Euler Lagrange differential equation.

Question

The Lagrange equation is a second order differential equation. However is it an ordinary or partial differential equation? Looking at wikipedia it says it is both, here it is a PDE and here it is an ODE. So which is it? \begin{equation} \frac{d}{dt}\bigg(\frac{\partial \mathscr L}{\partial \dot q}\bigg)-\frac{\partial \mathscr L}{\partial q}=0 \end{equation} In classical mechanics?

Answer 1

Your question depends on the problem and the context. I'll try to answer your question in general below (but will also answer the specific question you had).

In classical mechanics we are usually interested in determining the time-evolution of particles, i.e. we are interested in finding functions $q(t)$ or more generally $\vec{q}(t) = \{q_1(t),\ldots,q_n(t)\}$. For this kind of problems the Lagrangian is a function of only $q$ and $\dot{q}$ and this always leads to second order ordinary differential equations in $q$. The reson we cannot have a PDE here is that we only have one coordinate (time). When you write the Euler-Lagrange equation as

$$\frac{dL(q,\dot{q})}{dq} - \frac{d}{dt}\frac{dL(q,\dot{q})}{d\dot{q}} = 0$$

you are in this class of problems so for your situation the answer is ODE.

In general this is not the case. For example, when working with fields (e.g. the electromagnetic field, quantum fields, the gravitational field) instead of particles, one usually ends up with partial differential equations. The Euler-Lagrange equation in these cases is more complicated. If the Lagrangian is a function of the field $q$ and the derivatives of the fields $\frac{dq}{dt}$, $\frac{dq}{dx}$, $\ldots$ then we have

$$\frac{dL(q,\nabla q)}{dq} - \nabla_\mu\left(\frac{dL(q,\nabla q)}{d\nabla_\mu q}\right) = 0$$

where the last term is a sum over the different coordinates $\mu = t,x,y,z, $ and so on.

"The Lagrange equation is a second order differential equation."

In general the Euler-Lagrange equation does not have to be a second order ODE (or PDE). For example if the Lagrangian depends on $q,\dot{q}$ and $\ddot{q}$ then the Euler-Lagrange equation becomes

$$\frac{dL}{dq} - \frac{d}{dt} \frac{dL}{d{\dot q}}+ \frac{d^2}{dt^2}\frac{dL}{d{\ddot q}} = 0$$

which can lead to a third order equation (and so on). However for physics we almost always work with Lagrangians that lead to second order ODEs (or second order in time for the PDEs). The reason for this is that Lagrangians that leads to higher order equations of motion are plagued by instabillities and it seems that nature hates instabillities just as much as physicists.

Answer 2

Let's look at a very simple one-dimensional example. Suppose we have a particle in a potential field $U(x)$. We know that the kinetic energy $T(v)$ of that particle is given by

$$T(v)=\frac12 mv^2$$

where $m$ is the mass and $v$ is the velocity along the positive $x$ axis. The Lagrangian $\mathscr{L}(x,v)$ is

$$\begin{align} \mathscr{L}(x,v)&=T(v)-U(x)\\\\ &=\frac12 mv^2-U(x) \end{align}$$

Note that we view the Lagrangian as a function of two variables here, $v$ and $x$. We can determine the partial derivatives of $\mathscr{L}(x,v)$ with respect to the two independent variables as

$$\frac{\partial \mathscr{L}(x,v)}{\partial v}=mv$$

$$\frac{\partial \mathscr{L}(x,v)}{\partial x}=-\frac{d U(x)}{d x}$$

Lagrange's equation now views $x$ and $v$ as functions of time $x(t)$ and $v(t)$ with the relationship $v(t)=\frac{dx(t)}{dt}$. We then write

$$\begin{align} \frac{d}{dt}\left(\frac{\partial \mathscr{L}(x,v)}{\partial v}\right)-\frac{\partial \mathscr{L}(x,v)}{\partial x}&=\frac{d(mv(t))}{dt}+\frac{d U(x)}{d x}\\\\ &=0\end{align}$$

which yields Newton's second law of motion

$$\frac{dp}{dt}=-\frac{d U(x)}{d x}$$

where $p=mv$ is the momentum of the particle.