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Let

  • $d\in\left\{2,3\right\}$ with
  • $\Lambda\subseteq\mathbb R^d$ be bounded and open with $\partial\Lambda\in C^1$

In Lemma 6.1 of Navier-Stokes Equations and Nonlinear Functional Analysis by Roger Temam, the author is stating that if $u,v\in H^2(\Lambda,\mathbb R^d)$, then $$\frac\partial{\partial x_i}(u\cdot\nabla)v=\left(\frac{\partial u}{\partial x_i}\cdot\nabla\right)v+(u\cdot\nabla)\frac{\partial v}{\partial x_i}\tag 1$$ would belong to $L^2(\Lambda,\mathbb R^d)$, because $u,v$ belong to $L^6(\Lambda,\mathbb R^d)$ and $u,v$ are continuous on $\overline\Lambda$ (which is clear by the Sobolev inequalities as they can be found, for example, in the book of Evans).

Can the same statement be proved, if $\Lambda$ is just bounded and open and $u,v\in H_0^2(\Lambda,\mathbb R^d)$?

Let

  • $d\in\left\{2,3\right\}$ with
  • $\Lambda\subseteq\mathbb R^d$ be bounded and open with $\partial\Lambda\in C^1$

In Lemma 6.1 of Navier-Stokes Equations and Nonlinear Functional Analysis by Roger Temam, the author is stating that if $u,v\in H^2(\Lambda,\mathbb R^d)$, then $$\frac\partial{\partial x_i}(u\cdot\nabla)v=\left(\frac{\partial u}{\partial x_i}\cdot\nabla\right)v+(u\cdot\nabla)\frac{\partial v}{\partial x_i}\tag 1$$ would belong to $L^2(\Lambda,\mathbb R^d)$, because $u,v$ belong to $L^6(\Lambda,\mathbb R^d)$ and $u,v$ are continuous on $\overline\Lambda$ (which is clear by the Sobolev inequalities).

Can the same statement be proved, if $\Lambda$ is just bounded and open and $u,v\in H_0^2(\Lambda,\mathbb R^d)$?

Let

  • $d\in\left\{2,3\right\}$ with
  • $\Lambda\subseteq\mathbb R^d$ be bounded and open with $\partial\Lambda\in C^1$

In Lemma 6.1 of Navier-Stokes Equations and Nonlinear Functional Analysis by Roger Temam, the author is stating that if $u,v\in H^2(\Lambda,\mathbb R^d)$, then $$\frac\partial{\partial x_i}(u\cdot\nabla)v=\left(\frac{\partial u}{\partial x_i}\cdot\nabla\right)v+(u\cdot\nabla)\frac{\partial v}{\partial x_i}\tag 1$$ would belong to $L^2(\Lambda,\mathbb R^d)$, because $u,v$ belong to $L^6(\Lambda,\mathbb R^d)$ and $u,v$ are continuous on $\overline\Lambda$ (which is clear by the Sobolev inequalities as they can be found, for example, in the book of Evans).

Can the same statement be proved, if $\Lambda$ is just bounded and open and $u,v\in H_0^2(\Lambda,\mathbb R^d)$?

added 111 characters in body
Source Link
0xbadf00d
0xbadf00d
  • 167
  • 1
  • 5
  • 16

Let

  • $d\in\left\{2,3\right\}$ with
  • $\Lambda\subseteq\mathbb R^d$ be bounded and open with $\partial\Lambda\in C^1$

In Lemma 6.1 of Navier-Stokes Equations and Nonlinear Functional Analysis by Roger Temam, the author is stating that if $u,v\in H^2(\Lambda,\mathbb R^d)$, then $$\frac\partial{\partial x_i}(u\cdot\nabla)v=\left(\frac{\partial u}{\partial x_i}\cdot\nabla\right)v+(u\cdot\nabla)\frac{\partial v}{\partial x_i}\tag 1$$ would belong to $L^2(\Lambda,\mathbb R^d)$, because $u,v$ belong to $L^6(\Lambda,\mathbb R^d)$ and $u,v$ are continuous on $\overline\Lambda$ (which is clear by the Sobolev inequalities) and $u,v$ are continuous on $\overline\Lambda$.

I don't understand why $u,v$ need to be continuous on $\overline\Lambda$ (I know that they are Hölder continuous inCan the casesame statement be proved, if $d=1$)$\Lambda$ is just bounded and I don't understand how he's using this to obtain the claim.open and $u,v\in H_0^2(\Lambda,\mathbb R^d)$?

Let

  • $d\in\left\{2,3\right\}$ with
  • $\Lambda\subseteq\mathbb R^d$ be bounded and open with $\partial\Lambda\in C^1$

In Lemma 6.1 of Navier-Stokes Equations and Nonlinear Functional Analysis by Roger Temam, the author is stating that if $u,v\in H^2(\Lambda,\mathbb R^d)$, then $$\frac\partial{\partial x_i}(u\cdot\nabla)v=\left(\frac{\partial u}{\partial x_i}\cdot\nabla\right)v+(u\cdot\nabla)\frac{\partial v}{\partial x_i}\tag 1$$ would belong to $L^2(\Lambda,\mathbb R^d)$, because $u,v$ belong to $L^6(\Lambda,\mathbb R^d)$ (which is clear by the Sobolev inequalities) and $u,v$ are continuous on $\overline\Lambda$.

I don't understand why $u,v$ need to be continuous on $\overline\Lambda$ (I know that they are Hölder continuous in the case $d=1$) and I don't understand how he's using this to obtain the claim.

Let

  • $d\in\left\{2,3\right\}$ with
  • $\Lambda\subseteq\mathbb R^d$ be bounded and open with $\partial\Lambda\in C^1$

In Lemma 6.1 of Navier-Stokes Equations and Nonlinear Functional Analysis by Roger Temam, the author is stating that if $u,v\in H^2(\Lambda,\mathbb R^d)$, then $$\frac\partial{\partial x_i}(u\cdot\nabla)v=\left(\frac{\partial u}{\partial x_i}\cdot\nabla\right)v+(u\cdot\nabla)\frac{\partial v}{\partial x_i}\tag 1$$ would belong to $L^2(\Lambda,\mathbb R^d)$, because $u,v$ belong to $L^6(\Lambda,\mathbb R^d)$ and $u,v$ are continuous on $\overline\Lambda$ (which is clear by the Sobolev inequalities).

Can the same statement be proved, if $\Lambda$ is just bounded and open and $u,v\in H_0^2(\Lambda,\mathbb R^d)$?

Source Link
0xbadf00d
0xbadf00d
  • 167
  • 1
  • 5
  • 16

Is $(u\cdot\nabla)v\in H^1$, if $u,v\in H^2$?

Let

  • $d\in\left\{2,3\right\}$ with
  • $\Lambda\subseteq\mathbb R^d$ be bounded and open with $\partial\Lambda\in C^1$

In Lemma 6.1 of Navier-Stokes Equations and Nonlinear Functional Analysis by Roger Temam, the author is stating that if $u,v\in H^2(\Lambda,\mathbb R^d)$, then $$\frac\partial{\partial x_i}(u\cdot\nabla)v=\left(\frac{\partial u}{\partial x_i}\cdot\nabla\right)v+(u\cdot\nabla)\frac{\partial v}{\partial x_i}\tag 1$$ would belong to $L^2(\Lambda,\mathbb R^d)$, because $u,v$ belong to $L^6(\Lambda,\mathbb R^d)$ (which is clear by the Sobolev inequalities) and $u,v$ are continuous on $\overline\Lambda$.

I don't understand why $u,v$ need to be continuous on $\overline\Lambda$ (I know that they are Hölder continuous in the case $d=1$) and I don't understand how he's using this to obtain the claim.