= Solution
Equality of the mixed derivatives of the auxiliary vector $\Psi$ gives
$$
\Psi_{xt}=U_t\Psi+UV\Psi,
\qquad
\Psi_{tx}=V_x\Psi+VU\Psi.
$$
Thus the <zero-curvature condition> for this convention is
$$
\boxed{U_t-V_x+[U,V]=0}.
$$
Substituting the displayed matrices and collecting powers of the spectral parameter $\lambda$ makes every diagonal entry and every $\lambda$-dependent term cancel. The remaining matrix is
$$
U_t-V_x+[U,V]
=
\begin{pmatrix}
0&iq_t-q_{xx}-2rq^2\\
ir_t+r_{xx}+2qr^2&0
\end{pmatrix}.
$$
Consequently compatibility is equivalent to
$$
\boxed{
ir_t+r_{xx}+2qr^2=0,
\qquad
iq_t-q_{xx}-2rq^2=0
},
$$
so the requested constant is
$$
\boxed{a=2}.
$$
This is the <AKNS Lax pair for the nonlinear Schrodinger equation>.
The reduction
$$
q=\overline r
$$
is preserved by the two equations, which become complex conjugates and reduce to the <Focusing nonlinear Schrodinger equation>
$$
\boxed{ir_t+r_{xx}+2|r|^2r=0}.
$$
The analogous reduction
$$
q=-\overline r
$$
gives the <Defocusing nonlinear Schrödinger equation>
$$
\boxed{ir_t+r_{xx}-2|r|^2r=0}.
$$
For a complex field $r$ and rapidly decreasing boundary conditions, define
$$
H_{\mathrm{foc}}[r,\overline r]
=\int_{\mathbb R}\left(|r_x|^2-|r|^4\right)dx,
\qquad
H_{\mathrm{def}}[r,\overline r]
=\int_{\mathbb R}\left(|r_x|^2+|r|^4\right)dx.
$$
Their <variational derivatives>, after <integration by parts>, are
$$
\frac{\delta H_{\rm foc}}{\delta\overline r}
=-r_{xx}-2|r|^2r,
\qquad
\frac{\delta H_{\rm def}}{\delta\overline r}
=-r_{xx}+2|r|^2r.
$$
Both equations therefore have the <Hamiltonian field equation>
$$
\boxed{
ir_t=\frac{\delta H}{\delta\overline r}
},
$$
or, including the conjugate equation,
$$
\begin{pmatrix}r_t\\ \overline r_t\end{pmatrix}
=
\begin{pmatrix}0&-i\\ i&0\end{pmatrix}
\begin{pmatrix}\delta H/\delta r\\ \delta H/\delta\overline r\end{pmatrix}.
$$
This is the <Hamiltonian form of the cubic nonlinear Schrodinger equations>.
Now put
$$
r(x,t)=e^{-iEt}f(x),
$$
where $E$ and $f$ are real and $f$ is smooth and rapidly decreasing. For the focusing equation,
$$
f''+Ef+2f^3=0.
$$
Multiplication by $f'$ and use of the decay at infinity gives the first integral
$$
(f')^2=-Ef^2-f^4.
$$
A nonzero solution requires $E<0$. Writing $E=-\kappa^2$ with $\kappa>0$, separation of variables, or direct substitution, gives
$$
\boxed{
f(x)=\pm\kappa\operatorname{sech}\bigl(\kappa(x-x_0)\bigr)
}.
$$
Hence
$$
\boxed{
r(x,t)=\pm\kappa e^{i\kappa^2t}
\operatorname{sech}\bigl(\kappa(x-x_0)\bigr)
}
$$
is the <Bright standing soliton of the focusing nonlinear Schrodinger equation>.
For the defocusing equation the profile instead satisfies
$$
f''+Ef-2f^3=0,
$$
with first integral
$$
(f')^2=f^4-Ef^2=f^2(f^2-E).
$$
If a nonzero rapidly decreasing profile existed, $|f|$ would attain a positive maximum $A$. At that point $f'=0$, so the identity forces $E=A^2>0$. But along either tail, where $0<|f|<A$, the right-hand side $f^2(f^2-E)$ is negative, which is impossible. Thus there is no nonzero solution of the prescribed form, as recorded by <No rapidly decaying standing wave for the defocusing cubic nonlinear Schrodinger equation>.
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