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+++ b/content/know/concept/calculus-of-variations/index.pdc
@@ -234,3 +234,107 @@ $$\begin{aligned}
0 = \pdv{L}{f} - \sum_{n} \dv{x_n} \Big( \pdv{L}{f_{x_n}} \Big)
}
\end{aligned}$$
+
+
+## Constraints
+
+So far, for multiple functions $f_1, ... f_N$,
+we have been assuming that all $f_n$ are independent, and by extension all $\eta_n$.
+Suppose that we now have $M < N$ constraints $\phi_m$
+that all $f_n$ need to obey, introducing implicit dependencies between them.
+
+Let us consider constraints $\phi_m$ of the two forms below.
+It is important that they are **holonomic**,
+meaning they do not depend on any derivatives of any $f_n(x)$:
+
+$$\begin{aligned}
+ \phi_m(f_1, ..., f_N, x) = 0
+ \qquad
+ \int_{x_0}^{x_1} \phi_m(f_1, ..., f_N, x) \dd{x} = C_m
+\end{aligned}$$
+
+Where $C_m$ is a constant.
+Note that the first form can also be used for $\phi_m = C_m \neq 0$,
+by simply redefining the constraint as $\phi_m^0 = \phi_m - C_m = 0$.
+
+To solve this constrained optimization problem for $f_n(x)$,
+we introduce [Lagrange multipliers](/know/concept/lagrange-multiplier/) $\lambda_m$.
+In the former case $\lambda_m(x)$ is a function of all $x$, while in the
+latter case $\lambda_m$ is constant:
+
+$$\begin{aligned}
+ \int \lambda_m(x_i) \: \phi_m(\{f_n\}, x) \dd{x} = 0
+ \qquad
+ \lambda_m \int \phi_m(\{f_n\}, x) \dd{x} = \lambda_m C_m
+\end{aligned}$$
+
+The reason for this distinction in $\lambda_m$
+is that we need to find the stationary points with respect to $\varepsilon$
+of both constraint types. Written in the variational form, this is:
+
+$$\begin{aligned}
+ \delta \int \lambda_m \: \phi_m \dd{x} = 0
+\end{aligned}$$
+
+From this, we define a new Lagrangian $\Lambda$ for the functional $J$,
+with the contraints built in:
+
+$$\begin{aligned}
+ J[f_n]
+ &= \int \Lambda(f_1, ..., f_N; f_1', ..., f_N'; \lambda_1, ..., \lambda_M; x) \dd{x}
+ \\
+ &= \int L + \sum_{m} \lambda_m \phi_m \dd{x}
+\end{aligned}$$
+
+Then we derive the Euler-Lagrange equation as usual for $\Lambda$ instead of $L$:
+
+$$\begin{aligned}
+ 0
+ &= \delta \int \Lambda \dd{x}
+ = \int \pdv{\Lambda}{\varepsilon} \dd{x}
+ = \int \sum_n \Big( \pdv{\Lambda}{f_n} \pdv{f_n}{\varepsilon} + \pdv{\Lambda}{f_n'} \pdv{f_n'}{\varepsilon} \Big) \dd{x}
+ \\
+ &= \int \sum_n \Big( \pdv{\Lambda}{f_n} \eta_n + \pdv{\Lambda}{f_n'} \eta_n' \Big) \dd{x}
+ \\
+ &= \Big[ \sum_n \pdv{\Lambda}{f_n'} \eta_n \Big]_{x_0}^{x_1}
+ + \int \sum_n \eta_n \bigg( \pdv{\Lambda}{f_n} - \dv{x} \Big( \pdv{\Lambda}{f_n'} \Big) \bigg) \dd{x}
+\end{aligned}$$
+
+Using the same logic as before, we end up with a set of Euler-Lagrange equations with $\Lambda$:
+
+$$\begin{aligned}
+ 0
+ = \pdv{\Lambda}{f_n} - \dv{x} \Big( \pdv{\Lambda}{f_n'} \Big)
+\end{aligned}$$
+
+By inserting the definition of $\Lambda$, we then get the following.
+Recall that $\phi_m$ is holonomic, and thus independent of all derivatives $f_n'$:
+
+$$\begin{aligned}
+ \boxed{
+ 0
+ = \pdv{L}{f_n} - \dv{x} \Big( \pdv{L}{f_n'} \Big) + \sum_{m} \lambda_m \pdv{\phi_m}{f_n}
+ }
+\end{aligned}$$
+
+These are **Lagrange's equations of the first kind**,
+with their second-kind counterparts being the earlier Euler-Lagrange equations.
+Note that there are $N$ separate equations, one for each $f_n$.
+
+Due to the constraints $\phi_m$, the functions $f_n$ are not independent.
+This is solved by choosing $\lambda_m$ such that $M$ of the $N$ equations hold,
+i.e. solving a system of $M$ equations for $\lambda_m$:
+
+$$\begin{aligned}
+ \dv{x} \Big( \pdv{L}{f_n'} \Big) - \pdv{L}{f_n}
+ = \sum_{m} \lambda_m \pdv{\phi_m}{f_n}
+\end{aligned}$$
+
+And then the remaining $N - M$ equations can be solved in the normal unconstrained way.
+
+
+
+## References
+1. G.B. Arfken, H.J. Weber,
+ *Mathematical methods for physicists*, 6th edition, 2005,
+ Elsevier.