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Deflection of a membrane

Authors: Hans Petter Langtangen and Anders Logg.

Modified for DOLFINx by Jørgen S. Dokken

In the first FEniCSx program, we solved a simple problem which we could easily use to verify the implementation. In this section, we will turn our attentition to a physically more relevant problem with solutions of a somewhat more exciting shape.

We would like to compute the deflection $D(x,y)$ of a two-dimensional, circular membrane of radius $R$, subject to a load $p$ over the membrane. The appropriate PDE model is \begin{align} -T \nabla^2D&=p \quad\text{in }\quad \Omega={(x,y)\vert x^2+y^2\leq R^2 }. \end{align} Here, $T$ is the tension in the membrane (constant), and $p$ is the external pressure load. The boundary of the membrane has no deflection. This implies that $D=0$ is the boundary condition. We model a localized load as a Gaussian function: \begin{align} p(x,y)&=\frac{A}{2\pi\sigma}e^{-\frac{1}{2}\left(\frac{x-x_0}{\sigma}\right)^2-\frac{1}{2}\left(\frac{y-y_0}{\sigma}\right)^2}. \end{align} The parameter $A$ is the amplitude of the pressure, $(x_0, y_0)$ the location of the maximum point of the load, and $\sigma$ the "width" of $p$. We will take the center $(x_0,y_0)$ to be $(0,R_0)$ for some $0<R_0<R$. Then we have \begin{align} p(x,y)&=\frac{A}{2\pi\sigma}e^{-\frac{1}{2}\left(\left(\frac{x}{\sigma}\right)^2 +\left(\frac{y-R_0}{\sigma}\right)^2\right)}. \end{align}

Scaling the equation

There are many physical parameters in this problem, and we can benefit from grouping them by means of scaling. Let us introduce dimensionless coordinates $\bar{x}=\frac{x}{R}$, $\bar{y}=\frac{y}{R}$, and a dimensionless deflection $w=\frac{D}{D_e}$, where $D_e$ is a characteristic size of the deflection. Introducing $\bar{R}_0=\frac{R_0}{R}$, we obtain \begin{align} -\frac{\partial^2 w}{\partial \bar{x}^2} -\frac{\partial^2 w}{\partial \bar{y}^2} &=\frac{R^2A}{2\pi\sigma TD_e}e^{-\frac{R^2}{2\sigma^2}\left(\bar{x}^2+(\bar{y}-\bar{R}_0)^2\right)}\ &=\alpha e^{-\beta^2(\bar{x}^2+(\bar{y}-\bar{R}_0)^2} \end{align} for $\bar{x}^2+\bar{y}^2<1$ where $\alpha = \frac{R^2A}{2\pi\sigma TD_e}$ and $\beta=\frac{R}{\sqrt{2}\sigma}$.

With an appropriate scaling, $w$ and its derivatives are of size unity, so the left-hand side of the scaled PDE is about unity in size, while the right hand side has $\alpha$ as its characteristic size. This suggests choosing alpha to be unity, or around unity. In this particular case, we choose $\alpha=4$. (One can also find the analytical solution in scaled coordinates and show that the maximum deflection $D(0,0)$ is $D_e$ if we choose $\alpha=4$ to determine $D_e$.) With $D_e=\frac{AR^2}{8\pi\sigma T}$ and dropping the bars we obtain the scaled problem \begin{align} -\nabla^2 w = 4e^{-\beta^2(x^2+(y-R_0)^2)} \end{align} to be solved over the unit disc with $w=0$ on the boundary. Now there are only two parameters which vary the dimensionless extent of the pressure, $\beta$, and the location of the pressure peak, $R_0\in[0,1]$. As $\beta\to 0$, the solution will approach the special case $w=1-x^2-y^2$. Given a computed scaled solution $w$, the physical deflection can be computed by \begin{align} D=\frac{AR^2}{8\pi\sigma T}w. \end{align}