Laplace equation with u(x,∞) = sin(nπx/l)

Reading of the problem. The conditions are u(0, y) = u(l, y) = 0 and u(x, 0) = 0 with u(x, ∞) = sin(nπx/l). Taken literally this cannot hold: the solution that vanishes at y = 0 is proportional to sinh(nπy/l), which tends to ∞ rather than to a finite profile. The usual consistent reading in the strip 0 ≤ x ≤ l, y ≥ 0 is: u is bounded as y → ∞, with u = 0 on x = 0 and x = l, and the non-zero data sin(nπx/l) on the edge y = 0. I solve that problem, and also give the finite-top case y = b that the literal wording leads to.

Step 1: separate variables. u = X(x)Y(y): X″/X = −Y″/Y = −p².

X″ + p² X = 0  →  X = A cos px + B sin px
Y″ − p² Y = 0  →  Y = C e^{py} + D e^{−py}

Step 2: x-conditions. X(0) = 0 gives A = 0; X(l) = 0 gives p = mπ/l. So Xₘ = sin(mπx/l).

Step 3: y-condition at infinity. u must stay bounded as y → ∞, so C = 0:

u(x, y) = Σ_{m=1}^{∞} Bₘ sin(mπx/l) e^{−mπy/l}

Step 4: data on the edge y = 0. u(x, 0) = sin(nπx/l) = Σ Bₘ sin(mπx/l). By orthogonality, Bₙ = 1 and all other Bₘ = 0.

Answer (bounded strip):

u(x, y) = sin(nπx/l) e^{−nπy/l}

If the data sin(nπx/l) is on a finite top edge y = b with u(x, 0) = 0, the solution is

u(x, y) = sin(nπx/l) sinh(nπy/l) / sinh(nπb/l)

which has the same form as the earlier rectangle problem.