Find stationary point of sum of the square of integral

Question

Let $x \in [0, 1]^{10}$ be a ten dimensional vector. I want to maximize the functional with the form: $$G(f) = (\int_{x} f(x) x_1 dx)^2 + (\int_{x} f(x) e^{x_2} dx)^2$$ with constraints $$\int_{x} f(x) dx = 0 \text{ and } \int_{x} f^2(x) dx = 1,$$ where $x_1, x_2$ are the first and second element of $x$. The problem is easy to solve if $G(f) = (\int_{x} f(x) x_1 dx)^2$, since we can reduce that problem to maximize $T(f)=\int_{x} f(x) x_1 dx$ with the same constraints. Note that $T(f)$ has standard form in calculus of variation. We can solve it with Euler-Lagrange method and the maximizer is $f(x) = \sqrt{12}(x_1 - 1/2)$. Any hints or reference for this problem? thanks.

Answer 1

Hint: Using the method of Lagrange multipliers, you can replace the problem of maximizing $G[y]$ subject to the given constraints with the problem of finding the stationary points of the the modified functional $(\Omega=[0,1]^{10})$ $$ \tilde{G}[f;\lambda,\mu]=G[f]+\lambda\int_{\Omega}f(x)\,dx +\mu\left(\int_{\Omega}f^2(x)\,dx-1\right). \tag{1} $$ Thus \begin{align} \frac{\delta\tilde{G}}{\delta f(y)}=0 &\implies 2Ay_1+2Be^{y_2}+\lambda+2\mu f(y)=0 \\ &\implies f(y)=-\frac{1}{2\mu}\left(2Ay_1+2Be^{y_2}+\lambda\right)\quad(\mu\neq 0), \tag{2} \end{align} where $$ A:=\int_{\Omega}f(x)x_1\,dx, \qquad B:=\int_{\Omega}f(x)e^{x_2}\,dx. \tag{3} $$ To determine the constants $\lambda, \mu, A$ and $B$, you need four conditions. They are provided by the original constraints: $$ \frac{\partial \tilde{G}}{\partial \lambda}=0\implies \int_{\Omega}f(x)\,dx=0 \implies \int_{\Omega}(2Ax_1+2Be^{x_2}+\lambda)\,dx=0, \tag{4} $$ $$ \frac{\partial \tilde{G}}{\partial \mu}=0\implies \int_{\Omega}f^2(x)\,dx=1 \implies \int_{\Omega}(2Ax_1+2Be^{x_2}+\lambda)^2\,dx=4\mu^2, \tag{5} $$ and by the self-consistency conditions $(3)$: $$ \int_{\Omega}f(x)x_1\,dx=A \implies \int_{\Omega}(2Ax_1+2Be^{x_2}+\lambda)x_1\,dx=-2\mu A, \tag{6} $$ $$ \int_{\Omega}f(x)e^{x_2}\,dx=B \implies \int_{\Omega}(2Ax_1+2Be^{x_2}+\lambda)e^{x_2}\,dx=-2\mu B. \tag{7} $$ Now all you have to do is to solve the system of equations $(4)$-$(7)$.