The question is long because of the demonstrations I give, but the problem is simple, so bear with me for a minute.

I am trying to derive the dispersion relation of a semi-infinite system using Euler-Lagrange equations, and I started with the simplest case of a semi-infinite string. However, the result, as I show below, is $\omega^2 = -c_0^2 \gamma^2$ instead of the correct $\omega^2 = c_0^2 \gamma^2$.

My question is: can I use Euler-Lagrange equations to derive the dispersion relation of infinite and semi-infinite vibrating systems? I've seen it being used only for finite systems to calculate their natural frequencies. If the method is correct for semi-infinite and infinite systems, what did I do wrong in the string case?

I can't think of a reason why I shouldn't use it, but since it doesn't work for a string, how could it possibly be correct for more complicated media and geometries? And yet, it does work for a beam in pure bending, and this makes me think that either I did something wrong in the string case, or that it never works and the correct result for the beam is just a coincidence.

Considering transverse displacement $y = y(x,t) = A(t)X(x)$ of a point in a taut string, the kinetic energy $K$ of the string is $K = \frac{1}{2}\int{\rho\dot y^2}\text dx$, where $\rho$ is the string's linear mass density. As for the potential energy $U$, Graff [1, p.20] considers it as the work done by the tension $T$ as the string changes its length from $l$ to $l'$ when it vibrates. Therefore,

$$ l' = \int \text d s = \int \sqrt{1 + y_x^2} \, \text dx \approx \int1 + \frac 1 2 y_x^2 \, \text dx = l + \frac 1 2 \int y_x^2 \,\text dx $$ $$ U = T\Delta l = T(l' - l) = \frac T 2 \int y_x^2 \, \text dx $$

Making the Lagrangian $L = K - U$, and taking $A(t)$ as the generalized coordinate, the Euler-Lagrange equation becomes

$$ \frac{\text d}{\text dt}\left(\frac {\partial L}{\partial \dot q}\right) = \frac {\partial L}{\partial q} $$ $$ \frac{\text d}{\text dt}\left(\frac {\partial K}{\partial \dot A}\right) = -\frac {\partial U}{\partial A} $$ $$ \rho \ddot A(t) \int X(x) \, \text dx = - A(t) T \int (X'(x))^2 \, \text dx \tag{1} $$

Now, differentiating w.r.t $x$, and making $y$ an harmonic vibration, i.e. $\ddot A(t) = -\omega^2 A(t)$, $X(x) = \exp(-\text i \gamma x)$, and $c_0^2 = T/\rho$, the relationship between the frequency $\omega$ and the wavenumber $\gamma$ then comes out of $(1)$ as

$$ \omega^2 = -c_0^2 \gamma^2, $$ which is incorrect and can be verified by substituting $y(x,t)$ into the wave equation and making the same assumptions as above.

Now, if I try a Euler-Bernoulli beam in pure bending, taking the potential energy as $$ U = \frac 1 2 \int \frac{M^2}{EI} \, \text dx = \frac 1 2 \int \frac{(-EI y_{xx})^2}{EI} \, \text dx = \frac 1 2 \int EI y_{xx}^2 \, \text dx $$ and supposing again that $y = A(t) \exp(-\text i \gamma x)$, the dispersion relation is $$ \omega^2 = a^2 \gamma^4, \ \ \ a^2 = EI/\rho $$ which is correct, with $E$ and $I$ being the beam's Young modulus and cross section's second moment of area, respectively.

[1] Karl F. Graff. "Wave motion in elastic solids". 1st ed. Oxford Press, 1975.

  • 1
    $\begingroup$ Can you be a little more clear on what you would want to be answered? $\endgroup$ – GK A May 20 '19 at 1:18
  • $\begingroup$ @GKA Right, the question itself was buried somewhere in there. I tried to make it more explicit. $\endgroup$ – luispauloml May 20 '19 at 3:23

The correct way to use the Lagrangian formalism to predict equations of motion is to consider the lagrangian to be a functional of $y(x,t)$, given by

$$ L[y]=\frac{\rho}{2}\int\dot{y}^2\ dx-\frac{T}{2}\int y_x^2\ dx $$

Then the Euler-Lagrange equations come from minimizing $L$ with respect to all possible variations of $y(x,t)$ that fix the boundary conditions $y(x,t_0)$ and $y(x,t_1)$. In essence, we regard the function $y(x,t)$ to be an infinite collection of independent variables indexed by $x$, each of with varies with time. The Euler-Lagrange equations are

$$ \frac{\partial}{\partial t}\frac{\delta L}{\delta \dot y(x,t)}=\frac{\delta L}{\delta y(x,t)} $$ $$ \frac{\partial}{\partial t}[\rho \dot y(x,t)]=Ty_{xx}(x,t) $$ $$ \ddot y(x,t)=c_0^2y_{xx}(x,t) $$ Then the solution $y(x,t)=Ae^{i(\omega t-\gamma x)}$ gives

$$ \omega^2=\gamma^2c_0^2 $$ which is what you wanted.

The problem with your method is in plugging in the separable "guess" $y(x,t)=X(x)A(t)$ before doing the variational minimization. The Euler-Lagrange equations come from allowing the solution to vary over all possible configurations compatible with the boundary conditions, not just all separable solutions. You've essentially found a constrained minimum of the action, which is not the global minimum of the action. This is analogous to the elementary mistake many students make when first encountering the Lagrangian and try to "simplify" it by using conservation laws or equations of motion before applying Euler-Lagrange.

EDIT: I did this calculation using functional derivatives, to make it analagous to the case of $N$ coupled masses in the limit $N\rightarrow\infty$. To get, e.g., the right hand side, you have to take the functional derivative of $L$ with respect to $y(x,t)$, while treating $\dot y(x,t)$ as an independent variation. In other words, you compute $\frac{\delta L}{\delta y(x,t)}$ by computing $L[y+\delta y]$ and finding whatever is in front of $\delta y(x,t)$ in the resulting expression. In our case,

$$ L[y+\delta y]=\frac{\rho}{2}\int(\dot{y}+\dot{\delta y})^2\ dx-\frac{T}{2}\int (y_x+\delta y_x)^2\ dx\approx L[y]+\rho\int \dot y\dot{\delta y}\ dx-T\int y_x\delta y_x \ dx=L[y]+\rho\int \dot y\dot{\delta y}\ dx+T\int y_{xx}\delta y \ dx $$ where in the last step I integrated by parts. Now we can just read off that $\frac{\delta L}{\delta y(x,t)}=T y_{xx}(x,t)$.

If you're uncomfortable with this (maybe you haven't seen functional derivatives before) there is fortunately an alternative method to extract equations of motion from the Lagrangian, when the Lagrangian is a functional of some continuum function $y(x)$. This method doesn't require functional differentiation. The punchline is that you consider the Lagrangian to be the thing you integrate over $dt$ AND $dx$ to get the action (so for you, $L_{new}=\frac{\rho}{2}{\dot y}^2-\frac{T}{2}{y_x}^2$). Then the equations of motion become $$ \frac{\partial L}{\partial y}=\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot y}+\frac{\partial }{\partial x}\frac{\partial L}{\partial y_x} $$ which you can verify gives the same EOM as above.

| cite | improve this answer | |
  • $\begingroup$ Great! And thanks for pointing out the "elementary" mistake part. As an engineer, it shows that I need to understand this tool better. $\endgroup$ – luispauloml May 21 '19 at 2:21
  • $\begingroup$ I've just came back to this problem. How do you get the RHS of the second equation? Could you show me? $\endgroup$ – luispauloml May 23 '19 at 4:59
  • $\begingroup$ @luispauloml I edited, hopefully that helps. There are a few different ways to get the proper EOMs when your Lagrangian has an infinite number of coordinates, you should pick your favorite. If you want a more detailed explanation of either method, let me know. But it might make sense to put it in another question. $\endgroup$ – Jahan Claes May 23 '19 at 16:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.