# Why do we need projection in the definition of the Stokes operator?

$$\DeclareMathOperator{\div}{div}$$ $$\def\bu{\mathbf{u}}$$ 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?

• What are the extra exponents $2$ ($H^2(D)^2$, $L^2(D)^2$)?? – Ted Shifrin Mar 14 at 21:47
• @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, $\mathbf{u}\in H^2(D)^2$. – Oleg Mar 14 at 21:52
• 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. – Ted Shifrin Mar 14 at 21:53
• @TedShifrin Then why in the periodic case (periodic boundary conditions) the Stokes operator is equal to Laplacian? – Oleg Mar 14 at 22:01
• Hint: In the definition of $H$, in what sense is the divergence equal to zero? – maxmilgram Mar 15 at 17:42