1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

226 Chapter 3 The Wave Equation

Figure 4 Shapes of car antenna.

eachequalto256cyclespersecond.Thedifferenceinthesetoffrequen-

cies accounts for some of the difference between the sound of a stringed

instrument and that of a xylophone or glockenspiel.

13.My car’s antenna vibrates in the wind under various conditions in one of

the two shapes shown in Fig. 4. If the antenna is modeled as a uniform thin

beam with centerline displacementu(x,t),thenusatisfies the equation

∂^4 u

∂x^4 =−

1

c^2

∂^2 u

∂t^2 +f(x,t),^0 <x<a,^0 <t,

wherefis a “forcing function” that represents the effect of wind or other

distributed forces. Because the base of the antenna is built into the car, the

boundary conditions at the base are zero displacement and slope:

u( 0 ,t)= 0 , ∂∂xu( 0 ,t)= 0 , 0 <t.

The top of the antenna is free to move. There, the internal moment and

shear are both zero, leading to the conditions

∂^2 u

∂x^2 (a,t)=^0 ,

∂^3 u

∂x^3 (a,t)=^0 ,^0 <t.

(These four boundary conditions are standard for a cantilevered beam.)

It can be shown that the solution of the foregoing problem, together with

initial conditions onuandut, can be represented as a series of products

of the form

(

ancos

(

λ^2 nct

)

+bnsin

(

λ^2 nct

)

+Fn(t)

)

φn(x),

whereFn(t)comes from the forcing function andφn(x)andλnare related

through the eigenvalue problem

φ′′′′−λ^4 φ= 0 , 0 <x<a,

φ( 0 )= 0 ,φ′( 0 )= 0 ,φ′′(a)= 0 ,φ′′′(a)= 0.

This arises in the obvious way from the boundary conditions and the ho-

mogeneous partial differential equation.