Transversal stability of constraints

Often, constraint solvers get unstable, especially if they are working at velocity level. Actually the root cause is that they are explicit solvers, so both SHAKE and velocity time stepping (VTS) can suffer from similar problems. Interestingly enough, SHAKE right away explodes if it's stressed enough, while VTS manifests what is known as "jittering" (in the same conditions). And so far, nobody really knew what jittering really is and how to measure it. In what follows I will attempt to formulate jittering mathematically as oscillations transversal to the constraint directions. Actually, this fact was gradually hinted by a series of papers: [Servin et al. 2011], [Kaufman et al. 2014], [Tournier et al. 2015], [Andrews et al. 2017]. But what I'm trying to do is show that this is formally true, at least for bilateral constraints.

Another important thing to note is that instabilities don't come from the constraints themselves, but from how constraint forces are distributed to the other constraints and how these constraints are geometrically related to each other. This is is the essence of the geometric stiffness concept: that constraint forces and constraint orientations can act as stiff forces (e.g. angular springs) in the transversal directions. Take as example a thread made of particles and massless rigid constraints between them: the constraint forces that act along constraints to restrict half of the degrees of freedom (DOFs), also act on the rest of the DOFs (i.e. the angles of the multi-pendulum) and couples them together (like in the chaotic behavior of the double pendulum).

Consider we are given a dynamical system with $n$ degrees of freedom and $m$ constraints. I will consider the case $m<n$ here, and leave the over-constrained case for future treatment. The constraint are given implicitly by $m$ constraint functions: $\psi(q)=0$. In fact, these functions measure the displacement along the constraint direction, so they can be considered as a coordinate.

The trick I will use is to consider a coordinate change from $q$ to another set of coordinates $(\psi, \theta)$. As you can see, we already know $m$ coordinates given by $\psi$, and we need to pull out the rest of $n-m$ coordinates $\theta$ from somewhere. This is quite simple to do for a thread (multi-pendulum) where we have closed formulas for all the coordinate changes, but not so straightforward for the general case. Turns out we don't need to know these nonlinear transformation formulas, but rather their Jacobians. This approach is inspired from an older way of solving constraints, known as the local coordinates approach. The main reference used here will be [Potra and Rheinboldt 1991] (On the Numerical Solution of Euler-Lagrange Equations).

The Jacobian of $\psi(q)$ is denoted by $J=\nabla_q \psi(q)$ and is usually easy to compute and necessary for constraint solving. This is not true though for $L$, the Jacobian of $\theta(q)$. But [Potra and Rheinboldt 1991] show that $L$ exists locally and can be computed numerically (through QR factorization of J). Moreover, it can be chosen so that $L^TL=1$ (i.e. orthonormal). Once we have $L$ (even if locally), we can start looking at the equations of motion and how they would look like in $\theta$-space (i.e. transversal direction). So let's see the equations of motion in maximal coordinates $q$ with only constraints and no other forces:
\begin{equation}
M\ddot{q} = J^T \lambda.
\end{equation}
Move the mass on the right-hand side and multiply to the left by $L$:
\begin{equation}
L\ddot{q} = L M^{-1} J^T \lambda.
\end{equation}
We do this because we know the fact that $L=\nabla_q \theta(q)$ and so $\dot{\theta}=L\dot{q}$. Differentiate another time and we get $\ddot{\theta}=L\ddot{q} + \dot{L}\dot{q}$. Therefore we get:
\begin{equation}
\ddot{\theta} = L M^{-1} J^T \lambda + \dot{L}\dot{q}.
\end{equation}
I will call the last term a gyroscopic term, as it only appears because of the nonlinear change of coordinates and I will ignore it from now on, because I know it's not the culprit for instability. But I do know that the term  $L M^{-1} J^T \lambda$ is the one that's stiff. Most of this insight came (to me and others) from the first two papers mentioned in the first paragraph. So, in order to quantify the stiffness, we need to linearize this term (keep in mind we are already in a restricted vicinity of $q$ because of $L$). So let's differentiate the matrix $A = L M^{-1} J^T$ with respect to $\theta$:
\begin{equation}
\frac{\partial A}{\partial \theta}=\frac{\partial L}{\partial \theta} M^{-1} J^T + L M^{-1} \frac{\partial J^T}{\partial \theta}.
\end{equation}
But here's the catch: the transversal directions $L$ should not depend on the transversal coordinates $\theta$ but only on the other coordinates $\psi$. A similar choice of constraint directions $J$ can be made so that they only depend on $\theta$. What I cannot prove right now is that such a choice for both $J$ and $L$ always exists. However, I will assume it does, and continue based on that - namely, that $\frac{\partial L}{\partial \theta}=0$. Another thing we want to do, is express all terms as much as possible in terms of $\theta$, so we note that:
\begin{equation}
\frac{\partial J^T}{\partial \theta}= L^T \frac{\partial^2 \psi}{\partial \theta^2}.
\end{equation}
We can now finish our linearization, and after dumping the gyroscopic term, we get the following approximate local equation:
\begin{equation}
\ddot{\theta} = C + LM^{-1}L^T \left( \frac{\partial^2 \psi}{\partial \theta^2}\lambda \right) \theta = C + EK\theta,
\end{equation}
where $C$ is a constant term, $E$ is the inverse effective mass in transversal space and $K$ is the stiffness matrix as seen by the free degrees of freedom $\theta$. What we would like to show now is that there is a relation between this matrix $K=\frac{\partial^2 \psi}{\partial \theta^2}\lambda$ and the geometric stiffness $\tilde K = \frac{\partial^2 \psi}{\partial q^2}\lambda$. Given that $\frac{\partial^2 \psi}{\partial \psi^2} = \frac{\partial^2 \psi}{\partial \theta \partial \psi} = 0$, I was able to show that $\tilde K = L^T K L$. I may have to go again through the calculations, but it makes sense that the matrix $K$ is "pushed forward" in $q$-space and vice-versa.

Now, let's try and bring the last equation from transversal space back to $q$-space. Note that we linearized and approximated the original equations, so what we are bringing back now is basically just a portion of all the forces, but most importantly they are the stiff forces. This why we also leave out the constant term. Therefore, we multiply the equation to the left by $L^T$:
\begin{equation}
L^T\ddot{\theta} = L^T LM^{-1}L^T K \theta.
\end{equation}
We can now use the fact that $L^TL=1$:
\begin{equation}
\ddot{q} = M^{-1}L^T K \theta(q).
\end{equation}
We need again to linearize this equation (around a point $q^*$), so that stiffness shows its face:
\begin{equation}
\ddot{q} = M^{-1}L^T K L (q - q^*) + M^{-1}f^*.
\end{equation}
It is clear now to see that:
\begin{equation}
M\ddot{q} = \tilde K (q - q^*) + f^*.
\end{equation}
Again, we should ignore the constant terms and notice the fact that the stiffness matrix of the transverse forces in $q$-space is just $\tilde K$ - the geometric stiffness. Of course, this is just a linearization of the real force, but it is what matters locally, especially when doing numerical integration. Linear stability analysis is all we have, so the values of the $\tilde K$ matrix are very important for stability analysis in $q$-space (as the values of $K$ are in $\theta$-space). If the highest eigenvalue $\omega^2$ of $M^{-1}\tilde K$ does not satisfy $\omega < 2/h$, then the simulation becomes unstable. This is an idea that has already been proposed in [Andrews et al. 2017], but I was just not sure if it was true. Now that I've shown it is, it's time to focus on the modal analysis of $M^{-1}\tilde K$ as the proper way of studying stability in constrained based dynamics.

Implicit velocity projection

During the time I spent looking closer at PBD and what it really is I learned one clear thing: PBD is nothing else than a form of implicit integration (Backward Euler) of the constraint forces. Goldenthal calls it Implicit Constraint Directions (ICD). His optimized version he called Fast Projection (FP). FP solves a linear system at every nonlinear iteration. PBD runs just one Gauss-Seidel (or Jacobi) step at every iteration, resulting in what is known as Nonlinear Gauss-Seidel (NGS). 

The other thing I learned is that PBD is very stable just because it is fully implicit. And this is why you can do cloth with PBD, but not with velocity based methods, which I'll denote hereon as velocity time stepping (VTS). This is because articulated systems can get very unstable with VTS. The short explanation is that VTS is using the constraint direction at the beginning of the frame (i.e. explicit directions) and not the one at the end (implicit directions). Therefore a part of the system is integrated explicitly (the part transversal to constraints) and it is well known that explicit integrators can get easily unstable and even blow up. The good part is that VTS is very similar to SHAKE (pretty much a linearized version of it) which is symplectic. This means energy gets conserved better in the transverse direction - I will detail these things in another post.

Unfortunately conserving energy is not always what we want. Stability comes a lot of time for free from implicit integrators because they dissipate energy artificially. This may be bad for science projects, but not for games, graphics and real-time simulation. Therefore I've been looking for a way to make VTS stable without going really the PBD way. The differences between the two comes a lot from what we are trying to solve for: VTS solves for velocity constraints, PBD for position constraints. So how can we keep the VTS way without altering it too much, and also get the stability of PBD? The answer: integrate using implicit constraint directions but keep the velocity constraints.

Let's throw in some equations for this case:

\begin{align}
Mv^{l+1} &= Mv^l +hf_{ext} + hJ(x^{l+1})^T \lambda \\
x^{l+1} &= x^l + hv^{l+1} \\
0 &= \tfrac{\beta}{h} c(x^l) + J(x^l)v^{l+1}
\end{align}

We can express this as a system in the unknowns $v$ and $\lambda$:
\begin{align}
Mv &= Mv^l +hf_{ext} + hJ(x^l + hv)^T \lambda \\
0 &= \tfrac{\beta}{h} c(x^l) + J(x^l)v = G(v)
\end{align}

While the second equation is linear, the first one is not and we need to apply Newton's method to it. The Jacobian for the whole system then becomes:

\begin{equation}
\begin{bmatrix}
M-h^2 H(x)\lambda & -h J(x)^T \\
J(x^l) & 0
\end{bmatrix},
\end{equation}
where $x = x^l + hv$.We can proceed just like in the case of FP and ignore the geometric stiffness term $K=H\lambda$, where $H=\nabla J = \nabla^2 c$, and also set the residue of the first of the first equation to be always 0. We then get the following linear system:
\begin{equation}
\begin{bmatrix}
M & -h J(x_0)^T \\
J(x^l) & 0
\end{bmatrix}
\begin{pmatrix}
\Delta v_1 \\ \Delta \lambda_1
\end{pmatrix} =
\begin{pmatrix}
0 \\ -G(v_0)
\end{pmatrix}
\end{equation}

We can now take directly a Schur complement and obtain the usual constraint solver equation:

\begin{equation}
hJM^{-1}J^T\Delta\lambda_1 + G(v_0) = 0,
\end{equation}

where we can take by approximation $J$ to be the same on both side, i.e. $J(x_0)$. Then the change  in velocity is $\Delta v_1 = hM^{-1}J^T\Delta\lambda_1$ and the update step is given by:
\begin{align}
v_1 &= v_0 + \Delta v_1, \\
x_1 &= x^l + h v_1.
\end{align}
If you iterate this scheme you get implicit VTS and it's way more stable than VTS. I only tested this on a 2D thread/cable made of distance constraints and I haven't got to make measurements, but it looks just like PBD. But the algorithm looks a lot like VTS - there are two main differences though:

1. at every iteration, after you compute the new velocities, you compute new positions too, and
2. at every iteration you recompute the Jacobians of the constraints based on the new positions.

Otherwise, everything stays the same. You could use an exact solver for the Lagrange multiplier or better just use one Gauss-Seidel step, which results basically in NGS. The only trick is that you don't set $\beta$ too close to 1 or it explodes (why, I don't know yet).

Here's how it looks like for some extreme initial velocity conditions:

The same scenario, using VTS.

And for completeness, the PBD case.