and initial conditions $u(t=0,\cdot)=u_0(\cdot)$.
Claim: The pressure $p$ can be interpreted as Lagarange multiplicator that comes from incompressibility constraint $\nabla \cdot u=0$.
Basis: Lagrange multiplicator are associated to variational problem.
Problem: Navier-Stokes equation has no variational structure.
Alternative: Consider special cases: Stokes and Euler equation, which have a variational structure.
Then the Pressure $p$ is the Lagrange multiplicator coming from the incompressibility constraint.
Then the Pressure $p$ is the Lagrange multiplicator coming from the incompressibility constraint.
Then there exists $p:\Omega \to \R$ such that $g= - \nabla p$.
where the minimum is taken among all
Let us calculate the first variation of the functional in the minimization problem. Therefore take $v:\bar \Omega\to \R^d$ incompressible with zero Dirichlet boundary values, then we have
results By the previous Lemma, there exists $p:\Omega\to \R$ such that
whch is nothing else than the Stokes equation.
is a stationary point of the action functional
subject to $\Phi(t=0,\cdot)=\text{Id}$, $\Phi(t=T,\cdot)=\Phi_T(\cdot)$ and
Then there exists $p:\Omega \to \R$ such that $g= - \nabla p$.
Step 1: Weak form of the euler equation. We reformulate the Euler equation by starting from the auxilliary Lemma with Neumann boundary conditions. With the same reasoning as in the proof of the Claim for the Stokes equation, we obtain that $u$ satisfies the Euler equation if and only if
Furthermore, this characterization is equivalent to the time-integrated one
The boundary conditions allow for easy integratation by parts. Let us first investigate the nonlinear term
where we used the incompressibility condition. Hence, together with the term from the time derivative we arrive at
This we call the weak form of the Euler equation.
Step 2: From Arnold to Euler. Let $v: [0,T]\times \Omega\to \R^d$ be an admissible test vectorfield, i.e. incompressible, Neumann boundary and vanishing for $t=0,T$. Therewith, we want to construct a variation $\tilde \Phi(s,t,y)$ of $\Phi(t,y)$. Therefore, we first define the flow $\Psi(s,t,z)$ generated by $v(t,z)$ as usual
Therewith we set
Let us check, that this is an admissible variation
- $\tilde \Phi(0,t,y)= Phi(t,y)$ follows from $\Psi(0,t,z)=z$.
- $\tilde \Phi(s,t=\{0,T\},y)= \Phi(t=\{0,T\},y)$ is a consequence of $\Psi(s,t=\{0,T\},z)=z$ following from $v(t=\{0,T\},\cdot)=0$.
- $\det D\tilde \Phi(s,t,y)=1$ is a consequence of $D\Psi(s,t,z)=1$ following from $\nabla \cdot v=0$.
The first observation is that $\partial_t \tilde\Phi(0,t,y)=\partial_t \Phi(t,y)=u(t,\Phi(t,y))$. Let us calculate the second factor in the scalar product
Substituting everything back into the Euler-Lagrange equation of the Arnold functional leads to
which is nothing else than the weak form of the Euler equation derived in step 1.
Reference: Lecture notes: PDE's and applied analysis, F. Otto, Fall 2010 MPI MIS Leipzig.