Solving the Euler Lagrange equation for the functional $L(y)=\int_{-1}^7\sqrt{1+y'^2} dx$

Question

I want to solve this functional

\begin{equation} L(y)=\int_{-1}^7\sqrt{1+y'^2} \ dx \end{equation}

with IC: $y(0)=1,\ y(1)=2$

I start using the Euler Lagrange equation

\begin{equation} \frac{d}{dx}\frac{\partial F}{\partial y'}-\frac{\partial F}{\partial y}=0 \end{equation}

But I encounter immediately a problem, namely that

$\frac{d}{dx}\frac{\partial F}{\partial y'}=0$, since $\frac{\partial F}{\partial y'}=\frac{y'}{\sqrt{1+y'^2}}$ which has no x-variable. Also $\frac{\partial F}{\partial y}=0$, hence it does not make sense.

Alterantively, I set this equal to a constant instead, and get:

\begin{equation} \begin{array} f\frac{d}{dx}\frac{y'}{\sqrt{1+y'^2}}=C \\ \frac{y'}{\sqrt{1+y'^2}}=C_1x + C_3\\ \frac{y'}{\sqrt{1+y'^2}}=C_2x+C_4 \end{array} \end{equation}

How do I solve this?

Thanks

Answer 1

Method I:

Obviously, the functional $$ L(y)=\int_{a}^b \ \sqrt{1+y'^2} \ dx $$ represents the arc-length of a curve, and the variational problem reduces to finding a curve of shortest arc-length connecting two points on the plane. This is one of the classical problems in the Calculus of Variations.

Geometrically, the well-known answer is a straight line, $$y = m x + c.$$

The constants $m$ and $c$ can be found from the given boundary conditions. (we need two conditions basically)

You have given the conditions in the problem as $$ y(0) = 1, \ \ y(1) = 2 $$

Substituting, we get $$ 1 = m (0) + c \ \ \mbox{or} \ \ c = 1 $$ $$ 2 = m(1) + c \ \ \mbox{or} \ \ m = 2 - c = 2 - 1 = 1 $$

Hence, the integral of the given variational problem is $$ y = x + 1 $$ (Clearly, it satisfies the conditions $y(0) = 1, y(1) = 2$.

Method II:

Using the Euler-Lagrange equation $$ {\partial f \over \partial y} - {d \over dx}\left({\partial f \over \partial y'} \right) = 0 \tag{1} $$

Since $f$ is purely a function of $y'$, we have $$ {\partial f \over \partial y} = 0 $$

Thus, (1) simplifies to $$ {d \over dx}\left({\partial f \over \partial y'} \right) = 0 $$ or $$ {\partial f \over \partial y'} = \mbox{constant} $$

Thus, we get $$ {1 \over 2 \sqrt{1 + y'^2}} (2 y') = k_1 $$ or $$ {y' \over \sqrt{1 + y'^2}} = k_1 $$

Simplifying, we get $$ 1 + y'^2 = k_1 y'^2 $$ where $k_1 = {1 \over c_1^2} = $constant

Simplifying again, we get $$ y' = \mbox{constant} = m (\mbox{say}) $$

Integrating, we get $$ y = m x + c $$ which is a straight line.

The constants $m$ and $c$ can be found as in Method I.

Finally, we arrive at the optimal solution as $$ y = x + 1 $$

Answer 2

I found a second method:

Using the formula

\begin{equation} F-y'F_{y'}=C \end{equation}

I get:

\begin{equation} \begin{array} a\sqrt{1+y'^2}-y'\frac{y'}{\sqrt{1+y'^2}}=C \\ -y'^2+1+y'^2=C\sqrt{1+y'^2}\\ 1=C(1+y'^2)\\ Cy'^2+C=1\\ C(y'^2+1)=1\\ y'=\sqrt{C-1} \ set\ \sqrt{C-1}=A\\ \int y'=\int A \\ y(x)=Ax+B \end{array} \end{equation}

With initial conditions, this gives $y(x)=x+1$