Question Number 206273 by EmGent last updated on 10/Apr/24
$$\mathrm{Does}\:\mathrm{anyone}\:\mathrm{know}\:\mathrm{how}\:\mathrm{this}\:\mathrm{works}\:? \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{have}\:{d}\psi\:=\:\left({x}^{\mathrm{2}} -{cy}^{\mathrm{2}} \right){dy} \\ $$$$ \\ $$$$\mathrm{And}\:\mathrm{my}\:\mathrm{physics}\:\mathrm{teacher}\:\mathrm{says}\:\mathrm{it}\:\mathrm{is}\:\left(\mathrm{or}\:\mathrm{can}\right. \\ $$$$\left.\mathrm{be}\right)\:\mathrm{a}\:\mathrm{harmonic}\:\mathrm{function}\:\left(\Delta\psi\:=\:\mathrm{0}\right) \\ $$$$\mathrm{Can}\:\mathrm{anyone}\:\mathrm{explain}\:? \\ $$
Answered by aleks041103 last updated on 11/Apr/24
$${Since}\:\exists{d}\psi,\:{then} \\ $$$${d}\psi\:=\:\frac{\partial\psi}{\partial{x}}{dx}\:+\:\frac{\partial\psi}{\partial{y}}{dy}\:=\:\left({x}^{\mathrm{2}} −{cy}^{\mathrm{2}} \right){dy} \\ $$$$\Rightarrow\begin{cases}{\partial_{{x}} \psi\:=\:\mathrm{0}}\\{\partial_{{y}} \psi\:=\:{x}^{\mathrm{2}} −{cy}^{\mathrm{2}} }\end{cases} \\ $$$${it}\:{is}\:{obvious}\:{that}\:\partial_{{x}} \psi,\partial_{{y}} \psi\:{are}\:{continuous}. \\ $$$$\Rightarrow\psi\in{C}^{\mathrm{1}} \\ $$$${Also}\:{it}\:{is}\:{easily}\:{seen}\:{that}\:\exists\partial_{{x}} ^{\:\mathrm{2}} \psi,\partial_{{x}} \partial_{{y}} \psi,\partial_{{y}} \partial_{{x}} \psi,\partial_{{y}} ^{\:\mathrm{2}} \psi. \\ $$$${So},\:\psi\:{is}\:{twice}\:{differentiable}\:{and}\:{is}\:{once} \\ $$$${continously}\:{differentiable}. \\ $$$${Therefore}\:{Schwartz}\:{rule}\:{applies}: \\ $$$$\frac{\partial^{\mathrm{2}} \psi}{\partial{x}\partial{y}}\:=\:\frac{\partial^{\mathrm{2}} \psi}{\partial{y}\partial{x}} \\ $$$${BUT} \\ $$$$\frac{\partial^{\mathrm{2}} \psi}{\partial{x}\partial{y}}\:=\:\frac{\partial}{\partial{x}}\left(\frac{\partial\psi}{\partial{y}}\right)=\frac{\partial}{\partial{x}}\left({x}^{\mathrm{2}} −{cy}^{\mathrm{2}} \right)=\mathrm{2}{x} \\ $$$$\frac{\partial^{\mathrm{2}} \psi}{\partial{y}\partial{x}}\:=\:\frac{\partial}{\partial{y}}\left(\frac{\partial\psi}{\partial{x}}\right)=\frac{\partial}{\partial{y}}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${BUT}\:\:\mathrm{2}{x}\neq\mathrm{0} \\ $$$$\Rightarrow\:\nexists\psi:\:{d}\psi=\left({x}^{\mathrm{2}} −{cy}^{\mathrm{2}} \right){dy} \\ $$$$ \\ $$$${Therefore},\:{such}\:\psi\:{cannot}\:{be}\:{harmonic}, \\ $$$${since}\:{such}\:\psi\:{doesn}'{t}\:{exist}. \\ $$