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I am working on this problem

Consider the functional

$I=\int^a_b f (u'(x))dx$;

where $f \in C^1(\mathbb{R}; \mathbb{R})$ and $u \in \Phi =\{ u \in$ $D^1([a, b]; \mathbb{R})| u(a) = \alpha; u(b) = \beta \}$, for given
$\alpha; \beta \in \mathbb{R}$. ($D^1$: piecewise differentiable functions) Show that $\bar u(x) = \frac{\beta -\alpha}{b-a} (x-a)+ \alpha$ is a solution of the corresponding Euler-Lagrange equation and $u \in \Phi$.

Is the function $ \bar u$ a minimizer in $\Phi_2 = D^0_1([0; 1];$ $\mathbb{R})$ (i.e piecewise differentiable functions that vanish at the endpoints) of $I(u) = \int_0^1 e^{−u'^2(x)}dx$ ?

I have already shown that $\bar u(x) $ is a solution of the corresponding Euler-Lagrange equation and $\bar u \in \Phi$. I am stuck in the second part. Any advice on how to proceed would be much appreciated!

My idea is that there is no minimizer because that exponential $e^{-x^2}$ in the xy plane has asymptote in y=0, so I need a sequence to prove the inf of the integrand is 0. On the other hand $ \bar u$ does not seem to belong to $\Phi_2$, because at the endpoints of $[0, 1]$ I don't get 0 but instead $ \bar u(0) = \alpha; \bar u(1) = \beta$ because of the first part

some_math_guy
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    Take the admissible $\bar{u}$, break it in the middle as $\Lambda$-shape and let the top corner go to $+\infty$. What happens with the derivative $u'$ and the functional? – A.Γ. Feb 21 '23 at 07:22
  • @A.Γ. What do you mean with $\Lambda$. Is my idea that there is no minimizer correct? and isn't it a problem the fact that $\bar u (0)=0$ instead of $\alpha$ and $\bar u (1)=0 $instead of $\beta$ ? Or in other words, isn't that enough to say that there is no minimizer in $\Phi_2$? – some_math_guy Feb 21 '23 at 07:27
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    There is no problem in $\alpha=\beta=0$ as the first part works for all $\alpha,\beta\in\mathbb{R}$, doesn't it? The idea is just an intuition unless it is confirmed by a proper construction. Take the extremal, break it in the middle and consider a piecewise linear shape - first up then down (that looks like "$\Lambda$"). Try to figure out how to use these shapes to find a minimizing sequence. I would not count "just to say no minimizer" as a solution. – A.Γ. Feb 21 '23 at 07:39
  • @A.Γ. I think I can use $u_n(x)=n(x-1/2)^2-n/4$, so that $I(u_n)\to + \infty$ – some_math_guy Feb 21 '23 at 08:38
  • @A.Γ. I meant $I(u_n) \to 0$, but how do I conclude from the inf being 0 that there is no minimizer? – some_math_guy Feb 21 '23 at 08:49
  • Ok, this sequence works as well. No minimum because if the minimum existed it would give the same value as inf, that is zero, but it means the function $e^{-u'(x)^2}$ must be zero almost everywhere, which is impossible. – A.Γ. Feb 21 '23 at 11:16
  • @A.Γ. Couldn't I just argue that $I(\bar u)= \int_0^1e^{-o^2}dx =1$ and so u is not a minimizer because the functional is greater than its inf at that value? – some_math_guy Feb 21 '23 at 11:24

1 Answers1

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We can transcribe OP's variational problem into a related problem:

Consider the functional $$J[v]~:=~\int^a_b\! dx~f (v(x));$$ where $f \in C^1(\mathbb{R}; \mathbb{R})$ and $$v ~\in~ \Phi^{\prime} ~:=~\left\{ v \in D^0([a, b]; \mathbb{R})\left| \int^a_b\! dx~v(x)=\beta-\alpha\right.\right\},$$ for given $\alpha; \beta \in \mathbb{R}$. ($D^0$: piecewise continuous functions.)

It is not hard to see that the constant function $v(x)=\frac{\beta -\alpha}{b-a}$ is a minimum (maximum) for the functional $J$ if and only if $\frac{\beta -\alpha}{b-a}$ is a minimum (maximum) point for the function $f$, respectively.

There is a bijective map $$\Phi^{\prime}~\ni~v\quad \mapsto\quad \left[ x\mapsto\alpha+\int_a^x\! dx^{\prime}~v(x^{\prime}) \right] ~\in~\Phi.$$

It follows that the affine function $\bar{u}$ is a minimum (maximum) for the functional $I$ if and only if $\frac{\beta -\alpha}{b-a}$ is a minimum (maximum) point for the function $f$, respectively.

In particular, the affine function $\bar{u}$ happens to be a maximum for the functional $I$ in OP's title example.

Qmechanic
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