PDES: GENERAL AND PARTICULAR SOLUTIONS
20.1 Important partial differential equations
Most of the important PDEs of physics are second-order and linear. In order to
gain familiarity with their general form, some of the more important ones will
now be briefly discussed. These equations apply to a wide variety of different
physical systems.
Since, in general, the PDEs listed below describe three-dimensional situations,
the independent variables arerandt,whereris the position vector andtis
time. The actual variables used to specify the position vectorrare dictated by the
coordinate system in use. For example, in Cartesian coordinates the independent
variables of position arex,yandz, whereas in spherical polar coordinates they
arer,θandφ. The equations may be written in a coordinate-independent manner,
however, by the use of the Laplacian operator∇^2.
20.1.1 The wave equation
The wave equation
∇^2 u=
1
c^2
∂^2 u
∂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. The
equation also occurs in electromagnetism, whereumay be a component of the
electric or magnetic field in an elecromagnetic wave or the current or voltage
along a transmission line. The quantitycis the speed of propagation of the waves.
Find the equation satisfied by small transverse displacementsu(x, t)of a uniform string of
mass per unit lengthρheld under a uniform tensionT, assuming that the string is initially
located along thex-axis in a Cartesian coordinate system.
Figure 20.1 shows the forces acting on an elemental length ∆sof the string. If the tension
Tin the string is uniform along its length then the net upward vertical force on the
element is
∆F=Tsinθ 2 −Tsinθ 1.
Assuming that the anglesθ 1 andθ 2 are both small, we may make the approximation
sinθ≈tanθ. Since at any point on the string the slope tanθ=∂u/∂x,theforcecanbe
written
∆F=T
[
∂u(x+∆x, t)
∂x
−
∂u(x, t)
∂x
]
≈T
∂^2 u(x, t)
∂x^2
∆x,
where 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 the
mass of the element and its upward acceleration. The element has a massρ∆s,whichis
approximately equal toρ∆xif the vibrations of the string are small, and so we have
ρ∆x
∂^2 u(x, t)
∂t^2
=T
∂^2 u(x, t)
∂x^2
∆x.