How is pressure related to the energy cost Consider the following problem found in this webpage.
Consider a spherical bubble of radius $R$, of a certain fluid of density $\rho$, trapped inside of some other
fluid. The bubble is stabilized by the presence of surface tension. Namely, suppose that the bubble has a
nearly, but not perfectly, spherical surface, which we describe by a function $\zeta (\theta, \phi)$, denoting the difference
$\zeta = r - R$ between the actual radius $r$ and the original radius $R$. One can then write the energy cost of
this deformation as:
$$ E = \alpha \int d\theta d \phi \sin\theta(R+\zeta)^2\sqrt{1+ \left(
\frac{1}{R+\zeta} \frac{\partial \zeta}{\partial \theta} \right)^2
+ \left( \frac{1}{(R+\zeta)\sin \theta} \frac{\partial \zeta}{\partial \phi}\right)^2} .$$
The problem is then to argue that the pressure (difference from equilibrium pressure) at the surface of the bubble is: 
$$ P = \frac{2 \alpha \zeta}{R^2} + \frac{\alpha}{R^2} \nabla^2 \zeta $$
where $\nabla^2$ is spherical Laplacian.
I am not sure even in general case if one is given energy cost $E$ then how one would obtain pressure $P$. Is there a general definition or procedure that one would do? 
 A: I think one can get the expression for $P$ without that for $E$. Let $p_o$ be the pressure due to the fluid outside and $p_i$ be the pressure due to the fluid in the bubble. Since the bubble is initially spherical, 
\begin{equation}\tag{e1}\label{e1}
p_o - p_i = \frac{2\alpha}{R}.
\end{equation}
When the bubble is deformed its radius is given by $r = R + \zeta(\theta, \phi)$ so that the equation of its surface if $f(r, \theta, \phi) = 0$ where $f = r - R - \zeta(\theta, \phi)$. If the pressure inside the bubble is $p_f$,
\begin{equation}\tag{e2}\label{e2}
p_o - p_f = \alpha\Delta f,
\end{equation}
where $\Delta$ is the Laplace operator in $r, \theta, \phi$. Note that $R$ is a constant, so that
\begin{equation}\tag{e3}\label{e3}
\Delta f = \frac{2}{r} - \frac{\nabla^2\zeta}{r^2},
\end{equation}
where $\nabla^2$ is the Laplacian in $\theta, \phi$ alone. Now, 
\begin{equation}
r = R\left(1 + \frac{\zeta(\theta,\phi)}{R}\right).
\end{equation}
If $\zeta(\theta,\phi) \ll R$, we can approximate
\begin{equation}\tag{e4}\label{e4}
\frac{1}{r} = \frac{1}{R} - \frac{\zeta}{R^2}.
\end{equation}
Similarly,
\begin{equation}\tag{e5}\label{e5}
\frac{1}{r^2} = \frac{1}{R^2} - \frac{2\zeta}{R^3}.
\end{equation}
Substitution equations (e5) and (e4) in (e3) we get
\begin{equation}\tag{e6}\label{e6}
\Delta f = \frac{2}{R} - 2\frac{\zeta}{R^2} - \frac{\nabla^2\zeta}{R^2},
\end{equation}
where we have ignored the term $\zeta\nabla^2\zeta$ it being of a higher order in $\zeta$. From equation (e2) and (e6),
\begin{equation}\tag{e7}\label{e7}
p_o - p_f = \frac{2\alpha}{R} - \frac{2\alpha\zeta}{R^2} - \frac{\alpha\nabla^2\zeta}{R^2}
\end{equation}
Subtracting (e7) from (e1) we get
\begin{equation}\tag{e8}\label{e8}
p_f - p_i = \frac{2\alpha\zeta}{R^2} + \frac{\alpha\nabla^2\zeta}{R^2}.
\end{equation}
$p_f - p_i$ is the difference in the deformed bubble from its equilibrium pressure.
The integral in the expression for $E$ is just the area of a deformed sphere. You may want to refer to one of my questions for more details.
