# Euler-Lagrange Equations

Consider the integral

$\displaystyle I[L]=\int_{t_0}^{t_1}L(\bold{q}(t),\bold{\dot{q}}(t),t)dt$

The quantity $I[L]$ is called a functional. Just as a function takes a number and gives you a number back, a functional takes a function and gives you a number.

The integrand it takes is called the Lagrangian, which depends on a set of functions $\{q\}$ and their derivatives $\{\dot{q}\}$ with respect to the independent variable $t$.

The two limits of integration are taken to be fixed, so both $t_0$ and $t_1$ are constant.

Suppose we want to find the set $\{q(t)\}$ that extremises this integral, the one that makes it as big or as small as possible for a given Lagrangian. How do we go about doing this?

Let’s suppose $\{q\}$ is the set of variables that extremises the functional $I[L]$. Then if we perturb these variables by some small amount $\delta\bold{q}$, through

$\bold{q}\to\bold{q}+\delta\bold{q}$

$\bold{\dot{q}}\to\bold{\dot{q}}+\delta\bold{\dot{q}}$

the functional $I[L]$ does not change to first order. This is the same reasoning we use when we find the maximum or minimum of a normal, single-variable function $f$ – moving a small distance $\delta x$ along the curve from the extremum does not change the value of the function.

Since the endpoints of the integral are fixed, the perturbation $\delta\bold{q}$ must vanish there:

$\delta\bold{q}(t_0)=\delta\bold{q}(t_1)=0$

That is, all paths parametrised by the set $\{q(t)\}$ must start and end at the same place. The diagram above shows a schematic depiction of the perturbation of the extremising path.

The extremised integral $I$ plus the small perturbation $\delta I$ is

$\displaystyle I+\delta I=\int_{t_0}^{t_1} L(\bold{q}+\delta\bold{q},\bold{\dot{q}}+\delta\bold{\dot{q}},t)dt$

We can express the integrand in a more convenient form using a Taylor expansion:

$\displaystyle L(\bold{q}+\delta\bold{q},\bold{\dot{q}}+\delta\bold{\dot{q}},t)=L(\bold{q},\bold{\dot{q}},t)+\frac{\partial L}{\partial \bold{q}}\cdot\delta\bold{q}+\frac{\partial L}{\partial \bold{\dot{q}}}\cdot\delta\bold{\dot{q}}$

If this funny vector derivative notation is unfamiliar to you (as it was to me for a long time!), it is equivalent to the gradient in $\bold{q}$-space, so that the first inner product (for example) is

$\displaystyle \frac{\partial L}{\partial \bold{q}}\cdot\delta\bold{q}\equiv\frac{\partial L}{\partial q_1}\delta q_1+\frac{\partial L}{\partial q_2}\delta q_2+...$

Using this expansion, we have

$\displaystyle I+\delta I=\int_{t_0}^{t_1} \Big(L(\bold{q},\bold{\dot{q}},t)+\frac{\partial L}{\partial \bold{q}}\cdot\delta\bold{q}+\frac{\partial L}{\partial \bold{\dot{q}}}\cdot\delta\bold{\dot{q}}\Big)dt$

The first term in the integrand on the right-hand side can be extracted and made to cancel with $I$, leaving

$\displaystyle \delta I=\int_{t_0}^{t_1}\Big(\frac{\partial L}{\partial \bold{q}}\cdot\delta\bold{q}+\frac{\partial L}{\partial \bold{\dot{q}}}\cdot\delta\bold{\dot{q}}\Big)dt$

As stated before, this small change in the integral $I$ is 0 if we are ‘at’ an extremum. The expression above brings us a little closer, but is still pretty opaque. So we use a trick:

$\displaystyle \frac{d}{dt}\Big(\frac{\partial L}{\partial \bold{\dot{q}}}\cdot\delta\bold{q}\Big)=\frac{d}{dt}\Big(\frac{\partial L}{\partial \bold{\dot{q}}}\Big)\cdot\delta\bold{q}+\frac{\partial L}{\partial \bold{\dot{q}}}\cdot\delta\bold{\dot{q}}$

using the product rule for differentiation. Rearranging for the second term and substituting into the expression for $\delta I$, we find

$\displaystyle \delta I=\int_{t_0}^{t_1} \Big(\frac{\partial L}{\partial \bold{q}}\cdot\delta\bold{q}+\frac{d}{dt}\Big(\frac{\partial L}{\partial \bold{\dot{q}}}\cdot\delta\bold{q}\Big)-\frac{d}{dt}\Big(\frac{\partial L}{\partial \bold{\dot{q}}}\Big)\cdot\delta\bold{q}\Big)dt$

Splitting this integral up yields

$\displaystyle \delta I=\int_{t_0}^{t_1}\Bigg(\frac{\partial L}{\partial \bold{q}}-\frac{d}{dt}\Big(\frac{\partial L}{\partial \bold{\dot{q}}}\Big)\Bigg)\cdot\delta\bold{q}\ dt+\Bigg[\frac{\partial L}{\partial \bold{\dot{q}}}\cdot\delta\bold{q}(t)\Bigg]_{t_0}^{t_1}$

Recall the constraint we made earlier – the perturbation $\delta\bold{q}$ is equal to 0 at the limits in integration. This means the brackets vanish, leaving

$\displaystyle \delta I=\int_{t_0}^{t_1}\Bigg(\frac{\partial L}{\partial \bold{q}}-\frac{d}{dt}\Big(\frac{\partial L}{\partial \bold{\dot{q}}}\Big)\Bigg)\cdot\delta\bold{q}\ dt$

At this point we invoke the fundamental lemma of the calculus of variations. It allows us to state with certainty the intuitive result that if this integral is generally 0, then

$\displaystyle \frac{\partial L}{\partial \bold{q}}-\frac{d}{dt}\Big(\frac{\partial L}{\partial \bold{\dot{q}}}\Big) = 0$

This is the Euler-Lagrange equation. Really it is a set of equations, since the derivatives are really vectors, with as many components as there are functions in the set $\{q\}$:

$\displaystyle \frac{\partial L}{\partial q_i}=\frac{d}{dt}\Big(\frac{\partial L}{\partial \dot{q}_i}\Big)$

Though the Euler-Lagrange equations are often discussed in the context of Lagrangian mechanics, their full range of potential applications is vast. In a following post, we’ll use them to derive the shape of the catenary curve.