André Schlichting

Pressure as Lagrange multiplicator in Stokes and Euler equation

8 January 2013

Navier-Stokes equation in d=2,3d=2,3
u:[0,∞)×Ω→Rdvelocity (unknown) length / time p:[0,∞)×Ω→Rdpressure (unknown) force / area f:[0,∞)×Ω→Rdexternal volume force (data) force / volume ϱ∈[0,∞)mass density (data) mass / volume μ∈(0,∞)dynamic viscosity (data) pressure ⋅ time  \begin{aligned} u &: [0,\infty) \times \Omega \to \R^d &\text{velocity (unknown)} & \text{ length / time } \\ p &: [0,\infty) \times \Omega \to \R^d &\text{pressure (unknown)} & \text{ force / area } \\ f &: [0,\infty) \times \Omega \to \R^d &\text{external volume force (data)} & \text{ force / volume } \\ \varrho &\in [0,\infty) &\text{mass density (data)} & \text{ mass / volume } \\ \mu&\in (0,\infty) &\text{dynamic viscosity (data)} & \text{ pressure \(\cdot\) time } \end{aligned}

ϱ(∂tu+u⋅∇u)−μΔu+∇p=fin (0,∞)×Ω,∇⋅u=0in (0,∞)×Ω,u=0on (0,∞)×∂Ω\begin{aligned} \varrho \left( \partial_t u + u \cdot \nabla u \right) - \mu \Delta u + \nabla p &= f &&\text{in \((0,\infty)\times \Omega\)},\\ \nabla \cdot u &= 0 &&\text{in \((0,\infty)\times \Omega\)},\\ u &= 0 &&\text{on \((0,\infty)\times \partial \Omega\)} \end{aligned}

and initial conditions u(t=0,⋅)=u0(⋅)u(t=0,\cdot)=u_0(\cdot) .

Claim: The pressure pp can be interpreted as Lagarange multiplicator that comes from incompressibility constraint ∇⋅u=0\nabla \cdot u=0 .

Basis: Lagrange multiplicator are associated to variational problem.

Problem: Navier-Stokes equation has no known variational structure.

Alternative: Consider special cases: Stokes and Euler equation, which have a variational structure.

Claim 1 (Stokes equation)

The Stokes equation is the limit equation coming from the Navier-Stokes equation for ϱ→0\varrho\to 0

−μ∇u+∇p=fin Ω∇⋅u=0in Ωu=0on ∂Ω.\begin{aligned} -\mu \nabla u +\nabla p &= f && \text{in \(\Omega\)}\\ \nabla \cdot u &= 0 && \text{in \(\Omega\)}\\ u &= 0 && \text{on \(\partial \Omega\)}. \end{aligned}

Then the Pressure pp is the Lagrange multiplicator coming from the incompressibility constraint.

Claim 2 (Euler equation)

The Euler equation is the limit equation coming from the Navier-Stokes equation for μ→0\mu\to 0 and f≡0f\equiv 0

ϱ(∂tu+u⋅∇u)+∇p=0in Ω∇⋅u=0in Ωu⋅n=0on ∂Ω.\begin{aligned} \varrho\left(\partial_t u +u\cdot \nabla u\right) +\nabla p &= 0 && \text{in \(\Omega\)}\\ \nabla \cdot u &= 0 && \text{in \(\Omega\)}\\ u\cdot n &= 0 && \text{on \(\partial \Omega\)}. \end{aligned}

Then the Pressure pp is the Lagrange multiplicator coming from the incompressibility constraint.

To proof both claims, we first appeal to the following auxilliary result:

Auxilliary Lemma (Dirichlet boundary conditions)

Suppose g:Ω→Rdg :\Omega \to \R^d satisfies

∫Ωv⋅g=0∀v:Ωˉ→Rd with {∇⋅v=0in Ω,v=0on ∂Ω. \int_\Omega v\cdot g = 0 \quad \forall v: \bar \Omega\to \R^d \text{ with } \begin{cases} \nabla \cdot v &= 0 &\text{in \(\Omega\)},\\ v &= 0 &\text{on \(\partial \Omega\)}. \end{cases}

Then there exists p:Ω→Rp:\Omega \to \R such that g=−∇pg= - \nabla p .

Proof of Claim 1

Variational principle behind the Stokes equation: The stokes equation is the Euler-Lagrange equation of the minimazation problem

min⁡u12∫μ∣∇u∣2−∫f⋅u, \min_u \frac{1}{2} \int \mu |\nabla u|^2 - \int f \cdot u ,

where the minimum is taken among all

u:Ωˉ→Rd with {∇⋅u=0in Ω,u=0on ∂Ω. u: \bar \Omega\to \R^d \text{ with } \begin{cases} \nabla \cdot u &= 0 &\text{in \(\Omega\)},\\ u &= 0 &\text{on \(\partial \Omega\)}. \end{cases}

Let us calculate the first variation of the functional in the minimization problem. Therefore take v:Ωˉ→Rdv:\bar \Omega\to \R^d incompressible with zero Dirichlet boundary values, then we have

0=∫μ∇u:∇v−∫f⋅v=−∫(μΔu+f)⋅v. 0 = \int \mu \nabla u : \nabla v - \int f \cdot v = - \int \left(\mu \Delta u + f\right) \cdot v .

results
By the previous Lemma, there exists p:Ω→Rp:\Omega\to \R such that

−(μΔu+f)=−∇pin Ω, - \left(\mu \Delta u + f\right) = -\nabla p \quad \text{in \(\Omega\)} ,

whch is nothing else than the Stokes equation.

For the proof of Claim 2, we use Arnolds interpretation of the Euler equation. For simplicity we restrict to the case Ω=Rd\Omega=\R^d .

Arnolds interpretation of the Euler equation

uu is a solution of the Euler equation if and only if the associated flow Φ:(0,T)×Rd→Rd\Phi : (0,T)\times \R^d \to \R^d given by

∂tΦ(t,y)=u(t,Φ(t,y)),Φ(0,y)=y \partial_t \Phi(t,y) = u(t,\Phi(t,y)) , \quad \Phi(0,y)=y

is a stationary point of the action functional

∫0T∫Rdϱ2∣∂tΦ(t,y)∣2  dy  dt \int_0^T \int_{\R^d} \frac{\varrho}{2} |\partial_t \Phi(t,y) |^2 \;\text{d}y\;\text{d}t

subject to Φ(t=0,⋅)=Id\Phi(t=0,\cdot)=\text{Id} , Φ(t=T,⋅)=ΦT(⋅)\Phi(t=T,\cdot)=\Phi_T(\cdot) and

det⁡DΦ(t,y)=1∀(t,y)∈(0,T)×Rd. \det D\Phi(t,y) = 1 \quad \forall (t,y)\in (0,T)\times \R^d .

For the proof we again an auxillairy result, but with different boundary conditions.

Auxilliary Lemma (Neumann boundary conditions)

Suppose g:Ω→Rdg :\Omega \to \R^d satisfies

∫Ωv⋅g=0∀v:Ωˉ→Rd with {∇⋅v=0in Ω,v⋅n=0on ∂Ω. \int_\Omega v\cdot g = 0 \quad \forall v: \bar \Omega\to \R^d \text{ with } \begin{cases} \nabla \cdot v &= 0 &\text{in \(\Omega\)},\\ v \cdot n &= 0 &\text{on \(\partial \Omega\)}. \end{cases}

Then there exists p:Ω→Rp:\Omega \to \R such that g=−∇pg= - \nabla p .

Proof of Claim 2

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 uu satisfies the Euler equation if and only if

∀t∈[0,T]:∫(∂tu+u⋅∇u)⋅v  dx=0∀v:Ω→Rd with {∇⋅v=0in Ω,v⋅n=0on ∂Ω. \forall t\in[0,T] : \int (\partial_t u + u\cdot \nabla u) \cdot v \; \text{d}x = 0 \qquad \forall v:\Omega\to \R^d \quad\text{ with }\quad \begin{cases} \nabla \cdot v &= 0 &\text{in \(\Omega\)},\\ v \cdot n &= 0 &\text{on \(\partial \Omega\)}. \end{cases}

Furthermore, this characterization is equivalent to the time-integrated one

∫0T∫Ω(∂tu+u⋅∇u)⋅v  dx  dt=0∀v:[0,T]×Ω→Rd with {∇⋅v=0in Ω,v⋅n=0on ∂Ω,v(t=0,T,⋅)=0in Ω. \int_0^T \int_\Omega (\partial_t u + u\cdot \nabla u) \cdot v \; \text{d}x\; \text{d}t = 0 \quad \forall v:[0,T]\times \Omega\to \R^d \quad\text{ with } \quad\begin{cases} \nabla \cdot v &= 0 &\text{in \(\Omega\)},\\ v \cdot n &= 0 &\text{on \(\partial \Omega\)}, \\ v(t=0,T,\cdot) &= 0 &\text{in \(\Omega\)}. \end{cases}

The boundary conditions allow for easy integratation by parts. Let us first investigate the nonlinear term

∫Ω(u⋅∇u)⋅v=∫∫ujui,jvi=−∫Ω(ujvi),jui=−∫Ωuj,jviui+ujvi,jui=−∫Ω(∇⋅u)⏟=0(v⋅u)+(u⋅∇v)⋅u=−∫Ω(u⋅∇v)⋅u,\begin{split} \int_\Omega (u\cdot \nabla u) \cdot v &= \int\int u_j u_{i,j} v_i = - \int_\Omega (u_j v_i)_{,j} u_i = -\int_\Omega u_{j,j} v_i u_i + u_j v_{i,j} u_i \\ &= -\int_\Omega \underbrace{(\nabla \cdot u)}_{=0} ( v \cdot u) + (u\cdot \nabla v)\cdot u = -\int_\Omega (u\cdot \nabla v)\cdot u , \end{split}

where we used the incompressibility condition. Hence, together with the term from the time derivative we arrive at

0=∫0T∫Ω(∂tu+u⋅∇u)∇v  dx  dt=−∫0T∫Ωu⋅(∂tv+u⋅∇v)  dx  dt. 0 = \int_0^T \int_\Omega (\partial_t u + u\cdot\nabla u)\nabla v \;\text{d}x\;\text{d}t = - \int_0^T \int_\Omega u\cdot \left(\partial_t v +u\cdot \nabla v\right) \;\text{d}x\;\text{d}t .

This we call the weak form of the Euler equation.

Step 2: From Arnold to Euler.
Let v:[0,T]×Ω→Rdv: [0,T]\times \Omega\to \R^d be an admissible test vectorfield, i.e. incompressible, Neumann boundary and vanishing for t=0,Tt=0,T . Therewith, we want to construct a variation Φ~(s,t,y)\tilde \Phi(s,t,y) of Φ(t,y)\Phi(t,y) . Therefore, we first define the flow Ψ(s,t,z)\Psi(s,t,z) generated by v(t,z)v(t,z) as usual

∂sΨ(s,t,z)=v(t,Ψ(s,t,z)andΨ(0,t,z)=z. \partial_s \Psi(s,t,z) = v(t,\Psi(s,t,z) \quad\text{and}\quad \Psi(0,t,z)=z .

Therewith we set

Φ~(s,t,y):=Ψ(s,t,Φ(t,y)). \tilde \Phi(s,t,y) := \Psi\left(s,t,\Phi(t,y)\right) .

Let us check, that this is an admissible variation

  • Φ~(0,t,y)=Φ(t,y)\tilde \Phi(0,t,y)= \Phi(t,y) follows from Ψ(0,t,z)=z\Psi(0,t,z)=z .
  • Φ~(s,t={0,T},y)=Φ(t={0,T},y)\tilde \Phi(s,t=\{0,T\},y)= \Phi(t=\{0,T\},y) is a consequence of Ψ(s,t={0,T},z)=z\Psi(s,t=\{0,T\},z)=z following from v(t={0,T},⋅)=0v(t=\{0,T\},\cdot)=0 .
  • det⁡DΦ~(s,t,y)=1\det D\tilde \Phi(s,t,y)=1 is a consequence of DΨ(s,t,z)=1D\Psi(s,t,z)=1 following from ∇⋅v=0\nabla \cdot v=0 .

Hence Φ~\tilde \Phi is an admissible variation and therefore by assumption that Φ\Phi is a stationary point, we obtain

0=dds∣s=0∫0T∫Ωϱ2∣∂tΦ~(s,t,y)∣2  dy  dt=∫0T∫Ωϱ∂tΦ~(0,t,y)⋅∂t∂s∣s=0Φ~(s,t,y)  dy  dt.\begin{split} 0&= \left.\frac{\text{d}}{\text{d}s}\right|_{s=0} \int_0^T \int_\Omega \frac{\varrho}{2} |\partial_t \tilde \Phi(s,t,y)|^2 \;\text{d}y\;\text{d}t \\ &= \int_0^T \int_\Omega \varrho \partial_t \tilde\Phi(0,t,y) \cdot \partial_t \partial_s|_{s=0} \tilde \Phi(s,t,y)\;\text{d}y\;\text{d}t . \end{split}

The first observation is that ∂tΦ~(0,t,y)=∂tΦ(t,y)=u(t,Φ(t,y))\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

∂t∂s∣s=0Φ~(s,t,y)=∂t∂s∣s=0Ψ(s,t,Φ(t,y))=ddtv(t,Ψ(0,t,Φ(t,y)))=ddtv(t,Φ(t,y))=∂tv(t,Φ(t,y)+Dv(t,Φ(t,y)) ∂tΦ(t,y)=∂tv(t,Φ(t,y))+Dv(t,Φ(t,y)) u(t,Φ(t,y)). \begin{split} \partial_t \partial_s|_{s=0} \tilde \Phi(s,t,y) &= \partial_t \partial_s|_{s=0} \Psi(s,t,\Phi(t,y)) = \frac{\text{d}}{\text{d}t} v\left(t,\Psi(0,t,\Phi(t,y))\right) \\ &= \frac{\text{d}}{\text{d}t} v(t,\Phi(t,y)) = \partial_t v(t,\Phi(t,y) + D v(t,\Phi(t,y)) \ \partial_t \Phi(t,y) \\ &= \partial_t v(t,\Phi(t,y)) + D v(t,\Phi(t,y)) \ u(t,\Phi(t,y)) . \end{split}

Substituting everything back into the Euler-Lagrange equation of the Arnold functional leads to

0=∫0T∫Ω(u⋅(∂v+Dv u))(t,Φ(t,y))  dy  dt=∫0T∫Ωu⋅(∂v+u⋅∇v)(t,x)  dx  dt,\begin{split} 0 &= \int_0^T \int_\Omega \left(u\cdot (\partial v + Dv\ u)\right)(t,\Phi(t,y)) \;\text{d}y\;\text{d}t \\ &=\int_0^T \int_\Omega u\cdot(\partial v + u\cdot \nabla v)(t,x)\;\text{d}x\;\text{d}t , \end{split}

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.