Question 5 of 7: Geometric isotropy of polynomial field representations

Question 5: Geometric isotropy of polynomial field representations (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Three candidate polynomial expansions to audit, and a square bilinear element of side L whose field is written first in axes at node 1 and then in the $\xi,\eta$ axes centred at node 3.

Find. (a) a definition of geometric isotropy; (b) a verdict with justification for each of the three polynomials; (c) a demonstration that the bilinear square keeps its form under the change of axes.

Approach. State the definition, then apply the two operational tests it implies — completeness through some order, and symmetric pairing of any higher-order terms about the axis of Pascal's triangle — and finally substitute the coordinate transformation into the bilinear expansion to show that no new monomial is generated.

(a) Meaning of geometric isotropy. A polynomial field representation possesses geometric isotropy (also called spatial isotropy or geometric invariance) when the interpolated field is unchanged by a change of the reference coordinate axes — a translation of the origin to a different node, or an interchange or reversal of the axis directions. In practical terms the element must behave identically no matter which corner the analyst calls node 1 or which direction is called x: the polynomial basis is mapped into itself by the transformation, so only the coefficients change, never the set of monomials.

The requirement matters because a real mesh contains elements of the same type in every orientation. If the basis were not invariant, two identical elements rotated relative to one another would have different stiffness properties and the computed answer would depend on how the mesh happened to be numbered. Two working rules follow from the definition: the polynomial must be complete through some order p (every term of Pascal's triangle up to that row present), and any terms retained above order p must appear in symmetric pairs about the vertical axis of Pascal's triangle, that is, $x^{a}y^{b}$ must be accompanied by $x^{b}y^{a}$.

(b) Auditing the three expansions.

1. (i) One-dimensional quartic with a gap. The expansion $u = C_{1} + C_{2}x + C_{3}x^{2} + C_{4}x^{4}$ contains orders 0, 1, 2 and 4 but omits $x^{3}$, so it is not complete. Translating the origin, $x \to x^{\prime} + a$, expands $x^{4}$ into a term in $x^{\prime 3}$ that the basis cannot represent, so the interpolation changes when the axis origin moves. $$\text{(i): } \boxed{\text{no geometric isotropy}}$$
2. (ii) Complete two-dimensional cubic. The ten terms are precisely rows 0 to 3 of Pascal's triangle: $1$; $x, y$; $x^{2}, xy, y^{2}$; $x^{3}, x^{2}y, xy^{2}, y^{3}$. Completeness through order 3 is satisfied and nothing is retained above it, so every coordinate transformation maps the basis onto itself. $$\text{(ii): } \boxed{\text{geometric isotropy is possessed}}$$
3. (iii) Complete quadratic plus a symmetric cubic pair. The first six terms form the complete quadratic, and the two extra terms $x^{2}y$ and $xy^{2}$ are mirror images of one another about the axis of Pascal's triangle. The missing $x^{3}$ and $y^{3}$ are absent together, so no direction is privileged; interchanging x and y merely swaps $C_{7}$ and $C_{8}$. $$\text{(iii): } \boxed{\text{geometric isotropy is possessed}}$$ This is the eight-term basis of the serendipity quadratic quadrilateral, which is exactly why that element is usable in an arbitrarily oriented mesh.

(c) The bilinear square under axes at node 3. Take the square with node 1 at the origin of the $x,y$ frame, so the nodes are at $1(0,0)$, $2(L,0)$, $3(L,L)$ and $4(0,L)$. The figure places the new axes at node 3 with $\xi$ directed from node 3 towards node 4 and $\eta$ from node 3 towards node 2, which is a shift of the origin to node 3 combined with a reversal of both axis directions (equivalent to a 180° rotation of the reference frame).

1. Write the coordinate transformation. Measuring from node 3 backwards along each side:$$\xi = L - x \;\Longrightarrow\; x = L - \xi, \qquad\eta = L - y \;\Longrightarrow\; y = L - \eta$$
2. Substitute into the interpolation. Replacing $x$ and $y$ in $u = C_{1} + C_{2}x + C_{3}y + C_{4}xy$:$$u = C_{1} + C_{2}(L-\xi) + C_{3}(L-\eta) + C_{4}(L-\xi)(L-\eta)$$
3. Expand and collect. Multiplying out the product term gives $C_{4}\bigl(L^{2} - L\xi - L\eta + \xi\eta\bigr)$, and grouping by monomial:$$u = \underbrace{\bigl(C_{1} + C_{2}L + C_{3}L + C_{4}L^{2}\bigr)}_{C_{1}^{*}} + \underbrace{\bigl(-C_{2} - C_{4}L\bigr)}_{C_{2}^{*}}\xi + \underbrace{\bigl(-C_{3} - C_{4}L\bigr)}_{C_{3}^{*}}\eta + \underbrace{C_{4}}_{C_{4}^{*}}\,\xi\eta$$
4. Compare the two forms. The result is$$\boxed{\;u(\xi,\eta) = C_{1}^{*} + C_{2}^{*}\xi + C_{3}^{*}\eta + C_{4}^{*}\xi\eta\;}$$ which is the same four-term bilinear polynomial as the original, with only the constants relabelled. No quadratic or cubic monomial appears and no term is lost, so the element possesses geometric isotropy.

The mechanism is worth naming. The bilinear basis $\{1, x, y, xy\}$ is complete through order 1, and the single higher-order term $xy$ is its own mirror image about the Pascal axis, so the symmetric-pair rule is satisfied trivially. That is exactly the condition tested in part (b), and part (c) is a direct demonstration of it: a shift of origin can only redistribute coefficients among terms already in the basis. The same substitution with axes at node 2 or node 4 produces the same conclusion.