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I'm looking at the solution to a structural mechanics problem that is modeled with first-order elements and then as a comparison with second-order elements. It is clear that the first-order elements behave as very stiff elements and give very different displacements compared to the second-order elements. But shouldn't the solution still be exact at the nodes?

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3 Answers 3

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They should both converge to the same limit solution, but are expected to do so at different rates of convergence.

The best way to verify the convergence rates is to use a test-setting with known analytic solution. Then, when you compare the numerical solutions at different refinements with the analytical expression you can observe the actual convergence rate of your application.

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  • $\begingroup$ Maybe I have misunderstood something fundamental about the finite element method. So by convergence you mean refining the mesh until the solution is found? $\endgroup$
    – Tepa
    Oct 17, 2022 at 7:10
  • $\begingroup$ We do these simulations to find some behaviour as it is in the real world. Since we can only ever approximate the solutions with a finite set of points/elements, we will always be slightly off. This difference between numerical simulation and exact solution (if known) gets smaller the more points/elements you consider. The rate at which this difference gets smaller is the convergence rate and is different for first order finite element methods vs. higher order methods. So one method may approach the "true" solution faster than another one, but they both should meet at the same solution. $\endgroup$
    – MPIchael
    Oct 17, 2022 at 8:34
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It is not true that the finite element solution is exact a nodes.

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  • $\begingroup$ Let me add that in some special cases it might be exact, such as 1D Poisson problem with a constant material parameter and loading. $\endgroup$
    – knl
    Oct 17, 2022 at 18:00
  • $\begingroup$ @knl I think your comment is incorrect, however worth mentioning. If the solution is linear, both first and second order elements are exact at the nodes. If the solution is quadratic, however, only the latter is exact. $\endgroup$
    – ConvexHull
    Oct 17, 2022 at 18:37
  • $\begingroup$ No, try it yourself. Solve $-u''=1$ with $u(0)=u(1)=0$ using linear elements. The exact solution is $u(x) = x (1 -x)/2$ which is obviously quadratic. Despite this, the finite element solution is exact at the nodes. This has nothing to do with the solution being linear and has a deeper connection to Sobolev embedding theorem which states that in 1D $H^1$ is continuously embedded in the set of continuous functions. Therefore, without any additional regularity assumptions, we are able to use the pointwise evaluation as a projection operator in the analysis. This way one can prove the exactness. $\endgroup$
    – knl
    Oct 18, 2022 at 6:32
  • $\begingroup$ It is true that for this very specific case, the discrete solution is exact at the nodes. But this is not true if (i) you are in 2d, or (ii) you have non-constant coefficients, or (iii) you have a variable right hand side that you can't integrate exactly, or (iv) pretty much any other effect. In particular, I cannot see why that very specific situation should apply to the original question. $\endgroup$ Oct 18, 2022 at 17:02
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    $\begingroup$ Perhaps something like $-u''=1$ could be the first equation you are trying finite elements on while learning and will cause such confusion. This is not the first time I've seen this question. $\endgroup$
    – knl
    Oct 19, 2022 at 5:13
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The finite element solution is not, in general, exact at nodes. Perhaps this misunderstanding is caused by the fact that in some rare special cases the finite element solution can, in fact, be exact at nodes.

Here is an example of the finite element solution to $-u'' = 1$ with the boundary conditions $u(0)=u(1)=0$ using two linear elements, compared against the exact solution $u(x)=\frac{x(1-x)}{2}$:

In some cases finite element solution is exact at nodes.

We now observe that the finite element solution is exact at the node $x=0.3$. Nevertheless, this property does not work in general, e.g., in 2D or 3D second-order elliptic problems or simply when approximating more complicated 1D differential equations.

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    $\begingroup$ I think it's good that you're substantiating your claim, but from a pedagogical perspective I think this answer and the prior comments are doing a disservice to the questioner/readers, who may incorrectly infer/insist/wish that some "exactness-at-nodes" property holds, based on a lucky example or two. I think instead one should stress that FEM only guarantees what the weak form says on the tin: that the residual is orthogonal to the space of approximation. $\endgroup$ Oct 18, 2022 at 13:18
  • $\begingroup$ Thank you for the comment. I modified the answer so that I explain upfront that this is not a general property of the finite element approximation. $\endgroup$
    – knl
    Oct 18, 2022 at 14:22

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