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Consider the Neumann problem:

$$U_{xx} + U_{yy} = 0, \qquad 0 < x < \pi, \quad -1< y < 1$$ with

$$U_x(0,y) = U_x(\pi,y) = 0$$ $$U_y(x,-1) = 0$$ $$ U_y(x,1) = \alpha + \beta \sin(x)$$

Does the problem admit solutions for the following?

  1. $\alpha = 0, \beta = 1$

  2. $\alpha = -1, \beta = \frac{\pi}{2}$

  3. $\alpha = 1, \beta = \frac{\pi}{2}$

  4. $\alpha = 1, \beta = -\pi$

Mark McClure
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user120386
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    Does my edit ask the right question? Also, what have you tried, where are you stuck, ...? – not all wrong Jan 29 '14 at 12:11
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    For PDEs, it is always important to specify what you mean by solutions. Are you trying to find classical solutions (in the sense that $U\in \mathcal{C}^2([0,\pi]\times [-1,1])$) or weak solutions (in the sens of distributions)? – user37238 Jan 29 '14 at 12:21

1 Answers1

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A solution to the problem must satisfy $$\oint \frac{\partial u}{\partial \nu} ds=0 $$ where the integral is taken over the boundary of the domain. This reduces to $$\int_0^\pi \left( \alpha+\beta \sin x \right) \mathrm{d}x=0. $$

user1337
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  • sir, I am not familiar with this result, I would be thanful for give me explanation of this result.Please explain this – user120386 Jan 29 '14 at 16:43
  • It follows from Green's first identity applied to the functions $\phi=u,\psi=1$. The notations are a bit different, but in any case $\mathbf{n} \cdot \nabla \phi$ means the derivative with respect to the outwards pointing normal vector. – user1337 Jan 29 '14 at 17:13