# Deflection of a membrane#

Authors: Hans Petter Langtangen and Anders Logg.

Modified to 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

(11)#\begin{align} -T \nabla^2D&=p \quad\text{in }\quad \Omega=\{(x,y)\vert x^2+y^2\leq R \}. \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:

(12)#\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 localization 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

(13)#\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

(14)#\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

(15)#\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 to vary the dimensionless extent of the pressure, $$\beta$$, and the localization of the pressure peak, $$R_0\in[0,1]$$. As $$\beta\to 0$$, the solution will approach the special case $$1-x^2-y^2$$. Given a computed scaed solution $$w$$, the physical deflection can be computed by

(16)#\begin{align} D=\frac{AR^2}{8\pi\sigma T}w \end{align}