How to discretize $\frac{\partial^3 f}{\partial x\partial y^2}$ at mesh point $(i,j)$? We should use mesh points which are nearest to $(i,j)$.
1 Answer
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Yes. In my questions, I have figured it out for numerical simulations. $\frac{\partial^3 u}{\partial x\partial y^2}=\frac{u_{i+1,j+1}-u_{i-1,j+1}+u_{i+1,j-1}-u_{i-1,j-1}-2(u_{i+1,j}-u_{i-1,j})}{2\delta x(\delta y)^2}$
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$\begingroup$ Grr, I did not notice you had two derivatives in $y$. For fractions with multiple sup/superscripts, I suggest
\dfrac
or displayed formula... $\endgroup$– user147263Aug 19, 2014 at 13:12