# Lagrangian mechanics and initial conditions vs boundary conditions

It bothers me that many basic books on the classical mechanics don't discuss the following difference between "Newton's laws" and the "Principle of stationary action". Newton's laws can predict the behavior of a system if one sets initial positions and velocities. On the contrary, if we want to construct an action functional we have to set an initial and a final position. From the mathematical point of view, this is the difference between initial and boundary value problem, which have different qualitative behavior (BVPs may have many solutions or no solution).

My question: Which approach is more reasonable - the initial value problem or the boundary value problem?

• More reasonable for what? They are different types of problem and each has its uses. Jan 26 '15 at 12:50
• Possible duplicates: physics.stackexchange.com/q/38348/2451 and links therein. Jan 26 '15 at 12:53
• Thanks, I don't understand why I missed the other answer! Jan 26 '15 at 13:02

The Lagrangian boundary value problem is ill defined in general (it is partially well defined when the initial and final configurations are sufficiently close to each other also in time, see my ADDENDUM2 below). There is no existence and uniqueness property of the solutions. Conversely, the local and global problem with initial data, if the kinetic energy of the Lagrangian is positive-defined is always well-posed: The solution exists and is unique, if the Lagrangian is sufficiently regular.

Within this picture the variational principle has to be understood as a procedure to define just the equations of motion rather than the solutions.

ADDENDUM The existence and uniqueness theorem for the initial data problem -- the celebrated Cauchy problem-- of first order systems of differential equations can be stated as follows (the proof is essentially based on Banach's fixed point theorem).

Theorem. Let $$D\subset \mathbb{R} \times \mathbb{R}^n$$ be an open set and consider the 1st order ODE in normal form $$\frac{d\vec{x}}{dt}= \vec{F}(t, \vec{x}(t))\tag{0}$$ for the unknown given by the differentiable function $$\vec{x}= \vec{x}(t)$$ for $$t \in I$$, where $$I\subset \mathbb{R}$$ is an open interval to be determined and with the initial condition $$\vec{x}(t_0) = \vec{x}_0$$ where $$(t_0, \vec{x}_0) \in D$$ and $$t_0 \in I$$.

If $$\vec{F}$$ is

(i) jointly continuous and

(ii) locally Lipschitz with respect to the variable $$\vec{x}$$, then

(a) there exist a unique solution $$I \ni t \mapsto \vec{x}(t) \in \mathbb{R}^n$$ which is maximal (in the sense that there is no solution with the same initial contidion defined on la larger domain that extends the said solution).

(b) This solution is $$C^1$$.

(c) Every other solution of the same problem is a restriction of the above one.

A pair of crucial comments are in order

(A) If $$\vec{F}=\vec{F}(t,\vec{x})$$ is $$C^1(D)$$, then the regularity conditions (i) and (ii) are largely satisfied. However the existence part is valid with the only requirment that the $$\vec{F}$$ is jointly continuous, higher regularity is necessary for the uniqueness part, but it also simplify the proof of the existence part.

(B) The theorem applies to 2nd order differential systems in normal form $$\frac{d^2\vec{y}}{dt^2}= \vec{G}\left(t, \vec{y}(t), \frac{d\vec{y}}{dt}\right)\tag{1'}$$ with the following argument. Define the new 1st order system simply by rewriting the previous equation as $$\frac{d\vec{v}}{dt}= \vec{G}\left(t, \vec{y}(t), \vec{v}(t)\right)\tag{1}$$ and by adding the other 1st order equation $$\frac{d\vec{x}}{dt}= \vec{v}(t) \tag{2}\:.$$ Collecting (1) and (2) and using the variable $$\vec{x}:= (\vec{y}, \vec{v})$$ we have a system of the form (0) in $$\mathbb{R} \times \mathbb{R}^{2n}$$ and we can use the previous theorem. Obviously the number of initial contitions now are two $$\vec{y}(t_0) = \vec{y}_0\:; \quad \vec{v}(t_0) = \vec{v}_0\:.$$ Assuming $$\vec{G}$$ of class $$C^1$$ jointly in all variables is largely sufficient to guarantee the validity of the existence and uniqueness theorem.

(C) The procedure in (B) extends to a system of differential equations in normal form of every finite order through an obvious re-adaptation.

(D) All the results are valid as they stand when working with differential equations on manifolds, just by dealing with in local charts and finally gluing the solutions.

Let now pass to the Euler-Lagrange equations. For the sake of semplicity I will work in a local chart (the natural ambient would be the first jet bundle over the spacetime of configurations viewed as a fiber bundle over the time axis).

$$\frac{d}{dt}\left( \frac{\partial T(t,q(t),\dot{q}(t))}{\partial \dot{q}^k}\right) -\frac{\partial T(t,q(t),\dot{q}(t))}{\partial q^k} = Q_k(t, q(t), \dot{q}(t))\quad k=1,\ldots, n\tag{3}$$ $$\frac{dq^k}{dt}= \dot{q}^k(t)\quad k=1,\ldots, n \tag{4}$$

This is a first order system of differential equations for a curve $$I \ni t \mapsto (t,q^1(t), \ldots, q^n(t),\dot{q}^1(t), \ldots, \dot{q}^n(t))$$ for given initial conditions $$(t_0,q^1(t_0), \ldots, q^n(t_0),\dot{q}^1(t_0), \ldots, \dot{q}^n(t_0) =(t_0,q^1_0, \ldots, q^n_0,\dot{q}^1_0, \ldots, \dot{q}^n_0)\:.$$

The main problem is that the system (3)+(4) is not in normal form (only the higer order derivatives should take place on the left hand side and (3) violates this condition).

However the kinetic energy, for mechanical systems, has this form $$T(t,q,\dot{q}) = \sum_{h,k=1}^n a_{hk}(t,q) \dot{q}^h\dot{q}^k + \sum_{k=1}^n b_{k}(t,q)\dot{q}^k + c(t,q)\:,\tag{5}$$ where the matrix $$A= [a_{hk}(t,q)]$$ is non-singular and positive.

Taking advantage of this fact, a direct but tedious computation proves that (3)+(4) can be equivalently re-written as $$\frac{d\dot{q}^k}{dt}= \sum_{h=1}^n(A^{-1})_{kh}\left(G^h(t,q(t),\dot{q}(t))+ Q_h(t, q(t), \dot{q}(t))\right)\quad k=1,\ldots, n$$ $$\frac{dq^k}{dt}= \dot{q}^k(t)\quad k=1,\ldots, n$$ We have therefore reached a system of the form (0) and the theorem of existence and uniqueness is valid provided the requirments about regularity of the right-hand side are satisfied.

To this end, it is not difficult to prove that, if $$T(t,q,\dot{q})$$ of the said form is $$C^2$$ jointly in its arguments and $$Q_k(t,q,\dot{q})$$ is $$C^1$$ jointly in its arguments for every $$k$$, then also the regularity conditions sufficient to guarantee the validity of the existence ad uniqueness theorem are met.

If one has a Lagrangian $$L(t,q,\dot{q})= T(t,q,\dot{q}) + V(t,q,\dot{q})$$ encapsulating the $$Q_k(t,q,\dot{q)} = -\frac{d}{dt}\left( \frac{\partial V(t,q,\dot{q})}{\partial \dot{q}^k}\right) +\frac{\partial V(t,q,\dot{q})}{\partial q^k}$$

and where the Kinetic part $$T(t,q,\dot{q})$$ has the form (5) and $$V$$ is at most a 1st order polynomial (e.g., inertial forces or Lorentz forces) in the $$\dot{q}^k$$ with coefficients functions of $$t$$ and $$q^k$$, the situation is essentially identitcal. Existence and uniqueness are guaranteed by assuming that the Lagrangian is $$C^2$$ jointly in its arguments.

After more then 20 years (!), I eventually managed to interpret a comment in Arnold's book and to produce a proof of a statement he seems to state.

That is my version.

STATEMENT. Let us consider a $$C^4$$ Lagrangian in $$TQ$$, $$L(q, \dot{q})= \sum_{i,j=1}^n a_{ij}(q)\dot{q}^i\dot{q}^j - U(q)$$ with $$\det[a_{ij}(q)]\neq 0$$ everywhere.

Given $$q_0\in Q$$, if $$t_1>0$$ is sufficiently small, then there is an open ball $$B_r(q_0)$$ centered on $$q_0$$ of sufficiently small radius $$r>0$$ such that the problem $$\frac{d}{dt}\frac{\partial L(q(t), \dot{q}(t)}{\partial \dot q^k}- \frac{\partial L(q(t) \dot{q}(t))}{\partial q^k}=0\quad q(0)=q_0\:,\quad q(t_1)= q_1$$ admits a solution (not necessarily unique) for every given $$q_1 \in B_r(q_0)$$.

(I have also a statement fot the non-autonomous case where $$L$$ explicitly depends on time, but it is a bit cumbersome to write down here.)

Proof. If $$L$$ is $$C^4$$ then, when writing down the EL equations in normal form, we see that the right-hand side is $$C^2$$ and this, on the one hand, implies that the initial data problem is well posed (globally) and, on the other hand it enatail that the local one-paramter group of diffeomorphism $$\phi_t : TQ \to TQ$$ describing the solutions of the equaztions is of class $$C^2$$.

As is well known the domain of $$\phi$$ is an open subset $$A \subset \mathbb{R} \times TQ$$, $$I \ni t \mapsto \phi_t(q_0,\dot{q}_0)= \left(q(t, q_0, \dot{q}_0), \dot{q}(t, q_0, \dot{q}_0)\right)\in TQ$$ is the unique maximal solution of the initial-value problem with initial conditions $$(q(0), \dot{q}(0)) = (q_0,\dot{q}_0)$$, and $$I\ni 0$$ is the maximal open interval of definition of this solution.

Let us fix $$q_0 \in Q$$ and consider that map obtained as the first component of the map above, $$B \ni (t,\dot{q}_0) \mapsto (t, q(t, q_0, \dot{q}_0))\tag{1}$$ where the open subset $$B\subset A$$ is the obvious projection of $$A$$ at $$q=q_0$$ fixed.

I prove that if $$t=t_1>0$$ is sufficiently small, then the said map has Jacobian matrix non-singular at $$(t_1,q_0, \dot{q}_0=0)$$ (corresponding to $$(t_1,q_0,q_0)$$ in its image) and therefore it defines a local diffeomorphysm around that point. This way, we can invert the map around $$(t_1,q_0,q_0)$$ and writing there $$(t_1, \dot{q}_0) = (t_1, \dot{q}_0(t,q_0,q))\:.$$ In other words, if $$q=q_1$$ stays a neigborhhod of $$q_0$$, i.e. the ball $$B_r(q_0)$$ said above, we can find an inital vector such that the unique solution exiting $$q_0$$ with that initial vector achieves $$q$$ at the time $$t=t_1$$. This is what we wanted. Because it means that there is a solution joining $$q_0$$ at $$t=0$$ with $$q=q_1$$ at $$t=t_1$$.

Let us prove that the said Jacobian matrix is not singular at the said piint.

Since, in (1), the first entry $$t$$ is trivially transformed, it is sufficient to compute the Jacobian matrix of $$q^k$$ with respect to $$\dot{q}_0^h$$ and to prove that it is not singular if $$t>0$$ is sufficiently small when $$\dot{q}_0=0$$.

$$\frac{\partial q^k}{\partial \dot{q}^h}|_{(t,q_0,\dot{q}_0)}=\frac{\partial q^k}{\partial \dot{q}_0^h}|_{(0,q_0,\dot{q}_0)} + \frac{\partial^2 q^k}{\partial t\partial \dot{q}_0^h}|_{(0,q_0,\dot{q}_0)}t + O^k_{_h\:(q_0,\dot{q}_0)}(t^2)\:.$$ Above $$\frac{\partial q^k}{\partial \dot{q}_0^h}|_{(0,q_0,\dot{q}_0)}=0$$ evidently, because at $$t=0$$ every solutions passes through $$q_0$$ independently from its initial tangent vector $$\dot{q}$$.

Furthermore (here we use $$C^4$$ that implies that the group of diffeomorphisms is $$C^2$$), $$\frac{\partial^2 q^k}{\partial t\partial \dot{q}_0^h}|_{(0,q_0,\dot{q}_0)} = \frac{\partial }{\partial \dot{q}_0^h} \frac{\partial q^k}{\partial t}|_{(0,q_0,\dot{q}_0)} = \frac{\partial \dot{q}_0^k}{\partial \dot{q}_0^h} = \delta^k_h\:.$$ Putting all together $$\frac{\partial q^k}{\partial \dot{q}_0^h}|_{(t,q_0,\dot{q}_0)}= \delta^k_h t + O^k_{_h\:(q_0,\dot{q}_0)}(t^2)$$ In particular, for $$\dot{q}=0$$, $$\frac{\partial q^k}{\partial \dot{q}^h}|_{(t,q_0,0)}= \delta^k_h t + O^k_{_h\:(q_0,0)}(t^2)\:.$$ The determinant produces $$\det \left[\frac{\partial q^k}{\partial \dot{q}_0^h}|_{(t,q_0,0)} \right] = n t + O(t^2)\:.$$ It is now obvious that if $$t=t_1>0$$ is sufficiently small, the right-hand side cannot vanish as wanted. QED

Uniqueness generally fails. A trivial counterexample is provided by a Lagrangian for a free particle on $$S^2$$ whose kinitic energy is constructed out of the standard metric on $$S^2$$. Here the solutions of the EL equations are geodesics and time is an affine parameter. If $$q_0$$ and $$q$$ are respectively the south and the north pole, you see that there are infinitely many geodesics joining the two points within the same interval of time.

Also the condition of small $$t_1>0$$ cannot be relaxed in general regarding the existence issue as already stressed in Arnold's book by referring to the harmonic oscillator.

• For instance, if the Lagrangina is made of the only kinetic part only, which determines a Riemannian metric, sufficiently close means that the two points must belong to a geodesically convex neighborhood of that metric. Mar 20 '18 at 11:08
• Every textbook on ODEs. It is sufficiently to notice that the equations can be written into normal form an the right hand side is $C^1$ so that the global existence and uniqueness theorem is valid. Nov 29 '20 at 20:10
• @ RandomXYZ I added a long ADDENDUM concerning your issue. I hope it is sufficient. Nov 29 '20 at 22:07
• @RandomXYZ I added a statement about the problem with boundary conditions. Thank you for having insisten on that issue. Dec 16 '20 at 17:43
• Sorry, not yet. Now the book is only in Italian, also online, published by Springer. Next year I will publish it in English (always Springer). Online you may find a number of books of mine and also lecture notes, all in English, but nothing about analytical mechanics, since it is a subject of our BSc... Dec 16 '20 at 19:24