# Not enough degrees of freedom in Euler Lagrange equation

I have the following minimization problem $$\min_{p,r} \int_0^\infty (r^2 + p^2) dt$$ subject to $$\dot r = - r + \dot p,\quad p(0)=p_0,\quad r(0)=r_0, \quad p(\infty)=0, \quad r(\infty)=0.$$ This problem is feasible I can easily find admisble $$p(t)$$ fast converging to zero and then $$r(t)$$ such that this intergal is finite. So I built a Lagrangian: $$\cal L(p_0,r_0)= \int_0^\infty r^2+p^2 + \lambda ( \dot r+r - \dot p) dx$$ and derived Euler-Lagrange equations: \begin{align} p: 0=-2p + \dot \lambda \\ r: 0 = 2r +\lambda - \dot\lambda\\ \lambda: 0 = \dot r + r -\dot p\\ \end{align} combining second and third equation I am getting: $$\dot p = \dot r +r = 1/2(\ddot \lambda - \dot \lambda + \dot \lambda - \lambda) = 1/2(\ddot \lambda - \lambda).$$ On the other hand from the first equation: $$\dot p = -1/2(\ddot \lambda)$$ combining them I am getting: $$2\ddot \lambda -\lambda = 0.$$ The boundary conditions for $$\lambda$$ are the following \begin{align} \dot \lambda(0)&= 2p_0, \dot \lambda(\infty)=0 \text { from the first equation } \\ \dot \lambda(0)-\lambda(0) &= 2r_0, \lambda(\infty)=0 \text { from the second equation} \end{align} Simplifying: \begin{align} \dot \lambda(0)&= 2p_0, \dot \lambda(\infty)=0 \text { from the first equation } \\ \lambda(0) &= 2p_0-2r_0, \lambda(\infty)=0 \text { from the second equation} \end{align} The problem is that the second order ode has only 2 constants and I cannot satisfy 4 boundary conditions for arbirary $$p_0,r_0$$. So I am a bit confused , it seems there is no optimal solution for this system since I cannot resolve lagrange equations.However if I remove constrains for $$p_0$$ and solve the system, I will get $$\lambda^*(t),p^*(t),r^*(t)$$ . Then I can built a sequience $$p_\epsilon(t)\to p^*(t)$$ such that $$p_\epsilon(0)=p_0$$ in the limit it will be a function with inifnite derivative at 0.So $$p_\epsilon(t)$$ will be somehow $$\epsilon$$-suboptimal. My main question to the community: am I right? If yes why actually the above system does not have a solution and only $$\epsilon$$-solution.The problem is convex and well defined.

SOME UPDATE: $$\lambda(t) = c_1 e^{-\frac{t}{\sqrt{2}}} +c_2 e^{\frac{t}{\sqrt{2}}}.$$ The only case when all 4 conditions are satisfied is when $$c_2=0,-\frac{1}{\sqrt{2}} c_1 = 2p_0, c_1 =2p_0 -2 r_0$$ Namely
$$2p_0-2r_0 = -2 \sqrt{2} p_0 \to (1+\sqrt{2})p_0 = r_0.$$

• Something is off: denoting $\mu = 2\lambda,$ you have the system $\dot\mu = p; \dot \mu = r+ \mu, \dot r + r = \dot p$. Using $r = p-\mu,$ and differentiating, you can eliminate $r$ to get the system $p = \mu + \dot \mu, p = \dot \mu$, which imply that $\mu = 0$. Commented Jul 12 at 15:49
• Looking at the problem in its own right: note that the constraint can be written as $\dot s = e^{t}p,$ where $s = e^{t} r$. Now take $p$ to be a smooth function that is supported on $[0,\delta)$. Presumably $s$ will also be zero beyond $\delta$. Assuming that $p$ is nice enough that $s$ doesn't go wild, the objective would then be $O(\delta)$ (lots of ifs), suggesting that the optimal value is $0$ (and it is not attained, of course, if even one of $p_0, r_0$ is nonzero). This leads to the question of whether such nice bump-type solutions exist for any IVP of this form. Commented Jul 12 at 15:50
• Actually, wait, that was silly of me: it suffices to take $s$ to be bump-type. So the question just becomes if for every $a,b, \delta$, there is a map $s$ such that $s(0) = a, \dot s(0) = b$, $s$ is supported on $(-\infty, \delta)$ and $\sup_{t \ge 0} s(t)$ is bounded as a function of $(a,b)$. This seems like something one can just look up. Commented Jul 12 at 16:07
• Yes, I am thinking in similar direction. Also if I add to the functional $\epsilon \dot x^2$ solve it and make $\epsilon \to 0$ I should see how the solution converges to something not differenitable ( my hypothesis) but then I have oDE of the fourth order, not sure I can solve it anlytitvcally. I shluld get something like $e^{-t/\sqrt{\epsilon}}$ somewhere Commented Jul 12 at 17:08
• Actually, I don't think the vanishing cost business works, at least in the way I was thinking - under the parametrisation I proposed, the cost is $\int e^{-2t} (s^2 + \dot s^2)$, and if $s$ drops from $a \neq 0$ to $0$ in a window of size $\delta,$ then the typical $\dot s$ in that range has to look like $a/\delta$, so the integral of $\dot s^2$ would blow up as $\delta \to 0$. Also if we just set up E-L equations for $\int e^{-2t} (s^2 + \dot s^2),$ then again the same issue arises, i.e., that the initial values can't be met unless $r_0 = c p_0$ for a fixed $c$. Commented Jul 13 at 15:09

Considering the ode system

$$\cases{ -\lambda'(t)-2 p(t)=0\\ \lambda'(t)-\lambda(t)-2 r(t)=0\\ -p'(t)+r'(t)+r(t)=0 }$$

after addition of the first and second we have

$$-2p(t)-\lambda(t)-2r(t) = 0$$

so $$p(t),r(t),\lambda(t)$$ are not independent hence we will obtain only two general conditions instead of three.

Now solving the ode system we get

$$\left\{ \begin{array}{rcl} \lambda(t)&=&(34 c_1-28 c_2) \cosh \left(\frac{t}{\sqrt{2}}\right)+4 \sqrt{2} (6 c_1-5 c_2) \sinh \left(\frac{t}{\sqrt{2}}\right) \\ p(t)&=&2 (5 c_2-6 c_1) \cosh \left(\frac{t}{\sqrt{2}}\right)+\frac{(14 c_2-17 c_1) \sinh \left(\frac{t}{\sqrt{2}}\right)}{\sqrt{2}} \\ r(t)&=&(4 c_2-5 c_1) \cosh \left(\frac{t}{\sqrt{2}}\right)+\frac{(6 c_2-7 c_1) \sinh \left(\frac{t}{\sqrt{2}}\right)}{\sqrt{2}} \\ \end{array} \right.$$

• and you cannot satisfy with 2 constant c_1 c_2 both initial conditions and zero on infinity . That what I try to say. The system as seems to me does not have the extremum point, though I dont understand precisely what does it mean . Commented Jul 11 at 22:04
• Clearly the variational problem is ill posed. Commented Jul 11 at 22:22
• Could you elaborate please what is wrong here, and why there is no extreme solution Commented Jul 12 at 0:06
• A lagrangian with the structure $L=f(r,p)+\lambda(\dot r-\dot p+r)$ will require at most two conditions, because of dependency on $\{\lambda,r,p\}$ Commented Jul 12 at 7:13
• It is clear from derivation, but it is not clear initially. Also, it is not clear to me what it means in terms of optimal solutions. There are admissible $C^1$ even $C^2$ trajectories satisfying the constraints. But there is no best one? Also the discrete time version of this problem has nice analytical solution. Commented Jul 12 at 13:33

Here is one approach:

1. Define $$q:=p-r$$. Then the constraint becomes $$r=\dot{q}$$. The Lagrangian then becomes an inverted harmonic oscillator $$L~=~r^2+p^2~=~r^2+(r+q)^2~=~\dot{q}^2+(\dot{q}+q)^2~\sim~2\dot{q}^2+q^2,$$ where we have dropped a total time derivative term.

2. The EL equation is $$2\ddot{q}=q$$. The solution is $$q(t)~=~Ae^{t/\sqrt{2}}+ Be^{-t/\sqrt{2}}.$$ With OP's boundary conditions, it becomes $$q(t)~=~q_0e^{-t/\sqrt{2}}, \qquad q_0~:=~p_0-r_0.$$ Hence $$r(t)~=~\dot{q}(t)~=~r_0e^{-t/\sqrt{2}}, \qquad r_0~=~-\frac{q_0}{\sqrt{2}},$$ and $$p(t)~=~q(t)+r(t)~=~p_0e^{-t/\sqrt{2}}, \qquad p_0~=~q_0(1-\frac{1}{\sqrt{2}}).$$

3. Note that a stationary solution only exists if the boundary conditions satisfy $$p_0~=~(1-\sqrt{2})r_0,$$ i.e. generically the system is overconstrained, as OP already mentions. If the system is overconstrained then admissible configurations are never stationary.

TL;DR: OP's original system has 4 variables $$r$$, $$p$$, $$\dot{r}$$, $$\dot{p}$$, minus 1 constraint, so effectively 3 variables, but 4 boundary conditions, and is hence generically overconstrained.

• Of course, I know this solution and we get the same problem here. My question was in general: what does it mean “over constrained” in terms of optimal solution, because admissible trajectories satisfying all constraints do exist. My impression is the on the class of C^1 ( or may be C^2) the solution does not exist. If I add $\epsilon \dot p^2$to the functional - I will get equation of 4th order and will satisfy all constraints. Now taking $\epsilon\to 0$ I will see how the solution has a part with derivative converging to $\infty$. How to know this a priory? Commented Jul 15 at 22:10