Deriving it
Step 1 — differentiate the candidate quantity. Form
L - y'L_{y'} and take its total derivative along the curve. The
product rule on the second piece gives:
\frac{d}{dx}\left( L - y'L_{y'} \right) = \frac{dL}{dx} - y''L_{y'} - y'\frac{d}{dx}L_{y'}.
Step 2 — expand dL/dx by the chain rule. Since
L = L(y, y') has no explicit x, only the
y and y' slots carry the
x-dependence:
\frac{dL}{dx} = L_y\,y' + L_{y'}\,y''.
Step 3 — substitute and cancel. The two
L_{y'}y'' terms cancel, leaving
\frac{d}{dx}\left( L - y'L_{y'} \right) = L_y\,y' - y'\frac{d}{dx}L_{y'} = y'\left( L_y - \frac{d}{dx}L_{y'} \right).
Step 4 — invoke Euler–Lagrange. The bracket is exactly the Euler–Lagrange
expression, which is zero on any extremal. So the whole derivative is zero, and the quantity it
differentiates is constant:
\frac{d}{dx}\left( L - y'L_{y'} \right) = 0 \;\Longrightarrow\; L - y'L_{y'} = \text{const}.
Every cancellation above leaned on one assumption: L has
no explicit x. If x does
appear explicitly — say L = L(x, y, y') — then Step 2's chain rule
picks up an extra term:
\frac{dL}{dx} = \frac{\partial L}{\partial x} + L_y\,y' + L_{y'}\,y''.
That stray \partial L/\partial x survives the cancellation in Step 3,
so the honest result is \dfrac{d}{dx}(L - y'L_{y'}) = -\partial L/\partial x
— constant only when that partial derivative is zero. Reach for the Beltrami shortcut on a
Lagrangian like L = x\,y'^2, where x is
plainly there, and L - y'L_{y'} is not constant along the
extremal — you'd be smuggling in an invalid simplification of the full, second-order
Euler–Lagrange equation. Always check for explicit x first.
If the Lagrangian carries no explicit x (so
L = L(y, y')), then along any extremal:
-
L - y'\,\dfrac{\partial L}{\partial y'} = C for a constant
C;
-
this is a first integral — a first-order equation, easier than the
second-order Euler–Lagrange equation it replaces;
-
it is the variational echo of energy conservation: no explicit
x means a conserved quantity, just as no explicit time gives a
conserved energy.
The brachistochrone: the fastest slide is a cycloid
Here is the problem Beltrami was made for. A bead slides from
(0,0) down a frictionless wire y(x) to a
lower point. The descent time is the
functional
T[y] = \int \frac{\sqrt{1 + y'^2}}{\sqrt{2gy}}\,dx, \qquad L = \frac{\sqrt{1 + y'^2}}{\sqrt{2gy}}.
There's no explicit x in L — Beltrami
applies. Dropping the constant \sqrt{2g} and computing
L - y'L_{y'}, the algebra collapses neatly:
L - y'L_{y'} = \frac{1}{\sqrt{y}\,\sqrt{1 + y'^2}} = \text{const}.
Squaring and rearranging gives y\,(1 + y'^2) = k for a constant
k. That separable equation is solved by a clever substitution
y = \tfrac{k}{2}(1 - \cos\phi), and integrating the matching
dx produces the partner coordinate. With
r = k/2 the solution is the parametric cycloid:
x = r(\phi - \sin\phi), \qquad y = r(1 - \cos\phi).
The fastest descent is the curve traced by a point on the rim of a rolling wheel — turned
upside down. It dips below the straight ramp, trading a longer path for a steeper
early drop that builds speed fast.
A quick check. The trick works on simpler Lagrangians too. Take
L = \sqrt{1 + y'^2}, the arc-length integrand for the shortest path
between two points — again no explicit x. Since
L_{y'} = y'/\sqrt{1+y'^2}, Beltrami gives
L - y'L_{y'} = \frac{1 + y'^2 - y'^2}{\sqrt{1+y'^2}} = \frac{1}{\sqrt{1+y'^2}} = \text{const}.
A constant denominator forces y' itself to be constant — and a curve
with constant slope is a straight line, exactly as expected for the shortest
path in the plane.
In June 1696 Johann Bernoulli flung down a public gauntlet to "the sharpest mathematicians
in all the world": find the curve of fastest descent. Newton, the story goes, received the
problem in the evening, stayed up through the night, and mailed back the cycloid the next
morning — anonymously. Bernoulli recognised the author instantly, declaring "I know the lion
by his claw." Leibniz, l'Hôpital and Jacob Bernoulli also solved it. The contest is widely
taken as the birth of the calculus of variations, the moment optimising over shapes
became a discipline of its own.