Appendix 12 — Derivation of the Euler–Lagrange Equation

Appendix 12.1 — Definition of the Functional

We start with a function \( f_{1} \) that depends on three variables: \(t\), \(x_{1}(t)\) and \( \dot{x}_{1}(t) = \frac{dx_{1}}{dt} \):

Here \(x_{1}(t)\) is a function of \(t\), so \(\dot{x}_{1}(t)\) is not zero. In fact, \(t\) is the only independent variable; \(f_{1}\) is therefore a function of a function.

We now consider the functional:

We seek the extremal value (minimum, maximum, or saddle point) of \(I_{1}\). This requires:

Appendix 12.2 — Variation of the Path

To prove that \(I_{1}\) is extremal, consider a slightly displaced curve:

where:

• \(\lambda\) is a small parameter, independent of \(t\),
• \(\xi(t)\) is an arbitrary smooth function.

Because the variation cannot change the endpoints, we have:

Thus, the function \(x_{2}(t,\lambda)\) differs only between \(t_{1}\) and \(t_{2}\), but coincides with \(x_{1}(t)\) at the endpoints.

Appendix 12.3 — Variation of the Functional

The integral \(I_{2}\) for the nearby curve is:

With: \[ f_{2} = f\!\left(t,\; x_{2}(t,\lambda),\; \frac{d{x}_{2}(t,\lambda)}{dt}\right). \]

Substituting (6) into equation (4) gives:

Because \(I_{1}\) is an extremum, \(I_{2}\) must also be extremal for \(\lambda = 0\):

The extremum is found by taking the derivative with respect to \(\lambda\) and setting it to zero:

In combination with equation (6):

or equivalently:

Appendix 12.4 — Differentiation with Respect to the Variation Parameter

We had: \[ \lim_{\lambda \to 0} \frac{d}{d\lambda} \int_{t_{1}}^{t_{2}} f_{2}\, dt = 0. \] Since this is a product of two functions, we apply the rule of partial differentiation:

Because \(t\) and \(\lambda\) are independent: \[ \frac{dt}{d\lambda} = 0, \] so the second term vanishes:

Appendix 12.5 — Expanding \( \frac{d f_{2}}{d\lambda} \)

As previously stated, \(f_2\) is a function of three variables: \[ f_{2} = f\!\left(t,\; x_{2},\; \dot{x}_{2}\right). \]

We now apply the chain rule:

Thus:

Since \(t\) and \(\lambda\) are independent, the first term vanishes:

Since: \[ \frac{d\dot{x}_{2}}{d\lambda} = \frac{d^2 x_2}{dt\, d\lambda} = \frac{d}{dt} \left( \frac{dx_{2}}{d\lambda} \right) \tag{15} \] equation (14), together with (15), leads to:

\[ \lim_{\lambda \to 0} \int_{t_{1}}^{t_{2}} \left( \frac{\partial f_{2}}{\partial x_{2}}\frac{dx_{2}}{d\lambda} + \frac{\partial f_{2}}{\partial \dot{x}_{2}} \frac{d}{dt}\left(\frac{dx_{2}}{d\lambda}\right) \right) dt = 0 \tag{16} \]

\[ \lim_{\lambda \to 0} \left( \int_{t_{1}}^{t_{2}} \frac{\partial f_{2}}{\partial x_{2}} \frac{dx_{2}}{d\lambda} dt + \int_{t_{1}}^{t_{2}} \frac{\partial f_{2}}{\partial \dot{x}_{2}} \frac{d}{dt} \left( \frac{dx_{2}}{d\lambda} \right) dt \right) = 0 \]

Appendix 12.6 — Integration by Parts

We now partially integrate the second term:

The derivative of \(x_2\) with respect to \(\lambda\) is found from equation (4):

Since \(\xi(t)\) vanishes at the boundaries of the integral (see equation (5)), the first term on the right-hand side of equation (17) disappears:

So this gives:

Combining this result with equation (16) yields:

\[ \lim_{\lambda \to 0} \left( \int_{t_{1}}^{t_{2}} \frac{\partial f_{2}}{\partial x_2}\frac{dx_2}{d\lambda}\, dt + \int_{t_{1}}^{t_{2}} \frac{\partial f_{2}}{\partial \dot x_2}\, \frac{d\left(\frac{dx_2}{d\lambda}\right)}{dt}dt \right) = 0 \tag{20} \]

\[ \lim_{\lambda \to 0} \left( \int_{t_{1}}^{t_{2}} \frac{\partial f_{2}}{\partial x_2}\frac{dx_2}{d\lambda}\, dt - \int_{t_{1}}^{t_{2}} \frac{dx_{2}}{d\lambda}\, \frac{d}{dt}\left( \frac{\partial f_{2}}{\partial \dot{x}_{2}}\right)dt \right) = 0 \]

\[ \lim_{\lambda \to 0} \int_{t_{1}}^{t_{2}} \left( \frac{\partial f_{2}}{\partial x_2}\frac{dx_2}{d\lambda} - \frac{dx_{2}}{d\lambda}\, \frac{d}{dt} \left(\frac{\partial f_{2}}{\partial \dot{x}_{2}}\right)\right)dt = 0 \]

\[ \lim_{\lambda \to 0} \int_{t_{1}}^{t_{2}} \left( \frac{\partial f_{2}}{\partial x_2} - \, \frac{d}{dt} \left(\frac{\partial f_{2}}{\partial \dot{x}_{2}}\right)\right)\frac{dx_2}{d\lambda}dt = 0 \]

Factoring out \(dx_2/d\lambda\) and taking the limit \(\lambda \to 0\) gives the Euler–Lagrange condition:

\[ \lim_{\lambda \to 0} \left( \frac{\partial f_{2}}{\partial x_2} - \, \frac{d}{dt} \left(\frac{\partial f_{2}}{\partial \dot{x}_{2}}\right)\right) = 0 \]

Thus, \(\lambda\) has completely disappeared and we have obtained a general expression for the condition a function must satisfy for the integral \(I\) to have an extremum.

We started with equation (1) for our derivation, but we could generalize further by taking a function of multiple variables:

This leads to the generalized form of equation (21):

Or equivalently:

Equation (24) is the Euler–Lagrange equation. It gives the condition that a function must satisfy so that the integral \(I\) has an extremum.

This forms the foundation of variational calculus, classical mechanics, field theory, and general relativity.