In our textbook, we have the problem:
$u_{xx}+u_{yy}=0$ where $u(x,0)=g(x)$
We are told that the transformation $v=u_y$ transforms this PDE into:
$v_{xx}+v_{yy}=0$ where $v(x,0)=g(x)$
I am having trouble understanding how the textbook got this transformed PDE. I understand how we transformed the initial condition. Any guidance or tips would be much appreciated!
Thank you in advance.
At first glance I thought there must be a typo, but it turns out this works fine. Letting $v = u_{y}$ and computing partial derivatives of $v$ gives us
$$\begin{align*} v_{xx} &= u_{yxx}\\ v_{yy} &= u_{yyy} \end{align*} $$
Assuming symmetry of partial derivatives $$u_{yxx} = u_{xxy},$$
and so we have $$v_{xx} + v_{yy} = u_{xxy} + u_{yyy} = \frac{\partial}{\partial y}\left(u_{xx} + u_{yy}\right) = 0.$$