In this section, we will introduce the process for using Finite Different Method of the discrete Laplacian in different domain. There are rectangle and disc in our interesting domain. We also edit some different boundary conditions in the disc. In the end, we also arrange some domains about holey disc and rectangle.
4.1. The Rectangle with Dirichlet Boundary Conditions of Laplacian
∆u = f (x, y)
u(x, c) = m(x) , y = c u(b, y) = n(y) , x = b u(x, d) = p(x) , y = d u(a, y) = q(y) , x = a.
(14)
Let the domain of rectangle be a ≤ x ≤ b and c ≤ y ≤ d where a, b, c, d ∈ constant. Suppose ν, ω are the partition numbers of the two sides. Then we can get that the step size h = (b − a)/ν, and k = (c − d)/ω. Therefore we define grid points as following
xi = a + i × h, for i = 0, 1, 2, · · · , ν, yj = c + j × k, for j = 0, 1, 2, · · · , ω.
Now, we have ∆u = uxx+ uyy. The second differential item which is fixed respec-tively x and y can be obtained from Taylor series. They are
uxx = u(xi+1, yj) − 2u(xi, yj) + u(xi−1, yj) h2
uyy = u(xi, yj+1) − 2u(xi, yj) + u(xi, yj−1) k2
We rewrite u(xi, yj) = ui,j, f (xi, yj) = fi,j. We put above equations into (14), and use the finite difference method, we can get
∆u = uxx+ uyy
= u(xi+1, yj) − 2u(xi, yj) + u(xi−1, yj)
h2 + u(xi, yj+1) − 2u(xi, yj) + u(xi, yj−1) k2
= fi,j.
Let ρ = hk22, k = i + (j − 1)(ν − 1) with fk = fi,j , for i = 1, · · · , ν − 1, and j = 1, · · · , ω − 1. We suppose ν = 9 and ω = 4, then we can arrange some equations as following. We can see the discretization index from Figure 4.1.
0 2 4 6 8 10
−0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5
m(x) p(x)
n(x) q(x)
1 2 3 4 5 6 7 8
9 10 11 12 13 14 15 16
17 18 19 20 21 22 23 24
The discretization of a rectangle for ν = 9 and ω = 4.
x value
y value
Figure 4.1. It is the discretization of this laplacian for a rectangle. The partitions of two sides are ν = 9 and ω = 4.
(i, j) = (1, 1) : 1
h2(u2− 2u1+ u0,1) + 1
k2(u9− 2u1+ u1,0) = f1 (i, j) = (2, 1) : 1
h2(u3− 2u2+ u1) + 1
k2(u10− 2u2+ u2,0) = f2 (i, j) = (3, 1) : 1
h2(u4− 2u3+ u2) + 1
k2(u11− 2u3+ u3,0) = f3 (i, j) = (4, 1) : 1
h2(u5− 2u4+ u3) + 1
k2(u12− 2u4+ u4,0) = f4 (i, j) = (5, 1) : 1
h2(u6− 2u5+ u4) + 1
k2(u13− 2u5+ u5,0) = f5 (i, j) = (6, 1) : 1
h2(u7− 2u6+ u5) + 1
k2(u14− 2u6+ u6,0) = f6 (i, j) = (7, 1) : 1
h2(u8− 2u7+ u6) + 1
k2(u15− 2u7+ u7,0) = f7 (i, j) = (8, 1) : 1
h2(u9,1− 2u8+ u7) + 1
k2(u16− 2u8+ u8,0) = f8
================================================
(i, j) = (1, 2) : 1
h2(u10− 2u9+ u0,2) + 1
k2(u17− 2u9+ u1) = f9 (i, j) = (2, 2) : 1
h2(u11− 2u10+ u9) + 1
k2(u18− 2u10+ u2) = f10 (i, j) = (3, 2) : 1
h2(u12− 2u11+ u10) + 1
k2(u19− 2u11+ u3) = f11 (i, j) = (4, 2) : 1
h2(u13− 2u12+ u11) + 1
k2(u20− 2u12+ u4) = f12 (i, j) = (5, 2) : 1
h2(u14− 2u13+ u12) + 1
k2(u21− 2u13+ u5) = f13 (i, j) = (6, 2) : 1
h2(u15− 2u14+ u13) + 1
k2(u22− 2u14+ u6) = f14 (i, j) = (7, 2) : 1
h2(u16− 2u15+ u14) + 1
k2(u23− 2u15+ u7) = f15 (i, j) = (8, 2) : 1
h2(u9,2− 2u16+ u15) + 1
k2(u24− 2u16+ u8) = f16
================================================ We will solve the linear system question for Au = b.
Furthermore, the matrix
The right hand side b is
b =
and b3 =
The right hand side b is
b =
4.2. The Disc with Dirichlet Boundary Conditions of Laplacian ( ∆u = f (x, y)
u(x, y) = c(x, y) , (x, y) ∈ Ω (15) Let ∆r = 2ν+12 be the radial mesh width and ∆θ = 2πω be the azimuthal mesh width, where ν and ω is positive integer. Then we define grid points as following
ri = (i − 1
2)∆r, for i = 1, 2, · · · , ν, θj = (j − 1)∆θ, for j = 1, 2, · · · , ω.
Now, we have ∆u = uxx + uyy = (urr + 1rur) + r12uθθ, f (x, y) = f (r, θ), and c(x, y) = c(r, θ). Then we use the formula of ux and uxx to compute ur, urr, and uθθ.
urr = u(ri+1, θj) − 2u(ri, θj) + u(ri−1, θj)
∆r2 1
riur = u(ri+1, θj) − u(ri−1, θj) ri(2∆r)
1
ri2uθθ = u(ri, θj+1) − 2u(ri, θj) + u(ri, θj−1)
r2i∆θ2 .
We rewrite u(ri, θj) to ui,j and f (ri, θj) = fi,j. We put above equations into (15), and use the finite difference method, we can get
1
∆r2(ui+1,j−2ui,j+ui−1,j)+ 1
2ri∆r(ui+1,j−ui−1,j)+ 1
r2i∆θ2(ui,j+1−2ui,j+ui,j−1) = fi,j. To multiply ∆r2 to both right and left hand side, and to put ri = (i − 12)∆r into them. We can get
(ui+1,j−2ui,j+ui−1,j)+ 1
2(i − 12)(ui+1,j−ui−1,j)+ 1
(i −12)2∆θ2(ui,j+1−2ui,j+ui,j−1) = ∆r2fi,j.
Let λi = 2(i−11
2) = 2i−11 , βi = (i−11
2)2∆θ2, and k = ω(i − 1) + j with fk = fi,j, for i = 1, · · · , ν, and j = 1, · · · , ω. We suppose ν = 3, and ω = 8 then we can arrange some equations as following. We can consult Figure 2.1 in the Section 2.
(i, j) = (1, 1) : (1 + λ1)u9+ (−2 − 2β1)u1+ (1 − λ1)u5+ β1u2+ β1u8 = ∆r2f1 (i, j) = (1, 2) : (1 + λ1)u10+ (−2 − 2β1)u2+ (1 − λ1)u6+ β1u3+ β1u1 = ∆r2f2 (i, j) = (1, 3) : (1 + λ1)u11+ (−2 − 2β1)u3+ (1 − λ1)u7+ β1u4+ β1u2 = ∆r2f3 (i, j) = (1, 4) : (1 + λ1)u12+ (−2 − 2β1)u4+ (1 − λ1)u8+ β1u5+ β1u3 = ∆r2f4 (i, j) = (1, 5) : (1 + λ1)u13+ (−2 − 2β1)u5+ (1 − λ1)u1+ β1u6+ β1u4 = ∆r2f5 (i, j) = (1, 6) : (1 + λ1)u14+ (−2 − 2β1)u6+ (1 − λ1)u2+ β1u7+ β1u5 = ∆r2f6 (i, j) = (1, 7) : (1 + λ1)u15+ (−2 − 2β1)u7+ (1 − λ1)u3+ β1u8+ β1u6 = ∆r2f7 (i, j) = (1, 8) : (1 + λ1)u16+ (−2 − 2β1)u8+ (1 − λ1)u4+ β1u1+ β1u7 = ∆r2f8
=====================================================
(i, j) = (2, 1) : (1 + λ2)u17+ (−2 − 2β2)u9+ (1 − λ2)u1 + β2u10+ β2u16 = ∆r2f9 (i, j) = (2, 2) : (1 + λ2)u18+ (−2 − 2β2)u10+ (1 − λ2)u2+ β2u11+ β2u9 = ∆r2f10
(i, j) = (2, 3) : (1 + λ2)u19+ (−2 − 2β2)u11+ (1 − λ2)u3 + β2u12+ β2u10 = ∆r2f11 (i, j) = (2, 4) : (1 + λ2)u20+ (−2 − 2β2)u12+ (1 − λ2)u4 + β2u13+ β2u11 = ∆r2f12
(i, j) = (2, 5) : (1 + λ2)u21+ (−2 − 2β2)u13+ (1 − λ2)u5 + β2u14+ β2u12 = ∆r2f13 (i, j) = (2, 6) : (1 + λ2)u22+ (−2 − 2β2)u14+ (1 − λ2)u6 + β2u15+ β2u13 = ∆r2f14
(i, j) = (2, 7) : (1 + λ2)u23+ (−2 − 2β2)u15+ (1 − λ2)u7 + β2u16+ β2u14 = ∆r2f15 (i, j) = (2, 8) : (1 + λ2)u24+ (−2 − 2β2)u16+ (1 − λ2)u8+ β2u9+ β2u15 = ∆r2f16
=====================================================
(i, j) = (3, 1) : (1 + λ3)u4,1+ (−2 − 2β3)u17+ (1 − λ3)u9+ β3u18+ β3u24 = ∆r2f17 (i, j) = (3, 2) : (1 + λ3)u4,2+ (−2 − 2β3)u18+ (1 − λ3)u10+ β3u19+ β3u17 = ∆r2f18 (i, j) = (3, 3) : (1 + λ3)u4,3+ (−2 − 2β3)u19+ (1 − λ3)u11+ β3u20+ β3u18 = ∆r2f19 (i, j) = (3, 4) : (1 + λ3)u4,4+ (−2 − 2β3)u20+ (1 − λ3)u12+ β3u21+ β3u19 = ∆r2f20 (i, j) = (3, 5) : (1 + λ3)u4,5+ (−2 − 2β3)u21+ (1 − λ3)u13+ β3u22+ β3u20 = ∆r2f21 (i, j) = (3, 6) : (1 + λ3)u4,6+ (−2 − 2β3)u22+ (1 − λ3)u14+ β3u23+ β3u21 = ∆r2f22 (i, j) = (3, 7) : (1 + λ3)u4,7+ (−2 − 2β3)u23+ (1 − λ3)u15+ β3u24+ β3u22 = ∆r2f23 (i, j) = (3, 8) : (1 + λ3)u4,8+ (−2 − 2β3)u24+ (1 − λ3)u16+ β3u17+ β3u23 = ∆r2f24
After computing, we can find 1 − λ1 = 1 − 1 = 0. Put boundary conditions u(r, θ) = c(r, θ), we have u4,1 = c4,1. Use the same step for u18, · · · , u24. We will solve the linear system question for Au = b.
Furthermore, the matrix
A =
−2I − β1Ψ (1 + λ1)I
(1 − λ2)I −2I − β2Ψ (1 + λ2)I (1 − λ3)I −2I − β3Ψ
∈ R24×24,
where
Ψ =
2 −1 −1
−1 2 . ..
. .. ... −1
−1 −1 2
∈ R8×8.
The right hand side b is
The right hand side b is
b =
4.3. The Disc with Neumann Boundary Conditions of Laplacian ( ∆u = f (x, y)
∂u
∂n(x, y) = c(x, y) , (x, y) ∈ Ω (16) The process is the same with Section 4.2. After computing, we can find 1 − λ1 = 1 − 1 = 0. Put boundary conditions ∂u∂n(r, θ) = c(r, θ), we use them by Taylor expansions. For example u4,1:
∂u4,1
∂n = u4,1− u17
∆r = c4,1
imply
u4,1 = c4,1∆r + u17, then we can get
(1 + λ3)u4,1+ (−2 − 2β3)u17 = (1 + λ3)(c4,1∆r + u17) + (−2 − 2β3)u17
The right hand side b is
b =
The right hand side b is
4.4. The Disc with Robin Boundary Conditions of Laplacian ( ∆u = f (x, y)
∂u
∂n(x, y) + αu(x, y) = c(x, y) , (x, y) ∈ Ω (17) The process is the same with Section 4.2. After computing, we can find 1 − λ1 = 1 − 1 = 0. Put boundary conditions ∂u∂n(r, θ) = c(r, θ), we use them by Taylor expansions. For example u4,1:
∂u4,1 then we can get
(1 + λ3)u4,1+ (−2 − 2β3)u17 = (1 + λ3)( ∆rc4,1
The right hand side b is
The right hand side b is
b =
4.5. The Ring with Dirichlet Boundary Conditions of Laplacian
∆u = f (x, y)
u(x, y) = c(x, y) , (x, y) ∈ inside circle.
u(x, y) = d(x, y) , (x, y) ∈ outside circle.
(18)
Let ∆r = 2ν+12 be the radial mesh width and ∆θ = 2πω be the azimuthal mesh width, where ν and ω is positive integer. Then we define grid points as following
ri = (i − 1
2)∆r, for i = 1, 2, · · · , ν, θj = (j − 1)∆θ, for j = 1, 2, · · · , ω.
Now, we have ∆u = uxx+ uyy = (urr+ 1rur) + r12uθθ, f (x, y) = f (r, θ), c(x, y) = ci(r, θ), and d(x, y) = co(r, θ). Then we use the formula of ux and uxx to compute ur, urr, and uθθ.
urr = u(ri+1, θj) − 2u(ri, θj) + u(ri−1, θj)
∆r2 1
riur = u(ri+1, θj) − u(ri−1, θj) ri(2∆r)
1
ri2uθθ = u(ri, θj+1) − 2u(ri, θj) + u(ri, θj−1)
r2i∆θ2 .
We rewrite u(ri, θj) to ui,j and f (ri, θj) = fi,j. We put above equations into (18), and use the finite difference method, we can get
1
∆r2(ui+1,j−2ui,j+ui−1,j)+ 1
2ri∆r(ui+1,j−ui−1,j)+ 1
r2i∆θ2(ui,j+1−2ui,j+ui,j−1) = fi,j. Put ri = (i − 12)∆r into them. We can get
ui+1,j − 2ui,j+ ui−1,j
∆r2 + ui+1,j− ui−1,j
2(i −12)∆r2 +ui,j+1− 2ui,j+ ui,j−1
(i −12)2∆r2∆θ2 = fi,j. Let m be the number of inside circle (m < ν), λi = 2(i−11
2) = 2i−11 , βi = (i−11 2)2∆θ2, and k = ω(i − 1 − m) + j with fk = fi,j, for i = m + 1, · · · , ν, and j = 1, · · · , ω. We suppose m = 1, ν = 4, and ω = 8, then we can arrange some equations as following.
We can see the discretization index from Figure 4.2.
(i, j) = (2, 1) : (1 + λ2)u9+ (−2 − 2β2)u1+ (1 − λ2)u1,1+ β2u2+ β2u8 = ∆r2f1
(i, j) = (2, 2) : (1 + λ2)u10+ (−2 − 2β2)u2+ (1 − λ2)u1,2+ β2u3+ β2u1 = ∆r2f2 (i, j) = (2, 3) : (1 + λ2)u11+ (−2 − 2β2)u3+ (1 − λ2)u1,3+ β2u4+ β2u2 = ∆r2f3
(i, j) = (2, 4) : (1 + λ2)u12+ (−2 − 2β2)u4+ (1 − λ2)u1,4+ β2u5+ β2u3 = ∆r2f4 (i, j) = (2, 5) : (1 + λ2)u13+ (−2 − 2β2)u5+ (1 − λ2)u1,5+ β2u6+ β2u4 = ∆r2f5
(i, j) = (2, 6) : (1 + λ2)u14+ (−2 − 2β2)u6+ (1 − λ2)u1,6+ β2u7+ β2u5 = ∆r2f6 (i, j) = (2, 7) : (1 + λ2)u15+ (−2 − 2β2)u7+ (1 − λ2)u1,7+ β2u8+ β2u6 = ∆r2f7
(i, j) = (2, 8) : (1 + λ2)u16+ (−2 − 2β2)u8+ (1 − λ2)u1,8+ β2u1+ β2u7 = ∆r2f8
================================================
(i, j) = (3, 1) : (1 + λ3)u17+ (−2 − 2β3)u9+ (1 − λ3)u1 + β3u10+ β3u16 = ∆r2f9 (i, j) = (3, 2) : (1 + λ3)u18+ (−2 − 2β3)u10+ (1 − λ3)u2+ β3u11+ β3u9 = ∆r2f10
(i, j) = (3, 3) : (1 + λ3)u19+ (−2 − 2β3)u11+ (1 − λ3)u3 + β3u12+ β3u10 = ∆r2f11 (i, j) = (3, 4) : (1 + λ3)u20+ (−2 − 2β3)u12+ (1 − λ3)u4 + β3u13+ β3u11 = ∆r2f12
(i, j) = (3, 5) : (1 + λ3)u21+ (−2 − 2β3)u13+ (1 − λ3)u5 + β3u14+ β3u12 = ∆r2f13 (i, j) = (3, 6) : (1 + λ3)u22+ (−2 − 2β3)u14+ (1 − λ3)u6 + β3u15+ β3u13 = ∆r2f14
(i, j) = (3, 7) : (1 + λ3)u23+ (−2 − 2β3)u15+ (1 − λ3)u7 + β3u16+ β3u14 = ∆r2f15 (i, j) = (3, 8) : (1 + λ3)u24+ (−2 − 2β3)u16+ (1 − λ3)u8+ β3u9+ β3u15 = ∆r2f16
================================================
(i, j) = (4, 1) : (1 + λ4)u5,1+ (−2 − 2β4)u17+ (1 − λ4)u9+ β4u18+ β4u24 = ∆r2f17 (i, j) = (4, 2) : (1 + λ4)u5,2+ (−2 − 2β4)u18+ (1 − λ4)u10+ β4u19+ β4u17 = ∆r2f18 (i, j) = (4, 3) : (1 + λ4)u5,3+ (−2 − 2β4)u19+ (1 − λ4)u11+ β4u20+ β4u18 = ∆r2f19 (i, j) = (4, 4) : (1 + λ4)u5,4+ (−2 − 2β4)u20+ (1 − λ4)u12+ β4u21+ β4u19 = ∆r2f20 (i, j) = (4, 5) : (1 + λ4)u5,5+ (−2 − 2β4)u21+ (1 − λ4)u13+ β4u22+ β4u20 = ∆r2f21 (i, j) = (4, 6) : (1 + λ4)u5,6+ (−2 − 2β4)u22+ (1 − λ4)u14+ β4u23+ β4u21 = ∆r2f22 (i, j) = (4, 7) : (1 + λ4)u5,7+ (−2 − 2β4)u23+ (1 − λ4)u15+ β4u24+ β4u22 = ∆r2f23 (i, j) = (4, 8) : (1 + λ4)u5,8+ (−2 − 2β4)u24+ (1 − λ4)u16+ β4u17+ β4u23 = ∆r2f24
================================================
−1 −0.5 0 0.5 1
−1
−0.8
−0.6
−0.4
−0.2 0 0.2 0.4 0.6 0.8 1
1 2 3 4
5
6 7
8 9 10 11
12
13
14
15 16
17 18 19
20
21
22
23
24 The discretization of a ring for m = 1, ν = 3 and ω = 8.
x value
y value
Figure 4.2. It is the discretization of this laplacian for a ring. The partitions of two sides are m = 1, ν = 4 and ω = 8.
After computing, we can find 1 − λ1 = 1 − 1 = 0. Put boundary conditions u(r, θ) = c(r, θ) when (r, θ) ∈ inside circle, and u(r, θ) = d(r, θ) when (r, θ) ∈ outside circle. We use them by Taylor expansions. For example u1,1:
u1,1 = c1,1, then we can get (1 − λ2)u1,1 = (1 − λ2)c1,1.
The right hand side b is
b = general case for any m, ν and ω. The matrix
A =
The right hand side b is
4.6. The Ring with Neumann Boundary Conditions of Laplacian
The process is the same with Section 4.5. After computing, we can find 1 − λ1 = 1 − 1 = 0. Put boundary conditions ∂u∂n(r, θ) = c(r, θ) when (r, θ) ∈ inside circle, and then we can get
(−2 − 2β2)u1+ (1 − λ2)u1,1 = (−2 − 2β2)u1+ (1 − λ2)(u1− ∆rc1,1)
where
The right hand side b is
b = general case for any m, ν and ω. The matrix
A =
The right hand side b is
b =
4.7. The Ring with Robin Boundary Conditions of Laplacian
∆u = f (x, y)
∂u
∂n(x, y) + α1u(x, y) = c(x, y) , (x, y) ∈ inside circle.
∂u
∂n(x, y) + α2u(x, y) = d(x, y) , (x, y) ∈ outside circle.
(20)
The process is the same with Section 4.5. After computing, we can find 1 − λ1 = 1 − 1 = 0. Put boundary conditions ∂u∂n(r, θ) + α1u(x, y) = c(r, θ) when (r, θ) ∈ inside circle, and ∂u∂n(r, θ) + α2u(x, y) = d(r, θ) when (r, θ) ∈ outside circle. We use them by Taylor expansions. For example u1,1 and u5,1:
∂u1,1
∂n + α1u1,1 = u1− u1,1
∆r + α1u1,1= c1,1,
∂u5,1
∂n + α2u5,1= u5,1− u17
∆r + α2u5,1 = d5,1, imply
u1,1 = ∆rc1,1
α1∆r − 1− u1 α1∆r − 1, u5,1 = ∆rd5,1
1 + α2∆r + u17 1 + α2∆r, then we can get
(−2 − 2β2)u1+ (1 − λ2)u1,1 = (−2 − 2β2)u1+ (1 − λ2)( ∆rc1,1
α1∆r − 1 − u1 α1∆r − 1)
= c1,1∆r(1 − λ2)
α1∆r − 1 + (−(1 − λ2)
α1∆r − 1 − 2 − 2β2)u1. and
(1 + λ4)u5,1+ (−2 − 2β4)u17 = (1 + λ4)( ∆rd5,1
1 + α2∆r + u17
1 + α2∆r) + (−2 − 2β4)u17
= d5,1∆r(1 + λ4)
1 + α2∆r + ( 1 + λ4
1 + α2∆r− 2 − 2β4)u17. Use the same step for u2, · · · , u8, and u17, · · · , u24. Let η1 = −(1−λα1∆r−1m+1) and η2 =
1+λν
1+α2∆r. We will solve the linear system question for Au = b.
Furthermore, the matrix
A =
(η1− 2)I − β2Ψ (1 + λ2)I
(1 − λ3)I −2I − β3Ψ (1 + λ3)I (1 − λ4)I (η2− 2)I − β4Ψ
∈ R24×24,
where
Ψ =
2 −1 −1
−1 2 . ..
. .. ... −1
−1 −1 2
∈ R8×8.
The right hand side b is general case for any m, ν and ω. The matrix
A =
The right hand side b is
b =
Appendix
References
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[2] Tai-Chia Lin, Juncheng Wei (2005). ”Ground state of N coupled Nonlinear Schrodinger Equations in Rn , n ≤ 3.” Comm. Math. Phys. 255: 629-653.
[3] Ming-Chih Lai, Wen-Wei Lin, and Weichung Wang (2002). ”A fast spectral/difference method without pole condition for Poisson-type equation in cylindrical and spherical geometries,” IMA Journal of Numerical Analysis, 22(4): 537-548. (SCI)
[4] Ming-Chih. Lai (2001). ”A note on finite difference discretizations for Poisson equation on a disk , Numerical Methods for Partial Differential Equations,” vol 17, issue 3, 199-203.
[5] Tai-Chia Lin and Tsung-Fang. Wu. ”A two coupled nonlinear Schrodinger equa-tions in finite strip with holes, preprint.”
[6] Ming-Chih Lai. ”A Note on Finite Difference Discretizations for Poisson Equation on a Disk.”
[7] Yuen-Cheng Kuo, Wen-Wei Lin, Shih-Feng Shieh, and Weichung Wang, ”A Hyperplane-Constrained Continuation Method for Bound States of Coupled Nonlinear Schrodinger equations” (Submitted)
[8] Weichung Wang, Tsung-fang Wu, and Chien-Hsiang Liu, ”Existence of Multiple Positive Spike Solutions for Two Coupled Nonlinear Schr¨odinger Equations” (Sub-mitted)