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

Question

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?

Answer 1

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 $$ |b-a| $$

Answer 2

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 $[a,b]$.