2
$\begingroup$

I'm currently taking a course in Fluid Mechanics which I'm struggling quite a bit with. I don't have the best background in maths. In particular, I can't seem to deduce the following energy estimate for the Navier Stokes equation. If anyone could help me to understand where I'm going wrong, I would appreciate it.

Consider the incompressible Navier–Stokes equations on a bounded domain $\mathbb{T}_d$ with periodic boundary conditions. We know the Navier Stokes equations can be written as

$$\frac{\partial\mathbf{u}}{\partial t}+\mathbf{\omega}\times\mathbf{u}=\nu\Delta\mathbf{u}-\nabla(p+\frac{1}{2}|\mathbf{u}|^2)+\mathbf{f}$$

We can then find $\frac{d}{dt}\|\mathbf{u}\|^2_{L^2}$ and use Holder's inequality along with Young's and Poincare's inequality to deduce that

$$\frac{d}{dt}\|\mathbf{u}\|^2_{L^2}+2\nu\|\nabla\mathbf{u}\|_{L^2}^2\leq\delta\|\mathbf{u}\|_{L^2}^2+\frac{1}{\delta}\|\mathbf{f}\|_{L^2}^2$$

From here we can integrate the differential equation to find.

$$\|\mathbf{u}(\cdot,t)\|^2_{L^2}\leq \|\mathbf{u}(\cdot,0)\|^2_{L^2}\exp(-\nu t/c^2)+\bigg(\frac{c^2}{\nu}\bigg)^2\|\mathbf{f}(\cdot,t)\|_{L^2}^2(1-\exp(-\nu t/c^2))$$

But what I don't understand is that from here we can establish that for $T>0$

$$\mathbf{u}\in L^{\infty}([0,T];L^2(\mathbb{T}^d,\mathbb{R}^d))$$

Also, If we integrate the differential equation before applying Poincare's inequality we can find

$$\|\mathbf{u}(\cdot,t)\|^2_{L^2}+\frac{\nu}{c^2}\int_0^T\|\nabla\mathbf{u}(\cdot,\tau)\|_{L^2}^2d\tau\leq \nu \int_0^T\|\mathbf{u}(\cdot,\tau)\|_{L^2}^2d\tau+\frac{1}{\nu}\int_0^T\|\mathbf{f}(\cdot,\tau)\|_{L^2}^2d\tau$$

Again I don't follow why this means that for any $T>0$

$$\mathbf{u} \in L^2([0,T];H^1(\mathbb{T}^d,\mathbb{R}^d))$$ and thus

$$\mathbf{u} \in L^{\infty}([0,T];L^2(\mathbb{T}^d,\mathbb{R}^d))\cap L^2([0,T];H^1(\mathbb{T}^d,\mathbb{R}^d))$$

Many thanks.

$\endgroup$

1 Answer 1

2
$\begingroup$

First of all the equation that you writed is not the well-known Navier-Stokes equations since you included the term $\omega$, the term $\omega$ represent the vorticity and you get it when you take the curl of the velocity $u$. But, no problem let is return to the famous Navier Stokes equation which can be written as \begin{align} \frac{\partial u}{\partial t}+(u\cdot\nabla)u&= \Delta u -\nabla p\\ \nabla\cdot u &=0. \end{align} Before one pass to the energy inequality, one can get rid of the term $p$ which represent the pressure by applying the Helmholtz-Leray projection or by any other mathematical tool. Also, you have to know that there is a whole analysis one should make before talking about the energy estimates and here's a sketsh of this analysis: A global Lipschitz condition leads to global existence and uniqueness, whereas (since one have not a global Lipschitz condition), one needs to establish a local Lipschitz condition generally leads only to local existence and uniqueness. However, a local Lipschitz condition supplemented with some additional a priori bounds, if appropriate bounds exist and can be found, leads to global existence and uniqueness. And finally, without a local Lipschitz condition our construction of a continuous solution via Picard iteration is suspect and moreover, even if a continuous solution can be found by some other means, it is not necessarily unique. Now, since Navier-Stokes equation is not an ordinary differential equation but rather a PDE you have to approximate the equation by using as for example the Galerkin approximations sheme. Now, formally we return to the energy estimate:

To obtain the energy estimate you have to take the inner product in $L^2(\mathbb{T}^d)$ of the aformentioned Navier-Stokes equation against $u$, one infers that \begin{align} \langle\frac{\partial u}{\partial t}, u\rangle_{L^2}+\langle(u\cdot\nabla)u,u\rangle_{L^2}&= \langle\Delta u,u\rangle_{L^2}. \end{align} Now by using the fact that $\langle\frac{\partial u}{\partial t}, u\rangle_{L^2}=\frac{1}{2}\frac{d}{dt}\|u(t)\|_{L^2}^2$ and by integrating by parts $\langle\Delta u,u\rangle_{L^2}$, one obtains: \begin{align} \frac{d}{dt}\|u(t)\|_{L^2}^2+2\|\nabla u(t)\|_{L^2}^2\leq |\langle(u\cdot\nabla)u,u\rangle_{L^2}|. \end{align} As you mentioned in your question you can apply the Holder's and Young's product inequalities several times and use the cancellation law to get rid of the term $|\langle(u\cdot\nabla)u,u\rangle_{L^2}|$ to get the energy inequality \begin{align} \|u(s)\|_{L^2}^2+\int_0^s\|\nabla u(t)\|_{L^2}^2dt\leq \|u(0)\|_{L^2}^2. \end{align} The answer of your question consists in informing you that for any function $f\in \mathbb{R}_+\times\mathbb{T}^d$, one has: \begin{align} \|f\|_{L^\infty(\mathbb{R_+};L^2)}^2&=\sup\limits_{t\in \mathbb{R}_+}\|f(t)\|_{L^2}^2\\ \|f\|_{L^2(\mathbb{R_+};\dot{H}^1)}^2&=\int_{\mathbb{R}_+}\|\nabla f(t)\|_{L^2}^2dt. \end{align} Adding the information that $\|u(0)\|_{L^2}$ is constant. I hope that my answer was clear.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .