8.2. SEPARATION OF VARIABLES:
Lecture 11
How do we find the bn ?
Observe that we have a new type of eigenvalue problem subject to
X(0) = 0 X(L) = 0. Just as in the case with matrices we obtain sequence
of eigenvalues which in this case is infinite:
λn =
nπ L
n = 1, 2, . . .
(8.11)
and corresponding eigenfunctions
nπx xn (x) = sin λn x = sin
L
2πx
3πx
πx
, sin
, sin
,... .
sin
L
L
L
(8.12)
Recall that for symmetric matrices the eigenvectors form a basis.
Aside: How do we expand a vector?
Express f in terms of the basis vectors v1 , v2 , v3
f = α1 v1 + α2 v2 + α3 v3
f · ⎡vk = α1 v1 · vk + α2 v2 · vk ⎤
+ α3⎡v3 · v⎤
k
⎡
⎤
v1 · v1 v1 · v2 v1 · v3
α1
f · v1
⎣ v1 · v2 v2 · v2 v2 · v3 ⎦ ⎣ α2 ⎦ = ⎣ f · v2 ⎦
v1 · v3 v2 · v3 v3 · v3
α3
f · v3
(8.13)
If vk ⊥ v , k = i.e. the vk are orthogonal
αk =
f · vk
vk · vk
(8.14)
But functions are just infinite dimensional vectors:
53
Separation of Variables
f
[f1 , f2 , . . . , fN ]
g [g1 , g2 , . . . , gN ]
f · g = f1 g1 + f2 g2 + · · · + fN gN
=
N
Δx =
L
N
(8.15)
f (xk )g(xk ).
k=1
Now
N
L
f (xk )g(xk )Δx k=1
f (x)g(x) dx = f, g.
(8.16)
0
Back to finding bn :
f (x) =
∞
bn sin
nπx L
f (x) sin
0
kπx
L
dx =
sin
0
=
Inn
bn
sin
nπx 0
L
sin
cos(n − k)
L
sin
kπx
L
dx.
0
kπx
L
dx
πx
πx
− cos(n + k)
dx
L
L
n = k
1 sin(n − k)πx/L sin(n + k)πx/L L
=
−
2
(n − k)π/L
(n + k)π/L
0
= 0
(8.18)
L
L
nπx 2nπx
1
dx
=
sin2
1 − cos
dx =
L
2
L
0
= L/2
L
kπx
2
=
f (x) sin
dx.
L
L
0
54
nπx L
0
Therefore bk
L
n=1
L
1
2
∞
1
{cos(A − B) − cos(A + B)}. Therefore
2
Recall sin(A) sin B =
Ink =
(8.17)
L
n=1
8.2. SEPARATION OF VARIABLES:
Example 8.1
2x
0 < x < 12
L=1
2(1 − x) 12 < x < 1
⎧ 1
⎫
⎪
⎪
1
⎨2
⎬
= 2
2x sin(nπx) dx + 2(1 − x) sin(nπx) dx
⎪
⎪
⎩
⎭
1
f (x) =
bn
0
2
sin(nπ/2)
n2 π 2
∞
8 (−1)k
= 8
Therefore u(x, t) =
π2
k=0
(2k +
sin
1)2
3 4 5
nπn = 1 2
1
0
−1
0 1
L
2 2
sin (2k + 1)πx e−(2k+1) π t . (8.19)
• Observe as t → ∞ u(x, t) → 0 (all the heat leaks out).
• u(x, 0) =
∞
8 (−1)k
sin (2k + 1)πx .
2
2
π
(2k + 1)
k=0
•
π2
=
8
∞
k=0
1
1
(2k + 1)2
by letting x =
1 terms of the Fourier Series
1
f(x)
f(x)
2 terms of the Fourier Series
0.5
0.5
0
−0.5
−1
−2
1
⇒ f (x) = 1.
2
0
−0.5
−1
0
x
1
2
−1
−2
−1
0
x
1
2
Example 8.2
f (x) = x
0<x<1 L=1
1
x sin(nπx) dx = −2
bn = 2
0
Therefore u(x, t) =
cos(nπ)
(−1)n+1
=2
nπ
nπ
∞
2 (−1)n+1
2
sin(nπx)e−(nπ) t .
π
n
(8.20)
n=1
55
Separation of Variables
• As t → ∞ u(x, t) → 0.
• u(x, 0) =
u
1
,0
2
∞
2 (−1)n+1
sin(nπx).
π
n
n=1
=
1
2
•
π
4
1−
=
=
2
π
=
2
π
1 1
+ − ....
3 5
∞
n=1
∞
k=0
(−1)n+1
n
sin(nπ/2)
(−1)k
(2k + 1)
k n sin nπ
2
0 1
1
2
0
1 3
−1
4
0
2 5
1
1
1 terms of the Fourier Series
1
f(x)
f(x)
0
0
−0.5
−0.5
56
2 terms of the Fourier Series
0.5
0.5
−1
−2
(8.21)
−1
0
x
1
2
−1
−2
−1
0
x
1
2