34.4. LAPLACE EQUATION 685
First find the solution which has siny on the right and zero on the other edges. This wasdone in the above. It is
u1 (x,y) =∞
∑n=1
bn sinh(nπx)sin(nπy)
where bn is as given above. Thus
bn =2
sinh(nπa)
∫ 1
0sin(t)sin(nπt)dt =
2sinhnπa
nπ sin1(−1)n
n2π2−1
Hence this partial solution is
u1 (x,y) =∞
∑n=1
2sinhnπa
nπ sin1(−1)n
n2π2−1sinh(nπx)sin(nπy)
Next find the solution to the equation which has 1− (x−1)2 on the top and zero on theother sides. This is just like what was done earlier except that you would switch a and b.You find the eigenfunctions for the two opposite zero boundary conditions. These are
sin(nπ
2x), n = 1,2, · · ·
with eigenvalues n2π2
4 . Next you look for solutions to the equation which involve
a(y)sin(nπ
2x)
Thus
a′′ (y)sin(nπ
2x)+a(y)
(−π2
4n2 sin
π
2nx)= 0
Hence
a′′ (y)− π2
4n2a(y) = 0
and soa(y) = an cosh
(nπ
2y)+bn sinh
(nπ
2y)
Then the general solution is
u2 (x,y) =∞
∑n=1
(an cosh
(nπ
2y)+bn sinh
(nπ
2y))
sin(nπ
2x)
When y = 0, you are supposed to get 0 for the boundary condition. Hence an = 0. Wheny = b you need
1− (1− x)2 =∞
∑n=1
(bn sinh
(nπ
2
))sin(nπ
2x)
Therefore, you need
bn sinh(nπ
2
)=∫ 2
0
(1− (1− s)2
)sin(nπ
2s)
ds = 82−2(−1)n
n3π3