Mathematical Methods for Physics and Engineering - Matematica.NET

PDES: GENERAL AND PARTICULAR SOLUTIONS20.1 Important partial differential equationsMost of the important PDEs of physics are second-order and linear. In order togain familiarity with their general form, some of the more important ones willnow be briefly discussed. These equations apply to a wide variety of differentphysical systems.Since, in general, the PDEs listed below describe three-dimensional situations,the independent variables are r and t, wherer is the position vector and t istime. The actual variables used to specify the position vector r are dictated by thecoordinate system in use. For example, in Cartesian coordinates the independentvariables of position are x, y and z, whereas in spherical polar coordinates theyare r, θ and φ. The equations may be written in a coordinate-independent manner,however, by the use of the Laplacian operator ∇ 2 .The wave equation20.1.1 The wave equation∇ 2 u = 1 ∂ 2 uc 2 ∂t 2 (20.1)describes as a function of position and time the displacement from equilibrium,u(r,t), of a vibrating string or membrane or a vibrating solid, gas or liquid. Theequation also occurs in electromagnetism, where u may be a component of theelectric or magnetic field in an elecromagnetic wave or the current or voltagealong a transmission line. The quantity c is the speed of propagation of the waves.◮ Find the equation satisfied by small transverse displacements u(x, t) of a uniform string ofmass per unit length ρ held under a uniform tension T , assuming that the string is initiallylocated along the x-axis in a Cartesian coordinate system.Figure 20.1 shows the forces acting on an elemental length ∆s of the string. If the tensionT in the string is uniform along its length then the net upward vertical force on theelement is∆F = T sin θ 2 − T sin θ 1 .Assuming that the angles θ 1 and θ 2 are both small, we may make the approximationsin θ ≈ tan θ. Since at any point on the string the slope tan θ = ∂u/∂x, theforcecanbewritten[ ∂u(x +∆x, t)∆F = T−∂x]∂u(x, t)∂x≈ T ∂2 u(x, t)∆x,∂x 2where we have used the definition of the partial derivative to simplify the RHS.This upward force may be equated, by Newton’s second law, to the product of themass of the element and its upward acceleration. The element has a mass ρ ∆s, whichisapproximately equal to ρ ∆x if the vibrations of the string are small, and so we haveρ ∆x ∂2 u(x, t)= T ∂2 u(x, t)∆x.∂t 2 ∂x 2676