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Let $\mathcal{A}$ be the set of continuously differentiable functions at the interval $[a,b]$. Let $J$ be the functional

$$J(y)=\int_a^b \sqrt{1+y'(x)^2}dx$$

Find $\min_{y \in \mathcal{A}} J(y)$, if it exists.

I have tried the following:

$$J'(y)=\int_a^b \frac{\partial}{\partial{y}}(\sqrt{1+y'(x)^2})dx=\int_a^b \frac{y'(x)y''(x)}{\sqrt{1+y'(x)}}dx$$

Is it right so far? If so, how could we continue?

evinda
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  • I think that this question is false,because$$J(y)=\int_a^b \sqrt{1+y'(x)^2}dx$$ and true is$$J(c)=\int_a^c \sqrt{1+y'(x)^2}dx$$ – amir bahadory May 06 '15 at 23:44
  • @amirbahadory $J$ is a functional, i.e. it takes functions as an input, not scalars. – snar May 07 '15 at 00:04

2 Answers2

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We have

$$1 \leq \sqrt{1 + y'(x)^2}$$ with equality attained when $y'(x) = 0$.

Hence

$$\int_a^b \sqrt{1 + y'(x)^2} \ dx \geq \int_a^b 1 \ dx = b - a$$

with equality attained for any constant function $y$ on $[a,b]$.

Simon S
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    I see..The equality is attained when $\sqrt{1+y'(x)}=1 \Rightarrow y'(x)=0 \Rightarrow y(x)=c$, right? $$$$ Could you maybe also explain me the other approach, with the arc length? – evinda May 07 '15 at 16:53
  • Yes. On the other approach, see here http://math.stackexchange.com/questions/181110/the-shortest-distance-between-any-two-distinct-points-is-the-line-segment-joinin – Simon S May 07 '15 at 16:56
  • Small correction to your comment: $\sqrt{1 + y'(x)^2}$ – Simon S May 07 '15 at 17:15
  • Oh yes, right... I will look at the link that you sent me and ask you if I have questions.. Could you make also take a look at my other question: http://math.stackexchange.com/questions/1271719/the-functional-take-its-maximal-value-for-yt-t ? – evinda May 07 '15 at 17:30
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If you know the calculus of variations, this is not too hard a problem. But a much easier way to approach it is to note that $$\int_a^b \sqrt{1+y'(x)^2}dx$$

is the arc length along the curve $y(x)$ from the point where $x=a$ to the point where $x=b$. For any specific boundary conditions, $y(a)$ and $y(b)$ might be any arbitrary values, but in each case the minimal curve is the straight line from $(a,y(a))$ to $(b,y(b))$.

The answer to the min over all $y$ will come for the case where $y(a) = y(b)$ and the answer is $$ |b-a| $$

Mark Fischler
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  • Could you explain me further how we deduce that the answer is $|b-a|$? I haven't really understood it... – evinda May 07 '15 at 16:47