MATH0043 Handout: Fundamental lemma of the calculus of variations

In the proof of the Euler-Lagrange equation, the final step invokes a lemma known as the fundamental lemma of the calculus of variations (FLCV).

Lemma 1 (FLCV)   . Let y(x) be continuous on [a, b], and suppose that for all η(x)∈C2[a, b] such that η(a) = η(b) = 0 we have

$\displaystyle \int_{a}^{b}$ y(x)η(x) dx = 0.

Then y(x) = 0 for all axb.

Here is a sketch of the proof. Suppose, for a contradiction, that for some a < α < b we have y(α) > 0 (the case when α = a or α = b can be done similarly, but let's keep it simple). Because y is continuous, y(x) > 0 for all x in some interval (α0, α1) containing α.

Consider the function η : [a, b]→$\mathbb {R}$ defined by

η(x) = \begin{displaymath}
% latex2html id marker 1883
\begin{cases}(x-\alpha_0)^4 (x...
...& \alpha_0 < x < \alpha_1 \\
0 & \text{otherwise.}
\end{cases}\end{displaymath}    

η is in C2[a, b] -- it's difficult to give a formal proof without using a formal definition of continuity and differentiability, but hopefully the following plot shows what is going on:
Figure: The function η defined above, if a = 0, b = 1, α0 = 0.2, α1 = 0.4.
Image eta1

By hypothesis, $\int_{0}^{1}$y(x)η(x)  dx = 0. But y(x)η(x) is continuous, zero outside (α0, α1), and strictly positive for all x∈(α0, α1). A strictly positive continuous function on an interval like this has a strictly positive integral, so this is a contradiction. Similarly we can show y(x) never takes values < 0, so it is zero everywhere on [a, b].