--- title: "Lagrange and serendipity elements compared" subject: "Finite Element Method" chapter: 7 key: "semester_8/finite_element_method:lagrange-serendipity-elements-comparison" source: https://www.bctengineer.com/q/semester_8/finite_element_method/lagrange-serendipity-elements-comparison site: BCT Engineering --- # Lagrange and serendipity elements compared Finite Element Method · Chapter 7: Higher order Elements - Asked in: year not given, 5 marks - Video: [Patch Test in FEM | Shape Functions, Completeness & Serendipity vs Lagrange Elements](https://www.youtube.com/watch?v=iZZMZdZQU3k) — Reza Rafiee - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/finite_element_method/lagrange-serendipity-elements-comparison - Whole chapter: https://www.bctengineer.com/md/semester_8/finite_element_method/chapter-7 ## Question 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. ## 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). ## Answer 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 | ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which pair gives the quadratic element of each family? A. Both: 9-node B. Lagrange: 9-node; serendipity: 8-node C. Lagrange: 8-node; serendipity: 9-node D. Lagrange: 12-node; serendipity: 16-node Answer: B. The Lagrange quadratic has a centre node (9 nodes); the serendipity has only boundary nodes (8 nodes). (Key point: Quadratic elements) 2. What is a main difference in construction? A. They are identical for all orders B. Serendipity needs interior nodes for order 2 C. Lagrange has only boundary nodes D. Lagrange uses products of 1D Lagrange polynomials; serendipity has only boundary nodes and no interior nodes Answer: D. Lagrange elements are tensor products and include interior nodes from order 2, while serendipity elements have none. (Key point: Construction) 3. Which term is in the 9-node element but missing from the 8-node element? A. xi^2 eta^2 B. xi eta^2 C. xi^2 eta D. xi eta Answer: A. The 9-node element is a complete biquadratic including xi^2 eta^2; the 8-node omits this term. (Key point: Polynomial terms) 4. Why is the serendipity element usually preferred? A. It has an internal node that can be condensed B. It is exact for all problems C. The interior node adds degrees of freedom but little accuracy, so fewer DOF are needed D. It has more nodes Answer: C. Serendipity gives good accuracy with fewer nodes and DOF, making it cheaper. 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