Which pair gives the quadratic element of each family?

Lagrange: 9-node; serendipity: 8-node

Lagrange: 8-node; serendipity: 9-node

Lagrange: 12-node; serendipity: 16-node

What is a main difference in construction?

Lagrange uses products of 1D Lagrange polynomials; serendipity has only boundary nodes and no interior nodes

They are identical for all orders

Serendipity needs interior nodes for order 2

Lagrange has only boundary nodes

Why is the serendipity element usually preferred?

The interior node adds degrees of freedom but little accuracy, so fewer DOF are needed

It has an internal node that can be condensed

Practice · 5 marks

Lagrange and serendipity elements compared

Short answer

Lagrange elements have interior nodes and use tensor-product shape functions; serendipity elements have nodes only on boundary. Linear: both are the 4‑node bilinear. Quadratic: 9‑node Lagrange (terms 1,ξ,η,ξ²,ξη,η²,ξ²η,ξη²,ξ²η²) and 8‑node serendipity (same but omits ξ²η²).

Differentiate between Lagrange and serendipity families of two-dimensional elements. Name the elements of each family up to second order and list the polynomial terms of the 8-node and 9-node elements.

Both are higher-order rectangular (or quadrilateral) elements built on a $\xi$ - $\eta$ parent square ( $-1 \le \xi, \eta \le 1$ ).

Comparison

Point	Lagrange family	Serendipity family
Construction	Product of 1D Lagrange polynomials in $\xi$ and $\eta$	Nodes only on the boundary; functions built by inspection/combination
Interior nodes	Present for order 2 and above	None
Linear element	4-node bilinear	4-node bilinear (same)
Quadratic element	9-node (centre node)	8-node (4 corners + 4 mid-sides)
Cubic element	16-node (4 internal nodes)	12-node
Polynomial terms	Contains all terms of the tensor product	Complete up to its order, but misses the highest tensor-product terms
Number of DOF	More, so more costly	Fewer, so cheaper
Accuracy	Slightly better, and less sensitive to element distortion	Nearly as good with fewer nodes; no internal node to condense

 9-node Lagrange        8-node serendipity
 4--7--3                4--7--3
 |     |                |     |
 8  9  6                8     6
 |     |                |     |
 1--5--2                1--5--2

Polynomial terms (Pascal's triangle)

9-node Lagrange: $1,\ \xi,\ \eta,\ \xi^2,\ \xi\eta,\ \eta^2,\ \xi^2\eta,\ \xi\eta^2,\ \xi^2\eta^2$ .
8-node serendipity: the same list without $\xi^2\eta^2$ .

The 9-node element is a complete biquadratic. The 8-node element is a complete quadratic with two additional cubic terms, $\xi^2\eta$ and $\xi\eta^2$ , which keep the polynomial symmetric.

The serendipity element is preferred in most practice because the interior node adds degrees of freedom but little accuracy.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Lagrange and serendipity elements compared

Comparison

Polynomial terms (Pascal's triangle)

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