Just starting a course on Lagrangian Mechanics and I'm just wondering what about the Euler-Lagrange equation, and more specifically what I'm meant to be trying to do ..

One of the questions from my textbook reads :

Solve the Euler-Lagrange equation for the following function \begin{align*} f(y,y') = y^2+y'^2 \end{align*} Looks simple enough... But where should I be headed to start?

The Euler-Lagrange Equation as we have it is \begin{align*} \frac{\partial f}{\partial y} = \frac{d}{dx} \frac{\partial f}{\partial y'} \end{align*}

Do I just find the partial derivatives of f treating y and y' as independent variables as follows:

$$\frac{\partial f}{\partial y} = 2y,$$ $$\frac{\partial f}{\partial y'} = 2y$$ so then we have : $$2y = \frac{d}{dx}(2y')$$ $$y = \frac{d}{dx}(y')$$ $$yx+c = y'$$ $$\frac{yx^2}{2}+cx+d = y$$

And then simplify with the y isolated on the right so it looks nice .. or at least thats what I thought..

Any comments ? right track, wrong track? Also, what is this euler equation actually solving for? Is it just a nice way of solving differential equations in this form ?

When we did this in class, we did a monster proof showing it minimizes the path between 2 points or finds the extrema for the function $I(x) = \int^a_b \ F(y(x),y'(x),x) dx,$ which is all great but I'm not too sure how this relates to the equation we solve in the question .. Thanks for all

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    $\begingroup$ you seem to be 'losing' derivatives. $\frac{d}{dx} y' = y^{''}$. At least if $y=y(x)$. Then $y=y^{''}$ is the ODE you have to solve. Also, $\frac{df}{dy'} = 2 y'$, but that you fixed later on somehow, so it was possibly only a typo. $\endgroup$ – user20266 Jul 25 '12 at 12:23
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    $\begingroup$ How did you go from $y=\frac{d}{dx}y\,'$ to $yx+c=y\,'$? Did you perhaps do $\int ydx=yx+c$ (which is incorrect)? $\endgroup$ – anon Jul 25 '12 at 12:26
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    $\begingroup$ $y$ and $y'$ are not independent of $x$, so you cannot solve $y = \frac{d}{dx}y'$ this way. $\endgroup$ – N.U. Jul 25 '12 at 12:27
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    $\begingroup$ Though you are correct that we keep $y$ and $y\,'$ independent when we take partial derivatives. $\endgroup$ – anon Jul 25 '12 at 12:31
  • $\begingroup$ ok, so the problem line seems to be $y = \frac{d}{dx}(y')$ .. from here I should've gone to $y = y''$ which is easy enough to solve.. The reason I can't just integrate both sides with respect to $x$ is because the $y$ on the LHS is dependent of $x$ .. Can somebody confirm these thoughts for me ? thanks thanks $\endgroup$ – Rawb Jul 25 '12 at 13:21

The function $f(y,y') = {y'}^2+y^2$ can be interpreted as a Lagrangian $L = T-V$, where $T={y'}^2$ is the kinetic energy and $V = -y^2$ the potential energy of some object.${}^\dagger$ Among other things, this is the Lagrangian of pencil that is balanced vertically on its tip. $y$ is then interpreted as the angle the pencil makes with the vertical, and must be assumed to be small. $x$ should be interpreted as time. We expect at least one unstable solution since the pencil can fall over.

The action is extremized by solutions to the Euler-Lagrange equation. The equations of motion found by the principle of least action are equivalent to Newton's laws of motion---we are going to find out how the pencil drops.

First, note that $$\begin{eqnarray*} \frac{d}{dx} \frac{\partial f}{\partial y'} &=& \frac{d}{dx} 2 y' \\ &=& 2y'' \\ \frac{\partial f}{\partial y} &=& 2 y. \end{eqnarray*}$$ The Euler-Lagrange equation gives $$y'' = y.$$ The solutions are $$y(x) = A e^x + B e^{-x},$$ where $A$ and $B$ are some constants determined by the initial conditions. For example, if $y(0)=0$ and $y'(0)=1$ we find $$y(x) = \sinh x.$$ This solution is indeed unstable, the angle increases without bound. Of course, this can only be taken seriously for small $y$.

The principal advantage of finding equations of motion in this way is it allows the simple use of generalized coordinates. With practice it becomes automatic to write the total kinetic and potential energy of a system in terms of some convenient coordinates and to find the equations of motion via the Euler-Lagrange equations, which have the same form for any system of generalized coordinates. As a cultural side note, be aware that variational methods extend far beyond their application to the Lagrangian formulation of classical mechanics.

${}^\dagger$The force corresponding to this potential is $F = -\frac{d}{dy}V = 2y$. This is Hooke's law with the wrong sign. What happens if $f(y,y') = {y'}^2 - y^2$?

  • $\begingroup$ Thank you so much ! I have a hopeless lecturer (who is indeed very smart but who can't seem to teach and is very impatient) so I am badly in need of a better approach... I may be putting down a lot of questions in the next few months :/ $\endgroup$ – Rawb Jul 30 '12 at 11:19
  • $\begingroup$ So in the final y(x)=sinhx , does the x then represent time (oh wait, you told me that).. as for the †, would $f(y, y')=y'^2 - y^2$ then be the Lagrangian of a spring ? $\endgroup$ – Rawb Jul 30 '12 at 11:23
  • $\begingroup$ @Ronald: Glad to help. Yes, if you flip the sign you can also think of it as the Lagrangian of a pendulum for small oscillations. You'll get simple harmonic motion. $\endgroup$ – user26872 Jul 30 '12 at 16:42

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