Consider the variational problem (P)

J(y(x)) = \(\rm\displaystyle\int_0^1\)[(y')² − y|y| y' + xy] dx, y(0) = 0, y(1) = 0.

Which of the following statements is correct?

Concept:

Euler-Lagrange Equation: The extremal of the functional J[y] = \(\int_a^b\)f(x, y, y')dx satisfies \(\frac{\partial f}{\partial y}\) - \(\frac{d}{dx}\)(\(\frac{\partial f}{\partial y'}\)) = 0

Explanation:

J(y(x)) = \(\rm\displaystyle\int_0^1\)[(y')2 − y|y| y' + xy] dx, y(0) = 0, y(1) = 0

If y > 0 then

f(x, y, y') = (y')2 − y²y' + xy

So using  \(\frac{\partial f}{\partial y}\) - \(\frac{d}{dx}\)(\(\frac{\partial f}{\partial y'}\)) = 0 we get

-2yy' + x - \(\frac{d}{dx}\)(2y' - y²) = 0

- 2yy' + x - 2y'' +2yy' = 0

y'' = x/2....(i)

If y < 0 then

f(x, y, y') = (y')2 + y2y' + xy

So using  \(\frac{\partial f}{\partial y}\) - \(\frac{d}{dx}\)(\(\frac{\partial f}{\partial y'}\)) = 0 we get

2yy' + x - \(\frac{d}{dx}\)(2y' + y2) = 0

 2yy' + x - 2y'' - 2yy' = 0

y'' = x/2....(ii)

Hence in both case we get

y'' = x/2

Integrating 

y' = \(\frac{x^2}{4}+c_1\)

Integrating again

y = \(\frac{x^3}{12}+c_1x+c_2\)

Using  y(0) = 0, y(1) = 0 we get

c₂ = 0 and 0 = \(\frac{1}{12}\) + c₁ ⇒ c1 = - \(\frac{1}{12}\)

Hence solution is

y = \(\frac{x^3}{12}-\frac{x}{12}\)

Option (3) is correct