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I came across the following formulation of the problem.

Minimize the functional $L[u]$ given by

$L[u] = \int^b_a \sqrt{(1+(u'(x))^2}$ over $U = \{u\in C([a,b])\cap C^1((a,b)):u(a)=\alpha, u(b) = \beta\}$

What I am wondering is, shouldn't we require $u(x)\in C([a,b])\cap C^1([a,b])$ instead of $u(x)\in C([a,b])\cap C^1((a,b))$? That is, don't we need $u(x)$ to be continuously differentiable up to the boundary?

If a function is continuolsy differentiable only in the open interval, how do we know that $L[u]$ is well defined?

Does the continuity of u in the closed interval $[a,b]$ somehow gurantee that?

I am curious about this because I know that the function $f(x) = 1/x$ is in $C^1((0,1))$, but $\int^1_0f'(x)dx $ is not well defined.

mononono
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  • Can't you extend $u'$ by continuity on the border of $(a,b)$ using that $u$ is bounded on $[a,b]$? – Surb Feb 09 '15 at 23:01
  • @Surb : $:$ One is not necessarily able to do that. $;;;;$ –  Feb 09 '15 at 23:02
  • @RickyDemer Ok... I was not sure anymore... Do you maybe have an example of such function (where $u'$ is not extendable)? – Surb Feb 09 '15 at 23:04
  • @RickyDemer I would like to see an example too, if you know one. – mononono Feb 09 '15 at 23:05
  • Anyway note that for the definition of $L$, you don't need continuity of $u'$ on $[a,b]$ since $[a,b]=(a,b)\cup{a,b}$ and ${a,b}$ is of measure $0$. So whatever the value of $u'(a)$ and $u'(b)$ are, the value of $L$ stays unchanged. – Surb Feb 09 '15 at 23:06
  • @Surb Yes, but how do you know $\int^a_bu'dx$ is well defined? – mononono Feb 09 '15 at 23:09
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    @Surb : $:$ http://math.stackexchange.com/a/423279/57159 $;;;;$ –  Feb 09 '15 at 23:12
  • @RickyDemer Nice, thank you! – Surb Feb 09 '15 at 23:16

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Some people would consider $u\in C^1(a,b)$ to mean that $u$ is once continuously differentiable in $(a,b)$ and $$ \sup_{(a,b)} |u(x)| + \sup_{(a,b)} |u'(x)| <\infty. $$ In this case clearly $L$ is defined.

If no assumption is made on the derivative, you're right that $L$ will not be well defined: Take $a=0$, $b=1$ and $u(x)=x\sin(x^s)$, then for $s< -10$ say, we have $u$ is in your space with $\alpha=0$ but $u'$ is not integrable.

Jose27
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  • but even if $u'$ is not integralbe, isn't it possible though for $\sqrt{1+(u')^2}$ to be integrable? – mononono Feb 09 '15 at 23:18
  • No, since we have $\frac{1}{\sqrt{2}}(1+|u'|)\leq\sqrt{1+(u')^2}\leq 1+|u'|$, so that $u'$ is integrable in $(a,b)$ if and only if $ \sqrt{1+(u')^2}$ is. – Jose27 Feb 09 '15 at 23:22
  • In your definition of $u(x)$, your define $u(x)$ piecewisely so that $u(0)=0$ because $xsin(x^s)$ is undefined at x = 0? – mononono Feb 09 '15 at 23:29
  • @Dejon: Yes, exactly. – Jose27 Feb 09 '15 at 23:34