# An MCQ for finding the extremal of the functional $J = \int_{a}^{b} F(x, y, y^{'})$

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).

• Possible duplicate of Extremals of functional Oct 31 '15 at 12:15
• @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.
– user147263
Oct 31 '15 at 22:47
• @NormalHuman, I picked this as duplicate because the other was older, and had some personal effort show there, which is not there here! Nov 1 '15 at 12:32

## 1 Answer

Hints:

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