\\
\tilde{Y}_{15} & =&\frac{1}{2}(\sigma_{z}^{1}-\sigma_{z}^{2}),\nonumber \\
\tilde{Y}_{16} & =&\frac{1}{2}(\sigma_{x}^{1}\sigma_{y}^{2}-\sigma_{y}^{1}\sigma_{x}^{2}).\end{aligned}$$ The total Hamiltonian can now be rewritten as $H=\tilde{H}_{0}+\tilde{H}'$, in which $$\begin{aligned}
\tilde{H}_{0}=\tilde{W}_{1}\tilde{Y}_{1}+\tilde{W}_{2}\tilde{Y}_{2}+\cdots +\tilde{W}_{10}\tilde{Y}_{10} \end{aligned}$$ and $$\begin{aligned}
\tilde{H}'=\tilde{W}_{11}\tilde{Y}_{11}+\cdots +\tilde{W}_{16}\tilde{Y}_{16}. \end{aligned}$$

It is then evident that if we apply the following control Hamiltonian, i.e., $$\begin{aligned}
\label{control2}
\tilde{H}_{c}&=&\sum_{j=1}^{N}\pi \delta(t-T_j)\frac{\tilde{Y}_1}{2} \nonumber \\
&=& \sum_{j=1}^{N}\frac{\pi}{4}
\delta(t-T_j)(-I+\sigma_{x}^{1}\sigma_{x}^{2}+\sigma_{y}^{1}\sigma_{y}^{2}-\sigma_{z}^{1}\sigma_{z}^{2}),\end{aligned}$$ the time evolution operator of the controlled total system becomes entirely parallel to Eqs. (\[Ut\]) and (\[un\]) (with an arbitrary operator $O$ replaced by $\tilde{O}$). Hence, using the $N$ control pulse described by Eq. (\[control2\]), the quantum coherence defined by the expectation value of ${\cal P}_{|\tilde{0}\rangle}$ can be preserved up to $1+O(T^{N+1})$, for an arbitrary initial state. If the initial state is already the Bell state $|\tilde{0}\rangle$ (i.e., coincides with the $|\Psi\rangle$ that defines our coherence measure ${\cal P}_{|\Psi\rangle}$), then our UDD control sequence locks the system on this Bell state with a fidelity $1+O(T^{N+1})$, no matter how the system is coupled to its environment.

The constant term in the control Hamiltonian $\tilde{H}_c$ can be dropped because it only induces an overall phase of the evolving state. All other terms in $\tilde{H}_c$ represent two-body and hence nonlocal control. This confirms our initial expectation that suppressing the decoherence of entangled two-qubit 