My notes motivate Lax-Milgram theorem by using the membrane problem.
We take $E[u] = \int_{\Omega} \Big(\frac \sigma 2|\nabla u|^2+\frac \alpha 2u^2-f(x)u\Big)dx$ the functional associated to the problem and we make variations on it. So we define $g(s) = E[u+s\phi]$ with $\phi \in H_0^1(\Omega)$ and $u$ a potential minimum of $E$. Then $g'(0) = 0$ and $\int_{\Omega} (\sigma \nabla u \nabla \phi + \alpha u \phi - f \phi) dx = 0$. Then, my notes claim that $\int_{\Omega} (- \Delta u +\alpha v -f)\phi dx = 0$.
I don't understand this last equality. I have tried to do it with integration by parts. But if I write $\nabla u \nabla \phi = \frac{du}{dt}\frac{d\phi}{dt}+\frac{du}{dx}\frac{d\phi}{dx}$ the argument works well with the second part but for the first part I have to integrate $\int \frac{d\phi}{dt}dx$ and I get the derivative $\frac{du}{dxdt}$ not a addend of the laplacian.
When does the integration make sense?
As suggested by @Uskebasi, I was not integrating with respect to $t$.
Now my question is, do I have to impose anything to $\Omega$ so that the integration makes sense?
I treated a general setting where $\Omega$ was a bounded and "regular" domain of $\mathbb{R}^2$ (this is the case my notes seem to refer) and a more specific case where $x \in [0,L]$ and $t \in [0,\infty[$. In this case I wonder if the integration still makes sense.
For time-dependent problems, it really depends on how you define your functional $E$. There seems to have two different approaches: the classical approach where you treat space and time separately, and the modern approach where you treat space-time as a whole.
For the classical approach, your functional $E$ is no longer the potential energy, but action instead. Take your membrane $u=u(t,x)$ for instance. Its total potential energy reads $$ E_p[u]=\int_{x\in\left[0,L\right]}\left(\frac{\sigma}{2}\left|\frac{\partial u}{\partial x}\right|^2+\frac{\alpha}{2}\left|u\right|^2-fu\right){\rm d}x, $$ whereas its kinetic energy reads $$ E_k[u]=\int_{x\in\left[0,L\right]}\frac{\rho}{2}\left|\frac{\partial u}{\partial t}\right|^2{\rm d}x, $$ where $\rho$ stands for the surface mass density of this membrane. Obviously, this kinetic energy expression is exactly in analogy with $\left(1/2\right)mv^2$.
Now, there is a principle of least action in classical physics, stating that the physically true behavior of $u$ is such that the time integral of the kinetic energy minus the total potential is stationary, i.e., $$ J[u]=\int_{t\in\left[0,\infty\right)}\left(E_k[u]-E_p[u]\right){\rm d}t $$ is locally minimized. Thus in this approach, you need to make use of this $J$ instead of $E$, and Lax-Milgram would lead to $$ \left(\frac{\rm d}{{\rm d}s}J[u+s\phi]\right)\Bigg|_{s=0}=0. $$ You may check that this yields the desired $$ \rho\frac{\partial^2u}{\partial t^2}=\sigma\frac{\partial^2u}{\partial x^2}-\alpha u+f. $$
Alternatively, you may choose the modern approach by treating $\Omega\subseteq\mathbb{R}^2$ as $$ \left(t,x\right)\in\Omega=\left[0,\infty\right)\times\left[0,L\right]. $$ You may also put the coordinate as $\left(x,t\right)$. The order of time $t$ and space $x$ in the tuple does not make any difference. In this case, however, you should be very careful about the definition of a "vector" in space-time. You need to adopt the Lorentz metric instead of the standard Euclidean metric, i.e., $$ \left\|\left(T,X\right)\right\|^2=-T^2+X^2, $$ or in the three-dimensional-space case, $$ \left\|\left(T,X,Y,Z\right)\right\|^2=-T^2+X^2+Y^2+Z^2. $$ You may put $c$, the speed of light, in front of $T$ as well; mathematically the expressions read the same up to a coupling constant.
Now, get back to your problem, and $$ \nabla u:=\left(\sqrt{\rho}\frac{\partial u}{\partial t},\sqrt{\sigma}\frac{\partial u}{\partial x}\right) $$ becomes a "vector" in your space-time (with only one dimension in space). In this case, you could similarly define $$ I[u]=\int_{\Omega}\left(\frac{1}{2}\left\|\nabla u\right\|^2+\frac{\alpha}{2}u^2-fu\right){\rm d}V, $$ where the norm $\left\|\cdot\right\|$ follows the definition of Lorentz metric.
You could check that $$ I[u]=-J[u], $$ for which these two approaches are somewhat equivalent. Nevertheless, as time gets involved, it would be hard to say if the functional is minimized or maximized. And this is why I put it in above as that this functional is stationary.