shape of membrane on circular frame with pressure difference This is a problem I thought about recently, but I have no idea how to go about it: Consider a membrane evenly stretched across a round frame. What shape does this membrane take, when you have different air pressures on each side of the membrane? 
Basically we have a constant force in normal direction for every small piece of the same area. 
 A: The equation describing membranes separating two fluids with a pressure differential is the Young-Laplace equation first considered in the early 1800s. In modern geometric language the equation states that the mean curvature of the membrane is proportional to the pressure differential. 
In the case of the constant pressure differential and a circular boundary, just as the case of the meniscus in capillary surfaces, the solution is a portion of the round sphere. The radius of the sphere depends on the elasticity constant of the membrane and the pressure differential. 

For general membranes (idealized to 2D sheet):
Suppose the configuration "at rest" is that of a unit disc $D = \{x\in \mathbb{R}^2: |x| \leq 1\}$. The deformation map is $$\Phi: D \to \mathbb{R}^3$$ with the condition $\Phi |_{\partial D}$ is the expected embedding of the unit circle into the $x_1-x_2$ plane in $\mathbb{R}^3$. The local elastic energy density is (under isotropic assumption)
$$ e[\Phi] = f(|\nabla \Phi|^2, |\partial_1 \Phi \wedge \partial_2 \Phi|^2)$$
where the function $f$ models the material properties of the membrane
The case where $f(a,b) = \sqrt{b}$ is the soap-film model (minimizing area under constraint). By changing the function $f$ you get different models of elastic membranes. 
Once you have specified a function $f$, the situation you are interested in is described by thinking in terms of calculus of variations:


*

*Let $E[\Phi] = \int_D e[\Phi]~\mathrm{d}A$ be the total energy. We take its $L^2$ gradient $\nabla_\Phi E$ by defining it as the function satisfying 
$$ \int_D (\nabla_\Phi E)\psi ~\mathrm{d}A = \delta E[\Phi][\psi] $$
for every $\psi:D\to\mathbb{R}^3$ that vanishes on the boundary. Here $\delta E$ is the variation of $E$. 

*The pressure gives force per unit area; but the area is the stretched area. So the force density when translated into the parametrisation by the rest state is in fact $p(\Phi) |\partial_1\Phi \wedge \partial_2\Phi|$, the pressure times the area density. 

*Force balance requires then 
$$ p | \partial_1\Phi \wedge \partial_2\Phi| = \nabla_\phi E $$
(basically the Euler-Lagrange equation with constraint); I removed the dependence of the pressure $p$ on position $\Phi$ since you've stated that we are facing a constant pressure differential. This equation is in general a nonlinear system of elliptic partial differential equations. The precise form of this equation depends, again, on the function $f$ that you choose. 


In general probably this system of equations cannot be solved explicitly. 
A: This is classic  minimal area soap film surface , it is assumed that the property of surface tension does not change with pressure, or that that the membrane is very thin.
Normal curvature $k_n$ is constant everywhere and is given by mathematical treatment as Constant Mean Curvature  (CMC) surfaces.
For a circular boundary film a spherical bubble grows in proportion to the  pressure ( force per unit area). It is given in physics also under surface tension.
$$ p = 2 T k_n  = 2 T /R $$ 
A book by Cyril Isenberg ( Science of Soap films) is insightful.
A: To be honest, I am not really sure, but i think the equation that describes this situation is just $\Delta f(x,y)=k$, where $\Delta$ is the Laplace operator and $k$ is the constant pressure. For a circular membrane the solution would be a simple paraboloid.
