### Author Topic: General extremisation of functionals of arc length  (Read 133 times)

#### Aziz S.

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##### General extremisation of functionals of arc length
« on: November 28, 2016, 02:30:05 PM »
Functionals of form $\Phi = \int_{x_0}^{x_1} f(u)\sqrt{1 + (u')^2}$, where $u = u(x)$ may be extremised by solving the Euler-Lagrangian equation $\frac{\partial L}{\partial u} = \frac{\partial}{\partial x}\frac{\partial L}{\partial u'}$, where $L = f(u)\sqrt{1 + (u')^2}$.

$$\frac{\partial L}{\partial u} = f'\sqrt{1 + (u')^2}$$
$$\frac{\partial L}{\partial u'} = \frac{u'f}{{(1 + (u')^2)}^{1/2}}$$
And using product rule and recalling that $u$ and $u'$ are functions of x:
$$\frac{\partial }{\partial x} \frac{\partial L}{\partial u'} = \frac{\partial }{\partial x}\left(f \cdot u' \cdot {(1 + (u')^2)}^{-1/2}\right)$$
We obtain the Euler-Lagrangian equation
$$\frac{f' {u'}^2}{{(1 + (u')^2)}^{1/2}} +\frac{f u''}{{(1 + (u')^2)}^{1/2}} - \frac{f {u'}^2 u''}{{(1 + (u')^2)}^{3/2}} = f'{(1 + (u')^2)}^{1/2}$$
And multiply through by ${(1 + (u')^2)}^{3/2}$, expand, and simplify to obtain
$$f'u'' = f'(1 + (u')^2)$$
Rearranging and multiplying both sides by $u'$ we get
$$\frac{u'u''}{1 + (u')^2} = \frac{u'f'}{f}$$
Which we integrate $\mathrm{d}x$ to get the differential equation
\begin{align}
\frac{1}{2}\ln{(1 + (u')^2)} &= \ln f + C\\
1 + (u')^2 &= B {f(u)}^2
\end{align}

So the functions $u$ that extremise $\Phi$ solve the differential equation $1 + (u')^2 = B{f(u)}^2$ where $B$ is the constant of integration. When $f(u)$ is linear in u, consider $1 + \sinh^{2}x = \cosh^{2}x$ knowing that $(\cosh x)' = \sinh x$.
« Last Edit: November 28, 2016, 02:36:13 PM by Aziz S. »

#### Victor Ivrii

Correct. As it was mentioned both in the lectures and in textbook, if $L=L(u,u')$, so does not depend explicitly of $x$ then one does not need to write Euler-Lagrange equation, but a simpler one $u'L_{u'}=L=C$.It is equivalent to "multiply by $u'$ and integrate".
If $f(u)=u$ it is an area of the surface of the revolution of $x=f(u)$ around $x$''