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+---
+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 $f(x, \alpha\!=\!0)$ is the optimal route, then a slightly
+different (and therefore worse) path between the same two points can be expressed
+using the parameter $\alpha$:
+
+$$\begin{aligned}
+    f(x, \alpha) = f(x, 0) + \alpha \eta(x)
+    \qquad \mathrm{or} \qquad
+    \delta f = \alpha \eta(x)
+\end{aligned}$$
+
+Where $\eta(x)$ is an arbitrary differentiable deviation.
+Since $f(x, \alpha)$ 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 $\alpha$,
+a condition often written as $\delta J = 0$.
+In the following, the integration limits have been omitted:
+
+$$\begin{aligned}
+    0
+    &= \delta J
+    = \pdv{J}{\alpha} \Big|_{\alpha = 0}
+    = \int \pdv{L}{\alpha} \dd{x}
+    = \int \pdv{L}{f} \pdv{f}{\alpha} + \pdv{L}{f'} \pdv{f'}{\alpha} \dd{x}
+    \\
+    &= \int \pdv{L}{f} \eta + \pdv{L}{f'} \eta' \dd{x}
+    = \Big[ \pdv{L}{f'} \eta \Big]_{x_0}^{x_1} + \int \pdv{L}{f} \eta - \frac{d}{dx} \Big( \pdv{L}{f'} \Big) \eta \dd{x}
+\end{aligned}$$
+
+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 $\eta_n(x_0) = \eta_n(x_1) = 0$:
+
+$$\begin{aligned}
+    f_n(x, \alpha) = f_n(x, 0) + \alpha \eta_n(x)
+\end{aligned}$$
+
+The derivation procedure is identical to the case $N = 1$ from earlier:
+
+$$\begin{aligned}
+    0
+    &= \pdv{J}{\alpha} \Big|_{\alpha = 0}
+    = \int \pdv{L}{\alpha} \dd{x}
+    = \int \sum_{n} \Big( \pdv{L}{f_n} \pdv{f_n}{\alpha} + \pdv{L}{f_n'} \pdv{f_n'}{\alpha} \Big) \dd{x}
+    \\
+    &= \int \sum_{n} \Big( \pdv{L}{f_n} \eta_n + \pdv{L}{f_n'} \eta_n' \Big) \dd{x}
+    \\
+    &= \Big[ \sum_{n} \pdv{L}{f_n'} \eta_n \Big]_{x_0}^{x_1}
+    + \int \sum_{n} \eta_n \bigg( \pdv{L}{f_n} - \frac{d}{dx} \Big( \pdv{L}{f_n'} \Big) \bigg) \dd{x}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+    0
+    &= \pdv{J}{\alpha} \Big|_{\alpha = 0}
+    = \int \pdv{L}{\alpha} \dd{x}
+    = \int \pdv{L}{f} \pdv{f}{\alpha} + \sum_{n} \pdv{L}{f^{(n)}} \pdv{f^{(n)}}{\alpha} \dd{x}
+    \\
+    &= \int \pdv{L}{f} \eta + \sum_{n} \pdv{L}{f^{(n)}} \eta^{(n)} \dd{x}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+    f(x, y; \alpha) = f(x, y; 0) + \alpha \eta(x, y)
+\end{aligned}$$
+
+The derivation procedure starts in the exact same way as before:
+
+$$\begin{aligned}
+    0
+    &= \pdv{J}{\alpha} \Big|_{\alpha = 0}
+    = \iint \pdv{L}{\alpha} \dd{x} \dd{y}
+    \\
+    &= \iint \pdv{L}{f} \pdv{f}{\alpha} + \pdv{L}{f_x} \pdv{f_x}{\alpha} + \pdv{L}{f_y} \pdv{f_y}{\alpha} \dd{x} \dd{y}
+    \\
+    &= \iint \pdv{L}{f} \eta + \pdv{L}{f_x} \eta_x + \pdv{L}{f_y} \eta_y \dd{x} \dd{y}
+\end{aligned}$$
+
+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 $x_1, x_2, ..., x_N$:
+
+$$\begin{aligned}
+    \boxed{
+        0 = \pdv{L}{f} - \sum_{n} \dv{x_n} \Big( \pdv{L}{f_{x_n}} \Big)
+    }
+\end{aligned}$$