这是一份liverpool利物浦大学MATH334的成功案例

$$
D^{2} a_{1}-\frac{a_{1}\left(D a_{1}\right)}{2 a}=0
$$
The condition $D a=0$ as a consequence of has been used. Also simplifies to
$$
D b_{2}-\frac{a_{1}}{a}-\frac{3 b\left(D a_{1}\right)}{2 a}=0
$$
and integration gives the expression for $b_{2}$.
Another possibility is $G_{u}=0$, which gives
$$
\frac{9 b(D a)}{2 a}=\frac{D b-3 b a_{1}}{2 a}
$$
The free term transforms as
$$
u \frac{D b}{2 a}+\mu\left(\frac{D b}{a}-\frac{3 b D a}{2 a^{2}}\right)=\frac{3 b\left(D^{2} a_{1}\right)}{2 a}
$$

MATH334 COURSE NOTES ：

Let us consider a pair of the Zakharov-Shabat type equations,
$$
\psi_{t}(x, y, t)=\left(V_{0}+V_{1} T\right) \psi
$$
and
$$
\psi_{y}(x, y, t)=\left(U_{0}+U_{-1} T^{-1}\right) \psi .
$$
It differs from that used in the previous section by the change $T \rightarrow T^{-1}$ on the right-hand side .
In a $t$-lattice version with $j \in \mathbb{Z}$ we go to
$$
f(x, j+1)=\sum_{m=-M}^{N} U_{m} T^{m} f(x, j)
$$