# Prove differential form of Lagrangian

How to derive the Lagrangian differential force?

$$\frac{d}{dt}(\frac{\partial L}{\partial \dot{x}})+\frac{\partial L}{\partial x}=0$$

I was trying to do something.

$$L=T-U=\frac{1}{2} m\dot{x}^2-\frac{1}{2}kx^2$$ $$\frac{\partial L}{\partial \dot{x}}=m\dot{x}$$ $$\frac{\partial L}{\partial x}=-kx$$

$$\frac{d}{dt}\frac{\partial L}{\partial \ddot{x}}=m\ddot{x} (t)$$ So, I can write that $$m\ddot{x}(t)+kx$$

But, how can I prove that it's equal to $0$. I should find the equation from somewhere else. What's that?

~ https://people.umass.edu/~bvs/601_Lagrange.pdf

Which Euler's equation?

## 1 answer

Let's first remember how you check that you are at an extremum (minimum, maximum, saddle point) of a normal differentiable function. You do so by checking that the first derivative is zero (plus further tests that are irrelevant for the current question).

Now what does the derivative being zero mean? Well, it means that if you move a very small distance away from the point, the function value in first order doesn't change, that is, if $x_0$ is a minimum of $f$, then $$f(x+h) = f(x) + O(h^2)$$

Now when we seek an extremum of $S(q) = \int_{t_0}^{t_1} L(q, \dot q, t)\mathrm dt$, over *paths* q(t), we do essentially the same. One important thing to know is that the paths we consider all begin and end at the same point.

So let's evaluate the function at a path $\tilde q(t) = q(t) + \delta q(t)$ where $\delta q(t)$ is a small deviation, and of course in order to start and end at the same point, we must have $\delta q(t_0) = \delta q(t_1) = 0$.

We will also need the time derivative of $\tilde q(t)$, but that's easy: The derivative of a sum is just the sum of derivatives, that is $$\dot{\tilde q} = \dot q + \delta\dot q$$ (here and in the following I just write $q$ instead of $q(t)$ and so on; just remember that every $q$ has an implicit $(t)$).

So let's insert that into the action integral: $$S(\tilde q) = \int_{t_0}^{t_1} L(q+\delta q, \dot q+\delta\dot q, t)\mathrm dt$$ Now we can do a tailor expansion of $L$ in both arguments to first order (that is, basically ignore anything having a product of $\delta $ quantities): $$S(\tilde q)= \int_{t_0}^{t_1} \big( L(q,\dot q,t) + \frac{\partial L}{\partial q}(q\dot q,t)\delta q + \frac{\partial L}{\partial\dot q}(q\dot q,t)\delta\dot q\big) \mathrm dt$$ Now the integral of the sum can be split into a sum of the integral of the terms. Then we find that the first integral is just $S(q)$, and we get: $$S(\tilde q) = S(q) + \int_{t_0}^{t_1} \frac{\partial L}{\partial q}\delta q\mathrm dt + \int_{t_0}^{t_1} \frac{\partial L}{\partial \dot q}\delta\dot q\mathrm dt$$ Now we'd like to get rid of the $\delta\dot q$. But that's easy by using the partial integration rule

$$\int_a^b u\dot v\mathrm dt = [uv]_a^b - \int_a^b \dot uv\mathrm dt$$

In this case, $a=t_0$, $b=t_1$, $u=\frac{\partial L}{\partial\dot q}$ and $v=\delta q$. Nor remember that $\delta q(t_0)=\delta q(t_1) = 0$, therefore the first term on the right hand side vanishes, and we get $$\int_{t_0}^{t_1} \frac{\partial L}{\partial \dot q}\delta\dot q\mathrm dt = -\int_{t_0}^{t_1} \left(\frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial \dot q}\right)\delta\dot q\mathrm dt\delta q\,\mathrm dt$$ Inserting this into the formula for $S(\tilde q)$ and combining the two remaining integrals, we therefore get to first order in $\delta q$: $$S(\tilde q) = S(q) + \int_{t_0}^{t_1}\left(\frac{\partial L}{\partial q}-\frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial\dot q}\right)\delta q\,\mathrm dt$$ However out demand is that, to first order, $S(\tilde q)=S(q)$, which means that the integral above must be zero. And since $\delta q$ is supposed to be arbitrary (except at the end points), this means that $$\frac{\partial L}{\partial q}-\frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial\dot q}=0$$ But that is exactly the Euler-Lagrange equation.

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