Given an ODE then regard it as an Euler-Lagrange equation, how to find a functional relative to it?

Question

Recently, I have been studied about Lp Minkowski problem. I met some confusion. The equation is an ODE like that $$u_{xx}+u=\frac{g(x)}{u^{p+1}}, \quad x\in\mathbb{R} , \quad p\geq 0.$$ $g(x)$ is a positive function and $u>0, u(k)=u(2\pi+k),k\in\mathbb{R}$ . Then we regard it as an Euler-lagrange equation of a functional. The papers gave this one: $$J[u]=(\int_0^{2\pi}\frac{g(x)}{u^p}dx)^{\frac{2}{p}}(\int_0^{2\pi}(u^2-u_x^2)dx).$$ In other word, a solution of this ODE is essentially a critical point of the functional. My question is how to find this functional? I think it is strange and maybe relative to eigenvalue of ODE? But why it has the power $\frac{2}{p}$? Thanks in advance.

Answer 1

1. OP's functional is of the form $$ J[u]~= F[u]^r G[u]^s, \qquad r,s~\in~\mathbb{R},\tag{1}$$ where $$ F[u]~:=~ \int_0^{2\pi}\!\mathrm{d}x(u^2-u_x^2) \qquad\text{and}\qquad G[u]~:=~ \int_0^{2\pi}\!\mathrm{d}x\frac{g(x)}{u^p}. \tag{2}$$

2. The functional derivatives are $$ \frac{\delta F[u]}{\delta u(x)}~=~2u(x)+2u_{xx}(x) \qquad\text{and}\qquad \frac{\delta G[u]}{\delta u(x)}~=~ -p \frac{g(x)}{u^{p+1}(x)}.\tag{3} $$

3. The sought-for functional derivative becomes$^1$ $$\begin{align} \frac{1}{J[u]}\frac{\delta J[u]}{\delta u(x)} ~=~& \frac{r}{F[u]}\frac{\delta F[u]}{\delta u(x)}+ \frac{s}{G[u]}\frac{\delta G[u]}{\delta u(x)}\cr ~=~&\frac{2r}{F[u]}(u(x)+u_{xx}(x))- \frac{sp}{G[u]}\frac{g(x)}{u^{p+1}(x)}\end{align}\tag{4}$$

4. The stationary condition for $J$ is hence an ODE of the form $$ u(x)+u_{xx}(x)~= k\frac{g(x)}{u^{p+1}(x)}, \qquad k~\in~\mathbb{R}\backslash\{0\}.\tag{5}$$

5. Multiplying the ODE (5) with u(x) and integrating over $x$ implies that $$F[u]~=~kG[u]\tag{6}$$ on-shell.

6. Moreover, if the powers satisfy $$ 2r~=~sp,\tag{7}$$ then the ODE (5) is a stationary path for $J$. This essentially answers OP's question about the power $\frac{2}{p}$.

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$^1$ Strictly speaking $J[u]$ could be zero, so the factor $\frac{1}{J[u]}$ on the LHS of eq. (4) is better kept as a factor $J[u]$ on the RHS of eq. (4). We leave it to the reader to improve this.