Prof. Dr. R. Wulkenhaar WS 08/09 Dr. R. Brüske

Übungen zur Mathematik für Physiker III

Abgabe: Donnerstag, 27.11.08, bis 14h00 in den Briefkästen Blatt 6

Aufgabe 1. Für eine symmetrische Matrix A = A^{t} ∈ M(n, ℝ) werde durch F (x) := x^{t}Ax (x ∈ ℝ^{n} als Spaltenvektor aufgefaßt) eine Funktion F : ℝ^{n} → ℝ definiert. Zeige:

a) F besitzt auf der Einheitssphäre S^{n−1} := {y ∈ ℝ^{n} : ∑_{i=1}^{n} y_{i}^{2} = 1} ein Minimum λ, etwa in v ∈ ℝ^{n}.

b) Der Vektor v aus a) ist dann Eigenvektor von A zum Eigenwert v^{t}Av.

Aufgabe 2. Es sei F : ℝ^{n} → ℝ definiert durch F (x_{1}, …, x_{n}) = ∏_{i=1}^{n} x_{i}.

a) Man zeige: Eingeschränkt auf die Teilmenge

M := {(x_{1}, …, x_{n}) : x_{ν} ≥ 0 für alle 1 ≤ ν ≤ n , ∑_{ν=1}^{n} x_{ν} = 1}

besitzt F genau in x_{1} = … = x_{n} = \frac{1}{n} ein absolutes Maximum.

b) Man folgere: Für beliebige positive x_{1}, …, x_{n} gilt

∏_{i=1}^{n} x_{i} ≤ (\frac{1}{n} ∑_{i=1}^{n} x_{i})^{n} .

“=” gilt genau dann, wenn x_{1} = … = x_{n} gilt.

Aufgabe 3. Es sei p ∈ ℝ, p > 1, a ∈ ℝ^{n}. Betrachtet werde die Funktion F : ℝ^{n} → ℝ mit F (x) := ⟨a, x⟩ = ∑_{i=1}^{n} a_{i}x_{i} und die durch f(x) = 1 − ∑_{i=1}^{n} |x_{i}|^{p} = 0 gegebene Teilmenge M ⊂ ℝ^{n}.

a) Zeigen Sie, daß die Einschränkung von F auf M das absolute Maximum (∑_{i=1}^{n} |a_{i}|^{q})^{1/q} und das absolutes Minimum − (∑_{i=1}^{n} |a_{i}|^{q})^{1/q} hat, wobei \frac{1}{p} + \frac{1}{q} = 1 ist.

b) Man folgere:

| ∑_{i=1}^{n} a_{i}x_{i}| ≤ (∑_{i=1}^{n} |a_{i}|^{q})^{1/q} (∑_{i=1}^{n} |x_{i}|^{p})^{1/p} für alle x ∈ ℝ^{n} .

Aufgabe 4 ist auf der 2. Seite!

Aufgabe 4. Es seien 0 < r < R und K = {(x, 0, z) | (x − R)^{2} + z^{2} = r^{2}}.

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|M L  ϑ P = (x, y, z)

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0 φ  x y  × M L

Durch Drehen dieses Kreises um die z-Achse entsteht eine Fläche, die Torusfläche T. Jeder Punkt P von T ist eindeutig bestimmt durch den Drehwinkel φ (0 ≤ φ < 2π) und den Winkel ϑ (0 ≤ ϑ < 2π), der wie folgt entsteht: Man zeichne einen Querschnitt durch P und 0 parallel zur z-Achse. Das ist ein Kreis mit Radius r und Mittelpunkt M. Sei L der Punkt auf dem Kreis mit z = 0 und Abstand R − r zur z-Achse. Dann sei ϑ = ∡LMP. Man zeige:

a) T wird durch die Gleichung (\sqrt{x^{2} + y^{2}} − R)^{2} + z^{2} − r^{2} = 0 als zweidimensionale Fläche erkannt.

b) Man beschreibe x, y, z als Funktionen von φ, ϑ (lokales Koordinatensystem) und berechne den Tangentialraum von T in (\frac{π}{4}, \frac{π}{4}).