# Arnold's holonomic constraints being limits of potential energy

The following quote comes from Arnold's "Mathematical methods in mechanics" book:

"We consider potential energy $$U_N = Nq_2^2 + U_0(q_1, q_2)$$, depending on parameter $$N$$ (which we will tend to infinity). We consider the initial conditions on $$\gamma$$ [supposed to be path in $$q_1, q_2$$ coordinates]: $$q_1(0) = q_1^0, \dot{q}_1(0) = \dot{q}_1^0, q_2(0) = 0, \dot{q}_2 (0)= 0$$. Denote by $$q_1 = \phi(t,N)$$ the evolution of $$q_1$$ under a motion with these initial conditions in the field $$U_N$$."

Then he mentions a theorem but without a proof. I would be happy if someone could provide argument for believing the theorem.

"The following limit exists as $$N \to \infty$$: $$\lim \limits_{N \to \infty}\phi(t,N) = \psi(t)$$. The limit $$q_1 = \psi(t)$$ satisfies Lagrange equation [...] where [new Lagrangian] $$L_*(q_1, \dot{q}_1) = T|_{q_2=0, \dot{q}_2=0} - U_0|_{q_2=0}$$".

In the previous quote, $$T$$ is kinetic energy term.

• The 2nd particle $$q_2$$ is effectively attached to a spring with coupling constant $$2N$$. From mechanical energy conservation, $$|q_2|\leq\sqrt{E/N}$$. In the stiff spring limit $$N\to\infty$$, the 2nd particle $$q_2$$ becomes confined to stay in the origin, thereby enforcing the holonomic constraint $$q_2= 0$$.
• Meanwhile the 1st particle $$q_1$$ will go about doing its business $$m_1\ddot{q}_1~=~-\frac{\partial U_0(q_1,q_2)}{\partial q_1},\tag{1}$$ and interact with the confined 2nd particle at $$q_2=0$$. (Formally, we need continuity of the solution $$q_1$$ to the ODE (1) wrt. the parameter $$q_2$$. This is guaranteed by imposing certain regularity conditions on $$U_0(q_1,q_2).$$)
• thanks as always! I understand now that $q_2$ can indeed be bounded by taking $N$ sufficiently large as necessary. But I am still not sure about $\dot{q}_2$ being bounded. For example, intuitively (which is not an argument but still), when I have just harmonic oscillator it has property that no matter what kind of spring coupling $N$ I have then even though amplitude might become smaller, but if total energy is fixed, then the velocity at the origin in general is not arbitrarily small (but depends on the energy and is independent of $N$). – Daniels Krimans Sep 27 '19 at 18:36
• Note that the ODE (1) only couples to $q_2$ not $\dot{q}_2$. – Qmechanic Sep 27 '19 at 20:11
• You are absolutely right. Thank you for the help. Just in case you have some free time - do I understand correctly that the same algorithm for velocity dependent potentials cannot be used as we cannot secure (in general) that $\dot{q_2}$=0? – Daniels Krimans Sep 27 '19 at 20:14