# Lagrangian density for a Piano String

So I'm trying to do this problem where I'm given the Lagrangian density for a piano string which can vibrate both transversely and longitudinally. $\eta(x,t)$ is the transverse displacement and $\xi(x,t)$ is the longitudinal one. (So a point at $[x,0]$ at some later time t would be at $[x+\xi(x,t),\eta(x,t)]$). The Lagrangian density is given by $$\mathcal{L} = \frac{\rho_0}{2}[\ddot{\xi}^2+\ddot{\eta}^2]-\frac{\lambda}{2}[\frac{\tau_0}{2}+\xi'+\frac{1}{2}(\eta')^2]^2$$ where the dot is a partial time derivative and a prime is a partial x derivative. Also $\lambda$, $\tau_0$ and $\rho_0$ are parameters such as the Young's modulus, density and tension of the string respectively. So applying the action principle in the case of a continuous system I get the following PDEs for $\xi$ and $\eta$: $$\ddot{\xi} = \frac{\lambda}{\rho_0}(\xi''+\eta'\eta'')$$ and $$\ddot{\eta} = \frac{\lambda}{\rho_0}[\eta''(\frac{\tau_0}{\lambda} + \xi'+\frac{3}{2}(\eta')^2)+\eta'\xi'']$$.

Linearization of the equations yields two simple wave equations, the $\eta$ wave traveling at a speed $c_T^2 = \frac{\tau_0}{\rho_0}$ and the $\xi$ wave at $c_L^2 = \frac{\lambda}{\rho_0}$.

Now the part that is stumping me is that the problem asks us to show that if a given pulse of the form $\eta(x,t) = \eta(x-c_Tt)$ propagates along the string then that induces a concurrent longitudinal pulse of the form $\xi(x-c_Tt)$. Obviously I can't use the linearized equations to prove this. I would have to use some sort of iterative process by using exact PDE's to get the next step. But I tried doing this and I just get a mess. Any ideas where to go from here? (for reference this is problem 1.10 in Stone and Goldbart's book).

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Take you first EOM and use method of Fourier transform. Expand $\xi$ in terms of $exp(ikx)$ (coeff. will then depend on t and k) and expand $\eta'\eta''$ in terms of $exp(ik(x-c_Tt))$ (coefficients here will only depend on k). Then for each k you will get equation for a forced harmonic oscillator whose solution will necessarily consist of frequency imposed on it by forcing term. – user10001 Sep 6 '12 at 22:28

Let's look for solutions of the PDEs in a form $\eta(x, t)= \eta(z)$ and $\xi(x, t)= \xi(z)$, where $z= x-c_T t$. If we substitute these solution to the PDEs, they are reduced to ODEs. But to answer you question, we only need the second equation, which has the form: $$c_T^2 \eta_{zz} = {\lambda \over \rho_0} [\eta_{zz} ({\tau_0 \over \lambda} + \xi_z + {3 \over 2} (\eta_z)^2) + \eta_z \xi_{zz} ],$$ where index $z$ means derivative on $z$. Now, the first term in () on the right-hand side cancels with the left-hand-side term, because $\tau_0/\rho_0 = c_T^2$. The remaining terms are in fact the full derivative, therefore, we have: $$[\eta_z + \xi_z + (\eta_z)^3/2]_z = 0, \quad \mbox{or} \quad \eta_z + \xi_z + (\eta_z)^3/2= C,$$ where C is a constant (does not depend on $z$). Since $\eta(z)$ has a pulse shape, then $\eta_z= 0$ at $z \to \pm \infty$. We assume also that $\xi_z= 0$ at $z \to \pm \infty$. Then $C=0$. So, finally we get that $$\xi_z= -\eta_z[1+(\eta_z)^2]. \qquad (*)$$ We know that $\eta(z)$ has a pulse shape. Therefore, it's derivative $\eta_z$ is zero at $z= -\infty$, then it decreases to some minimum, then it increases, crossing zero, goes to maximum and decreases again to zero at $z= +\infty$. Since exprssion in [ ] in Eq.(*) is always positive, $\xi_z$ has behavour similar to $\eta_z$, but with signs flipped. This means that $\xi(z)$ has also a pulse shape, but this pulse is flipped. I.e. if $\eta(z)$ is a hill-shaped, then $\xi(z)$ is a well-shaped.