usumdelphini
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 Feb25 comment Polar coordinates for vector difference in $\mathbb{R}^2$ up up up up up hp Dec13 comment Finite differences and conservation law I edited the question to avoid getting too much into the detail of my specific problem, I hope that increases the interest. Dec13 comment Finite differences and conservation law I edited the question to avoid getting too much into the detail of my specific problem, I hope that rises the interest. Dec13 comment Finite differences and conservation law It seems to me that here the stiff term will have trouble, because there is no implicit part in your scheme. Dec12 comment Finite differences and conservation law I tried to impose conservation on the whole r.h.s, by discretising it in conserved form, i.e. as a first derivative, but that gives me an unstable thing (I see sawtooth modes entering the grid) Jul8 comment Space-dependent diffusivity and finite-differences Found the mistake, in the explicit part I used $u_{j+1}-u_j$ instead of $[u_{j+1}-u_{j-1}]/2$! Jul8 comment Space-dependent diffusivity and finite-differences Usually, to test it in the simplest case, I use $D(x)=x$ or $D(x)=1-x$, to see how it behaves going to both boundaries. Jun5 comment Multiplying a vector and a matrix's rows Could you please explicit that? Oct25 comment Vectorial derivative @Nick Could you help me trying to explicit the calculation in components of with the index notation? Because I cannot see it, sorry. Oct25 comment Vectorial derivative Why should I do that, mate? Feb21 comment Orthogonal Part Operator $(\boldsymbol{p}\cdot\nabla\boldsymbol{p})_\alpha=p_\beta\partial_\beta p_\alpha$ Feb21 comment Orthogonal Part Operator No, I simply meant the application of $P$ to those terms: $P(\boldsymbol{p}\cdot\nabla\boldsymbol{p})$ Feb21 comment Orthogonal Part Operator Yes, I need to be clearer. $\boldsymbol{p}$ is a column vector of $\mathbb R^3$, function of $\boldsymbol{x}$, respect to which the differentiation in $\nabla$ is carried. The $P$ operator is exactly the matrix $P_{\alpha\beta}=\delta_{\alpha\beta}-p_\alpha p_\beta$ Jun22 comment Closed form with of a series Mathematica Yes, but what about the closed form if $x$ can be every value, so not necessarily $0x/x_{0}}\frac{1}{k-x/x_{0}}\right)=\frac{1}{x_{‌​0}}\left(\sum_{-k'x/x_{0}}\frac{1}{k-x/x_{0‌​}}\right)=\frac{1}{x_{0}}\left(\sum_{k'>-x/x_{0}}\frac{1}{x/x_{0}+k'}+\sum_{k>x/x‌​_{0}}\frac{1}{k-x/x_{0}}\right)\leq\frac{1}{x_{0}}\sum_{k>x/x_{0}}\left(\frac{1}{‌​x/x_{0}+k}+\frac{1}{k-x/x_{0}}\right) =\frac{1}{x_{0}}\sum_{k>x/x_{0}}\left(\frac{1}{k^{2}-x^{2}/x_{0}^{2}}\right) $$Jun17 comment Closed form of \sum_{k=-\infty}^{+\infty}\frac{1}{|x-kx_0|} This is my friend's proof of convergence:$$S=\sum_{k=-\infty}^{+\infty} \frac{1}{|x-kx_0|} = \frac{1}{x_0}\left(\sum_{kx/x_0} \frac{1}{k-x/x_0}\right) = \frac{1}{x_0}\left(\sum_{-k'x/x_0} \frac{1}{k-x/x_0}\right)= \frac{1}{x_0}\left(\sum_{k'>-x/x_0} \frac{1}{x/x_0+k'}+\sum_{k>x/x_0} \frac{1}{k-x/x_0}\right)\leq \frac{1}{x_0}\sum_{k>x/x_0}\left(\frac{1}{x/x_0+k}+\frac{1}{k-x/x_0}\right)= \frac{1}{x_0} \sum_{k>x/x_0} \left( \frac{1}{k^2-x^2/x_0^2} \right)$\$