Minimization problem $\min \int_{0}^{1}\sqrt{1+y'(x)^2}dx \\s.t. y(0)=1$

Question

I have some problems with the following exercise:

I have to determine the candidate for a weak extremum to the following problem:

$\min \int_{0}^{1}\sqrt{1+y'(x)^2}dx \\s.t. y(0)=1$

What I have done is the following:

I have used the Euler-Lagrange equation:

I obtained: $L_y=\frac{d}{dx}L_{y'} =>0=\frac{d}{dx}(\frac 1 2(1+y'(x))^{-\frac1 2})=-\frac1 4y''(x)(1+y'(x))^{-\frac3 2}$

Furthermore we know that

$y(0)=1$

and

$L_{y'}(1,y(1),y'(1))=0=\frac 1 2(1+y'(1))^{-\frac1 2}$

My first question is: is everything correct up to here?

If everything is correct my second question would be: how do I solve the differential equation $0=-\frac1 4y''(x)(1+y'(x))^{-\frac3 2}$?

I have very little experience with differential equations and until now I have only had homogeneous cases or 1st degree inhomogeneous cases. I have recently started studying inhomogeneous 2nd grade cases but I still can't understand them well.

Furthermore, this case seems to be a bit more complicated than a simple inhomogeneous case of second degree.

Can anyone explain me how to solve it?

Answer 1

You forgot the square on $y'$ in $L$ while computing $\frac d{dx}L_{y'}.$ Moreover, it was useless to compute it: $$\begin{align}0=\frac d{dx}L_{y'}&\Longleftrightarrow C=L_{y'}=\frac{y'}{\sqrt{1+y'^2}}\\&\Longleftrightarrow y'=\frac C{\sqrt{1-C^2}}=:D\\&\Longleftrightarrow y=Dx+y(0)\\&\Longleftrightarrow y=Dx+1\end{align}$$ and $D$ would be determined by a boundary condition for $y(1).$ If there is none, just minimize $\int_0^1\sqrt{1+D^2}\,\mathrm dx.$ The solution is $D=0$ i.e. $y'=0,$ which was obvious from the very beginning.

Answer 2

Don't let your reasoning be too complicated. Note $$ \int^1_0 \sqrt{1+y'(x)^2}~\mathrm{d}x \geq \int^1_0 \sqrt{1}~\mathrm{d}x=1 $$ This is obtained exactly if $y'(x)=0$ almost everywhere. So $y$ is constant. Because of $y(0)=1$ it must hold that $y(t)=1$ for all $t$.