# Pressure as Lagrange multiplicator in Stokes and Euler equation

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Navier-Stokes equation in (d=2,3)
\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} \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 $$u(t=0,\cdot)=u_0(\cdot)$$.

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 known variational structure.

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

# weak L¹-convergence

$$\newcommand{\R}{\mathbb{R}} \newcommand{\dx}{\mathrm{d}}$$ Let $$\Omega\subset \mathbb{R}^n$$, a sequence $$(u_n)$$ converges weakly to $$u\in L^p(\Omega)$$ if $\int_\Omega u_n v\;\dx{x} = \int_\Omega u v \;\dx{x} , \qquad \forall v\in L^\infty(\Omega) .$ As usual the convergence is denoted by $$u_n \rightharpoonup u$$ in $$L^1(\Omega)$$. Continue reading

# Classic results for porous medium equation

### Selfsimilar solutions

$$\newcommand{\R}{\mathbb{R}} \newcommand{\dx}{\mathrm{d}}$$ Let us seek for solutions of $$\partial \varrho = \Delta \varrho^m$$ satisfying the scaling hypothesis $(x,t)\mapsto (y,\tau):=\left(\frac{x}{s(t)},\tau(t)\right) \qquad \varrho(x,t) = \frac{1}{s(t)^N} u\left(\frac{x}{s(t)},\tau(t)\right),$ where $$s: \R_+ \to \R_+$$ and $$\tau:\R_+ \to \R_+$$ are reparametrization of time and $$u:\R_+\times \R^N \to \R$$. The prefactor $$\frac{1}{s(t)^N}$$ ensures that $$u$$ conserves mass, i.e. $1 = \int_{\R^N} \varrho(x,t) \; \dx{x} \stackrel{x\mapsto ys}{=} \int_{\R^N} \frac{1}{s(t)^N} \varrho(y s(t),t)\;\dx{y} = \int_{\R^N} u(y,\tau(t)) \; \dx{y}.$ Continue reading

# Fokker Planck equation

### Description

$$\newcommand{\R}{\mathbb{R}} \newcommand{\dx}{\mathrm{d}}$$ The Fokker Planck equation has the form $\partial_t \varrho(x,t) = \nabla \cdot \bigl(\beta^{-1} \nabla \varrho(x,t) + \varrho(x,t) \nabla H(x)\bigr) ,\qquad \varrho(x,0)=\varrho_0(x) ,$ where $$H:\R^n\to \R$$ is a smooth function, $$\beta>0$$ some parameter and $$\varrho_0$$ a probability density on $$\R^n$$. The partial differential equation is in divergence form and conserves mass. Hence, also $$\varrho(\cdot,t)$$ is a probability density on $$\R^n$$. In the case, where $$H$$ has some growth at $$\infty$$, the equilibrium solutions $$\varrho_\infty:\R^n \to \R$$ are characterized by $\beta \nabla \varrho_\infty + \varrho_\infty \nabla H =0$ leading to the solution $\varrho_\infty(x) = Z^{-1} e^{-\beta H(x)} \quad \text{with} \quad Z = \int e^{-\beta H(x)} \; \dx x$ Continue reading