My query is related to conversion of PDE into bilinear form using some transformation. The detailed description about bilinear forms can be seen in book of Yoshimasa Matsuno . The bilinear operator may be defined as :
$D_{x}^{m}D_{t}^{n}\{f\cdot g\}=(\partial_{x}-\partial_{x^{\prime}})^{m}(\partial_{t}-\partial_{t^{\prime}})^{n}\Big|_{x=x^{\prime},\,t=t^{\prime}}$
For KdV equation
$u_{t}+6uu_{x}+u_{xxx}=0$
one can use transformation
$u=2\frac{\partial^{2}}{\partial x^{2}}(\log f)$
which transforms KdV into
$f_{xt}f-f_{x}f_{t}+f_{xxxx}f-4f_{xxx}f_{x}+3f_{x}^{2}=0$
which further can be written as
$D_{x}(D_{t}+D_{x}^{3})\left\{f\cdot g\right \}=0$
Now this we call a bilinear form for KdV equation. This seems straight but let take some difficult version as given in article at pp.4, section 3.1:
$\int u_{t}\,dx+\frac{1}{147}u^{4}+\frac{1}{7}uu_{x}^{2}+\frac{2}{7}u^{2}u_{2x}+u_{2x}^{2}+u_{x}u_{3x}+uu_{4x}+u_{6x}=0$.
Using following transformation
$$u=42\frac{\partial^{2}}{\partial x^{2}}(\log f)$$
above PDE reduce to bilinear form
$\left(D_{x}D_{t}+\frac{3}{10}D_{x}^{8}+\frac{7}{10}D_{x}^{3}D_{z}\right)\left(f\cdot f\right)=0$
$\left(D_{x}^{6}-D_{x}D_{z}\right)\left(f\cdot f\right)=0$
where additional variable $z$ has been introduced to obtain bilinear form. How in algorithmic manner one can predict such a difficult bilinear form? Although there is package PDEBellII(catalogue identifier AEQP_v1_0) for this type of conversion but that too work only for straightforward bilinear forms.