# Illustrating Simple Variation Problem

### Example

For example, consider the functional

$v[f(x)]=\int_0^1 f'(x)^2+12 x f(x)dx; f(0)=0,f(1)=1$

Euler-Lagrange Equation is given by

${\partial F\over\partial f} - {d\over dx} {\partial F\over\partial f'} =0$

$\implies {\partial ( f'(x)^2+12 x f(x) )\over\partial f} - {d\over dx} {\partial ( f'(x)^2+12 x f(x) )\over\partial f'} =0$

$\implies 12 x - {d\over dx} 2f'(x)=0$

$\implies 12 x - f''(x)^2 =0$

Solving by integrating twice, we get $f(x) = A x^3+Bx+C$, applying boundary conditions f(0)=0, f(1)=1, we get extrema as $f(x)=x^3$.

That is, if you put $f(x)=x^3\implies f'(x)=3x^2$ and hence, the functioal becomes

$v[x^3]=\int_0^1 [(3x^2)^2+12x^4]dx=4.2$

That is, if you take any other $f(x)$ that satisfies the boundary condition, the functional value for that function will be definitely greater than $v[x^3]=4.2$. For example, suppose we choose $f(x)=x$, then $v[x]=5$, if $f(x)=x^4$, then $v[x^4]=4.2857...$, which is greater than 4.2. Similarly, $v[sin({\pi\over2}x]=6.0971...$

That is, functional takes minimum value for the function $f(x)=x^3$

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