# Is the derivative in the Euler-Lagrange equation defined as an implicit function?

Consider the Euler-Lagrange equation in multiple dimensions (which is actually a system of equations): $$\frac{\partial L}{\partial q_i}(t, \mathbf{q}(t), \dot{\mathbf{q}}(t)) - \frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i}(t, \mathbf{q}(t),\dot{\mathbf{q}}(t)) =0.$$ My question is the following:

Is the relationship $$\dot{\mathbf{q}}(t)=\frac{d}{dt}\mathbf{q}(t)$$ an assumption of the Euler-Lagrange equation, or is it a conclusion?

My argument -- where are the mistakes? I have argued that it is a conclusion before, see in this answer, I don't know how to prove or disprove that this is correct.

Specifically, when evaluating $\displaystyle\frac{\partial L}{\partial q_i}$ no terms of the form $\displaystyle\frac{\partial \dot{q}_i}{\partial q_i}$ show up, and when evaluating $\displaystyle\frac{\partial L}{\partial \dot{q}_i}$ no terms of the form $\displaystyle\frac{\partial q_i}{\partial \dot q_i}$ show up, which wouldn't make sense if $\dot{q}_i$ was being treated as a function, for example, if $q_i = \frac{1}{2}t^2$ then $\dot{q}_i = t$ so then $\displaystyle\frac{\partial q_i}{\partial \dot{q}_i} = t \not=0$.

However, neglecting such terms would/does make sense if the $\dot{q}_i$ are being treated as independent variables, rather than functions. The only problem then is that one can't say that one independent variable is the derivative of another independent variable.

BUT, along the solution curves, the independent variable denoted by an abuse of notation $\dot{q}_i$ could/would be/is an implicit function of $t$, call it $f_i(t)$, as would the independent variable $q_i$ be an implicit function of $t$ along the solution curve call it $g_i(t)$, and then a consequence of the Euler-Lagrange equation might be then that $f_i(t) = \frac{d}{dt}g_i(t)$.

This seems to be a natural way to think of the situation/setup if one thinks about differential equations as "phase flows" in "phase space", e.g. as in the first section of the first chapter of Vladimir Arnold's excellent book on ODE's.

The answers to this question seem to suggest this might be a possibility: Derivative with respect to $y'$ in the Euler-Lagrange differential equation; however they don't state whether or not the relation $\dot{\mathbf{q}}(t)=\frac{d}{dt}\mathbf{q}(t)$ is a consequence or not, although that should be implied since it could not be an assumption if $y'$ were an independent variable, as the answers suggest.

Note: Before dismissing this possibility out of hand, I would ask one to at least consider that partial derivative notation is notorious for obfuscating and making more confusing issues surrounding implicit differentiation and implicit functions, see for example here, here, here, here, here, here, here, here, here: "A problem with Jacobi's notation is that... substitution of variables can lead to absurd formulae or, at least, to formulae that require careful interpretation", and here.

• @Bye_World Not really; I understand that it is thought of as a function of time only, but if it is a function, then how can you differentiate with respect to it? That would seem to be possible only if the generalized velocity was an independent variable. But if it is an independent variable, then you can't say that it has an "antiderivative". Sep 14, 2016 at 23:15
• @Bye_World But then if you use the chain rule those terms I mentioned which do disappear shouldn't disappear (I think). Or is one substituting into the independent variable "$y$" the generalized velocity, i.e. $\tilde{L}(t)= L(t, q(t), \dot{q}(t)$, where $L=L(t,x,y)$. I guess my question could also be "is it possible to formulate the EL equation in such a way that this relationship is a conclusion and not an assumption?" Sep 14, 2016 at 23:16
• Related and also helpful like the accepted answer math.stackexchange.com/a/697189/327486 Dec 31, 2021 at 3:30

Let $L: \Bbb R^3 \to \Bbb R: (x_1, x_2, x_3) \mapsto L(x_1, x_2, x_3)$ and $q: \Bbb R\to \Bbb R: t \mapsto q(t)$. Then denote by $\dot q$ the derivative of $q$. Let $\gamma: \Bbb R \to \Bbb R^3: t\mapsto (t,q(t),\dot q(t))$. Now consider the restriction of $L$ to $\gamma$:

$$(L\circ \gamma)(t):\Bbb R\to \Bbb R \\ t\mapsto L(t,q(t),\dot q(t))$$

This is the usual way of writing the Lagrangian.

Now the expressions $\frac{\partial L}{\partial q}$ and $\frac{\partial L}{\partial {\dot q}}$ in the EL equation are really just the partial derivatives of the above function $L$ wrt its second and third arguments: $$\frac{\partial L}{\partial q}:= \frac{\partial L}{\partial x_2} \qquad \frac{\partial L}{\partial \dot q}:= \frac{\partial L}{\partial x_3}$$

Thus the value of these partials does not depend on the fact that $\dot q = \frac{dq}{dt}$.

Now let's look at $\frac{d}{dt}\frac{\partial L}{\partial \dot q}$. This is really the derivative of the restriction of $\frac{\partial L}{\partial x_3}$ to the path $\gamma$

$$\frac{d}{dt}\frac{\partial L}{\partial \dot q} := \left[\frac{d}{dt}\left(\frac{\partial L}{\partial x_3}\circ \gamma\right)\right](t)$$

So the EL equation, written in this new notation is:

$$\left(\frac{\partial L}{\partial x_2}\circ \gamma\right)(t) - \left[\frac{d}{dt}\left(\frac{\partial L}{\partial x_3}\circ \gamma\right)\right](t) =0$$

or written slightly nicer (but also slightly less clear):

$$\left.\frac{\partial L}{\partial x_2}\right|_{\gamma(t)} - \frac{d}{dt}\left(\left.\frac{\partial L}{\partial x_3}\right|_{\gamma(t)}\right) = 0$$

Or equivalently, one could interpret the EL equation as saying that this particular function is the null function on $\Bbb R$:

$$\frac{\partial L}{\partial x_2}\circ \gamma - \frac{d}{dt}\left(\frac{\partial L}{\partial x_3}\circ \gamma\right) = 0$$

• "Maybe this is an abuse of notation and one should define new variables $\eta = q_i(t)$ and $\xi=\dot{q}_i(t)$..." So then by assumption, the inputs to $L$ are assumed to lie in the image of $\mathbb{R}$ under $q_i$ on the $\eta$-line and in the image of $\mathbb{R}$ under $\dot{q}_i$ on the $\xi$ line? My confusion stems from the fact that I am not sure if $q_i$ and $\dot{q}_i$ are given, or if they are implicit functions derived from how the the values of the points of solution curves on the $\eta$/$\xi$ lines respectively relate to their value on the $t$ line Sep 15, 2016 at 8:30
• $q_i$ is not given. The EL equation is a PDE whose solution is the $q_i$ (or $q_i$'s) that extremizes the classical action (the integral of $L$ wrt $t$).
– user137731
Sep 15, 2016 at 11:21
• I don't know that I'd call $q_i(t)$ an implicit function of $\eta$. There is an explicit relationship between them (equality). $\eta$ is an implicit function of $t$, however.
– user137731
Sep 15, 2016 at 11:32
• When we say $y = f(x)$ and $z=g(x,y)$ we're treating $y$ like a variable within $g$, but its values are always in the image of $f$. So we treat $y$ as both another name for the function $f$ (outside $g$) and as an independent variable (inside $g$). When we bring in partial derivatives the lines between variable and function become even more blurred.
– user137731
Sep 15, 2016 at 14:33
• @William Did you ever see my edited response? I think this should be correct.
– user137731
Sep 24, 2016 at 15:33