I have some trouble in understanding the origin of lagrange multipliers for minimization of a functional with some constraints. I would appreciate if anyone could clarify this in detail.
The functional in my expression is given by,
$$ \tag{1} F(\phi_1, \phi_2, \phi_3) = \int_{\Omega} \left(\Sigma_1 |\nabla \phi_1|^2 + \Sigma_2 |\nabla \phi_2|^2 + \Sigma_3 |\nabla \phi_3|^2 \right) d\Omega + \int_{\partial \Omega} \psi^s ds $$
with $\psi^s = \sigma_1 (3\phi_1^2 - 2\phi_1^3) + \sigma_2 (3\phi_2^2 - 2\phi_2^3) + \sigma_3 (3\phi_3^2 - 2\phi_3^3)$. Here, $\Sigma_i, \sigma_i \ \forall \ i = 1,2,3$ are constants. $F$ has to be minimized with the constraint $$ \tag{2} \phi_1 + \phi_2 + \phi_3 = 1 $$
Minimization of this functional is performed using variational derivatives and I am just interested in the boundary terms along $\partial \Omega$. This yields
$$ \tag{3} 2 \Sigma_1 \nabla \phi_1 \cdot \mathbf{n} + 6 \sigma_1 \phi_1 (1 - \phi_1) - \Gamma_1 = 0 \\ 2 \Sigma_2 \nabla \phi_2 \cdot \mathbf{n} + 6 \sigma_2 \phi_2 (1 - \phi_2) - \Gamma_2 = 0 \\ 2 \Sigma_3 \nabla \phi_3 \cdot \mathbf{n} + 6 \sigma_3 \phi_3 (1 - \phi_3) - \Gamma_3 = 0 $$
I do understand the procedure of this minimization, but the article I refer directly uses different lagrange multipliers $\Gamma_1, \Gamma_2, \Gamma_3$ in Eq. (3).
I do understand that Eq. (3) would satisfy the constraint (2) when
$$ \tag{4} 6 \frac{\sigma_1}{\Sigma_1} \phi_1 (1 - \phi_1) + 6 \frac{\sigma_2}{\Sigma_2} \phi_2 (1 - \phi_2) + 6 \frac{\sigma_3}{\Sigma_3} \phi_3 (1 - \phi_3) - \frac{\Gamma_1}{\Sigma_1} - \frac{\Gamma_2}{\Sigma_2} - \frac{\Gamma_3}{\Sigma_3} = 0 $$
An additional constraint which needs to be satisfied in my case is
$$ \tag{5} \Gamma_1|_{\phi_1 = 0} = 0, \Gamma_2|_{\phi_2 = 0} = 0, \Gamma_3|_{\phi_3 = 0} = 0. $$
I also understand that if the constraint (5) was absent, this problem could be easily solved by adding the term $\int_{\partial \Omega} \Gamma (\phi_1 + \phi_2 + \phi_3 - 1)$ to Eq. (1) such that $\Gamma_1 = \Gamma_2 = \Gamma_3 = \Gamma$. In that case, Eq. (3) could be easily manipulated to yield an expression of $\Gamma$. What I don't understand is how $\Gamma_1, \Gamma_2, \Gamma_3$ need to be added to the functional in (1) to yield Eq. (3).