Not enough degrees of freedom in Euler Lagrange equation

Question

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
$$ 2p_0-2r_0 = -2 \sqrt{2} p_0 \to (1+\sqrt{2})p_0 = r_0. $$

Answer 1

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

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

Answer 2

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.