Aufgabensammlung Physik: Brachistochronen Problem

Achtung, Baustelle!

Brachistchronen Problem.

Berechne den schnellsten Weg eines Teilchens von Punkt A zu Punkt B, dass sich reibungsfrei in einem Gravitationspotential bewegt.

1. Finde die Funktionenschar

Formuliere den allgemeinen Integranden $ds$ in eine für das Problem geeignete Form um.

Lösung Teil 1

$min{\stackrel {!}{=}}\int _{t_{0}}^{t_{1}}dt=\int _{1}^{2}{\frac {ds}{v}}$

Aus den bekannten Variablen für das System kann eine Form für $v$ gefunden werden. Dazu wird die Energiebilanz des Systems aufgestellt:

${\begin{aligned}E_{ges}=E_{pot}-E_{kin}\\=ymg-{\frac {1}{2}}my'^{2}\end{aligned}}$

Mit $y_{t_{0}}=0$ ist $E_{ges}=0$ . Nach Umformen ergibt sich also:

$v={\sqrt {2yg}}$

Und damit:

${\begin{aligned}\int _{t_{0}}^{t_{1}}dt=\int _{1}^{2}{\frac {ds}{v}}\\=\int {\frac {ds}{\sqrt {2gy}}}\end{aligned}}$

Mit: $ds={\sqrt {1+y'^{2}}}dx$

Ergibt sich:

$min{\stackrel {!}{=}}\int {\sqrt {\frac {1+y'^{2}}{2gy}}}dx$

2. Aufstellen der Eulerschen Gleichung

Stelle die Eulersche Gleichung auf und löse diese.

Lösung der Aufgabe 2

• $min{\stackrel {!}{=}}\int {\sqrt {\frac {1+y'^{2}}{2gy}}}dx={\frac {1}{\sqrt {2g}}}\int {\sqrt {\frac {1+y'^{2}}{y}}}dx$
• Die Eulersche Gleichung lautet:

${\frac {\partial f}{\partial y}}-{\frac {d}{dx}}{\frac {\partial f}{\partial y'}}=0$

Dabei ist:

${\begin{aligned}{\frac {\partial f}{\partial y}}&=-{\frac {1}{2}}{\sqrt {1+y'^{2}}}y^{-{\frac {3}{2}}}\\{\frac {\partial f}{\partial y'}}&={\frac {y'}{\sqrt {y(1+y'^{2})}}}\end{aligned}}$

Weiter ist dann:

${\begin{aligned}{\frac {d}{dx}}{\frac {\partial f}{\partial y'}}&={\frac {d}{dx}}{\frac {y'}{\sqrt {y(1+y'^{2})}}}\end{aligned}}$

Mit:

${\begin{aligned}{\frac {\partial }{\partial x}}{\frac {y'}{\sqrt {y(1+y'^{2})}}}=0\\{\frac {\partial }{\partial y}}{\frac {y'}{\sqrt {y(1+y'^{2})}}}\cdot y'=-{\frac {1}{2}}{\frac {y'^{2}}{\sqrt {1+y'^{2}}}}\cdot {\frac {1}{y^{\frac {3}{2}}}}\\{\frac {\partial }{\partial y'}}{\frac {y'}{\sqrt {y(1+y'^{2})}}}\cdot y''={\frac {y''}{\sqrt {y(1+y'^{2})}}}-{\frac {y'^{2}y''}{\sqrt {y}}}\cdot {\frac {1}{(1+y'^{2})^{\frac {3}{2}}}}\end{aligned}}$

Es ergibt sich also insgesamt:

${\begin{aligned}0={\frac {\partial f}{\partial y}}-{\frac {d}{dx}}{\frac {\partial f}{\partial y'}}\\=-{\frac {1}{2}}{\sqrt {1+y'^{2}}}y^{-{\frac {3}{2}}}+{\frac {1}{2}}{\frac {y'^{2}}{\sqrt {1+y'^{2}}}}\cdot {\frac {1}{y^{\frac {3}{2}}}}-{\frac {y''}{\sqrt {y(1+y'^{2})}}}+{\frac {y'^{2}y''}{\sqrt {y}}}\cdot {\frac {1}{(1+y'^{2})^{\frac {3}{2}}}}&&|\cdot {\sqrt {1+y'^{2}}}\\=-{\frac {1+y'^{2}}{2}}y^{-{\frac {3}{2}}}+{\frac {1}{2}}{\frac {y'^{2}}{y^{\frac {3}{2}}}}-{\frac {y''}{\sqrt {y}}}+{\frac {y'^{2}y''}{\sqrt {y}}}\cdot {\frac {1}{1+y'^{2}}}&&|\cdot y^{\frac {3}{2}}\\=-{\frac {1+y'^{2}}{2}}+{\frac {y'^{2}}{2}}-y''y+{\frac {yy'^{2}y''}{1+y'^{2}}}\\\Longrightarrow 1+y'^{2}=y'^{2}-2yy''+{\frac {2yy'^{2}y''}{1+y'^{2}}}\end{aligned}}$

• ${\begin{aligned}\end{aligned}}$