Fixing the Poisson equation to match the deformation of elastic sheet with experimental observation I am working on the calculation of the deformation of a circular elastic sheet with radius $R=1.2~m$ when a plate with mass $M$ and radius $r_0 = 4~cm$ is sitting in the center of the sheet.

I used the wave equation without the $z_{tt}$ so that $v_s^2 \Delta z = f(\mathbf{r})$ where $v_s$ is the wave speed on the sheet. Here $f=g= 9.81~m/s^2$ and for the region the plate is sitting, I scaled the load as $f=\tilde{g}=(1+(M/A)/\rho)g$ where $A$ is the plate area $A=\pi r_0^2$ and $\rho = 137~g/m^2$ is the area density of the sheet.
Using the polar form of the Laplacian $\Delta z = \frac{1}{r} \frac{\partial}{\partial r}(r \frac{\partial z}{\partial r}) + \frac{1}{r^2} \frac{\partial^2 z}{\partial \varphi^2}$ and apply the axial symmetry, the general solution to $v_s^2 \Delta z = g$ is found to be $z(r)=\frac{g}{4v_s^2}r^2 + C_1 \log{r} + C_2$. I patched the turning point $r=r_0$ with
$$z(r_0^-)=z(r_0^+)$$
$$z'(r_0^-)=z'(r_0^+)$$
and used a boundary condition $z(R)=0$. This gives me the depression $D \equiv 0-z(r_0)$ as a function of the plate mass $M$:
$$D(M) = \frac{g \log{(R/r_0)}}{2\pi \rho_m v_s^2} \cdot M + \frac{g}{4v_s^2} (R^2-r_0^2)$$
which is a very simple linear response.
The problem came after I plugged in the experimentally measured $v_s^{expt} = 4.4~m/s$. It turns out this value predicts a way large deformation $D \sim 40~cm$, while adopting $v_s^{fit} = 8.0~m/s$ gives a satisfactory prediction.
My doubt is did I missed a constant in the Poisson equation? Maybe it should be $v_s^2 \Delta z = f/4$ instead of $v_s^2 \Delta z = f$?

I'd really appreciate your kind help!
 A: Your derivation seems valid to me. Here is a derivation based on a more physical intuition yielding the same result. The weight P of the mass must be balanced at $r=R$, by a lineic reaction force $F_{y}$ such that $2\pi R F_{y}=-P$. This is actually true if we perform a virtual cut of the elastic sheet along any circumference at $r \geq r_{0}$, so if we introduce the tension of the elastic sheet $T$, for small deflections and neglecting the weight of the sheet for the time being:
$$ 2\pi r T \frac{\partial w}{\partial r} = -P \tag{1}$$
Substituting $T=\rho V_{s}^{2}$ and integrating we find:
$$    w=\frac{-P}{2 \pi \rho V_{s}^{2}} \ln(r) + C_{1} $$
The boundary condition at $r=R$ yields at once:
$$w=-\frac{P}{ 2 \pi \rho V_{s}^{2}} \ln (\frac{r}{R})$$
and finally:
$$d=w(r_{0})= -\frac{mg}{2\pi\rho V_{s}^{2}} \ln \left(\frac{r_{0}}{R}\right)$$
which is exactly the first term of your solution, the second term accounts for the deflection of the sheet due to ist own weight and is an offset in the deflection with mass. To explain the difference with experimental data:


*

*the $V_{s}$ measurement you have might be inconsistent (how was $V_{s}$ measured? If you measured the frequency of a fundamental vibration mode, are you sure you associated this frequency with the correct mode? )

*the sheet elastic behaviour significantly departs from that of a membrane and seems much stiffer, maybe this calls for an elastic behaviour in between a thin plate and a membrane (have you checked the deflection obtained if you assume your elastic sheet deforms as a thin plate under its own weigth?) 
