Hamilton's principle/Mathematics/The fundamental lemma

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

The fundamental lemma of the calculus of variations closes the gap left at the end of the first variation. It states that a continuous function whose weighted integral vanishes for every admissible weight must itself be zero everywhere. Without it, stationarity yields an integral identity and nothing more; with it, stationarity yields a differential equation.

Statement

A localised test function.

Lemma. Let $E$ be continuous on $[a,b]$. Suppose

${\int }_a^{b}E(x)\eta (x)dx=0\mathrm{(}\mathrm{1}\mathrm{)}$

for every $\eta \in C^2[a,b]$ with $\eta (a)=\eta (b)=0$. Then $E(x)=0$ for every $x\in [a,b]$.

The strength of the lemma lies in how little it asks of the test class. It is enough that (1) hold for smooth $\eta$ vanishing at the ends; the conclusion is then pointwise and holds on the closed interval.

Proof

The argument is by contradiction, and its shape is worth naming before the details: if $E$ is nonzero somewhere, build an $\eta$ that is supported exactly where $E$ keeps its sign, and the integral cannot cancel.

Suppose $E(x_0)\ne 0$ for some $x_0\in (a,b)$. Without loss of generality take $E(x_0)>0$; otherwise replace $E$ by $-E$, which changes nothing in (1).

Step 1. Isolate a subinterval where $E$ keeps its sign. $E$ is continuous, so there is $\delta >0$ with $[x_0-\delta ,x_0+\delta ]\subset (a,b)$ and

$E(x)>\frac{1}{2}E(x_0)>0\text{for}|x-x_0|<\delta \mathrm{(}\mathrm{2}\mathrm{)}$

Step 2. Build a test function supported there. Put

$\eta (x)=\{\begin{array}{cc}[{\delta }^2-(x-x_0{)}^2{]}^3,& |x-x_0|\le \delta \\ 0,& \text{otherwise}\end{array}\mathrm{(}\mathrm{3}\mathrm{)}$

This $\eta$ is admissible. It vanishes at $a$ and $b$, being zero outside a subinterval of $(a,b)$. It is non-negative, and strictly positive on $(x_0-\delta ,x_0+\delta )$. The cube is what buys the smoothness: the bracket vanishes to first order at $x=x_0\pm \delta$, so the cube vanishes to third order there and $\eta \in C^2[a,b]$, as the test class requires.

Step 3. Evaluate. Because $\eta$ vanishes outside the subinterval,

${\int }_a^{b}E\eta dx={\int }_{x_0-\delta }^{x_0+\delta }E(x)\eta (x)dx>\frac{1}{2}E(x_0){\int }_{x_0-\delta }^{x_0+\delta }\eta (x)dx>0\mathrm{(}\mathrm{4}\mathrm{)}$

using (2) on the first factor and $\eta \ge 0$, not identically zero, on the second. This contradicts (1).

Therefore $E$ vanishes on the open interval $(a,b)$, and by continuity at the endpoints as well. $◼$

Where each hypothesis is used

hypothesis	used in	what fails without it
$E$ continuous	Step 1	sign cannot be propagated from a point to a neighbourhood; the conclusion drops to "almost everywhere"
$\eta$ arbitrary in the class	Step 2	a fixed or finite family of test functions cannot localise, and cancellation is not excluded
$\eta$ vanishes at the endpoints	Step 2	not needed for the lemma, but required for $\eta$ to be an admissible variation
$\eta$ in $C^2$	Step 2	the bracket must be cubed rather than squared to meet the class

The middle row is the load-bearing one. The lemma is a statement about the richness of the test class, and every proof of it is a construction of a sufficiently localised bump.

Variants

Du Bois-Reymond form. If instead

${\int }_a^{b}F(x){\eta }^{\prime }(x)dx=0\mathrm{(}\mathrm{5}\mathrm{)}$

for every admissible $\eta$, with $F$ continuous, then $F$ is constant. This variant avoids assuming in advance that $L_{y^{\prime }}$ is differentiable along the extremal, and it is the honest route to the Euler-Lagrange equation when the smoothness of $y$ is not given. It is used on the Weierstrass excess function, where extremals with corners are admitted.

Higher order. If (1) holds with $\eta$ replaced by ${\eta }^{(k)}$, then $E$ is a polynomial of degree less than $k$. This is the form needed for integrands containing higher derivatives.

Several dimensions. On a bounded domain $\Omega \subset {ℝ}^n$, if ${\int }_{\Omega }E\eta d^{n}x=0$ for every smooth $\eta$ of compact support in $\Omega$, then $E\equiv 0$. The proof is the same, with the bump (3) built from ${\delta }^2-\Vert x-x_0{\Vert }^2$. This is the version used on Several independent variables.

Remark on generality

The lemma is the point at which the calculus of variations makes contact with distribution theory. In modern language it says that a continuous function which is zero as a distribution is zero as a function. Nothing in this volume requires that language, but it explains why the lemma feels like an identity rather than a theorem: it is the statement that testing against enough smooth functions determines a continuous function completely.

Next

the Euler-Lagrange equation applies the lemma to equation (7) of the first variation.