Smoothly Filling Holes in 3D meshes using Variational Calculus and Surface Fairing

$

ewcommand{\mvec}[1]{\mathbf{#1}}

ewcommand{\gvec}[1]{\boldsymbol{#1}}\definecolor{eqcol2}{RGB}{114,0,172}\definecolor{eqcol1}{RGB}{172,0,114}\definecolor{eqcol3}{RGB}{0,172,114}\definecolor{eqcol4}{RGB}{230,190,120}$In this article, we describe an approach to smoothly filling holes in broken meshes that is based on variational calculus. However, do note that it is not assumed that the reader has experience in variational calculus, and we will instead introduce the necessary concepts from this topic when they are needed. Most of the techniques described in this article are based the description of Surface Fairing in section 4.3 of [1]. In the below image, it is shown how our described algorithm can be used to smoothly and naturally fill holes in a broken mesh.

Our Approach

First, we shall describe our general approach to solving the problem. Let us say we have some geometry with a hole, and from the side, it looks like this

As can be observed, we describe the height at every point with a function $f$, so that $f(1)$ is the height of the first vertex, and so on. At $x=1$, $x=2$, $x=7$ and $x=8$ there are vertices, but in between, there is a hole at $x=3$, $x=4$, $x=5$, $x=6$. We wish to add a patch of four vertices at these four points that fills the hole, and we also want to ensure the height of these vertices follows the general curvature around the hole. We will find these four heights by formulating a system of equations, and then solving it, so that the obtained solution is the values of $f(3)$, $f(4)$, $f(5)$ and $f(6)$.

What kind of solution would be a good and desirable solution? We claim a solution like the below is a good solution

While the below solution is a bad solution

In the bad solution, the patch that fills the hole, is simply a linear interpolation from $x=2$ to $x=7$. The issue with such a solution is that it poorly preserves the slopes(that is, the derivatives) of the surroundings of the patch. The derivative at some point $x$ can be approximated as \[ f_x(x) = \frac{f(x + h) - f(x)}{h} = f(x + 1) - f(x) \] where we have let the step size be $h = 1$, and where we are using the notation $f_x(x) = \frac{\partial f}{\partial x}$ to denote the first derivative with respect to $x$, and then the second derivative is denoted $f_{xx}(x)$. Observe that $f_x(1) = 5 - 2 = 3$. In the good solution, we have that $f_x(2) = 2.1$, but in the bad solution we have $f_x(2) = 0.2$. The derivative at $f_x(2)$ is much closer to the derivative at $f_x(1)$ in the good solution compared to the bad. The change in derivative is smaller for the good solution, and this results in the transition from $f(2)$ to $f(3)$ being a much more natural transition, since this transition is more consistent with the derivatives of the surrounding vertices. To summarize, we desire a small change in the value of the derivative, as we go from one vertex to its neighbour. The change of derivative is simply the second derivative, and we wish to minimize this quantity over the patch.

This point is very important and the key of the technique, so let us clarify it even more: the first derivative, $f_x(x)$, measures how the function $f(x)$ changes as we go from $x$ to $x+1$. Then the definition of the second derivative, $f_{xx}(x)$ , is obvious: it measures the change of $f_x(x)$ as we go from $x$ to $x+1$. In order to create a smooth patch that naturally follows the curvature around the hole, we will create a patch that minimizes the second derivative.

Minimizing the Second-Derivative with Variational Calculus

In this section, we will implement the approach described in the previous section. Our objective is to find some function $f(x)$, that describes a patch that has the requirements we described in the previous section. For the case of our example, we already know the values of this function for $x=1$, $x=2$, $x=7$ and $x=8$, but we need to solve and find the values for $x=3$, $x=4$, $x=5$ and $x=6$. The example function we have been dealing with up until this point has been a discrete function $f(x)$, that is only defined at the integers. We will now temporarily move from the discrete domain into the continuous domain, since this allows us to use the powerful tools of variational calculus. Henceforth, $f$ is not only defined for values such as $x=2$ and $x=6$, but also for values such as $x=1.2$ and $x=1.5$.

We wish to find a function $f(x)$, such that the second derivative is minimal at all points in the patch. Then it is natural to define the energy function \[ E(f) = \int_a^b (f_{xx})^2\ dx \] (This integral would not have been possible in the discrete domain!)The above integral can be seen as a sum. Put in a simplified way, it evaluates the value of $(f_{xx})^2$ at many, many points in the interval $[a, b]$, and then adds all these evaluations together into one large sum. If this integral is small, then the second derivatives in the interval $[a, b]$ will be small, and this means $f$ is a good patch by our requirements. Note that $a$ and $b$ are the boundary points of the patch, and the boundary is $x=2$ and $x=7$ for the example function.

In order to find the $f$ that minimizes $E(f)$, we utilize a common approach of variational calculus: First, we assume that $f$ is the minimizer of $E(f)$. Further, let $u(x)$ be any function that is defined in the interval $[a,b]$, and is once differentiable in this interval(this means that we can compute its first derivative in this interval), and that further satisfies $u(a) = u(b) = 0$ and $u_x(a) = u_x(b) = 0$. It will soon become clear why we are doing this.

Consider now the function $E(f(x) + u(x)\lambda)$. This is a function where we are adding a certain amount of $u(x)$ to $f(x)$ and then giving this sum to $E()$. How much of $u(x)$ that we add to $f(x)$ is controlled by the scalar $\lambda$. We assumed that $f(x)$ is the minimizer of $E(f)$, and this implies that $\lambda=0$ is the minimizer of $E(f(x) + u(x)\lambda)$. But this means that the derivative of $E(f(x) + u(x)\lambda)$ with respect to $\lambda$ must be $0$ when $\lambda = 0$(since the derivative is zero at the minimum point). That is \[ \frac{\partial E(f(x) + u(x)\lambda)}{\partial\lambda} \bigg\rvert_{\lambda = 0}= 0 \] We shall now expand and simplify the left-hand side. For convenience, we will onward write $u(x)$ as $u$ and $f(x)$ as $f$. First observe that \[ E(f + u\lambda) = \int_b^a ( (f + u\lambda)_{xx} )^2\ dx = \int_b^a (f_{xx} + u_{xx}\lambda)^2\ dx \] Next we compute the derivative with respect to $\lambda$. \[ \frac{\partial E(f + u\lambda)}{\partial\lambda} = \frac{\partial}{\partial\lambda}\left(\int_b^a (f_{xx} + u_{xx}\lambda)^2\ dx\right) = \int_a^b 2(f_{xx} + u_{xx}\lambda)u_{xx}\ dx \] (we have here performed differentiation under the integral, and this is completely acceptable by the Leibniz's rule). When $\lambda=0$, the right-hand side must equal zero, and thus it follows \[ \int_a^b f_{xx}u_{xx}\ dx = 0 \] Recall that $u(a) = u(b) = 0$ and $u_x(a) = u_x(b) = 0$. By using these facts and integration by parts, it is possible to simplify the above even further. Integration by parts yields \[ \int_a^b f_{xx}u_{xx} dx = [f_{xx}u_x]^b_a - \int_a^b f_{xxx} u_x\ dx = 0 \] However, $u_x(a) = u_x(b) = 0$, and thus it simplifies to \[ \int_a^b f_{xxx} u_x\ dx = 0 \] Integrate by parts again \[ \int_a^b f_{xxx} u_x dx = [f_{xxx}u]^b_a - \int_a^b f_{xxxx}u\ dx = 0 \] But $u(a) = u(b) = 0$, so we can simplify as \[ \int_a^b f_{xxxx}u\ dx = 0 \] Finally, here comes the coup de grace. Recall our definition of $u(x)$: it is any function that satisfies $u(a) = u(b) = 0$ and $u_x(a) = u_x(b) = 0$, and is once differentiable. How can we ensure that the above equality is satisfied, while still letting $u(x)$ be any function? The only possible way to satisfy such a strong condition, is by requiring that \[ f_{xxxx} = 0 \] That is, $f_{xxxx}$ must always be zero within our domain $[a,b]$. This implies that the minimizer of $E(f)$ is a function whose fourth derivative is always zero within the domain. We will use this fact to solve for the minimizer $f$ in the following section.

However, note that instead of solving for the $f$ whose fourth derivative with respect to $x$ is zero, we will find the $f$ whose bi-laplacian is zero, which is of course equivalent to the former. We are in this example dealing with only one dimension, and the laplacian in one dimension is simply $\Delta f = f_{xx}$. This means that our condition $f_{xxxx} = 0$ is equivalent to \[ \Delta\Delta f = \Delta^2 f = 0 \] where $\Delta^2$ is often called the bi-laplacian or the bi-harmonic operator, which is just the laplacian of the laplacian. The laplacian is more commonly used in notation than the fourth derivative, so this is why we are formulating our condition in terms of the laplacian.

Discretizing and Solving for the Minimizer

In the beginning, we showed a function with bar charts with a hole in it, and now we will show how to find the minimizer that fills this hole, and at the same time make sure that the second derivatives are minimal. The minimizer $f$ must satisfy the condition \[ \Delta\Delta f = \Delta^2 f = 0 \] However, this condition is in the continuous domain, and our bar chart is in the discrete domain, and so is a 3D mesh. In practical applications, discrete domains are very common, and so the bi-laplacian must be discretized. We will demonstrate how this can be done in the one-dimensional case. The bi-laplacian is the fourth derivative in one dimension. The first step is to discretize the first derivative, using central differences. \[ f_x(x) = \frac{f(x + 0.5h) - f(x - 0.5h)}{h} = f(x+0.5) - f(x-0.5) \] Where we used a step length of $h=1$, the smallest possible step length in our discrete problem. The second derivative, is the derivative of the first derivative \[ \Delta f(x) = f_{xx}(x) = f_x(x+0.5) - f_x(x-0.5) = f(x-1) - 2f(x) + f(x+1) \] Finally, the bi-laplacian is just the laplacian of the laplacian \[ \Delta^2 f(x) = f_{xxxx}(x) = f_{xx}(x-1) - 2f_{xx}(x) + f_{xx}(x+1) = f(x-2) - 4f(x-1) + 6f(x) - 4f(x+1) + f(x+2) \] Now we can finally solve for $f$ in our bar chart problem. First, observe that we desire to have $\Delta^2 f(3) = 0$. Using our discretization scheme, this can be simplified to \[ f(1) - 4f(2) + 6f(3) - 4f(4) + f(5) = 0 \] And same type of linear equation also holds for $f(4)$, $f(5)$, and $f(6)$. We summarize all these linear equations into a single matrix equation \[ \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & -4 & 6 & -4 & 1 & 0 & 0 & 0 \\ 0 & 1 & -4 & 6 & -4 & 1 & 0 & 0 \\ 0 & 0 & 1 & -4 & 6 & -4 & 1 & 0 \\ 0 & 0 & 0 & 1 & -4 & 6 & -4 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} f(1) \\ f(2) \\ f(3) \\ f(4) \\ f(5) \\ f(6) \\ f(7) \\ f(8) \\ \end{bmatrix} = \begin{bmatrix} 2 \\ 5 \\ 0 \\ 0 \\ 0 \\ 0 \\ 6 \\ 3 \\ \end{bmatrix} \] Note that rows 1, 2, 7 and 8 simply specify the values at the boundaries of the bar chart. These are the boundary conditions of our problem, and must be specified in order to create a solvable equation. The remaining rows specify that the bi-laplacians of the unknown variables $f(3)$, $f(4)$, $f(5)$, $f(6)$ must equal zero. However, this matrix equation can be simplified. For row 3, we have \[ f(1) - 4f(2) + 6f(3) - 4f(4) + f(5) = 0 \] But since the values of $f(1)=2.0$ and $f(2)=5.0$ are known, we can move some constant values to the right-hand side, keeping only the variables on the left side: \[ 6f(3) - 4f(4) + f(5) = 18 \] We do the same for the other three linear equations, and remove rows 1,2, 7 and 8, since these rows are now redundant. Our simplified matrix equation is \[ \begin{bmatrix} 6 & -4 & 1 & 0 \\ -4 & 6 & -4 & 1 \\ 1 & -4 & 6 & -4 \\ 0 & 1 & -4 & 6 \\ \end{bmatrix} \begin{bmatrix} f(3) \\ f(4) \\ f(5) \\ f(6) \\ \end{bmatrix} = \begin{bmatrix} 18 \\ -5 \\ -6 \\ 21 \\ \end{bmatrix} \] If we solve, we obtain our solution below, which is also the good solution that we showed earlier.

Smooth 3D Hole-Filling Surface Patches

We have shown how to smoothly fill holes in the one-dimensional case. We will now briefly discuss how to smoothly fill holes in 3D meshes. Most of the reasoning from the one-dimensional case carries over to the 3D case, so this discussion will be more brief.

First, we locate the hole in the mesh. The below image illustrates this hole

and then compute the center position of the vertices in the hole boundary(this is the average of the vertex positions). The vertices in the boundary are connected with a newly created vertex at the center position, thus creating a surface patch that fills the hole. The triangles in this patch are also upsampled(subdivided), so that we have more vertices to manipulate for when we will be making the patch more smooth. The result can be seen below

Note that we also upsampled the triangles of the original bunny mesh, since this makes it very easy to fuse together the vertices of the original patch and the original geometry. By "fuse", we mean that we ensure that the boundary of the original mesh, and the boundary of the patch share the same vertices.

Next, we create a smooth patch by solving the equation $\Delta^2 f = 0$ on the patch. We need to discretize the laplace operator to achieve this, and the discretization of the laplace operator on the surface of a mesh is called the Laplace-Beltrami operator. This is a topic that is too large to be covered in this article, and will probably be the subject of a future article. Instead, we mention that libigl provides code that creates the discretized Laplace-Beltrami operator, and then solves the equation $\Delta^2 f = 0$. This is what we used in our demo implementation of the technique. We solve for $f$, and the results can be seen below

Finally, we downsample the mesh, using a mesh decimation algorithm.

This concludes the article. In a github repository, we provide a commented implementation of the technique implemented with libigl and Eigen.

References

[1] Mario Botsch, Leif Kobbelt, Mark Pauly, Pierre Alliez, Bruno Lévy., "Polygon Mesh Processing".