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Let $f\in \mathcal C^2([a,b]\times \mathbb R\times \mathbb R)$. I have to show that if $\bar u\in \mathcal C^2([a,b])\cap X$ is a minimizer of $$\inf_{u\in X}\left\{\int_a^b f(x,u(x),u'(x))dx\right\},$$ for $X=\{u\in \mathcal C^1([a,b])\mid u(a)=\alpha \}$, then $$\begin{cases}\frac{d}{dx}f_\xi=f_u\\ f_\xi(b,\bar u(b),\bar u'(b))=\alpha \end{cases}.$$

We suppose WLOG that $u(a)=0$. The proof goes as the proof of Euler-Lagrange Equation. Except that at the end, I get $$\int_a^b \left(f_u-\frac{d}{dx}f_\xi\right)v+f_{\xi}(b, \bar u(b),\bar u'(b))v(b)=0.$$ for all $v\in X$. Now, it's written that using fundamental theorem of calculs of variation, we get $$\begin{cases}\frac{d}{dx}f_\xi=f_u\\ f_\xi(b,\bar u(b),\bar u'(b))=0\end{cases}.$$

I would agree if $$\int_a^b \left(f_u-\frac{d}{dx}f_\xi\right)v=0,$$ but it's not what we have... so how can it work ?

user330587
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1 Answers1

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The fundamental lemma says this (see : https://en.wikipedia.org/wiki/Fundamental_lemma_of_calculus_of_variations )
If : $\int_a^b f(x)h(x) dx = 0 $ for all compactly supported smooth functions $h(x) \text{ on } (a,b)$ , then this implies $\implies f(x)=0$.

Your $'h(x)'=v$ , and your $'f(x)'=f_u-\frac{d}{dx}f_\xi $ .

Your question was why can we assume $\int_a^b \left(f_u-\frac{d}{dx}f_\xi\right)v dx$ must be $0$ ?

"Compactly supported" means "vanishes outside" $(c,d)$ for some $c,d$ such that $a<c<d<b$. So in that case the fundamental theorem applies.

When you say : $ \forall \enspace v \in X : \int_a^b \left(f_u-\frac{d}{dx}f_\xi\right)v dx +f_{\xi}(b, \bar u(b),\bar u'(b))v(b)=0 $ that means that also this is valid for all compactly supported $v \in X$. But for all compactly supported $v$ we have $v(b)=0 $ implying that : $f_{\xi}(b, \bar u(b),\bar u'(b))v(b)$ must be zero for all compactly supported $v$. Then we see that $\int_a^b \left(f_u-\frac{d}{dx}f_\xi\right)v dx$ must also be zero for all compactly supported $v$ and we can apply the fundamental theorem.

It follows that $f_u-\frac{d}{dx}f_\xi=0$ and then also : $f_{\xi}(b, \bar u(b),\bar u'(b))=0$.

Rutger Moody
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  • why $v(b)=0$ ? $b$ can be on the support of $v$... – user330587 Mar 26 '17 at 15:49
  • @user330587 $v(b)=0$ for compactly supported $v(x)$ which is a subset of all smooth $v$. So if statement is valid for all $v$ it is certainly also valid for all compactly supported $v$ which is what we need for the fundamental theorem and we have $a<c<d<b$ according to wikipedia at least, but they also can be wrong of course.. – Rutger Moody Mar 26 '17 at 15:54
  • @user330587 Looked it up here, but no I don't think $b$ can be on the support of $v$ in variational calculus. – Rutger Moody Mar 26 '17 at 15:59
  • Just a samll question more : why can we apply this fundamental lemma ? Indeed, it says that if $\int fh=0$ for all test function $h$, then $f=0$. But here $v$ is not a test function (it's only $C^1$). Thank you – user330587 Apr 09 '17 at 14:33
  • @user330587 Thx. The point is that the lemma applies if the condition is true for all $C^1$ functions with $v(a)=v(b)=0$. You had already assumed $v(a)=0$. So $v(b)$ doesn't have to be $0$ , but there is certainly a subset of all $v()$ that do have $v(b)=0$. The condition must apply for those functions as well so we can apply the lemma. And once we know $f_u-\frac{d}{dx}f_\xi=0$ then it follows that also the other part must be zero : $f_{\xi}(b, \bar u(b),\bar u'(b))=0$ . Or is that not what you mean? – Rutger Moody Apr 09 '17 at 15:30