5—Fourier Series 125
No solutions there, so tryλ= 0
u(x) =A+Bx, then u(0) =A= 0 and so u(L) =BL= 0⇒B= 0
No solutions here either. Tryλ < 0 , settingλ=−k^2.
u(x) =Asinkx+Bcoskx, then u(0) = 0 =B and so u(L) =AsinkL= 0
Now there are many solutions becausesinnπ= 0allowsk=nπ/Lwithnany integer. But,sin(−x) =−sin(x)
so negative integers just reproduce the same functions as do the positive integers; they are redundant and you
can eliminate them. The complete set of solutions to the equationu′′=λuwith these boundary conditions have
λn=−n^2 π^2 /L^2 with
un(x) = sin
(nπx
L
)
n= 1, 2 , 3 ,... and
〈
un,um
〉
=
∫L
0
dxsin
(nπx
L
)
sin
(mπx
L
)
= 0 if n 6 =m (14)
There are other choices of boundary condition that will make the bilinear concomitant vanish. For example
u(0) = 0, u′(L) = 0 gives un(x) = sin
(
n+^1 / 2
)
πx/L n= 0, 1 , 2 , 3 ,...
and you have the orthogonality integral for non-negative integersnandm
∫L
0
dxsin
(
(n+^1 / 2 )πx
L
)
sin
(
(m+^1 / 2 )πx
L
)
= 0 if n 6 =m (15)
A very common choice of boundary conditions is
u(a) =u(b), u′(a) =u′(b) (periodic boundary conditions) (16)
It’s often more convenient to use complex exponentials in this case (though of course not necessary). On
0 < x < L
u(x) =eikx, where k^2 =−λ and u(0) = 1 =u(L) =eikL