Question 7 of 7: Isoparametric quadrilateral — Jacobian and the validity of a mapping

Question 7: Isoparametric quadrilateral — Jacobian and the validity of a mapping (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.

(a) Isoparametric element. An element is isoparametric when the same shape functions, written in the natural coordinates, are used both to interpolate the field variable and to map the geometry: u = ΣN_iu_i together with x = ΣN_ix_i and y = ΣN_iy_i. That single choice lets a simple parent shape represent curved or irregular physical elements while automatically satisfying the completeness and compatibility requirements.

Given. The four-node quadrilateral with the shape functions printed on the paper — note that they place node 3 at (−1, +1) and node 4 at (+1, +1) in the parent domain, so the boundary is traversed in the order 1–2–4–3 — and two candidate global elements built on the same four points (2, 2), (5, 2), (2, 4) and (5, 5) with different node numbering.

Find. The general Jacobian matrix of the element, the Jacobian determinant of each of the two numbered elements, and the conclusion those two expressions support.

Approach. Differentiate the isoparametric map to obtain [J] in terms of the nodal coordinates, substitute each set of coordinates, and examine the sign of the determinant over the whole parent square.

1. (b) Differentiate the mapping. With $x=\sum N_ix_i$ and $y=\sum N_iy_i$, the Jacobian matrix is$$[J]=\begin{bmatrix}\dfrac{\partial x}{\partial\xi}&\dfrac{\partial y}{\partial\xi}\\[4pt]\dfrac{\partial x}{\partial\eta}&\dfrac{\partial y}{\partial\eta}\end{bmatrix}=\frac14\begin{bmatrix}-(1-\eta)&(1-\eta)&-(1+\eta)&(1+\eta)\\-(1-\xi)&-(1+\xi)&(1-\xi)&(1+\xi)\end{bmatrix}\begin{bmatrix}x_1&y_1\\x_2&y_2\\x_3&y_3\\x_4&y_4\end{bmatrix}$$Carrying out the product gives the compact form used below:$$\boxed{\begin{aligned}J_{11}&=\tfrac14\big[(1-\eta)(x_2-x_1)+(1+\eta)(x_4-x_3)\big],&J_{12}&=\tfrac14\big[(1-\eta)(y_2-y_1)+(1+\eta)(y_4-y_3)\big],\\J_{21}&=\tfrac14\big[(1-\xi)(x_3-x_1)+(1+\xi)(x_4-x_2)\big],&J_{22}&=\tfrac14\big[(1-\xi)(y_3-y_1)+(1+\xi)(y_4-y_2)\big]\end{aligned}}$$Each entry is an average of two edge vectors, weighted by position along the element.
2. (c) Element (i). With $1(2,2)$, $2(5,2)$, $3(2,4)$, $4(5,5)$ the differences are $x_2-x_1=x_4-x_3=3$, $y_2-y_1=0$, $y_4-y_3=1$, $x_3-x_1=x_4-x_2=0$, $y_3-y_1=2$, $y_4-y_2=3$, so$$[J]_{(i)}=\begin{bmatrix}\tfrac32&\tfrac14(1+\eta)\\[2pt]0&\tfrac14(5+\xi)\end{bmatrix}\qquad\Longrightarrow\qquad\boxed{|J|_{(i)}=\tfrac38\,(5+\xi)}$$Over the parent square this runs from $\tfrac38(4)=1.5$ at $\xi=-1$ to $\tfrac38(6)=2.25$ at $\xi=+1$ — strictly positive. Integrating it, $\iint|J|\,d\xi\,d\eta=7.5$, which matches the physical area of the quadrilateral computed by the shoelace rule.
3. Element (ii). The same four points are now numbered $1(2,2)$, $2(2,4)$, $3(5,5)$, $4(5,2)$, giving $x_2-x_1=0$, $x_4-x_3=0$, $y_2-y_1=2$, $y_4-y_3=-3$, $x_3-x_1=x_4-x_2=3$, $y_3-y_1=3$, $y_4-y_2=-2$, so$$[J]_{(ii)}=\begin{bmatrix}0&-\tfrac14(1+5\eta)\\[2pt]\tfrac32&\tfrac14(1-5\xi)\end{bmatrix}\qquad\Longrightarrow\qquad\boxed{|J|_{(ii)}=\tfrac38\,(1+5\eta)}$$This runs from $\tfrac38(-4)=-1.5$ at $\eta=-1$ to $\tfrac38(6)=2.25$ at $\eta=+1$, vanishing on the line $\eta=-\tfrac15$.
4. (d) Interpret the two results. The determinant of the Jacobian is the local area ratio between the parent and the global domain, so the mapping is one-to-one and invertible only where $|J|\gt 0$. Element (i) satisfies this everywhere, so its mapping is valid, its inverse $[J]^{-1}$ exists at every Gauss point and the element can be integrated reliably. Element (ii) contains points where $|J|=0$ and a whole region where $|J|\lt 0$: the mapping folds the parent square over on itself. The cause is purely the node numbering — taking the parent traversal 1–2–4–3 into the global points (2,2), (2,4), (5,2), (5,5) produces a self-intersecting "bow-tie" instead of a simple quadrilateral, as the figure shows. The conclusion is therefore:$$\boxed{|J|\gt 0\ \text{throughout} \Rightarrow \text{valid element (i)};\quad |J|\ \text{changing sign} \Rightarrow \text{invalid element (ii)}}$$Nodes must be numbered consecutively around the element boundary in a consistent (here counter-clockwise) sense; a mesh generator that reports a negative or zero Jacobian is reporting this error, not a modelling subtlety.