1. Can you help me explain the basic difference between FDM, FEM and FVM?

  2. What is the best method and why?

  3. Advantage and disadvantage of them?

  • 23,901
  • 3
  • 44
  • 75
Anh-Thi DINH
  • 463
  • 1
  • 6
  • 15
  • FEM is the most powerful and flexible yet mathematically sound method of the methods you listed. – Shuhao Cao Jun 24 '13 at 01:21
  • There is also the "Material point method", where a continuum body is described by a number of small Lagrangian elements. It's not a mesh based method and is instead categorized as a meshless/meshfree or continuum-based particle method – skan Aug 07 '19 at 22:27

7 Answers7


This is a difficult question to answer.

"The FDM is the oldest and is based upon the application of a local Taylor expansion to approximate the differential equations. The FDM uses a topologically square network of lines to construct the discretization of the PDE. This is a potential bottleneck of the method when handling complex geometries in multiple dimensions. This issue motivated the use of an integral form of the PDEs and subsequently the development of the finite element and finite volume techniques." (http://www2.imperial.ac.uk/ssherw/spectralhp/papers/HandBook.pdf)

Here are two references to review so you can get a better feel for these methods.

  • 749
  • 7
  • 19
  • 55,120
  • 25
  • 75
  • 108
  • 2
    Nice post! Sometimes OP's don't know that their questions are difficult to answer...+1 – amWhy Jun 26 '13 at 00:06
  • 1
    Wow, Amzoti, when you literally copy something, you should put quotation marks around the sentence, so that other people would not think that these wise words belong to you. If there are no quotation marks, then it can be considered to be an instance of a plagiarism. I hope you will edit this post properly. – Artem May 16 '14 at 00:41

Here is an old scicomp.SE question that answered some of your question: What are criteria to choose between finite-differences and finite-elements?

In my humble opinion, FEM is the most flexible one in terms of dealing with complex geometry and complicated boundary conditions. FEM also allows the adaptive/local procedure to get higher order local approximation or battling singularities. FEM's basis can be discontinuous and not well-defined pointwisely, which is a nice heritage from the Hilbert space framework. For computational fluid dynamics and electromagnetism, FEM is the way to incorporate the intrinsic geometrical properties of the solutions.

For FVM: partly you can refer to my answer here: How should a numerical solver treat conserved quantities? It is also worth noting that FVM can only have lower order of approximation.

In some recently development in FEM addresses the problem I mentioned in the answer above. For example, for convection-dominated pde, tradition continuous Galerkin framework for FEM doesn't work well, which introduces dissapation over time and oscillation over material-layers for the numerical solution. Now there are Discontinuous Galerkin FEM (higher order FVM) and hybrized DGFEM (see here: Unified hybridization of discontinuous Galerkin, mixed, and continuous Galerkin methods for second order elliptic problems) to remedy these two effects.

FDM and FVM are easy to implement, but you get trade-off from this convenience of implementation for limited usage for different PDEs.

Shuhao Cao
  • 18,320
  • 4
  • 51
  • 105
  • I see papers for finite difference method for non uniform grids. So if you can do this, why not use finite difference for everything? Is there not a way to deal with discontinuity with finite difference? – Frank Jul 30 '19 at 00:38
  • 1
    @Frank Yeah, there are many new developments in FDM like mimetic methods and virtual elements in recent years, which can deal with discontinuities, and their formulations are as complicated as DG... – Shuhao Cao Jul 30 '19 at 19:54

It shall be argued in this post that the whole idea of one Numerical Method being superior to another is merely a prejudice that rests upon insufficient in-depth analysis of the real thing.

The argumentation will proceed at hand of a two-dimensional example.
The reader is invited not to skip through but take notice of the details.

Numerical Analysis of Diffusion starts with a well known Partial Differential Equation (PDE). The problem will be restricted here to the simpler case of two space dimensions: $$ \frac{\partial Q_x}{\partial x} + \frac{\partial Q_y}{\partial y} = 0 $$ $ (x,y) = $ Cartesian coordinates. A possible interpretation of the vector $ (Q_x,Q_y) $ is the heat flux. The differential equation then follows from the law of conservation of energy. In case of pure diffusion of heat, also known as conduction, the components of the heat flux are related to temperature $T$ as follows: $$ Q_x = - \lambda \frac{\partial T}{\partial x} \qquad \qquad Q_y = - \lambda \frac{\partial T}{\partial y} $$ Where $ \lambda = $ thermal conductivity. Hence the final differential equation for the temperature field is actually of the second degree. In order to make the PDE amenable for numerical treatment, an integration procedure has to be resorted to. At this point, there occurs a splitting into several distinct roads, all leading to a numerical solution, more or less efficiently.

Triangle isoparametrics

The simplest Finite Element in two dimensions - and my absolute favorite - is the linear triangle:
enter image description here
Let's summarize the isoparametrics (= affine transformation) in the first place: $$ \begin{cases} x = x_1 + (x_2-x_1)\xi + (x_3-x_1)\eta \\ y = y_1 + (y_2-y_1)\xi + (y_3-y_1)\eta \\ f = f_1 + (f_2-f_1)\xi + (f_3-f_1)\eta \end{cases} $$ It will be shown next how partial differentiation at such a linear triangle takes place. First do the chain rules with global $(x,y)$ and local $(\xi,\eta)$ coordinates: $$ \begin{cases} \Large \frac{\partial f}{\partial \xi} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial \xi} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial \xi} \\ \Large \frac{\partial f}{\partial \eta} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial \eta} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial \eta} \end{cases} \quad \Longleftrightarrow \quad \Large \begin{bmatrix} \frac{\partial f}{\partial \xi} \\ \frac{\partial f}{\partial \eta} \end{bmatrix} = \begin{bmatrix} \frac{\partial x}{\partial \xi} & \frac{\partial y}{\partial \xi} \\ \frac{\partial x}{\partial \eta} & \frac{\partial y}{\partial \eta} \end{bmatrix} \begin{bmatrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \end{bmatrix} $$ But what we need is the inverse, with determinant / Jacobian $\;\Delta = (\partial x / \partial \xi)(\partial y / \partial \eta) - (\partial x / \partial \eta)(\partial y / \partial \xi)$ : $$ \Large \begin{bmatrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \end{bmatrix} = \begin{bmatrix} \frac{\partial x}{\partial \xi} & \frac{\partial y}{\partial \xi} \\ \frac{\partial x}{\partial \eta} & \frac{\partial y}{\partial \eta} \end{bmatrix}^{-1} \begin{bmatrix} \frac{\partial f}{\partial \xi} \\ \frac{\partial f}{\partial \eta} \end{bmatrix} = \begin{bmatrix} \frac{\partial y}{\partial \eta} & -\frac{\partial y}{\partial \xi} \\ -\frac{\partial x}{\partial \eta} & \frac{\partial x}{\partial \xi} \end{bmatrix} / \Delta \begin{bmatrix} \frac{\partial f}{\partial \xi} \\ \frac{\partial f}{\partial \eta} \end{bmatrix} $$ Giving full discretization of the function derivatives, with determinant / Jacobian $\;\Delta = (x_2-x_1)(y_3-y_1)-(x_3-x_1)(y_2-y_1)$ : $$ \Delta \begin{bmatrix} \partial f / \partial x \\ \partial f / \partial y \end{bmatrix} = \begin{bmatrix} (y_3-y_1) & -(y_2-y_1) \\ -(x_3-x_1) & (x_2-x_1) \end{bmatrix} \begin{bmatrix} f_2-f_1 \\ f_3-f_1 \end{bmatrix} $$ Here $\Delta$ is the area of a vector parallelogram, which is twice the area of the triangle. The above can also be written as: $$ \Delta \left[ \begin{array}{c} \partial f / \partial x \\ \partial f / \partial y \end{array} \right] = \left[ \begin{array}{ccc} +(y_2 - y_3) & +(y_3 - y_1) & +(y_1 - y_2) \\ -(x_2 - x_3) & -(x_3 - x_1) & -(x_1 - x_2) \end{array} \right] \left[ \begin{array}{c} f_1 \\ f_2 \\ f_3 \end{array} \right] $$ The matrix in this last formula should be memorized; it is called a differentiation matrix.

Finite Element Method

When using a Finite Element method, the differential equation may be multiplied at first with an arbitrary (test)function. Subsequently the PDE is integrated over the domain of interest. Let the test function be called $f$, then: $$ \iint f . \left[ \frac{\partial Q_x}{\partial x} + \frac{\partial Q_y}{\partial y} \right] \, dx dy = 0 $$ It can be shown that this integral formulation is (more or less) equivalent with the original partial differential equation. This is due to the fact that $f$ is an arbitrary function. It should be non-zero, continuous and integrable, though.
Partial integration, or applying Green's theorem (which is the same), results in an expression with line-integrals over the boundaries and an area integral over the bulk field. The latter is given by: $$ - \iint \left[ \frac{\partial f}{\partial x}.Q_x + \frac{\partial f}{\partial y}.Q_y \right] \, dx dy $$ Mind the minus sign. The advantage accomplished herewith is a reduction of the difficulty of the problem: only derivatives of the first degree are left. As a next step, the domain of interest is split up into "elements" $E$. Due to this, also the integral will split up into separate contributions, each contribution corresponding with an element: $$ - \sum_E \iint \left[ \frac{\partial f}{\partial x}.Q_x + \frac{\partial f}{\partial y}.Q_y \right] \, dx dy $$ It is clear that $\partial f / \partial x$ and $\partial f / \partial y$ are constants. While considering only 2-D diffusion, $Q_x$ and $Q_y$ are also partial derivatives of the first degree, hence constants. Herewith the bulk Finite Element formulation, for one triangle, is given by: $$ - \left[ \frac{\partial f}{\partial x}.Q_x + \frac{\partial f}{\partial y}.Q_y \right] \iint dx dy = - \left[ \frac{\partial f}{\partial x}.Q_x + \frac{\partial f}{\partial y}.Q_y \right] \Delta/2 $$ The remaining integral is equal, namely, to de area of the triangle. Applying now the differentiation matrix, we find: $$ = - \frac{1}{2} \left[ \begin{array}{ccc} f_1 & f_2 & f_3 \end{array} \right] \left[ \begin{array}{cc} y_2 - y_3 & x_3 - x_2 \\ y_3 - y_1 & x_1 - x_3 \\ y_1 - y_2 & x_2 - x_1 \end{array} \right] \left[ \begin{array}{c} Q_x \\ Q_y \end{array} \right] = $$ $$ = \frac{1}{2} \left[ \begin{array}{ccc} f_1 & f_2 & f_3 \end{array} \right] \left[ \begin{array}{c} (y_3 - y_2) Q_x - (x_3 - x_2) Q_y \\ (y_1 - y_3) Q_x - (x_1 - x_3) Q_y \\ (y_2 - y_1) Q_x - (x_2 - x_1) Q_y \end{array} \right] $$ Actually, we don't want to subdivide the Finite Element domain into triangular elements, but rather into quadrilateral elements. However, any quad element, in turn, can be subdivided yet into triangles, even in two different ways:
enter image description here
In addition, what we want is a configuration in which all quad vertices play an equally important role. In order to accomplish this, all of the four triangles must be present in our formulation, simultaneously. For just one quadrilateral, it boils down to renumbering vertices in the formulation for a single triangle, according to the following permutations:

   1  2  3       2  4  1       3  1  4       4  3  2
Also an upper label (not a power) will be attached to the values $(Q_x,Q_y)$, because it must be denoted at which triangle the discretization takes place. Any contributions are summed now over the four triangles (and the whole is divided by a factor two again): $$ \frac{1}{4} \left[ \begin{array}{ccc} f_1 & f_2 & f_3 \end{array} \right] \left[ \begin{array}{c} (y_3 - y_2) Q^1_x - (x_3 - x_2) Q^1_y \\ (y_1 - y_3) Q^1_x - (x_1 - x_3) Q^1_y \\ (y_2 - y_1) Q^1_x - (x_2 - x_1) Q^1_y \end{array} \right] + $$ $$ \frac{1}{4} \left[ \begin{array}{ccc} f_2 & f_4 & f_1 \end{array} \right] \left[ \begin{array}{c} (y_1 - y_4) Q^2_x - (x_1 - x_4) Q^2_y \\ (y_2 - y_1) Q^2_x - (x_2 - x_1) Q^2_y \\ (y_4 - y_2) Q^2_x - (x_4 - x_2) Q^2_y \end{array} \right] + $$ $$ \frac{1}{4} \left[ \begin{array}{ccc} f_3 & f_1 & f_4 \end{array} \right] \left[ \begin{array}{c} (y_4 - y_1) Q^3_x - (x_4 - x_1) Q^3_y \\ (y_3 - y_4) Q^3_x - (x_3 - x_4) Q^3_y \\ (y_1 - y_3) Q^3_x - (x_1 - x_3) Q^3_y \end{array} \right] + $$ $$ \frac{1}{4} \left[ \begin{array}{ccc} f_4 & f_3 & f_2 \end{array} \right] \left[ \begin{array}{c} (y_2 - y_3) Q^4_x - (x_2 - x_3) Q^4_y \\ (y_4 - y_2) Q^4_x - (x_4 - x_2) Q^4_y \\ (y_3 - y_4) Q^4_x - (x_3 - x_4) Q^4_y \end{array} \right] \mbox{ } $$ The four overlapping triangles at the corners of the quadrilateral are a small but significant twist to the standard Finite Element Method, which is motivated by the end-result.

Finite Volume Method

In order to save unnecessary paperwork, the following shorthand notation has been adopted. It may be interpreted as an outer product: $$ r_{ij} \times Q_k = (y_i - y_j) Q^k_x - (x_i - x_j) Q^k_y = (x_j - x_i) Q^k_y - (y_j - y_i) Q^k_x $$ Terms belonging to $f_k, k=1 ... 4$ are collected together. By doing so, the standard Finite Element assembly procedure is demonstrated at a small scale. What else is the Finite Element matrix than just an incomplete system of equations? $$ \frac{1}{4} \left[ \begin{array}{cccc} f_1 & f_2 & f_3 & f_4 \end{array} \right] \left[ \begin{array}{c} r_{32} \times Q_1 + r_{42} \times Q_2 + r_{34} \times Q_3 + 0 \\ r_{13} \times Q_1 + r_{14} \times Q_2 + 0 + r_{34} \times Q_4 \\ r_{21} \times Q_1 + 0 + r_{41} \times Q_3 + r_{42} \times Q_4 \\ 0 + r_{21} \times Q_2 + r_{13} \times Q_3 + r_{23} \times Q_4 \end{array} \right] $$ Subsequently use: $$ r_{32} = r_{34} + r_{42} \qquad r_{14} = r_{13} + r_{34} \qquad r_{41} = r_{42} + r_{21} \qquad r_{23} = r_{21} + r_{13} $$ To put the above in a more handsome form: $$ \left[ \begin{array}{cccc} f_1 & f_2 & f_3 & f_4 \end{array} \right] \left[ \begin{array}{c} \frac{1}{2} r_{42} \times \frac{1}{2} (Q_1+Q_2) + \frac{1}{2} r_{34} \times \frac{1}{2} (Q_1+Q_3) \\ \frac{1}{2} r_{13} \times \frac{1}{2} (Q_1+Q_2) + \frac{1}{2} r_{34} \times \frac{1}{2} (Q_2+Q_4) \\ \frac{1}{2} r_{21} \times \frac{1}{2} (Q_1+Q_3) + \frac{1}{2} r_{42} \times \frac{1}{2} (Q_3+Q_4) \\ \frac{1}{2} r_{21} \times \frac{1}{2} (Q_2+Q_4) + \frac{1}{2} r_{13} \times \frac{1}{2} (Q_3+Q_4) \end{array} \right] $$ It's a triviality, but nevertheless: a picture says more that a thousand words.
enter image description here
enter image description here
It is seen that the four pieces-of-equations correspond with four pieces of line-integrals, each of them belonging to one of the vertices. Midpoints of triangle sides are connected by lines at which the integration takes place. The heat flux at a midpoint is the average of values at the vertices.
Let's adopt another point of view now and no longer concentrate on elements but on vertices. Instead of arranging vertices around an element, elements are arranged around a vertex. Label triangle side midpoints as $a,b,c,d,e,f,g,h$.
It is immediately noted that the lines connecting the midsides of the triangles around a vertex, when tied together, neatly delineate a closed area, which can be interpreted as a kind of 2-D Finite Volume. Expressed in the outer product formalism, we find: $$ r_{ba} \times Q_a + r_{cb} \times Q_c + r_{dc} \times Q_c + r_{ed} \times Q_e + r_{fe} \times Q_e + r_{gf} \times Q_g + r_{hg} \times Q_g + r_{ah} \times Q_a $$ Which is the content of one equation in the Finite Element global matrix. All terms together represent a discretization of the following circular integral: $$ \sum r \times Q = \oint Q_y dx - Q_x dy $$ With help of Green's theorem, however, such a circular integral can be converted into a "volume" integral, over the area indicated in the above figure: $$ \oint Q_y dx - Q_x dy = + \iint \left[ \frac{\partial Q_x}{\partial x} + \frac{\partial Q_y}{\partial y} \right] \, dx dy $$ Conservation of heat is integrated over a finite volume, which is wrapped around a vertex. So we have arrived at sort of a Finite Difference method. To be more precise: at a Finite Volume Method. It is remarked that this F.V. procedure has been applicable for curvilinear grids from the start.
Apply a Finite Element (Galerkin) method to a mesh of quadrilaterals. Subdivide each of the quads into four (overlapping) triangles, in the two ways that are possible. Then such a method is equivalent to a Finite Volume method: midsides of the triangles, around the vertex of interest, are neatly connected together, to form the boundary of a 2-D finite volume, and the conservation law is integrated over this volume.
Unification of a Finite Element and a Finite Volume method has been accomplished herewith, for a restricted class of 2-D diffusion problems.

Han de Bruijn
  • 16,250
  • 2
  • 41
  • 77


FDM is created from basic definition of differentiation that is $$ \frac{df}{dx}=\frac{f(x+h)-f(x)}{h}$$ here "h" tends to zero.

In numerical analysis, its not possible to divide a number by "0" so "zero" means a small number. So FDM is similar to differential calculus but it has killed the heart that is limit tenda to "zero". So in most of the cases accuracy of FDM increases with refining grid. Easy method but not reliable for conservative differential equations and solutions having shocks. Tough to implement in complex geometry where it needs complex mapping and mapping makes governing equation even tougher. Extending to higher order accuracy is very simple


It is a numerical tool that is borrowed from calculus of variation. There are lot of types of FEM like point collocation method, sub-domain method etc. Here they assume some trial function and multiply that trial function with weighting function . In Galerkins method the trial function itself weighting function. Different methods follow different ways in weighting. Then this weighting function is multiplied with trial function then integrated over the control volume ( weak form) and equated to zero (This procedure will differ for different types of FEM but theme is same). Then we get one set of algebraic equations. Solving that will give solution. Here we are working only in error and differential equation some times conservative law may be violated. This method is more accurate than FVM and FDM. Ideal for linear PDEs, expensive and complex for non-linear PDEs. Here higher order accuracy is achieved by using higher order basis (i.e) shape functions. Extending to higher order accuracy is relatively complex than FVM and FDM. Higher order accurate calculations are expensive in computation and Mathematical formulation especially for non-linear PDEs. Mostly suitable for Heat transfer, Structural mechanics, vibrational analysis etc.

FVM: This is similar to FDM but. It didn't kill the theme of differentiation because we are integrating the differential equation over a control volume and discretizing the domain. Since we have integrated the differential equation discetization is mathematically a valid one. It can be loosely viewed as FEM but weight here used is 1. Here fluxes are integrated and resultant is set to zero, so flux is conserved. Can handle almost any PDEs and complex domain. Interpolation is done from face to centre will reduce the accuracy of this process. Here accuracy is based on order of polynomial used. FVM can also produce any order accurate numerical solution similar to FDM but more expensive than FDM Aero acoustic problems use FVM about $11^{th}$ order schemes such schemes are rarely used even in DNS and LES. Ideal for Fluid mechanics.


Isn't it a pity that one has to choose between these methods, while they all have their advantages and disadvantages? Wouldn't it be better trying to take the best of the worlds and mix ingredients together? Perhaps a decent research effort of the kind will result in just one numerical method for solving PDE's instead of two or three distinct ones. Here is my attempt:

Where it should be emphasized that more than one life will be needed to really accomplish things. So, where is my backup?
Han de Bruijn
  • 16,250
  • 2
  • 41
  • 77

Labrujère's Problem

In Februari 1976, Dr. Th.E. Labrujère, at the National Aerospace Laboratory NLR, the Netherlands, wrote a memorandum which is titled, when translated in English: The "Least Squares - Finite Element" Method [L.S.FEM] applied to the 2-D Incompressible Flow around a Circular Cylinder. To be more precise: incompressible and irrotational flow.
In this memorandum, it was firmly established that a straightforward application of the Least Squares Method, using linear triangular Finite Elements, quite unexpectedly, does not work well. Herewith, Labrujère's report is demonstrating a scientific integrity which is rarely seen these days. With our own software we have been able to reproduce the poor results as obtained by NLR:

enter image description here

Improving on these results has been a non-trivial task. On the side of NLR, it could only be accomplished by introducing highly complicated elements. On the side of myself, it could only be accomplished by adopting an approach which is quite deviant from the common Finite Element methodology. It has to be decided by Occam's Razor which of the two approaches is to be preferred.

The Calgary Solution

In December 1976, Labrujère's problem was "solved" by G. de Vries, T.E. Labrujère himself and D.H. Norrie, at the mechanical Engineering Department of The University of Calgary, Alberta, Canada. The result is written down in their Report no.86: A Least Squares Finite Element Solution for Potential Flow. The abstract of this report is copied here:
enter image description here
It seems to me that the above solution is of pure academical interest, though. The apparent need for fifth-order trial functions shall make this method unworkable in practice. Even if attention is restricted to the simple case at hand, it's way too complicated. What's worse, generalization is likely to be hard. In the end, 2-D and 3-D Navier Stokes equations (at a curvilinear grid, preferably) need to be solved. So the point of departure must be something which is much more simple. Especially the number of unknowns at each nodal point should not exeed the absolute minimum, the number of degrees of freedom: two. I have never been in doubt that an alternative least squares finite element solution, having such desirable properties, must be possible. I have a dream ..

Incompressible irrotational (ideal) flow of an inviscid fluid is described by the following system of linear first-order (!) Partial Differential Equations (PDE's): $$ \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0 \quad \mbox{: incompressible} \\ \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} = 0 \quad \mbox{: irrotational} $$ Here: $(x,y) =$ coordinates , $(u,v) =$ velocity-components.
There does not exist a kind of "natural" variational principle for the above differential equations. Conventional Finite Element Methods, however, are very much dependent upon the existence of such principles. There must be something to minimize (or to "make stationary"). In cases like the above, it seems, at first sight, that L.S.FEM offers a possible solution. That is because Least Squares Finite Element Methods proceed by constructing an alternative minimum principle: square the equations just as-they-are (!) , add these squares together, integrate their sum over the area of interest and minimize the result as a function of the unknowns. This is the approach as described in O.C. Zienkiewicz "The Finite Element Method" (1977) chapter 3.14.2. In our case: $$ \iint \left\{ \left[ \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} \right]^2 + \left[ \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right]^2 \right\} \, dx.dy = \mbox{minimum}(u,v) $$ Simple as it sounds, but watch out! People (including myself) have wasted very much time trying to get this method to work. After many years of frustration, I even had to give up for a while. Appearance is highly deceptive here: Least Squares may be the most tricky Finite Element Method that has ever been invented. We have already seen that Least Squares does not work well for linear triangles, that is iff the method is applied in a straightforward Finite Element manner. Which is the bare essence of Labrujère's Problem. Start of personal motivation. It is our purpose to show, in the end, that Labrujère's problem can be solved in a proper manner. Herewith I mean: a simple and straightforward manner. However, to that end, we must look at the problem from a different, or should I rather say a "difference" perspective. As if it were essentially a Finite Difference problem, namely, instead of the Finite Element problem that it only appears to be. With other words:
the Least Squares Finite Element Method is a Finite Difference Method in disguise.

A Difference Perspective

Let's look at the details. At first, the global F.E. integral is split up into separate contributions, from all finite elements $(E)$ in the mesh: $$ \sum_E \iint \left\{ \left[ \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} \right]^2 + \left[ \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right]^2 \right\} \, dx.dy = \mbox{minimum} $$ It is often advantageous to carry out a Numerical Integration, instead of an "exact" one (see e.g. Zienkiewicz chapter 8.8). This means that function values are to be determined at so-called integration points $p$. With each integration point $p$ a certain weight factor $w_p$ is associated: $$ \sum_E \sum_p w_p \left\{ \left[ \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} \right]_p^2 + \left[ \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right]_p^2 \right\} \, J_p = \mbox{minimum} $$ Here $J_p$ is the Jacobian (determinant), which is the result of a transformation from global to local F.E. coordinates. The jacobians $J_p$ as well as the weighting factors $w_p$ are positive real-valued numbers.
What follows is a small step for man: unify the summations over the elements and the integration points, resulting in one global summation over all integration points $(i=E,p)$, where $(i)$ becomes the global index of any "integration point". This merely says that summing over elements, together with their integration points, is equivalent with summing over all the integration points in the whole domain of interest, in one big sweep. In this way, integration points can be interpreted as more elementary than the elements themselves. And an element with more than one integration point can be considered as a superposition of elementary integrated elements, with only one integration point $(i)$ in each of them: $$ \sum_i w_i \left\{ \left[ \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} \right]_i^2 + \left[ \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right]_i^2 \right\} \, J_i = \mbox{minimum} = 0 $$ In order for L.S.FEM to work properly, the minimum required must be a small number, rapidly approximating zero, as the size of the elements becomes less. Thus maybe it would be not such a weird idea to demand that the minimum value should merely be zero from the start. But in that case the above "variational integral" would have been equivalent to an non-squared system of linear equations. Because when a sum of squares can possibly be zero? If and only if each of the separate terms in the sum is equal to zero: $$ \left[ \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} \right]_i = 0 \quad ; \quad \left[ \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right]_i = 0 \quad \mbox{: for each integration point } (i) $$ Let's go one more step further. It is realized that each 'integration point' in the grid does in fact nothing else than creating two independent equations. All integration points together contribute to the fact that a whole system of linear equations emerges in this way. Nothing prevents us from calling this a "Finite Difference" system of equations. Let's therefore, at last, replace the notion of 'an integration point' simply by: 'an F.D. equation'. And here we are!

Any feasible Least Squares Finite Element Method is equivalent with forcing to zero the sum of squares of all equations emerging from some Finite Difference Method.
L.S.FEM gives rise to the same solution as an equivalent system of finite difference equations.

We are ready now to look at Labrujère's problem in the following way. Let it be required that the Least Squares Finite Element Method always leads to an acceptable solution, with moderate mesh sizes. Then, of course, in the associated Finite Difference system, the number of unknowns $N$ should be equal to the number of independent equations $M$. If such is not the case, namely, then the system is likely to be overdetermined. And it is doubtful if the Least Squares minimum can still approach zero, fast enough. A simple count of the triangles involved with Labrujère's problem reveals that such kind of a delicate balance between unknowns and equations is definitely not achieved there: the number of elements outweights the number of nodal points by a factor $2$! This means that there are roughly twice as many "unsquared" F.D. equations as there are unknowns. Apart from of any more complicated kind of argument, like higher order continuity, this surely throws up a basic question.

I am not qualified to check out whether Norrie and DeVries implicitly adressed that question, in their report. They first kept the triangular shapes. I guess that, in order to compensate for an abundance of elementary equations, they had to introduce even so many additional variables. Now it becomes clear what kind of different approach may be feasible here. For the only thing that has to be accomplished is: a perfect balancing between the number of equations and the number of unknowns. Instead of increasing both these numbers by some complicated mechanism.


Th.E. Labrujère,
'DE "EINDIGE ELEMENTEN - KLEINSTE KWADRATEN" METHODE TOEGEPAST OP DE 2D INCOMPRESSIBELE STROMING OM EEN CIRKEL CYLINDER', Memorandum WD-76-030, Nationaal Lucht- en Ruimtevaartlaboratorium (NLR), Noordoostpolder, 23 februari 1976.

G. de Vries, T.E. Labruj`ere, D.H. Norrie,
'A LEAST SQUARES FINITE ELEMENT SOLUTION FOR POTENTIAL FLOW', Report No.86, Department of Mechanical Engineering, The University of Calgary, Alberta, Canada, December 1976.

O.C. Zienkiewicz,
'The Finite Element Method', 3th edition, Mc.Graw-Hill U.K. 1977, ISBN 0-07-084072-5

To be continued as:

Any employment for the Varignon parallelogram?

Take a good look at these (Least Squares) Finite Difference Elements:

  • No continuity requirements, at all, on the components of velocity
  • First order trial functions (linear) for both components of velocity
  • Numerical results in close agreement with the theoretical solution

Han de Bruijn
  • 16,250
  • 2
  • 41
  • 77

finite difference method is the oldest method to find the limited region or close region and FEM is the structural method to solve the partial deferential equation.