= Solution
The spectral equation $(S)$ is
$$
-\phi_{xx}+u\phi=\lambda\phi,
\qquad\text{equivalently}\qquad
\phi_{xx}=(u-\lambda)\phi.
$$
This is the <KdV Schrodinger spectral problem>. Differentiating it with respect to $t$ gives
$$
\phi_{xxt}=(u_t-\dot\lambda)\phi+(u-\lambda)\phi_t.
$$
For
$$
Q=\phi_t+u_x\phi-2(u+2\lambda)\phi_x,
$$
a direct differentiation, followed by substitution of the two preceding identities, yields
$$
Q_{xx}=(u-\lambda)Q+
\bigl(u_t+u_{xxx}-6uu_x-\dot\lambda\bigr)\phi.
$$
Therefore
$$
\begin{aligned}
\partial_x(\phi_xQ-\phi Q_x)
&=\phi_{xx}Q-\phi Q_{xx}\\
&=\phi^2\bigl(\dot\lambda-u_t-u_{xxx}+6uu_x\bigr).
\end{aligned}
$$
Comparison with $(*)$ gives
$$
\boxed{\alpha=-1.}
$$
This is the reusable <KdV Schrodinger spectral problem> <Wronskian>[Wronskian identity].
Now take $u$ to satisfy the <Korteweg-De Vries equation> and $\phi=\varphi_n$, where $\lambda_n=-\kappa_n^2$. Both $\varphi_n$ and its derivative decay at infinity, as does $u$, so integration of $(*)$ over the real line gives
$$
0=\dot\lambda_n\int_{-\infty}^{\infty}\varphi_n^2\,dx
=\dot\lambda_n.
$$
The normalization was used in the last equality. Hence
$$
\boxed{\lambda_n(t)=\lambda_n(0)},
$$
which is the <Isospectrality of the KdV discrete spectrum>.
With the KdV equation and $\dot\lambda_n=0$, the Wronskian identity says
$$
\partial_x(\varphi_{n,x}Q-\varphi_nQ_x)=0.
$$
Decay at infinity makes the constant zero. Thus $(Q/\varphi_n)_x=0$ between zeros, and continuation across the isolated zeros gives
$$
Q(x,t)=h_n(t)\varphi_n(x,t).
$$
Multiplying by $\varphi_n$ and integrating,
$$
h_n=\int\varphi_n\varphi_{n,t}\,dx
+\int u_x\varphi_n^2\,dx
-2\int(u+2\lambda_n)\varphi_n\varphi_{n,x}\,dx.
$$
The first integral is zero by differentiating the normalization. Integration by parts turns the last integral into $\int u_x\varphi_n^2dx$, so
$$
h_n=2\int u_x\varphi_n^2\,dx.
$$
To evaluate this, differentiate $(S)$ in $x$ and use it to write
$$
u_x\varphi_n^2
=\varphi_n\varphi_{n,xxx}-\varphi_{n,x}\varphi_{n,xx}.
$$
Its integral is $-2\int\varphi_{n,x}\varphi_{n,xx}dx=0$ by decay. Hence
$$
\boxed{h_n=0,\qquad Q=0.}
$$
As $x\to+\infty$, rapid decay of $u$ and $u_x$, together with
$$
\varphi_n\sim c_n(t)e^{-\kappa_nx},
\qquad
\varphi_{n,x}\sim-\kappa_nc_n(t)e^{-\kappa_nx},
$$
reduces $Q=0$ to
$$
c_n'-4\kappa_n^3c_n=0.
$$
Since $\kappa_n$ is constant,
$$
\boxed{c_n(t)=c_n(0)e^{4\kappa_n^3t},}
$$
as stated by the <Evolution of a KdV discrete norming constant>.
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