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The spectral equation $(S)$ isThis is the KdV Schrodinger spectral problem. Differentiating it with respect to $t$ givesFora direct differentiation, followed by substitution of the two preceding identities, yieldsThereforeComparison with $(\ast )$ givesThis is the reusable KdV Schrodinger spectral problem Wronskian identity.

Now take $u$ to satisfy the Korteweg-De Vries equation and $\varphi ={\phi }_n$, where ${\lambda }_n=-{\kappa }_n^2$. Both ${\phi }_n$ and its derivative decay at infinity, as does $u$, so integration of $(\ast )$ over the real line givesThe normalization was used in the last equality. Hencewhich is the Isospectrality of the KdV discrete spectrum.

With the KdV equation and ${\dot{\lambda }}_n=0$, the Wronskian identity saysDecay at infinity makes the constant zero. Thus $(Q/{\phi }_n{)}_x=0$ between zeros, and continuation across the isolated zeros givesMultiplying by ${\phi }_n$ and integrating,The first integral is zero by differentiating the normalization. Integration by parts turns the last integral into $\int u_x{\phi }_n^{2}dx$, soTo evaluate this, differentiate $(S)$ in $x$ and use it to writeIts integral is $-2\int {\phi }_{n,x}{\phi }_{n,xx}dx=0$ by decay. Hence

As $x\to +\infty$, rapid decay of $u$ and $u_x$, together withreduces $Q=0$ toSince ${\kappa }_n$ is constant,as stated by the Evolution of a KdV discrete norming constant.