Understanding traveling waves as critical points of the constrained energy I am trying to understand a (straightforward I guess) statement of an old (but outstanding) paper on stability of solitary waves. Let us consider the following functionals: 
$$
V(u)=\dfrac{1}{2}\int_\mathbb{R} u^2dx \quad \hbox{and}\quad E(u)=\int_\mathbb{R}(\tfrac{1}{2}uMu-\tfrac{1}{2}u^2-F(u))dx.
$$
As you can guess, these functionals are conserved quantities of a certain one-dimensional PDE. This PDE has some special kind of solution called traveling wave solutions of speed $c$ (that is, $u(t,x):=\phi_c(x-ct)$ is a solution of the PDE). Note that the fixed profile also depends on $c$ (I mean, not only the speed of the wave but also the "scale" of the profile depends on $c$). Traveling waves of speed $c$ satisfy the following equation:
$$
\qquad \qquad M\phi_c+c\phi_c-\phi_c-f(\phi_c)=0. \qquad \qquad (*)
$$
Here $M$ denotes a non-specified differential operator and $f$ denotes the non-linearity (both of them quite general, but satisfying some regularity and growth assumptions). From now on we shall denote the profile $\phi_c$ just by $\phi$ and we will assume that $c>0$ is fixed. Then, it is clear from the $(*)$ that $\phi$ satisfies $E'(\phi)+cV'(\phi)=0$. Nevertheless, (here is where I got confused) the authors say that from this latter identity we can see $\phi$ as a critical point of $E$ subject to the constraint $V(u)=V(\phi)$. 
I don't really understand this last statement, in particular, it bothers me the fact that on the identity $E'(\phi)+cV'(\phi)=0$ there is a factor $c$ appearing, which I don't know how to connect with the statement "$\phi$ is a critical point of $E$ subject to the constraint $V(u)=V(\phi)$". Could anyone explain me how to see this equivalence? I am particularly confused because my knowledge of optimization on functionals is almost zero, I was thinking in a kind of "Lagrange multipliers" but I don't really understand how to implement this method for functionals.
PS: Sorry, I forgot to say, in the definition of $E(u)$ the function $F$ denotes a primitive of the nonlinearity $f(u)$, that is, $F$ satisfies that $F'(x)=f(x)$ and $F(0)=0$. If it helps, $f$ also satisfies that $f(0)=0$.
Edit: I understand that Lagrange multipliers have a $\lambda$ factor on the constraint, but on our equation we don't have any $\lambda$ but $c$, that is, exactly the speed of the traveling wave (that is specifically my doubt). My problem is that $c$ is not a parameter, it is FIXED (positive, arbitrary, but fixed). So it cannot play the role of $\lambda$ on the Lagrange multipliers method.
 A: The problem is precisely that you should not fix a priori the parameter $c$, because it will be exactly the Lagrange multiplier (or rather, $\lambda=-c$ will).
Think of it like this: Let me introduce a new parameter $R>0$, which I think of as a prescribed $L^2$ energy level.
Then the minimization problem
$$
\min\limits_{u\in(\dots)}\Bigg\{E(u):\qquad V(u)=R\Bigg\}
$$
is a constrained minimization problem. Presumably under structural assumptions on your operator $M$ this problem is well-behaved (convex, coercive, and what have you), hence there should be at least a solution $u$.
If you work out the Lagrange multipliers setting, the outcome is that the (Fréchet) gradient $E'(u)$ should be colinear (in $L^2$) with the gradient $V'(u)$. In other words, there is a real number $\lambda$ such that
$$
E'(u)=\lambda V'(u)
$$
Note carefully that I'm purposedly omitting the dependence of both $c,u$ on $R$ (and in fact they need not be unique, more on this later).
Writing $c:=-\lambda$ and given your explicit expressions of $V,E$ we get the PDE
$$
E'(u)+cV'(u)=0
\qquad 
\Leftrightarrow\qquad
Mu-u-f(u)+cu=0.
$$
Usually your PDE is of the form $\partial_tu=-Mu+u+f(u)$, hence you can interpret $-cu$ as a convective derivative $\partial_t(u(x-ct))$ and you get indeed a travelling wave.
For any "$V$-energy" level $R>0$ that you chose you get solutions $(u_c,c)$, the typical case being that a value of $R$ gives rise a finitely many solutions. In your partcular case, varying continuously $R>0$ is what gives the whole family of travelling waves.
NB I am fully aware that this is just a purely heuristic explanation, but given that the OP is not very specific I cannot really "prove" anything here.
Note on the non-uniqueness: very often the orientation $c>0$, corresponding to waves moving to the right $u(x-ct)$, is completely arbitrary. In this case one usually also gets left-moving waves $u(x+ct)$, so for any solution $(c,u_c)$ you get another (reflected) solution $(-c,u_{-c})$, which means that the pairs of (constrained minimizers, Lagrange multipliers) are not unique. But this strongly depends on the operator $M$ (and its invariance under symmetries such as $x\to-x$).
PS: sorry for using $u$ instead of $\phi$, but I guess old habits die hard.
