$\DeclareMathOperator{\div}{div}$ $\def\bu{\mathbf{u}}$ Let $D$ be the square $[0,1]^2$ and consider the following space: $$ V:=\{\bu: \bu\in H^2(D)^2, \div \bu=0, u|_{\partial D}=0 \}. $$

Introduce also $$ H:=\{\bu: \bu\in L^2(D)^2, \div \bu=0\}. $$

Then the Stokes operator is defined as an operator $V\to H$ as $$ A\bu:=-P_L\Delta\bu, $$ where $P_L$ is the projection on $H$.

I don't understand, why the Stokes operator is not equal to minus Laplacian? Indeed, if $\bu\in V$, then $\div \bu=0$ and thus $$ \div \Delta\bu=u^1_{xxx}+u^1_{yyx}+u^2_{xxy}+u^2_{yyy}=(u^1_x+u^2_y)_{xx}+(u^1_x+u^2_y)_{yy}=\Delta\div\bu=0. $$ In this case, $\Delta \bu\in H$ and thus the projection is not needed.

Question: what is wrong in my reasoning? Why the Stokes operator is not equal to Laplacian? Related question: Can one write explicitly the eigenfunctions of Stokes operator in this simple situation? At least can maybe one bound $\lambda_1$, the smallest eigenvalue, from below?

  • $\begingroup$ What are the extra exponents $2$ ($H^2(D)^2$, $L^2(D)^2$)?? $\endgroup$ – Ted Shifrin Mar 14 at 21:47
  • $\begingroup$ @TedShifrin $\mathbf{u}=(u_1,u_2)$ is a vector. Each of its 2 coordinates $u_1$ and $u_2$ belongs to $H^2(D)$. Thus, $\mathbf{u}\in H^2(D)^2$. $\endgroup$ – Oleg Mar 14 at 21:52
  • $\begingroup$ Oh, I see ... So, offhand, my answer to your question is that you need $C^3$ to interchange partial derivatives, and you don't have $C^3$ functions. $\endgroup$ – Ted Shifrin Mar 14 at 21:53
  • $\begingroup$ @TedShifrin Then why in the periodic case (periodic boundary conditions) the Stokes operator is equal to Laplacian? $\endgroup$ – Oleg Mar 14 at 22:01
  • $\begingroup$ Hint: In the definition of $H$, in what sense is the divergence equal to zero? $\endgroup$ – maxmilgram Mar 15 at 17:42

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