Mathematics Asked by Nicole Douglas on December 2, 2020

Find a possible solution for the minimization of the functional:

$$J[x]=int_0^1 (tdot{x}+dot{x}^2) , dttag1$$

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 $x(1)= c,cinmathbb{R}$ 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

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$^1$.

Instead we can complete the square in OP's functional: $$begin{align}J[x]~:=~&int_0^1! mathrm{d}t~ (dot{x}^2+tdot{x})cr ~=~& int_0^1! mathrm{d}t~ (dot{x}+frac{t}{2})^2-int_0^1! mathrm{d}t~frac{t^2}{4}cr ~=~& int_0^1! mathrm{d}t~ (dot{x}+frac{t}{2})^2- frac{1}{12}.end{align}tag{1} $$ 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 $x(t)=Asin t$.

Correct answer by Qmechanic on December 2, 2020

Using the Euler-Lagrange equations we arrive to

$$ 2ddot x+1 = 0 $$

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

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

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