Consider a functional

$$J = \int_{a}^{b} F(x, y, y^{'}),$$ where $F(x, y, y^{'}) = \frac{1 + y^{2}}{(y^{'})^2}$ for admissible function $y(x).$ Which of the following are extremals for $J$?

  1. $y(x) = A\sin x $

  2. $y(x) = A\sinh x + B\cosh x$

  3. $y(x) = A\sinh(Ax + B)$

  4. $y(x) = A\sin x + B\cos x$

I use Cauchy Euler equation. But it does not give me any direction to justify the option(s).

  • 2
    $\begingroup$ Possible duplicate of Extremals of functional $\endgroup$ Oct 31 '15 at 12:15
  • 3
    $\begingroup$ @JessePFrancis I think the answer to the present one is better; it just needs an upvote so that the question can be closed in the opposite direction. $\endgroup$
    – user147263
    Oct 31 '15 at 22:47
  • 2
    $\begingroup$ @NormalHuman, I picked this as duplicate because the other was older, and had some personal effort show there, which is not there here! $\endgroup$ Nov 1 '15 at 12:32


  1. Strictly speaking, to have a well-posed variational problem, one should impose boundary conditions. We will assume Dirichlet boundary conditions in what follows.

  2. Note that the Lagrangian $$F(x, y, y^{'}) = \frac{y^2+1}{(y^{\prime})^2}$$ does not depend explicit on the independent variable $x$.

  3. We can therefore use the Beltrami identity to get a first integral. $$\exists C:~~C~=~ \left(y^{\prime}\frac{\partial }{\partial y^{\prime}} -1\right)F~=~-3F. $$

  4. In other words, $$ y^{\prime}~=~K \sqrt{y^2+1} , \qquad K~:=~\sqrt{-\frac{3}{C}}. $$

  5. Separation of variables: $$ \frac{\mathrm{d} y}{\sqrt{y^2+1}}~=~K\mathrm{d} x.$$

  6. Integrate: $~~{\rm arsinh} (y) ~=~ K (x-x_0).$

  7. Isolate: $~~y~=~\sinh\left(K (x-x_0)\right).$


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