I'm working to understand how the partition of unity method is used to build a global PUM space from local approximation spaces. Can someone please explain the mechanics of gluing the local spaces together and what type of object it results in (e.g matrix or vector)? I would really appreciate any help solidifying the main ideas here.

Let $\{ \Omega_i \}$ (for $i = 1,...,m$ ) be an open over of $\Omega \subset \mathbb{R}^n$ and let $\{ \varphi_i\}$ be a $(M,C_{\infty},C_G)$ partition of unity subordinate to $\{ \Omega_i \}$. Let $V_i \subset H^1(\Omega_i \cap\Omega)$ (the local approximation spaces) be given. In this paper (statement 2.2 with slight notation change) they state that the global PUM space is given by the definition: $ S(\Omega) = \{\psi : \psi = \sum _{i = 1} ^m \sum _{j=1}^{N_i} \varphi_i v_i ^{[j]} \mbox{ where } v_i ^{[j]} \in V_i \}$. Note here that $N_i$ is the size of each local approximation space's basis.

I'm having trouble visualizing what would pop out of these sums. In this definition local spaces $V_i$ are matrices whose columns form a basis of the corresponding element of the open cover and partition of unity functions could be piecewise linear shape (hat) functions, for example. Then how would I interpret these sums? Does it give me a single vector? Or a number? Or is it a matrix?

My current thoughts: Each local basis vector $v_i ^{[j]} \in V_i$ has length $N$ (for every local space this length will be the same). The inner sum $\sum _{j=1}^{N_i} \varphi_i v_i ^{[j]}$ is like weighting each local basis vector by the hat function value (each element of the cover has one $\varphi_i(x,y)$). I think $\varphi_i(x,y)$ needs to be evaluated at the same $(x,y)$ values that $v_i ^{[j]}$ is defined on. In which case I think the inner sum is the sum of a $(1,N)$ vector (meaning $\varphi_i(x,y)'$) component wise times each $(N,1)$ vector in the local basis (meaning $v_i ^{[j]}$).This gives us a weighted matrix. And then we use the direct sum to add them up and get a block matrix.

Any help making these ideas more concrete is greatly appreciated. Thanks!

  • $\begingroup$ What’s a PUM space? $\endgroup$
    – Deane
    Commented May 18 at 14:59
  • $\begingroup$ @Deane the the global space created from using the partition of unity method to glue together the local spaces $\endgroup$
    – k12345
    Commented May 18 at 14:59
  • $\begingroup$ Could you give a rigorous mathematical definition of a PUM? Normally, no partition of unity is used to build a space. They are typically used to build smooth functions on a space. $\endgroup$
    – Deane
    Commented May 18 at 15:04
  • $\begingroup$ @Deane I'm really sorry I'm still very new to this topic and I don't know how to give many more details on this. As far as I understand the definition I provided above is the main definition. These concepts are also discussed in the paper "The partition of unity finite element method: Basic theory and applications" by Babuska and Melenk. There (in Def 2.2) it is referred to as PUFEM space and it is defined in the same way. $\endgroup$
    – k12345
    Commented May 18 at 15:14
  • $\begingroup$ OK. I misunderstood. When you say "space" you mean a space of functions. Alas, I don't have the time to wade through all the terminology and notation. It might be helpful if you try to provide a more detailed definition, break it down into simpler pieces, and identify where exactly you get stuck. $\endgroup$
    – Deane
    Commented May 18 at 16:06

1 Answer 1


It seems as though your general question is how to represent a global function that is the sum of local basis functions.

The global representation you have defined seems slightly askew, as the fuctions $v_i^j$ are the basis functions, there should be a coefficient for each $i,j$, so that $\psi = \sum_{i=1}^n\sum_{j=1}^{N_i}\phi_i^jv_i^j$.

If all of the $N_i=N$ for some fixed integer, then the set of coefficients can be represented by a matrix $\Phi\in\mathbb R^{n\times N}$, or equivalently as a vector $\Phi\in\mathbb R^{nN}$. Notice that the vector representation is more flexible, as if the $N_i$ values are not fixed, it is not clear how to represent the coefficients as a matrix, but they can still be represented as a vector $\Phi\in\mathbb R^K$ with $K=\sum_{i=1}^nN_i$.

We can use the summation $\psi = \sum_{i=1}^K\phi_iv_i$ which expresses $\psi$ as a function (which would evaluate to a scalar), but is also uniquely represented by the vector $\Psi$.

Using the vector representation is particularly useful in finite element methods, as the whole problem can be boiled down the a linear system of the form $A\Psi = b$ (for linear problems at least).

Taking a specific example, consider the continuous piecewise linear finite element method for the Poisson problem with homogeneous Dirichlet BCs, over a uniform mesh on $[0,1]$.

The weak formulation is stated: Find $u_h\in U_h$ such that $$\int_0^1u_h'v_h' = \int_0^1fv_h\quad\forall v_h\in U_h.$$ Expressing $u_h=\sum_{i=1}^nU_iv_i$, the above is equivalent to $$\int_0^1\sum_{i=1}^nU_iv_i'v_j'=\int_0^1 fv_j\quad \forall j=1,\ldots,n.$$ Now each basis function satisfies $v_j(x_i)=1 $ if $i=j$, and $0$ otherwise (to visualise it think of a tent function, it is only non zero in the two subintervals on each side of $x_j$). This tells us that $v_j' = 1/h$ on $[x_{j-1}, x_j]$, and $v_j'=-1/h$ on $[x_j,x_{j+1}]$, and is zero everywhere else.

The matrix $A$ for the linear system satisfies $$A_{ij}=\int_0^1v_i'v_j'.$$

If $i=j$ we get $2h(1/h)^2=2/h$, and if $|i-j|=1$ we get $h(1/h)(-1/h)=-1/h$, and otherwise we get $0$. So $A$ is a tradiagonal matrix with $2/h$ on the diagonal entries, and $-1/h$ on the off diagonal entries. Interestingly this is exactly the same as the linear system for the central finite difference method.

  • $\begingroup$ Thank you very much for your response! So you end up with $\Psi$ as the coefficients we would solve for in finite element methods. I thought it would be more like the stiffness matrix, A. I think at the end we need to add up what we get from each space in a block matrix to get A. But maybe your work showing $\Psi$ can be connected to that a bit more? Can you please help me connect these ideas a bit better? $\endgroup$
    – k12345
    Commented May 29 at 15:43
  • $\begingroup$ I've added a bit more info to the answer $\endgroup$
    – Ellya
    Commented May 30 at 9:31

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