Weak formulation, Variational formulation, Solution of a PDE.

Question

Ok, I'm struggling with some basic stuff. My question is: given a PDE are the concepts of a variational formulation and weak formulation the same?

Take a PDE (letting $\phi : \mathbb{R}\to \mathbb{R}$ say convex) $$ \partial_t u(t,x)=-\nabla_x \phi(u(t,x)),\quad u(0,x)=u_{(0)}(x)\label{1}\tag{1} $$ A strong solution $u:([0,T] \times \mathbb{R}^d) \to \mathbb{R}$ satisfies the above equation for all $t$ and $x$.

Upon multiplication by a test function $\psi$ and integration of \eqref{1} and moving the derivatives onto $\psi$ via integration by parts one can obtain the weak formulation.

$\textbf{Question :}$ My question is sometimes instead of writing down the weak formulation of a PDE an author will claim it has an associated variational formulation, for instance see this book chapter, where Eq. $(1)$ is the PDE, and eq. $(5)$ is its variational formulation. What does a variational formulation mean? And where does it come from?

Answer 1

a sequence x_n in hilbert space is said to converge in the weak sense if for every y in space <x_n,y> -> <x,y> strong convergence means

||x_n-x|| -> 0

strong implies weak, but not the other way around. take for example: the space l_2 of sequences which are bounded in norm. take the sequence of x_n = 1 on the nth place and 0 otherwise. then this sequence is not even cauchy.

||x_n-x_m|| = sqrt(2)

for every n,m.

so it has no convergent subsequence.

but, take <x_n-x_m,y> for all y in l_2: cauchy-schwartz tells us:

<x_n-x_m,y>  <  ||x_n-x_m||•||y|| = sqrt(2)||y||  -> 0

as y is a bounded sequence in l_2.

so there is a weakly convergent subsequence.

variational is exactly the weak sense convergence.