Aufgabensammlung Mathematik: Kettenregeln für die partielle Ableitung

Seien $x,y,z,w$ Zustandsgrößen, von denen jede von jeweils zwei anderen abhängt. Für $x$ gibt es also Funktionen $x(y,z)$, $x(z,w)$ sowie $x(y,w)$. Analoges gilt für die anderen drei Zustandsgrößen. Außerdem sei ${\left(\frac{\partial y}{\partial x}\right)}_z\ne 0$. Man beweise:

1. ${\left(\frac{\partial x}{\partial y}\right)}_z=\frac{1}{{\left(\frac{\partial y}{\partial x}\right)}_z}$
2. ${\left(\frac{\partial x}{\partial y}\right)}_z{\left(\frac{\partial y}{\partial z}\right)}_x{\left(\frac{\partial z}{\partial x}\right)}_y=-1$
3. ${\left(\frac{\partial x}{\partial y}\right)}_z={\left(\frac{\partial x}{\partial y}\right)}_w+{\left(\frac{\partial x}{\partial w}\right)}_y{\left(\frac{\partial w}{\partial y}\right)}_z$

Lösung zur 1. Teilaufgabe

Es ist

$\begin{array}{cc}\mathrm{d}x& ={\left(\frac{\partial x}{\partial y}\right)}_z\mathrm{d}y+{\left(\frac{\partial x}{\partial z}\right)}_y\mathrm{d}z\\ \mathrm{d}y& ={\left(\frac{\partial y}{\partial x}\right)}_z\mathrm{d}x+{\left(\frac{\partial y}{\partial z}\right)}_x\mathrm{d}z\end{array}$

Nun kann die Gleichung $\mathrm{d}y={\left(\frac{\partial y}{\partial x}\right)}_z\mathrm{d}x+{\left(\frac{\partial y}{\partial z}\right)}_x\mathrm{d}z$ in die Gleichung für $\mathrm{d}x$ eingesetzt werden. So erhält man

$\begin{array}{cc}\mathrm{d}x& ={\left(\frac{\partial x}{\partial y}\right)}_z\mathrm{d}y+{\left(\frac{\partial x}{\partial z}\right)}_y\mathrm{d}z\\ [5px]& \downarrow \mathrm{d}y={\left(\frac{\partial y}{\partial x}\right)}_z\mathrm{d}x+{\left(\frac{\partial y}{\partial z}\right)}_x\mathrm{d}z\\ [5px]& ={\left(\frac{\partial x}{\partial y}\right)}_z\cdot ({\left(\frac{\partial y}{\partial x}\right)}_z\mathrm{d}x+{\left(\frac{\partial y}{\partial z}\right)}_x\mathrm{d}z)+{\left(\frac{\partial x}{\partial z}\right)}_y\mathrm{d}z\\ & ={\left(\frac{\partial x}{\partial y}\right)}_z{\left(\frac{\partial y}{\partial x}\right)}_z\mathrm{d}x+{\left(\frac{\partial x}{\partial y}\right)}_z{\left(\frac{\partial y}{\partial z}\right)}_x\mathrm{d}z+{\left(\frac{\partial x}{\partial z}\right)}_y\mathrm{d}z\end{array}$

Also ist

$0=({\left(\frac{\partial x}{\partial y}\right)}_z{\left(\frac{\partial y}{\partial x}\right)}_z-1)\mathrm{d}x+({\left(\frac{\partial x}{\partial y}\right)}_z{\left(\frac{\partial y}{\partial z}\right)}_x+{\left(\frac{\partial x}{\partial z}\right)}_y)\mathrm{d}z$

Da $\mathrm{d}x$ und $\mathrm{d}z$ linear unabhängig sind, muss gelten:

${\left(\frac{\partial x}{\partial y}\right)}_z{\left(\frac{\partial y}{\partial x}\right)}_z-1=0$

und

${\left(\frac{\partial x}{\partial y}\right)}_z{\left(\frac{\partial y}{\partial z}\right)}_x+{\left(\frac{\partial x}{\partial z}\right)}_y=0$

Aus der ersten Gleichung folgt

$\begin{array}{c}\begin{array}{ccc}& {\left(\frac{\partial x}{\partial y}\right)}_z{\left(\frac{\partial y}{\partial x}\right)}_z-1& =0\\ \iff & {\left(\frac{\partial x}{\partial y}\right)}_z{\left(\frac{\partial y}{\partial x}\right)}_z& =1\\ \iff & {\left(\frac{\partial x}{\partial y}\right)}_z& =\frac{1}{{\left(\frac{\partial y}{\partial x}\right)}_z}\end{array}\end{array}$

Lösung zur 2. Teilaufgabe

In der Lösung zu zweiten Teilaufgabe wurde gezeigt, dass

${\left(\frac{\partial x}{\partial y}\right)}_z{\left(\frac{\partial y}{\partial z}\right)}_x+{\left(\frac{\partial x}{\partial z}\right)}_y=0$

Daraus folgt, dass

$\begin{array}{c}\begin{array}{ccc}& {\left(\frac{\partial x}{\partial y}\right)}_z{\left(\frac{\partial y}{\partial z}\right)}_x+{\left(\frac{\partial x}{\partial z}\right)}_y& =0\\ \iff & {\left(\frac{\partial x}{\partial y}\right)}_z{\left(\frac{\partial y}{\partial z}\right)}_x& =-{\left(\frac{\partial x}{\partial z}\right)}_y\\ [5px]& & \downarrow {\left(\frac{\partial x}{\partial z}\right)}_y=\frac{1}{{\left(\frac{\partial z}{\partial x}\right)}_y}\\ [5px]\iff & {\left(\frac{\partial x}{\partial y}\right)}_z{\left(\frac{\partial y}{\partial z}\right)}_x& =-\frac{1}{{\left(\frac{\partial z}{\partial x}\right)}_y}\\ \iff & {\left(\frac{\partial x}{\partial y}\right)}_z{\left(\frac{\partial y}{\partial z}\right)}_x{\left(\frac{\partial z}{\partial x}\right)}_y& =-1\end{array}\end{array}$

Lösung zur 3. Teilaufgabe

Es ist

$\begin{array}{cc}\mathrm{d}x& ={\left(\frac{\partial x}{\partial y}\right)}_z\mathrm{d}y+{\left(\frac{\partial x}{\partial z}\right)}_y\mathrm{d}z\\ \mathrm{d}x& ={\left(\frac{\partial x}{\partial y}\right)}_w\mathrm{d}y+{\left(\frac{\partial x}{\partial w}\right)}_y\mathrm{d}w\\ \mathrm{d}w& ={\left(\frac{\partial w}{\partial y}\right)}_z\mathrm{d}y+{\left(\frac{\partial w}{\partial z}\right)}_y\mathrm{d}z\end{array}$

Wenn man nun die ersten beiden Gleichungen gleichsetzt und in dieser Gleichung $\mathrm{d}w$ durch den Term ${\left(\frac{\partial w}{\partial y}\right)}_z\mathrm{d}y+{\left(\frac{\partial w}{\partial z}\right)}_y\mathrm{d}z$ ersetzt, dann erhält man

$\begin{array}{c}\begin{array}{ccc}& {\left(\frac{\partial x}{\partial y}\right)}_z\mathrm{d}y+{\left(\frac{\partial x}{\partial z}\right)}_y\mathrm{d}z& ={\left(\frac{\partial x}{\partial y}\right)}_w\mathrm{d}y+{\left(\frac{\partial x}{\partial w}\right)}_y\mathrm{d}w\\ [5px]& & \downarrow dw={\left(\frac{\partial w}{\partial y}\right)}_z\mathrm{d}y+{\left(\frac{\partial w}{\partial z}\right)}_y\mathrm{d}z\\ [5px]\Rightarrow & {\left(\frac{\partial x}{\partial y}\right)}_z\mathrm{d}y+{\left(\frac{\partial x}{\partial z}\right)}_y\mathrm{d}z& ={\left(\frac{\partial x}{\partial y}\right)}_w\mathrm{d}y+{\left(\frac{\partial x}{\partial w}\right)}_y\cdot ({\left(\frac{\partial w}{\partial y}\right)}_z\mathrm{d}y+{\left(\frac{\partial w}{\partial z}\right)}_y\mathrm{d}z)\\ \Rightarrow & {\left(\frac{\partial x}{\partial y}\right)}_z\mathrm{d}y+{\left(\frac{\partial x}{\partial z}\right)}_y\mathrm{d}z& ={\left(\frac{\partial x}{\partial y}\right)}_w\mathrm{d}y+{\left(\frac{\partial x}{\partial w}\right)}_y{\left(\frac{\partial w}{\partial y}\right)}_z\mathrm{d}y+{\left(\frac{\partial x}{\partial w}\right)}_y{\left(\frac{\partial w}{\partial z}\right)}_y\mathrm{d}z\\ \Rightarrow & {\left(\frac{\partial x}{\partial y}\right)}_z\mathrm{d}y+{\left(\frac{\partial x}{\partial z}\right)}_y\mathrm{d}z& =({\left(\frac{\partial x}{\partial y}\right)}_w+{\left(\frac{\partial x}{\partial w}\right)}_y{\left(\frac{\partial w}{\partial y}\right)}_z)\mathrm{d}y+{\left(\frac{\partial x}{\partial w}\right)}_y{\left(\frac{\partial w}{\partial z}\right)}_y\mathrm{d}z\\ \Rightarrow & 0& =({\left(\frac{\partial x}{\partial y}\right)}_w+{\left(\frac{\partial x}{\partial w}\right)}_y{\left(\frac{\partial w}{\partial y}\right)}_z-{\left(\frac{\partial x}{\partial y}\right)}_z)\mathrm{d}y+({\left(\frac{\partial x}{\partial w}\right)}_y{\left(\frac{\partial w}{\partial z}\right)}_y-{\left(\frac{\partial x}{\partial z}\right)}_y)\mathrm{d}z\end{array}\end{array}$

Da $\mathrm{d}y$ und $\mathrm{d}z$ linear unabhängig sind, ist

${\left(\frac{\partial x}{\partial y}\right)}_w+{\left(\frac{\partial x}{\partial w}\right)}_y{\left(\frac{\partial w}{\partial y}\right)}_z-{\left(\frac{\partial x}{\partial y}\right)}_z=0$

und damit

${\left(\frac{\partial x}{\partial y}\right)}_z={\left(\frac{\partial x}{\partial y}\right)}_w+{\left(\frac{\partial x}{\partial w}\right)}_y{\left(\frac{\partial w}{\partial y}\right)}_z$