A Comparison Between FEM and FVM via the Method of Weighted Residuals

Abstract

The Method of Weighted Residuals (MWR) is used to compare the finite element method (FEM) with the finite volume method (FVM) through nodal recursion relations. Both methods reside under the general MWR structure, with the underlying switch between the two methods established through the weighting function. Both methods yield comparable spatial accuracy for steady-state conditions. However, the flexibility of the FEM permits additional options that can increase accuracy, but generally at the expense of additional time and resource constraints.

Appendix

1D FEM Formulation with Advection

For simplicity, we define $\phi (x,t)$ as a 1D scalar transport variable. Linear shape functions are used for two consecutive adjacent linear elements. After assembly (creating a 3 × 3 global matrix), the recursion relation for the i^th node can be established.

Setting W(x) = N(x) and using the integral relation for the 1D element spanning nodes i-1 to i, the governing equation for time-dependent transport and diffusion,

$$\frac{\partial \phi }{\partial t} + U\frac{\partial \phi }{\partial x} - K\frac{{\partial^{2} \phi }}{{\partial x^{2} }} = 0$$

can be written numerically as

$$\int {N_{i} } N_{i - 1} dx\left\{ {\begin{array}{*{20}c} {\dot{\phi }_{i - 1} } \\ {\dot{\phi }_{i} } \\ \end{array} } \right\} + \int {UN_{j} N_{i} } \frac{{dN_{i - 1} }}{dx}dx\left\{ {\begin{array}{*{20}c} {\phi_{i - 1} } \\ {\phi_{i} } \\ \end{array} } \right\} + K\int {\frac{{dN_{i} }}{dx}\frac{{dN_{i - 1} }}{dx}} dx\left\{ {\begin{array}{*{20}c} {\phi_{i - 1} } \\ {\phi_{i} } \\ \end{array} } \right\} = 0$$

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where $\dot{\phi }(x,t)$ denotes time dependence and U(x) is velocity. The integral terms are similarly evaluated for elements spanning i to i + 1. Thus, for the two adjacent linear elements,

For $\Delta x_{ - } = x_{i} - x_{i - 1}$

$$\frac{{\Delta x_{ - } }}{6}\left[ {\begin{array}{*{20}c} 2 & 1 \\ 1 & 2 \\ \end{array} } \right]\left\{ {\begin{array}{*{20}c} {\dot{\phi }_{i - 1} } \\ {\dot{\phi }_{i} } \\ \end{array} } \right\} - \frac{U}{2}\left[ {\begin{array}{*{20}c} { - 1} & 1 \\ { - 1} & 1 \\ \end{array} } \right]\left\{ {\begin{array}{*{20}c} {\phi_{i - 1} } \\ {\phi_{i} } \\ \end{array} } \right\} - \frac{K}{{\Delta x_{ - } }}\left[ {\begin{array}{*{20}c} 1 & { - 1} \\ { - 1} & 1 \\ \end{array} } \right]\left\{ {\begin{array}{*{20}c} {\phi_{i - 1} } \\ {\phi_{i} } \\ \end{array} } \right\} = 0$$

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For $\Delta x_{ + } = x_{i + 1} - x_{i}$

$$\frac{{\Delta x_{ + } }}{6}\left[ {\begin{array}{*{20}c} 2 & 1 \\ 1 & 2 \\ \end{array} } \right]\left\{ {\begin{array}{*{20}c} {\dot{\phi }_{i} } \\ {\dot{\phi }_{i + 1} } \\ \end{array} } \right\} - \frac{U}{2}\left[ {\begin{array}{*{20}c} { - 1} & 1 \\ { - 1} & 1 \\ \end{array} } \right]\left\{ {\begin{array}{*{20}c} {\phi_{i} } \\ {\phi_{i + 1} } \\ \end{array} } \right\} - \frac{K}{{\Delta x_{ + } }}\left[ {\begin{array}{*{20}c} 1 & { - 1} \\ { - 1} & 1 \\ \end{array} } \right]\left\{ {\begin{array}{*{20}c} {\phi_{i} } \\ {\phi_{i + 1} } \\ \end{array} } \right\} = 0$$

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For simplicity, the velocity is assumed constant. The use of hypermatrix formulation is needed to deal with variable velocity—more details are described in [19].

Assembling the two adjacent elements into a 3 × 3 global matrix,

$$\begin{aligned} \frac{1}{6}\left[ {\begin{array}{*{20}c} {2\Delta x_{ - } } & {\Delta x_{ - } } & 0 \\ {\Delta x_{ - } } & {2\Delta x_{ - } + 2\Delta x_{ + } } & {\Delta x_{ + } } \\ 0 & {\Delta x_{ + } } & {2\Delta x_{ + } } \\ \end{array} } \right]\left\{ {\begin{array}{*{20}c} {\dot{\phi }_{i - 1} } \\ {\dot{\phi }_{i} } \\ {\dot{\phi }_{i + 1} } \\ \end{array} } \right\} + \frac{U}{2}\left[ {\begin{array}{*{20}c} { - 1} & 1 & 0 \\ { - 1} & 0 & 1 \\ 0 & { - 1} & 1 \\ \end{array} } \right]\left\{ {\begin{array}{*{20}c} {\phi_{i - 1} } \\ {\phi_{i} } \\ {\phi_{i + 1} } \\ \end{array} } \right\} \hfill \\ + \frac{K}{6}\left[ {\begin{array}{*{20}c} {\frac{1}{{\Delta x_{ - } }}} & {\frac{ - 1}{{\Delta x_{ - } }}} & 0 \\ {\frac{ - 1}{{\Delta x_{ - } }}} & {\frac{1}{{\Delta x_{ - } }} + \frac{1}{{\Delta x_{ + } }}} & {\frac{ - 1}{{\Delta x_{ + } }}} \\ 0 & {\frac{ - 1}{{\Delta x_{ + } }}} & {\frac{1}{{\Delta x_{ + } }}} \\ \end{array} } \right]\left\{ {\begin{array}{*{20}c} {\phi_{i - 1} } \\ {\phi_{i} } \\ {\phi_{i + 1} } \\ \end{array} } \right\} = 0 \hfill \\ \end{aligned}$$

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Setting Δx_- = Δx₊ = Δx, the recursion relation in 1D is obtained by stripping out the central expression within the global matrix, i.e.,

$$\frac{1}{6}\left( {\dot{\phi }_{i - 1} + 4\dot{\phi }_{i} + \dot{\phi }_{i + 1} } \right) + \frac{U}{2\Delta x}(\phi_{i + 1} - \phi_{i - 1} ) - \frac{K}{{\Delta x^{2} }}(\phi_{i + 1} - 2\phi_{i} + \phi_{i - 1} ) = 0$$

(37)

which appears similar to the central difference FDM (or FVM). The improved accuracy of the FEM lies in the time-dependent terms, $\dot{\phi }$. When employing Crank–Nicolson time averaging, Eq. (37) yields O(Δx⁴) versus O(Δx²) in space.

2D FEM Formulation with Advection

The general transport equation in 2D can be discretized using either 2D triangular elements or quadrilateral elements and then assembled over eight triangular elements, four quadrilateral elements, or one quadratic Lagrangian quadrilateral element, consecutively, to establish a set of recursion relations. The resulting set of recursion relations can then be solved using the Strongly Implicit Procedure (SIP), or with a Modified SIP [20, 22]. A more detailed discussion on the formulation of FEM and FVM methods that includes advection and upwinding is provided in Idelsohn and Onate [23].

The matrix equivalent equation of the transport relation (assume $\phi (x,y) \equiv T(x,y)$ for simplicity) can be written as

$$[M]\left\{ {\dot{T}} \right\} + [A]\left\{ T \right\} - [K]\left\{ T \right\} = 0$$

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where [M] is the mass matrix, [A(V)] is the advection matrix, and [K] is the diffusion matrix

$$[M] = \iint {N_{i} }N_{j} dxdy$$

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$$[A(V(u,v))] = \iint {\left( {u \cdot N_{i} N_{j} \frac{{\partial N_{k} }}{\partial x} + v \cdot N_{i} N_{j} \frac{{\partial N_{k} }}{\partial y}} \right)dxdy}$$

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$$[K] = \iint {K\left( {\frac{{\partial N_{i} }}{\partial x}\frac{{\partial N_{j} }}{\partial x} + \frac{{\partial N_{i} }}{\partial y}\frac{{\partial N_{j} }}{\partial y}} \right)}dxdy$$

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where the i, j, k indices denote row and column vectors [20]. After establishing the global matrices for a patch consisting of 8 bilinear triangular elements, 4 bilinear quadrilateral elements, and one Lagrangian quadratic quadrilateral element, a set of recursion relations for the T_{i, j} node can be established.

2D BASIS Functions

Bilinear triangular and quadratic elements are used for the 2D trial functions [19, 20]. The general 2D bilinear element configurations are shown in Figs. 11 and 12. Notice that the triangular element array permits either a 9-node configuration or a 5-node configuration.

Fig. 11

Basis function-triangular element

Fig. 12

Basis function-quadrilateral element

The set of finite element expressions, defined over a rectangular subspace and formulated using chapeau basis functions, can be interpreted as integrated averaged difference approximations. After applying Galerkin’s method, integrating by parts, and using isoparametric transformations, we obtain the equations as follows:

2D bilinear triangular elements:

$$\begin{aligned} & \frac{1}{36}\left[ {\left( {3\left( {\dot{T}_{i + 1,j + 1} + \dot{T}_{i - 1,j - 1} + \dot{T}_{i + 1,j} + \dot{T}_{i - 1,j} + \dot{T}_{i,j + 1} + \dot{T}_{i,j - 1} } \right) + 18\dot{T}_{i,j} } \right)} \right] \\ & \quad + \frac{K}{{\Delta x^{2} }}\left[ {T_{i - 1,j} - 2T_{i,j} + T_{i + 1,j} } \right] + \frac{K}{{\Delta y^{2} }}\left[ {T_{i,j + 1} - 2T_{i,j} + T_{i,j - 1} } \right] \\ & \quad + \frac{u}{6\Delta x}\left[ { + T_{i,j + 1} - 2T_{i + 1,j} + 2T_{i - 1,j} - T_{i,j - 1} } \right] \\ & \quad + \frac{v}{6\Delta y}\left[ {T_{i - 1,j} - T_{i - 1,j - 1} + 2T_{i,j + 1} - 2T_{i,j - 1} + T_{i + 1,j + 1} - T_{i + 1,j} } \right] = 0 \\ \end{aligned}$$

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2D bilinear quadrilateral elements:

$$\begin{aligned} & \frac{1}{9}\left[ {\left( \begin{aligned} \left( {\dot{T}_{i - 1,j + 1} + \dot{T}_{i + 1,j + 1} + \dot{T}_{i - 1,j - 1} + \dot{T}_{i + 1,j - 1} } \right) \hfill \\ + 4\left( {\dot{T}_{i - 1,j} + \dot{T}_{i + 1,j} + \dot{T}_{i,j + 1} + \dot{T}_{i,j - 1} } \right) + 16\dot{T}_{i,j} \hfill \\ \end{aligned} \right)} \right] \\ & \quad + \frac{K}{{6\Delta x^{2} }}\left[ \begin{aligned} - T_{i - 1,j + 1} - 4T_{i - 1,j} - T_{i - 1,j - 1} + 2T_{i,j + 1} \hfill \\ + 8T_{i,j} + 2T_{i,j - 1} - T_{i + 1,j + 1} - 4T_{i + 1,j} - T_{i + 1,j - 1} \hfill \\ \end{aligned} \right] \\ & \quad + \frac{K}{{6\Delta y^{2} }}\left[ \begin{aligned} - T_{i - 1,j + 1} + 2T_{i - 1,j} - T_{i - 1,j - 1} - 4T_{i,j + 1} \hfill \\ + 8T_{i,j} - 4T_{i,j - 1} - T_{i + 1,j + 1} + 2T_{i + 1,j} - T_{i + 1,j - 1} \hfill \\ \end{aligned} \right] \\ & \quad + \frac{u}{6\Delta x}\left[ \begin{aligned} - T_{i - 1,j + 1} - 4T_{i - 1,j} - T_{i - 1,j - 1} \hfill \\ + T_{i + 1,j + 1} + 4T_{i + 1,j} + T_{i + 1,j - 1} \hfill \\ \end{aligned} \right] \\ & \quad + \frac{v}{6\Delta y}\left[ \begin{aligned} T_{i - 1,j + 1} - T_{i - 1,j - 1} \hfill \\ + 4T_{i,j + 1} - 4T_{i,j - 1} + T_{i + 1,j + 1} - T_{i + 1,j - 1} \hfill \\ \end{aligned} \right] = 0 \\ \end{aligned}$$

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2D Quadratic Lagrangian quadrilateral element:

$$\begin{gathered} \frac{1}{{225}}\left[ {\left( \begin{gathered} 4\left( {\dot{T}_{{i - 1,j + 1}} + \dot{T}_{{i - 1,j - 1}} + \dot{T}_{{i + 1,j + 1}} + \dot{T}_{{i + 1,j - 1}} } \right) \hfill \\ + 32\left( {\dot{T}_{{i - 1,j}} + \dot{T}_{{i,j + 1}} + \dot{T}_{{i,j - 1}} + \dot{T}_{{i + 1,j}} } \right) + 256\dot{T}_{{i,j}} \hfill \\ \end{gathered} \right)} \right] \hfill \\ + \frac{K}{{45\Delta x^{2} }}\left[ \begin{gathered} - 8T_{{i - 1,j + 1}} - 64T_{{i - 1,j}} - 8T_{{i - 1,j - 1}} + 16T_{{i - 1,j}} + 128T_{{i,j}} \hfill \\ + 16T_{{i,j - 1}} - 8T_{{i + 1,j + 1}} - 64T_{{i + 1,j}} - 8T_{{i + 1,j - 1}} \hfill \\ \end{gathered} \right] \hfill \\ + \frac{K}{{45\Delta y^{2} }}\left[ \begin{gathered} - 8T_{{i - 1,j + 1}} + 16T_{{i - 1,j}} - 8T_{{i - 1,j - 1}} - 64T_{{i - 1,j}} \hfill \\ + 128T_{{i,j}} - 64T_{{i,j - 1}} - 8T_{{i + 1,j + 1}} + 16T_{{i + 1,j}} - 8T_{{i + 1,j - 1}} \hfill \\ \end{gathered} \right] \hfill \\ + \frac{u}{{45\Delta x}}\left[ \begin{gathered} - 4T_{{i - 1,j + 1}} - 32T_{{i - 1,j}} - 4T_{{i - 1,j - 1}} \hfill \\ + 4T_{{i + 1,j + 1}} + 32T_{{i + 1,j}} + 4T_{{i + 1,j - 1}} \hfill \\ \end{gathered} \right] \hfill \\ + \frac{V}{{45\Delta y}}\left[ \begin{gathered} 4T_{{i - 1,j + 1}} - 4T_{{i - 1,j - 1}} + 32T_{{i - 1,j}} \hfill \\ - 32T_{{i,j - 1}} + 4T_{{i + 1,j + 1}} - 4T_{{i + 1,j - 1}} \hfill \\ \end{gathered} \right] = 0 \hfill \\ \end{gathered}$$

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