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+---
+title: "Beltrami identity"
+firstLetter: "B"
+publishDate: 2022-09-17
+categories:
+- Physics
+- Mathematics
+
+date: 2022-09-10T13:39:06+02:00
+draft: false
+markup: pandoc
+---
+
+# Beltrami identity
+
+Consider a general functional $J[f]$ of the following form,
+with $f(x)$ an unknown function:
+
+$$\begin{aligned}
+    J[f] = \int_{x_0}^{x_1} L(f, f', x) \dd{x}
+\end{aligned}$$
+
+Where $L$ is the Lagrangian.
+To find the $f$ that maximizes or minimizes $J[f]$,
+the [calculus of variations](/know/concept/calculus-of-variations/)
+states that the Euler-Lagrange equation must be solved for $f$:
+
+$$\begin{aligned}
+    0 = \pdv{L}{f} - \dv{x} \Big( \pdv{L}{f'} \Big)
+\end{aligned}$$
+
+We now want to know exactly how $L$ depends on the free variable $x$,
+since it is a function of $x$, $f(x)$ and $f'(x)$.
+Using the chain rule:
+
+$$\begin{aligned}
+    \dv{L}{x}
+    = \pdv{L}{f} \dv{f}{x} + \pdv{L}{f'} \dv{f'}{x} + \pdv{L}{x}
+\end{aligned}$$
+
+Substituting the Euler-Lagrange equation into the first term gives us:
+
+$$\begin{aligned}
+    \dv{L}{x}
+    &= f' \dv{x} \Big( \pdv{L}{f'} \Big) + \dv{f'}{x} \pdv{L}{f'} + \pdv{L}{x}
+    \\
+    &= \dv{x} \bigg( f' \pdv{L}{f'} \bigg) + \pdv{L}{x}
+\end{aligned}$$
+
+Although we started from the "hard" derivative $\dv*{L}{x}$,
+we arrive at an expression for the "soft" derivative $\pdv*{L}{x}$,
+describing the *explicit* dependence of $L$ on $x$:
+
+$$\begin{aligned}
+    - \pdv{L}{x}
+    = \dv{x} \bigg( f' \pdv{L}{f'} - L \bigg)
+\end{aligned}$$
+
+What if $L$ does not explicitly depend on $x$, i.e. $\pdv*{L}{x} = 0$?
+In that case, the equation can be integrated to give the **Beltrami identity**:
+
+$$\begin{aligned}
+    \boxed{
+        f' \pdv{L}{f'} - L
+        = C
+    }
+\end{aligned}$$
+
+Where $C$ is a constant.
+This says that the left-hand side is a conserved quantity in $x$,
+which could be useful to know.
+If we insert a concrete expression for $L$,
+the Beltrami identity might be easier to solve for $f$ than the full Euler-Lagrange equation.
+The assumption $\pdv*{L}{x} = 0$ is justified;
+for example, if $x$ is time, it means that the potential is time-independent.
+
+
+## Higher dimensions
+
+Above, a 1D problem was considered, i.e. $f$ depended only on a single variable $x$.
+Consider now a 2D problem, such that $J[f]$ is given by:
+
+$$\begin{aligned}
+    J[f] = \iint_{(x_0, y_0)}^{(x_1, y_1)} L(f, f_x, f_y, x, y) \dd{x} \dd{y}
+\end{aligned}$$
+
+In which case the Euler-Lagrange equation takes the following form:
+
+$$\begin{aligned}
+    0 = \pdv{L}{f} - \dv{x} \Big( \pdv{L}{f_x} \Big) - \dv{y} \Big( \pdv{L}{f_y} \Big)
+\end{aligned}$$
+
+Once again, we calculate the hard $x$-derivative of $L$ (the $y$-derivative is analogous):
+
+$$\begin{aligned}
+    \dv{L}{x}
+    &= \pdv{L}{f} \dv{f}{x} + \pdv{L}{f_x} \dv{f_x}{x} + \pdv{L}{f_y} \dv{f_y}{x} + \pdv{L}{x}
+    \\
+    &= \dv{f}{x} \bigg( \dv{x} \Big( \pdv{L}{f_x} \Big) + \dv{y} \Big( \pdv{L}{f_y} \Big) \bigg)
+    + \pdv{L}{f_x} \dv{f_x}{x} + \pdv{L}{f_y} \dv{f_y}{x} + \pdv{L}{x}
+    \\
+    &= \dv{x} \Big( f_x \pdv{L}{f_x} \Big) + \dv{y} \Big( f_x \pdv{L}{f_y} \Big) + \pdv{L}{x}
+\end{aligned}$$
+
+This time, we arrive at the following expression for the soft derivative $\pdv*{L}{x}$:
+
+$$\begin{aligned}
+    - \pdv{L}{x}
+    &= \dv{x} \Big( f_x \pdv{L}{f_x} - L \Big) + \dv{y} \Big( f_x \pdv{L}{f_y} \Big)
+\end{aligned}$$
+
+Due to the derivatives, this cannot be cleanly turned into an analogue of the 1D Beltrami identity,
+and therefore we use that name only in the 1D case.
+
+However, if $\pdv*{L}{x} = 0$, this equation is still useful.
+For an off-topic demonstration of this fact,
+let us choose $x$ as the transverse coordinate, and integrate over it to get:
+
+$$\begin{aligned}
+    0
+    &= - \int_{x_0}^{x_1} \pdv{L}{x} \dd{x}
+    \\
+    &= \int_{x_0}^{x_1} \dv{x} \Big( f_x \pdv{L}{f_x} - L \Big) + \dv{y} \Big( f_x \pdv{L}{f_y} \Big) \dd{x}
+    \\
+    &= \Big[ f_x \pdv{L}{f_x} - L \Big]_{x_0}^{x_1} + \dv{y} \int_{x_0}^{x_1} \Big( f_x \pdv{L}{f_y} \Big) \dd{x}
+\end{aligned}$$
+
+If our boundary conditions cause the boundary term to vanish (as is often the case),
+then the integral on the right is a conserved quantity with respect to $y$.
+While not as elegant as the 1D Beltrami identity,
+the above 2D counterpart still fulfills the same role.
+
+
+
+## References
+1.  O. Bang,
+    *Nonlinear mathematical physics: lecture notes*, 2020,
+    unpublished.