Mathematical Methods for Physics and Engineering : A Comprehensive Guide

20.3 GENERAL AND PARTICULAR SOLUTIONS

the terms ready for substitution into (20.20), we obtain

∂u

∂x

=a

df(p)

dp

,

∂u

∂y

=b

df(p)

dp

,

∂^2 u

∂x^2

=a^2

d^2 f(p)

dp^2

,

∂^2 u

∂x∂y

=ab

d^2 f(p)

dp^2

,

∂^2 u

∂y^2

=b^2

d^2 f(p)

dp^2

,

which on substitution give

(

Aa^2 +Bab+Cb^2

)d^2 f(p)

dp^2

=0. (20.21)

This is the form we have been seeking, since now a solution independent of

the form offcan be obtained if we require thataandbsatisfy

Aa^2 +Bab+Cb^2 =0.

From this quadratic, two values for the ratio of the two constantsaandbare

obtained,
b/a=[−B±(B^2 − 4 AC)^1 /^2 ]/ 2 C.

If we denote these two ratios byλ 1 andλ 2 thenanyfunctions of the two variables

p 1 =x+λ 1 y, p 2 =x+λ 2 y

will be solutions of the original equation (20.20). The omission of the constant

factorafromp 1 andp 2 is of no consequence since this can always be absorbed

into the particular form of any chosen function; only therelativeweighting ofx

andyinpis important.

Sincep 1 andp 2 are in general different, we can thus write the general solution

of (20.20) as

u(x, y)=f(x+λ 1 y)+g(x+λ 2 y), (20.22)

wherefandgare arbitrary functions.

Finally, we note that the alternative solutiond^2 f(p)/dp^2 = 0 to (20.21) leads

only to the trivial solutionu(x, y)=kx+ly+m, for which all second derivatives

are individually zero.

Find the general solution of the one-dimensional wave equation

∂^2 u

∂x^2

−

1

c^2

∂^2 u

∂t^2

=0.

This equation is (20.20) withA=1,B=0andC=− 1 /c^2 , and so the values ofλ 1 andλ 2
are the solutions of

1 −

λ^2

c^2

=0,

namelyλ 1 =−candλ 2 =c. This means that arbitrary functions of the quantities

p 1 =x−ct, p 2 =x+ct