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Acceleration-level contact complementarity decides whether a smooth, closed contact with zero normal velocity must carry a compressive reaction or begin to separate. At $d=0$ and $\dot{d}=0$, the normal conditions areThey do not constrain the sign of acceleration at a positive gap, and they do not resolve an impact with negative incoming normal velocity. The distinction between smooth constrained dynamics and time-stepped contact/impact models is explicit in the primary dynamics treatment.^[ref]

For a frictionless multibody model with generalized velocity $v$, invertible mass matrix $M$ and contact Jacobian $J$ defined by $\dot{d}=Jv$, writewhere $b=J{M}^{-1}f+\dot{J}v$ is the free normal acceleration and $W=J{M}^{-1}{J}^{\mathsf{T}}$. The normal reactions then solve the linear complementarity problemThe matrix couples contacts: choosing each reaction from a separate scalar formula generally fails when it changes acceleration at the other contacts. This formulation assumes the displayed velocity-coordinate kinematics and active closed contacts; additional friction forces modify the dynamics.

For an original coupled example, take free normal acceleration $b=(-3,-3{)}^{\mathsf{T}}\mathrm{m}{\mathrm{s}}^{-2}$ andThe reaction $N=(1,1{)}^{\mathsf{T}}\mathrm{N}$ gives zero normal acceleration at both contacts. Ignoring the off-diagonal coupling would instead choose $1.5\mathrm{N}$ at each; inserted into the full system, that produces $1.5\mathrm{m}{\mathrm{s}}^{-2}$ of outward acceleration at both loaded contacts and violates complementarity. Positive reaction and positive separating acceleration cannot coexist in this smooth active-contact model.

Because these are acceleration-level conditions, a numerical trajectory can still accumulate position or velocity drift. They must be paired with an integration/contact-update procedure that maintains the assumed state. During separated free flight, a negative acceleration can be perfectly admissible until the next contact event. At impact, an impulse and an appropriate velocity-level law are needed instead of applying the smooth equations across a discontinuity.

A linear complementarity problem asks for real vectors $z,w\in {\mathbb{R}}^n$ satisfyingfor a prescribed real matrix $M$ and vector $q$. The inequalities hold componentwise. Since every product $z_i{w}_i$ is nonnegative, the final equality means $z_i{w}_i=0$ for every index: each paired component has at least one zero. The linear complementarity problem is a feasibility problem; it need not come with an objective function. Its affine equation does not make the product condition linear.^[ref]

For an original two-variable example, takeThe pair $z=(1/2,0{)}^T$ and $w=(0,1/2{)}^T$ satisfies all three conditions. Guessing which member of each pair vanishes reduces the remaining equations to a candidate linear system, but each candidate must still satisfy nonnegativity. Choosing $z=0$ here gives $w=q$ and fails that check. Some data admit no solution: in one dimension, $M=-1,q=-1$ gives $w=-z-1<0$ for every $z\ge 0$.

The connection with quadratic programming explains one important tractable case. For a symmetric matrix $M$, considerThe Karush-Kuhn-Tucker conditions set its gradient equal to a nonnegative constraint multiplier $w$ and impose complementary slackness, giving precisely the displayed linear complementarity problem. If $M$ is a positive-definite matrix, the objective function is strictly convex and grows without bound as $\Vert z\Vert \to \infty$, so a unique minimizer exists. This conclusion uses that additional matrix property, not merely the word linear.

There is also a short uniqueness proof. For two complementary pairs,Positive definiteness then forces $z=z^{\prime }$. This argument gives a directly checkable reason for uniqueness in the stipulated case.

In unilateral contact complementarity, a nonnegative separation pairs with a nonnegative normal contact force: an open gap has zero contact reaction, while a positive reaction requires closed contact. A time-discretized rigid body model can relate its next-step gap affinely to the unknown reaction, producing a linear complementarity problem. The particular discretization and impact law determine that affine relation; the algebraic form alone does not select the physical contact model.^[ref]