Help with Euler Lagrange equations: neighboring curves of the extremum

[Reuben_Leib]

TL;DR Summary

Euler Lagrange equations

I tried writing this out but I think there is a bug or something as its not always displaying the latex, so sorry for the image.
I have gone through various sources and it seems that the reason for u being small varies. Sometimes it is needed because of the taylor expansion, this time (below) is "needed" for equations(2.18) and (2.21). I provided all the information I think you may need. So in this case, why must u be small?

 

[pasmith]

The unsoken assumption is that you are truncating the taylor series of [itex]x(t,u)[/itex] with respect to [itex]u[/itex] about [itex]u = 0[/itex]: [tex]
x(t, u) = x(t,0) + u\left.\frac{\partial x}{\partial u}\right|_{u = 0} + \frac12 u^2 \left.\frac{\partial^2 x}{\partial u^2}\right|_{u = 0} + \dots[/tex] where here [tex]\eta = \left.\frac{\partial x}{\partial u}\right|_{u=0}.[/tex] The neglect of the higher order terms is only justified if [itex]u[/itex] is small. But really what you are looking for to derive the Euler-Lagrange equations is a functional derivative at [itex]x[/itex] in the direction of [itex]\eta[/itex], [tex]
\delta I = \lim_{u \to 0} \frac{I[x + u\eta] - I[x]}{u} = \left(\left.\frac{d}{du}I[x + u\eta]\right|_{u = 0}\right)
[/tex] and in this formulation [itex]x(t,u) = x(t,0) + u \eta(t)[/itex] is exact, not approximate.

 

[Reuben_Leib]

Thank you pasmith for your reply, From my understanding I believe that equations(2.18) and (2.21) should hold regardless of the size of ##u##? Maybe I am not understanding taylor series correctly, but I thought that for an example a first order taylor expansion: $$I(u) = I(0) + uI'(0) + \frac{I''(\epsilon)}{2}u^2$$ where ##\epsilon \in [0,u]##, is exact?

standardly let: $$I(u) = I(0) + uI'(0) + O(u^2)$$

The I can go on to prove the(the necessary condition) that ##\delta I = 0## :

Suppose $I(0)$ is the extremum is a minimum, then proof by contradiction: let $I'(0) > 0$.
Let ##C## be chosen so that ##C \geq \frac{I''(\epsilon)}{2}## for all ##u \in [-U,U]##, thus we get ##Cu^2 \geq \frac{I''(\epsilon)}{2}u^2 = O(u^2)##.
Now we can get ##|I(u)-I(0)-uI'(0)| = O(u^2) \leq Cu^2##
Now let ##u<0##
As ##I'(0)## is a constant(not dependent on ##u##),chose ##u## such that ##C|u|=\frac{I'(0)}{4} < \frac{I'(0)}{2}##
Letting ##C|u| \leq \frac{I'(0)}{2}##, thus ##-\frac{I'(0)}{2}\leq Cu \leq \frac{I'(0)}{2}##, thus ##-u\frac{I'(0)}{2}\geq Cu^2 \geq \frac{I'(0)}{2}u##, because ##u<0##.
Now as ##|I(u) -I(0) - uI'(0)|\leq Cu^2##, thus ##-Cu^2\leq I(u) -I(0) - uI'(0)\leq Cu^2##. Taking ##I(u)\leq I(0) + uI'(0)+Cu^2##. and ##Cu^2\leq-u\frac{I'(0)}{2}## we get:##I(u)\leq I(0) + uI'(0)+Cu^2\leq f(0) + u\frac{I'(0)}{2}< I(0)##
Thus ##I(0)## can't be minimum, hence ##I'(0) \leq 0##, similar to prove for maximum ##I'(0) \geq 0##, thus ##I'(0) = 0##
So this proves that ##I'(0) = 0##.

So from this I believe that I proved then necessary condition without the use of ##u## being small?

 

[vanhees71]

It's much easier to argue just with the finding of extrema of functions of a real variable. You simply define
$$F(u)=\int_{t_1}^{t_2} \mathrm{d} t [L(\vec{x} + u \vec{\eta},\dot{\vec{x}}+u \dot{\vec{\eta}}].$$
Then for this to get extremal at ##\eta=0## you must have
$$F'(u)|_{u=0}=0.$$
Evaluate the derivative
$$F'(u)=\int_{t_1}^{t_2} \mathrm{d}t \left [\vec{\eta} \cdot \frac{\partial L}{\partial \vec{x}} + \dot{\vec{\eta}} \cdot \frac{\partial L}{ \partial \dot{\vec{x}}} \right].$$
Next you use that in the Hamilton principle (in Lagrangian form) the endpoints of the trajectories under consideration are held fixed, i.e., you have ##\vec{\eta}(t_1)=\vec{\eta}(t_2)=0##. Then you can integrate the 2nd contribution by parts and get
$$F'(u)=\int_{t_1}^{t_2} \mathrm{d}t \vec{\eta} \cdot \left [\frac{\partial L}{\partial \vec{x}} - \frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{ \partial \dot{\vec{x}}} \right].$$
Now set ##u## to 0 and then, because ##F'(u=0)=0## must hold for all trajectories ##\vec{\eta}(t)##, the bracket in the integral must vanish (this is known as the "fundamental theorem of variational calculus"), which leads to the Euler-Lagrange equations,
$$\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{ \partial \dot{\vec{x}}}=\frac{\partial L}{\partial \vec{x}}.$$

 

[Reuben_Leib]

It's much easier to argue just with the finding of extrema of functions of a real variable. You simply define
$$F(u)=\int_{t_1}^{t_2} \mathrm{d} t [L(\vec{x} + u \vec{\eta},\dot{\vec{x}}+u \dot{\vec{\eta}}].$$
Then for this to get extremal at ##\eta=0## you must have
$$F'(u)|_{u=0}=0.$$
Evaluate the derivative
$$F'(u)=\int_{t_1}^{t_2} \mathrm{d}t \left [\vec{\eta} \cdot \frac{\partial L}{\partial \vec{x}} + \dot{\vec{\eta}} \cdot \frac{\partial L}{ \partial \dot{\vec{x}}} \right].$$
Next you use that in the Hamilton principle (in Lagrangian form) the endpoints of the trajectories under consideration are held fixed, i.e., you have ##\vec{\eta}(t_1)=\vec{\eta}(t_2)=0##. Then you can integrate the 2nd contribution by parts and get
$$F'(u)=\int_{t_1}^{t_2} \mathrm{d}t \vec{\eta} \cdot \left [\frac{\partial L}{\partial \vec{x}} - \frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{ \partial \dot{\vec{x}}} \right].$$
Now set ##u## to 0 and then, because ##F'(u=0)=0## must hold for all trajectories ##\vec{\eta}(t)##, the bracket in the integral must vanish (this is known as the "fundamental theorem of variational calculus"), which leads to the Euler-Lagrange equations,
$$\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{ \partial \dot{\vec{x}}}=\frac{\partial L}{\partial \vec{x}}.$$

This is the problem I have, I 100% understand this method(also the one from wikipedia), but the one my studyguide gives me I can't understand.

 

[fresh_42]

So from this I believe that I proved then necessary condition without the use of ##u## being small?

Sure? I don't follow your calculation in detail, but you set
$$
\dfrac{I'(0)}{2}\geq C\cdot |u| \geq \dfrac{I''(\varepsilon )}{2}|u|\text{ and so } \dfrac{I'(0)}{I''(\varepsilon }\geq |u|>0
$$
so you already assumed the size of ##u##.