# Why is the Euler-Lagrange equation a necessary condition? (without counter example)

I'm currently studying Arfken's Mathematical Methods for Physicists, and I'm specifically focusing on the calculus of variations. While studying, I came across the derivation of the Euler-Lagrange equation as a necessary condition for extremizing arbitrary functionals. However, I have some doubts regarding this process. Typically, in other books, functionals are treated as functions that depend on a specific variation, and the extremum of that function is sought in Arfken's book consider the functional $$J$$ as $$$$J[y]=\int_{x_1}^{x_2} f(y,y_x,x) dx$$$$ and define $$$$y(x,\alpha)=y(x,0) + \alpha \eta(x)$$$$ where $$y(x,0)$$ is minimize path. Then $$J$$ is now a function of $$\alpha$$: $$$$J=J[Y]=J(\alpha).$$$$ Since $$J$$ is a function of $$\alpha$$, then we can minimize with respect to $$\alpha$$ and ......

Here is a my question. The variation of $$y$$ is $$\alpha \eta(x)$$. But we only consider $$\alpha$$. Although there emerge the $$\eta(x)$$ from almost end of step, it is just arbitrary $$\eta(x)$$. I think it is different minimize with respect to $$\eta(x)$$.

In my opinion, I believe that the optimization should be performed with respect to both variables $$\alpha$$ and function $$\eta(x)$$, and by considering only the variation with respect to $$\alpha$$, we might be neglecting the information related to $$\eta(x)$$. Therefore, I think the Euler-Lagrange equation becomes a necessary condition.

Using an analogy, let's consider a function $$f(x, y)$$, and suppose $$f(x_0, y_0)$$ is the minimum value of $$f$$. In this case, $$\nabla f$$ would be zero at $$(x_0, y_0)$$, but just because $$\frac{\partial f}{\partial x}$$ is zero at $$(x_0, y_0)$$ doesn't guarantee that $$f(x_0, y_0)$$ is the minimum value. I think in this case, Euler-Lagrange equation and $$\frac{\partial f}{\partial x}$$ is same character. I'm curious to know if there are any flaws in my thinking. Additionally, I'm curious why the functional $$J$$ is not considered as a function solely dependent on $$\alpha$$. I have tried my best to summarize my questions, and I appreciate your patience in reading through my lengthy inquiry. Have a great day!

• Just wanted to clarify that the Euler-Lagrange equations only give an extremal point of the integral, and do not necessarily give the minimum. And in most physics cases an extremal point is enough - the Action just needs to be extremized, not minimized. If you do want a minimum for the integral you will need other checks (second order variational derivative), and even then it may not be a global minimum. Again, that is usually not a concern, but I guess it depends on your application. Commented Jun 23, 2023 at 15:50

The $$\eta(x)$$ function that you are referring to is a helper function, and the whole point is that the shape of that helper function is immaterial, as long as it is zero at the designated points.
The only reason for introducing the helper function $$\eta(x)$$ is to facilitate some wriggle room. The multiplication factor $$\alpha$$ executes the variation sweep. At the point where $$\alpha$$ is zero the variation coincides with the curve that you are in the process of solving for.
For the end result it is the derivative at the point where $$\alpha$$ is zero that is relevant.
(Again, at the point where $$\alpha$$ is zero the variation coincides with the curve you are in the process of solving for.)
Another way of looking at it: the derivation is organized in such a way that towards the end the helper function $$\eta(x)$$ drops out. It would be really bad if the helper function would not drop out, that would mean you are stuck.