Proof of Euler–Lagrange Equation
Proof / Derivation of Euler–Lagrange Equation
Let all possible paths between \(A\) and \(B\) such as
\(y_1, y_2, y_3, y_4, \dots\) be represented by a family of curves or paths:
\( y(x,\alpha) = y(x) + \alpha \, \eta(x) \)
Then:
-
\( \dfrac{\partial y}{\partial \alpha} =
\dfrac{\partial}{\partial \alpha} \big(y(x)+\alpha \eta(x)\big)
= \eta(x) \)
-
\( \dfrac{\partial y'}{\partial \alpha} =
\dfrac{\partial}{\partial \alpha} \left(\dfrac{\partial y}{\partial x}\right)
= \eta'(x) \)
- \( \dfrac{\partial x}{\partial \alpha} = 0 \)
-
At \(x_1:\; y(x_1,\alpha) = y(x_1)+\alpha \eta(x_1)\;\Rightarrow\; \eta(x_1)=0\).
-
At \(x_2:\; y(x_2,\alpha) = y(x_2)+\alpha \eta(x_2)\;\Rightarrow\; \eta(x_2)=0\).
Derivation
Now let \( f = f(y(x,\alpha), y'(x,\alpha), x) \).
\( I = \int_{x_1}^{x_2} f(y(x,\alpha), y'(x,\alpha), x) \, dx \)
Since \(I\) depends on \(\alpha\), for extremum values we require:
\[
\frac{\partial I}{\partial \alpha}
= \int_{x_1}^{x_2}
\left[
\frac{\partial f}{\partial y}\frac{\partial y}{\partial \alpha} +
\frac{\partial f}{\partial y'}\frac{\partial y'}{\partial \alpha} +
\frac{\partial f}{\partial x}\frac{\partial x}{\partial \alpha}
\right] dx = 0
\]
Substituting the results:
\[
\frac{\partial I}{\partial \alpha} =
\int_{x_1}^{x_2}
\left[
\frac{\partial f}{\partial y}\eta(x) +
\frac{\partial f}{\partial y'} \eta'(x)
\right] dx
\]
Integrating the second term by parts:
\[
\int_{x_1}^{x_2} \frac{\partial f}{\partial y'} \eta'(x) \, dx
= \left[ \frac{\partial f}{\partial y'} \eta(x) \right]_{x_1}^{x_2}
- \int_{x_1}^{x_2} \frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)\eta(x)\, dx
\]
Since \(\eta(x_1) = \eta(x_2) = 0\):
\[
\frac{\partial I}{\partial \alpha} =
\int_{x_1}^{x_2} \left[
\frac{\partial f}{\partial y} -
\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)
\right] \eta(x) \, dx = 0
\]
Because \(\eta(x)\) is arbitrary, the integrand must vanish:
\[
\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right) -
\frac{\partial f}{\partial y} = 0
\]
This is the Euler–Lagrange equation.