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When looking up the tag description of (finite-element-method), we saw the following :

[ .. ] It consists of a method of discretization splitting the domain
into disjoint subdomains [ .. ]
Take a look at the picture below. Suppose that we have two Finite Element meshes, covering the same domain of interest. Can they be united together, to form a mesh of overlapping triangular elements that also works?

enter image description here

Suppose that discretization at the mesh on the Left leads to the equations : $\mathbf{S_L x_L} = \mathbf{b_L}$ .
Suppose that discretization at the mesh on the Right leads to the equations : $\mathbf{S_R x_R} = \mathbf{b_R}$ .
Putting these two meshes together then corresponds to : $\left(\mathbf{S_L} + \mathbf{S_R}\right) \mathbf{x_U} = \left(\mathbf{b_L}+ \mathbf{b_R}\right)\,$ .
This surely is fulfilled if $\mathbf{x_U} = \mathbf{x_L}$ and $\mathbf{x_U} = \mathbf{x_R}$. But all we know is that Left and Right are decent meshes, that lead to the "true" solution upon sufficient refinement. Thus we know that $\mathbf{x_L}$ and $\mathbf{x_R}$ are not too different in the end: $$ \mathbf{x_R} = \mathbf{x_L} + \mathbf{\Delta} \quad \mbox{with} \quad \mathbf{\Delta} \to \mathbf{0} $$ Can we conclude herefrom that the United mesh also leads to a decent approximate solution $\mathbf{x_U}$ ?

EDIT. Suppose that $\mathbf{x_U} \approx \frac{1}{2}\left(\mathbf{x_L}+ \mathbf{x_R}\right)$ and calculate the residual in some suitable norm $\left\|\right\|$ : $$ \left\| \left(\mathbf{S_L} + \mathbf{S_R}\right) \mathbf{x_U} - \left(\mathbf{b_L}+ \mathbf{b_R}\right)\right\| = \\ \left\|\mathbf{S_L}\frac{1}{2}\mathbf{x_L} + \mathbf{S_L}\frac{1}{2}\left(\mathbf{x_L}+\mathbf{\Delta}\right) + \mathbf{S_R}\frac{1}{2}\left(\mathbf{x_R}-\mathbf{\Delta}\right) + \mathbf{S_R}\frac{1}{2}\mathbf{x_R} - \left(\mathbf{b_L} + \mathbf{b_R}\right)\right\|= \\ \frac{1}{2}\left\|\left(\mathbf{S_L}-\mathbf{S_R}\right)\mathbf{\Delta}\right\| \le \frac{1}{2}\left\|\mathbf{S_L}-\mathbf{S_R}\right\|\left\|\Delta\right\| \quad \mbox{with} \quad \left\|\Delta\right\| \to 0 $$ In fact, the residual is even smaller than one might think, because only the difference of the quadrilateral element diagonal contributions ( $\color{red}{\mbox{Left}}$ minus $\color{green}{\mbox{Right}}$ ) is still present in $\,\left(\mathbf{S_L}-\mathbf{S_R}\right)$ :
enter image description here

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  • $\begingroup$ I don't think they have to be. If not then we probably get over determined equations. $\endgroup$ Commented May 30, 2018 at 14:42
  • $\begingroup$ After "interleaving" you still get disjoint elements. The point is that each element has its own "simple" local function (usually a low degree polynomial). It makes no sense to ask for it to be one local function due to the presence of one element and simultaneously another one due to the presence of another element. This would require all the parameters to simultaneously be continuous (otherwise you would spoil regularity inside of an element), which is the same as just having one bigger element. $\endgroup$
    – Ian
    Commented May 30, 2018 at 15:09
  • $\begingroup$ @Ian: Yes and no, I think. See my answer below, where the bigger element is a quadrilateral with integration points at its vertices. $\endgroup$ Commented Jun 1, 2018 at 20:38

1 Answer 1

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Also in the description of the (finite-element-method) tag it is said that

It is closely tied to the calculus of variations.
So let us consider the following variational integral: $$ \iint \left[ \left(\frac{\partial \phi}{\partial x}\right)^2 + \left(\frac{\partial \phi}{\partial y}\right)^2 \right] dx\,dy = \mbox{minimum}(\phi) $$ In Finite Element methodology, as a rule, two coordinate systems are distinguished: global coordinates $\,(x,y)$ - applicable to the whole domain of interest - and local coordinates $\,(\xi,\eta)$ - applicable to one element only. In this way a mapping is induced between local and global coordinates. That mapping is called isoparametric if every function $\,f\,$ at an element - global coordinates being no exception - is expressed in the local coordinates, in the same way. Thus we have $\,\phi(\xi,\eta)$ , $x(\xi,\eta)$ , $y(\xi,\eta)$ . And they all are similar expressions; examples following soon.

Of course we are mainly interested in the global derivatives $\,\partial \phi / \partial x\,$ and $\,\partial \phi / \partial y$ . How can they be expressed in local coordinates? As follows. Apply the chain rule for partial deriviatives, starting in the local coordinate system: $$ \begin{cases} \large \frac{\partial \phi}{\partial \xi} = \frac{\partial \phi}{\partial x}\frac{\partial x}{\partial \xi} + \frac{\partial \phi}{\partial y}\frac{\partial y}{\partial \xi} \\ \large \frac{\partial \phi}{\partial \eta} = \frac{\partial \phi}{\partial x}\frac{\partial x}{\partial \eta} + \frac{\partial \phi}{\partial y}\frac{\partial y}{\partial \eta} \end{cases} $$ Two equations with two unknowns. They can be solved with Cramer's rule : $$ \frac{\partial \phi}{\partial x} = \large \frac{\frac{\partial \phi}{\partial \xi}\frac{\partial y}{\partial \eta} - \frac{\partial y}{\partial \xi}\frac{\partial \phi}{\partial \eta}} {\frac{\partial x}{\partial \xi} \frac{\partial y}{\partial \eta} - \frac{\partial y}{\partial \xi}\frac{\partial x}{\partial \eta}} $$ $$ \frac{\partial \phi}{\partial y} = \large \frac{\frac{\partial x}{\partial \xi}\frac{\partial \phi}{\partial \eta} - \frac{\partial \phi}{\partial \xi}\frac{\partial x}{\partial \eta}} {\frac{\partial x}{\partial \xi} \frac{\partial y}{\partial \eta} - \frac{\partial y}{\partial \xi}\frac{\partial x}{\partial \eta}} $$ The simplest Finite Element in two dimensions is a linear triangle :

The vertex values of an arbitrary function $f$ at a linear triangle are interpolated by: $$ f(\xi,\eta) = (1-\xi-\eta).f_1+\xi.f_2+\eta.f_3 $$ The very meaning of an isoparametric transformation is that we thus have: $$ \begin{cases} x(\xi,\eta) = (1-\xi-\eta).x_1+\xi.x_2+\eta.x_3 \\ y(\xi,\eta) = (1-\xi-\eta).y_1+\xi.y_2+\eta.y_3 \\ \phi(\xi,\eta) = (1-\xi-\eta).\phi_1+\xi.\phi_2+\eta.\phi_3 \end{cases} $$ So there is no need to calculate the local derivatives separately. Just substitute for $\,f\,$ the "true" function, which is $\,x\,$ or $\,y\,$ or $\,\phi$ : $$ \frac{\partial f}{\partial \xi} = f_2-f_1 \quad ; \quad \frac{\partial f}{\partial \eta} = f_3-f_1 \quad \mbox{with} \quad f = x,y,\phi $$ Next take a look at Quadrilateral Interpolation :

If the parent element of the quadrilateral on the left is $\left[-\frac{1}{2},+\frac{1}{2}\right]\times\left[-\frac{1}{2},+\frac{1}{2}\right]$ then we have for an arbitrary function $f$ at the element on the right: $$ f(\xi,\eta) = \left(\frac{1}{2}-\xi\right)\left(\frac{1}{2}-\eta\right)f_1 + \left(\frac{1}{2}+\xi\right)\left(\frac{1}{2}-\eta\right)f_2 \\ + \left(\frac{1}{2}-\xi\right)\left(\frac{1}{2}+\eta\right)f_3 + \left(\frac{1}{2}+\xi\right)\left(\frac{1}{2}+\eta\right)f_4 $$ And for its derivatives, at vertex $(1)$ in particular: $$ \frac{\partial f}{\partial \xi} = \left(\frac{1}{2}-\eta\right)(f_2-f_1) + \left(\frac{1}{2}+\eta\right)(f_4-f_3) \quad \Longrightarrow \quad \frac{\partial f}{\partial \xi}(\eta = -1/2) = f_2-f_1\\ \frac{\partial f}{\partial \eta} = \left(\frac{1}{2}-\xi\right)(f_3-f_1) + \left(\frac{1}{2}+\xi\right)(f_4-f_2) \quad \Longrightarrow \quad \frac{\partial f}{\partial \eta}(\xi = -1/2) = f_3-f_1 $$ Where again we must replace our dummy $\,f\,$ by the true functions $\,x,y,\phi\,$ in order to take care of the real thing.

It is readily observed now that all of the local derivatives at vertex $(1)$ of the quadrilateral are exactly the same as all of the local derivatives at the linear triangle $(1,2,3)$. By symmetry, we can even conclude that the local derivatives at any corner of the quadrilateral are exactly the same as the local derivatives at the triangle corresponding with that corner. But then the same holds for the global derivatives $\,\partial \phi / \partial x\,$ and $\,\partial \phi / \partial y$ , because the analytical formula that expresses them into the local derivatives is the same for all (isoparametric) elements. Last but not least, the analytical as well as the numerical formulation for the area elements does not make any difference too: $$ dx\,dy = \frac{1}{4}\left[\frac{\partial x}{\partial \xi} \frac{\partial y}{\partial \eta} - \frac{\partial y}{\partial \xi}\frac{\partial x}{\partial \eta}\right]d\xi\,d\eta $$ When specified for corner $(1)$, quadrilateral as well as triangle (each overlapping triangle counted half its weight/area): $$ dx\,dy = \frac{1}{4}\left[(x_2-x_1)(y_3-y_1)-(y_2-y_1)(x_3-x_1)\right]d\xi\,d\eta $$ Consequently, integration points at the vertices of a quadrilateral cannot be distinguished from four overlapping triangles at the corners of that quadrilateral:
enter image description here
All mathematical components of both discretizations are the same in this variational integral: $$ \iint \left[ \left(\frac{\partial \phi}{\partial x}\right)^2 + \left(\frac{\partial \phi}{\partial y}\right)^2 \right] dx\,dy = \mbox{minimum}(\phi) $$ So indeed: Finite Elements need not to be disjoint per se. Concerning the integration points:
Are there any two-dimensional quadrature that only uses the values at the vertices of triangles?

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