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}, throughlagrange



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:


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.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s