APM3711 ASSIGNMENT 4 Solutions 2026
UNISA
Numerical Methods II
,QUESTION 1
We are given the boundary value problem
∇2 𝑢 = −4, 0 < 𝑥 < 2, 0 < 𝑦 < 2
with
𝑢=0
on all four boundaries.
The mesh spacing is
ℎ = 1.
These are the conditions given in the assignment.
1(a) Sketch the computational domain
1(b) Five-point finite difference equation
The five-point approximation to the Laplace operator is
𝑢𝑖+1,𝑗 + 𝑢𝑖−1,𝑗 + 𝑢𝑖,𝑗+1 + 𝑢𝑖,𝑗−1 − 4𝑢𝑖,𝑗
∇2 𝑢 ≈ .
ℎ2
Since
∇2 𝑢 = −4
we have
, 𝑢𝑖+1,𝑗 + 𝑢𝑖−1,𝑗 + 𝑢𝑖,𝑗+1 + 𝑢𝑖,𝑗−1 − 4𝑢𝑖,𝑗
= −4.
ℎ2
Given
ℎ = 1,
we obtain
𝑢𝑖+1,𝑗 + 𝑢𝑖−1,𝑗 + 𝑢𝑖,𝑗+1 + 𝑢𝑖,𝑗−1 − 4𝑢𝑖,𝑗 = −4.
At the only interior node (1, 1),
𝑢2,1 + 𝑢0,1 + 𝑢1,2 + 𝑢1,0 − 4𝑢1,1 = −4.
All four neighbouring points lie on the boundary, so
𝑢2,1 = 0,
𝑢0,1 = 0,
𝑢1,2 = 0,
and
𝑢1,0 = 0.
Therefore,
0 + 0 + 0 + 0 − 4𝑢1,1 = −4.
Hence
−4𝑢1,1 = −4.
Dividing both sides by −4,
𝑢1,1 = 1.
Thus the finite difference equation is
−4𝑢1,1 = −4 .
UNISA
Numerical Methods II
,QUESTION 1
We are given the boundary value problem
∇2 𝑢 = −4, 0 < 𝑥 < 2, 0 < 𝑦 < 2
with
𝑢=0
on all four boundaries.
The mesh spacing is
ℎ = 1.
These are the conditions given in the assignment.
1(a) Sketch the computational domain
1(b) Five-point finite difference equation
The five-point approximation to the Laplace operator is
𝑢𝑖+1,𝑗 + 𝑢𝑖−1,𝑗 + 𝑢𝑖,𝑗+1 + 𝑢𝑖,𝑗−1 − 4𝑢𝑖,𝑗
∇2 𝑢 ≈ .
ℎ2
Since
∇2 𝑢 = −4
we have
, 𝑢𝑖+1,𝑗 + 𝑢𝑖−1,𝑗 + 𝑢𝑖,𝑗+1 + 𝑢𝑖,𝑗−1 − 4𝑢𝑖,𝑗
= −4.
ℎ2
Given
ℎ = 1,
we obtain
𝑢𝑖+1,𝑗 + 𝑢𝑖−1,𝑗 + 𝑢𝑖,𝑗+1 + 𝑢𝑖,𝑗−1 − 4𝑢𝑖,𝑗 = −4.
At the only interior node (1, 1),
𝑢2,1 + 𝑢0,1 + 𝑢1,2 + 𝑢1,0 − 4𝑢1,1 = −4.
All four neighbouring points lie on the boundary, so
𝑢2,1 = 0,
𝑢0,1 = 0,
𝑢1,2 = 0,
and
𝑢1,0 = 0.
Therefore,
0 + 0 + 0 + 0 − 4𝑢1,1 = −4.
Hence
−4𝑢1,1 = −4.
Dividing both sides by −4,
𝑢1,1 = 1.
Thus the finite difference equation is
−4𝑢1,1 = −4 .
