Question 1 of 7: Geometric isotropy of polynomial interpolations

Question 1: Geometric isotropy of polynomial interpolations (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. An eight-node square element of side $L$, carrying the eight-term interpolation above written in Cartesian axes $x,y$ with their origin at node 1 ($x$ to the right along side 1–2, $y$ upward along side 1–4). A second frame $\xi,\eta$ has its origin at node 4, with $\xi$ to the right along side 4–3 and $\eta$ pointing downward along side 4–1. Part (b) supplies three further trial polynomials.

Find. A definition of geometric isotropy; a reasoned verdict on each of the three interpolations in (b); and, for (c), the eight-term polynomial re-expressed in the $\xi,\eta$ frame, exhibiting a term that the original does not contain.

Approach. Define geometric isotropy through invariance of the term set, test (b) with the completeness plus symmetric-pair criteria, then in (c) perform the explicit coordinate substitution $x=\xi$, $y=L-\eta$ and collect like powers, which is the only test that also detects loss of invariance under a translation of the origin.

1. Define geometric isotropy (part a). An interpolation possesses geometric isotropy — equivalently geometric invariance or spatial isotropy — when the field it can represent is independent of the position and orientation of the reference axes used to write it. Formally, if the polynomial is re-expressed in any other admissible Cartesian frame obtained by translating, rotating or reflecting the original axes, the set of monomials that appears must be unchanged, so that the same family of functions is spanned. Physically this means an element gives the same answer whichever corner the analyst chooses as the local origin and whichever way the local axes are drawn, which is essential because a mesh generator numbers and orients elements arbitrarily.
2. State the working criteria used in part (b). A polynomial has geometric isotropy when both of the following hold: (1) completeness — every monomial of Pascal’s triangle up to some order $p$ is present; and (2) symmetry — any higher-order terms retained beyond that complete order occur in symmetric pairs, i.e. $x^{a}y^{b}$ appears together with $x^{b}y^{a}$. Completeness is what guarantees that constant and linear fields (rigid-body motion and constant strain) survive an axis change; symmetry is what guarantees invariance when $x$ and $y$ are interchanged.
3. Test interpolation (i). The terms are $\{1,\ x,\ y\}$, the complete linear polynomial of Pascal’s triangle, with no higher-order terms at all. Both criteria are met trivially, and any translation or rotation of the axes maps a linear function to another linear function. $$u=C_1+C_2x+C_3y\ \Longrightarrow\ \boxed{\text{(i) HAS geometric isotropy}}$$ This is the classic constant-strain triangle interpolation, and its invariance is precisely why the CST passes the patch test in any orientation.
4. Test interpolation (ii). The terms are $\{1,\ x,\ y,\ x^2,\ xy,\ y^2,\ x^2y,\ xy^2\}$. The first six are the complete quadratic set, so the polynomial is complete to order $p=2$. The two remaining cubic terms $x^2y$ and $xy^2$ are the mirror image of one another under the interchange $x \leftrightarrow y$, so they form a symmetric pair; the missing cubic terms $x^3$ and $y^3$ are absent together, which preserves the symmetry. $$\{x^2y,\ xy^2\}\ \text{symmetric},\quad \text{complete to order }2\ \Longrightarrow\ \boxed{\text{(ii) HAS geometric isotropy}}$$ This is the familiar eight-node Serendipity quadrilateral interpolation.
5. Test interpolation (iii). The terms are $\{1,\ x^2,\ y^2,\ x^2y^2\}$. The first-order monomials $x$ and $y$ are missing, so the polynomial is complete only to order zero: it cannot represent a uniform gradient, and therefore cannot represent a state of constant strain. The completeness criterion already fails, and a direct rotation confirms it. Rotating the axes by $45^\circ$ through $x=\tfrac{1}{\sqrt{2}}(x'-y')$, $y=\tfrac{1}{\sqrt{2}}(x'+y')$ turns the pair $C_2x^2+C_3y^2$ into $$\tfrac{1}{2}(C_2+C_3)(x'^2+y'^2)+(C_3-C_2)\,x'y',$$ so a cross term $x'y'$ appears that the original set does not contain (unless $C_2=C_3$). $$\text{incomplete at order }1\ \Longrightarrow\ \boxed{\text{(iii) does NOT have geometric isotropy}}$$
6. Set up the coordinate change for part (c). Node 4 lies at $(x,y)=(0,L)$. The $\xi$ axis runs in the same sense as $x$, and the $\eta$ axis points from node 4 toward node 1, i.e. opposite to $y$. Hence a point has $$\xi = x, \qquad \eta = L - y \qquad\Longleftrightarrow\qquad x=\xi,\quad y = L-\eta .$$ Sanity check on the corners: node 1 $(0,0)\to(\xi,\eta)=(0,L)$; node 3 $(L,L)\to(L,0)$; node 4 $(0,L)\to(0,0)$ as required.
7. Substitute and expand. Replacing $x$ by $\xi$ and $y$ by $L-\eta$ term by term, $$C_3y = C_3L-C_3\eta,\qquad C_5y^2 = C_5\left(L^2-2L\eta+\eta^2\right),$$ $$C_6x^2y = C_6L\xi^2 - C_6\xi^2\eta,\qquad C_7xy^2 = C_7L^2\xi - 2C_7L\,\xi\eta + C_7\xi\eta^2,$$ $$C_8x^2y^2 = C_8L^2\xi^2 - 2C_8L\,\xi^2\eta + C_8\xi^2\eta^2 .$$ The single term that matters is $C_7xy^2$: because $y^2$ expands to $L^2-2L\eta+\eta^2$, the product with $x=\xi$ manufactures a $\xi\eta$ contribution.
8. Collect like powers in the new frame. Gathering all of the above, $$\begin{aligned} u(\xi,\eta) =\ & \left(C_1+C_3L+C_5L^2\right) + \left(C_2+C_7L^2\right)\xi - \left(C_3+2C_5L\right)\eta \\[2pt] &+ \left(C_4+C_6L+C_8L^2\right)\xi^2 + C_5\,\eta^2 \;\underbrace{-\;2C_7L\,\xi\eta}_{\text{new term}} \\[2pt] &- \left(C_6+2C_8L\right)\xi^2\eta + C_7\,\xi\eta^2 + C_8\,\xi^2\eta^2 . \end{aligned}$$
9. Compare the two term sets and conclude. In the original frame the monomials are $\{1,\ x,\ y,\ x^2,\ y^2,\ x^2y,\ xy^2,\ x^2y^2\}$ — note that $xy$ is absent. In the frame at node 4 the monomials are $\{1,\ \xi,\ \eta,\ \xi^2,\ \eta^2,\ \boldsymbol{\xi\eta},\ \xi^2\eta,\ \xi\eta^2,\ \xi^2\eta^2\}$, one term richer. The extra term carries the coefficient $$\boxed{\;\text{coefficient of }\xi\eta \;=\; -2C_7L \;\neq\; 0\ \text{ whenever } C_7\neq 0\;}$$ so the same element, described from node 4, spans a different function space than it does when described from node 1. The interpolation is therefore not geometrically isotropic.
10. Identify the underlying defect and contrast with a bilinear element. The eight terms do satisfy the symmetric-pair rule — $(x^2,y^2)$, $(x^2y,xy^2)$ and the self-symmetric $x^2y^2$ — which is why the element survives a pure reflection: setting $L=0$ above (origin unmoved, $\eta=-y$) reproduces the original term set exactly. What defeats it is the translation of the origin from node 1 to node 4, which the missing $xy$ term cannot absorb. By contrast the four-term bilinear set $u=C_1+C_2x+C_3y+C_4xy$ under the same substitution becomes $(C_1+C_3L)+(C_2+C_4L)\xi-C_3\eta-C_4\,\xi\eta$, whose term set is unchanged — the bilinear quadrilateral is geometrically isotropic. Completeness to order 1 plus the symmetric pair is what the eight-term set lacks at the quadratic level, since it keeps $x^2$ and $y^2$ but drops their companion $xy$.