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Computing a discrete surface integral of a scalar function

  1. Oct 24, 2013 #1
    Consider a triangulated discrete manifold (a polyhedron) with known vertices (i.e. each vertex is given in terms of its $$(x,y,z)$$ coordinates ).

    Assign scalar values (some kind of potentials) to each vertex (i.e. at each vertex, a $$k_t(\mathbf{v})$$ is known through its value, no analytical expression is given!).

    For any triangular face of this discrete manifold, one is required to compute the surface integral of the $$k_t$$ scalar discrete field:
    $$ \int_{\Delta}{k_t dS} $$

    I have a small sketch for a solution, but I do not feel it is mathematically sound. Could anyone follow it briefly and see what goes wrong (if anything?)?

    \mathbf{v}(s,t) = (1-t)\left[ (1-s)\mathbf{v}_k + s\mathbf{v}_j \right] + t \mathbf{v}_i,
    with $ (s,t) \in [0,1]^2 $ be a parametrization for $ \Delta $ . If $f \equiv k_t $ is the energy function defined over a surface patch, then
    f(\mathbf{v}(s,t)) = (1-t)[(1-s) f(\mathbf{v}_k) + s f(\mathbf{v}_j)] + t f(\mathbf{v}_i).
    Hence, the surface integral can be transformed into the following double integral:
    \int_{\Delta_{\mathbf{v}_i\mathbf{v}_j\mathbf{v}_k}} {f(x,y,z) dS} = \int_0^1 \int_0^1 {g(s,t) \left\lVert \frac{\partial \mathbf{v}}{\partial s} \times \frac{\partial \mathbf{v}}{\partial t} \right\rVert ds dt },
    where $$g(s,t) = f(\mathbf{v}(s,t)) $$. This last expression is directly computable and leads to the following discrete bending energy
    formulation over the 1-ring of a $$\mathbf{v}_i$$ vertex:
    E_{bending}(\mathbf{v}_i) = \sum_{\Delta_{\mathbf{v}_i\mathbf{v}_j\mathbf{v}_k} \in \mathcal{N}^1(\mathbf{v}_i)}
    \frac{3f(\mathbf{v}_j) + 3f(\mathbf{v}_k) + 4f(\mathbf{v}_i)}{15} \cdot \\
    \cdot \left\lVert \mathbf{v}_j\times\mathbf{v}_i - \mathbf{v}_j\times \mathbf{v}_k - \mathbf{v}_k\times\mathbf{v}_i \right\rVert .
  2. jcsd
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