Existence and uniqueness of stationary Navier-Stokes equation

By | January 8, 2018

Let \Omega be a bounded domain in \mathbb{R}^2 or \mathbb{R}^3. We consider the Dirichlet boundary value problem for the stationary Navier-Stokes equation:

    \begin{align*} -\nu\triangle u+\left(u\cdot\nabla\right)u+\nabla p & =f\qquad \text{in } \Omega \\ \Div u & =0\qquad \text{in } \Omega \\ u & =0 \qquad \text{on } \partial \Omega \end{align*}

where \nu>0 is a viscosity constant. We call this problem as (NS).

Assume f\in \Sob{-1}{2} (\Omega). Now we consider a weak formulation of (NS).

A function u\in \Sob{1}{2}_0(\Omega) is called a weak solution of (NS) if \Div u =0 in \Omega and u satisfies

    \[ \nu \int_\Omega \nabla u : \nabla v \intd{x} = \int_\Omega (u\otimes u) : \nabla v \intd{x} + \action{ f,v } \]

for any v\in C_{0,\sigma}^\infty(\Omega).

The definition is well established. Take inner product to the first equation in (NS) with v\in C_{0,\sigma}^\infty(\Omega) and integrate it by parts. For the first part,

(1)   \begin{equation*} -\int_\Omega \nu \triangle u v \intd{x} = \nu \int_\Omega \nabla u : \nabla v \intd{x}. \end{equation*}

For the second part,

    \begin{align*} \int_\Omega (u\cdot \nabla u) v \intd{x} &= \sum_{i,j=1}^3 \int_\Omega u^j \frac{\partial u^i}{\partial x_j} v^i \intd{x}\\ &= -\sum_{i,j=1}^3 \int_\Omega \frac{\partial (u^j v^i)}{\partial x_j} u^i \intd{x} \\ &=- \int_\Omega \left(\sum_{j=1}^3 \frac{\partial u^j}{\partial x_j}\right)\sum_{i=1}^3 v^i u^i \intd{x}-\sum_{i,j=1}^3 \int_\Omega u^i u^j \frac{\partial v^i}{\partial x_j} \intd{x} \\ & =- \int_\Omega (u\otimes u) : \nabla v \intd{x}. \end{align*}

For the thrid part, it vanishes since \Div v =0. Also, for u\in \Sob{1}{2}(\Omega), for n=2, 3, by the Sobolev embedding theorem and the boundedness of \Omega, u \in \Leb{4} (\Omega). Hence \eqref{eq:weak-sol-st-NS} makes sense.

From now, we assume \nu =1. In 1933, J. Leray proved the existence of weak solution of (NS). To present the theorem of Leray, we define some terminology and the fundamental fixed point theorem proved by Leray and Schauder.

Theorem (Leray-Schauder’s fixed point theorem). Let X be a Banach space. Suppose that the compact operator A:X\rightarrow X satisfies the following: there exists a constant M>0 such that if u\in X is a solution of u=\sigma Au, 0\leq \sigma < 1 and

    \[ \norm{u}{X}<M \]

then there exists u in X satisfying u=Au.

If X is reflexive and T: X\rightarrow X is completely continuous, then T is compact. So in this case, we can apply the Leray-Schauder principle for such operator. See this article for proof.

Theorem. Let \Omega be a bounded Lipschitz domain. Then there exists at least one weak solution of (NS).

Proof. Since \Omega is a bounded Lipschitz domain, due to Poincar\’e’s inequality, one can check that [u,v] = \int_\Omega \nabla u : \nabla v \intd{x} is an inner product on \Sob{1}{2}_{0,\sigma} (\Omega).

Also, for any w\in \Sob{1}{2}_{0,\sigma} (\Omega), \Div w\otimes w \in \Sob{-1}{2}(\Omega). Indeed, for any v\in \Sob{1}{2}_{0,\sigma} (\Omega),

    \begin{align*} \action{\Div (w\otimes w) , v} = -\action{w\otimes w,\nabla v}&\le \norm{w\otimes w}{2;\Omega}\norm{\nabla v}{2;\Omega}\\ &\leq C\norm{w}{4;\Omega}\norm{w}{4;\Omega} \norm{\nabla v}{2;\Omega} \intertext{By the Sobolev embedding $\Sob{1}{2}(\Omega) \hookrightarrow \Leb{6}(\Omega)\hookrightarrow\Leb{4}(\Omega)$, we get} &\leq C\norm{w}{1,2;\Omega}^2 \norm{v}{1,2;\Omega}. \end{align*}

So \Div(w \otimes w)\in \Sob{-1}{2}(\Omega).

Now for each w\in \Sob{1}{2}_{0,\sigma}(\Omega) by the existence result on Stokes equation, there exists a unique weak solution u \in \Sob{1}{2}_{0,\sigma}(\Omega) satisfying

(2)   \begin{equation*} \int_{\Omega} \nabla u : \nabla v \intd{x} = \action{-\Div(w\otimes w),v} + \action{f,v} \end{equation*}

for all v\in C_{0,\sigma}^\infty (\Omega). By Lemma ??, the above identity holds for any v\in \Sob{1}{2}_{0,\sigma} (\Omega).

Since -\Div(w\otimes w) \in \Sob{-1}{2}(\Omega), by the Riesz representation theorem, there exists a unique A(w) in \Sob{1}{2}_{0,\sigma}(\Omega) such that

    \[ [A(w) , v] = -\action{\Div(w\otimes w) , v} \]

for all v\in \Sob{1}{2}_{0,\sigma} (\Omega). By the uniqueness, the operator A:\Sob{1}{2}_{0,\sigma}(\Omega) \rightarrow \Sob{1}{2}_{0,\sigma}(\Omega) is well-defined. Similarly, there exists a unique F\in \Sob{1}{2}_{0,\sigma}(\Omega) such that

    \[ [F,v] = \action{f,v} \]

for all v\in \Sob{1}{2}_{0,\sigma} (\Omega) with \norm{\nabla F}{2;\Omega}= \norm{f}{-1,2;\Omega}. So \eqref{eq:perturb-Stokes} can be written as

    \[ [u,v ] =[A(w),v] +[F,v],\]

i.e., u=A(w)+F in operator form. Note that the existence of weak solution of (NS) is equivalent to the existence of a fixed point of the above operator equation. If there exists u such that u=A(u)+F, then [A(u),v]=\int_\Omega u\times u :\nabla v \intd{x}. So u becomes a weak solution of (NS).

To use the Leray-Schauder fixed point theorem, we need to show that the operator T:\Sob{1}{2}_{0,\sigma} (\Omega) \rightarrow \Sob{1}{2}_{0,\sigma} (\Omega) defined by T(w) =A(w)+F is a compact operator and a priori estimate.

We show that T is completely continuous on \Sob{1}{2}_{0,\sigma}(\Omega). Let u_n \rightharpoonup u in \Sob{1}{2}_{0,\sigma}(\Omega). Then there exists a constant C>0 such that \norm{w_k}{1,2;\Omega} \leq C for all k. For simplicity, let u_k = T(w_k). Since \Omega is bounded Lipschitz domain, the Rellich-Kondrashov theorem shows that \Sob{1}{2}_{0,\sigma}(\Omega) is compactly embedded in \Leb{4}(\Omega). Thus, w_k \rightarrow w strongly in \Leb{4}(\Omega). Now

    \begin{align*} &\relphantom{=} \norm{w_k \otimes w_k - w\otimes w}{\Leb{2}}\\ & = \norm{(w_k-w)\otimes w_k +w \otimes w_k -w\otimes w}{\Leb{2}}\\ & \leq C\norm{w_k-w}{\Leb{4}} \norm{w_k}{\Leb{4}} +\norm{w}{\Leb{4}} \norm{w_k-w}{\Leb{4}}. \end{align*}

So as k\rightarrow \infty, w_k \otimes w_k \rightarrow w\otimes w in \Leb{2}(\Omega). Note

    \[ [u_k-u_m,v] =\int_{\Omega} w_k\otimes w_k -w_m \otimes w_m : \nabla v \intd{x}\qquad \text{for all } v\in \Sob{1}{2}_{0,\sigma}(\Omega). \]

Put v=u_k -u_m. Then we get

    \[ \norm{\nabla u_k -\nabla u_m}{2;\Omega}^2 \leq \norm{\nabla u_k -\nabla u_m}{2;\Omega} \norm{w_k\otimes w_k -w_m\otimes w_m}{2;\Omega} \]

So

    \[ \norm{\nabla u_k -\nabla u_m}{2;\Omega} \leq \norm{w_k\otimes w_k -w_m\otimes w_m}{2;\Omega}\rightarrow 0 \]

as k,m\rightarrow \infty. This shows that T is completely continuous.

Now we left to show a priori estimate.

First, note that for any v\in C_{0,\sigma}^\infty(\Omega),

    \begin{align*} &\relphantom{=} \int_\Omega v \otimes v : \nabla v \intd{x} \\ &=\sum_{i,j=1}^3 \int_\Omega v^i v^j \frac{\partial v^i }{\partial x_j} \intd{x} \\ & =-\sum_{i,j=1}^3 \int_{\Omega} \frac{\partial v^j}{\partial x_j} v^i v^i \intd{x} -\sum_{i,j=1}^3 \int_\Omega v^i v^j \frac{\partial v^i }{\partial x_j} \intd{x} \\ & =-\sum_{i,j=1}^3 \int_\Omega v^i v^j \frac{\partial v^i }{\partial x_j} \intd{x}=-\int_\Omega v \otimes v : \nabla v \intd{x}. \end{align*}

The last identity holds because of divergence free condition. By density, \int_\Omega v \otimes v : \nabla v \intd{x} =0 for any v\in \Sob{1}{2}_{0,\sigma}(\Omega).

Now suppose u\in \Sob{1}{2}_{0,\sigma}(\Omega) satisfies u=\sigma (Au +F), 0\leq \sigma <1. So

    \begin{align*} [u,u] &= [\sigma (A(u)+F),u] = \sigma [A(u),u] + \sigma [F,u] \\ &=\sigma \int_\Omega (u\otimes u) :\nabla u \intd{x} + \sigma \action{f,u} \\ & \leq \norm{f}{-1,2;\Omega}\norm{\nabla u}{2;\Omega}. \end{align*}

Here we used

    \[ \int_\Omega v \otimes v : \nabla v \intd{x} =0 \qquad \text{for all } v\in \Sob{1}{2}_{0,\sigma} (\Omega). \]

So

(3)   \begin{equation*} \norm{\nabla u}{2;\Omega} \leq \norm{f}{-1,2;\Omega} <(\norm{f}{-1,2;\Omega} +1). \end{equation*}

Here the constant does not depend on \sigma. Therefore, by the Leray-Schauder fixed point theorem, T has a fixed point u. This u is the desired weak solution of (NS).

Now we prove the uniqueness.

Theorem. Let \Omega be a bounded Lipschitz domain and suppose that

    \[ 2c_0^2(n,\Omega) \norm{f}{-1,2;\Omega} <1, \]

where c_0 (n,\Omega) is a constant in the inequality

(4)   \begin{equation*} \norm{v}{4;\Omega} \leq c_0 (n,\Omega) \norm{\nabla v}{2;\Omega} \end{equation*}

for any v\in \Sob{1}{2}_0 (\Omega). Then (NS) has a unique weak solution.

Remark. The inequality (4) holds since \Omega is bounded and the Sobolev embedding theorem.

Proof. Suppose u_1 and u_2 are two different solutions to (NS). Then

    \begin{align*} \norm{\nabla u_1 -\nabla u_2}{2;\Omega}^2&= \int_{\Omega} (u_1\otimes u_1 - u_2\otimes u_2): \nabla (u_1-u_2) \intd{x} \\ & = \int_{\Omega} (u_1\otimes (u_1-u_2)): \nabla (u_1-u_2) \intd{x} \\ & \relphantom{=}+ \int_{\Omega} (u_1-u_2)\otimes u_2: \nabla (u_1-u_2) \intd{x}\\ & \leq (\norm{u_1}{4;\Omega} +\norm{u_2}{4;\Omega}) \norm{u_1-u_2}{4;\Omega} \norm{\nabla u_1 - \nabla u_2}{2;\Omega}. \end{align*}

Apply (3) and (4)}. Then we get

    \begin{align*} & \leq 2c_0^2 \norm{f}{-1,2;\Omega} \norm{\nabla u_1- \nabla u_2}{2;\Omega}^2. \end{align*}

Now by assumption, we get

    \begin{align*} \norm{\nabla u_1 -\nabla u_2}{2;\Omega}^2& <\norm{\nabla u_1 -\nabla u_2}{2;\Omega}^2, \end{align*}

which is a contradiction. Therefore, the solution of (NS) is unique.

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