Find possible solution for minimization of a functional with free boundary condition

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

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 $$x(t)~=~1-\frac{t^2}{4}.$$

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$$^1$$ 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

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

• Thank you and do we have adequate boundary conditions to use the Euler-Lagrange equations? Sep 5, 2020 at 14:30
• 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 $C_0$. Sep 5, 2020 at 16:28