Numerical Methods for Partial Differential Equations

3. Triangular domain
Given;
∆ u=f
Let Ω=Ω∩ R+¿2 ¿ amd Ω :=Ω ∪0< x , y <1
Taking i=0, we discretize as follows;
∂n
2 u
∂ x2 ( 0 ,1−x ) = 1
h2 ( u−1 , j−2 u0 , j +ui , j ) +O ( h2 ) ... ... ... ... ... (i)
Applying the boundary conditions;
−γ0 , j= ∂n u
∂ x ( 0 , 1−x )= 1
2 h ( u1 , j −u−1 , j ) +O ( h2 ) ... ... ... ... ... ... .(ii)
Combining (i) and (ii);
∂n
2 u
∂ x2 ( t ) = 1
h2 ( 2ui , j−2u0 , j ) + 2 γ o , j
h +O(h2)
Whereγ =0;
∂n
2 u
∂ x2 ( t ) = 1
h2 ( 2ui , j−2u0 , j ) +O(h2)
From the solution obtained we observe that the consistency level is 2.
From the solution we obtain that an alternating sequence is formed meaning that north point is
equal to the east point as well as west point being equal to the south point as described in the
figure below (Khasminskii, 2011).

From the details in the diagram we come up with a matrix as below;
(
−4 1 0
1 −4 1
0 1 −4
1 0 0
0 1 0
0 0 1
0 0 0
0 0 0
0 0 0
1 0 0
0 1 0
0 0 1
−4 1 0
1 −4 1
0 1 −4
1 0 0
0 1 0
0 0 1
0 0 0
0 0 0
0 0 0
1 0 0
0 1 0
0 0 1
−4 1 0
1 −4 1
0 1 −4
)( u1,1
u2,1
u3,1
u1,2
u2,2
u3,2
u 1,3
u 2,3
u 3,3
) =
( h2 f 1,1−u1,0−u 0,1
h2 f 2,1−u 2,0
h2 f 3,1−u 3,0−u 4,1
h2 f 1,2−u 0,2
h2 f 2,2
h2 f 3,2−u 4,2
h2 f 1,3−u 0,3−u 1,4
h 62 f 2,3−u2,4
h2 f 3,3−ui 4,3−u 3,4
)4. Rothe’s method
uk +1=uk + τ ∆ uk+1 ,(x , y)Ω
Rearranging we have;
uk +1−uk =τ ∆ uk +1
Using the 5-point finite difference formula to discrete the PDE in space we obtain;
dui , j
dt ( x , y ) = D
h2 ( ui+1 , j +ui−1 , j +ui , j+1 +ui , j−1−4 ui , j ) +O(h2 )
Therefore,
uk+1 −uk
h =ui+1 , j
k+1 + ui−1 , j
k+1 −4 ui , j
k+1 +ui , j+1
k +1 +ui , j−1
k+1
h2 +O (h2)
Multiplying both sides by h;
uk +1−uk =ui +1 , j
k+1 +ui−1, j
k+1 −4 ui , j
k+1 +ui , j+1
k +1 +ui , j−1
k +1
h +O(h2)
The convergence order is 2.
To make the linear stable; τ ≤ h2
Comparing the 5 point finite difference formula to the implicit method, the implicit method is
inefficient.