Proof of the Euler-Lagrange Equation

The Euler-Lagrange equation is the foundational result of the calculus of variations, providing necessary conditions for a function to be an extremal of a functional. Its derivation combines integration by parts with the fundamental lemma of the calculus of variations.

Statement

Theorem7.3Euler-Lagrange Equation

Let $L(x,y,p) \in C^2$ and suppose $y^* \in C^2([a,b])$ is a stationary point of the functional $J[y] = \int_a^b L(x,y(x),y'(x))\,dx$ subject to fixed boundary conditions $y(a) = A$ , $y(b) = B$ . Then $y^*$ satisfies the Euler-Lagrange equation: $\frac{\partial L}{\partial y}(x,y^*,(y^*)') - \frac{d}{dx}\frac{\partial L}{\partial p}(x,y^*,(y^*)') = 0$ for all $x \in [a,b]$ .

Proof

Step 1: Setup the variation. Let $\eta \in C^2([a,b])$ satisfy $\eta(a) = \eta(b) = 0$ (admissible variation). For $\varepsilon \in \mathbb{R}$ , define $\Phi(\varepsilon) = J[y^* + \varepsilon\eta]$ . Since $y^*$ is a stationary point, $\Phi'(0) = 0$ .

Step 2: Compute the derivative. Differentiating under the integral: $\Phi'(\varepsilon) = \int_a^b \left[\frac{\partial L}{\partial y}(x, y^*+\varepsilon\eta, (y^*)'+\varepsilon\eta')\cdot\eta + \frac{\partial L}{\partial p}(x, y^*+\varepsilon\eta, (y^*)'+\varepsilon\eta')\cdot\eta'\right]dx$

Evaluating at $\varepsilon = 0$ : $0 = \Phi'(0) = \int_a^b\left[L_y(x,y^*,(y^*)')\eta + L_p(x,y^*,(y^*)')\eta'\right]dx$

where we write $L_y = \partial L/\partial y$ and $L_p = \partial L/\partial p$ for brevity.

Step 3: Integration by parts. Apply integration by parts to the second term: $\int_a^b L_p\,\eta'\,dx = \left[L_p\,\eta\right]_a^b - \int_a^b\frac{d}{dx}(L_p)\,\eta\,dx$

The boundary term $[L_p\,\eta]_a^b = L_p(b)\eta(b) - L_p(a)\eta(a) = 0$ since $\eta(a) = \eta(b) = 0$ . Substituting: $0 = \int_a^b\left[L_y - \frac{d}{dx}L_p\right]\eta\,dx$

Step 4: Fundamental lemma. We invoke the fundamental lemma of the calculus of variations: if $f \in C([a,b])$ and $\int_a^b f(x)\eta(x)\,dx = 0$ for every $\eta \in C^2([a,b])$ with $\eta(a) = \eta(b) = 0$ , then $f(x) = 0$ for all $x \in [a,b]$ .

Proof of the lemma: Suppose $f(x_0) > 0$ for some $x_0 \in (a,b)$ . By continuity, $f > 0$ on $(x_0-\delta, x_0+\delta)$ for some $\delta > 0$ . Choose $\eta(x) = [(x-x_0+\delta)(x_0+\delta-x)]^3$ on $(x_0-\delta,x_0+\delta)$ and $\eta = 0$ outside. Then $\eta \geq 0$ , $\eta \in C^2$ , and $\int f\eta\,dx > 0$ , contradicting the hypothesis. Similarly for $f(x_0) < 0$ . Therefore $f \equiv 0$ .

Step 5: Conclusion. Since $L_y - \frac{d}{dx}L_p$ is continuous (by the $C^2$ hypotheses on $L$ and $y^*$ ), the fundamental lemma gives: $\frac{\partial L}{\partial y}(x,y^*,(y^*)') - \frac{d}{dx}\frac{\partial L}{\partial p}(x,y^*,(y^*)') = 0 \quad \text{for all } x \in [a,b]$

which is the Euler-Lagrange equation. $\square$

■

ExampleExpanded Form of the Euler-Lagrange Equation

Expanding $\frac{d}{dx}L_p$ by the chain rule: $L_{px} + L_{py}\,y' + L_{pp}\,y'' = L_y$ . When $L_{pp} \neq 0$ (the Legendre condition, also called regularity), this is a second-order ODE: $y'' = \frac{1}{L_{pp}}[L_y - L_{px} - L_{py}\,y']$ . The condition $L_{pp} > 0$ (strict convexity in $p$ ) ensures that the extremal is a local minimum (Jacobi's strengthened condition).

RemarkWeak and Strong Extremals

The proof above yields a necessary condition for weak extremals (variations small in the $C^1$ norm). For strong extremals (small in $C^0$ norm), additional conditions are needed: the Weierstrass excess function $\mathcal{E}(x,y,p,q) = L(x,y,q) - L(x,y,p) - (q-p)L_p(x,y,p) \geq 0$ for all $q$ (Weierstrass necessary condition), and the Jacobi condition (no conjugate points). Together, the Euler-Lagrange equation, Legendre condition, Weierstrass condition, and Jacobi condition form the complete set of sufficient conditions for a strong minimum.