The d d d -dimensional wave equation is given by the form:
u t t = a 2 ∇ 2 u + f ( x , t ) u_{tt} = a^2 \nabla^2 u + f(\mathbf{x},t) u tt = a 2 ∇ 2 u + f ( x , t ) where u ( x , t ) : R d × R → R u(\mathbf{x,t}): \mathbb{R}^d\times \mathbb{R}\to \mathbb{R} u ( x , t ) : R d × R → R . The one dimensional wave equation u t t = a 2 u x x u_{tt}=a^2u_{xx} u tt = a 2 u xx very naturally arise in modeling of a vibrating string .
The Vibrating String ¶ Consider a horinzontally placed string with length L L L . We make the following assumptions
Suppose at equilibrium the tension is a constant T T T . As the string vibrate at small amplitude, we can assume at arbitrary position the tension is a constant and takes the value of the equilibrium tension T T T .
The string vibrates in the vertical direction, i.e., no horizontal stretching
Let ρ ( x ) \rho(x) ρ ( x ) be the density distribution of the string.
Now to model the vibration mathematically, we place the string along the x x x -axis, let u ( x , t ) u(x,t) u ( x , t ) be the vertical displacement of the string at position x x x from equilibrium position at time t t t .
The Lagrangian Way ¶ It’s clear that the kinetic energy of the string should be given by:
K = 1 2 ∫ 0 L m u t d x K=\frac{1}{2}\int_0^L mu_tdx K = 2 1 ∫ 0 L m u t d x