Find a possible solution for the minimization of the functional: $$J[x]=\int_0^1 (t\dot{x}+\dot{x}^2) \, dt\tag1$$

with $x(0)=1$ and $x(1)$ is a free variable.

I am trying to solve the above, but not sure whether I am going the right way!

My attempt:

Try to find the function $x$, perhaps the line passing from points $x(0)$ and $x(1)$? If $x(0)=1$, and $x(1)= c,c\in\mathbb{R}$ then $$x(t)= (c-1)t+1. \tag2$$

Then replace $(2)$ into the equation $(1)$ and now $$J[x]=\int_0^1 (c-1)t+(c-1)^2 \, dt$$

Doing the calculations $$=c^2 -\frac{3}{2}c+\frac{1}{2}\tag3$$

So the problem becomes minimizing $(3)$.

Let $F(c)=c^2 -\frac{3}{2}c+\frac{1}{2}$. Then $F(c)$ has a critical point at $c=\frac{3}{4}$ and it is a minimum for $F(c)$ based on the second order condition. Hence, one possible minimization of the functional is:

$$x(t)= \frac{-1}{4}t+1 $$

But not sure if it is a correct approach or I should use Euler-Lagrange like in this example Minimization of functional using Euler-Lagrange


2 Answers 2

  1. We cannot assume the Euler-Lagrange (EL) equation since we don't have adequate boundary conditions (BCs). Recall that the proof of the EL equation uses BCs to get rid of boundary terms$^1$.

  2. Instead we can complete the square in OP's functional: $$\begin{align}J[x]~:=~&\int_0^1\! \mathrm{d}t~ (\dot{x}^2+t\dot{x})\cr ~=~& \int_0^1\! \mathrm{d}t~ (\dot{x}+\frac{t}{2})^2-\int_0^1\! \mathrm{d}t~\frac{t^2}{4}\cr ~=~& \int_0^1\! \mathrm{d}t~ (\dot{x}+\frac{t}{2})^2- \frac{1}{12}.\end{align}\tag{1} $$ Clearly, the minimum configuration satisfies $$ \dot{x}~=~-\frac{t}{2}.$$ Given the initial condition $x(0)=1$, the unique solution is $$ x(t)~=~1-\frac{t^2}{4}.$$


$^1$ Perhaps a counterexample is in order. Consider the functional $$K[x]~:=~ \int_0^2\! \mathrm{d}t~ (\dot{x}^2-x^2)$$ with $x(0)=0$ and $x(2)$ is a free variable. It is not hard to prove that $K$ is unbounded from below e.g. by considering configurations of the form $x(t)=A\sin t$.


Using the Euler-Lagrange equations we arrive to

$$ 2\ddot x+1 = 0 $$

and after solving with the initial condition $x(0) = 1$ we arrive to

$$ x=1-\frac{t^2}{4}+ tC_0 $$

To determine $C_0$ we substitute the found $x$ into the integral obtaining

$$ \int_0^1 (t \dot x+\dot x^2) dt = -\frac{1}{12}+C_0^2 $$

and to have a minimum we set $C_0 = 0$ so the solution is

$$ x = 1-\frac{t^2}{4} $$

  • 1
    $\begingroup$ Thank you and do we have adequate boundary conditions to use the Euler-Lagrange equations? $\endgroup$ Sep 5, 2020 at 14:30
  • 1
    $\begingroup$ The boundary problem associated to the E-L conditions, will have an undetermined constant: here $C_0$ but the optimality condition suffices to give us the value for $C_0$. $\endgroup$
    – Cesareo
    Sep 5, 2020 at 16:28

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