Change of variables for obtaining a unitary group

Question

Consider the following NLS:

$$i u_t + \Delta u- 2 \operatorname{Re} u = F(u),$$

where $F(u):=(u + \bar{u} + |u|^2)u.$

In Scattering for the Gross–Pitaevskii equation, the authors S. Gustafson, K. Nakanishi, and T.-P. Tsai used a change of variables to get the free evolution as a unitary group:

$$u \mapsto v:=V^{-1}u:= U^{-1} \operatorname{Re} u + i \operatorname{Im}u,$$

where $U:=\sqrt{- \Delta (2-\Delta)^{-1}}$.

Then $v$ satisfies the equation

$$i v_t - \sqrt{- \Delta (2-\Delta)} v = - i V^{-1} i F(V v).$$

I have been trying to understand the motivation of this change of variable but no result! The main obstacles for me is the squire root, I can not see where the square root came from? Any help is appreciated.

Updates

I tried to write the equation as a dynamical system of two PDEs, real and imaginary parts, and tried to diagonalized the operator but I couldn't get the same result.

Answer 1

Let $u=u_1+iu_2$, $v=v_1+iv_2$ with $u_1,u_2,v_1,v_2$ being real valued. What one would like to do is find a $2\times 2$ matrix of real symmetric operators $$ \begin{pmatrix} A & B\\ C & D \end{pmatrix} $$ such that if one lets $$ \begin{pmatrix}v_1\\ v_2\end{pmatrix}:=\begin{pmatrix} A & B\\ C & D \end{pmatrix} \begin{pmatrix}u_1\\ u_2\end{pmatrix}\ , $$ the free equation $i\partial_t u=-\Delta u+2\Re u$ is equivalent to an evolution for $v$ which is of the form $i\partial_t v=H v$ where $H$ is a real symmetric operator, or rather its complexification.

One could start with general $A,B,C,D$ and see what happens, but with the hindsight provided by the authors of the mentioned article, we know we will be able to arrange for $B,C=0$, and $D={\rm Id}$. Let's go ahead and rename $A=U^{-1}$, for some $U$ TBA.

In terms of real and imaginary components, the equation for $u$ is equivalent to the system $$ \left\{ \begin{array}{ccc} \partial_t u_1 &=& -\Delta u_2\\ \partial_t u_2 &=& \Delta u_1 -2u_1 \end{array} \right. $$ while for the components of $v$ we must have $$ \left\{ \begin{array}{ccc} \partial_t v_1 &=& H u_2\\ \partial_t v_2 &=& -H u_1 \end{array} \right. $$ Since we assumed the relations $v_1=U^{-1}u_1$, $v_2=u_2$, we immediately get $$ \partial_t v_1=-U^{-1}\Delta v_2 $$ which suggests $H=-U^{-1}\Delta$. But we also get $$ \partial_t v_2=\partial_t u_2=(\Delta-2)u_1=(\Delta-2)Uv_1 $$ which requires also $H=-(\Delta-2)U$. So consistency for the choice of $H$ needs the equation $$ -U^{-1}\Delta=(2-\Delta)U $$ to hold. Assuming $U$ commutes with $\Delta$, we obtain the desired square root formula.

Answer 2

The problem is to show that the equation $$ iu_t+\Delta u-2\,{\rm Re}\,u=F(u),\label{1}\tag{1} $$ with $u=Vv$, is equivalent to $$i v_t - \sqrt{- \Delta (2-\Delta)} v = - i V^{-1} i F(V v).\label{2}\tag{2}$$

Let me decompose $u=u_1+iu_2$ into real and imaginary parts. By definition $$v= [- \Delta (2-\Delta)^{-1}]^{-1/2} u_1 + i u_2,\label{3}\tag{3}$$ hence $$\sqrt{- \Delta (2-\Delta)} v=(2-\Delta)u_1+i\sqrt{- \Delta (2-\Delta)}u_2.\label{4}\tag{4}$$

I now start from equation (2) and multiply both sides by $Vi$, $$-u_t+V[i(\Delta-2)u_1+\sqrt{- \Delta (2-\Delta)}u_2]=iF(u).\label{5}\tag{5}$$ I note that $$V^{-1}(i\Delta u-2iu_1)=i(\Delta-2)u_1+\sqrt{- \Delta (2-\Delta)}u_2.\label{6}\tag{6}$$ Substitution of equation \eqref{6} into equation \eqref{5} gives equation \eqref{1}, so indeed, equations \eqref{1} and \eqref{2} are equivalent.