I have tried to solve the following problem with Lagrange multipliers, but I have trouble getting started.
I want to minimize $$F(\alpha) = \frac12\int_0^1 [u(x) - 1]^2\,dx$$ with the constraints $f(x) = -u''(x), \; u(0) = 0, \; u'(1) = \alpha$.
$u(x)$ is the solution of $\frac{d^2u(x)}{dx^2} = 0,$ so $u(x) = \alpha x$.
I don't know if I need this.
Does anyone know how to solve this problem? Thank you!
Off-hand, I'd say you could just do \begin{align*}\frac12\int_0^1[\alpha x-1]^2\,dx&=\frac12\int_0^1[\alpha^2 x^2-2\alpha x+1]\,dx \\ &=\frac12\left[\frac{\alpha^2 x^3}{3}-\alpha x^2+x\right]_0^1 \\ &=\frac12\left[\frac{\alpha^2}{3}-\alpha+1\right]. \end{align*} You can complete the square or take the derivative (I opt for the latter): $$\frac{dF(\alpha)}{d\alpha}=\frac{\alpha}{3}-\frac12\overset{\text{set}}{=}0\quad\implies\quad \alpha=\frac32\quad\implies\quad F(3/2)=\frac12\left[\frac34-\frac32+1\right]=\frac18.$$ But this completely ignores the condition $f(x)=-u''(x).$ For this condition to be consistent with your other conditions will require that $f(x)=0.$