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I have my pde 2D problem with robin condition (form: du/dn +ku=g) to solve with matlab. i have the exact function u and I want to find the function g in robin condition. How can i do it? thanks for the help :)

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  • $\begingroup$ As you can see with the benefit of the Answer below, there is missing information needed to explain the normal derivative, namely what shape is the boundary? The normal derivative is the directional derivative of u with respect to an outward facing unit vector perpendicular to the curve defining the boundary of your domain. $\endgroup$ – hardmath Dec 28 '13 at 14:50
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You can compute $\nabla u$ and the outward normal $n$ to get $\frac{\partial u}{\partial n}$. Add that and $\kappa u$ on the Robin boundary and get $g$. Is that what you're looking for?

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  • $\begingroup$ Yes in particular my problem is how to calculate the outward normal.I don't understand how to do it..Then the normal is outward my domain? $\endgroup$ – Betelgeuse Dec 14 '13 at 22:48
  • $\begingroup$ @Betelgeuse: What is the shape of your domain? Does it have curved boundaries? $\endgroup$ – Paul Dec 15 '13 at 2:58
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    $\begingroup$ @Betelgeuse: the normal vector is simply a vector $\mathbf b=\mathbf n(\mathbf x)$ that at every point $\mathbf x$ of the boundary points perpendicular to the boundary, outward of the domain, with length one. For example, if your domain is a box, $\mathbf n$ is either $(\pm 1, 0)^T$ or $(0, \pm 1)^T$, depending on which of the four parts of the domain you're on. The normal derivative $\partial u/\partial n=\mathbf n \cdot \nabla u$ is then just the dot product of the gradient and this normal vector. $\endgroup$ – Wolfgang Bangerth Dec 15 '13 at 8:12
  • $\begingroup$ Of course, $\mathbf b$ in the first line should have been $\mathbf n$. $\endgroup$ – Wolfgang Bangerth Dec 15 '13 at 12:48
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    $\begingroup$ If your domain is a box, the normal on a side of the box is just the unit vector orthogonal to that box side. The outward unit normal is the one that points outside the box. $\endgroup$ – Jesse Chan Dec 15 '13 at 16:47

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