What is the shrinkage formula in image denoising?

Question

The minimizing for $ \omega$ is equivalent to solving,for i=1,2,...,$n^2$, $$\min_{w_i}=\|w_i\|_2+\frac{\beta}{2}\|w_i-D_iu\|_2^2 \; \; \; \;\;\;\;(1)$$ Which we first introduce two auxiliary vectors $\omega_1,\omega_2\in \mathcal{R}^{n^2}$ to approximate $D^{(1)}u$ and $D^{(2)}u$,respectively,where we recall that $D^{(1)},D^{(2)}\in\mathcal{R}^{n^2\times n^2}$ and $u \in\mathcal{R}^{n\times n}$.We denote $\omega =(\omega_1;\omega_2) \in \mathcal{R}^{2n^2}$ and $D=(D^{(1)};D^{(2)})\in\mathcal{R}^{2n^2\times n^2}$,for i=1,2,...,$n^2$,we let $\omega_i=((\omega_1)_i;(\omega_2)_i) \in \mathcal{R}^2$,which is an approximation of $D_iu=[(D^{(1)}u)_i;(D^{(2)}u)_i]\in \mathcal{R}^2$.To keep $\omega_i$ close to $D_iu$,we apply a quadratic penalt term $\beta$ .In my opinion,to solve the (1): $$when\; w_i\neq 0$$ $$\min_{w_i}=\sqrt{(\omega_1)_i^2+(\omega_2)_i^2}+\frac{\beta}{2}([(\omega_1)_i-(D^{(1)}u)_i)^2+((\omega_2)_i-(D^{(2)}u)_i)^2)]$$ By derivation then we can get: $$\frac{(\omega_1)_i}{\left \|\omega_i \right \|}+\beta((\omega_1)_i-(D^{(1)}u)_i)=0$$ $$\frac{(\omega_2)_i}{\left \|\omega_i \right \|}+\beta((\omega_2)_i-(D^{(2)}u)_i)=0$$ So it means: $$\frac{\omega_i}{\left \|\omega_i \right \|}+\beta(\omega_i-D_iu)=0$$ $$\omega_i=D_iu-\frac{\omega_i}{\beta\left \|\omega_i \right \|}$$ What let me feel confuse is the paper said the solution for the (1):for which the unique minimizer is given by the following two-demensional(2D) shrinkage fomula: $$\omega_i=max\left \{{\left \|D_iu \right \|-\frac{1}{\beta}},0 \right \}\frac{D_iu}{\left \|D_iu \right \|},i=1,2,...,n^2\; \; \; \;\;\;\;(2)$$ So what is shrinkage formula doing for (1) to (2)?And why I get the different answer and where if I made mistakes?Thanks for your answers.