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 )


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

$Uhhhh .$ $can you guys solve Partial differantial equation$ $▽^{2} ϕ = 0$ $Cylinderical Laplacian case$ $▽^{2} = \frac{1}{ρ} \cdot \frac{\partial}{\partial ρ} (ρ \frac{\partial}{\partial ρ}) + {(\frac{1}{ρ})}^{2} \frac{\partial^{2}}{\partial ϕ^{2}} + \frac{\partial^{2}}{\partial z^{2}}$ $Spherical Laplacian case$ $▽^{2} = {(\frac{1}{r})}^{2} \frac{\partial}{\partial r} (r^{2} \frac{\partial}{\partial r}) + \frac{1}{r^{2} \sin (θ)} \cdot \frac{\partial}{\partial θ} (\sin (θ) \frac{\partial}{\partial θ}) + \frac{1}{r^{2} \sin^{2} (θ)} \cdot \frac{\partial^{2}}{\partial φ^{2}}$
	Answered by MrGaster last updated on 30/Oct/24

	$▽^{2} ϕ = 0$ $\frac{1}{ρ} \frac{\partial}{\partial ρ} (ρ \frac{\partial ϕ}{\partial ϕ}) + \frac{1}{ρ^{2}} \frac{\partial^{2} ϕ}{\partial ϕ^{2}} + \frac{\partial^{2} ϕ}{\partial z^{2}} = 0$ $\frac{1}{r^{2}} \frac{\partial}{\partial r} (r^{2} \frac{\partial ϕ}{\partial r}) + \frac{1}{r^{2} \sin θ} \frac{\partial}{\partial θ} (\sin θ \frac{\partial ϕ}{\partial θ}) + \frac{1}{r^{2} \sin^{2} θ} \frac{\partial^{2} ϕ}{\partial φ^{2}} = 0$
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