Cell[BoxData[ \(t\ \[Rule] \ \(+\[Infinity]\)\)]], ". \n\n\tThe steady state solution ", Cell[BoxData[ \(\(U\_p\) \((t)\)\)]], " can be obtained by representing ", Cell[BoxData[ \(\(U\_p\) \((t)\)\)]], " by its Fourier series and substituting ", Cell[BoxData[ \(\(U\_p\%\(\(\ \)\(''\)\)\) \((t)\), \(U\_p\%\(\(\ \)\('\)\)\) \ \((t)\), \ and\ \ \(U\_p\) \((t)\)\)]], " into the nonhomogeneous differential equation and solving the resulting \ system for the Fourier coefficients of ", Cell[BoxData[ \(\(U\_p\) \((t)\)\)]], ". The ", StyleBox["general solution", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " to the nonhomogeneous linear differential equation is \n\n\t", Cell[BoxData[ \(U \((t)\)\ = \ \(U\_h\) \((t)\)\ + \ \(U\_p\) \((t)\)\)]], "." }], "Text"], Cell[TextData[{ StyleBox["\n", FontWeight->"Bold"], StyleBox["Example 11.4, Page 504.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Find the general solution to ", Cell[BoxData[ \(U'' \((t)\) + 2\ U' \((t)\) + U \((t)\) = F \((t)\)\)]], ", where F(t) is given by the Fourier series ", Cell[BoxData[ \(F[ t] = \[Sum]\+\(k = 1\)\%\[Infinity] Cos[\((2 k\ - \ 1)\) t]\/\((2 \ k\ - \ 1)\)\^2\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 11.4.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text"], Cell["\<\ First, find the transient solution to the differential \ equation.\ \>", "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[c, U, t];\)\ \), "\n", \(\(de = \(U''\)[t] + 2\ \(U'\)[t] + U[t] \[Equal] 0;\)\ \), "\n", \(\(solset = DSolve[de, U[t], t];\)\ \), "\[IndentingNewLine]", \(\(Uh[t_] = ReplaceAll[ solset\_\(\(\[LeftDoubleBracket]\)\(1, 1, \ 2\)\(\[RightDoubleBracket]\)\), {C[1] \[Rule] c\_1, C[2] \[Rule] c\_2}];\)\ \), "\[IndentingNewLine]", \(\(Print[de];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[solset];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\"];\)\ \), "\ \[IndentingNewLine]", \(\(Print[\*"\"\<\!\(U\_h\)[t] = \>\"", Uh[t]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell["\<\ Next, use Fourier series to construct the steady state solution to the \ differential equation. Enter the formula for the n-th terms T[n,t] and \ F[n,t] of the Fourier Series for U[t] and F[t], respectively. Then form the \ equation relating the n-terms and the equations involving Cos[n t] and Sin[n \ t]. Then solve these equations for the Fourier coefficients of U[t].\ \>", \ "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[a, b, A, B, S, t, T, U];\)\ \), "\[IndentingNewLine]", \(\(Clear[EqA, EqB, f, F, k, L, m, n];\)\ \), "\n", \(\(T[n_, t_]\ = \ A\_n\ Cos[n\ t]\ + \ B\_n\ Sin[n\ t];\)\ \), "\n", \(\(F[0, t_]\ = \ 0;\)\ \), "\n", \(\(F[n_, t_]\ = \ \(\(1 - \((\(-1\))\)\^n\)\/2\) Cos[n\ t]\/n\^2;\)\ \), "\n", \(\(Print["\", T[n, t]];\)\ \), "\n", \(\(Print["\", F[n, t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[ Eq\ = \ \[PartialD]\_\(t, t\)T[n, t]\ + \ 2\ \[PartialD]\_t\ T[n, t]\ + \ T[n, t]\ \[Equal] \ F[n, t]];\)\ \), "\n", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[ EqA\ = \ ReplaceAll[ Eq, {Cos[n\ t]\ \[Rule] \ 1, Sin[n\ t]\ \[Rule] \ 0}]];\)\ \), "\n", \(\(Print[ EqB\ = \ ReplaceAll[ Eq, {Sin[n\ t]\ \[Rule] \ 1, Cos[n\ t]\ \[Rule] \ 0}]];\)\ \), "\n", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(solset\ = \ Solve[{EqA, EqB}, \ {A\_n, B\_n}];\)\ \), "\n", \(\(a\_n_\ = \ solset\_\(\(\[LeftDoubleBracket]\)\(1, 1, \ 2\)\(\[RightDoubleBracket]\)\);\)\ \), "\n", \(\(b\_n_\ = \ solset\_\(\(\[LeftDoubleBracket]\)\(1, 2, \ 2\)\(\[RightDoubleBracket]\)\);\)\ \), "\n", \(\(a\_0\ \ = \ 2\ F[0, t];\)\ \), "\n", \(\(Print[\*"\"\<\!\(a\_0\) = \>\"", a\_0];\)\ \), "\n", \(\(Print[\*"\"\<\!\(a\_n\) = \>\"", a\_n];\)\ \), "\n", \(\(Print[\*"\"\<\!\(b\_n\) = \>\"", b\_n];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell["\<\ Use the Fourier coefficients of U[t] and form the n-th term of the series \ U[n,t], and verify that it satisfies the given D.E.\ \>", "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(U[n_, t_]\ = \ a\_n\ Cos[n\ t]\ + \ b\_n\ Sin[n\ t];\)\ \), "\n", \(\(U[0, t]\ \ \ = \(\(\ \)\(a\_0\)\)\/2;\)\ \), "\n", \(\(Print["\", U[0, t]];\)\ \), "\n", \(\(Print["\", U[n, t]];\)\ \), "\n", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\"];\)\ \), "\n", \(\(L\ = \ \[PartialD]\_\(t, t\)U[n, t]\ + \ 2 \[PartialD]\_t U[n, t]\ + \ U[n, t];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[PartialD]\_\(t, t\)\)U[n,t] + \ 2\!\(\[PartialD]\_t\)U[n,t] + U[n,t] = \>\"", F[n, t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[L, "\< = \>", F[n, t]];\)\ \), "\n", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[MapAll[Together, L], "\< = \>", F[n, t]];\)\ \), "\n", \(\(Print[ MapAll[Together, L \[Equal] F[n, t]]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\nConstruct the Trigonometric Polynomial ", Cell[BoxData[ \(S\_5[t]\)], AspectRatioFixed->True], " of degree n = 5 and verify that it satisfies the D.E." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(S[ t_]\ = \ \[Sum]\+\(n = 0\)\%5 U[n, t];\)\ \), "\n", \(\(L\ = \ \(S''\)[t]\ + \ 2\ \(S'\)[t]\ + \ S[t];\)\ \), "\n", \(\(Print[\*"\"\<\!\(S\_5\)[t] = \!\(\[Sum]\+\(n = 0\)\%5\)U[n,t] = \ \>\"", S[t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(S\_5\)'[t] = \>\"", \(S'\)[t]];\)\ \), "\n", \(\(Print[\*"\"\<\!\(S\_5\)''[t] = \>\"", \(\(S'\)'\)[t]];\)\ \), "\n", \(\(Print["\< \>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(S\_5\)''[t] + 2 \!\(S\_5\)'[t] + \!\(S\_5\)[t] = \ \>\"", Expand[L]];\)\ \), "\n", \(\(f[t_]\ = \ \[Sum]\+\(n = 0\)\%5 F[n, t];\)\ \), "\n", \(\(Print[\*"\"\<\!\(f\_5\)[t] = \!\(\[Sum]\+\(n = 0\)\%5\)F[n,t] = \ \>\"", f[t]];\)\ \), "\n", \(\(Print["\< \>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(S\_5\)''[t] + 2 \!\(S\_5\)'[t] + \!\(S\_5\)[t] \ \[Equal] \!\(f\_5\)[t]\>\""];\)\ \), "\n", \(\(Print[Expand[L], "\< \[Equal] \>", f[t]];\)\ \), "\n", \(\(Print[Expand[L]\ \[Equal] \ f[t]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\nLook at the size of the coefficients ", Cell[BoxData[ \(a\_n\ \ and\ \ b\_n\)]], " and observe that\n", Cell[BoxData[{ \(lim\_\(n \[Rule] \[Infinity]\)\ a\_n = \(lim\_\(n \[Rule] \[Infinity]\)\ \(\((\(-1\)\ + \ \ \((\(-1\))\)\^n)\)\ \((\(-1\)\ + \ n\^2)\)\)\/\(2\ n\^2\ \((1\ + \ \ n\^2)\)\^2\)\ = \ 0\)\), "\n", \(lim\_\(n \[Rule] \[Infinity]\)\ b\_n = \(lim\_\(n \[Rule] \[Infinity]\)\ \(1\ - \ \ \((\(-1\))\)\^n\)\/\(n\ \((1\ + \ n\^2)\)\^2\)\ = \ 0\)\)}]], " \nWe can investigate these sequences." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(m = 5;\)\ \), "\[IndentingNewLine]", \(\(as = Table[a\_k, {k, 0, m}];\)\ \), "\[IndentingNewLine]", \(\(bs = Table[b\_k, {k, 1, m}];\)\ \), "\[IndentingNewLine]", \(\(as = Append[as, "\<...\>"];\)\ \), "\[IndentingNewLine]", \(\(bs = Append[bs, "\<...\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<{\!\(a\_n\)} = \>\"", as];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<{\!\(b\_n\)} = \>\"", bs];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<{\!\(a\_n\)} = \>\"", NumberForm[N[as], 3]];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<{\!\(b\_n\)} = \>\"", NumberForm[N[bs], 3]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ "\nMaybe ", Cell[BoxData[ \(S\_5[t]\)]], " is a close approximation to the solution. A graph of ", Cell[BoxData[ \(S\_5[t]\)]], " is given below." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(Plot[S[t], {t, \(-2\) Pi, 2 Pi}, PlotStyle \[Rule] Magenta, \[IndentingNewLine]PlotRange \[Rule] {{\(-2\) \[Pi], 2 \[Pi]}, {\(-0.5\), 0.5}}, AxesLabel \[Rule] {"\", "\"}, \[IndentingNewLine]Ticks \ \[Rule] {Range[\(-2\) \[Pi], 2 \[Pi], \[Pi]], Range[\(-0.5\), 0.5, 0.1]}];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\\""];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\\ \""];\)\ \), "\n", \(\(Print[\*"\"\<\!\(S\_5\)[t] = \>\"", S[t]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\nFor illustration purposes, use the approximate solution ", Cell[BoxData[ \(U[t] = U\_h[t] + S\_5[t]\)], AspectRatioFixed->True], " and use it so solve ", Cell[BoxData[ \(U[0] = 1, \(U'\)[0] = 0\)]], ", and plot the result." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(eqns = {Uh[0] + S[0] \[Equal] 1, \(Uh'\)[0] + \(S'\)[0] \[Equal] 0};\)\ \), "\n", \(\(solset = Solve[eqns, {c\_1, c\_2}];\)\ \), "\n", \(\(U[t_] = S[t] + ReplaceAll[Uh[t], solset\_\(\(\[LeftDoubleBracket]\)\(1\)\(\[RightDoubleBracket]\)\)\ ];\)\ \), "\[IndentingNewLine]", \(\(Plot[{U[t], S[t]}, {t, 0, 4 Pi}, PlotStyle \[Rule] {Blue, Magenta}, \[IndentingNewLine]PlotRange \[Rule] {{0, 4 \[Pi]}, {\(-0.5\), 1.05}}, AxesLabel \[Rule] {"\", "\"}, \[IndentingNewLine]Ticks \ \[Rule] {Range[0, 4 \[Pi], \[Pi]], Range[\(-0.5\), 1.0, 0.25]}];\)\ \), "\n", \(\(Print[\*"\"\\""];\)\ \), "\n", \(\(Print[\*"\"\\""];\)\ \), "\n", \(\(Print["\"];\)\ \ \), "\n", \(\(Print["\", U[t]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"] }, Closed]], Cell[TextData[{ StyleBox["\n", FontWeight->"Bold"], StyleBox["Exercise 1 (a), Page 505.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Find the general solution to ", Cell[BoxData[ \(U'' \((t)\) + 3\ U' \((t)\) + U \((t)\) = F \((t)\)\)]], ", where F(t) is given by the Fourier series ", Cell[BoxData[ \(F[ t] = \[Sum]\+\(n = 1\)\%\[Infinity]\((\(-1\))\)\^\(n - 1\)\/n\ Sin[ n\ t]\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 1 (a).", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text"], Cell["\<\ First, find the transient solution to the differential \ equation.\ \>", "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[c];\)\ \), "\[IndentingNewLine]", \(\(Clear[U, t];\)\ \), "\n", \(\(de = \(U''\)[t] + 3\ \(U'\)[t] + U[t] \[Equal] 0;\)\ \), "\n", \(\(solset = DSolve[de, U[t], t];\)\ \), "\[IndentingNewLine]", \(\(Uh[t_] = ReplaceAll[ solset\_\(\(\[LeftDoubleBracket]\)\(1, 1, \ 2\)\(\[RightDoubleBracket]\)\), {C[1] \[Rule] c\_1, C[2] \[Rule] c\_2}];\)\ \), "\[IndentingNewLine]", \(\(Print[de];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[solset];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\"];\)\ \), "\ \[IndentingNewLine]", \(\(Print[\*"\"\<\!\(U\_h\)[t] = \>\"", Uh[t]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell["\<\ Next, use Fourier series to construct the steady state solution to the \ differential equation. Enter the formula for the n-th terms T[n,t] and \ F[n,t] of the Fourier Series for U[t] and F[t], respectively. Then form the \ equation relating the n-terms and the equations involving Cos[n t] and Sin[n \ t]. Then solve these equations for the Fourier coefficients of U[t].\ \>", \ "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[a, b, A, B, S, t, T, U];\)\ \), "\[IndentingNewLine]", \(\(Clear[EqA, EqB, f, F, k, L, m, n];\)\ \), "\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", \(\(Print["\", T[n, t]];\)\ \), "\n", \(\(Print["\", F[n, t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[ Eq\ = \ \[PartialD]\_\(t, t\)T[n, t]\ + \ 3\ \[PartialD]\_t\ T[n, t]\ + \ T[n, t]\ \[Equal] \ F[n, t]];\)\ \), "\n", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[ EqA\ = \ ReplaceAll[ Eq, {Cos[n\ t]\ \[Rule] \ 1, Sin[n\ t]\ \[Rule] \ 0}]];\)\ \), "\n", \(\(Print[ EqB\ = \ ReplaceAll[ Eq, {Sin[n\ t]\ \[Rule] \ 1, Cos[n\ t]\ \[Rule] \ 0}]];\)\ \), "\n", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(solset\ = \ Solve[{EqA, EqB}, \ {A\_n, B\_n}];\)\ \), "\n", \(\(a\_n_\ = \ solset\_\(\(\[LeftDoubleBracket]\)\(1, 1, \ 2\)\(\[RightDoubleBracket]\)\);\)\ \), "\n", \(\(b\_n_\ = \ solset\_\(\(\[LeftDoubleBracket]\)\(1, 2, \ 2\)\(\[RightDoubleBracket]\)\);\)\ \), "\n", \(\(a\_0\ \ = \ 2\ F[0, t];\)\ \), "\n", \(\(Print[\*"\"\<\!\(a\_0\) = \>\"", a\_0];\)\ \), "\n", \(\(Print[\*"\"\<\!\(a\_n\) = \>\"", a\_n];\)\ \), "\n", \(\(Print[\*"\"\<\!\(b\_n\) = \>\"", b\_n];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell["\<\ Use the Fourier coefficients of U[t] and form the n-th term of the series \ U[n,t], and verify that it satisfies the given D.E.\ \>", "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(U[n_, t_]\ = \ a\_n\ Cos[n\ t]\ + \ b\_n\ Sin[n\ t];\)\ \), "\n", \(\(U[0, t]\ \ \ = \(\(\ \)\(a\_0\)\)\/2;\)\ \), "\n", \(\(Print["\", U[0, t]];\)\ \), "\n", \(\(Print["\", U[n, t]];\)\ \), "\n", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\"];\)\ \), "\n", \(\(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]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[L, "\< = \>", F[n, t]];\)\ \), "\n", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[MapAll[Together, L], "\< = \>", F[n, t]];\)\ \), "\n", \(\(Print[ MapAll[Together, L \[Equal] F[n, t]]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\nConstruct the Trigonometric Polynomial ", Cell[BoxData[ \(S\_5[t]\)], AspectRatioFixed->True], " of degree n = 5 and verify that it satisfies the D.E." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(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]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(S\_5\)'[t] = \>\"", \(S'\)[t]];\)\ \), "\n", \(\(Print[\*"\"\<\!\(S\_5\)''[t] = \>\"", \(\(S'\)'\)[t]];\)\ \), "\n", \(\(Print["\< \>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(S\_5\)''[t] + 3 \!\(S\_5\)'[t] + \!\(S\_5\)[t] = \ \>\"", Expand[L]];\)\ \), "\n", \(\(f[t_]\ = \ \[Sum]\+\(n = 0\)\%5 F[n, t];\)\ \), "\n", \(\(Print[\*"\"\<\!\(f\_5\)[t] = \!\(\[Sum]\+\(n = 0\)\%5\)F[n,t] = \ \>\"", f[t]];\)\ \), "\n", \(\(Print["\< \>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(S\_5\)''[t] + 3 \!\(S\_5\)'[t] + \!\(S\_5\)[t] \ \[Equal] \!\(f\_5\)[t]\>\""];\)\ \), "\n", \(\(Print[Expand[L], "\< \[Equal] \>", f[t]];\)\ \), "\n", \(\(Print[Expand[L]\ \[Equal] \ f[t]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\nLook at the size of the coefficients ", Cell[BoxData[ \(a\_n\ \ and\ \ b\_n\)]], " and observe that\n", Cell[BoxData[{ \(lim\_\(n \[Rule] \[Infinity]\)\ a\_n = \(lim\_\(n \[Rule] \[Infinity]\)\ \(3\ \ \((\(-1\))\)\^n\)\/\(1 + 7\ n\^2 + n\^4\)\ = \ 0\)\), "\n", \(lim\_\(n \[Rule] \[Infinity]\)\ b\_n = \(lim\_\(n \[Rule] \[Infinity]\)\ \(\((\(-1\))\)\^n\ \((\(-1\ \) + n\^2)\)\)\/\(n\ \((1 + 7\ n\^2 + n\^4)\)\)\ = \ 0\)\)}]], " \nWe can investigate these sequences." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(m = 5;\)\ \), "\[IndentingNewLine]", \(\(as = Table[a\_k, {k, 0, m}];\)\ \), "\[IndentingNewLine]", \(\(bs = Table[b\_k, {k, 1, m}];\)\ \), "\[IndentingNewLine]", \(\(as = Append[as, "\<...\>"];\)\ \), "\[IndentingNewLine]", \(\(bs = Append[bs, "\<...\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<{\!\(a\_n\)} = \>\"", as];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<{\!\(b\_n\)} = \>\"", bs];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<{\!\(a\_n\)} = \>\"", NumberForm[N[as], 3]];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<{\!\(b\_n\)} = \>\"", NumberForm[N[bs], 3]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ "\nMaybe ", Cell[BoxData[ \(S\_5[t]\)]], " is a close approximation to the solution. 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" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 2 (a).", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text"], Cell["\<\ First, find the transient solution to the differential \ equation.\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[U, t];\)\ \), "\n", \(\(de = 2 \( U''\)[t] + 2\ \(U'\)[t] + U[t] \[Equal] 0;\)\ \), "\n", \(\(solset = DSolve[de, U[t], t];\)\ \), "\[IndentingNewLine]", \(\(Uh[t_] = ReplaceAll[ solset\_\(\(\[LeftDoubleBracket]\)\(1, 1, \ 2\)\(\[RightDoubleBracket]\)\), {C[1] \[Rule] c\_1, C[2] \[Rule] c\_2}];\)\ \), "\[IndentingNewLine]", \(\(Print[de];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[solset];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\"];\)\ \), "\ \[IndentingNewLine]", \(\(Print[\*"\"\<\!\(U\_h\)[t] = \>\"", Uh[t]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell["\<\ Next, use Fourier series to construct the steady state solution to the \ differential equation. Enter the formula for the n-th terms T[n,t] and \ F[n,t] of the Fourier Series for U[t] and F[t], respectively. Then form the \ equation relating the n-terms and the equations involving Cos[n t] and Sin[n \ t]. Then solve these equations for the Fourier coefficients of U[t].\ \>", \ "Text"] }, Open ]], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[a, b, A, B, S, t, T, U];\)\ \), "\[IndentingNewLine]", \(\(Clear[EqA, EqB, f, F, k, L, m, n];\)\ \), "\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", \(\(Print["\", T[n, t]];\)\ \), "\n", \(\(Print["\", F[n, t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[ Eq\ = \ 2\ \[PartialD]\_\(t, t\)T[n, t]\ + \ 2\ \[PartialD]\_t\ T[n, t]\ + \ T[n, t]\ \[Equal] \ F[n, t]];\)\ \), "\n", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[ EqA\ = \ ReplaceAll[ Eq, {Cos[n\ t]\ \[Rule] \ 1, Sin[n\ t]\ \[Rule] \ 0}]];\)\ \), "\n", \(\(Print[ EqB\ = \ ReplaceAll[ Eq, {Sin[n\ t]\ \[Rule] \ 1, Cos[n\ t]\ \[Rule] \ 0}]];\)\ \), "\n", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(solset\ = \ Solve[{EqA, EqB}, \ {A\_n, B\_n}];\)\ \), "\n", \(\(a\_n_\ = \ solset\_\(\(\[LeftDoubleBracket]\)\(1, 1, \ 2\)\(\[RightDoubleBracket]\)\);\)\ \), "\n", \(\(b\_n_\ = \ \ solset\_\(\(\[LeftDoubleBracket]\)\(1, 2, 2\)\(\ \[RightDoubleBracket]\)\);\)\ \), "\n", \(\(a\_0\ \ = \ 2\ F[0, t];\)\ \), "\n", \(\(Print[\*"\"\<\!\(a\_0\) = \>\"", a\_0];\)\ \), "\n", \(\(Print[\*"\"\<\!\(a\_n\) = \>\"", a\_n];\)\ \), "\n", \(\(Print[\*"\"\<\!\(b\_n\) = \>\"", b\_n];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell["\<\ Use the Fourier coefficients of U[t] and form the n-th term of the series \ U[n,t], and verify that it satisfies the given D.E.\ \>", "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(U[n_, t_]\ = \ a\_n\ Cos[n\ t]\ + \ b\_n\ Sin[n\ t];\)\ \), "\n", \(\(U[0, t]\ \ \ = \(\(\ \)\(a\_0\)\)\/2;\)\ \), "\n", \(\(Print["\", U[0, t]];\)\ \), "\n", \(\(Print["\", U[n, t]];\)\ \), "\n", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\"];\)\ \), "\n", \(\(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]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[L, "\< = \>", F[n, t]];\)\ \), "\n", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[MapAll[Together, L], "\< = \>", F[n, t]];\)\ \), "\n", \(\(Print[ MapAll[Together, L \[Equal] F[n, t]]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\nConstruct the Trigonometric Polynomial ", Cell[BoxData[ \(S\_5[t]\)], AspectRatioFixed->True], " of degree n = 5 and verify that it satisfies the D.E." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(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]];\)\ \), 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\[Rule] \[Infinity]\)\ a\_n = \(lim\_\(n \[Rule] \[Infinity]\)\ \(2\ \ \((\(-1\))\)\^n\)\/\(1 + 4\ n\^4\)\ = \ 0\)\), "\n", \(lim\_\(n \[Rule] \[Infinity]\)\ b\_n = \(lim\_\(n \[Rule] \[Infinity]\)\ \(\((\(-1\))\)\^n\ \((\(-1\ \) + 2\ n\^2)\)\)\/\(n\ \((1 + 4\ n\^4)\)\)\ = \ 0\)\)}]], " \nWe can investigate these sequences." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(m = 5;\)\ \), "\[IndentingNewLine]", \(\(as = Table[a\_k, {k, 0, m}];\)\ \), "\[IndentingNewLine]", \(\(bs = Table[b\_k, {k, 1, m}];\)\ \), "\[IndentingNewLine]", \(\(as = Append[as, "\<...\>"];\)\ \), "\[IndentingNewLine]", \(\(bs = Append[bs, "\<...\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<{\!\(a\_n\)} = \>\"", as];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<{\!\(b\_n\)} = \>\"", bs];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<{\!\(a\_n\)} = \>\"", NumberForm[N[as], 3]];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<{\!\(b\_n\)} = \>\"", 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CounterAssignments->{{"Subsubsection", 0}}, FontSize->14, FontWeight->"Bold"], Cell[StyleData["Subsection", "Presentation"], CellMargins->{{36, 10}, {11, 32}}, LineSpacing->{1, 0}, FontSize->22], Cell[StyleData["Subsection", "Condensed"], CellMargins->{{16, Inherited}, {6, 12}}, FontSize->12], Cell[StyleData["Subsection", "Printout"], CellMargins->{{9, 0}, {7, 22}}, FontSize->12] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Subsubsection"], CellDingbat->"\[FilledSmallSquare]", CellMargins->{{22, Inherited}, {8, 18}}, CellGroupingRules->{"SectionGrouping", 50}, PageBreakBelow->False, CounterIncrements->"Subsubsection", FontWeight->"Bold"], Cell[StyleData["Subsubsection", "Presentation"], CellMargins->{{34, 10}, {11, 26}}, LineSpacing->{1, 0}, FontSize->18], Cell[StyleData["Subsubsection", "Condensed"], CellMargins->{{17, Inherited}, {6, 12}}, FontSize->10], Cell[StyleData["Subsubsection", "Printout"], CellMargins->{{9, 0}, {7, 14}}, FontSize->11] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Styles for Body Text", "Section"], Cell[CellGroupData[{ Cell[StyleData["Text"], CellMargins->{{12, 10}, {7, 7}}, LineSpacing->{1, 3}, CounterIncrements->"Text"], Cell[StyleData["Text", "Presentation"], CellMargins->{{24, 10}, {10, 10}}, LineSpacing->{1, 5}], Cell[StyleData["Text", "Condensed"], CellMargins->{{8, 10}, {6, 6}}, LineSpacing->{1, 1}], Cell[StyleData["Text", "Printout"], CellMargins->{{2, 2}, {6, 6}}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["SmallText"], CellMargins->{{12, 10}, {6, 6}}, LineSpacing->{1, 3}, CounterIncrements->"SmallText", FontFamily->"Helvetica", FontSize->9], Cell[StyleData["SmallText", "Presentation"], CellMargins->{{24, 10}, {8, 8}}, LineSpacing->{1, 5}, FontSize->12], Cell[StyleData["SmallText", "Condensed"], CellMargins->{{8, 10}, {5, 5}}, LineSpacing->{1, 2}, FontSize->9], Cell[StyleData["SmallText", "Printout"], CellMargins->{{2, 2}, {5, 5}}, FontSize->7] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Styles 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If a cell's FormatType matches the name of one of the styles \ defined below, then that style is applied between the cell's style and its \ own options.\ \>", "Text"], Cell[StyleData["CellExpression"], PageWidth->Infinity, CellMargins->{{6, Inherited}, {Inherited, Inherited}}, ShowCellLabel->False, ShowSpecialCharacters->False, AllowInlineCells->False, AutoItalicWords->{}, StyleMenuListing->None, FontFamily->"Courier", Background->GrayLevel[1]], Cell[StyleData["InputForm"], AllowInlineCells->False, StyleMenuListing->None, FontFamily->"Courier"], Cell[StyleData["OutputForm"], PageWidth->Infinity, TextAlignment->Left, LineSpacing->{1, -5}, StyleMenuListing->None, FontFamily->"Courier"], Cell[StyleData["StandardForm"], LineSpacing->{1.25, 0}, StyleMenuListing->None, FontFamily->"Courier"], Cell[StyleData["TraditionalForm"], LineSpacing->{1.25, 0}, SingleLetterItalics->True, TraditionalFunctionNotation->True, DelimiterMatching->None, StyleMenuListing->None], Cell["\<\ The style defined below is mixed in to any cell that is in an \ inline cell within another.\ \>", "Text"], Cell[StyleData["InlineCell"], TextAlignment->Left, ScriptLevel->1, StyleMenuListing->None], Cell[StyleData["InlineCellEditing"], StyleMenuListing->None, Background->RGBColor[1, 0.749996, 0.8]] }, Closed]] }, Open ]] }] ] (******************************************************************* Cached data follows. 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