Question

Example. Laplace equation in polar coordinates $\frac{\partial^2 u}{\partial r^2} + \frac{1}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2} = 0$ $u = u(r, \theta), 0 < r < 1, 0 < \theta < 2\pi$ $u(1, \theta) = \theta$ Prove that: $u(r, \theta) = 2 - 2 \sum_{n=0}^{\infty} \frac{1}{n}r^n \sin n\theta$

          Example.
Laplace equation in polar coordinates
$\frac{\partial^2 u}{\partial r^2} + \frac{1}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2} = 0$
$u = u(r, \theta), 0 < r < 1, 0 < \theta < 2\pi$
$u(1, \theta) = \theta$
Prove that:
$u(r, \theta) = 2 - 2 \sum_{n=0}^{\infty} \frac{1}{n}r^n \sin n\theta$

Example.
Laplace equation in polar coordinates
(∂^2 u)/(∂ r^2) + (1)/(r)(∂ u)/(∂ r) + (1)/(r^2)(∂^2 u)/(∂θ^2) = 0
u = u(r, θ), 0 < r < 1, 0 < θ < 2π
u(1, θ) = θ
Prove that:
u(r, θ) = 2 - 2 ∑n=0^∞(1)/(n)r^n sin nθ

Added by Anne W.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Narayan Hari

Step 1

First, let's assume that the solution to the Laplace equation in polar coordinates is of the form u(r,θ) = R(r)Θ(θ), where R(r) is a function of r only and Θ(θ) is a function of θ only. Show more…

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Laplace equation in polar coordinates âˆ‚^2u/âˆ‚r^2 + (1/r)âˆ‚u/âˆ‚r + (1/r^2)âˆ‚^2u/âˆ‚Î¸^2 = 0 u = u(r,Î¸), 0 < r < 1, 0 < Î¸ < 2Ï€ u(1,Î¸) = 0 Prove that: u(r,Î¸) = 2 - 2r^n sin(nÎ¸), n = 0