5—Fourier Series 107Apply the Theorem
As an example, carry out a full analysis of the case for whicha= 0andb=L, and for the boundary
conditionsu(0) = 0andu(L) = 0. The parameterλis positive, zero, or negative. Ifλ > 0 , then set
λ=k^2 and
u(x) =Asinhkx+Bcoshkx, then u(0) =B= 0
and so u(L) =AsinhkL= 0⇒A= 0
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, so u(L) =AsinkL= 0
Now there are many solutions becausesinnπ = 0 allows k = nπ/Lwithn any 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′′=λu
with these boundary conditions hasλn=−n^2 π^2 /L^2 and reproduces the result of the explicit integration
as in Eq. (5.6).
un(x) = sin
(nπx
L
)
n= 1, 2 , 3 ,... and
〈
un,um
〉
=
∫L
0dxsin
(nπx
L
)
sin(mπx
L
)
= 0 if n 6 =m (5.17)
There are other choices of boundary condition that will make the bilinear concomitant vanish.
(Verify these!) For example
u(0) = 0, u′(L) = 0 gives un(x) = sin
(
n+^1 / 2
)
πx/L n= 0, 1 , 2 , 3 ,...
and without further integration you have the orthogonality integral for non-negative integersnandm
〈
un,um
〉
=
∫L
0dxsin
(
(n+^1 / 2 )πx
L
)
sin(
(m+^1 / 2 )πx
L
)
= 0 if n 6 =m (5.18)
A very common choice of boundary conditions isu(a) =u(b), u′(a) =u′(b) (periodic boundary conditions) (5.19)
It is 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
The periodic behavior of the exponential implies thatkL= 2nπ. The condition that the derivatives
match at the boundaries makes no further constraint, so the basis functions are