Illustrating Simple Variation Problem


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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