Compact stencil Last updated September 23, 2025 A 2D compact stencil using all 8 adjacent nodes, plus the center node (in red). In mathematics , especially in the areas of numerical analysis called numerical partial differential equations , a compact stencil is a type of stencil that uses only nine nodes for its discretization method in two dimensions. It uses only the center node and the adjacent nodes. For any structured grid utilizing a compact stencil in 1, 2, or 3 dimensions the maximum number of nodes is 3, 9, or 27 respectively. Compact stencils may be compared to non-compact stencils . Compact stencils are currently implemented in many partial differential equation solvers, including several in the topics of CFD, FEA, and other mathematical solvers relating to PDE's [ 1] [ 2]
Two Point Stencil Example The two point stencil for the first derivative of a function is given by:
f ′ ( x 0 ) = f ( x 0 + h ) − f ( x 0 − h ) 2 h + O ( h 2 ) . {\displaystyle f'(x_{0})={\frac {f{\left(x_{0}{+}h\right)}-f{\left(x_{0}{-}h\right)}}{2h}}+{\mathcal {O}}{\left(h^{2}\right)}.}
This is obtained from the Taylor series expansion of the first derivative of the function given by:
f ′ ( x 0 ) = f ( x 0 + h ) − f ( x 0 ) h − f ″ ( x 0 ) 2 ! h − f ( 3 ) ( x 0 ) 3 ! h 2 − f ( 4 ) ( x 0 ) 4 ! h 3 + ⋯ . {\displaystyle f'(x_{0})={\frac {f{\left(x_{0}{+}h\right)}-f(x_{0})}{h}}-{\frac {f''(x_{0})}{2!}}h-{\frac {f^{(3)}(x_{0})}{3!}}h^{2}-{\frac {f^{(4)}(x_{0})}{4!}}h^{3}+\cdots .}
Replacing h {\displaystyle h} with − h {\displaystyle -h} , we have:
f ′ ( x 0 ) = − f ( x 0 − h ) − f ( x 0 ) h + f ″ ( x 0 ) 2 ! h − f ( 3 ) ( x 0 ) 3 ! h 2 + f ( 4 ) ( x 0 ) 4 ! h 3 + ⋯ . {\displaystyle f'(x_{0})=-{\frac {f{\left(x_{0}{-}h\right)}-f(x_{0})}{h}}+{\frac {f''(x_{0})}{2!}}h-{\frac {f^{(3)}(x_{0})}{3!}}h^{2}+{\frac {f^{(4)}(x_{0})}{4!}}h^{3}+\cdots .}
Addition of the above two equations together results in the cancellation of the terms in odd powers of h {\displaystyle h} :
2 f ′ ( x 0 ) = f ( x 0 + h ) − f ( x 0 ) h − f ( x 0 − h ) − f ( x 0 ) h − 2 f ( 3 ) ( x 0 ) 3 ! h 2 + ⋯ f ′ ( x 0 ) = f ( x 0 + h ) − f ( x 0 − h ) 2 h − f ( 3 ) ( x 0 ) 3 ! h 2 + ⋯ = f ( x 0 + h ) − f ( x 0 − h ) 2 h + O ( h 2 ) . {\displaystyle {\begin{aligned}2f'(x_{0})&={\frac {f{\left(x_{0}{+}h\right)}-f(x_{0})}{h}}-{\frac {f{\left(x_{0}{-}h\right)}-f(x_{0})}{h}}-2{\frac {f^{(3)}(x_{0})}{3!}}h^{2}+\cdots \\[1ex]f'(x_{0})&={\frac {f{\left(x_{0}{+}h\right)}-f{\left(x_{0}{-}h\right)}}{2h}}-{\frac {f^{(3)}(x_{0})}{3!}}h^{2}+\cdots \\&={\frac {f{\left(x_{0}{+}h\right)}-f{\left(x_{0}{-}h\right)}}{2h}}+{\mathcal {O}}{\left(h^{2}\right)}.\end{aligned}}}
Three Point Stencil Example For example, the three point stencil for the second derivative of a function is given by:
f ″ ( x 0 ) = f ( x 0 + h ) + f ( x 0 − h ) − 2 f ( x 0 ) h 2 + O ( h 2 ) . {\displaystyle f''(x_{0})={\frac {f{\left(x_{0}{+}h\right)}+f{\left(x_{0}{-}h\right)}-2f(x_{0})}{h^{2}}}+{\mathcal {O}}{\left(h^{2}\right)}.}
This is obtained from the Taylor series expansion of the first derivative of the function given by: f ′ ( x 0 ) = f ( x 0 + h ) − f ( x 0 ) h − f ″ ( x 0 ) 2 ! h − f ( 3 ) ( x 0 ) 3 ! h 2 − f ( 4 ) ( x 0 ) 4 ! h 3 + ⋯ . {\displaystyle f'(x_{0})={\frac {f{\left(x_{0}{+}h\right)}-f(x_{0})}{h}}-{\frac {f''(x_{0})}{2!}}h-{\frac {f^{(3)}(x_{0})}{3!}}h^{2}-{\frac {f^{(4)}(x_{0})}{4!}}h^{3}+\cdots .}
Replacing h {\displaystyle h} with − h {\displaystyle -h} , we have: f ′ ( x 0 ) = − f ( x 0 − h ) − f ( x 0 ) h + f ″ ( x 0 ) 2 ! h − f ( 3 ) ( x 0 ) 3 ! h 2 + f ( 4 ) ( x 0 ) 4 ! h 3 + ⋯ . {\displaystyle f'(x_{0})=-{\frac {f{\left(x_{0}{-}h\right)}-f(x_{0})}{h}}+{\frac {f''(x_{0})}{2!}}h-{\frac {f^{(3)}(x_{0})}{3!}}h^{2}+{\frac {f^{(4)}(x_{0})}{4!}}h^{3}+\cdots .}
Subtraction of the above two equations results in the cancellation of the terms in even powers of h {\displaystyle h} : 0 = f ( x 0 + h ) − f ( x 0 ) h + f ( x 0 − h ) − f ( x 0 ) h − 2 f ( 2 ) ( x 0 ) 2 ! h − 2 f ( 4 ) ( x 0 ) 4 ! h 3 + ⋯ . f ( 2 ) ( x 0 ) = f ( x 0 + h ) + f ( x 0 − h ) − 2 f ( x 0 ) h 2 − 2 f ( 4 ) ( x 0 ) 4 ! h 2 + ⋯ . f ( 2 ) ( x 0 ) = f ( x 0 + h ) + f ( x 0 − h ) − 2 f ( x 0 ) h 2 + O ( h 2 ) . {\displaystyle {\begin{aligned}0&={\frac {f{\left(x_{0}{+}h\right)}-f(x_{0})}{h}}+{\frac {f{\left(x_{0}{-}h\right)}-f(x_{0})}{h}}-2{\frac {f^{(2)}(x_{0})}{2!}}h-2{\frac {f^{(4)}(x_{0})}{4!}}h^{3}+\cdots .\\[1ex]f^{(2)}(x_{0})&={\frac {f{\left(x_{0}{+}h\right)}+f{\left(x_{0}{-}h\right)}-2f(x_{0})}{h^{2}}}-2{\frac {f^{(4)}(x_{0})}{4!}}h^{2}+\cdots .\\[1ex]f^{(2)}(x_{0})&={\frac {f{\left(x_{0}{+}h\right)}+f{\left(x_{0}{-}h\right)}-2f(x_{0})}{h^{2}}}+{\mathcal {O}}{\left(h^{2}\right)}.\end{aligned}}}
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