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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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I wonder if you can interpolate the derivatives, instead of derivating the interpolation.
POSTED BY: Gianluca Gorni
POSTED BY: Gianluca Gorni
SALIOU TELLY
SALIOU TELLY
Posted 9 years ago
POSTED BY: SALIOU TELLY
SALIOU TELLY
SALIOU TELLY
Posted 9 years ago
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POSTED BY: SALIOU TELLY
POSTED BY: Daniel Lichtblau
SALIOU TELLY
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

I doubt it will give a usable result. I transformed to polar because that made the expressions more tractable; this involves also making new variables for sine and cosine, and adding the usual identity to link those algebraically. Found there are four solutions and they are huge. For better control over the process I used GroebnerBasis directly, then used Roots to solve for the radial variable, and back-substituted to get the sine and cosine solution for the angular variable (there is one per radial solution).

e1 = {- xa + 
    1/2*(\[Rho]*xd + \[Alpha]* 
        Sqrt[(xd^2) + (
         yd^2) ] + (\[Lambda]^2*(xd^2 + 
          yd^2)*(\[Rho]*xd + \[Alpha]*  
           Sqrt[(xd^2) + (yd^2) ]))/((\[Rho]* xd + \[Alpha]*  
           Sqrt[(xd^2) + (yd^2) ])^2 + (\[Rho]* yd + \[Beta]* 
           Sqrt[(xd^2) + (yd^2) ])^2)),
   -ya + 1/
     2*(\[Rho]*yd + \[Beta]* 
        Sqrt[(xd^2) + (
         yd^2) ] - (\[Lambda]^2*(xd^2 + 
          yd^2)*(\[Rho]*yd + \[Beta]*  
           Sqrt[(xd^2) + (yd^2) ]))/((\[Rho]* xd + \[Alpha]*  
           Sqrt[(xd^2) + (yd^2) ])^2 + (\[Rho]* yd + \[Beta]* 
           Sqrt[(xd^2) + (yd^2) ])^2))};
e2 = e1 /. {(xd^2 + yd^2) -> r^2, Sqrt[(xd^2) + (yd^2) ] -> r};

e3 = Join[
   Together[
    e2 /. {xd -> r*ctheta, yd -> r*stheta}], {ctheta^2 + stheta^2 - 
     1}];

Timing[gb = 
   GroebnerBasis[e3, {ctheta, stheta, r}, 
    CoefficientDomain -> RationalFunctions];]

(* Out[57]= {2.448772, Null} *)

radii = Solve[gb[[1]] == 0, r, Quartics -> False];
gb2 = Rest[gb] /. radii[[1]];

sin = Solve[gb2[[1]] == 0, stheta];
cos = Solve[gb2[[2]] == 0, ctheta];

As I wrote, this is probably too large to be useful. And there are three others; which is "correct" could vary with parameter values, including the specific values of xa and ya.

In[82]:= radii[[1]]

(* Out[82]= {r -> 
  Root[16 xa^4 \[Alpha]^4 + 32 xa^2 ya^2 \[Alpha]^4 + 
     16 ya^4 \[Alpha]^4 + 32 xa^4 \[Alpha]^2 \[Beta]^2 + 
     64 xa^2 ya^2 \[Alpha]^2 \[Beta]^2 + 
     32 ya^4 \[Alpha]^2 \[Beta]^2 + 16 xa^4 \[Beta]^4 + 
     32 xa^2 ya^2 \[Beta]^4 + 16 ya^4 \[Beta]^4 - 
     16 xa^4 \[Alpha]^2 \[Rho]^2 - 32 xa^2 ya^2 \[Alpha]^2 \[Rho]^2 - 
     16 ya^4 \[Alpha]^2 \[Rho]^2 - 16 xa^4 \[Beta]^2 \[Rho]^2 - 
     32 xa^2 ya^2 \[Beta]^2 \[Rho]^2 - 
     16 ya^4 \[Beta]^2 \[Rho]^2 + (-32 xa^3 \[Alpha]^5 - 
        32 xa ya^2 \[Alpha]^5 - 32 xa^2 ya \[Alpha]^4 \[Beta] - 
        32 ya^3 \[Alpha]^4 \[Beta] - 64 xa^3 \[Alpha]^3 \[Beta]^2 - 
        64 xa ya^2 \[Alpha]^3 \[Beta]^2 - 
        64 xa^2 ya \[Alpha]^2 \[Beta]^3 - 
        64 ya^3 \[Alpha]^2 \[Beta]^3 - 32 xa^3 \[Alpha] \[Beta]^4 - 
        32 xa ya^2 \[Alpha] \[Beta]^4 - 32 xa^2 ya \[Beta]^5 - 
        32 ya^3 \[Beta]^5 - 32 xa^3 \[Alpha]^3 \[Lambda]^2 - 
        32 xa ya^2 \[Alpha]^3 \[Lambda]^2 + 
        32 xa^2 ya \[Alpha]^2 \[Beta] \[Lambda]^2 + 
        32 ya^3 \[Alpha]^2 \[Beta] \[Lambda]^2 - 
        32 xa^3 \[Alpha] \[Beta]^2 \[Lambda]^2 - 
        32 xa ya^2 \[Alpha] \[Beta]^2 \[Lambda]^2 + 
        32 xa^2 ya \[Beta]^3 \[Lambda]^2 + 
        32 ya^3 \[Beta]^3 \[Lambda]^2 + 48 xa^3 \[Alpha]^3 \[Rho]^2 + 
        48 xa ya^2 \[Alpha]^3 \[Rho]^2 + 
        48 xa^2 ya \[Alpha]^2 \[Beta] \[Rho]^2 + 
        48 ya^3 \[Alpha]^2 \[Beta] \[Rho]^2 + 
        48 xa^3 \[Alpha] \[Beta]^2 \[Rho]^2 + 
        48 xa ya^2 \[Alpha] \[Beta]^2 \[Rho]^2 + 
        48 xa^2 ya \[Beta]^3 \[Rho]^2 + 48 ya^3 \[Beta]^3 \[Rho]^2 + 
        16 xa^3 \[Alpha] \[Lambda]^2 \[Rho]^2 + 
        16 xa ya^2 \[Alpha] \[Lambda]^2 \[Rho]^2 - 
        16 xa^2 ya \[Beta] \[Lambda]^2 \[Rho]^2 - 
        16 ya^3 \[Beta] \[Lambda]^2 \[Rho]^2 - 
        16 xa^3 \[Alpha] \[Rho]^4 - 16 xa ya^2 \[Alpha] \[Rho]^4 - 
        16 xa^2 ya \[Beta] \[Rho]^4 - 
        16 ya^3 \[Beta] \[Rho]^4) #1 + (24 xa^2 \[Alpha]^6 + 
        8 ya^2 \[Alpha]^6 + 32 xa ya \[Alpha]^5 \[Beta] + 
        56 xa^2 \[Alpha]^4 \[Beta]^2 + 40 ya^2 \[Alpha]^4 \[Beta]^2 + 
        64 xa ya \[Alpha]^3 \[Beta]^3 + 
        40 xa^2 \[Alpha]^2 \[Beta]^4 + 56 ya^2 \[Alpha]^2 \[Beta]^4 + 
        32 xa ya \[Alpha] \[Beta]^5 + 8 xa^2 \[Beta]^6 + 
        24 ya^2 \[Beta]^6 + 48 xa^2 \[Alpha]^4 \[Lambda]^2 + 
        16 ya^2 \[Alpha]^4 \[Lambda]^2 + 
        32 xa^2 \[Alpha]^2 \[Beta]^2 \[Lambda]^2 - 
        32 ya^2 \[Alpha]^2 \[Beta]^2 \[Lambda]^2 - 
        16 xa^2 \[Beta]^4 \[Lambda]^2 - 
        48 ya^2 \[Beta]^4 \[Lambda]^2 + 
        24 xa^2 \[Alpha]^2 \[Lambda]^4 + 
        8 ya^2 \[Alpha]^2 \[Lambda]^4 - 
        32 xa ya \[Alpha] \[Beta] \[Lambda]^4 + 
        8 xa^2 \[Beta]^2 \[Lambda]^4 + 
        24 ya^2 \[Beta]^2 \[Lambda]^4 - 52 xa^2 \[Alpha]^4 \[Rho]^2 - 
        20 ya^2 \[Alpha]^4 \[Rho]^2 - 
        64 xa ya \[Alpha]^3 \[Beta] \[Rho]^2 - 
        72 xa^2 \[Alpha]^2 \[Beta]^2 \[Rho]^2 - 
        72 ya^2 \[Alpha]^2 \[Beta]^2 \[Rho]^2 - 
        64 xa ya \[Alpha] \[Beta]^3 \[Rho]^2 - 
        20 xa^2 \[Beta]^4 \[Rho]^2 - 52 ya^2 \[Beta]^4 \[Rho]^2 - 
        40 xa^2 \[Alpha]^2 \[Lambda]^2 \[Rho]^2 - 
        8 ya^2 \[Alpha]^2 \[Lambda]^2 \[Rho]^2 + 
        8 xa^2 \[Beta]^2 \[Lambda]^2 \[Rho]^2 + 
        40 ya^2 \[Beta]^2 \[Lambda]^2 \[Rho]^2 - 
        4 xa^2 \[Lambda]^4 \[Rho]^2 - 4 ya^2 \[Lambda]^4 \[Rho]^2 + 
        32 xa^2 \[Alpha]^2 \[Rho]^4 + 16 ya^2 \[Alpha]^2 \[Rho]^4 + 
        32 xa ya \[Alpha] \[Beta] \[Rho]^4 + 
        16 xa^2 \[Beta]^2 \[Rho]^4 + 32 ya^2 \[Beta]^2 \[Rho]^4 + 
        8 xa^2 \[Lambda]^2 \[Rho]^4 - 8 ya^2 \[Lambda]^2 \[Rho]^4 - 
        4 xa^2 \[Rho]^6 - 4 ya^2 \[Rho]^6) #1^2 + (-8 xa \[Alpha]^7 - 
        8 ya \[Alpha]^6 \[Beta] - 24 xa \[Alpha]^5 \[Beta]^2 - 
        24 ya \[Alpha]^4 \[Beta]^3 - 24 xa \[Alpha]^3 \[Beta]^4 - 
        24 ya \[Alpha]^2 \[Beta]^5 - 8 xa \[Alpha] \[Beta]^6 - 
        8 ya \[Beta]^7 - 24 xa \[Alpha]^5 \[Lambda]^2 - 
        8 ya \[Alpha]^4 \[Beta] \[Lambda]^2 - 
        16 xa \[Alpha]^3 \[Beta]^2 \[Lambda]^2 + 
        16 ya \[Alpha]^2 \[Beta]^3 \[Lambda]^2 + 
        8 xa \[Alpha] \[Beta]^4 \[Lambda]^2 + 
        24 ya \[Beta]^5 \[Lambda]^2 - 24 xa \[Alpha]^3 \[Lambda]^4 + 
        8 ya \[Alpha]^2 \[Beta] \[Lambda]^4 + 
        8 xa \[Alpha] \[Beta]^2 \[Lambda]^4 - 
        24 ya \[Beta]^3 \[Lambda]^4 - 8 xa \[Alpha] \[Lambda]^6 + 
        8 ya \[Beta] \[Lambda]^6 + 24 xa \[Alpha]^5 \[Rho]^2 + 
        24 ya \[Alpha]^4 \[Beta] \[Rho]^2 + 
        48 xa \[Alpha]^3 \[Beta]^2 \[Rho]^2 + 
        48 ya \[Alpha]^2 \[Beta]^3 \[Rho]^2 + 
        24 xa \[Alpha] \[Beta]^4 \[Rho]^2 + 
        24 ya \[Beta]^5 \[Rho]^2 + 
        32 xa \[Alpha]^3 \[Lambda]^2 \[Rho]^2 - 
        32 ya \[Beta]^3 \[Lambda]^2 \[Rho]^2 + 
        8 xa \[Alpha] \[Lambda]^4 \[Rho]^2 + 
        8 ya \[Beta] \[Lambda]^4 \[Rho]^2 - 
        24 xa \[Alpha]^3 \[Rho]^4 - 
        24 ya \[Alpha]^2 \[Beta] \[Rho]^4 - 
        24 xa \[Alpha] \[Beta]^2 \[Rho]^4 - 
        24 ya \[Beta]^3 \[Rho]^4 - 
        8 xa \[Alpha] \[Lambda]^2 \[Rho]^4 + 
        8 ya \[Beta] \[Lambda]^2 \[Rho]^4 + 8 xa \[Alpha] \[Rho]^6 + 
        8 ya \[Beta] \[Rho]^6) #1^3 + (\[Alpha]^8 + 
        4 \[Alpha]^6 \[Beta]^2 + 6 \[Alpha]^4 \[Beta]^4 + 
        4 \[Alpha]^2 \[Beta]^6 + \[Beta]^8 + 
        4 \[Alpha]^6 \[Lambda]^2 + 
        4 \[Alpha]^4 \[Beta]^2 \[Lambda]^2 - 
        4 \[Alpha]^2 \[Beta]^4 \[Lambda]^2 - 
        4 \[Beta]^6 \[Lambda]^2 + 6 \[Alpha]^4 \[Lambda]^4 - 
        4 \[Alpha]^2 \[Beta]^2 \[Lambda]^4 + 
        6 \[Beta]^4 \[Lambda]^4 + 4 \[Alpha]^2 \[Lambda]^6 - 
        4 \[Beta]^2 \[Lambda]^6 + \[Lambda]^8 - 
        4 \[Alpha]^6 \[Rho]^2 - 12 \[Alpha]^4 \[Beta]^2 \[Rho]^2 - 
        12 \[Alpha]^2 \[Beta]^4 \[Rho]^2 - 4 \[Beta]^6 \[Rho]^2 - 
        8 \[Alpha]^4 \[Lambda]^2 \[Rho]^2 + 
        8 \[Beta]^4 \[Lambda]^2 \[Rho]^2 - 
        4 \[Alpha]^2 \[Lambda]^4 \[Rho]^2 - 
        4 \[Beta]^2 \[Lambda]^4 \[Rho]^2 + 6 \[Alpha]^4 \[Rho]^4 + 
        12 \[Alpha]^2 \[Beta]^2 \[Rho]^4 + 6 \[Beta]^4 \[Rho]^4 + 
        4 \[Alpha]^2 \[Lambda]^2 \[Rho]^4 - 
        4 \[Beta]^2 \[Lambda]^2 \[Rho]^4 - 2 \[Lambda]^4 \[Rho]^4 - 
        4 \[Alpha]^2 \[Rho]^6 - 
        4 \[Beta]^2 \[Rho]^6 + \[Rho]^8) #1^4 &, 1]} *)
POSTED BY: Daniel Lichtblau

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
SALIOU TELLY
Posted 9 years ago
POSTED BY: SALIOU TELLY

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
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

POSTED BY: SALIOU TELLY
SALIOU TELLY
SALIOU TELLY
Posted 9 years ago
POSTED BY: SALIOU TELLY

It would be more easy to test this if Mathematica code were provided.
POSTED BY: Daniel Lichtblau
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