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Given $\mathbf X(s_1, s_2, v) = \Delta t\mathbf v+\sigma s_1(\hat{\mathbf n}_1+\mathbf v)+\tau s_2(\hat{\mathbf n}_2+\mathbf v)$, is it possible to express $\hat{\mathbf n}_1\cdot\nabla_{\mathbf X}$ and $\hat{\mathbf n}_2\cdot\nabla_{\mathbf X}$ in terms of the $\partial_{s_1}$, $\partial_{s_2}$, and/or $\partial_{\mathbf v}$? I need to put this operator on a function in $\mathbf X$ and then do a double integral with respect to $s_1$ and $s_2$. Expressing it in terms of the derivatives of these two will simplify the integrals.

$\Delta t, \sigma$ and $\tau$ are fixed constants. $\hat{\mathbf n}_1$ and $\hat{\mathbf n}_2$ are also fixed directions and $\mathbf v$ are independent vectors from the two, not necessarily orthogonal.

I used directional derivatives and obtained $\hat{\mathbf n}_1\cdot\nabla_{\mathbf X}=\partial_{s_1}-\frac{\sigma}{\tau}\partial_{s_2}$ and $\hat{\mathbf n}_2\cdot\nabla_{\mathbf X}=\partial_{s_2}-\frac{\tau}{\sigma}\partial_{s_1}$

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  • $\begingroup$ Try to provide some background on the problem, and also tell us what you have tried. I suggest you see this article How to ask a good question $\endgroup$ Commented Oct 18 at 20:11
  • $\begingroup$ $\nabla_{\mathbf X + f\mathbf Y} = \nabla_{\mathbf X} + f\nabla_{\mathbf Y}$. But your notation is far from clear. $\endgroup$ Commented Oct 18 at 20:22
  • $\begingroup$ I added some details about the variables. $\endgroup$ Commented Oct 18 at 20:33
  • $\begingroup$ This seems quite wrong. $\nabla_{s_1\hat{\mathbf n}_1} = s_1\nabla_{\hat{\mathbf n}_1}$. The $\mathbf n_1$ certainly can never pull out as you suggested. $\endgroup$ Commented Oct 18 at 21:11

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