Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Why are advanced finite-element methods not more popular?

  1. Apr 21, 2014 #1
    I'm an engineer who uses a lot of finite-element software for doing multiphyics modeling. I've been learning a lot about the foundations of the finite-element method, but I have a particulary nagging question that I havn't been able to get a satisfying answer to: Why do most finite-element programs use relatively low order lagrange elements?

    From my understanding, are more advanced elements, with higher order that have many advantages like continous spatial derivatives. For some applications like solid-mechanics or heat transfer, shouldn't more advanced elements be desirable? If the field variable is the displacements, than the strains are formed from the derivatives. If the field variable is temperature, then the heat flux is determined according to the gradient. Since Lagrange elements are only C0 continuous (derivatives are discontinous between elements) strain and heat flux may be discontinous between elements. Shouldn't we use fancier C1 continous elements? Like Argyris elements? Wouldn't this also benifit the accuracy of convective terms?

    I know that they can be more computationally expensive because higher order elements require more quadurature points and shape funtions to evaluate. Obviously there must be some significant downside or else they would be common.
  2. jcsd
  3. Apr 21, 2014 #2
    Discontinuous fluxes aren't necessarily a bad thing if you have discontinuous media. We know there will be jump conditions on interior boundaries so the derivatives of the dependent variables aren't always expected to be continuous at these interfaces.

    It depends what you want. If you are a mathematician and don't like modelling errors then you might want higher order elements, but when used in practice on a machine with finite RAM the user will soon learn to make practical compromises.
  4. Apr 21, 2014 #3


    User Avatar
    Science Advisor
    Homework Helper

    High order continuity elements don't play well with strongly nonlinear problems. For example if you have a thermal problem where you need to track the moving boundary of a phase change between solid and liquid, with different thermal hehavior (e.g. the convection only happens in the liquid) and the latent heat effects at the interface, a fine mesh of low order elements works better at capturing the discontinuities on the solution that are physically real. Smoothing the discontinuities away with high order elements is physically wrong.

    Even for appropriate problems, specifying boundary conditions for high order elements is non trivial and error prone, especially at locations in the model like sharp corners. (There was once an infamous element for structural analysis called the Dupuis quadrilateral, which had 13 degrees of freedom per node including derivatives like ##\partial^3u/\partial^2x\partial y##. Thankfully, it's now extinct.)

    Also, any numerical method that is of higher order accuracy than the underlying PDE it is solving is liable to produce "solutions" that are an artefact of the solution process, and have nothing to do with the physics. Of course you can design methods that suppress the spurious solutions for nicely behaved problems, but if the model is strongly nonlinear, you can usually find some way to break the numerical solution.
    Last edited: Apr 21, 2014
  5. Apr 21, 2014 #4
    Interesting! It isn't surprising that a slope discontinuty could cause problems with elements designed to prevent those. The same must apply to multiphysics problems since each element can have different material properties than the neighbours?

    I will have to look into the Dupuis element. That sounds like an interesting story.

    Do these problems occur for the 'hp'-fem methods where the solution of the linear system is no longer nodal values?
  6. Apr 21, 2014 #5


    User Avatar
    Science Advisor
    Homework Helper

    IIRC it was in the MARC finite element program back in about 1975-1980. This story is a bit ironic, since MARC were doing multiphysics long before the name was invented.

    MARC Analysis Inc has now been bought out by MSC (the NASTRAN company). The current sales literature on Marc doesn't even mention what element types it uses.

    If you want to try a literature search, two of the senior MARC guys from that time were Joop Nachtegaal and IIRC Cees Gelten. There should be some papers by them in journals like Int J Num Eng. Dave Hibbitt (the "H" of HKS, the Abaqus company, later bought by Dassault) was also at MARC round about then, before he left to do his own thing.

    I don't know much about hp methods (and the only experience I had with them was poor, but that could have been the fault of the software not the method).

    The exponential growth in CPU speed and memory size in personal computers have put many techniques in FEA into the category of "it seemed like a good idea at the time". Tne engineering problems to be solved haven't changed much, but not many people can remember the best ways of solving them on computers with clock speeds of a few MHz (not a few GHz), and less than 1Mb of RAM. Even the first Cray supercomputers were an order of magnitude less powerful than a today's cheap notebooks.
    Last edited: Apr 21, 2014
  7. Apr 22, 2014 #6
    I work with high order spectral elements. The problems that I study have extreme anisotropies, and here the spectral convergence properties of these elements is incredibly handy. However, there is a lot of over head involved in using spectral (or hp) elements. In general they are harder to code. The matrices they produce are denser, and harder to invert.

    In our case, where we need extreme resolution there is a huge payoff to using spectral elements. But its hard to justify the effort of developing spectral elements in less demanding cases where standard finite elements work just fine,

    Yes, when using spectral elements there are similar advantages to using C0 elements. Also, you can still use a nodal basis (like Gauss-Labatto-Legendre polynomials) with spectral elements. Many proofs use a modal basis (like Legendre polynomials) because the math is simpler. The two methods are equivalent, and both representations have different advantages.

    Finally, I should mention that there are advantages to using C1 elements. For example, C1 elements are useful for systems of equations that contain 4th order derivatives. You can still use C0 elements, but you have to introduce an auxiliary field.
  8. Apr 22, 2014 #7
    It makes sense that before sparse matrix solvers were refined, higher order methods producing denser matrices offered better accuracy for the same size of matrix (which used to be a significant limiting factor).

    When you say harder to invert/solve? What exactly do you mean? Do you just mean less sparse or are you also talking about the numerical accuracy? It seems to me that a non-node based solution would have a larger range of numbers, possibly leading to numerical problems.

    I have not formally read about hp and spectral elements, it is tough to find a resource at the appropriate math level (not too simple, not too advanced). Are they very closely related? I was under the impression that hp-method was similar to the h-method with the addition of extra shape functions, not using nodal values. I also thought spectral spectral elements took a Fourier type approach using sines/cosines as shape functions (to exploit orthogonality)?
  9. Apr 22, 2014 #8
    Try the book of Sherwin and Karniadakis on spectral/hp finite element methods. I programmed my first hp method using this book. The p in this case simply means that you can have p-th order basis functions. Originally, Chebyshev polynomials were used in spectral finite element methods as well, like for classic spectral methods. Other basis functions have proven more effective, like Lagrange polynomials through the Gauss-Lobatto-Legendre points.

    The reason people are not using very high order methods in FEM is the same reason it's not used in finite difference methods: higher than say 6th order doesn't give you a more accurate solution for the same degrees of freedom in practical problems. If the local (Sobolev) regularity of the solution is smaller than the polynomial degree of your basis functions, you are basically trying to approximate a linear line with a high order polynomial.
  10. Apr 22, 2014 #9
    Thanks bigfooted, I will take a look at the Sherwin and Karniadakis book.

    Don't high order methods allow for larger elements? Obviously if you continue to increase the element order, when already converged you gain nothing. For some reason I have it in my head that large finite-elements with appropriately high order should be 'better' than smaller elements of a lower order (assuming the problem is not over meshed).
  11. Apr 22, 2014 #10
    Sherwin and Karniadakis is a great book!

    They do to a point. But remember how there are advantages to allowing for discontinues in derivatives across elements. We lose that advantage if we use to large of an element. So the ability to both reduce the element size and increase the order of the elements basis functions are advantageous.

    Both, Spectral elements matrices are less sparse. I also believe that they typically have worse conditions numbers, so accuracy is a concern. There are tricks that help, and modern linear algebra packages have really advance over the last decade.

    The p refers to the fact that you can increase the order of the shape function. For example the simplest FEM codes often use linear elements. A -p method would allow you to increase the order of the elements. So you could use linear, quadratic, cubic, or even higher order elements.

    The difference between high order finite elements and spectral elements is blurry. I usually refer to hp element methods as high order finite element methods if they converge algebraically as p is increased (error goes like p^-(constant)). I use spectral methods to refer to methods that can achieve spectral convergence (error ~ e^-p) for a sufficiently smooth solutions.

    I caution you. It sound like you equate nodal basis with low finite element and modal basis with spectral element. This is wrong. Both methods can use both types of basis functions. You could for instance do a low order finite element method using sin x as your basis function. Likewise you can, and many people do, use Lagrange Polynomials (high order tent polynomials) in spectral element methods. However, you have to be care in how you chose you node locations if you want to achieve spectral convergence.

    I strongly disagree with this statement. I do agree that if your solution is represented by a low order polynomial, then you don't gain accuracy by increasing polynomial agree. However many simple problems aren't accurately represented by low order polynomials. For example heat conduction in a cylinder.

    I recently did a benchmark case that involved solving a simple 2D elliptical equation. To illustrate the power of spectral elements I compared a mesh using low order elements (representative of standard FEM) with a mesh that uses high order elements (spectral element). For the FEM mesh I get an error of 1e-8 using 128x128 mesh with 2nd order polynomials. This mesh has approximately 100,000 degrees of freedom. For the same problem I get an error of 1e-9 on a 2x2 mesh with 10th order polynomials. This mesh has about 500 degrees of freedom.
    Thus using spectral elements I can get a solution that is 10 times more accurate using 200 times fewer degrees of freedom over traditional finite elements. This reduction in the problem size also reduced the computation time by a factor of 400.
  12. Sep 9, 2014 #11
    A lot depends on the actual application area. For example, bending of plates and shells require, for the sake of accuracy, high-order elements: bending moments are the (scaled) 2nd derivatives of the displacements; shear forces are even "worse", since they require accurate calculations of the 3rd derivatives.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook