How to solve $L(u,v)=\int_{-1}^8\sqrt{u'^2+v'^2}dx$

Question

I am trying to solve this functional:

$L(u,v)=\int_{-1}^8\sqrt{u'^2+v'^2}dx$

where the points $(u,v)$ on BC: $-1<x<8$

I use the Euler Lagrange formula and get two equations, considering $u$ and $v$ to represent two different dimensions. That is:

\begin{cases} \frac{d}{dx}\frac{\partial F}{\partial u'}-\frac{\partial F}{\partial u}=C_1\\ \frac{d}{dx}\frac{\partial F}{\partial v'}-\frac{\partial F}{\partial v}=C_2 \end{cases}

obtaining with $F=\sqrt{u'^2+v'^2}$,

\begin{cases} \frac{d}{dx}\frac{u'}{\sqrt{u'^2+v'^2}}=C_1\\ \frac{d}{dx}\frac{v'}{\sqrt{u'^2+v'^2}}=C_2 \end{cases}

Integrating on both sides I get:

\begin{cases} \frac{u'}{\sqrt{u'^2+v'^2}}=C_1x\\ \frac{v'}{\sqrt{u'^2+v'^2}}=C_2x \end{cases}

which leaves the relation:

\begin{cases} \frac{1}{x\sqrt{u'^2+v'^2}}=\frac{C_1}{u'}\\ \frac{1}{x\sqrt{u'^2+v'^2}}=\frac{C_2}{v'} \end{cases}

So now we can equate the RHS:

\begin{equation} \frac{C_1}{u'}=\frac{C_2}{v'}\\ C_1u'=C_2v' \end{equation}

The we get:

\begin{equation} u'=\frac{C_2}{C_1}v' \rightarrow \\ u=\frac{C_2}{C_1}v+A \end{equation}

But what is $v$ ? Is it simply the inverse of $u$?

Thanks

Answer 1

Something like $$ u =Cv + D$$ for some constants $C, D$ is probably the best you can get. Your functional is the length functional (thinking of $\gamma (x) = (u(x), v(x))$ as a curve in $\mathbb R^2$) and in $\mathbb R^2$ only the straight lines minimizes the length.

You cannot get further (that is, you cannot find $u, v$ explicitly, even with specific boundary conditions) since the length functional is invariant under re-parametrization. That is, if $\sigma : [-1 ,8]\to [-1, 8]$ is any $C^1$ bijective maps with $\sigma ' >0$ for all $t\in [-1, 8]$, then

$$ L(\gamma) = L ( \gamma\circ \sigma).$$