Laplace equation solutions; u(x,0)=sin x

Part 1: All possible solutions of u_xx + u_yy = 0

Assume u = X(x)Y(y). Then X″Y + XY″ = 0, so X″/X = −Y″/Y = k (a constant). Three cases arise.

Case 1: k = 0. X″ = 0 and Y″ = 0:

u = (c₁x + c₂)(c₃y + c₄)

Case 2: k = p² > 0. X″ − p²X = 0 and Y″ + p²Y = 0:

u = (c₅ e^{px} + c₆ e^{−px})(c₇ cos py + c₈ sin py)

Case 3: k = −p² < 0. X″ + p²X = 0 and Y″ − p²Y = 0:

u = (c₉ cos px + c₁₀ sin px)(c₁₁ e^{py} + c₁₂ e^{−py})

Which solution is used depends on the boundary conditions.

Part 2: The given problem

Conditions (taking the intended form): u(0, y) = 0, u(π, y) = 0, u is bounded as y → ∞, and u(x, 0) = sin x.

Choice of solution. The x-boundary conditions are homogeneous at x = 0 and x = π, so X must be oscillatory. This is Case 3:

u = (A cos px + B sin px)(C e^{py} + D e^{−py})

• u(0, y) = 0 gives A = 0.
• u(π, y) = 0 gives B sin pπ = 0, so p = n (integer).
• Bounded as y → ∞ gives C = 0 (e^{ny} would blow up).

u(x, y) = Σ_{n=1}^{∞} bₙ e^{−ny} sin nx

Condition u(x, 0) = sin x:

Σ bₙ sin nx = sin x

Comparing coefficients: b₁ = 1, and bₙ = 0 for n ≥ 2.

Answer:

u(x, y) = e^{−y} sin x

Check: u_xx = −e^{−y} sin x and u_yy = e^{−y} sin x, which sum to 0 ✓. u(0, y) = u(π, y) = 0 ✓. u(x, 0) = sin x ✓. u → 0 as y → ∞ ✓.