Hey I have following two PDEs: \begin{align} \frac{\partial^2 u_1}{\partial t \partial x_3}+\frac{\partial U}{\partial x_3} & = \nu\left(\frac{\partial^3 u_1}{\partial x_3 \partial x_2^2}+\frac{\partial^3 u_1}{\partial x_3 \partial x_3^2}\right) \\[2mm] \frac{\partial^2 u_1}{\partial t \partial x_2}+\frac{\partial U}{\partial x_2} & = \nu\left(\frac{\partial^3 u_1}{\partial x_2 \partial x_2^2}+\frac{\partial^3 u_1}{\partial x_2 \partial x_3^2}\right) \end{align} Since the first PDE contains $\frac{\partial}{\partial x_{3}}$ and the second term contains $\frac{\partial}{\partial x_{2}}$, is it possible to rewrtite each PDE by extracting the derivatives, leading to:
$$\frac{\partial u_1}{\partial t } + U = \nu\left(\frac{\partial^2 u_1}{ \partial x_2^2}+\frac{\partial^2 u_1}{\partial x_3^2}\right) {\Large ?} $$ Or do I need to add the first two PDES leading to: \begin{align} &\frac{\partial}{\partial t}\left(\frac{\partial u_1}{\partial x_3}+\frac{\partial u_1}{\partial x_3}\right)+\frac{\partial U}{\partial x_2}+\frac{\partial U}{\partial x_3} \\[2mm] = &\ \nu\left[\frac{\partial}{\partial x_2^2}\left(\frac{\partial u_1}{\partial x_3}+\frac{\partial u_1}{\partial x_3}\right) + \frac{\partial}{\partial x_3^2}\left(\frac{\partial u_1}{\partial x_3}+\frac{\partial u_1}{\partial x_3}\right)\right] \end{align}
I want to solve for $u_{1}$ while I already know $U$.