Poisson: ∇²u = -10(x²+y²+10)

Poisson's equation ∇²u = f with f(x,y) = −10(x² + y² + 10) is solved with the five-point formula (h = 1):

u(i−1,j) + u(i+1,j) + u(i,j−1) + u(i,j+1) − 4u(i,j) = h²f(i,j)

so u(i,j) = ¼[ sum of four neighbours − f(i,j) ].

Set up

The square 0 ≤ x, y ≤ 3 with h = 1 has four interior points: u₁ = u(1,2), u₂ = u(2,2), u₃ = u(1,1), u₄ = u(2,1). All boundary values are 0.

f(1,2) = −10(1+4+10) = −150      f(2,2) = −10(4+4+10) = −180
f(1,1) = −10(1+1+10) = −120      f(2,1) = −10(4+1+10) = −150

u₁ = ¼(0 + 0 + u₂ + u₃ + 150) = ¼(150 + u₂ + u₃)
u₂ = ¼(180 + u₁ + u₄)
u₃ = ¼(120 + u₁ + u₄)
u₄ = ¼(150 + u₂ + u₃)

(u₁ and u₄ have the same equation, so u₁ = u₄.)

Iteration (Gauss–Seidel)

Start with u = 30 for all points.

Converged to 4 decimals after 12 iterations.

Direct solution (check)

Put u₁ = u₄ = a, u₂ = b, u₃ = c: a = ¼(150 + b + c), b = ¼(180 + 2a), c = ¼(120 + 2a). Then 4a = 150 + (300 + 4a)/4 → 3a = 225 → a = 75.0000, b = 82.5000, c = 67.5000.

Answer: u₁ = 75.0000, u₂ = 82.5000, u₃ = 67.5000, u₄ = 75.0000