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Sid June 27, 2001 14:24

Inverse Map for Bilinear Elements
Hello all,

I want to compute (\xi,\eta) for given (x,y). I know there is a simple map from the former to the latter: the famous bilinear map. But can I do the inverse for general quadrilateral? How is this usually done?

( I need to interpolate some function value within a quadrilateral using bilinear interpolation. So, I want to compute a function value at some point (x0,y0) in the PHYSICAL plane. I don't know where this point is mapped in the \xi-\eta plane..... So I cannot use N1=(1-\xi)*(1-\eta), etc)


Adrin Gharakhani July 1, 2001 15:01

Re: Inverse Map for Bilinear Elements
There is nothing magical about the transformed equations. If you go through the derivation process in any finite element book you can see how you can evaluate the function at any point within the sub-domain.

The transformation equation for the quadrilateral is nothing but the parametric version of:

f(x,y) = a + b*x + c*y + d*x*y

You can obtain a,b,c, and d by substituting the function values at the four vertices and solving the 4 eqn. 4 unknown system:

f(x_i,y_i) = a + b*x_i + c*y_i + d*x_i*y_i

where (x_i,y_i) are the coordinates at the 4 corners and f() is its functional value. Once you have obtained, a, b, c and d, you can evaluate f(x_0,y_0) easily.

Adrin Gharakhani

Sid July 1, 2001 15:48

Re: Inverse Map for Bilinear Elements
Wow! It looks so easy if I do it numerically. Thanks!


Adrin Gharakhani July 1, 2001 19:05

Re: Inverse Map for Bilinear Elements
You don't do it numerically. It's a 4x4 matrix and you can invert it analytically. If you go through the math and are careful with simplification and factorization you will see terms very similar to the paramteric form of the interpolations emerging. So, you'll end up with just a quadratic polynomial (with known a,b,c and d) that you can use. No need to solve the matrix numerically.

Adrin Gharakhani

Sid July 2, 2001 11:20

Re: Inverse Map for Bilinear Elements
Thank you again for your comments. I haven't been able to figure out the formula. It is messy. Well, I'll keep trying.....


jebediahkerman April 30, 2016 15:55

Although this thread is almost 15 years old now, I found a paper with an explicit formula that may be useful for anyone else who lands on this thread from a google. Surprisingly simple, but I also had trouble getting there on my own. Some of the coefficients might change depending how elements are defined, but the formula is pretty simple.

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