The principle of stationary action?

Question

In proving that the action $$S\equiv \int^{t_2}_{t_1}L(x, x',t)dt$$ has a has a stationary point $x_0$ that satisfies the following: $$\frac{d}{dt}(\frac{\partial L}{\partial x'_0})=\frac{\partial L}{\partial x_0}$$ My textbook finds the action of $$x_a(t)\equiv x_0(t)+a\beta (t)$$ differentiated with respect a to the first order to reach the expression: $$\frac{\partial}{\partial a} S[x_a(t)]=\int^{t_2}_{t_1}(\frac{\partial L}{\partial x_a} - \frac{d}{dt}(\frac{\partial L}{\partial x'_a}))\beta dt$$ It then says that the only way this can be 0 for all $\beta$ is if the expression in the parentheses evaluated at $a=0$ is identically equal to zero. I don't understand why we can evaluate it at 0, please could someone explain? (I would understand if the thing in the parentheses had to be 0 for all $a$ but why does been it been 0 for $a=0$ imply it is 0 for all $a$?)

Answer 1

Comment to the question (v1): Consider the composed function

$$s(a)~:=~ S[x_a] , $$

where

$$x_a(t) ~=~ x_0(t) + a \beta(t),$$

for fixed $\beta$.

It should be stressed that the function $a\mapsto s(a)$ is not necessarily independent of $a$, or equivalently, the derivative $s^{\prime}(a)$ is not necessarily zero for all $a$, even if $x_0(t)$ is a stationary path.

However, if $x_0(t)$ is a stationary path, then $s^{\prime}(0)=0$ by definition.

The full derivation of Euler-Lagrange equations from the stationary action principle is done in many textbooks and websites, e.g. Wikipedia. For more information, see also e.g. this Phys.SE post.