Euler–Lagrange equation

Question

Does this P.D.E: $$\nabla\cdot\left( \frac{ \nabla u}{u} \right)+a\, \Delta u+b\,u=0 \hspace{3cm} (*)$$ have a variational structure? Here $a$ and $b$ are constants.

In other words, the question I am asking is:

Does there exist a functional such that the corresponding Euler–Lagrange equation is (*)?

Answer 1

I)

Answer: Yes, choose functional $$\tag{1} S[u]~:=~\int_{\Omega\subseteq \mathbb{R}^n} d^nx~{\cal L}(u,\nabla u),\qquad d^nx~:=~dx_1 \wedge \ldots \wedge dx_n,$$
where $$\tag{2} {\cal L}(u,\nabla u)~:=~\frac{1}{2}\left(a+\frac{1}{u}\right)^2(\nabla u)^2 -\frac{ab}{2}u^2 -bu. $$

Sketched proof: The functional derivative is

$$ \frac{\delta S[u]}{\delta u} ~=~\frac{\partial {\cal L}(u,\nabla u)}{\partial u} -\nabla \cdot \frac{\partial {\cal L}(u,\nabla u)}{\partial (\nabla u)}$$ $$\tag{3}~=~-\left(a+\frac{1}{u}\right) \left[-\frac{1}{u^2}(\nabla u)^2 +\left(a+\frac{1}{u}\right)\nabla^2u +bu\right],$$

which leads to the Euler-Lagrange equation

$$\tag{4}\nabla\cdot\left(\frac{1}{u}\nabla u\right) +a\nabla^2u +bu~=~0\quad \vee\quad u~=~-\frac{1}{a}.$$

II)

More generally, let us for fun consider the normalized equation $$\tag{5} \nabla^2u+ f(u)(\nabla u)^2+g(u)~=~0, $$ where $f$ and $g$ are given functions of $u$. The question is now whether we can find a variational principle for eq. (5)? The answer is again Yes.

Let $F$ and ${\cal V}$ be antiderivatives (aka. primitive or indefinite integrals) of $f$ and $ge^{2F}$, respectively, i.e.

$$\tag{6}F^{\prime}(u)~=~f(u) \quad\text{and}\quad {\cal V}^{\prime}(u)~=~g(u)e^{2F(u)}.$$

We now multiply the normalized equation (5) with an integrating factor

$$\tag{7}\lambda(u)~:=~ e^{2F(u)}. $$

Define the Lagrangian density $$\tag{8}{\cal L}(u,\nabla u) ~:=~\frac{1}{2}e^{2F(u)}(\nabla u)^2-{\cal V}(u).$$ It is straightforward to show that the Euler-Lagrange equation for the Lagrangian density (8) is eq. (5).

Sketched proof:

$$ \frac{\delta S[u]}{\delta u} ~=~\frac{\partial {\cal L}(u,\nabla u)}{\partial u} -\nabla \cdot \frac{\partial {\cal L}(u,\nabla u)}{\partial (\nabla u)}$$ $$\tag{9}~=~-e^{2F(u)}\left[ \nabla^2u+ f(u)(\nabla u)^2+g(u) \right].$$