In this question, you are asked to find the solution u(x, y) that satisfies the partial differential equation below using the spectral collocation method (with Gauss-Lobatto and also evenly spaced points), 8² u 8²u + მ2 дуг f(x, y). In order to do that, you will need to first find the derivative matrices and use them to construct the system of algebraic equation that looks something like [A] {u} = {f}. To extend the one-dimensional spectral collocation method with Gauss-Lobatto points into two dimensions, we can tile the one dimensional mesh along the second direction. Suppose that {X} and {Y} are arrays containing Nth order (N + 1 point) one-dimensional meshes, then we generate our two-dimensional mesh by repeating {X} at a spacing determined by the points in {Y}. The idea is akin to how MATLAB's meshgrid() function works. To illustrate this consider a two-dimensional mesh generated from the third order Gauss--Lobatto points.
In this question, you are asked to find the solution u(x, y) that satisfies the partial differential equation below using the spectral collocation method (with Gauss-Lobatto and also evenly spaced points), 8² u 8²u + მ2 дуг f(x, y). In order to do that, you will need to first find the derivative matrices and use them to construct the system of algebraic equation that looks something like [A] {u} = {f}. To extend the one-dimensional spectral collocation method with Gauss-Lobatto points into two dimensions, we can tile the one dimensional mesh along the second direction. Suppose that {X} and {Y} are arrays containing Nth order (N + 1 point) one-dimensional meshes, then we generate our two-dimensional mesh by repeating {X} at a spacing determined by the points in {Y}. The idea is akin to how MATLAB's meshgrid() function works. To illustrate this consider a two-dimensional mesh generated from the third order Gauss--Lobatto points.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![Y
1
0.5
0
-0.5
−1
213,0
= 2413
212,0 219
=
211,0 = 215
−1
240,0 = 241
213,1 = 244
212,1 = 240
211,1 = 216
240,1 = 212
-0.5
u(x, y) ≈
Hence the second derivatives are given by
0
N N
X
243,2 = 2415
i=0 j=0
212,2 = 2411
241,2 = 214
20,2 = 213
0.5
ΣΣυ.L;(2)L;(y).
Here the nodes follow the lexicographic ordering from left to right, bottom to top.
The advantage of this type of construction is that it essentially decouples the axes, meaning that
we can take the one dimensional Lagrange basis functions and multiply them together to produce
the two-dimensional interpolants
1
Uxx (x, y)
_ouj, L{ (æ)Lj(y),
Uyy (x, y) = U₁L₁(x) L (y).
થરૂ,3 = 16
212,3 = 242
211,3 = 218
240,3 = 24](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F12fdac56-4b52-4434-8b00-5772085cd8ad%2F43cecc25-b25c-4007-9826-0ae7983c96c8%2F2nqwat_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Y
1
0.5
0
-0.5
−1
213,0
= 2413
212,0 219
=
211,0 = 215
−1
240,0 = 241
213,1 = 244
212,1 = 240
211,1 = 216
240,1 = 212
-0.5
u(x, y) ≈
Hence the second derivatives are given by
0
N N
X
243,2 = 2415
i=0 j=0
212,2 = 2411
241,2 = 214
20,2 = 213
0.5
ΣΣυ.L;(2)L;(y).
Here the nodes follow the lexicographic ordering from left to right, bottom to top.
The advantage of this type of construction is that it essentially decouples the axes, meaning that
we can take the one dimensional Lagrange basis functions and multiply them together to produce
the two-dimensional interpolants
1
Uxx (x, y)
_ouj, L{ (æ)Lj(y),
Uyy (x, y) = U₁L₁(x) L (y).
થરૂ,3 = 16
212,3 = 242
211,3 = 218
240,3 = 24
![In this question, you are asked to find the solution u(x, y) that satisfies the partial differential
equation below using the spectral collocation method (with Gauss-Lobatto and also evenly spaced
points),
8² u
8² u
əx² მყ2
+
=
= f(x, y).
In order to do that, you will need to first find the derivative matrices and use them to construct the
system of algebraic equation that looks something like
[A]{u} = {f}.
To extend the one-dimensional spectral collocation method with Gauss-Lobatto points into two
dimensions, we can tile the one dimensional mesh along the second direction. Suppose that {X}
and {Y} are arrays containing Nth order (N + 1 point) one-dimensional meshes, then we
generate our two-dimensional mesh by repeating {X} at a spacing determined by the points in
{Y}. The idea is akin to how MATLAB's meshgrid() function works.
To illustrate this consider a two-dimensional mesh generated from the third order Gauss--Lobatto
points.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F12fdac56-4b52-4434-8b00-5772085cd8ad%2F43cecc25-b25c-4007-9826-0ae7983c96c8%2F6e7387c_processed.jpeg&w=3840&q=75)
Transcribed Image Text:In this question, you are asked to find the solution u(x, y) that satisfies the partial differential
equation below using the spectral collocation method (with Gauss-Lobatto and also evenly spaced
points),
8² u
8² u
əx² მყ2
+
=
= f(x, y).
In order to do that, you will need to first find the derivative matrices and use them to construct the
system of algebraic equation that looks something like
[A]{u} = {f}.
To extend the one-dimensional spectral collocation method with Gauss-Lobatto points into two
dimensions, we can tile the one dimensional mesh along the second direction. Suppose that {X}
and {Y} are arrays containing Nth order (N + 1 point) one-dimensional meshes, then we
generate our two-dimensional mesh by repeating {X} at a spacing determined by the points in
{Y}. The idea is akin to how MATLAB's meshgrid() function works.
To illustrate this consider a two-dimensional mesh generated from the third order Gauss--Lobatto
points.
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Follow-up Questions
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Follow-up Question
![For convenience, we write the coefficients uj,¿ in the vector
{u} = (u₁, U₂,.
where l
-
1 + i + (N + 1)j.
Q5
Ulg...
... 9
Then the second derivative matrix may be written as
where is some operation on the one-dimensional second derivative matrix [D²] and the identity
matrix [I] that produces the correct two-dimensional derivative matrix.
[D²].x
Urr (xi, Yi) = [I] [D²]{u},
®
x x
Uyy (Xi, Yi) = [D²] [I]{u},
‚ U(N+1)²)™ = (u0,0, U0,1, . . . , Wi‚j:
For the moment, assume a 2 × 2 mesh. Show that the two dimensional derivative matrices are
given by
[D²] y
=
=
[A] [B]
D₁₁ D12
D² D2
11
21
D₁₁
D²₁
D%2
The result above can be extended to an N x N mesh. The operation
the Kronecker product which is implemented in MATLAB using the kron()
matrix [A] and a p × q matrix [B], it is defined as
D²₁
D²21
11
D²₁₁ D₁2
D21 D22
D112
D 222
a₁1[B]
⠀
am1[B]
‚UN,N)T
D²12
ain [B]
:
amn [B]
on two matrices is called
function. For an m × n](https://content.bartleby.com/qna-images/question/12fdac56-4b52-4434-8b00-5772085cd8ad/9372dda9-8b0b-4df9-9995-dfbdabcc0fe2/ffhwibq_thumbnail.jpeg)
Transcribed Image Text:For convenience, we write the coefficients uj,¿ in the vector
{u} = (u₁, U₂,.
where l
-
1 + i + (N + 1)j.
Q5
Ulg...
... 9
Then the second derivative matrix may be written as
where is some operation on the one-dimensional second derivative matrix [D²] and the identity
matrix [I] that produces the correct two-dimensional derivative matrix.
[D²].x
Urr (xi, Yi) = [I] [D²]{u},
®
x x
Uyy (Xi, Yi) = [D²] [I]{u},
‚ U(N+1)²)™ = (u0,0, U0,1, . . . , Wi‚j:
For the moment, assume a 2 × 2 mesh. Show that the two dimensional derivative matrices are
given by
[D²] y
=
=
[A] [B]
D₁₁ D12
D² D2
11
21
D₁₁
D²₁
D%2
The result above can be extended to an N x N mesh. The operation
the Kronecker product which is implemented in MATLAB using the kron()
matrix [A] and a p × q matrix [B], it is defined as
D²₁
D²21
11
D²₁₁ D₁2
D21 D22
D112
D 222
a₁1[B]
⠀
am1[B]
‚UN,N)T
D²12
ain [B]
:
amn [B]
on two matrices is called
function. For an m × n
Solution
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