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on a cube:
∆ 3 u = u xx + u yy + u zz = 0 in D,
D = {0 < x < π, 0 < y < π, 0 < z < π},

u(π, y, z) = g(y, z),
u(0, y, z) = u(x, 0, z) = u(x, π, z) = u(x, y, 0) = u(x, y, π) = 0.

To solve this problem we separate variables and use the five homogeneous boundary
conditions:
X 00 Y 00 Z 00
u = X(x)Y (y)X(z), + + = 0,
X Y Z
X(0) = Y (0) = Z(0) = Y (π) = Z(π) = 0.
.Each quotient X”/X, Y ”/Y , and Z”/Z must be a constant. In the familiar way, we
find
Y (y) = sin my (m = 1, 2, . . .)
and
Z(z) = sin nz(n = 1, 2, . . . ),

so that
2
2
X” = (m + n )X, X(0) = 0.
Therefore,
√
2
2
X(x) = A sinh( m + n x).
Summing up, our complete solution is
∞ ∞
X X √
2
2
u(x, y, z) = A mn sinh( m + n x) sin my sin nz. (10.22)
n=1 m=1
Finally, we plug in our inhomogeneous condition at x = π :
X X √
2
2
g(y, z) = A mn sinh( m + n π) sin my sin nz.
This is a double Fourier sine series in the variables y and z! Its theory is similar
to that of the single series. In fact, the eigenfunctions {sin my sin nz} are mutually
orthogonal on the square {0 < y < π, 0 < z < π}. Their normalizing constants are
Z π Z π π 2
2
(sin my sin nz) dydz = .
0 0 4
Therefore,
Z π Z π
4
A mn = √ g(y, z) sin my sin nzdydz. (10.23)
2
2
2
π sinh( m + n π) 0 0
Hence the solutions can be expressed as the doubly infinite series (10.22) with the
coefficients A mn . The complete solution to this example is (10.22) and (10.23). With
such a series, as with a double integral, one has to be careful about the order of
summation, although in most cases any order will give the correct answer.

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