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Question Number 187433 by Mastermind last updated on 17/Feb/23

$$\mathrm{Prove}\mathrm{that}▽\times ▽\varphi =0.\mathrm{If}\varphi \mathrm{is}$$$$\mathrm{assumed}\mathrm{to}\mathrm{be}\mathrm{continous}\mathrm{and}$$$$\mathrm{having}\mathrm{continous}\mathrm{first}\mathrm{and}\mathrm{second}$$$$\mathrm{order}\mathrm{derivatives}.$$$$$$$$\mathrm{M}.\mathrm{m}$$

Answered by aleks041103 last updated on 17/Feb/23

$$▽\times ▽f=\left|\begin{array}{ccc}i& j& k\\ {\partial }_x& {\partial }_y& {\partial }_z\\ {\partial }_{x}f& {\partial }_{y}f& {\partial }_{z}f\end{array}\right|=$$$$=i({\partial }_{yz}f-{\partial }_{zy}f)+j({\partial }_{zx}f-{\partial }_{xz}f)+k({\partial }_{xy}f-{\partial }_{yx}f)=$$$$=0i+0j+0k=0$$$$sincebyshwartzrule:$$$${\partial }_{ab}f=\frac{{\partial }^{2}f}{\partial a\partial b}=\frac{{\partial }^{2}f}{\partial b\partial a}={\partial }_{ba}f$$$$foralltwotimesdifferentiablef$$

Commented by Mastermind last updated on 17/Feb/23

$$\mathrm{Thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{my}\mathrm{BOSS}$$

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