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One-Dimensional Heat Equation
.

√
Y (0) sinh sx 1 ∫ x √
′
= √ − √ f(u) sinh s(x − u)du.
s s 0
Now, Y (1) = 0, implying

∫ 1
1 √
′
Y (0) = √ f(u) sinh s(1 − u)du.
sinh s 0
Thus,

1 sinh sx sin s(1 − u) x sinh s(x − u)
∫ √ √ ∫ √
U(x, s) = f(u) √ √ du − f(u) √ du.
0 s sinh s 0 s

∫ 1 ∫ x ∫ 1
We can write = + and use the fact that
0 0 x
sinh(z ± w) = sinh z cosh w ± cosh z sinh w.

Then

x sinh sx sinh s(1 − u) sinh s(x − u)
∫ [ √ √ √ ]
U(x, s) = f(u) √ √ − √ du+
0 s sinh s s
1 sinh sx sinh s(1−u) x sinh s(1−x) sinh su
∫ √ √ ∫ √ √
+ f(u) √ √ du= f(u) √ √ du+
x s sinh s 0 s sinh s
∫ √ √
x sinh sx sinh s(1 − u)
+ f(u) √ √ du.
1 s sinh s

To determine the inverse we use the complex inversion formula. When it
is applied to the first integral we have

∫ x 0 +i∞ { √ √ }
1 sinh s(1 − x) sinh su ∑
e ts f(u) √ √ ds = Res .
2πi s sinh s
x 0 −i∞
2 2
There are simple poles in this case at s = 0 and s = −n π , n = 1, 2,
0
n
. . .
√ √
∫
x sinh s(1 − x) sinh su
Res(0) = lim f(u) √ √ du = 0.
s→0 s sinh s
0
√
√
∫ x sinh s(1 − x) sinh su
2 2
2 2
Res(−n π ) = lim (s + n π )e ts f(u) √ √ du =
2 2
s→−n π 0 s sinh s
√
√
2 2
s + π n ∫ x sinh s(1 − x) sinh su
lim √ · lim e ts f(u) √ du =
s→−n π sinh s s→−n π 0 s
2 2
2 2
∫
x sinh[(πni)(1 − x)] sinh(πni)u
2 2
= 2e −n π t f(u) du =
cosh(πni)
0
∫
x sin[πn(1 − x)] sin πnu
2 2
= 2e −n π t f(u) du,
. . . . 0 − cos πn
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