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Second week without fest
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23
S3/MaPhyIII/VL/MaPhIIIVL2.typ
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23
S3/MaPhyIII/VL/MaPhIIIVL2.typ
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// Main VL template
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#import "../preamble.typ": *
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// Fix theorems to be shown the right way in this document
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#import "@preview/ctheorems:1.1.3": *
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#show: thmrules
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// Main settings call
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#show: conf.with(
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// May add more flags here in the future
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num: 5,
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type: 0, // 0 normal, 1 exercise
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date: datetime.today().display(),
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//date: datetime(
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// year: 2025,
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// month: 5,
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// day: 1,
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//).display(),
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)
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= Uebersicht
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Diese Vorlesung ist ausgefallen.
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108
S3/MaPhyIII/VL/MaPhIIIVL3.typ
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S3/MaPhyIII/VL/MaPhIIIVL3.typ
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// Main VL template
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#import "../preamble.typ": *
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// Fix theorems to be shown the right way in this document
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#import "@preview/ctheorems:1.1.3": *
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#show: thmrules
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// Main settings call
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#show: conf.with(
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// May add more flags here in the future
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num: 3,
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type: 0, // 0 normal, 1 exercise
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date: datetime.today().display(),
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//date: datetime(
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// year: 2025,
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// month: 5,
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// day: 1,
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//).display(),
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)
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= Uebersicht
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Wiederholung der *Bessel'schen Ungleichung*. Betrachte $f in RR [- pi, pi] \, space c^(k) = 1/(2 pi) integral _(- pi) ^(pi) f (x) e ^(i k x) dif x$
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$
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==> sum abs(c^(k) )^2 <= norm(f)_(L^2 ) ^2 \
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norm(f)_(L^2 ) ^2 = lr(angle.l f, f angle.r) \
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lr(angle.l f, g angle.r) = 1/(2 pi) integral _(- pi) ^(pi) overline(f (x)) g (x) dif x
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$
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Im Beweis
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$
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norm(f - sum _(- n) ^(n) c_(k) e_(k) )^2 = lr(angle.l f - sum c_(k) e_(k) , f - sum c_(k) e_(k) angle.r) \
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= norm(f)^2 - sum abs(c_(k))^2 -> 0 .. (n -> oo) "wenn" sum c_(k) e_(k) "gegen" f "bezueglich" norm(*)_(L^2 ) \
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e_(k) (x) = e ^(i k x).
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$
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Es folgt die *Parseval'sche Gleichung*
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$
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norm(f)_(2) ^(2) = lim_(n -> oo) sum_(- n)^(n) abs(c_(k) )^2 \
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==> sum abs(c_(k) )^2 < oo.
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$
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$==>$ Eindeutigkeit der Fourierkoeffizienten.
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#proof[
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Sei $f = sum c_(k) e_(k) \, space g = sum d_(k) e_(k) $. Mit $c_(k) = lr(angle.l e_(k) , f angle.r) \, space d_(k) = lr(angle.l e_(k) , g angle.r)$.
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Ist $f != g$ so gilt
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$
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0 != norm(f - g)_(L^2 ) ^2 = sum abs(lr(angle.l e_(k) , f - g angle.r))^2 = sum abs(c_(k) - d_(k) )^2.
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$
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Angenommen $c_(k) = d_(k) forall k ==> 0 != 0 $. Widerspruch.
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$f ~ g$ bezueglich der $norm(*)_(L_(1) ) :<==> f "und" g "auf Nullmenge verschieden sein duerfen" $.
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Also $f, g in C [- pi, pi] ==>( f != g ==> exists x_0 in [- pi, pi]: f (x_0 ) != g (x_0 ))$.
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Parseval $==>$ $sum_(k=1)^(oo) 1/k^2 = pi^2 /6$.
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#figure(
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image("typst-assets/drawing-2025-11-07-10-46-32.rnote.svg"),
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)
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$
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f (x) = x "auf" [- pi, pi] \, space c_(k) = (- 1)^(k) ) i/k \, space k != 0, c_0 = 0 \
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==> sum_(- oo)^(oo) abs(c_(k) )^2 = 2 sum_(k=1)^(oo) 1/k^2 =^("Parseval") lr(angle.l f, f angle.r) = 1/(2 pi) integral _(- pi) ^(pi) x^2 dif x = pi^2 /3
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$
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]
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Die Schwingende Rechtecksmembran durch den Tayloransatz loesen
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$
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u (t, x, y) = sum _(n = 1) ^(oo) sum_(m = 1)^(oo) (a_(m, n) cos (sqrt(lambda_(n, m) ) c t) + b_(n, m) sin (sqrt(lambda_(n, m) ) c t)) nu_(n, m).
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$
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Cladni-Figuren in Wolframalpha.
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= Kreisfoernige Membran
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Wellengleichung
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$
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arrow(nabla) ^2 u = 1/v^2 partial _(t) ^2 u.
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$
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Produktansatz $==>$ $u = v (t) w (x, y)$.
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Wenn man auf Polarkoordinaten transformiert, dann muessen die Randbedingungen alle erfuellt sein, was man durch den Limes prueft.
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$
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(partial _(x) ^2 + partial _(y) ^2 ) v = - lambda v
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$
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Es wird dann wieder ein Ansatz gemacht
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$
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V (r, theta) = R (r) A (theta) \
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==> (r^2 R'' + v R')1/R + lambda r^2 = - (A'') / (A).
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$
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$
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1/r partial _(r) (r partial _(r) (R (r) A (theta))) + 1 / r^2 partial _(theta) ^2 (R (r) A (theta)) + lambda R (r) A (theta) = ^(!) 0
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$
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Potenzreihenansatz als
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$
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f (rho ) = sum a_(k) rho^(k).
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$
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Es wird gleichmaessige Konvergenz angennommen. Es folgt
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$
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rho^2 f'' (rho) = rho^2 (sum a_(k) k rho ^(k - 1) )' = rho^2 sum a_(k) k (k - 1) rho ^(k - 2).
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$
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Nachdem die anderen Teile ausgerechnet wurden kann ein Koeffizientenvergleich gemacht werden.
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BIN
S3/MaPhyIII/VL/typst-assets/drawing-2025-11-07-10-46-32.rnote
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S3/MaPhyIII/VL/typst-assets/drawing-2025-11-07-10-46-32.rnote
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<?xml version="1.0" standalone="no"?>
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</svg>
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