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 Hamilton's principle/Mathematics/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 Hamilton's principle/Mathematics/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 Hamilton's principle/Mathematics/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

Hamilton's principle/Mathematics/The Euler-Lagrange equation applies the lemma to equation (7) of Hamilton's principle/Mathematics/The first variation.