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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.

For a continuous closed path $\gamma :[0,1]\to \mathbb{C}\setminus \{0\}$, choose a continuous argument liftIts winding number of a continuous closed path about zero isFor a piecewise smooth path this equals $(2\pi i{)}^{-1}{\int }_{\gamma }dz/z$.

If $|\gamma (t)|>|\varphi (t)|$, thennever vanishes, since $|H(s,t)|\ge |\gamma (t)|-s|\varphi (t)|>0$. Thus $H$ is a homotopy through closed paths avoiding zero. By homotopy invariance of winding number, or directly by the dominated-perturbation lemma,

More generally, ${\gamma }_0$ and ${\gamma }_1$ are homotopic by paths in $\Gamma$ when there is a continuous functionwith $H(0,t)={\gamma }_0(t)$, $H(1,t)={\gamma }_1(t)$, and $H(s,0)=H(s,1)$ for every $s$. The winding-number theorem states that such a homotopy implies

For the Fundamental theorem of algebra, let $P(z)=a_n{z}^n+\cdots +a_0$ with $n\ge 1$. For sufficiently large $R$,for every $t$. The dominated-perturbation lemma shows that $t\mapsto P(R{e}^{2\pi it})$ has the same winding number as $t\mapsto a_n{R}^n{e}^{2\pi int}$, namely $n$. If $P$ had no zero, however,would be a homotopy in $\Gamma$ from that loop to the constant loop $P(0)$, whose winding number is zero. This contradiction proves that $P$ has a complex root. This is the winding-number proof of the fundamental theorem of algebra.

Finally suppose that a continuous retraction $r:D^2\to S^1$ existed. The boundary loop $\eta (t)=e^{2\pi it}$ has winding number one, while $(1-s)\eta (t)$ contracts it to zero inside the disc. Composing this contraction with $r$ gives a homotopy through loops in $S^1$ from $r\circ \eta =\eta$ to the constant loop $r(0)$. Their winding numbers are respectively one and zero, contradicting homotopy invariance. Hence there is no such retraction, as in the winding-number proof of the no-retraction theorem.