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Question Number 213097 by issac last updated on 30/Oct/24

Uhhhh. can you guys solve Partial differantial equation ▽^2 𝛗=0 Cylinderical Laplacian case ▽^2 =(1/ρ)∙((∂ )/∂ρ)(ρ((∂ )/∂ρ))+((1/ρ))^2 (∂^2 /∂φ^2 )+((∂^2 )/∂z^2 ) Spherical Laplacian case ▽^2 =((1/r))^2 ((∂ )/∂r)(r^2 ((∂ )/∂r))+(1/(r^2 sin(θ)))∙((∂ )/∂θ)(sin(θ)((∂ )/∂θ))+(1/(r^2 sin^2 (θ)))∙(∂^2 /∂ϕ^2 )

$$\mathrm{Uhhhh}. \\ $$$$\mathrm{can}\:\mathrm{you}\:\mathrm{guys}\:\mathrm{solve}\:\mathrm{Partial}\:\mathrm{differantial}\:\mathrm{equation} \\ $$$$\bigtriangledown^{\mathrm{2}} \boldsymbol{\phi}=\mathrm{0} \\ $$$$\mathrm{Cylinderical}\:\mathrm{Laplacian}\:\mathrm{case} \\ $$$$\bigtriangledown^{\mathrm{2}} =\frac{\mathrm{1}}{\rho}\centerdot\frac{\partial\:\:}{\partial\rho}\left(\rho\frac{\partial\:\:}{\partial\rho}\right)+\left(\frac{\mathrm{1}}{\rho}\right)^{\mathrm{2}} \frac{\partial^{\mathrm{2}} \:}{\partial\phi^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} \:\:}{\partial{z}^{\mathrm{2}} } \\ $$$$\mathrm{Spherical}\:\mathrm{Laplacian}\:\mathrm{case} \\ $$$$\bigtriangledown^{\mathrm{2}} =\left(\frac{\mathrm{1}}{{r}}\right)^{\mathrm{2}} \frac{\partial\:\:}{\partial{r}}\left({r}^{\mathrm{2}} \frac{\partial\:\:}{\partial{r}}\right)+\frac{\mathrm{1}}{{r}^{\mathrm{2}} \mathrm{sin}\left(\theta\right)}\centerdot\frac{\partial\:\:}{\partial\theta}\left(\mathrm{sin}\left(\theta\right)\frac{\partial\:\:}{\partial\theta}\right)+\frac{\mathrm{1}}{{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\theta\right)}\centerdot\frac{\partial^{\mathrm{2}} \:}{\partial\varphi^{\mathrm{2}} } \\ $$

Answered by MrGaster last updated on 30/Oct/24

▽^2 φ=0 (1/ρ) (∂/∂ρ)(ρ(∂φ/∂φ))+(1/ρ^2 ) (∂^2 φ/∂φ^2 )+(∂^2 φ/∂z^2 )=0 (1/r^2 ) (∂/∂r)(r^2 (∂φ/∂r))+(1/(r^2 sin θ)) (∂/∂θ)(sin θ(∂φ/∂θ))+(1/(r^2 sin^2 θ)) (∂^2 φ/∂ϕ^2 )=0

$$\bigtriangledown^{\mathrm{2}} \phi=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\rho}\:\frac{\partial}{\partial\rho}\left(\rho\frac{\partial\phi}{\partial\phi}\right)+\frac{\mathrm{1}}{\rho^{\mathrm{2}} }\:\frac{\partial^{\mathrm{2}} \phi}{\partial\phi^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} \phi}{\partial{z}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\:\frac{\partial}{\partial{r}}\left({r}^{\mathrm{2}} \frac{\partial\phi}{\partial{r}}\right)+\frac{\mathrm{1}}{{r}^{\mathrm{2}} \mathrm{sin}\:\theta}\:\frac{\partial}{\partial\theta}\left(\mathrm{sin}\:\theta\frac{\partial\phi}{\partial\theta}\right)+\frac{\mathrm{1}}{{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \theta}\:\frac{\partial^{\mathrm{2}} \phi}{\partial\varphi^{\mathrm{2}} }=\mathrm{0} \\ $$

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