Laplace equation with u(x,a) = sin(nπx/L)

Problem. u_xx + u_yy = 0 in the rectangle 0 ≤ x ≤ L, 0 ≤ y ≤ a, with

• u(0, y) = 0, u(L, y) = 0
• u(x, 0) = 0
• u(x, a) = sin(nπx/L)

Step 1: separate variables. u = X(x)Y(y): X″/X = −Y″/Y = −p² (the negative sign is chosen so that X is oscillatory, since X must vanish at both x = 0 and x = L).

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

Step 2: conditions at x = 0 and x = L. A = 0 and B sin pL = 0, so p = mπ/L.

Step 3: condition at y = 0. Y(0) = 0 means C + D = 0, so Y ∝ sinh(py).

General solution:

u(x, y) = Σ_{m=1}^{∞} Bₘ sin(mπx/L) sinh(mπy/L)

Step 4: condition at y = a.

Σ Bₘ sinh(mπa/L) sin(mπx/L) = sin(nπx/L)

By orthogonality of sines, Bₘ sinh(mπa/L) = 1 for m = n and 0 for m ≠ n. So only the n-th term survives:

Bₙ = 1 / sinh(nπa/L)

Answer:

u(x, y) = sin(nπx/L) · sinh(nπy/L) / sinh(nπa/L)

Check: u(x, a) = sin(nπx/L) ✓; u = 0 on the other three sides ✓.