## Master 4

(If you are interested check the previous posts: Part 1, Part 2, Part 3

We consider now the case where we have three fluids in the container ${\Omega}$. Then the energy functional has the form

$\displaystyle \mathcal{E}(E)=\sum_{1\leq i

where ${\Sigma_{ij}=\partial^* E_i \cap \partial^* E_j \cap \Omega}$

Since we have

$\displaystyle \mathcal{H}^{N-1}(\Sigma_{ij})+\mathcal{H}^{N-1}(\Sigma_{ik})=\text{Per}_\Omega(E_i),$

we can rewrite the energy functional in the following form

$\displaystyle \mathcal{E}(E)=\sum_{i=1}^3 \gamma_i\text{Per}_\Omega(E_i)+\sum_{i=1}^3 \beta_i \mathcal{H}^{N-1}(\partial^* E_i \cap \partial \Omega)+\sum_{i=1}^3 g\rho_i \int_\Omega x_N\chi_{E_i}dx, \ \ \ \ \ (1)$

where ${\gamma_i, \ i=1..3}$ satisfy the relations ${\gamma_i+\gamma_j=\sigma_{ij}}$ for ${1\leq i < j \leq 3}$. Indeed, we can calculate that

$\displaystyle \gamma_i=\frac{\sigma_{ij}+\sigma_{ik}-\sigma_{kj}}{2},$

and by the physical considerations on the interfacial tensions ${\sigma_{ij}}$ it is reasonable to assume that ${\gamma_i\geq 0}$. In fact, the condition ${\gamma_i \geq 0}$ is a necessary condition for the existence of a lower bound for the functional ${\mathcal{E}}$.

Remark 1 We cannot obtain a formulation similar to (1) for a number of four or more fluids, because the system of equations

$\displaystyle \gamma_i+\gamma_j=\sigma_{ij}, 1 \leq i

is overdetermined, and has solution if and only if the following obvious condition holds on ${\sigma_{ij}}$

$\displaystyle \sigma_{ij}+\sigma_{kl}=\sigma_{ik}+\sigma_{jl},$

for all indexes ${1 \leq i,j,k,l \leq n}$ which are pairwise distinct. These conditions are not physically justified. Take for example the values of the interfacial tensions (measured in dynes) given in the Smithsonian Physical Tables for air, water, olive oil and mercury.

$\begin{tabular}{|c || c | c| c| c |} \hline & Water & Air & Mercury & Olive Oil\\ \hline \hline Water &0 & 75 &392 & 18.6 \\ \hline Air & 75 & 0 &513 &34.6 \\ \hline Mercury &392 &513&0& 317\\ \hline Olive Oil &18.6&34.6&317&0\\ \hline \end{tabular}$

$\displaystyle \sigma_{\text{Air-Water}}+\sigma_{\text{Mercury-Oil}}= 392$

$\displaystyle \sigma_{\text{Air-Mercury}}+\sigma_{\text{Water-Oil}}= 531.6$

The lower bound for the energy functional can be deduced from the general form of the energy if ${\sigma_{ij}=\gamma_i+\gamma_j \geq 0}$, because then we trivially have

$\displaystyle \mathcal{E}(E) \geq -\sum_{i=1}^3 \left(|\beta_i|\text{Per}(\Omega)+|g|\rho_i |\Omega|\max_\Omega |x_N|\right).$

Of course, the condition ${\sigma_{ij} \geq 0}$ is necessary, because otherwise, we can make the boundary between ${E_i}$ and ${E_j}$ to be very rough, making ${\mathcal{H}^{N-1}(\Sigma_{ij}) \rightarrow \infty}$ and therefore ${\mathcal{E}(E) \rightarrow -\infty}$.

This means that one necessary condition for the existence of the minimum is ${\sigma_{ij} \geq 0}$ for all ${1\leq i < j \leq 3}$. If one ${\sigma_{ij}}$ is zero, then the fluid ${E_i}$ can spread freely in ${E_j}$, because the cost of the boundary of ${E_i}$ and ${E_j}$ becomes zero. This contradicts the immiscibility of the fluids ${E_i}$ and ${E_j}$. Therefore, regarding this fact, and the physical considerations, we will assume that ${\sigma_{ij}>0}$ for all ${1 \leq i.

Let’s now turn to the compactness property. Denote ${0<\sigma_i=\min\limits_{j \neq i} \sigma_{ij}}$. Then we have the following inequality:

$\displaystyle \sigma_i \text{Per}_\Omega(E_i)\leq \sum_{ j\neq i} \sigma_{ij}\mathcal{H}^{N-1}(\Sigma_{ij})\leq \mathcal{E}(E)+\sum_{i=1}^3 \left( |\beta_i|\text{Per}(\Omega)+|g|\rho_i |\Omega|\max_\Omega |x_N|\right).$

If we choose ${(E^h)}$ a minimizing sequence for ${\mathcal{E}}$, then ${\mathcal{E}(E^h)}$ is bounded, and by the inequality above it follows that ${\text{Per}_\Omega(E_i^h)}$ is bounded. Since this happens for every ${i=1..3}$ and ${|E_i^h|=c_i,\ i=1..3}$, by the compactness property and a diagonal argument it follows that ${(E^h)}$ has a convergent subsequence in ${(L^1(\Omega))^3}$ topology to some element ${(E)=(E_1,E_2,E_3)}$ such that ${\text{Per}_\Omega(E_i)<\infty,\ i=1..3}$.

To be able to prove the lower semicontinuity of the functional ${\mathcal{E}}$ in the ${(L^1(\Omega))^3}$ topology, in the case when ${\Omega}$ has smooth boundary, we have the necessary condition ${|\beta_i-\beta_j| \leq \sigma_{ij}=\gamma_i+\gamma_j}$, a condition which is similar with the condition we had in the case of two fluids. If these conditions do not hold, we have the following counterexample.

Example 1 In this example we suppose that ${\beta_1 > \beta_2+ \sigma_{12}}$ and we will see that the functional ${\mathcal{E}}$ is not lower semicontinuous. This example can be extended to more dimensions, or to multiple fluids to see that the condition ${|\beta_i-\beta_i|\leq \sigma_{ij}}$ is really necessary for the lower semicontinuity of the energy functional. The main idea is that if ${\beta_1 > \beta_2+\sigma_{12}}$ if we replace the contact region of ${E_1}$ with the container by a thin layer of ${E_2}$ then we get a strictly lower energy value.

Consider the unit square ${[0,1]^2}$ and define ${E_3=[0,1]\times[0,1/2]}$ (this is the fixed position of ${E_3}$). Consider ${E_1=[0,1/2]\times [1/2,1]}$ and ${E_2=[1/2,1]\times[1/2,1]}$ (see the figure).

Define ${E_1^h=[0,1/2]\times [1/2,1-1/h],\ E_2^h=[0,1]^2 \setminus (E_3 \cup E_1^h)}$ (we take a thin strip of ${E_1}$ and pass it to ${E_2}$). Notice that all terms concerning ${E_3}$ are constant, so we just look at the terms containing ${E_1,E_2,E_1^h,E_2^h}$.

$\displaystyle \lim_{h \rightarrow \infty} [\sigma_{12}\mathcal{H}^{N-1}(\Omega \cap \partial E_1^h \cap \partial E_2^h)+\beta_1 \mathcal{H}^{N-1}(\partial\Omega \cap \partial E_1^h)+\beta_2 \mathcal{H}^{N-1}(\partial\Omega \cap \partial E_2^h)]=$

$\displaystyle = \sigma_{12}+\beta_1/2+3\beta_2/2.$

On the other hand

$\displaystyle \sigma_{12}\mathcal{H}^{N-1}(\Omega \cap \partial E_1 \cap \partial E_2)+\beta_1 \mathcal{H}^{N-1}(\partial\Omega \cap \partial E_1)+\beta_2 \mathcal{H}^{N-1}(\partial\Omega \cap \partial E_2)=$

$\displaystyle = \sigma_{12}/2 +\beta_1+\beta_2,$

which proves that

$\displaystyle \limsup_{h \rightarrow \infty} \mathcal{E}(E)-\mathcal{E}(E^h)=\frac{\beta_1-\beta_2-\sigma_{12}}{2}>0$

and ${\mathcal{E}}$ is not lower semicontinuous.

We are now ready to state the main result of this section.

Theorem 1 Consider ${\Omega \subset \Bbb{R}^N}$ an open set with ${C^1}$ boundary. If ${\gamma_i>0, \gamma_i+\gamma_j>0}$ and ${\gamma_i+\gamma_j > |\beta_i-\beta_j|}$ for every ${1\leq i,j \leq 3}$ with ${i \neq j}$ then the functional

$\displaystyle \mathcal{E}(E)=\sum_{i=1}^3 \gamma_i\text{Per}_\Omega(E_i)+\sum_{i=1}^3 \beta_i \mathcal{H}^{N-1}(\partial^* E_i \cap \partial \Omega)+\sum_{i=1}^3 g\rho_i \int_\Omega x_N\chi_{E_i}dx$

is lower semicontinuous with respect to ${(L^1(\Omega))^3}$ convergence of characteristic functions.

Proof: First let’s notice that the gravity term is continuous, and this reduces the problem to proving that

$\displaystyle \mathcal{S}(E)=\sum_{i=1}^3 \gamma_i\text{Per}_\Omega(E_i)+\sum_{i=1}^3 \beta_i \mathcal{H}^{N-1}(\partial^* E_i \cap \partial \Omega)$

is lower semicontinuous.

Pick ${E \subset \mathcal{K}}$ and consider ${(E^h) \subset \mathcal{K}}$ such that ${E^h \rightarrow E}$ in the sense of ${L^1(\Omega)}$ convergence of characteristic functions. Without loss of generality we can assume that ${\beta_1\leq \beta_2\leq \beta_3}$. We notice that ${\chi_{E_1}+\chi_{E_2}+\chi_{E_3}=\chi_{E_1^h}+\chi_{E_2^h}+\chi_{E_3^h}=1}$ almost everywhere in ${\Omega}$. Therefore, we can replace ${\chi_{E_2}}$ and ${\chi_{E_2^h}}$ in terms of ${E_1,E_3,E_1^h,E_3^h}$ in the trace term from the expression of ${\mathcal{S}(E)}$. We also take ${\varepsilon,\rho >0}$ sufficiently small and use the trace inequality. Therefore we have

$\displaystyle \mathcal{S}(E)-\mathcal{S}(E^h)=\sum_{i=1}^3 \bigg[\gamma_i\left( \int_\Omega |D \chi_{E_i}|-\int_\Omega |D \chi_{E_i^h}|\right)+$

$\displaystyle +\beta_i \int_{\partial \Omega}(\chi_{E_i}-\chi_{E_i^h})d\mathcal{H}^{N-1}\bigg] \leq$

$\displaystyle \leq \sum_{i=1}^3 \gamma_i\left( \int_{\Omega \setminus \Omega_\rho} |D \chi_{E_i}|-\int_{\Omega \setminus \Omega_\rho} |D \chi_{E_i^h}|\right)+$

$\displaystyle +\sum_{i=1}^3 \gamma_i \int_{\Omega_\rho} | D \chi_{E_i}| +\sum_{j=1,3} (1+\varepsilon)|\beta_j-\beta_2| \int_{\Omega_\rho} |D \chi_{E_j}|+$

$\displaystyle + \sum_{j=1,3} \left((1+\varepsilon)|\beta_j-\beta_2|-\gamma_j \right)\int_{\Omega_\rho} |D \chi_{E_j^h}|-\gamma_2\int_{\Omega_\rho} |D \chi_{E_2^h}|+$

$\displaystyle +k\sum_{j=1,3}|\beta_j-\beta_2| \int_{\Omega_\rho} |\chi_{E_i}-\chi_{E_i^h}|dx$

Denote

$\displaystyle G(E^h)= \sum_{j=1,3} \left((1+\varepsilon)|\beta_j-\beta_2|-\gamma_j \right)\int_{\Omega_\rho} |D \chi_{E_j^h}|-\gamma_2\int_{\Omega_\rho} |D \chi_{E_2^h}|.$

If we prove that ${\limsup\limits_{h \rightarrow 0} G(E^h) \leq 0}$ then we are done, since

$\displaystyle \limsup_{h \rightarrow 0}\mathcal{F}(E)-\mathcal{F}(E^h) \leq \sum_{i=1}^3 \gamma_i \int_{\Omega_\rho} |D \chi_{E_i}|+\sum_{j=1,3} |\beta_j-\beta_2|\int_{\Omega_\rho}| D \chi_{E_j}|,$

and the right hand side term tends to zero as ${\rho \rightarrow 0}$. So let’s prove that ${\limsup\limits_{h \rightarrow 0} G(E^h) \leq 0}$. If ${\gamma_i \geq (1+\varepsilon)|\beta_j-\beta_2|,\ j=1,3}$ then we are done. If ${\gamma_1 <(1+\varepsilon)|\beta_1-\beta_2|}$, we obtain

$\displaystyle G(E^h) \leq\left((1+\varepsilon) |\beta_1-\beta_2|-\gamma_1\right)\left(\int_{\Omega_\rho}|D \chi_{E_2^h}|+\int_{\Omega_\rho} |D \chi_{E_3^h}|\right)+$

$\displaystyle +\left((1+\varepsilon)(\beta_3-\beta_2)-\gamma_3\right)\int_{\Omega_\rho} |D \chi_{E_3^h}|-\gamma_2\int_{\Omega_\rho}|D \chi_{E_2^h}|=$

$\displaystyle =((1+\varepsilon)(\beta_2-\beta_1)-\gamma_1-\gamma_2)\int_{\Omega_\rho}|D \chi_{E_2^h}|+$

$\displaystyle +((1+\varepsilon)(\beta_3-\beta_1)-\gamma_1-\gamma_3)\int_{\Omega_\rho}|D \chi_{E_3^h}|\leq 0$

for ${\varepsilon >0}$ chosen small enough.

If ${\gamma_3 <(1+\varepsilon)|\beta_3-\beta_2|}$ then a similar inequality takes place. It is easy to see that if both ${\gamma_1 <(1+\varepsilon)|\beta_1-\beta_2|}$ and If ${\gamma_3 <(1+\varepsilon)|\beta_3-\beta_2|}$ hold then we contradict the hypothesis for ${\varepsilon>0}$ small enough.

Therefore the functional ${\mathcal{S}}$ is lower semicontinuous, and the theorem is proved.\hfill ${\square}$

As in the case of two fluids, we can formulate a similar result for ${\Omega}$ with Lipschitz boundary by strengthening the inequalities between ${\gamma_i}$ and ${\beta_i}$ such that the trace inequality for Lipschitz continuous domains can be applied.

Theorem 2 Consider ${\Omega \subset \Bbb{R}^N}$ an open set with Lipschitz continuous boundary. If ${\gamma_i>0, \gamma_i+\gamma_j>0}$ and ${\gamma_i+\gamma_j > \sqrt{1+L^2} |\beta_i-\beta_j|}$ for every ${1\leq i,j \leq 3}$ with ${i \neq j}$ (${L}$ is the Lipschitz constant of ${\partial \Omega}$) then the functional

$\displaystyle \mathcal{E}(E)=\sum_{i=1}^3 \gamma_i\text{Per}_\Omega(E_i)+\sum_{i=1}^3 \beta_i \mathcal{H}^{N-1}(\partial^* E_i \cap \partial \Omega)+\sum_{i=1}^3 g\rho_i \int_\Omega x_N\chi_{E_i}dx$

is lower semicontinuous with respect to ${(L^1(\Omega))^3}$ convergence of characteristic functions.

We will see in the next section that the lower semicontinuity of the functionals presented in this sections work for domains ${\Omega}$ with Lipschitz continuous boundary with the same conditions as for the ${C^1}$ boundary, but the tools used in proving the lower semicontinuity are completely different.