= Solution
Compatibility of the auxiliary system, computed on a fundamental <matrix> of solutions, gives the <zero-curvature condition>
$$
\boxed{U_t-V_x+[U,V]=0.}
$$
Indeed equality of the mixed <derivatives> gives $(U_t+UV-V_x-VU)\Psi=0$. A single possibly zero vector solution would not justify the <matrix> identity; compatibility means a full fundamental family.
Put $H=W_t-VW$. Using the <zero-curvature condition> gives $H_x=UH$. At $x=0$, $W_t=0$ and hence $H(0)=-V(0)$. Uniqueness of the initial-value problem gives $H=-WV(0)$, or $W_t=VW-WV(0)$. Periodicity implies $V(2\pi)=V(0)=v$, so the <monodromy matrix> $w=W(2\pi)$ obeys the <Lax equation>
$$
\boxed{w_t=[v,w].}
$$
Cyclicity of the trace yields $\frac d{dt}\operatorname{tr}(w^k)=k\operatorname{tr}(w^{k-1}[v,w])=0$ for every $k\geq1$.
There is a genuine factor-of-two error in the printed sine-Gordon pair. Assign the second printed <matrix> to $U$ and the first to $V_{\rm print}$:
$$
U=\frac i2\begin{pmatrix}2\lambda&u_x\\u_x&-2\lambda\end{pmatrix},\qquad
V_{\rm print}=\frac1{2i\lambda}\begin{pmatrix}\cos u&-i\sin u\\i\sin u&-\cos u\end{pmatrix}.
$$
Direct multiplication gives
$$
U_t-(V_{\rm print})_x+[U,V_{\rm print}]
=\frac i2\begin{pmatrix}0&u_{xt}-2\sin u\\u_{xt}-2\sin u&0\end{pmatrix}.
$$
Thus \b[the printed <matrices> encode $u_{xt}=2\sin u$].
For the stated <Sine-Gordon equation>, explicitly repair the first prefactor by taking $V=V_{\rm print}/2$, that is $1/(4i\lambda)$. Its <zero-curvature condition> becomes $u_{xt}=\sin u$. For this corrected pair, $\operatorname{tr}(w(t,\lambda)^k)$ is conserved for every $\lambda\ne0$. Since $U$ is entire in $\lambda$, so are $W$ and these traces. Their Taylor coefficients in $\lambda$ supply an infinite family of <first integrals>, by the identity theorem extending conservation through $\lambda=0$. At a fixed spectral parameter the higher traces of a $2\times2$ <matrix> are related by the <Cayley-Hamilton theorem>; \b[no independence of the resulting <first integrals> is claimed].
Back to article page
