---
tag: RB08381QYR
kind: definition
name: Hilbert's invariant integral
msc: [49K05, 49L99]
defines: D1EBAAFEWV
uses: [K4DFT3H9BJ, R8TWRPSPNV, 0H5MR9B6GT, 0ZV14XRNF0]
sources:
  - url: https://en.wikipedia.org/wiki/Calculus_of_variations
---

## Definition

Let $L(x, y, p)$ be a $C^2$ function on $\R \times \R \times \R$, and let $J(y) = \int_a^b L(x, y, y') \, dx$.
Let $p(x, y)$ be the slope function of a [[field of extremals|K4DFT3H9BJ]] of $J$ on a simply connected open set $D \subseteq \R^2$.
For a piecewise $C^1$ curve $\gamma$ in $D$, *Hilbert's invariant integral* is the [[line integral|0H5MR9B6GT]]
$$I^*(\gamma) = \int_\gamma \Big( L(x, y, p) - p \, L_p(x, y, p) \Big) \, dx + L_p(x, y, p) \, dy,$$
where $p = p(x, y)$ inside the integrand.

## Remarks

The integrand is a [[closed differential form|0ZV14XRNF0]] on $D$ because the field consists of extremals.
Since $D$ is simply connected, the form is exact, so $I^*(\gamma)$ depends only on the end points of $\gamma$.

Along a curve of the field one has $dy = p \, dx$, so $I^*(\gamma) = \int L(x, y, p) \, dx = J$ on that curve.
For the graph of any piecewise $C^1$ function $y$ in $D$, subtracting $I^*$ from $J$ gives
$$J(y) - I^*(y) = \int_a^b E\big(x, y(x), p(x, y(x)), y'(x)\big) \, dx,$$
with $E$ the [[Weierstrass excess function|R8TWRPSPNV]].
If $y_0$ is a field curve and $y$ has the same end points, then $I^*(y) = J(y_0)$, which gives the Weierstrass formula $J(y) - J(y_0) = \int_a^b E \, dx$.

Example: for $L = p^2$ and the field $y = x + c$ with $p = 1$, the integrand is $(1 - 2) \, dx + 2 \, dy = d(2y - x)$, which is exact.

Non-example: with the slope function $p = 1$ and $L = p^2 + y^2$, the curves $y = x + c$ are not extremals, and the form $(y^2 - 1) \, dx + 2 \, dy$ is not closed, since $\partial_y (y^2 - 1) = 2y$ is not identically equal to $\partial_x 2 = 0$.