The Chemistry Maths Book, Second Edition

(Grace) #1

14.8 Exercises 413


The general solution of the wave equation that satisfies the boundary conditions is


therefore a superposition of normal modes


(14.103)


(the constantsc


1

, 1 c


2

, 1 c


3

,=in (14.101) have been absorbed in the A’s and B’s in


(14.103)). This solution satisfies the initial conditions when the A’s and B’s are


determined from the equations


(14.104)


The two series in (14.104) are examples of the Fourier series discussed in Chapter 15.


We return to this problem in Section 15.5 to determine the A’s and B’s for typical


initial functionsf(x)andg(x).


0 Exercises 16 –18


14.8 Exercises


Section 14.2


1.Show that the functionf(x, t) 1 = 1 a 1 sin 1 (bx) 1 cos 1 (vbt)(i)satisfies the one-dimensional wave


equation (14.1), (ii)has the formf(x, t) 1 = 1 F(x 1 + 1 vt) 1 + 1 G(x 1 − 1 vt).


2.The diffusion equation


provides a model of, for example, the transfer of heat from a hot region of a system to a


cold region by conduction whenf(x, 1 t)is a temperature field, or the transfer of matter


from a region of high concentration to one of low concentration whenfis the


concentration. Find the functionsV(x)for whichf(x, t) 1 = 1 V(x)e


ct

is a solution of the


equation.



  1. (i) It is shown in Example 14.2 that the functionf(x, t) 1 = 1 a exp[−b(x 1 − 1 vt)


2

]is a


solution of the wave equation (14.1). Sketch graphs off(x, t)as a function of xat times


t 1 = 1 0, t 1 = 122 v, t 1 = 142 v(use, for example,a 1 = 1 b 1 = 11 ), to demonstrate that the function


represents a wave travelling to the right (in the positive x-direction) at constant speed


v. (ii)Verify thatg(x, t) 1 = 1 a exp[−b(x 1 + 1 vt)


2

]is also a solution of the wave equation,


and hence that every superpositionF(x, t) 1 = 1 f(x, t) 1 + 1 g(x, t)is a solution. (iii) Sketch


appropriate graphs off(x, t) 1 + 1 g(x, t)to demonstrate how this function develops


in time.




=




f


t


D


f


x


2

2

t

n

nn

y


t


gx B


nx


l


=

=









==



0

1

() sin



ω


π


yx f x A


nx


l


n

n

() (),= = sin


=


0


1


π


yxt


nx


l


AtBt


n

nnnn

( ),=sin cos +sin








=


1


π


ωω

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