Checknig the DF condition in the DF-SEM

   \begin{equation*}  \mathbf{u'}(x)=\sqrt{\frac{V_{b}k}{N\sigma^{3}}}\sum_{k=1}^{N}-\frac{q(\frac{|\mathbf{x}|}{\sigma})}{|\mathbf{x}|^{3}}\mathbf{x}\times\alpha^{k}\end{equation*}  (1)

Using a general function $ q(x,y,z) $, we may calculate the divergence in this case, which of course in a DF method must be imposed to be 0.

  \begin{equation*}  \begin{array}{lll}  \nabla\cdot\mathbf{u'=} & \sigma^{2}\{ & [\alpha_{3}(x-x^{k})-\alpha_{1}(z-z^{k})][+\frac{1}{r^{3}}\frac{\partial q}{\partial y}]+...\\ & & [\alpha_{2}(x-x^{k})-\alpha_{1}(y-y^{k})][-\frac{1}{r^{3}}\frac{\partial q}{\partial z}]+...\\ & & [\alpha_{3}(y-y^{k})-\alpha_{2}(z-z^{k})][-\frac{1}{r^{3}}\frac{\partial q}{\partial x}]\}\end{array}  \end{equation*} (2)

where $ r = \sqrt{(x - x^k)^2+(y - y^k)^2+(z-z^k)^2} $


 \begin{equation*}  \begin{cases}  \frac{\partial q}{\partial x} - \frac{3 (x - x^k) q}{ r^2} = 0 \\ \frac{\partial q}{\partial y} - \frac{3 (y - y^k) q}{ r^2} = 0 \\ \frac{\partial q}{\partial z} - \frac{3 (z - z^k) q}{ r^2} = 0  \end{cases} \end{equation*}(3)

The final function is: $ q = A_0 r^3 $. This is basically a cubic function that, when used, simplifies with the denominator in equation (1), leaving the i-th velocity component independent from the i-th axis (u independent of x, v independent of y and w independent of z).

Further investigation showed that the system (3) is not the only solution of our problem. In fact basically in equation (2) may be grouped in a different way and lead us to a different system of solving equations. Anyway, in the Matlab provided at the bottom of the page, I tried (using a symbolic calculator) to manage a $ q(\sqrt{(x - x^k)^2+(y - y^k)^2+(z-z^k)^2}) $ general function, and for each function depending only on the distance leads us to a divergence free velocity field [see matlab script].

shape_function_winckelmans.jpg

After all these considerations I wanted to check everything in the DF-SEM. I calculate then the divergence in any simulations and I plotted the results in fig. 2.I calculate the derivative with a central difference 2nd order scheme. This picture shows that the method is divergence free everywhere but near the eddy edge.

problem_divergence_eddy_boundary.jpg

After all these considerations I wanted to check everything in the DF-SEM. I calculate then the divergence in any simulations and I plotted the results in fig. 2.

   \begin{equation*}  \begin{array}{l}  u = \frac{q(\mathbf{r})}{\mathbf{r}^3} (\alpha_{3}(y-y^{k})-\alpha_{2}(z-z^{k}))\\ v=\frac{q(\mathbf{r})}{\mathbf{r}^{3}}(\alpha_{1}(z-z^{k})-\alpha_{3}(x-x^{k}))\\ w=\frac{q(\mathbf{r})}{\mathbf{r}^{3}}(\alpha_{2}(x-x^{k})-\alpha_{1}(y-y^{k}))  \end{array}  \end{equation*}  (4)

   \begin{equation*}  \begin{array}{l}  \frac{\partial u}{\partial x}=[\frac{\partial q(\mathbf{r})}{\partial\mathbf{r}}\frac{\partial\mathbf{r}}{\partial x}\mathbf{r}^{3}-q(\mathbf{r})3\mathbf{r}^{2}\frac{\partial\mathbf{r}}{x}](\alpha_{3}(y-y^{k})-\alpha_{2}(z-z^{k}))\\  \frac{\partial v}{\partial y}=[\frac{\partial q(\mathbf{r})}{\partial\mathbf{r}}\frac{\partial\mathbf{r}}{\partial y}\mathbf{r}^{3}-q(\mathbf{r})3\mathbf{r}^{2}\frac{\partial\mathbf{r}}{y}](\alpha_{1}(z-z^{k})-\alpha_{3}(x-x^{k})) \\  \frac{\partial w}{\partial z}=[\frac{\partial q(\mathbf{r})}{\partial\mathbf{r}}\frac{\partial\mathbf{r}}{\partial z}\mathbf{r}^{3}-q(\mathbf{r})3\mathbf{r}^{2}\frac{\partial\mathbf{r}}{z}](\alpha_{2}(x-x^{k})-\alpha_{1}(y-y^{k}))  \end{array}  \end{equation*}  (5)

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elsem symcal2cart.m manage 1.2 K 2010-04-07 - 12:42 RuggeroPoletto  
Topic revision: r10 - 2010-04-12 - 15:55:39 - RuggeroPoletto
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