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Bei mehr als 8 \ Punkten gilt die Klausur als bestanden, f\[UDoubleDot]r eine 1,0 ben\ \[ODoubleDot]tigen Sie 20 Punkte, die vier zus\[ADoubleDot]tzlichen Punkte \ dienen dazu eventuelle Rechenfehler oder Ungenauigkeiten zu kompensieren. \ \>", "Text"], Cell["Die L\[ODoubleDot]sungen sind jeweils grau unterlegt.", "Answer"], Cell[CellGroupData[{ Cell["1. Aufgabe", "Section"], Cell[TextData[{ "a) Was versteht man unter einem Fundamentalsystem einer \ Differentialgleichung ?", StyleBox[" (1 Punkt)", FontWeight->"Bold"] }], "Text"], Cell[TextData[{ "Ein Fundamentalsystem sind alle linear unabh\[ADoubleDot]ngigen L\ \[ODoubleDot]sungen ", Cell[BoxData[ \(TraditionalForm\`\(y\_i\)(x)\)]], " einer Differentialgleichung ", Cell[BoxData[ \(TraditionalForm\`n\)]], "-ter Ordnung, die Funktionen sind linear unabh\[ADoubleDot]ngig wenn die \ Wronski-Determinate nicht verschwindet" }], "Answer"], Cell[BoxData[ FormBox[ RowBox[{\(W(\(y\_1\)(x), .. , \(y\_n\)(x))\), "=", RowBox[{"|", GridBox[{ {\(y\_1\), \(y\_2\), "..."}, {\(y\_1'\), \(y\_2'\), "\[CenterEllipsis]"}, {"\[VerticalEllipsis]", "\[VerticalEllipsis]", "\[DescendingEllipsis]"}, {\(y\_1\^\((n - 1)\)\), \(y\_2\^\((n - 1)\)\), "\[CenterEllipsis]"} }], "|", \(\(\[NotEqual]\)\(0\)\)}]}], TraditionalForm]], "Answer"], Cell[TextData[{ "b) Man l\[ODoubleDot]se die gew\[ODoubleDot]hnlichen \ Differentialgleichungen ", StyleBox["(5 Punkte)", FontWeight->"Bold"], ":" }], "Text"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{ SuperscriptBox["y", "\[Prime]", MultilineFunction->None], "(", "x", ")"}], "+", \(\[Lambda]\ \(y(x)\)\)}], "=", "0"}], TraditionalForm]], "DisplayFormula"], Cell[BoxData[ \(TraditionalForm[ y[x] /. \(DSolve[\(y'\)[x] + \[Lambda]\ y[x] \[Equal] 0, y[x], x]\)[\([1]\)]]\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{\(\[ExponentialE]\^\(\(-x\)\ \[Lambda]\)\), " ", SubscriptBox[ TagBox["c", C], "1"]}], TraditionalForm]], "Answer"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{ SuperscriptBox["y", "\[Prime]", MultilineFunction->None], "(", "x", ")"}], "+", \(x\/\[Sigma]\^2\ \(y(x)\)\)}], "=", "0"}], TraditionalForm]], "DisplayFormula"], Cell[BoxData[ \(TraditionalForm[ y[x]\ /. \ \(DSolve[\(y'\)[x] + x\ y[x]/\[Sigma]\^2 \[Equal] 0, y[x], x]\)[\([1]\)]]\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{\(\[ExponentialE]\^\(-\(x\^2\/\(2\ \[Sigma]\^2\)\)\)\), " ", SubscriptBox[ TagBox["c", C], "1"]}], TraditionalForm]], "Answer"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{ SuperscriptBox["y", "\[Prime]\[Prime]", MultilineFunction->None], "(", "x", ")"}], "+", \(\(\[Omega]\^2\) \(y(x)\)\)}], "=", "0"}], TraditionalForm]], "DisplayFormula"], Cell[BoxData[ \(TraditionalForm[ y[x]\ /. \ \(DSolve[\(y''\)[x] + \(\[Omega]\^\(\(2\)\(\ \)\)\) y[x] \[Equal] 0, y[x], x]\)[\([1]\)]]\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ SubscriptBox[ TagBox["c", C], "1"], " ", \(cos(x\ \[Omega])\)}], "+", RowBox[{ SubscriptBox[ TagBox["c", C], "2"], " ", \(sin(x\ \[Omega])\)}]}], TraditionalForm]], "Answer"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{ SuperscriptBox["y", "\[Prime]", MultilineFunction->None], "(", "x", ")"}], "+", \(y(x)\)}], "=", "x"}], TraditionalForm]], "DisplayFormula"], Cell[BoxData[ \(TraditionalForm[ y[x]\ /. \ \(DSolve[\(y'\)[x] + y[x] \[Equal] x, y[x], x]\)[\([1]\)]]\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{"x", "+", RowBox[{\(\[ExponentialE]\^\(-x\)\), " ", SubscriptBox[ TagBox["c", C], "1"]}], "-", "1"}], TraditionalForm]], "Answer"], Cell[BoxData[ \(TraditionalForm\`\(y\^\((3)\)\)(x) + y(x) = 0\)], "DisplayFormula"], Cell[BoxData[ \(TraditionalForm[ y[x]\ /. \ \(DSolve[\(y'''\)[x] + y[x] \[Equal] 0, y[x], x]\)[\([1]\)]]\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{\(\[ExponentialE]\^\(-x\)\), " ", SubscriptBox[ TagBox["c", C], "1"]}], "+", RowBox[{\(\[ExponentialE]\^\(x/2\)\), " ", SubscriptBox[ TagBox["c", C], "3"], " ", \(cos(\(\@3\ x\)\/2)\)}], "+", RowBox[{\(\[ExponentialE]\^\(x/2\)\), " ", SubscriptBox[ TagBox["c", C], "2"], " ", \(sin(\(\@3\ x\)\/2)\)}]}], TraditionalForm]], "Answer"] }, Open ]], Cell[CellGroupData[{ Cell["2. Aufgabe", "Section"], Cell[" a) Man forme die Differentialgleichung 2. Ordnung:", "Text"], Cell[BoxData[ \(TraditionalForm\`\(x\^2\) y'' \((x)\) + \ x\ y' \((x)\) + \((x\^2 - 1)\)\ \(y(x)\) = 0\)], "DisplayFormula"], Cell[TextData[{ "in ein System von Gleichungen 1. Ordnung um. Welche Probleme ergeben \ sich, wenn die L\[ODoubleDot]sung ", StyleBox["numerisch", FontSlant->"Italic"], " f\[UDoubleDot]r die Anfangsbedingungen ", Cell[BoxData[ \(TraditionalForm\`y(0) = 0\)]], " und ", Cell[BoxData[ \(TraditionalForm\`y'\ \((0)\) = 1/2\)]], " ", "berechnet werden soll ? Man mache Vorschl\[ADoubleDot]ge die Probleme zu \ umgehen.", StyleBox[" (2 Punkte)", FontWeight->"Bold"] }], "Text"], Cell[TextData[{ "Das System erh\[ADoubleDot]lt man mit ", Cell[BoxData[ \(TraditionalForm\`y(x) = u(x)\)]], ", ", Cell[BoxData[ \(TraditionalForm\`y' \((x)\) = v(x)\)]] }], "Answer"], Cell[BoxData[{ \(TraditionalForm\`u' \((x)\) = v(x)\), "\[IndentingNewLine]", \(TraditionalForm\`v' \((x)\) = \(-\ \(\((x\ \ \(v( x)\) + \((x\^2 - 1)\)\ \(u( x)\))\)\/x\^2\)\)\)}], "DisplayFormula", Background->GrayLevel[0.900008]], Cell[TextData[{ "Die Anfangsbedingungen lauten dann ", Cell[BoxData[ \(TraditionalForm\`u(0) = 0\)]], " und ", Cell[BoxData[ \(TraditionalForm\`v(0) = 1/2. \)]] }], "Answer"], Cell[TextData[{ "Die Anfangsbedingungen sind f\[UDoubleDot]r den ", StyleBox["singul\[ADoubleDot]ren Punkt", FontSlant->"Italic"], " der Gleichung bei ", Cell[BoxData[ \(TraditionalForm\`x = 0\)]], " gegeben, eine numerische L\[ODoubleDot]sung l\[ADoubleDot]uft Gefahr, \ mit einer Division durch 0 zu enden" }], "Answer"], Cell[TextData[{ "Entweder man verwendet statt ", Cell[BoxData[ \(TraditionalForm\`\(\(x\^2\)\(\ \)\)\)]], " im Nenner ", Cell[BoxData[ \(TraditionalForm\`\((x\^2 + \[Epsilon])\)\)]], " mit ", Cell[BoxData[ \(TraditionalForm\`\[Epsilon] << \ 1\)]], " oder man macht einen Reihenansatz" }], "Answer"], Cell[BoxData[ \(TraditionalForm\`y( x) = \(x\^\[Lambda]\) \(\[Sum]\+\(k\ = \ 0\)\%\[Infinity]\ \(a\_k\) x\^k\)\)], "DisplayFormula", Background->GrayLevel[0.900008]], Cell["um von der singul\[ADoubleDot]ren Stelle zu entkommen.", "Answer"], Cell["b) Zum numerischen L\[ODoubleDot]sen des Gleichungssystems", "Text"], Cell[BoxData[{ \(TraditionalForm\`u' \((x)\) = \ 998\ \(u(x)\) + 1998\ \(v(x)\)\), "\[IndentingNewLine]", \(TraditionalForm\`v' \((x)\) = \(-999\)\ \(u(x)\)\ - \ 1999\ \(v(x)\)\)}], "DisplayFormula"], Cell[TextData[{ "mit den Anfangsbedingungen ", Cell[BoxData[ \(TraditionalForm\`u(0) = 1\)]], "und ", Cell[BoxData[ \(TraditionalForm\`v(0) = 0\)]], " stehen ein verbessertes Euler-Verfahren" }], "Text"], Cell[BoxData[{ \(TraditionalForm\`\(y\^*\) = y\_n + h\ \(f(x, y\_n)\)\), "\[IndentingNewLine]", \(TraditionalForm\`y\_\(n + 1\) = y\_n + h\/2\ \((f(x + h, \(y\^*\)) + f(x, y\_n))\)\)}], "DisplayFormula"], Cell["und eine Trapez-Regel", "Text"], Cell[BoxData[ \(TraditionalForm\`y\_\(n + 1\) = y\_n + \(\(\(h\)\(\ \)\)\/2\) \((f(x + h, y\_\(n + 1\)) + f(x, y\_n))\)\)], "DisplayFormula"], Cell[TextData[{ "zu Auswahl. Welches Verfahren verwenden Sie ? Begr\[UDoubleDot]nden Sie \ ihre Entscheidung.", StyleBox[" (3 Punkte)", FontWeight->"Bold"] }], "Text"], Cell[TextData[{ "Das System von Gleichungen ist steif und mu\[SZ] mit dem ", StyleBox["impliziten", FontSlant->"Italic"], " Verfahren (also der Trapez-Regel) gel\[ODoubleDot]st werden. Ein \ explizites Verfahren ben\[ODoubleDot]tigt f\[UDoubleDot]r die Stabilit\ \[ADoubleDot]t der Integration einen um den Faktor 1000 kleinere \ Schrittweite. " }], "Answer"], Cell[CellGroupData[{ Cell["Hinweise", "Subsubsection"], Cell["Die Gleichung", "Text"], Cell[BoxData[ \(TraditionalForm\`\(x\^2\) y'' \((x)\) + \ x\ y' \((x)\) + \((x\^2 - n\^2)\)\ \(y(x)\) = 0\)], "DisplayFormula"], Cell[TextData[{ "ist eine Besselsche Differentialgleichung ", Cell[BoxData[ \(TraditionalForm\`n\)]], "-ter Ordnung und mu\[SZ] nicht numerisch gel\[ODoubleDot]st werden." }], "Text"], Cell["Die Matrix ", "Text"], Cell[BoxData[ FormBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"998", "1998"}, {\(-999\), \(-1999\)} }, ColumnAlignments->{Decimal}], "\[NoBreak]", ")"}], TraditionalForm]], "DisplayFormula", FormatType->StandardForm], Cell[TextData[{ "hat die Eigenwerte ", Cell[BoxData[ FormBox[Cell[TextData[Cell[BoxData[ \(TraditionalForm\`\[Lambda]\_1 = \(-1\)\)]]]], TraditionalForm]]], " und ", Cell[BoxData[ \(TraditionalForm\`\[Lambda]\_2 = \(-1000\)\)]] }], "Text"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["3. Aufgabe", "Section"], Cell["Man l\[ODoubleDot]se das Randwertproblem", "Text"], Cell[BoxData[ \(TraditionalForm\`y'' \((x)\) = x\)], "DisplayFormula"], Cell[TextData[{ "f\[UDoubleDot]r ", Cell[BoxData[ \(TraditionalForm\`x \[Element] \([\(-\[Pi]\), \[Pi]]\)\)]], " mit den periodischen Randbedingungen, ", Cell[BoxData[ \(TraditionalForm\`y(\(-\[Pi]\)) = y(\[Pi])\)]], ". ", StyleBox["(8 Punkte)", FontWeight->"Bold"] }], "Text"], Cell["Als erstes wird das Eigenwertproblem", "Answer"], Cell[BoxData[{ \(TraditionalForm\`y'' \((x)\) = \(-\[Lambda]\)\ \(y( x)\)\), "\[IndentingNewLine]", \(TraditionalForm\`y(\(-\[Pi]\)) = y(\[Pi])\)}], "DisplayFormula", Background->GrayLevel[0.900008]], Cell["gel\[ODoubleDot]st. Die Differentialgleichung", "Answer"], Cell[BoxData[ \(TraditionalForm\`y'' \((x)\) + \[Lambda]\ \(y(x)\) = 0\)], "DisplayFormula", Background->GrayLevel[0.900008]], Cell["hat die L\[ODoubleDot]sungen:", "Answer"], Cell[BoxData[ \(TraditionalForm\`\[Lambda] = \(0\ \ \ \ \ \(y(x)\) = c\_1 + c\_2\ x\)\)], "DisplayFormula", Background->GrayLevel[0.900008]], Cell[BoxData[ \(TraditionalForm\`\[Lambda] \[NotEqual] 0\ \ \ \ \ \(y(x)\) = c\_1\ \(sin(\@\[Lambda]\ x)\) + c\_2\ \(cos(\(\@\[Lambda]\) x)\)\)], "DisplayFormula", Background->GrayLevel[0.900008]], Cell[TextData[{ "f\[UDoubleDot]r ", Cell[BoxData[ \(TraditionalForm\`\[Lambda] = 0\)]], " erf\[UDoubleDot]llt nur die Funktion ", Cell[BoxData[ \(TraditionalForm\`\(y\_0\)(x) = c\)]], " die Randbedingungen, f\[UDoubleDot]r ", Cell[BoxData[ \(TraditionalForm\`\[Lambda] \[NotEqual] 0\)]], " sind sowohl ", Cell[BoxData[ \(TraditionalForm\`\(y\_k\)(x) = \ c\ \(cos(k\ x)\)\)]], " und ", Cell[BoxData[ \(TraditionalForm\`\(\(y\&~\)\_k\)(x) = c\ \(sin(k\ x)\)\)]], " Eigenfunktionen wenn ", Cell[BoxData[ \(TraditionalForm\`k\)]], " eine ganze Zahl ist. Man erh\[ADoubleDot]lt also die Eigenwerte:" }], "Answer"], Cell[BoxData[ \(TraditionalForm\`\[Lambda]\_k = 0, 1, 4, \[Ellipsis], k\^2, \[Ellipsis]\)], "DisplayFormula", Background->GrayLevel[0.900008]], Cell["und die Eigenfunktionen", "Answer"], Cell[BoxData[ \(TraditionalForm\`\(y\_0\)(x) = 1\/\@\(2 \[Pi]\)\)], "DisplayFormula", Background->GrayLevel[0.900008]], Cell[BoxData[ \(TraditionalForm\`\(y\_k\)( x) = \(1\/\@\[Pi]\) \(cos(k\ x)\)\)], "DisplayFormula", Background->GrayLevel[0.900008]], Cell[BoxData[ \(TraditionalForm\`\(\(y\&~\)\_k\)( x) = \(1\/\@\[Pi]\) \(sin(k\ x)\)\)], "DisplayFormula", Background->GrayLevel[0.900008]], Cell[TextData[{ "Zum L\[ODoubleDot]sen des Randwertproblems mu\[SZ] die rechte Seite ", Cell[BoxData[ \(TraditionalForm\`f(x) = x\)]], " in eine Fourier-Reihe nach diesen Funktionen entwickelt werden. F\ \[UDoubleDot]r ", Cell[BoxData[ \(TraditionalForm\`f(x)\)]], " erh\[ADoubleDot]lt man" }], "Answer"], Cell[BoxData[ \(TraditionalForm\`f( x) = \(x = \(2 \[Sum]\)\+\(k\ = \ 1\)\%\[Infinity]\ \(\(\((\(-1\))\)\ \^\(k + 1\)\) \(\(\(1\)\(\ \)\)\/k\) \(sin(k\ x)\)\)\)\)], "DisplayFormula", Background->GrayLevel[0.900008]], Cell[TextData[{ "Die Koeffizienten zu den Eigenfunktionen ", Cell[BoxData[ \(TraditionalForm\`\(cos(k\ x)\)/\@\[Pi]\)]], " verschwinden, weil das Integral " }], "Answer"], Cell[BoxData[ \(TraditionalForm\`\(1\/\@\[Pi]\) \(\[Integral]\_\(-\[Pi]\)\%\[Pi] x\ \ \(cos(k\ x)\) \[DifferentialD]x\) = 0\)], "Text", Background->GrayLevel[0.900008]], Cell["\<\ verschwindet (ungerade Funktion \[UDoubleDot]ber ein symmetrisches \ Intervall). Mit dem Reihenansatz\ \>", "Answer"], Cell[BoxData[ \(TraditionalForm\`y( x) = \[Sum]\+\(k\ = \ 1\)\%\[Infinity]\ \(\(a\_k\)\(\ \)\(sin( k\ x)\)\(\ \)\)\)], "DisplayFormula", Background->GrayLevel[0.900008]], Cell["erh\[ADoubleDot]lt man", "Answer"], Cell[BoxData[ \(TraditionalForm\`y(\(''\) \(x\)) = \(\[Sum]\+\(k\ = \ 1\)\%\[Infinity]\ \ a\_k\ \((\(-k\^2\))\)\ \(sin(k\ x)\) = \(f( x) = \ \(2 \[Sum]\)\+\(k\ = \ 1\)\%\[Infinity]\ \(\((\(-1\))\)\^\ \(k + 1\)\) \(\(\(1\)\(\ \)\)\/k\) \(sin(k\ x)\)\)\)\)], "DisplayFormula", Background->GrayLevel[0.900008]], Cell["oder", "Answer"], Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(a\_k\ = \ 2\ \ \(\((\(-1\))\)\^k\) \(\(1\)\(\ \)\)\/k\^3\)\)\)], "DisplayFormula",\ Background->GrayLevel[0.900008]], Cell[BoxData[ \(TraditionalForm\`y( x) = \[Sum]\+\(k\ = \ 1\)\%\[Infinity]\ \ \ \(\(2\)\(\ \ \)\(\((\(-1\ \))\)\^k\) \(\(\(1\)\(\ \)\)\/k\^3\)\(\ \)\(sin( k\ x)\)\(\ \)\)\)], "DisplayFormula", Background->GrayLevel[0.900008]], Cell["\<\ Als Bemerkung zu 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