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I have a system of PDEs that switch between the Cartesian coordinate frame $(x,y)$ and the coordinates along an obstacle $(l,n)$, where $l$ is tangential to the obstacle and $n$ is normal to it. The equations to link these coordinates are given by: $$ x=l \cos{\delta} - n\sin{\delta}, \qquad y=l \sin{\delta} + n \cos{\delta} $$ or inversely, $$ l = x \cos{\delta} + y \sin{\delta}, \qquad n=-x\sin{\delta}+y\cos{\delta} $$ where $\delta=\arctan{\frac{ds}{dx}}$. My question is how do you systematically determine how a partial derivative would translate to the other coordinate axis? Is it correct that: $$ \frac{\partial}{\partial l} = \frac{\partial x}{\partial l}\frac{\partial}{\partial x}+\frac{\partial y}{\partial l}\frac{\partial}{\partial y} $$ So now I just need to find $\frac{\partial x}{\partial l}$ and $\frac{\partial y}{\partial l}$. To find the former, I take the first equation solved for $x$ and implicitly differentiate assuming everything but $x,l$ are constant. Therefore: $$ \partial x = \cos{\delta}\ \partial l \implies \frac{\partial x}{\partial l}=\cos{\delta} $$ Doing the same thing with the equation solved for $y$ yields: $$ \partial y = \sin{\delta}\ \partial l \implies \frac{\partial y}{\partial l}=\sin{\delta}$$ Putting this all together, then it would follow that $$ \frac{\partial}{\partial l} = \cos{\delta}\frac{\partial}{\partial x}+\sin{\delta}\frac{\partial}{\partial y} $$ Is this correct? Further, I am a bit concerned that $\delta$ actually has some $x$ dependence based on its definition. Should this affect the relationship between the partial derivatives?

EDIT: Based on the comment from Kurt, it seems that my concern is correct that $\delta=\delta(x)$. I believe the consequence of this is that when you implicitly differentiate the equation solved for $x$, you find: $$ \partial x=\cos{\delta}\ \partial l - l \sin{\delta} \ \partial x - n \cos{\delta} \ \partial x $$ and hence $$ \frac{\partial x}{\partial l} = \frac{\cos{\delta}}{1+l\sin{\delta}+n\cos{\delta}} $$ Is this now the correct procedure for continuing with the other relations?

EDIT 2: Assume further that there is no time dependence within the obstacle so $\delta=\delta(x)$ as stated above (i.e. ignore the old comment).

What would be the gradient and Laplacian in this new coordinate frame?

Mjoseph
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    This is not correct precisely for the reason you are concerned about. What is $\frac{ds}{dx}?$ – Kurt G. Aug 29 '24 at 17:10
  • Here, $s$ is an unknown function of $x,t$ that has to be solved for. – Mjoseph Aug 30 '24 at 13:01
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    Of $x,\color{red}t,?$ Welcome $t$ to the zoo of unexplained variables! – Kurt G. Aug 30 '24 at 13:10
  • $t$ represents time in this problem. $(x,z)$ and $(l,n)$ are two differential spatial coordinate frames in which I view the problem - none of these have time dependence although I suppose $\delta$ does. – Mjoseph Aug 30 '24 at 13:28
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    $\color{red}z,?$ Another one. Whatever you mean by those pairs, these are coordinates that depend on time as soon as $\delta$ does. – Kurt G. Aug 30 '24 at 13:33
  • Sorry I meant $y$ - there is no $z$. I see though thank you. So should I instead have an additional term in the equation for $\partial_l$ of the form $\frac{\partial t}{\partial l}\frac{\partial}{\partial t}$? And then similarly for all the other relations? – Mjoseph Aug 30 '24 at 13:36
  • Further, should I have terms that include $\frac{\partial \delta}{\partial x}$? Take for instance the second to last equation in the edited part. Is that correct or should the second and third terms on the RHS also include $\frac{\partial \delta}{\partial x}$? – Mjoseph Aug 30 '24 at 13:40
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    You have to view the coordinate tranformation as a differentiable mapping between $(x,y)$ and $(l,n),.$ Forget first about a specific form and apply the full chain rule. This will show how the partial derivatives transform. The method is the same as here. – Kurt G. Aug 30 '24 at 13:51
  • I guess you're studying a problem in hydrodynamics. Can you point out a reference? Your description confused me a bit, and maybe it's better to look at the original problem. – Quillo Apr 10 '25 at 17:26
  • The idea to change to this coordinate frame came from the paper: "Surface-tension-driven flow on a moving curved surface" by P.D. Howell from the Journal of Engineering Mathematics written in 2003. Although what they are doing there is much more general. – Mjoseph Apr 11 '25 at 13:38
  • I am just trying to solve the Stokes' equations rewritten in this coordinate frame (l,n). The first step to doing this obviously is transforming the gradient and Laplacian operators which is what I am asking for help with here. – Mjoseph Apr 11 '25 at 13:39

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