Consider a string clamped at the end points \(a\) and \(b\), say, with \(u(a,t)=0\) and \(u(b,t)=0\), where \(u(x,t)\) is the string’s deviation from the horizontal rest position. The kinetic energy of the string is \[ E_k(t)=\frac {1}{2}\int _a^b \rho u_t^2d s, \] where \(\rho (x,t)\) is the mass density and \(ds=\sqrt {1+u_x^2}d x\) is an infinitesimal arc length. The potential energy consists of the sum of the energy due to the stretching of the string, and the work done against a load: \[ E_p(t)=\int _a^b \left (d(\sqrt {1+u_x^2}-1)-lu\sqrt {1+u_x^2}\right )d x. \] Here \(d(x,t)\) is the string’s elastic coefficient and \(l(x,t)\) is the load on the string. The classical action for the continuum model is \[ J=\int _{t_1}^{t_2}[E_k(t)+E_p(t)]d t. \]
- Substitute the integral expressions for \(E_k\) and \(E_p\) into the action \(J\), and consider the variations \(u+\epsilon \psi \) such that \(\psi \) vanishes at the end points \(a,b\) and at the initial and final times \(t_1,t_2\). Calculate the first variation \(\delta J\).
- After integrating by parts terms of the form \(u_t\psi _t\) and \(u_x\psi _x\), and using the boundary conditions for the variation \(\psi \), show that Hamilton’s principle \(\delta J=0\) yields the following PDE for \(u\): \[ (\rho u_t)_t-(1+u_x^2)^{-1/2} \left [d(1+u_x^2)^{-1/2}u_x\right ]_x-l=0. \]
- Show that if \(\rho ,d\) are constants, there is no load (\(l=0\)), and the small slope approximation \(\abs {u_x}\ll 1\) holds, then the PDE of part (b) reduces to the classical one-dimensional wave equation. Write down an expression for the wave speed.