1.3 – Anwendung der Divergenz

 

Man zeige:

\frac{{{\partial ^2}{t_{ij}}}}{{\partial {x_i}\partial {x_j}}} = \nabla \cdot \left( {\nabla \cdot T} \right)

Lösung

Wir zeigen am Beispiel mit zwei Komponenten:

\nabla \cdot \left( {\nabla \cdot T} \right) = \left( {\begin{array}{*{20}{c}}{\frac{\partial }{{\partial x}}} \\{\frac{\partial }{{\partial y}}} \\ \end{array} } \right) \cdot \left( {\left( {\begin{array}{*{20}{c}}{\frac{\partial }{{\partial x}}} \\{\frac{\partial }{{\partial y}}} \\ \end{array} } \right) \cdot \left( {\begin{array}{*{20}{c}}{{t_{11}}} & {{t_{12}}} \\{{t_{21}}} & {{t_{22}}} \\ \end{array} } \right)} \right)

= \left( {\begin{array}{*{20}{c}}{\frac{\partial }{{\partial x}}} \\{\frac{\partial }{{\partial y}}} \\ \end{array} } \right)\left( {\begin{array}{*{20}{c}}{\frac{{\partial {t_{11}}}}{{\partial x}}+\frac{{\partial {t_{12}}}}{{\partial y}}} \\{\frac{{\partial {t_{21}}}}{{\partial x}}+\frac{{\partial {t_{22}}}}{{\partial y}}} \\ \end{array} } \right)

= \frac{{{\partial ^2}{t_{11}}}}{{\partial {x^2}}}+\frac{{{\partial ^2}{t_{12}}}}{{\partial x\partial y}}+\frac{{{\partial ^2}{t_{21}}}}{{\partial y\partial x}}+\frac{{{\partial ^2}{t_{22}}}}{{\partial {y^2}}}

= \frac{{{\partial ^2}{t_{ij}}}}{{\partial {x_i}\partial {x_j}}}