Quantum synchronization of the Schrödinger–Lohe model

Let U_i and $U_{i}^{+}$ be a unitary $d\times d$ matrix and its Hermitian conjugate, respectively, and let H_i be the $d\times d$ Hermitian matrix whose eigenvalues correspond to the natural frequencies of the oscillator at node i. Then, the Lohe model [18, 19] for quantum synchronization can be written as follows:

$$\begin{eqnarray}&&{\rm i}{{\dot{U}}_{j}}={{H}_{j}}{{U}_{j}}+\frac{{\rm i}K}{2N}\mathop{\sum }\limits_{k=1}^{N}\left( {{U}_{k}}-{{U}_{j}}U_{k}^{+}{{U}_{j}} \right),\quad \qquad 1\leqslant j\leqslant N.\end{eqnarray} \tag{ 2.1 }$$

For two special cases, the Lohe model (2.1) can be related to the Kuramoto model and the Schrödinger equation. For example, consider the one-dimensional case, d = 1. In this case, we write U_j and H_j as

$$\begin{eqnarray*}&&{{U}_{j}}={{{\rm e}}^{-{\rm i}{{\theta }_{j}}}},\quad \quad \quad {{H}_{j}}={{\Omega }_{j}}.\end{eqnarray*}$$

Then, it is easy to see that (2.1) reduces to the Kuramoto model

$$\begin{eqnarray*}&&{{\dot{\theta }}_{j}}={{\Omega }_{j}}+\frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}{\rm sin} ({{\theta }_{k}}-{{\theta }_{j}}),\quad \qquad 1\leqslant j\leqslant N.\end{eqnarray*}$$

However, for zero coupling, $K=0$, system (2.1) becomes the Schrödinger equation

$$\begin{eqnarray*}&&{\rm i}{{\dot{U}}_{j}}={{H}_{j}}{{U}_{j}},\quad \qquad {\rm i}{\rm .e}{\rm .,}\qquad {{U}_{i}}(t)={{{\rm e}}^{-{\rm i}{{H}_{i}}t}}{{U}_{i}}(0).\end{eqnarray*}$$

Thus, the Lohe model (2.1) is a quantum generation of the Kuramoto model and the Schrödinger equation with finite quantum states. A natural question to ask then is 'If we allow each quantum particle to take an infinite quantum state, what happens to the system (2.1)?' In this case, we need to use the standard Schrödinger equation instead of the finite-dimensional one. Note that our purpose is to present a quantum model exhibiting the synchronization property. Thus, in some sense, we need to introduce a synchronization enforcing term into the standard Schrödinger equation. In this paper, we employ the Lohe synchronization for this flocking mechanism

$$\begin{eqnarray}&&{\rm i}{{\partial }_{t}}{{\psi }_{j}}=-\Delta {{\psi }_{j}}+\frac{{\rm i}K}{N}\mathop{\sum }\limits_{k=1}^{N}\left( {{\psi }_{k}}-\frac{\langle {{\psi }_{j}},{{\psi }_{k}}\rangle }{\langle {{\psi }_{j}},{{\psi }_{j}}\rangle }{{\psi }_{j}} \right).\end{eqnarray} \tag{ 2.2 }$$

Note that, on the manifold $||{{\psi }_{i}}|{{|}_{2}}=1$, system (2.2) can be rewritten as

$$\begin{eqnarray*}&&{\rm i}{{\partial }_{t}}{{\psi }_{j}}=-\Delta {{\psi }_{j}}+\frac{{\rm i}K}{N}\mathop{\sum }\limits_{k=1}^{N}\left( {{\psi }_{k}}-\langle {{\psi }_{j}},{{\psi }_{k}}\rangle {{\psi }_{j}} \right)\end{eqnarray*}$$

which has structural similarity to (2.1). For additional details, we refer to sections 3 and 6 of [18].

In this subsection, we show that the solution to the S–L system (1.1) can be given by the successive composition of the free Schrödinger equation and the Lohe system.

Consider the Cauchy problem for the free Schrödinger equation

$$\begin{eqnarray}\begin{array}{rcl} {{\partial }_{t}}\psi & = & {\rm i}\Delta \psi ,\quad \qquad x\in {{\mathbb{R}}^{d}},\quad \qquad t\gt 0, \\ \psi (x,0) & = & {{\psi }_{0}}(x),\quad \qquad t=0. \\ \end{array}\end{eqnarray} \tag{ 2.3 }$$

Define a one-parameter family of solution operator S(t) to (2.3) with an L² initial datum ψ₀ as follows:

$$\begin{eqnarray}\begin{array}{rcl} S(t){{\psi }_{0}} & := & K(\cdot ,t){{*}_{x}}{{\psi }_{0}}=\frac{1}{{{(4\pi {\rm i}t)}^{{\rm d}/2}}}{{\int }_{{{\mathbb{R}}^{d}}}}{{{\rm e}}^{-\frac{{{(x-y)}^{2}}}{4{\rm i}t}}}{{\psi }_{0}}(y){\rm d}y,\quad \qquad t\gt 0, \\ S(0) & = & I{\rm d}. \\ \end{array}\end{eqnarray} \tag{ 2.4 }$$

Then, it is easy to see that $S(t){{\psi }_{0}},\;t\gt 0,$ is an analytic function. In the following two lemmas, we study the basic properties of the solution operator S(t).

Lemma 2.1. Let $\psi =S(t){{\psi }_{0}}$ be the solution to (2.3) with L²_x initial datum ψ₀. Then, we have

$$\begin{eqnarray*}&&||S(t){{\psi }_{0}}|{{|}_{2}}=||{{\psi }_{0}}|{{|}_{2}},\quad t\geqslant 0.\end{eqnarray*}$$

Proof. It is the result of the Plancherel theorem. By the Fourier transform, we have

$$\begin{eqnarray*}&&\widehat{S(t){{\psi }_{0}}}=\widehat{{{K}_{t}}{{*}_{x}}{{\psi }_{0}}}=\widehat{{{K}_{t}}}\widehat{{{\psi }_{0}}}={\rm exp} (-{\rm i}t|\bar{x}{{|}^{2}})\widehat{{{\psi }_{0}}}.\end{eqnarray*}$$

It follows from the above relation that

$$\begin{eqnarray*}&&||S(t){{\psi }_{0}}|{{|}_{2}}=||\widehat{S(t){{\psi }_{0}}}|{{|}_{2}}=||{\rm exp} (-{\rm i}t|\bar{x}{{|}^{2}})\widehat{{{\psi }_{0}}}|{{|}_{2}}=||\widehat{{{\psi }_{0}}}|{{|}_{2}}=||{{\psi }_{0}}|{{|}_{2}}.\end{eqnarray*}$$

Therefore, we have

$$\begin{eqnarray*}&&||S(t){{\psi }_{0}}|{{|}_{2}}=||{{\psi }_{0}}|{{|}_{2}}.\end{eqnarray*}$$

We next study the unitary property of the solution operator S(t).

Lemma 2.2.  (Unitary property of S(t)) Let S(t) be a solution operator given by (2.4). Then for any complex-valued $L_{x}^{2}({{\mathbb{R}}^{d}})$ functions ϕ₁₀ and ϕ₂₀, we have

$$\begin{eqnarray*}&&\langle S(t){{\phi }_{10}},S(t){{\phi }_{20}}\rangle =\langle {{\phi }_{10}},{{\phi }_{20}}\rangle ,\qquad t\geqslant 0.\end{eqnarray*}$$

Proof. Since equation (2.3) is linear, $S(t)({{\psi }_{10}}-{{\psi }_{20}})$ is the solution of (2.3) with initial datum ${{\psi }_{10}}-{{\psi }_{20}}$. Then, by lemma 2.1, we have

$$\begin{eqnarray*}&&||S(t)({{\psi }_{10}}-{{\psi }_{20}})|{{|}_{2}}=||{{\psi }_{10}}-{{\psi }_{20}}|{{|}_{2}},\quad \qquad t\geqslant 0.\end{eqnarray*}$$

We take the square of the above relation and invoke L² conservation to obtain the desired result.

We next consider L² conservation of the space-dependent Lohe system

$$\begin{eqnarray}&&\frac{{\rm d}{{\psi }_{j}}}{{\rm d}t}=\frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\left( {{\psi }_{k}}-\frac{\langle {{\psi }_{k}},{{\psi }_{j}}\rangle }{\langle {{\psi }_{j}},{{\psi }_{j}}\rangle }{{\psi }_{j}} \right),\quad \qquad t\gt 0.\end{eqnarray} \tag{ 2.5 }$$

Note that the spatial variable x acts as a parameter in (2.5). For a solution $\Psi =({{\psi }_{1}},...,\;{{\psi }_{N}})$ to (2.5) with initial data ${{\Psi }_{0}}={{\Psi }_{0}}(x)$, we set the solution operator L(t) to the system (2.5) as follows:

$$\begin{eqnarray*}&&\Psi (t)=L(t){{\Psi }_{0}},\quad \qquad t\gt 0,\quad \qquad L(0)=I{\rm d}.\end{eqnarray*}$$

Lemma 2.3. The solution operator L(t) conserves an L² norm:

$$\begin{eqnarray*}&&||L(t){{\psi }_{j0}}|{{|}_{2}}=||{{\psi }_{j0}}|{{|}_{2}},\qquad 1\leqslant i\leqslant N,\qquad t\geqslant 0.\end{eqnarray*}$$

Proof. Note that

$$\begin{eqnarray*}\begin{array}{rcl} \frac{{\rm d}}{{\rm d}t}\langle {{\psi }_{j}},{{\psi }_{j}}\rangle & = & \left\langle \frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\left( {{\psi }_{k}}-\frac{\langle {{\psi }_{j}},{{\psi }_{k}}\rangle }{\langle {{\psi }_{j}},{{\psi }_{j}}\rangle }{{\psi }_{j}} \right),{{\psi }_{j}} \right\rangle +\left\langle {{\psi }_{j}},\frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\left( {{\psi }_{k}}-\frac{\langle {{\psi }_{j}},{{\psi }_{k}}\rangle }{\langle {{\psi }_{j}},{{\psi }_{j}}\rangle }{{\psi }_{j}} \right) \right\rangle \\ {} & = & \frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\left( \langle {{\psi }_{k}},{{\psi }_{j}}\rangle -\frac{\langle {{\psi }_{j}},{{\psi }_{k}}\rangle }{\langle {{\psi }_{j}},{{\psi }_{j}}\rangle }\langle {{\psi }_{j}},{{\psi }_{j}}\rangle \right) \\ {} & {} & +\frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\left( \langle {{\psi }_{j}},{{\psi }_{k}}\rangle -\frac{\overline{\langle {{\psi }_{j}},{{\psi }_{k}}\rangle }}{\langle {{\psi }_{j}},{{\psi }_{j}}\rangle }\langle {{\psi }_{j}},{{\psi }_{j}}\rangle \right) \\ {} & = & 0. \\ \end{array}\end{eqnarray*}$$

We next state the essential key tool for our later analysis of the solution decomposition relation for the wave function that denotes the scattering property of system (1.1).

Proposition 2.1. Let $\Psi =({{\psi }_{1}},...,{{\psi }_{N}})$ be a solution to system (1.1) with L² initial data Ψ ₀ . Then, the solution ψ_j can be decomposed as the successive composition of S and L operators to initial datum ψ_j0 , i.e.,

$$\begin{eqnarray*}&&{{\psi }_{j}}(x,t)=S(t)\;\circ \;L(t){{\psi }_{j0}}(x),\quad \qquad x\in {{\mathbb{R}}^{d}},\quad \qquad t\gt 0.\end{eqnarray*}$$

Proof. Let $\Psi =({{\psi }_{1}},...,{{\psi }_{N}})$ be a solution to (1.1). Then, we first use $S{{(t)}^{-1}}=S(-t)$ to see that

$$\begin{eqnarray*}&&{{\psi }_{j}}(x,t)=S(t)\;\circ \;L(t){{\psi }_{j0}}(x)\quad \Longleftrightarrow \quad S(-t){{\psi }_{j}}(x,t)=L(t){{\psi }_{j0}}(x).\end{eqnarray*}$$

Thus, to prove the desired result, it suffices to check that $S(-t){{\psi }_{j}}(x,t)$ satisfies the Lohe system (2.5). For this, we apply the linear solution operator $S(-t)$ to both sides of (1.1) to obtain

$$\begin{eqnarray}&&S(-t)\left( {{\partial }_{t}}{{\psi }_{j}}-{\rm i}\Delta {{\psi }_{j}} \right)=S(-t)\left[ \frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\left( {{\psi }_{k}}-\frac{\langle {{\psi }_{j}},{{\psi }_{k}}\rangle }{\langle {{\psi }_{j}},{{\psi }_{j}}\rangle }{{\psi }_{j}} \right) \right].\end{eqnarray} \tag{ 2.6 }$$

• (Estimate of the left-hand side): Note that the representation formula (2.4) yields
  $$\begin{eqnarray*}&&S(-t){{\psi }_{j}}=K(\cdot ,-t){{*}_{x}}{{\psi }_{j}}(\cdot ,t).\end{eqnarray*}$$
  We differentiate the above relation with respect to t to obtain
  $$\begin{eqnarray}\begin{array}{rcl} {{\partial }_{t}}S(-t){{\psi }_{j}} & = & ({{\partial }_{t}}K(\cdot ,-t)){{*}_{x}}{{\psi }_{j}}(\cdot ,t)+K(\cdot ,-t){{*}_{x}}{{\partial }_{t}}{{\psi }_{j}}(\cdot ,t) \\ {} & = & -{\rm i}\Delta K(x,-t){{*}_{x}}{{\psi }_{j}}(\cdot ,t)+K(\cdot ,-t){{*}_{x}}{{\partial }_{t}}{{\psi }_{j}}(\cdot ,t) \\ {} & = & K(x,-t){{*}_{x}}(-{\rm i}\Delta {{\psi }_{j}}(\cdot ,t))+K(\cdot ,-t){{*}_{x}}{{\partial }_{t}}{{\psi }_{j}}(\cdot ,t) \\ {} & = & K(x,-t){{*}_{x}}\left( {{\partial }_{t}}{{\psi }_{j}}(\cdot ,t)-{\rm i}\Delta {{\psi }_{j}}(\cdot ,t) \right) \\ {} & = & S(-t)\left( {{\partial }_{t}}{{\psi }_{j}}(\cdot ,t)-{\rm i}\Delta {{\psi }_{j}}(\cdot ,t) \right), \\ \end{array}\end{eqnarray} \tag{ 2.7 }$$
  where we used the fact that $K(x,-t)$ satisfies the backward free Schrödinger equation
  $$\begin{eqnarray*}&&-{\rm i}{{\partial }_{t}}K(x,-t)+\Delta K(x,-t)=0.\end{eqnarray*}$$
• (Estimate of the rhs): Since the solution operator $S(-t)$ is linear and unitary, we have
  $$\begin{eqnarray}&&\begin{array}{l} S(-t)\left[ \frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\left( {{\psi }_{k}}-\frac{\langle {{\psi }_{j}},{{\psi }_{k}}\rangle }{\langle {{\psi }_{j}},{{\psi }_{j}}\rangle }{{\psi }_{j}} \right) \right] \\ \quad =\frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\left( S(-t){{\psi }_{k}}-\frac{\langle {{\psi }_{j}},{{\psi }_{k}}\rangle }{\langle {{\psi }_{j}},{{\psi }_{j}}\rangle }S(-t){{\psi }_{j}} \right) \\ \quad =\frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\left( S(-t){{\psi }_{k}}-\frac{\langle S(-t){{\psi }_{j}},S(-t){{\psi }_{k}}\rangle }{\langle S(-t){{\psi }_{j}},S(-t){{\psi }_{j}}\rangle }S(-t){{\psi }_{j}} \right). \\ \end{array}\end{eqnarray} \tag{ 2.8 }$$

Finally, we combine the estimates (2.6)–(2.8) to obtain

$$\begin{eqnarray*}&&{{\partial }_{t}}(S(-t){{\psi }_{j}})=\frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\left( S(-t){{\psi }_{k}}-\frac{\langle S(-t){{\psi }_{j}},S(-t){{\psi }_{k}}\rangle }{\langle S(-t){{\psi }_{j}},S(-t){{\psi }_{j}}\rangle }S(-t){{\psi }_{j}} \right).\end{eqnarray*}$$

However, since $S(-t){{\psi }_{i}}{{\mid }_{t=0}}={{\psi }_{i0}}$, by the uniqueness of (2.5), we have

$$\begin{eqnarray*}&&S(-t){{\psi }_{j}}=L(t){{\psi }_{j0}},\quad \qquad {\rm i}{\rm .e}{\rm .,}\quad {{\psi }_{j}}=S(t)\;\circ \;L(t){{\psi }_{j0}}.\end{eqnarray*}$$

We next study the relation between the S–L model (2.2) and the Kuramoto model which was observed in [18].

Consider normalized initial data with the same amplitude and different phases, i.e.,

$$\begin{eqnarray}&&{{\psi }_{j0}}(x)=\sqrt{\rho (x)}{{{\rm e}}^{{\rm i}{{\theta }_{j0}}}},\quad \qquad x\in {{\mathbb{R}}^{d}},\quad ||\rho |{{|}_{1}}=1.\end{eqnarray} \tag{ 2.9 }$$

Then, with initial phases ${{\Theta }_{0}}=({{\theta }_{10}},...,{{\theta }_{N0}})$, we solve the Cauchy problem for the Kuramoto model

$$\begin{eqnarray*}&&{{\dot{\theta }}_{j}}=\frac{2K}{N}\mathop{\sum }\limits_{k=1}^{N}{\rm sin} ({{\theta }_{k}}-{{\theta }_{j}}),\quad \qquad t\gt 0,\quad {{\theta }_{j}}(0)={{\theta }_{j0}},\quad j=1,...,N.\end{eqnarray*}$$

We then set

$$\begin{eqnarray}&&{{\psi }_{j}}(x,t):=S(t)\;\circ \;(\sqrt{\rho (x)}{{{\rm e}}^{{\rm i}{{\theta }_{j}}(t)}}).\end{eqnarray} \tag{ 2.10 }$$

Below, we verify that ${{\psi }_{j}}(x,t)$ constructed by relation (2.10) is a smooth solution to the S–L system. Because of proposition 2.1, it suffices to show that the functions ${{\tilde{\psi }}_{j}}(x,t)=\sqrt{\rho (x)}{{{\rm e}}^{{\rm i}{{\theta }_{j}}(t)}}$ satisfy the Lohe system (2.5).

It follows from direct calculations that ${{\tilde{\psi }}_{j}}$ satisfies

$$\begin{eqnarray}&&\frac{{\rm d}{{\tilde{\psi }}_{j}}}{{\rm d}t}=\frac{{\rm d}}{{\rm d}t}\left( \sqrt{\rho (x)}{{{\rm e}}^{{\rm i}{{\theta }_{j}}(t)}} \right)={\rm i}{{\tilde{\psi }}_{j}}{{\dot{\theta }}_{j}}(t)\end{eqnarray} \tag{ 2.11 }$$

and

$$\begin{eqnarray}&&\begin{array}{l} \frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\left( {{\tilde{\psi }}_{k}}-\frac{\langle {{\tilde{\psi }}_{j}},{{\tilde{\psi }}_{k}}\rangle }{\langle {{\tilde{\psi }}_{j}},{{\tilde{\psi }}_{j}}\rangle }{{\tilde{\psi }}_{j}} \right) \\ \quad =\frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\left( \sqrt{\rho (x)}{{{\rm e}}^{{\rm i}{{\theta }_{k}}(t)}}-\sqrt{\rho (x)}\langle \sqrt{\rho (x)}{{{\rm e}}^{{\rm i}{{\theta }_{j}}(t)}},\sqrt{\rho (x)}{{{\rm e}}^{{\rm i}{{\theta }_{k}}(t)}}\rangle {{{\rm e}}^{{\rm i}{{\theta }_{j}}(t)}} \right) \\ \quad =\frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\sqrt{\rho (x)}\left( {{{\rm e}}^{{\rm i}{{\theta }_{k}}(t)}}-{{{\rm e}}^{2{\rm i}{{\theta }_{j}}(t)-{\rm i}{{\theta }_{k}}(t)}}\langle \sqrt{\rho (x)},\sqrt{\rho (x)}\rangle \right) \\ \quad =\frac{K{{\tilde{\psi }}_{j}}}{N}\mathop{\sum }\limits_{k=1}^{N}\left( {{{\rm e}}^{{\rm i}({{\theta }_{k}}(t)-{{\theta }_{j}}(t))}}-{{{\rm e}}^{{\rm i}({{\theta }_{j}}(t)-{{\theta }_{k}}(t))}} \right), \\ \end{array}\end{eqnarray} \tag{ 2.12 }$$

where we used $||\rho |{{|}_{1}}=1$ in (2.9).

Finally, we combine (2.11) and (2.12) to obtain

$$\begin{eqnarray*}\begin{array}{rcl} \frac{{\rm d}{{\tilde{\psi }}_{j}}}{{\rm d}t} & - & \frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\left( {{\tilde{\psi }}_{k}}-\frac{\langle {{\tilde{\psi }}_{k}},{{\tilde{\psi }}_{j}}\rangle }{\langle {{\tilde{\psi }}_{j}},{{\tilde{\psi }}_{j}}\rangle }{{\psi }_{i}} \right) \\ {} & {} & ={{\tilde{\psi }}_{j}}\left( {\rm i}{{{\dot{\theta }}}_{j}}(t)-\frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\left( {{{\rm e}}^{{\rm i}({{\theta }_{k}}(t)-{{\theta }_{j}}(t))}}-{{{\rm e}}^{{\rm i}({{\theta }_{j}}(t)-{{\theta }_{k}}(t))}} \right) \right. \\ {} & {} & ={{\tilde{\psi }}_{j}}{\rm i}\left( {{{\dot{\theta }}}_{j}}(t)-\frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}{\rm sin} ({{\theta }_{k}}(t)-{{\theta }_{j}}(t)) \right) \\ {} & {} & =0. \\ \end{array}\end{eqnarray*}$$

Thus, ${{\psi }_{j}}(t)=S(t){{\tilde{\psi }}_{j}}$ satisfies (1.1).

Note that in the above calculation, the order of ψ_j and ψ_k in the coupling term in (2.2) plays a key role. If we take the coupling term as

$$\begin{eqnarray*}&&{{\psi }_{k}}-\frac{\langle {{\psi }_{k}},{{\psi }_{j}}\rangle }{\langle {{\psi }_{j}},{{\psi }_{j}}\rangle }{{\psi }_{j}}\end{eqnarray*}$$

then the rhs of (2.12) becomes zero. Thus, (1.1) are equivalent with ${{\dot{\theta }}_{i}}=0$, i.e., there is no synchronization.

We first introduce a Lyapunov functional measuring the degree of quantum synchronization. Given an ordered N-tuple of wave functions $\Psi =({{\psi }_{1}},...,{{\psi }_{N}})$, we set

$$\begin{eqnarray*}&&D(\Psi ):=\mathop{{\rm max} }\limits_{1\leqslant i,j\leqslant N}||{{\psi }_{i}}-{{\psi }_{j}}|{{|}_{2}}.\end{eqnarray*}$$

Then, it is easy to see that the functional $D(\Psi )$ is Lipschitz continuous, so it is almost everywhere differentiable.

Definition 3.1. Let $\Psi =({{\psi }_{1}},...,{{\psi }_{N}})$ be a solution to system (1.1). Then the system (1.1) exhibits time-asymptotic quantum synchronization if and only if the following estimate holds

$$\begin{eqnarray}&&\mathop{{\rm lim} }\limits_{t\to \infty }\;D(\Psi (t))=0.\end{eqnarray} \tag{ 3.1 }$$

Remark 3.1. Note that our quantum synchronization given by the relation (3.1) corresponds to the phase synchronization in classical case [13].

Before we proceed to estimate quantum synchronization, we make two remarks. First, in the absence of coupling, i.e., K = 0, the L² distance between solutions to the S–L model is constant, i.e.,

$$\begin{eqnarray*}&&||{{\psi }_{i}}(t)-{{\psi }_{j}}(t)|{{|}_{2}}=||{{\psi }_{i0}}-{{\psi }_{j0}}|{{|}_{2}},\quad \qquad t\geqslant 0.\end{eqnarray*}$$

Second, because of lemma 2.1 and proposition 2.1, we have

$$\begin{eqnarray*}&&||{{\psi }_{i}}(t)-{{\psi }_{j}}(t)|{{|}_{2}}=||S(-t){{\psi }_{i}}(t)-S(-t){{\psi }_{j}}(t)|{{|}_{2}}=||L(t){{\psi }_{i0}}-L(t){{\psi }_{j0}}|{{|}_{2}}.\end{eqnarray*}$$

Thus, quantum synchronization of the S–L model reduces to synchronization of the Lohe system (2.5).

As mentioned, in this section, we consider the space-homogeneous S–L system. We next derive a Gronwall-type inequality for $D(\Psi )$ using conservation of the L² norm along the flow (1.1).

Lemma 3.1. Suppose that the coupling strength and initial data satisfy

$$\begin{eqnarray*}&&K\gt 0\quad \qquad \;{\rm and}\;\quad \qquad ||{{\psi }_{i0}}|{{|}_{2}}=1.\end{eqnarray*}$$

Then, for any solution $\Psi =({{\psi }_{1}},...,{{\psi }_{N}})$ to (2.5), we have

$$\begin{eqnarray*}\begin{array}{rcl} \frac{{\rm d}}{{\rm d}t}||{{\psi }_{i}}-{{\psi }_{j}}||_{2}^{2} & \leqslant & \frac{2K}{N}\mathop{\sum }\limits_{k=1}^{N}\left( -||{{\psi }_{i}}-{{\psi }_{j}}||_{2}^{2}+||{{\psi }_{k}}-{{\psi }_{j}}|{{|}_{2}}\cdot ||{{\psi }_{i}}-{{\psi }_{j}}||_{2}^{2} \right. \\ {} & {} & -\left. |\langle {{\psi }_{i}}-{{\psi }_{j}},{{\psi }_{k}}\rangle {{|}^{2}}+||{{\psi }_{i}}-{{\psi }_{k}}|{{|}_{2}}\cdot ||{{\psi }_{i}}-{{\psi }_{j}}||_{2}^{2} \right). \\ \end{array}\end{eqnarray*}$$

Proof. It follows from lemma 2.3 that

$$\begin{eqnarray*}&&||{{\psi }_{i}}(t)|{{|}_{2}}=1,\quad \qquad i=1,...,N\quad ,\;\;t\gt 0.\end{eqnarray*}$$

Then ψ_i satisfies

$$\begin{eqnarray*}\begin{array}{rcl} \frac{{\rm d}}{{\rm d}t}||{{\psi }_{i}}-{{\psi }_{j}}||_{2}^{2} & = & \left\langle \frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\left( \langle {{\psi }_{j}},{{\psi }_{k}}\rangle {{\psi }_{j}}-\langle {{\psi }_{i}},{{\psi }_{k}}\rangle {{\psi }_{i}} \right),{{\psi }_{i}}-{{\psi }_{j}} \right\rangle \\ {} & {} & +\left\langle {{\psi }_{i}}-{{\psi }_{j}},\frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\left( \langle {{\psi }_{j}},{{\psi }_{k}}\rangle {{\psi }_{j}}-\langle {{\psi }_{i}},{{\psi }_{k}}\rangle {{\psi }_{i}} \right) \right\rangle \\ {} & = & \frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\left[ \langle {{\psi }_{j}},{{\psi }_{k}}\rangle \langle {{\psi }_{j}},{{\psi }_{i}}-{{\psi }_{j}}\rangle -\langle {{\psi }_{i}},{{\psi }_{k}}\rangle \langle {{\psi }_{i}},{{\psi }_{i}}-{{\psi }_{j}}\rangle \right] \\ {} & {} & +\frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\left[ \overline{\langle {{\psi }_{j}},{{\psi }_{k}}\rangle }\langle {{\psi }_{i}}-{{\psi }_{j}},{{\psi }_{j}}\rangle -\overline{\langle {{\psi }_{i}},{{\psi }_{k}}\rangle }\langle {{\psi }_{i}}-{{\psi }_{j}},{{\psi }_{i}}\rangle \right] \\ {} & = & \frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\left[ \langle {{\psi }_{j}},{{\psi }_{k}}\rangle \langle {{\psi }_{j}},{{\psi }_{i}}-{{\psi }_{j}}\rangle -\langle {{\psi }_{i}},{{\psi }_{k}}\rangle \langle {{\psi }_{i}},{{\psi }_{i}}-{{\psi }_{j}}\rangle \right] \\ {} & {} & +\frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\left[ \overline{\langle {{\psi }_{j}},{{\psi }_{k}}\rangle }\overline{\langle {{\psi }_{j}},{{\psi }_{i}}-{{\psi }_{j}}\rangle }-\overline{\langle {{\psi }_{i}},{{\psi }_{k}}\rangle }\overline{\langle {{\psi }_{i}},{{\psi }_{i}}-{{\psi }_{j}}\rangle } \right]. \\ \end{array}\end{eqnarray*}$$

Therefore, we have

$$\begin{eqnarray*}&&\frac{{\rm d}}{{\rm d}t}||{{\psi }_{i}}-{{\psi }_{j}}||_{2}^{2}=\frac{2K}{N}\mathop{\sum }\limits_{k=1}^{N}\operatorname{Re}\left[ \langle {{\psi }_{j}},{{\psi }_{k}}\rangle \langle {{\psi }_{j}},{{\psi }_{i}}-{{\psi }_{j}}\rangle -\langle {{\psi }_{i}},{{\psi }_{k}}\rangle \langle {{\psi }_{i}},{{\psi }_{i}}-{{\psi }_{j}}\rangle \right].\end{eqnarray*}$$

To estimate the rhs of the above relation, we divide it into four parts as follows:

$$\begin{eqnarray*}&&\begin{array}{l} \langle {{\psi }_{j}},{{\psi }_{k}}\rangle \langle {{\psi }_{j}},{{\psi }_{i}}-{{\psi }_{j}}\rangle -\langle {{\psi }_{i}},{{\psi }_{k}}\rangle \langle {{\psi }_{i}},{{\psi }_{i}}-{{\psi }_{j}}\rangle \\ \quad =\langle {{\psi }_{j}},{{\psi }_{k}}\rangle \langle {{\psi }_{j}}-{{\psi }_{i}},{{\psi }_{i}}-{{\psi }_{j}}\rangle +\langle {{\psi }_{j}},{{\psi }_{k}}\rangle \langle {{\psi }_{i}},{{\psi }_{i}}-{{\psi }_{j}}\rangle -\langle {{\psi }_{i}},{{\psi }_{k}}\rangle \langle {{\psi }_{i}},{{\psi }_{i}}-{{\psi }_{j}}\rangle \\ \quad =\langle {{\psi }_{j}},{{\psi }_{k}}\rangle \langle {{\psi }_{j}}-{{\psi }_{i}},{{\psi }_{i}}-{{\psi }_{j}}\rangle +\langle {{\psi }_{j}}-{{\psi }_{i}},{{\psi }_{k}}\rangle \langle {{\psi }_{i}},{{\psi }_{i}}-{{\psi }_{j}}\rangle \\ \quad =\langle {{\psi }_{j}},{{\psi }_{j}}\rangle \langle {{\psi }_{j}}-{{\psi }_{i}},{{\psi }_{i}}-{{\psi }_{j}}\rangle +\langle {{\psi }_{j}},{{\psi }_{k}}-{{\psi }_{j}}\rangle \langle {{\psi }_{j}}-{{\psi }_{i}},{{\psi }_{i}}-{{\psi }_{j}}\rangle \\ \qquad +\langle {{\psi }_{j}}-{{\psi }_{i}},{{\psi }_{k}}\rangle \langle {{\psi }_{i}},{{\psi }_{i}}-{{\psi }_{j}}\rangle \\ \quad =\langle {{\psi }_{j}},{{\psi }_{j}}\rangle \langle {{\psi }_{j}}-{{\psi }_{i}},{{\psi }_{i}}-{{\psi }_{j}}\rangle +\langle {{\psi }_{j}},{{\psi }_{k}}-{{\psi }_{j}}\rangle \langle {{\psi }_{j}}-{{\psi }_{i}},{{\psi }_{i}}-{{\psi }_{j}}\rangle \\ \qquad +\langle {{\psi }_{j}}-{{\psi }_{i}},{{\psi }_{k}}\rangle \langle {{\psi }_{k}},{{\psi }_{i}}-{{\psi }_{j}}\rangle +\langle {{\psi }_{j}}-{{\psi }_{i}},{{\psi }_{k}}\rangle \langle {{\psi }_{i}}-{{\psi }_{k}},{{\psi }_{i}}-{{\psi }_{j}}\rangle \\ \quad ={{\mathcal{I}}_{11}}+{{\mathcal{I}}_{12}}+{{\mathcal{I}}_{13}}+{{\mathcal{I}}_{14}}. \\ \end{array}\end{eqnarray*}$$

We next provide the estimates for ${{\mathcal{I}}_{1i}},\;i=1,...,4$ separately.

• (Estimates for ${{\mathcal{I}}_{11}}$ and ${{\mathcal{I}}_{13}})$: By direct calculation, we have
  $$\begin{eqnarray*}&&{{\mathcal{I}}_{11}}=-||{{\psi }_{i}}-{{\psi }_{j}}||_{2}^{2},\qquad \quad {{\mathcal{I}}_{13}}=-|\langle {{\psi }_{i}}-{{\psi }_{j}},{{\psi }_{k}}\rangle {{|}^{2}}.\end{eqnarray*}$$
• (Estimates for ${{\mathcal{I}}_{12}}$): Since $||{{\psi }_{i}}|{{|}_{2}}=1$, we have
  $$\begin{eqnarray*}\begin{array}{rcl} |{{\mathcal{I}}_{12}}| & \leqslant & ||{{\psi }_{j}}|{{|}_{2}}\cdot ||{{\psi }_{k}}-{{\psi }_{j}}|{{|}_{2}}\cdot ||{{\psi }_{j}}-{{\psi }_{i}}|{{|}_{2}}\cdot ||{{\psi }_{i}}-{{\psi }_{j}}|{{|}_{2}} \\ {} & = & ||{{\psi }_{k}}-{{\psi }_{j}}|{{|}_{2}}\cdot ||{{\psi }_{i}}-{{\psi }_{j}}||_{2}^{2}. \\ \end{array}\end{eqnarray*}$$
• (Estimates for ${{\mathcal{I}}_{14}}$): Similarly, we have
  $$\begin{eqnarray*}\begin{array}{rcl} |{{\mathcal{I}}_{14}}| & \leqslant & ||{{\psi }_{j}}-{{\psi }_{i}}|{{|}_{2}}\cdot ||{{\psi }_{k}}|{{|}_{2}}\cdot ||{{\psi }_{i}}-{{\psi }_{k}}|{{|}_{2}}\cdot ||{{\psi }_{i}}-{{\psi }_{j}}|{{|}_{2}} \\ {} & = & ||{{\psi }_{i}}-{{\psi }_{k}}|{{|}_{2}}\cdot ||{{\psi }_{i}}-{{\psi }_{j}}||_{2}^{2}. \\ \end{array}\end{eqnarray*}$$

Finally, we combine all these estimates to obtain

$$\begin{eqnarray*}\begin{array}{rcl} \frac{{\rm d}}{{\rm d}t}||{{\psi }_{i}}-{{\psi }_{j}}||_{2}^{2} & \leqslant & \frac{2K}{N}\mathop{\sum }\limits_{k=1}^{N}\left( -||{{\psi }_{i}}-{{\psi }_{j}}||_{2}^{2}+||{{\psi }_{k}}-{{\psi }_{j}}||\cdot ||{{\psi }_{i}}-{{\psi }_{j}}||_{2}^{2} \right. \\ {} & {} & -\left. |\langle {{\psi }_{i}}-{{\psi }_{j}},{{\psi }_{k}}\rangle {{|}^{2}}+||{{\psi }_{i}}-{{\psi }_{k}}|{{|}_{2}}\cdot ||{{\psi }_{i}}-{{\psi }_{j}}||_{2}^{2} \right). \\ \end{array}\end{eqnarray*}$$

□

Lemma 3.2. Suppose that the coupling strength and initial data satisfy

$$\begin{eqnarray*}&&K\gt 0\quad \qquad {\rm and}\quad ||{{\psi }_{j0}}|{{|}_{2}}=1,\quad 1\leqslant j\leqslant N.\end{eqnarray*}$$

Then, for any solution $\Psi =({{\psi }_{1}},...,{{\psi }_{N}})$ to (2.5), $D(\Psi )$ satisfies

$$\begin{eqnarray*}&&\dot{D}(\Psi )\leqslant K\left( -D(\Psi )+2D{{(\Psi )}^{2}} \right),\quad \qquad {\rm a}.{\rm e}.\;t\gt 0.\end{eqnarray*}$$

Proof. For $t\in {{\mathbb{R}}_{+}}$, we can take an index pair $({{i}_{t}},{{j}_{t}})$ such that

$$\begin{eqnarray*}&&D(\Psi )(t)=||{{\psi }_{{{i}_{t}}}}(t)-{{\psi }_{{{j}_{t}}}}(t)|{{|}_{2}}.\end{eqnarray*}$$

We now use lemma 3.1 to obtain

$$\begin{eqnarray*}\begin{array}{rcl} \frac{{\rm d}}{{\rm d}t}D{{(\Psi )}^{2}} & = & \frac{{\rm d}}{{\rm d}t}||{{\psi }_{{{i}_{t}}}}(t)-{{\psi }_{{{j}_{t}}}}(t)||_{2}^{2} \\ {} & \leqslant & \frac{2K}{N}\mathop{\sum }\limits_{k=1}^{N}\left( -||{{\psi }_{{{i}_{t}}}}-{{\psi }_{{{j}_{t}}}}||_{2}^{2}+||{{\psi }_{k}}-{{\psi }_{{{j}_{t}}}}|{{|}_{2}}\cdot ||{{\psi }_{{{i}_{t}}}}-{{\psi }_{{{j}_{t}}}}||_{2}^{2} \right. \\ {} & {} & -\left. |\langle {{\psi }_{{{i}_{t}}}}-{{\psi }_{{{j}_{t}}}},{{\psi }_{k}}\rangle {{|}^{2}}+||{{\psi }_{{{i}_{t}}}}-{{\psi }_{k}}|{{|}_{2}}\cdot ||{{\psi }_{{{i}_{t}}}}-{{\psi }_{{{j}_{t}}}}||_{2}^{2} \right). \\ \end{array}\end{eqnarray*}$$

It follows from the definition of $D(\Psi )$ that

$$\begin{eqnarray*}&&\frac{{\rm d}}{{\rm d}t}D{{(\Psi )}^{2}}\leqslant 2K\left( -D{{(\Psi )}^{2}}+2D{{(\Psi )}^{3}} \right)\quad {\rm a}.{\rm e}.\;t\gt 0.\end{eqnarray*}$$

Then, lemma 3.2 yields the quantum synchronization estimate as follows.

Theorem 3.1. Suppose that the coupling strength and initial data satisfy

$$\begin{eqnarray*}&&K\gt 0,\quad \qquad ||{{\psi }_{j0}}|{{|}_{2}}=1,\quad 1\leqslant j\leqslant N,\quad \qquad D({{\Psi }_{0}})\lt \frac{1}{2}.\end{eqnarray*}$$

Then, for any solution $\Psi =({{\psi }_{1}},...,{{\psi }_{N}})$ to (2.5), the diameter $D(\Psi )$ satisfies

$$\begin{eqnarray*}&&D(\Psi (t))\leqslant \frac{D({{\Psi }_{0}})}{D({{\Psi }_{0}})+(1-2D({{\Psi }_{0}})){{{\rm e}}^{Kt}}},\quad \qquad t\geqslant 0.\end{eqnarray*}$$

Proof. It follows from lemma 3.2 that we have

$$\begin{eqnarray*}&&\dot{D}(\Psi )\leqslant K\left( -D(\Psi )+2D{{(\Psi )}^{2}} \right).\end{eqnarray*}$$

By the comparison principle of ODEs, we have

$$\begin{eqnarray*}&&D(\Psi (t))\leqslant \frac{D({{\Psi }_{0}})}{D({{\Psi }_{0}})+(1-2D({{\Psi }_{0}})){{{\rm e}}^{Kt}}}.\end{eqnarray*}$$

Note that the condition $D({{\Psi }_{0}})\lt \frac{1}{2}$ is needed to exclude the finite-time blowup of $D(\Psi (t))$.

Remark 3.2. (1). In section 6 of [18], two finite-dimensional reductions to the S–L model were discussed. For example, for some special class of initial data with the common amplitudes, the S–L model can be reduced to the Kuramoto model which exhibit synchronization for large coupling strength (see section 2.2). However, theorem 3.1 can be applied to initial data with distinct amplitudes.

(2). For the Kuramoto model, exponential synchronization has been extensively studied in [5–7, 9, 8, 13, 14]. As can be seen in section 2.3, the quantum model (1.1) can be reduced to the Kuramoto model with zero natural frequency for some class of initial configurations given by (2.9). There are two important concepts of synchronization for the Kuramoto model. The first one is 'phase synchronization' which means that all relative phases go to zero asymptotically. Our definition 3.1 corresponds to this phase synchronization in classical case, and the second one is the 'complete synchronization' which means that all relative phase velocities (frequencies) go to zero asymptotically. For the Kuramoto model with zero natural frequencies, phase synchronization will occur only for some restricted class of initial configurations which can be confined to the half circle. For example, consider a bipolar configuration $({{\theta }_{1}},{{\theta }_{2}})=(0,\pi )$. This configuration is clearly an equilibrium for the Kuramoto model with zero natural frequency so that there will be no phase synchronization in this case. In contrast, the complete synchronization will occur for a generic initial configurations, although the complete proof for this is not yet available in literature (e.g., for identical oscillators, this can be found in [8]).

In this subsection, we study higher order synchronization of the wave functions. For this, we set

$$\begin{eqnarray*}&&{{M}_{\alpha }}[\Psi (t)]:=\mathop{{\rm max} }\limits_{1\leqslant i\leqslant N}||\partial _{x}^{\alpha }{{\psi }_{i}}(t)||_{2}^{2},\quad \qquad \alpha \in {{\mathbb{Z}}^{d}}.\end{eqnarray*}$$

Lemma 3.3. Suppose that $\Psi =({{\psi }_{1}},...,{{\psi }_{N}})$ be a global solution to (2.5) satisfying

$$\begin{eqnarray*}&&||{{\psi }_{i0}}|{{|}_{2}}=1,\quad 1\leqslant i\leqslant N,\qquad {{M}_{\alpha }}[{{\Psi }_{0}}]\lt \infty ,\qquad D(\Psi (t))\leqslant {{C}_{0}}{{{\rm e}}^{-{{C}_{1}}t}},\quad t\geqslant 0.\end{eqnarray*}$$

Then, we have

$$\begin{eqnarray*}&&\mathop{{\rm sup} }\limits_{0\leqslant t\lt \infty }{{M}_{\alpha }}[\Psi (t)]\leqslant {{M}_{\alpha }}[{{\Psi }_{0}}]{{{\rm e}}^{2K{{C}_{0}}/{{C}_{1}}}}.\end{eqnarray*}$$

Proof. We differentiate (2.5) with respect to t and obtain

$$\begin{eqnarray}&&\frac{{\rm d}}{{\rm d}t}\partial _{x}^{\alpha }{{\psi }_{i}}=\frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\left( \partial _{x}^{\alpha }{{\psi }_{k}}-\langle {{\psi }_{i}},{{\psi }_{k}}\rangle \partial _{x}^{\alpha }{{\psi }_{i}} \right),\end{eqnarray} \tag{ 3.2 }$$

where we used the fact that $\langle {{\psi }_{k}},{{\psi }_{i}}\rangle$ is only dependent on t. Then we use (3.2) to obtain

$$\begin{eqnarray*}&&\begin{array}{l} \frac{1}{2}\frac{{\rm d}}{{\rm d}t}||\partial _{x}^{\alpha }{{\psi }_{i}}(t)||_{2}^{2} \\ \quad =\operatorname{Re}\left\{ \frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\left\langle \partial _{x}^{\alpha }{{\psi }_{k}}(t)-\langle {{\psi }_{i}}(t),{{\psi }_{k}}(t)\rangle \partial _{x}^{\alpha }{{\psi }_{i}}(t),\partial _{x}^{\alpha }{{\psi }_{i}}(t) \right\rangle \right\} \\ \quad =\operatorname{Re}\left\{ \frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\left\langle \partial _{x}^{\alpha }{{\psi }_{k}}(t),\partial _{x}^{\alpha }{{\psi }_{i}}(t) \right\rangle \right\}-\frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\operatorname{Re}\left\{ \langle {{\psi }_{i}}(t),{{\psi }_{k}}(t)\rangle ||\partial _{x}^{\beta }{{\psi }_{i}}(t)|{{|}^{2}} \right\}. \\ \end{array}\end{eqnarray*}$$

For $t\in {{\mathbb{R}}_{+}}$, we can take an index i_t such that

$$\begin{eqnarray*}&&{{M}_{\alpha }}[\psi (t)]=||\partial _{x}^{\alpha }{{\psi }_{{{i}_{t}}}}(t)||_{2}^{2}.\end{eqnarray*}$$

By the above relation, we have

$$\begin{eqnarray*}&&\frac{1}{2}\frac{{\rm d}}{{\rm d}t}{{M}_{\alpha }}[\psi (t)]=\frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\operatorname{Re}\left\{ \left\langle \partial _{x}^{\alpha }{{\psi }_{k}}(t),\partial _{x}^{\alpha }{{\psi }_{i}}(t) \right\rangle \right\}-\frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\operatorname{Re}\left\{ \langle {{\psi }_{i}}(t),{{\psi }_{k}}(t)\rangle {{M}_{\alpha }}[\psi (t)] \right\}.\end{eqnarray*}$$

It follows from performing elementary calculations on the above equation that

$$\begin{eqnarray*}\begin{array}{rcl} \frac{1}{2}\frac{{\rm d}}{{\rm d}t}{{M}_{\alpha }}[\psi (t)] & \leqslant & K{{M}_{\alpha }}[\psi (t)]-\frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\operatorname{Re}\left\{ \left\langle {{\psi }_{i}}(t),{{\psi }_{k}}(t) \right\rangle \right\}{{M}_{\alpha }}[\psi (t)] \\ {} & = & \frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\operatorname{Re}\left\{ \left\langle {{\psi }_{i}}(t),{{\psi }_{i}}(t) \right\rangle \right\}{{M}_{\alpha }}[\psi (t)]-\frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\operatorname{Re}\left\{ \left\langle {{\psi }_{i}}(t),{{\psi }_{k}}(t) \right\rangle \right\}{{M}_{\alpha }}[\psi (t)] \\ {} & = & \frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\operatorname{Re}\left\{ \left\langle {{\psi }_{i}}(t),{{\psi }_{i}}(t)-{{\psi }_{k}}(t) \right\rangle \right\}{{M}_{\alpha }}[\psi (t)] \\ {} & \leqslant & \frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}||{{\psi }_{i}}(t)-{{\psi }_{k}}(t)|{{|}_{2}}{{M}_{\alpha }}[\psi (t)]. \\ \end{array}\end{eqnarray*}$$

This implies

$$\begin{eqnarray*}&&\frac{1}{2}\frac{{\rm d}}{{\rm d}t}{\rm log} {{M}_{\alpha }}[\psi (t)]\leqslant \frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}||{{\psi }_{i}}(t)-{{\psi }_{k}}(t)|{{|}_{2}}\leqslant KD(\Psi (t))\leqslant K{{C}_{0}}{{{\rm e}}^{-{{C}_{1}}t}}.\end{eqnarray*}$$

We integrate the above inequality to obtain the desired result.

Theorem 3.2. Suppose that the coupling strength K and initial data satisfy

$$\begin{eqnarray*}&&K\gt 0,\quad {{M}_{\alpha }}[{{\Psi }_{0}}]\lt \infty ,\quad D({{\Psi }_{0}})\lt \frac{1}{2}\quad {\rm and}\quad ||{{\psi }_{i0}}|{{|}_{2}}=1,\quad 1\leqslant i\leqslant N.\end{eqnarray*}$$

Then, for any solution $\Psi =({{\psi }_{1}},...,{{\psi }_{N}})$ to equation (2.5), we have

$$\begin{eqnarray*}&&||\partial _{x}^{\alpha }({{\psi }_{i}}(t)-{{\psi }_{j}}(t))|{{|}_{2}}\leqslant {{C}_{2}}{{{\rm e}}^{-\frac{K}{2}(2-D{{({{\Psi }_{0}})}^{2}})t}},\end{eqnarray*}$$

where C₂ is the positive constant depending on initial data ψ_i0 and ψ_j0.

Proof. Below, for notational simplicity, we suppress the x dependence in ψ_i, i.e.,

$$\begin{eqnarray*}&&{{\psi }_{i}}(t)\equiv {{\psi }_{i}}(x,t).\end{eqnarray*}$$

Note that $\partial _{x}^{\alpha }({{\psi }_{i}}(t)-{{\psi }_{j}}(t))$ satisfies

$$\begin{eqnarray}&&\frac{{\rm d}}{{\rm d}t}\partial _{x}^{\alpha }({{\psi }_{i}}(t)-{{\psi }_{j}}(t))=\frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\left( \langle {{\psi }_{j}}(t),{{\psi }_{k}}(t)\rangle \partial _{x}^{\alpha }{{\psi }_{j}}(t)-\langle {{\psi }_{i}}(t),{{\psi }_{k}}(t)\rangle \partial _{x}^{\alpha }{{\psi }_{i}}(t) \right).\end{eqnarray} \tag{ 3.3 }$$

We multiply (3.3) by $\partial _{x}^{\alpha }({{\psi }_{i}}(t)-{{\psi }_{j}}(t))$ and integrate the resulting relation with respect to x to obtain

$$\begin{eqnarray}&&\begin{array}{l} \frac{1}{2}\frac{{\rm d}}{{\rm d}t}||\partial _{x}^{\alpha }{{\psi }_{i}}(t)-\partial _{x}^{\alpha }{{\psi }_{j}}(t)||_{2}^{2} \\ \quad =\operatorname{Re}\left\{ \frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\left\langle \langle {{\psi }_{j}}(s),{{\psi }_{k}}(s)\rangle \partial _{x}^{\alpha }{{\psi }_{j}}(s)-\langle {{\psi }_{i}}(s),{{\psi }_{k}}(s)\rangle \partial _{x}^{\alpha }{{\psi }_{i}}(s),\partial _{x}^{\alpha }{{\psi }_{i}}(t)-\partial _{x}^{\alpha }{{\psi }_{j}}(t) \right\rangle \right\} \\ \quad =\operatorname{Re}\left\{ \frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\left\langle \langle {{\psi }_{j}}(s),{{\psi }_{k}}(s)\rangle (\partial _{x}^{\alpha }{{\psi }_{j}}(s)-\partial _{x}^{\alpha }{{\psi }_{i}}(s)),\partial _{x}^{\alpha }{{\psi }_{i}}(t)-\partial _{x}^{\alpha }{{\psi }_{j}}(t) \right\rangle \right\} \\ \qquad -\operatorname{Re}\left\{ \frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\left\langle \langle {{\psi }_{i}}(s)-{{\psi }_{j}}(s),{{\psi }_{k}}(s)\rangle \partial _{x}^{\alpha }{{\psi }_{i}}(s),\partial _{x}^{\alpha }{{\psi }_{i}}(t)-\partial _{x}^{\alpha }{{\psi }_{j}}(t) \right\rangle \right\} \\ \quad =-\operatorname{Re}\left\{ \frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\langle {{\psi }_{j}}(s),{{\psi }_{k}}(s)\rangle \right\}||\partial _{x}^{\alpha }{{\psi }_{i}}(t)-\partial _{x}^{\alpha }{{\psi }_{j}}(t)||_{2}^{2} \\ \qquad -\operatorname{Re}\left\{ \frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}\left\langle \langle {{\psi }_{i}}(s)-{{\psi }_{j}}(s),{{\psi }_{k}}(s)\rangle \partial _{x}^{\alpha }{{\psi }_{i}}(s),\partial _{x}^{\alpha }{{\psi }_{i}}(t)-\partial _{x}^{\alpha }{{\psi }_{j}}(t) \right\rangle \right\} \\ \quad ={{\mathcal{I}}_{21}}+{{\mathcal{I}}_{22}}. \\ \end{array}\end{eqnarray} \tag{ 3.4 }$$

We next estimate the terms ${{\mathcal{I}}_{2i}},\;i=1,2,$ as follows.

• (Estimate of ${{\mathcal{I}}_{21}}$): Note the relation
  $$\begin{eqnarray}\begin{array}{lll} \operatorname{Re}\langle {{\psi }_{j}}(s),{{\psi }_{k}}(s)\rangle & = & \frac{1}{2}\left( \langle {{\psi }_{k}}(s),{{\psi }_{j}}(s)\rangle +\langle {{\psi }_{j}}(s),{{\psi }_{k}}(s)\rangle \right) \\ {} & = & \frac{1}{2}\left( 2-||{{\psi }_{k}}(s)||_{2}^{2}+\langle {{\psi }_{j}}(s),{{\psi }_{k}}(s)\rangle +\langle {{\psi }_{k}}(s),{{\psi }_{j}}(s)\rangle -||{{\psi }_{j}}(s)||_{2}^{2} \right) \\ {} & = & \frac{1}{2}\left( 2-||{{\psi }_{k}}(s)-{{\psi }_{j}}(s)||_{2}^{2} \right) \\ {} & \geqslant & \frac{1}{2}\left( 2-D{{(\Psi )}^{2}} \right)\geqslant \frac{1}{2}\left( 2-D{{({{\Psi }_{0}})}^{2}} \right)=\left( 1-\frac{1}{2}D{{({{\Psi }_{0}})}^{2}} \right). \\ \end{array}\end{eqnarray} \tag{ 3.5 }$$

Here we used the assumption that $D({{\Psi }_{0}})\lt 1/2$ and lemma 3.1 to see the monotonicity of $D(\Phi )$. Relation (3.5) implies

$$\begin{eqnarray}&&{{\mathcal{I}}_{21}}\leqslant -\frac{K}{2}\left( 2-D{{({{\Psi }_{0}})}^{2}} \right)||\partial _{x}^{\alpha }{{\psi }_{i}}(t)-\partial _{x}^{\alpha }{{\psi }_{j}}(t)||_{2}^{2}.\end{eqnarray} \tag{ 3.6 }$$

• (Estimate of ${{\mathcal{I}}_{22}}$): In this case, we use theorem 3.1, lemma 3.3, and
  $$\begin{eqnarray*}&&||{{\psi }_{i}}(s)-{{\psi }_{j}}(s)|{{|}_{2}}\leqslant D(\Psi ),\quad ||{{\psi }_{k}}(s)|{{|}_{2}}=1,\quad ||\partial _{x}^{\alpha }{{\psi }_{i}}(s)|{{|}_{2}}\leqslant {{M}_{\alpha }}[{{\Psi }_{0}}]{{{\rm e}}^{C}}\end{eqnarray*}$$
  to derive
  $$\begin{eqnarray}\begin{array}{rcl} |{{\mathcal{I}}_{22}}| & \leqslant & \frac{K}{N}\mathop{\sum }\limits_{k=1}^{N}||{{\psi }_{i}}(s)-{{\psi }_{j}}(s)|{{|}_{2}}\cdot ||{{\psi }_{k}}(s)|{{|}_{2}}\cdot ||\partial _{x}^{\beta }{{\psi }_{i}}(s)|{{|}_{2}}\cdot ||\partial _{x}^{\alpha }{{\psi }_{i}}(t)-\partial _{x}^{\alpha }{{\psi }_{j}}(t)|{{|}_{2}} \\ {} & \leqslant & KD(\Psi (t)){{M}_{\alpha }}[{{\Psi }_{0}}]{{{\rm e}}^{2K{{C}_{0}}/{{C}_{1}}}}||\partial _{x}^{\alpha }{{\psi }_{i}}(t)-\partial _{x}^{\alpha }{{\psi }_{j}}(t)|{{|}_{2}} \\ {} & \leqslant & \frac{KD({{\Psi }_{0}}){{M}_{\alpha }}[{{\Psi }_{0}}]{{{\rm e}}^{2K{{C}_{0}}/{{C}_{1}}}}}{1-2D({{\Psi }_{0}})}{{{\rm e}}^{-Kt}}||\partial _{x}^{\alpha }{{\psi }_{i}}(t)-\partial _{x}^{\alpha }{{\psi }_{j}}(t)|{{|}_{2}}. \\ \end{array}\end{eqnarray} \tag{ 3.7 }$$

Finally, we combine the estimates (3.4), (3.6), and (3.7) to derive the first-order differential inequality for $||\partial _{x}^{\alpha }({{\psi }_{i}}(t)-{{\psi }_{j}}(t)|{{|}_{2}}$:

$$\begin{eqnarray*}\begin{array}{rcl} \frac{{\rm d}}{{\rm d}t}||\partial _{x}^{\alpha }({{\psi }_{i}}(t)-{{\psi }_{j}}(t))|{{|}_{2}} & \leqslant & \frac{KD({{\Psi }_{0}}){{M}_{\alpha }}[{{\Psi }_{0}}]{{{\rm e}}^{2K{{C}_{0}}/{{C}_{1}}}}}{1-2D({{\Psi }_{0}})}{{{\rm e}}^{-Kt}} \\ {} & {} & -\frac{K}{2}\left( 2-D{{({{\Psi }_{0}})}^{2}} \right)||\partial _{x}^{\alpha }{{\psi }_{i}}(t)-\partial _{x}^{\alpha }{{\psi }_{j}}(t)|{{|}_{2}}. \\ \end{array}\end{eqnarray*}$$

Then, Gronwallʼs lemma implies

$$\begin{eqnarray*}\begin{array}{rcl} ||\partial _{x}^{\alpha }({{\psi }_{i}}(t)-{{\psi }_{j}}(t))|{{|}_{2}} & \leqslant & ||\partial _{x}^{\alpha }({{\psi }_{i0}}-{{\psi }_{j0}})|{{|}_{2}}{{{\rm e}}^{-\frac{K}{2}(2-D{{({{\Psi }_{0}})}^{2}})t}} \\ {} & {} & +\frac{2{{M}_{\alpha }}[{{\Psi }_{0}}]{{{\rm e}}^{2K{{C}_{0}}/{{C}_{1}}}}}{D({{\Psi }_{0}})(1-2D({{\Psi }_{0}}))}\left( {{{\rm e}}^{-\frac{K}{2}(2-D{{({{\Psi }_{0}})}^{2}})t}}-{{{\rm e}}^{-Kt}} \right). \\ \end{array}\end{eqnarray*}$$

This yields the desired result.