Given $h(t),$ and $p(r,t)$ are both functions independent of a variable $z.$
Is it valid to solve the following equation by differentiating both sides with respect to $z$ twice:
$\left[z^2-zh(t)\right]\left[\frac{\partial^2 p(r,t)}{\partial r^2}+\frac{\partial p(r,t)}{\partial r} \frac{1}{r}\right] + \frac{dh(t)}{dt} \frac{1}{h(t)} = 0$ (1)
My thought process was that by doing this I could end up with the following PDE:
$\frac{\partial^2 p(r,t)}{\partial r^2}+\frac{\partial p(r,t)}{\partial r} \frac{1}{r} = 0$ (2)
Update/Edit:
I tried the technique in order to obtain the above differential equation (2) for the p(r,t) function.
Next, I was given the following boundary conditions:
$p(r_1,t) = 0$
$p(r_2,t) = 0$
Solving the PDE I got:
$p(r,t) = e^{c_1(t)}*ln(r) + c_2(t)$
Solving using the boundary conditions I ended up with
$c_2(t) = 0$
$c_1(0) = ln(0)$ which is very wrong...
Am I solving this PDE wrong or is my initial method of obtaining the PDE flawed?
Edit #2:
Regarding the origins of this PDE (note everything is in cylindrical coordinates)
$p(r,t)$ is a function representing pressure, where pressure doesn't depend on $\theta$ or $z$
$h(t)$ represents the height of a certain surface
The geometries are irrelevant and thus a diagram is not necessary.
It is also given that the components of the velocity field are
$v_r(r,z,t) = \frac{1}{2 \mu} * \frac{\partial{p(r,t)}}{\partial r} * [ z^2 - zh(t)]$
$v_\theta = 0$
$v_z(z,t) = \frac{\partial{h(t)}}{\partial t} * \frac{z}{h(t)}$
I got the PDE by applying the continuity equation (a consequence of conservation of mass): $\nabla \cdot \vec{v} $
For an incompressible fluid and cylindrical coordinate that becomes:
$ \frac{1}{r} \frac{\partial{(r v_r)}}{\partial r} + \frac{1}{r} \frac{\partial{(v_\theta)}}{\partial \theta} + \frac{\partial{(v_z)}}{\partial z} = 0$
No, generally speaking you cannot take the partial derivative of both sides of an equation while maintaining equality. For example, if $y=x^2$ then $$x^2-y = 0$$ but it does not follow that $$\frac{\partial}{\partial x}(x^2-y) = 2x = 0.$$
What is true is the following: if $f(x,r,t)$ is sufficiently smooth, and $f(x,r,t)=0$, then necessarily $\frac{d f}{d x}(x,r,t) = 0$. Notice that here we're taking the full derivative with respect to an independent variable. Not a partial derivative.
For your PDE, if you intend to find a single solution $h(t), p(r,t)$ which satisfy your equation for all values of $z$, then your approach of taking the second derivative with respect to $z$ is totally fine.
If you intend to find a one-parameter family of solutions $h_z(t), p_z(r,t)$ that vary with the choice of $z$, you can still take the derivative of both sides, but you will get additional chain rule terms.