Edit: Note that I am doing only the first variation, and I am not doing each and every step, mainly those pertinent in understanding how the general shape equation is determined. If you want to see the full derivation, you will need to understand the Geometric Mathematic Primer discussed in Sections 2 and 3 of the book.
- Geometric Methods in the Elastic Theory of Membranes in Liquid
Crystal Phases by Zhong-Can Ou-Yang, Ji-Xing Liu, Yu-Zhang Xie, Xie
Yu-Zhang
$c_{0}$: Spontaneous curvature of the membrane surface
$k_{c}$: Bending rigidity of the vesicle membrane
$H$: Mean curvature of the membrane surface at any point $P$
$K$: Gaussian curvature of the membrane surface at any point $P$
$dA$: Area element of the membrane
$dV$: Volume element enclosed by the closed bilayer
$\lambda$: Surface tension of the bilayer, or the tensile strength acting on the membrane
$\Delta p$: Pressure difference between the inside and outside of the membrane.
The shape energy of a vesicle is given by:
$$
F = F_{c} + \Delta p \int dV + \lambda \int dA
$$
Where
$$
F_{c}=\frac{k_{c}}{2}(2H-c_{0})^{2} = \frac{k_{c}}{2}(c_{1}+c_{2}-c_{0})^{2}
$$
The variation of $dA$ and $dV$ are needed, refer to the book to locate those.
Next we'll calculate the first variation of $F$. And we can break this into components by starting with the first variation $F_{c}$.
$$
\delta ^{(1)}F_{c} = \frac{k_{c}}{2}\oint (2H+c_{0})^{2} \delta ^{(1)}(dA) + \frac{k_{c}}{2}\oint 4(2H+c_{0})^{2}(\delta ^{(1)}H)dA
$$
Where the first order variation of $\psi$ gives us:
$$
\delta ^{(1)}dA = -2H\psi g^{1/2}dudv
$$
$$
\delta ^{(1)}dV = \psi g^{1/2}dudv
$$
$$
\delta ^{(1)}H = (2H^{2}-K))\psi + (1/2)g^{ij}(\psi _{ij}-\Gamma _{ij}^{k}\psi_{k})
$$
Note: $\Gamma_{ij}^{k}$ is the Christoffel symbol defined by (for reference):
$$
\Gamma_{ij}^{k} = \frac{1}{2}g^{kl}(g_{il,j} + g_{jl,i} - g_{ij,l})
$$
And we plug those into the variation of $F_{c}$:
$$
\delta ^{(1)}F_{c} = k_{c}\oint [(2H+c_{0})^{2}((2H^{2}-K)\psi + (1/2)g^{ij}(\psi_{ij}-\Gamma_{ij}^{k}\psi_{k}))]
$$
$$
= k_{c}\oint [(2H+c_{0})(2H^{2}-c_{0}H-2K)\psi + (1/2)g^{ij}(2H+c_{0})\psi_{ij} - g^{ij}\Gamma_{ij}^{k}(2H+c_{0})\psi_{k}]g^{1/2}dudv
$$
And there are two relations ($i,j = u,v$)
$$
\oint f\phi_{i}dudv = -\oint f_{i}\phi dudv
$$
$$
\oint f\phi_{ij}dudv = \oint f_{ij}\phi dudv
$$
So then we have:
$$
\delta ^{(1)}F_{c} = k_{c}\oint \left \{ (2H+c_{0})(2H^{2}-c_{0}H-2K)g^{1/2} + [g^{1/2}g^{ij}(2H+c_{0})]_{ij} + [g^{1/2}g^{ij}(2H+c_{0})\Gamma_{ij}^{k}]_{k} \right \}\psi dudv
$$
And we can re-write:
$$
[g^{1/2}g^{ij}(2H+c_{0})]_{ij} = [(g^{1/2}g^{ij})_{j}(2H+c_{0})]_{i} + [g^{1/2}g^{ij}(2H+c_{0})\Gamma_{ij}^{k}]_{k} \psi dudv
$$
And for functions $f(u,v)$, where $u,v = i,j$, we can directly expand:
$$
[(g^{1/2}g^{ij})_{j}f]_{i} = -(\Gamma_{ij}^{k}g^{1/2}g^{ij}f)_{k}
$$
A Laplacian operator for these surfaces is defined in the book, and is given as:
$$
\bigtriangledown^{2} = g^{1/2}\frac{\partial }{\partial i}(g^{1/2}g^{ij}\frac{\partial }{\partial j})
$$
So then we have:
$$
[g^{1/2}g^{ij}(2H+c_{0})_{j}]_{i} = g^{1/2}\bigtriangledown^{2}(2H+c_{0})
$$
Using these methods in the first variation:
$$
\delta^{(1)}F_{c} = k_{c}\oint [(2H+c_{0})(2H^{2}-c_{0}H-2K) + \bigtriangledown^{2}(2H+c_{0})]\psi g^{1/2}dudv
$$
And now we want the variation of $F$.
$$
\delta^{(1)}F = \delta^{(1)}F_{c} + \delta^{(1)}(\Delta p\int dV) + \delta^{(1)}(\lambda\int dA)
$$
Which gives us:
$$
\delta^{(1)}F = \oint [\Delta p-2\lambda H + k_{c}(2H+c_{0})(2H^{2}-c_{0}-2K) + k_{c}\bigtriangledown^{2}(2H+c_{0})]\psi g^{1/2}dudv
$$
And since $\psi$ is a very small, well smooth function of $u$ and $v$, the vanishing of the first variation of $F$ requires that:
$$
\Delta p = 2\lambda H + k_{c}(2H+c_{0})(2H^{2}-c_{0}H-2K) + k_{c}\bigtriangledown^{2}(2H+c_{0}) = 0
$$
Which is the general shape equation of the vesicle membrane. $c_{0}$ is a constant unless the symmetry effect of the membrane and its environment varies between each point (we assume it doesn't) otherwise $c_{0}$ becomes a function of $u$ and $v$. So we can reduce to:
$$
\Delta p = 2\lambda H + k_{c}(2H+c_{0})(2H^{2}-c_{0}H-2K) + 2k_{c}\bigtriangledown^{2}H = 0
$$
Hope this helps. Again I would locate that book to see the full derivations. I don't know if the visible section of the book on Google shows you everything that you need to know, but I surely hope this points you in the right direction to understanding the problem.
Best Answer
I see two ways in which the thermal de Broglie wavelength is defined. In both cases we can get it from the probability distribution and partition function of an ideal gas. We will consider a 3D ideal gas with non-relativistic dispersion.
First Way
Consider the partion function of an ideal gas:
\begin{align} Z &= \int_{p_x=-\infty}^{+\infty}\int_{p_y=-\infty}^{+\infty}\int_{p_z=-\infty}^{+\infty} e^{-\frac{1}{2mkT}(p_x^2+p_y^2+p_z^2)}dp_xdp_ydp_z\\ &= \left(2\pi mkT \right)^{\frac{3}{2}} \end{align}
Note that this has dimensions of momentum cubed. Noticing this, we can define the characteristic thermal momentum
\begin{align} p_T = \sqrt{2\pi mkT} \end{align}
We can consider the de Broglie wavelength of a particle with this characteristic momentum to get the first definition of the thermal de Broglie wavelength:
$$\lambda_T = \frac{h}{p_T} = \frac{h}{\sqrt{2\pi mkT}}$$
Second Way
Consider the average energy of an ideal 3D gas. This can be found from the equipartion theorem to be
\begin{align} \langle E \rangle = \frac{3}{2}kT \end{align}
A relation for the de Broglie wavelength is
$$ \lambda = \frac{h}{\sqrt{2mE}} $$
Considering the de Broglie wavelength of a particle with energy equal to the average thermal energy of a 3D ideal gas we get the second definition of the thermal de Broglie wavelength:
$$ \lambda_T = \frac{h}{\sqrt{3mkT}} $$
A Note on a Possible Third Way
A third way which would make sense would be to calculate the average de Broglie wavelength of all of the particles in an ideal gas:
$$ \langle \lambda \rangle = \iiint \frac{h}{p} e^{-p^2/2mkT} dp^3 = \frac{2h}{\sqrt{2\pi m kT}} $$
We see that this is within a factor of 2 of the first definition.
Summary
The three approaches differ by factors of order unity so they all refer to similar length scales. In the end the thermal de Broglie wavelength is largely a notational convenience so we don't need to carry around factors of $\frac{h^2}{mkT}$ all over the place so we shouldn't worry too much about the pre-factor. But it is nice to know where the different conventions come from. Though it is largely a notational convenience it does clearly have physical significance since it is related to $\langle \lambda \rangle$.
I have never really seen the third way presented. I have seen the first way presented by far the most often. I think this is because the partition function appears all over the place so very commonly the specific factor $2\pi mkT$ arises so it is nice to give this quantity a name. The second approach may be presented more often in introductory approaches to statistical mechanics. This convention is most problematic because it very clearly relies on the specific problem of a 3 dimensional ideal gas.