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

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

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

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.

Let me decompose $u=u_1+iu_2$ into real and imaginary parts, then $$v= [- \Delta (2-\Delta)^{-1}]^{-1/2} u_1 + i u_2,\qquad\qquad(1)$$ hence $$\sqrt{- \Delta (2-\Delta)} v=(2-\Delta)u_1+i\sqrt{- \Delta (2-\Delta)}u_2.\qquad\qquad(2)$$ The equation $$i v_t - \sqrt{- \Delta (2-\Delta)} v = - i V^{-1} i F(V v)\qquad\qquad(3)$$ with $Vv=u$ then takes the form, upon multiplication of both sides by $Vi$, $$-u_t-V[i(2-\Delta)u_1-\sqrt{- \Delta (2-\Delta)}u_2]=iF(u).\qquad\qquad(4)$$

I now note that $$i(2-\Delta)u_1-\sqrt{- \Delta (2-\Delta)}u_2=-V^{-1}(i\Delta u-2iu_1).\qquad\qquad(5)$$ Substitution of equation (5) into equation (4) gives $$iu_t+\Delta u-2u_1=F(u),\qquad\qquad(6)$$ which is the desired result: equations (3) and (6) are equivalent.

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

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

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

Let me decompose $u=u_1+iu_2$ into real and imaginary parts, then $$v= [- \Delta (2-\Delta)^{-1}]^{-1/2} u_1 + i u_2,$$ hence $$\sqrt{- \Delta (2-\Delta)} v=(2-\Delta)u_1+i\sqrt{- \Delta (2-\Delta)}u_2.$$ The equation $$i v_t - \sqrt{- \Delta (2-\Delta)} v = - i V^{-1} i F(V v)$$ with $Vv=u$ then takes the form, upon multiplication of both sides by $Vi$, $$-u_t-V[i(2-\Delta)u_1-\sqrt{- \Delta (2-\Delta)}u_2]=iF(u).$$ This equation should be equivalent to $$- u_t + i\Delta u- 2i u_1 = iF(u),$$ which is the case if $V^{-1}(i\Delta u-2iu_1)$ equals $-[i(2-\Delta)u_1-\sqrt{- \Delta (2-\Delta)}u_2]$. Let me check, $$V^{-1}(i\Delta u-2iu_1)=[- \Delta (2-\Delta)^{-1}]^{-1/2}(-\Delta u_2)+i(\Delta-2)u_1,$$ and it checks out.

Done.

Let me decompose $u=u_1+iu_2$ into real and imaginary parts, then $$v= [- \Delta (2-\Delta)^{-1}]^{-1/2} u_1 + i u_2,\qquad\qquad(1)$$ hence $$\sqrt{- \Delta (2-\Delta)} v=(2-\Delta)u_1+i\sqrt{- \Delta (2-\Delta)}u_2.\qquad\qquad(2)$$ The equation $$i v_t - \sqrt{- \Delta (2-\Delta)} v = - i V^{-1} i F(V v)\qquad\qquad(3)$$ with $Vv=u$ then takes the form, upon multiplication of both sides by $Vi$, $$-u_t-V[i(2-\Delta)u_1-\sqrt{- \Delta (2-\Delta)}u_2]=iF(u).\qquad\qquad(4)$$

I now note that $$i(2-\Delta)u_1-\sqrt{- \Delta (2-\Delta)}u_2=-V^{-1}(i\Delta u-2iu_1).\qquad\qquad(5)$$ Substitution of equation (5) into equation (4) gives $$iu_t+\Delta u-2u_1=F(u),\qquad\qquad(6)$$ which is the desired result: equations (3) and (6) are equivalent.