path: root/content/know/concept/calculus-of-variations/index.pdc

---
title: "Calculus of variations"
firstLetter: "C"
publishDate: 2021-02-24
categories:
- Mathematics
- Physics

date: 2021-02-24T18:50:06+01:00
draft: false
markup: pandoc
---

# Calculus of variations

The **calculus of variations** lays the mathematical groundwork
for Lagrangian mechanics.

Consider a **functional** $J$, mapping a function $f(x)$ to a scalar value
by integrating over the so-called **Lagrangian** $L$,
which represents an expression involving $x$, $f$ and the derivative $f'$:

$$\begin{aligned}
    J[f] = \int_{x_0}^{x_1} L(f, f', x) \dd{x}
\end{aligned}$$

If $J$ in some way measures the physical "cost" (e.g. energy) of
the path $f(x)$ taken by a physical system,
the **principle of least action** states that $f$ will be a minimum of $J[f]$,
so for example the expended energy will be minimized.

If $f(x, \varepsilon\!=\!0)$ is the optimal route, then a slightly
different (and therefore worse) path between the same two points can be expressed
using the parameter $\varepsilon$:

$$\begin{aligned}
    f(x, \varepsilon) = f(x, 0) + \varepsilon \eta(x)
    \qquad \mathrm{or} \qquad
    \delta f = \varepsilon \eta(x)
\end{aligned}$$

Where $\eta(x)$ is an arbitrary differentiable deviation.
Since $f(x, \varepsilon)$ must start and end in the same points as $f(x,0)$,
we have the boundary conditions:

$$\begin{aligned}
    \eta(x_0) = \eta(x_1) = 0
\end{aligned}$$

Given $L$, the goal is to find an equation for the optimal path $f(x,0)$.
Just like when finding the minimum of a real function,
the minimum $f$ of a functional $J[f]$ is a stationary point
with respect to the deviation weight $\varepsilon$,
a condition often written as $\delta J = 0$.
In the following, the integration limits have been omitted:

$$\begin{aligned}
    0
    &= \delta J
    = \pdv{J}{\varepsilon} \Big|_{\varepsilon = 0}
    = \int \pdv{L}{\varepsilon} \dd{x}
    = \int \pdv{L}{f} \pdv{f}{\varepsilon} + \pdv{L}{f'} \pdv{f'}{\varepsilon} \dd{x}
    \\
    &= \int \pdv{L}{f} \eta + \pdv{L}{f'} \eta' \dd{x}
    = \Big[ \pdv{L}{f'} \eta \Big]_{x_0}^{x_1} + \int \pdv{L}{f} \eta - \frac{d}{dx} \Big( \pdv{L}{f'} \Big) \eta \dd{x}
\end{aligned}$$

The boundary term from partial integration vanishes due to the boundary
conditions for $\eta(x)$. We are thus left with:

$$\begin{aligned}
    0
    = \int \eta \bigg( \pdv{L}{f} - \dv{x} \Big( \pdv{L}{f'} \Big) \bigg) \dd{x}
\end{aligned}$$

This holds for all $\eta$, but $\eta$ is arbitrary, so in fact
only the parenthesized expression matters:

$$\begin{aligned}
    \boxed{
        0 = \pdv{L}{f} - \dv{x} \Big( \pdv{L}{f'} \Big)
    }
\end{aligned}$$

This is known as the **Euler-Lagrange equation** of the Lagrangian $L$,
and its solutions represent the optimal paths $f(x, 0)$.


## Multiple functions

Suppose that the Lagrangian $L$ depends on multiple independent functions
$f_1, f_2, ..., f_N$:

$$\begin{aligned}
    J[f_1, ..., f_N] = \int_{x_0}^{x_1} L(f_1, ..., f_N, f_1', ..., f_N', x) \dd{x}
\end{aligned}$$

In this case, every $f_n(x)$ has its own deviation $\eta_n(x)$,
satisfying $\eta_n(x_0) = \eta_n(x_1) = 0$:

$$\begin{aligned}
    f_n(x, \varepsilon) = f_n(x, 0) + \varepsilon \eta_n(x)
\end{aligned}$$

The derivation procedure is identical to the case $N = 1$ from earlier:

$$\begin{aligned}
    0
    &= \pdv{J}{\varepsilon} \Big|_{\varepsilon = 0}
    = \int \pdv{L}{\varepsilon} \dd{x}
    = \int \sum_{n} \Big( \pdv{L}{f_n} \pdv{f_n}{\varepsilon} + \pdv{L}{f_n'} \pdv{f_n'}{\varepsilon} \Big) \dd{x}
    \\
    &= \int \sum_{n} \Big( \pdv{L}{f_n} \eta_n + \pdv{L}{f_n'} \eta_n' \Big) \dd{x}
    \\
    &= \Big[ \sum_{n} \pdv{L}{f_n'} \eta_n \Big]_{x_0}^{x_1}
    + \int \sum_{n} \eta_n \bigg( \pdv{L}{f_n} - \frac{d}{dx} \Big( \pdv{L}{f_n'} \Big) \bigg) \dd{x}
\end{aligned}$$

Once again, $\eta_n(x)$ is arbitrary and disappears at the boundaries,
so we end up with $N$ equations of the same form as for a single function:

$$\begin{aligned}
    \boxed{
        0 = \pdv{L}{f_1} - \dv{x} \Big( \pdv{L}{f_1'} \Big)
        \quad \cdots \quad
        0 = \pdv{L}{f_N} - \dv{x} \Big( \pdv{L}{f_N'} \Big)
    }
\end{aligned}$$


## Higher-order derivatives

Suppose that the Lagrangian $L$ depends on multiple derivatives of $f(x)$:

$$\begin{aligned}
    J[f] = \int_{x_0}^{x_1} L(f, f', f'', ..., f^{(N)}, x) \dd{x}
\end{aligned}$$

Once again, the derivation procedure is the same as before:

$$\begin{aligned}
    0
    &= \pdv{J}{\varepsilon} \Big|_{\varepsilon = 0}
    = \int \pdv{L}{\varepsilon} \dd{x}
    = \int \pdv{L}{f} \pdv{f}{\varepsilon} + \sum_{n} \pdv{L}{f^{(n)}} \pdv{f^{(n)}}{\varepsilon} \dd{x}
    \\
    &= \int \pdv{L}{f} \eta + \sum_{n} \pdv{L}{f^{(n)}} \eta^{(n)} \dd{x}
\end{aligned}$$

The goal is to turn each $\eta^{(n)}(x)$ into $\eta(x)$, so we need to
partially integrate the $n$th term of the sum $n$ times. In this case,
we will need some additional boundary conditions for $\eta(x)$:

$$\begin{aligned}
    \eta'(x_0) = \eta'(x_1) = 0
    \qquad \cdots \qquad
    \eta^{(N-1)}(x_0) = \eta^{(N-1)}(x_1) = 0
\end{aligned}$$

This eliminates the boundary terms from partial integration, leaving:

$$\begin{aligned}
    0
    &= \int \eta \bigg( \pdv{L}{f} + \sum_{n} (-1)^n \dv[n]{x} \Big( \pdv{L}{f^{(n)}} \Big) \bigg) \dd{x}
\end{aligned}$$

Once again, because $\eta(x)$ is arbitrary, the Euler-Lagrange equation becomes:

$$\begin{aligned}
    \boxed{
        0 = \pdv{L}{f} + \sum_{n} (-1)^n \dv[n]{x} \Big( \pdv{L}{f^{(n)}} \Big)
    }
\end{aligned}$$


## Multiple coordinates

Suppose now that $f$ is a function of multiple variables.
For brevity, we only consider two variables $x$ and $y$,
but the results generalize effortlessly to larger amounts.
The Lagrangian now depends on all the partial derivatives of $f(x, y)$:

$$\begin{aligned}
    J[f] = \iint_{(x_0, y_0)}^{(x_1, y_1)} L(f, f_x, f_y, x, y) \dd{x} \dd{y}
\end{aligned}$$

The arbitrary deviation $\eta$ is then also a function of multiple variables:

$$\begin{aligned}
    f(x, y; \varepsilon) = f(x, y; 0) + \varepsilon \eta(x, y)
\end{aligned}$$

The derivation procedure starts in the exact same way as before:

$$\begin{aligned}
    0
    &= \pdv{J}{\varepsilon} \Big|_{\varepsilon = 0}
    = \iint \pdv{L}{\varepsilon} \dd{x} \dd{y}
    \\
    &= \iint \pdv{L}{f} \pdv{f}{\varepsilon} + \pdv{L}{f_x} \pdv{f_x}{\varepsilon} + \pdv{L}{f_y} \pdv{f_y}{\varepsilon} \dd{x} \dd{y}
    \\
    &= \iint \pdv{L}{f} \eta + \pdv{L}{f_x} \eta_x + \pdv{L}{f_y} \eta_y \dd{x} \dd{y}
\end{aligned}$$

We partially integrate for both $\eta_x$ and $\eta_y$, yielding:

$$\begin{aligned}
    0
    &= \int \Big[ \pdv{L}{f_x} \eta \Big]_{x_0}^{x_1} \dd{y} + \int \Big[ \pdv{L}{f_y} \eta \Big]_{y_0}^{y_1} \dd{x}
    \\
    &\quad + \iint \eta \bigg( \pdv{L}{f} - \dv{x} \Big( \pdv{L}{f_x} \Big) - \dv{y} \Big( \pdv{L}{f_y} \Big) \bigg) \dd{x} \dd{y}
\end{aligned}$$

But now, to eliminate these boundary terms, we need extra conditions for $\eta$:

$$\begin{aligned}
    \forall y: \eta(x_0, y) = \eta(x_1, y) = 0
    \qquad
    \forall x: \eta(x, y_0) = \eta(x, y_1) = 0
\end{aligned}$$

In other words, the deviation $\eta$ must be zero on the whole "box".
Again relying on the fact that $\eta$ is arbitrary, the Euler-Lagrange
equation is:

$$\begin{aligned}
    0 = \pdv{L}{f} - \dv{x} \Big( \pdv{L}{f_x} \Big) - \dv{y} \Big( \pdv{L}{f_y} \Big)
\end{aligned}$$

This generalizes nicely to functions of even more variables $x_1, x_2, ..., x_N$:

$$\begin{aligned}
    \boxed{
        0 = \pdv{L}{f} - \sum_{n} \dv{x_n} \Big( \pdv{L}{f_{x_n}} \Big)
    }
\end{aligned}$$