I Help with Euler Lagrange equations: neighboring curves of the extremum

AI Thread Summary
The discussion centers on the necessity of keeping the variable u small when deriving the Euler-Lagrange equations, particularly in relation to Taylor series expansions. It is argued that neglecting higher-order terms in the expansion is justified only if u is small, as this affects the accuracy of the approximation. However, some participants believe that the equations can hold regardless of u's size, suggesting that the necessary conditions for extrema can be proven without this assumption. The conversation also touches on the fundamental theorem of variational calculus, which leads to the Euler-Lagrange equations when considering the functional derivative. Overall, the participants explore the implications of u's size on the derivation process and the validity of their approaches.
Reuben_Leib
Messages
7
Reaction score
1
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?

1682098811578.png
 
Physics news on Phys.org
The unsoken assumption is that you are truncating the taylor series of x(t,u) with respect to u about u = 0: <br /> 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 where here \eta = \left.\frac{\partial x}{\partial u}\right|_{u=0}. The neglect of the higher order terms is only justified if u is small. But really what you are looking for to derive the Euler-Lagrange equations is a functional derivative at x in the direction of \eta, <br /> \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) <br /> and in this formulation x(t,u) = x(t,0) + u \eta(t) is exact, not approximate.
 
Last edited:
  • Like
Likes Reuben_Leib and vanhees71
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?
 
Last edited:
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,
$$\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{ \partial \dot{\vec{x}}}=\frac{\partial L}{\partial \vec{x}}.$$
 
vanhees71 said:
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.
 
Reuben_Leib said:
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##.
 
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Hello! Let's say I have a cavity resonant at 10 GHz with a Q factor of 1000. Given the Lorentzian shape of the cavity, I can also drive the cavity at, say 100 MHz. Of course the response will be very very weak, but non-zero given that the Loretzian shape never really reaches zero. I am trying to understand how are the magnetic and electric field distributions of the field at 100 MHz relative to the ones at 10 GHz? In particular, if inside the cavity I have some structure, such as 2 plates...
Back
Top