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Mathematics Equation Solving Wolfram Language

Finding the inverse of a mapping of a disk onto an airfoil

SALIOU TELLY

Posted 9 years ago

POSTED BY: SALIOU TELLY

13 Replies

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Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 9 years ago

I wonder if you can interpolate the derivatives, instead of derivating the interpolation.

POSTED BY: Gianluca Gorni

Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 9 years ago

If a numerical inverse is enough for you, you can compute the direct transformation in the points in a grid, reverse the order and then use something like ListInterpolation.

POSTED BY: Gianluca Gorni

SALIOU TELLY

Posted 9 years ago

POSTED BY: SALIOU TELLY

SALIOU TELLY

Posted 9 years ago

Daniel, Thanks for the suggestion. Attached is the simplified notebook that I am using.... Regards, Saliou Telly Attachments:

POSTED BY: SALIOU TELLY

Daniel Lichtblau

Daniel Lichtblau, Wolfram Research

Posted 9 years ago

POSTED BY: Daniel Lichtblau

SALIOU TELLY

Posted 9 years ago

Thanks Daniel! I am actually interested in the inverse of the last map in the notebook ((xd, yd) ---> (xa, ya)), which maps the circular annulus to the airfoil annulus. So I would like to find an expression for the inverse map ((xa, ya) ---> (xd, yd)). What you have solved is the inverse mapping of a circular disk to a circular annulus, which was just a step that I use to generate a circular annulus before mapping it to an airfoil annulus. I will try your approach to see if it helps in finding the inverse map of interest..... Thanks again, Saliou

POSTED BY: SALIOU TELLY

Daniel Lichtblau

Daniel Lichtblau, Wolfram Research

Posted 9 years ago

POSTED BY: Daniel Lichtblau

Kay Herbert

Kay Herbert, Chief scientist, Sloan Valve Co.

Posted 9 years ago

I think it would be much easier to keep the map w=f[z] in the complex plane. If you want to invert the map you need to solve for z=f^-1[z] . Mathematica usually can handle it, for example: In[1]:= Solve[a z + b/z == w, z] In[2]:= Solve[a (z - z0) + b/(z - z0) == w, z] Out[2]= {{z -> (w - Sqrt[-4 a b + w^2] + 2 a z0)/(2 a)}, {z -> ( w + Sqrt[-4 a b + w^2] + 2 a z0)/(2 a)}}

POSTED BY: Kay Herbert

SALIOU TELLY

Posted 9 years ago

Thanks Kay! I will give it a try in the complex domain where it was first derived. However, the map in the complex form (w = f(z)) involved magnitude of z ( \|z\| ), which you can see in the algebraic form as sqrt (xd^2 + yd^2)... I was not sure would be easily inverted by Mathematica or any other tool. But it is worth a try and thanks for the suggestion.... Regards, Saliou

POSTED BY: SALIOU TELLY

Kay Herbert

Kay Herbert, Chief scientist, Sloan Valve Co.

Posted 9 years ago

All depends, but usually you get an answer if it exists. Example: In[9]:= Solve[w == (z - z0)/Abs[z] + 1/(z Abs[z]), z] Out[9]= {{z -> (z0 - Sqrt[-4 - 4 w + z0^2])/(2 (1 + w))}, {z -> ( z0 + Sqrt[-4 - 4 w + z0^2])/( 2 (1 + w))}, {z -> (-z0 - Sqrt[-4 + 4 w + z0^2])/( 2 (-1 + w))}, {z -> (-z0 + Sqrt[-4 + 4 w + z0^2])/(2 (-1 + w))}} You can also try Reduce

All depends, but usually you get an answer if it exists. Example:

In[9]:= Solve[w == (z - z0)/Abs[z] + 1/(z Abs[z]), z]

Out[9]= {{z -> (z0 - Sqrt[-4 - 4 w + z0^2])/(2 (1 + w))}, {z -> (
   z0 + Sqrt[-4 - 4 w + z0^2])/(
   2 (1 + w))}, {z -> (-z0 - Sqrt[-4 + 4 w + z0^2])/(
   2 (-1 + w))}, {z -> (-z0 + Sqrt[-4 + 4 w + z0^2])/(2 (-1 + w))}}

You can also try Reduce

POSTED BY: Kay Herbert

SALIOU TELLY

Posted 9 years ago

Thanks Kay! This seems promising as it does appear to be able to solve in the complex form. I will look into the solution in more detail for correctness and also see if I can re-express it in Cartesian form, which is really the ideal form that I would need to work with. I am more of a Matlab person but I am amazed by the power of Mathematica when it comes to symbolic calculations..... Thanks again for your time, Saliou