My Question is, if my solution is right.
I am trying to minimize $$ I[f]=\int_1^2 (x^2 f'(x)^2+2f(x)^2)\,\mathrm dx $$ with the conditions $f(1)=0$, $f(2)=1$ Assuming $f$ is the minimizer and $\epsilon \in \mathbb{R}$ I define $g(\epsilon):=I[f+\epsilon h]=\int_1^2 x^2(f'(x)+\epsilon h'(x))^2+2(f(x)+\epsilon h(x))^2dx$ Because I assumed f is a minimum of I[f], I[f+$\epsilon$ h] hast it's minimum for $\epsilon=0$, especially g'(0)=0
Calculating the derivative gives
$g'(\epsilon)=\int_1^2 2x(f'(x)+\epsilon h'(x))^2+2x^2(f'(x)+\epsilon h'(x))^2(f''(x)+\epsilon h''(x))+4(f(x)+\epsilon h(x))(f'(x)+\epsilon h'(x))) dx$
So as a result I get $0=g'(0)$=$\int_1^2 2x f'(x)^2+2x^2f'(x)^2f''(x)+4f(x)f'(x)dx$=$\int_1^2 f'(x)(2xf'(x)+2x^2f'(x)f''(x)+4f(x)) dx$
, so as a the answer I get $f'(x)=0$ and f(x)=c where $c \in \mathbb{R}$ which does not seem right considering the condition that f(1)=0, f(2)=1
Did I make a mistake? Thanks in advance
You can use "Calculus of Variations" for this. Given an equation of the form: $$J[y]=\int_{x_1}^{x_2}L(x,f,f')dx$$ the minimum is given by: $$\frac{\partial L}{\partial f}-\frac{d}{dx}\frac{\partial L}{\partial f'}=0$$
In your case you have: $$L=x^2f'(x)^2+2f(x)^2$$ and so: $$\frac{\partial L}{\partial f}=4f$$ $$\frac{d}{dx}\frac{\partial L}{\partial f'}=\frac{d}{dx}\left(2x^2f'(x)\right)=4xf'(x)+2x^2f''(x)$$ now combing this we get: $$4f(x)-4xf'(x)-2x^2f''(x)=0$$
To solve this we can first divide through by two: $$x^2f''(x)+2xf'(x)-2f(x)=0$$ now a form of substitution like $x^n$ should make this easy to solve