Elliptic: 0≤x,y≤3 boundary

With h = 1 the interior points are (1,1), (2,1), (1,2), (2,2). Name them u₁ = u(1,2), u₂ = u(2,2), u₃ = u(1,1), u₄ = u(2,1) (y increases upward). For Laplace's equation ∇²u = 0 on a square mesh, each interior value equals the mean of its four neighbours (standard five-point formula):

u(i,j) = ¼[u(i−1,j) + u(i+1,j) + u(i,j−1) + u(i,j+1)]

Boundary values: top (y = 3) = 90, bottom (y = 0) = 10, left (x = 0) = 70, right (x = 3) = 0.

Equations

u₁ = ¼(160 + u₂ + u₃)
u₂ = ¼(90 + u₁ + u₄)
u₃ = ¼(80 + u₁ + u₄)
u₄ = ¼(10 + u₂ + u₃)

Iteration (Gauss–Seidel / Liebmann)

Initial value 42 for every unknown (close to the mean of the boundary values); each new value is used at once in the next equation.

Converged to 4 decimals after 11 iterations.

Direct solution of the equations (check)

u₁ = 61.2500
u₂ = 43.7500
u₃ = 41.2500
u₄ = 23.7500

Answer: u₁ = 61.2500, u₂ = 43.7500, u₃ = 41.2500, u₄ = 23.7500