The Euler-Lagrange equation

Plain-language version: What the Euler-Lagrange equation says

The Euler-Lagrange equation is the differential equation satisfied by every stationary path of a functional. It is the central result of the machinery track, and everything in the applications track is an instance of it.

Statement

The two rates the Euler-Lagrange equation balances.

Theorem. Let $L(x,y,y^{\prime })$ have continuous partial derivatives up to second order, and let

$J[y]={\int }_a^{b}L(x,y(x),y^{\prime }(x))dx\mathrm{(}\mathrm{1}\mathrm{)}$

on the admissible set $𝒜$ of Functionals and the space of paths. If $y\in 𝒜$ is stationary for $J$, then

$\frac{\partial L}{\partial y}-\frac{d}{dx}\left(\frac{\partial L}{\partial y^{\prime }}\right)=0\mathrm{(}\mathrm{2}\mathrm{)}$

for every $x\in [a,b]$.

Derivation

Stationarity means $\delta J[y;\eta ]=0$ for every $\eta \in 𝒱$. By equation (6) of the first variation,

${\int }_a^b[L_y-\frac{d}{dx}{L}_{y^{\prime }}]\eta dx=0\mathrm{(}\mathrm{3}\mathrm{)}$

The bracket is continuous on $[a,b]$: $L$ has continuous second partials by hypothesis and $y\in C^2$, so $\frac{d}{dx}{L}_{y^{\prime }}$ exists and is continuous. Equation (3) holds for every admissible $\eta$. The fundamental lemma of the fundamental lemma therefore applies to the bracket, and it vanishes identically. That is (2). $◼$

The derivation is three lines because the work was done on the two preceding pages. That is the correct division: the integration by parts is where the endpoints are used, and the lemma is where the arbitrariness of $\eta$ is used.

Expanded form

Carrying out the total derivative in (2) with the chain rule,

$\frac{d}{dx}{L}_{y^{\prime }}=L_{y^{\prime }x}+L_{y^{\prime }y}{y}^{\prime }+L_{y^{\prime }{y}^{\prime }}{y}^{″}\mathrm{(}\mathrm{4}\mathrm{)}$

so that (2) becomes

$L_{y^{\prime }{y}^{\prime }}{y}^{″}+L_{y^{\prime }y}{y}^{\prime }+L_{y^{\prime }x}-L_y=0\mathrm{(}\mathrm{5}\mathrm{)}$

This is a second-order ordinary differential equation for $y$, generally nonlinear. Three consequences follow immediately from its form.

It is second order, so it needs two conditions. The admissible set supplies exactly two, $y(a)=y_a$ and $y(b)=y_b$. The count matches, which is why the fixed-endpoint problem is well posed. It is a boundary value problem, not an initial value problem, and that difference is the source of the non-uniqueness discussed on the Jacobi condition and conjugate points.

Degeneracy when $L_{y^{\prime }{y}^{\prime }}=0$. If the coefficient of $y^{″}$ vanishes, (5) is not a second-order equation at all and the problem is singular. The condition $L_{y^{\prime }{y}^{\prime }}\ne 0$ is called non-degeneracy, and its signed version is the Legendre condition of the Legendre condition.

Solutions are called extremals, not extrema. A solution of (2) is an extremal. The word records only that the first variation vanishes. Whether an extremal minimises anything is undecided by (2) and is settled, when it can be settled, on the second variation.

First integrals

Two cases reduce the order of (2) by one.

$L$ free of $y$. Then $L_y=0$ and (2) gives

$\frac{d}{dx}{L}_{y^{\prime }}=0⟹L_{y^{\prime }}=\text{const}\mathrm{(}\mathrm{6}\mathrm{)}$

The conserved quantity $L_{y^{\prime }}$ is the conjugate momentum. Where $x$ is time and $y$ a coordinate, (6) is conservation of momentum.

$L$ free of $x$. Then

$y^{\prime }{L}_{y^{\prime }}-L=\text{const}\mathrm{(}\mathrm{7}\mathrm{)}$

which is the Beltrami identity, derived on the Beltrami identity. Where $x$ is time, (7) is conservation of energy.

Both cases are instances of one theorem. A symmetry of $L$ — here, independence of a variable — produces a conserved quantity. The general statement is Noether's, on Noether's first theorem. Noether states it in the form that $\rho$ linearly independent combinations of the Lagrange expressions become divergences.

Worked example: the shortest plane curve

Minimise arc length between $(a,y_a)$ and $(b,y_b)$:

$J[y]={\int }_a^b\sqrt{1+y{\text{'}}^2}dx\mathrm{(}\mathrm{8}\mathrm{)}$

Here $L=\sqrt{1+y{\text{'}}^2}$ is free of both $x$ and $y$. Using (6),

$L_{y^{\prime }}=\frac{y^{\prime }}{\sqrt{1+y{\text{'}}^2}}=c\mathrm{(}\mathrm{9}\mathrm{)}$

Solving (9) for $y^{\prime }$ gives $y{\text{'}}^2(1-c^2)=c^2$, so $y^{\prime }$ is a constant wherever $|c|<1$, and $y$ is a straight line. The two endpoint conditions fix its two parameters.

Note what has not been shown: that the line is the shortest curve. Equation (2) is necessary, not sufficient. The line is here in fact the unique minimiser, but establishing that requires the sufficiency theory, and the geodesic problem on a sphere in the Jacobi condition and conjugate points is the case where the corresponding extremal is stationary and is not a minimum.

Historical note

Euler reached the equation in 1744 by a limiting argument on a polygonal approximation to the curve, in Methodus inveniendi lineas curvas maximi minimive proprietate gaudentes. Lagrange's variational derivation, close to the one above, followed in the 1750s correspondence with Euler and reached print in the Mécanique analytique of 1788.

Provenance: this paragraph is editorial summary, not transcription. The Euler and Lagrange source rows of the survey on Sources are not yet ESTABLISHED, and no equation on this page is attributed to either. The two attributed equations, (2) as a Lagrange expression and the Theorem I statement, come from Noether 1918, which is ESTABLISHED.

Next

Boundary terms and natural boundary conditions relaxes the fixed-endpoint condition and recovers the boundary term dropped at equation (5) of the first variation.