lecture4.tex

\section{Lecture 4: Lagrange Multipliers, Near Equilibrium Dynamics \& Oscillators}

\subsection{Lagrange Multipliers: Constrained Optimization}

Our standard Lagrangian approach requires expressing the positions of all particles in 
terms of generalized coordinates, $r_\alpha\left(q_i, t\right)$. This can be complex, 
making it difficult to obtain an explicit form for the Lagrangian, 
$L\left(q_i, \dot{q_i}, t\right)$. A natural question arises: can we formulate mechanics 
directly in terms of the particle positions, $\mathbf{r_\alpha}$, and impose constraints
directly at the level of the equations of motion, rather than at the level of the action?
This is where Lagrange multipliers become a valuable tool.

\subsubsection{Review of Lagrange Multipliers in Calculus}

Let's first recall how Lagrange multipliers work in standard calculus. Suppose we want to 
minimize a function $F\left(x, y\right)$. Without constraints, this involves simply 
finding where the partial derivatives vanish:

\begin{equation}
    \frac{\partial F}{\partial x} = \frac{\partial F}{\partial y} = 0
\end{equation}

However, if we want to minimize $F\left(x, y\right)$ subject to a constraint given by 
$g\left(x, y\right) = 0$, we introduce a Lagrange multiplier, $\lambda$.  We then 
minimize the modified function: $F\left(x, y\right) + \lambda \ g\left(x, y\right)$. This 
leads to the following set of equations:

\begin{align}
    \frac{\partial F}{\partial x} + \lambda \ \frac{\partial g}{\partial x} &= 0 \\
    \frac{\partial F}{\partial y} + \lambda \ \frac{\partial g}{\partial y} &= 0 \\
    g\left(x, y\right) &= 0 
\end{align}

Notice that we now have three equations for three unknowns: $x$, $y$, and $\lambda$. The 
first two equations define the stationary points of $F$ subject to the constraint, and 
the third equation ensures that the constraint is satisfied. This method provides a 
systematic way to solve constrained optimization problems.

\subsubsection{Lagrange Multipliers in the Calculus of Variations}

The same principle can be extended to the calculus of variations. Let's say we have $M$ 
functions $y_i\left(x\right), i = 1, 2, \cdots, M$, and we want to minimize an action 
integral:

\begin{equation}
    I[y_i\left(x\right)] = \int dx \ F\left(y_i, \dot{y_i}, x\right)
\end{equation}

Furthermore, suppose we have $M-N$ constraints of the form $G_k[y_i] = 0, k = 1, 2, \cdots, M - N$. 
The traditional approach would be to express all $M$ functions, $y_i$, in terms of $N$ 
independent degrees of freedom, however, we can avoid this by using Lagrange multipliers. 
To incorporate the constraints, we modify the action integral as follows:

\begin{equation}
    \tilde{I}[y_i, \lambda_k] = \int dx \ \left(F\left(y_i, \dot{y_i}, x\right) + \sum_k \lambda_k \left(x\right) G_k[y_i]\right)
\end{equation}

Here, $\lambda_k(x)$ are now functions of $x$, the independent variable of the integral. 
We treat $\tilde{I}$ as the functional we want to minimize, and obtain its Euler-Lagrange 
equations:

\begin{align}
    \frac{\delta \tilde{I}}{\delta y_i}  = \frac{\delta I}{\delta y_i} + \sum_k \lambda_k \frac{\delta G_k}{\delta y_i} &= 0 \\
    G_k[y_i] &= 0
\end{align}

This gives us a total of $M + (M - N)$ equations. We have $M$ equations from the 
functional derivative with respect to each $y_i$, plus $M-N$ equations from the 
constraints. These equations involve the $M$ unknowns  $y_i$ and the $M-N$ Lagrange 
multiplier functions, $\lambda_k(x)$. These can, in principle, be solved to obtain the 
solution to the constrained variational problem.

\begin{example}
    Compute the equations of motion in \ref{ex:2-1} again, but this time using Lagrange
    multipliers.
\end{example}

\begin{solution}
    Now we treat $x$ \& $y$ as independent variables, and they subject to $x^2+y^2-l^2=0$.
    
    \begin{equation}
        G\left(x, y\right) = x^2 + y^2 - l^2
    \end{equation}

    \begin{equation}
        L = \frac{1}{2} m \left(\dot{x} ^ 2 + \dot{y} ^ 2\right) - mgy
    \end{equation}

    \begin{equation}
        L + \lambda G = \frac{1}{2} m \left(\dot{x} ^ 2 + \dot{y} ^ 2\right) - mgy + \lambda \left(x^2 + y^2 - l^2\right)
    \end{equation}

    So by taking the derivative of $L + \lambda G$ with respect to $x$, we can get:

    \begin{align}
        \label{eq:4-12}
        m \ddot{x} - 2 \lambda x &= 0 \\
        \label{eq:4-13}
        m \ddot{y} + mg - 2 \lambda y &= 0 \\
        x ^ 2 + y ^ 2 &= l ^ 2
    \end{align}

    From \ref{eq:4-12} and \ref{eq:4-13} we get:

    \begin{equation}
        m \frac{\ddot{x}}{x} = m \left(\frac{\ddot{x}}{x} + \frac{g}{y}\right)
    \end{equation}

    \begin{equation}
        \label{eq:4-16}
        \ddot{x} y - \ddot{y} x - gx = 0
    \end{equation}

    Previously from \ref{eq:2-12} and \ref{eq:2-13} we get the time derivatives of the 
    coordinates are:

    \begin{align}
        \dot{x} &= l \cos{\theta} \cdot \dot{\theta} \\
        \dot{y} &= l \sin{\theta} \cdot \dot{\theta} 
    \end{align}

    So the second time derivatives of the coordinates are:

    \begin{align}
        \label{eq:4-19}
        \ddot{x} &= l\left(\cos \theta \ \ddot{\theta} - \sin \theta \ \dot{\theta}^2\right) \\
        \label{eq:4-20}
        \ddot{y} &= l\left(\sin \theta \ \ddot{\theta} + \cos \theta \ \dot{\theta}^2\right)
    \end{align}

    Take \ref{eq:4-19} and \ref{eq:4-20} into \ref{eq:4-16}, we can get:
\end{solution}