16. solution of elliptic partial differential...
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16. Solution of elliptic partial differential equation
Recall in the first lecture of this course ….
Assume…• you know how to use a computer to compute;
• but have not done any serious numerical computations,
• and only know some very basic numerics.
This course will teach you…
• FORTRAN programing language to compute,
• good habits to write structural Fortran programs,
• to write test driver to test the Fortran program written by you or others,
• to use well developed and optimized libraries, such as LAPACK, FFTW,
• to do parallel computing in multi-process computer using OpenMP
• to use the these skills to solve partial differential equations governing simple
physical systems, such as Poisson equation, wave equation, heat equation.
… and using these basic blocks plus some extra hard work, eventually you are able to develop your own numerical program to simulate the more realistic physical system, such as …
Numerical simulation of flow motion
• A popular tool due to the burst in power of computing.
• Substitute for experiment when measurement is inaccessible.
velocity: � = � �,�,�,�
= � �,�,�,� �̂+ � �,�,�,� �̂+ � �,�,�,� ��
pressure: � = �(�,�,�,�)
• Property variables to describe the flow:
• Equations governing the flow:
��
��+��
��+��
��= 0 � ⋅� =0
��
��= − � ⋅� � −
1
��� + �� ��
= − �� − �(�)
��
��= − �
��
��− �
��
��− �
��
��−1
�
��
��+ �
���
���+���
���+���
���
��
��= − �
��
��− �
��
��− �
��
��−1
�
��
��+ �
���
���+���
���+���
���
��
��= − �
��
��− �
��
��− �
��
��−1
�
��
��+ �
���
���+���
���+���
���
Governing equations for motion of an incompressible fluid
• 4 unknowns with 4 equations
incompressibility of fluid (solenoidal condition):
momentum conservation:
�
��
� ⋅� =0
��
��= − �� − �(�)
• Define velocity and pressure at discrete time instance �� = �Δ�: �� and ��
Numerical solution based on explicit schemeHarlow, F.H. & Welch, J.E., 1965. Numerical calculation of time-dependent viscous incompressible flow of fluid with free surface. Phys. Fluids, 8(12), 2182−2189.
��
��≈��
��=��� � − ��
Δ�≈ − ��� − �(��)
� ⋅�� = 0
i.e., after obtaining �� and �� at �� , marching towards ��� � to get ��� �
• But, 1. How is the solenoidal condition � ⋅�� = 0 satisfied?
2. What is the equation for the pressure ��� � ?
� ⋅��� � − ��
Δ�≈ − ��� − �(��)
• Take divergence of the discretized momentum equation:
� ⋅��� � − ��
Δ�≈ − ��� − � �� ⟹
� ⋅��� � − � ⋅��
Δ�≈ − � ⋅��� − � ⋅�(��)
• Ideally, � ⋅��� � = 0 and � ⋅�� = 0, since the solenoidal condition � ⋅� = 0 must be satisfied at
every time instance.
∴ − � ⋅��
Δ�≈ − � ⋅��� − � ⋅� �� ⟹ � ��� ≈
1
Δ�� ⋅�� − � ⋅�(��) ≡ �
���
���+���
���+���
���= � �,�,�
• This is an elliptic-type partial differential equation called Poisson equation:
• Take divergence of the time-discretized momentum equation:
• But due to numerical approximation, the solenoidal condition may not be satisfied exactly. Let say,
there is a small value of � ⋅�� at �= �� , i.e., � ⋅�� ≈ 0 but � ⋅�� ≠ 0.
• We then project that � ⋅��� � = 0 at �= ��� �.
• So the solenoidal condition at �= ��� � is satisfied by solving the pressure Poisson equation.
This means that pressure is a tuning property to make sure the flow remain solenoidal.
��,� = � ��,�� = � � ��� ,��� �� ��� ����
��� �
�� ���
��� �
� � ���
��� ,� =1
�
1
�� � ��,��
� � �� ��� ����
�� �
�� �
� � �
�� �
�� = � �� = � ��� �� �
����
��� �
� � ���
= � ��� �� �� ��
��� �
� � ���
��� =1
�� ���
� � �����
� � �
�� �
=1
�� ���
� � �� ��
� � �
�� �
Discrete Fourier Transform in Higher Dimension
• One-dimensional:
• Two-dimensional:
0 1 2� =
0 ��
�� − 1
�� = ��
�� =����
�� =��
��
�� =����
� = 0 1 2 �− 1 �� = 0
1
2
�− 1
�
�� =2�
���
�� =2�
���
��,��
• Elliptic-type partial differential equation for � = � �,�in a rectangular domain:
���
���+���
���= � �,�
Numerical Solution of Poisson Equation
• Numerical solution of the equation means finding discrete ��,� satisfies the discretized equation:
���
�����,�
+���
�����,�
≈ ��,�
• Discrete representation of the continuous function:
� = � �,� ⟹ ��,� = �(��,��)
� = � �,� ⟹ ��,� = �(��,��)
��
��
�
�
�� =����
�� =����
� = 0 1 2 �− 1 �� = 0
12
�− 1�
Discrete ��,� can be represented by a discrete Fourier series with the coefficients to be determined:
��,� = � � ��̂ ,��� �� ��� ����
��� �
�� ���
��� �
� � ���
Similarly,
��,� = � � ��� ,��� �� ��� ����
��� �
�� ���
��� �
� � ���
Since ��,� are given ⟹ ��� ,� are known.
� = 0,1,2,…,� − 1� = 0,1,2,…,� − 1
Rectangular domain with periodic conditions in both directions
� = 0 1 2 � − 1�� = 0
1
2
�
� − 1
�� =����
�� =����
�� =2�
��� �� =
2�
���
� = ��
� = ���
�
���
���+���
���= � �,�
periodicperiodic
periodic
periodic
(��,��)
• This is valid when � ≠ 0 and � ≠ 0. � = 0 and � = 0 corresponds to the constant mode.
• The solution of the Poisson equation subject to periodic boundary conditions in both direction is indeterminant, i.e. any constant, � �,� = �, can be the solution.
∴ For the well posedness of the problem, ��̂,� ≡ 0.
• Substitute ��,� and ��,� into Poisson equation:���
���+���
���= � �,�
��,� = � � ��̂ ,� �� ����� ����
��� �
�� ���
��� �
� � ���
��,� = � � ��� ,��� �� ��� ����
��� �
�� ���
��� �
� � ���
� � − �����̂ ,� − �����̂ ,� = ��� ,� �� �� ��� ����
��
� � ��� ���̂ ,� �� �� ��� ����
��� �
�� ���
+ � � ��� ���̂ ,��� �� ��� ����
��� �
�� ���
=
��� �
� � ���
� � ��� ,� �� ����� ����
��� �
�� ���
��� �
� � ���
��� �
� � ���
�� =2�
��� �� =
2�
���
�� =������ =
����
• Only need to consider � = 0~�/2 and � = 0~�/2 for real DFT.
∴ ��̂ ,� = −��� ,�
��� + ���⟹ ��,� = � � ��̂ ,� �
� ����� �̃��
��� �
�� ���
��� �
� � ���
Rectangular domain with periodic condition in one direction and Dirichlet conditions on the other boundaries
���
���+���
���= � �,�
periodicperiodic
�
�
� = �(�)= given
� = �(�)= given� = ��
� = ��
0
� �,� = � � + ��,�• Unknown:
� �,0 = � � = � � + �� = given
�(�,��)= � � = � � + �� = given
• Boundary conditions:
• Since � �,� is periodic in � direction only, it can be represented as:
Similarly,
�� � = � ��,� = � ��� � ���� ��
��� �
� � ���
���
���(��,�)= � − �����̂ � ������
��� �
� � ���
���
���(��,�)= �
����̂ �
���������
��� �
� � ���
⟹
� ��,0 = �� ⟹ � ��̂ 0 ������
��� �
� � ���
= � ��� ���� ��
��� �
� � ���
� ��,�� = �� ⟹ � ��̂ �� ������
��� �
� � ���
= � ��� ������
��� �
� � ���
�� � = � ��,� = � ��̂ � �������
��� �
� � ���
= � ��̂ � ������
��� �
� � ���
�� =2�
��� �� = �
���
�� = ����
� = 0 1 2 �� − 1
�� = ��
�� = ��
periodicperiodic
�
�
���
���+���
���= � ��,�
• This is an ordinary differential equation for ��̂ � subject to the
upper and lower boundary conditions of Dirichlet type for different � .
• Only need to consider � = 0 to �/2 for real DFT.
• Substitute the expansions of � ��,� and � ��,� into the Poisson equation:���
���+���
���= � ��,�
• Substitute the expansions of � ��,� ,� �� and � �� into the boundary conditions:
� − �����̂ � +����̂ �
��������� = � ��� � ������
��� �
� � ���
��� �
� � ���
⇒����̂ �
���− �����̂ � = ��� �
� ��,0 = �� ⇒ � ��̂ 0 ���� ��
��� �
� � ���
= � ��� ���� ��
��� �
� � ���
⇒��̂ 0 = ���
� ��,�� = �� ⇒ � ��̂ �� ���� ��
��� �
� � ���
= � ��� ���� ��
��� �
� � ���
⇒��̂ �� = ���
�� =2�
���
��̂ 0 = ���
��̂ �� = ���
����̂���
− �����̂ = ��� �
� = 0
� = ��
• If use 2nd-order finite-difference scheme to approximate the differentiation of �:
����̂ �����
− �����̂ �� = ��� �� ��̂ ,�� � − 2��̂ ,� + ��̂ ,�� �
△�− �����̂ ,� = ��� ,�
2
1
0
�
� + 1
� − 1
�
� − 1
� − 2
• Discretize ��̂ � at � = �� = �Δ, Δ = ��/�, � = 0~�.
The unknowns are ��̂ �� = ��̂ ,�, � = 0~�/2, � = 1~� − 1
��̂ (0)= ���
��̂ (��)= ��� ��̂ ,� = ���
− 2 + ���Δ� ��̂ ,�� � + ��̂ ,�� � = Δ���� ,�� � − ���
��̂ ,� − 2 + ���Δ� ��̂ ,� = Δ���� ,� − ���
�� =2�
���
��̂ ,�� � − 2 + ���Δ� ��̂ ,� + ��̂ ,�� � = Δ���� ,�
��̂ ,� = ���
�
� = 0
� + 1
� − 1
� − 1�
� − 2
12
− 2 + ���∆� 1
�
1 − 2 + ���∆� 1
1 − 2 + ���∆� 1
1 − 2 + ���∆� 1
�
1 − 2 + ���∆�
��� ,�
��� ,�
⋮��� ,�� �
��� ,�
��� ,�� �
⋮��� ,�� �
��� ,�� �
=
Δ���� ,� − ���⋮⋮
△� ��� ,�� �
△� ��� ,�
△� ��� ,�� �
⋮⋮
Δ���� ,�� � − ���
The discretized equation can be represented as a system of linear equation:
Discretized equation for ��̂ �� = ��̂ ,� :
− 2 + ���Δ� ��̂ ,�� � + ��̂ ,�� � = Δ���� ,�� � − ���
��̂ ,� − 2 + ���Δ� ��̂ ,� = Δ���� ,� − ���
��̂ ,�� � − 2 + ���Δ� ��̂ ,� + ��̂ ,�� � = Δ���� ,�
− 2 + ���∆� 1
�
1 − 2 + ���∆� 1
1 − 2 + ���∆� 1
1 − 2 + ���∆� 1
�
1 − 2 + ���∆�
��� ,�
��� ,�
⋮��� ,�� �
��� ,�
��� ,�� �
⋮��� ,�� �
��� ,�� �
=
Δ���� ,� − ���⋮⋮
△� ��� ,�� �
△� ��� ,�
△� ��� ,�� �
⋮⋮
Δ���� ,�� � − ���
In the above system of equation � � = � , � is real but � and � are complex.
So the real and imaginary parts of the unknown vector can be solved separately as:
� �� = �� and � �� = ��
The (�− 1)× �− 1 matrix is tridiagonal. The linear system can be solved efficiently for various �.
Only need to consider � = 0 to �/2 for real DFT.
Numerical implementation
• Given the source function of the Poisson equation ��,�, and the lower and upper boundary
values �� and ��, where � = 0~(� − 1)and � = 1~(� − 1)
• Call FFT to compute ��� and ���Loops of � for calling FFT to compute ��� ,�
• Loops of � = 0~�/2 for solving for ��̂ ,�
Loops of �=1~�− 1 to construct the tridiagonal matrix � for each �
end loops in �
Call suitable solver to solve � �� = �� and � �� = �� for ��̂ ,� of each �
end loops of �
• Loops of � for calling inverse FFT to get ��̂,�
Rectangular domain with periodic condition in one direction and Neumann conditions on the other boundaries
���
���+���
���= � �,�
periodicperiodic
�
�
��
��= �(�)= given
��
��= �(�)= given
� = ��
� = ��
0
� �,� = � � + ��,�• Unknown:
��
���,0 = � � = � � + �� = given
��
���,�� = � � = � � + �� = given
• Boundary conditions:
• Since � �,� is periodic in � direction only, it can be represented as:
Similarly,
�� � = � ��,� = � ��� � ���� ��
��� �
� � ���
���
���(��,�)= � − �����̂ � ������
��� �
� � ���
���
���(��,�)= �
����̂ �
���������
��� �
� � ���
⟹
��
����,0 = �� ⟹ �
���̂��
�� �
���� ��
��� �
� � ���
= � ��� ������
��� �
� � ���
��
����,�� = �� ⟹ �
���̂��
�� ��
���� ��
��� �
� � ���
= � ��� ���� ��
��� �
� � ���
�� � = � ��,� = � ��̂ � �������
��� �
� � ���
= � ��̂ � ������
��� �
� � ���
�� =2�
��� �� = �
���
�� = ����
� = 0 1 2 �� − 1
��
���
= ��
��
���
= ��
periodicperiodic
�
�
���
���+���
���= � ��,�
• Again, this is an ordinary differential equation for ��̂ � subject to the
upper and lower boundary conditions of Neumann type for different �.
• Substitute the expansions of � ��,� and � ��,� into the Poisson equation:���
���+���
���= � ��,�
• Substitute the expansions of � ��,� ,� �� and � �� into the boundary conditions:
� − �����̂ � +����̂ �
��������� = � ��� � ������
��� �
� � ���
��� �
� � ���
⇒����̂ �
���− �����̂ � = ��� �
��
����,0 = �� �
���̂��
������
��� �
� � ���
= � ��� ���� ��
��� �
� � ���
⇒���̂��
= ���
��
����,�� = �� �
���̂��
���� ��
��� �
� � ���
= � ��� ������
��� �
� � ���
⇒���̂��
= ���
• Only need to consider � = 0 to �/2 for real DFT.
�� =2�
���
���̂��
= ���
���̂��
= ���
����̂���
− �����̂ = ��� �
� = 0
� = ��
• If use 2nd-order finite-difference scheme to approximate the differentiation of �:
����̂ �����
− �����̂ �� = ��� ��
��̂ ,�� � − 2��̂ ,� + ��̂ ,�� �
△�− �����̂ ,� = ��� ,�
1
0
− 1
�
� + 1
� − 1
� + 1
�
� − 1
• Discretize ��̂ � at � = �� = �Δ, Δ = ��/�, � = 0~�, the unknowns are ��̂ �� = ��̂ ,� , � = 0 to �/2.
���̂��
= ���
���̂��
= ���
��̂ ,�� � − ��̂ ,�� �
2Δ= ��� ��̂ ,�� � = ��̂ ,�� � + 2Δ���
− 2 + ���Δ� ��̂ ,� + 2��̂ ,�� � = Δ���� ,� − 2Δ���
��̂ ,� − ��̂ ,� �
2Δ= ��� ��̂ ,� � = ��̂ ,� − 2Δ���
2��̂ ,� − 2 + ���Δ� ��̂ ,� = Δ���� ,� + 2Δ���
�� =2�
���
��̂ ,�� � − 2 + ���Δ� ��̂ ,� + ��̂ ,�� � = Δ���� ,�
�
� = 0
� + 1
� − 1
� − 1�
� − 2
12
The (�+ 1)× �+ 1 matrix is tridiagonal. The linear system can be solved efficiently for various �.
Only need to consider � = 0 to �/2 for real DFT.
− 2 + ���∆� 2
�
1 − 2 + ���∆� 1
1 − 2 + ���∆� 1
1 − 2 + ���∆� 1
�
2 − 2 + ���∆�
��� ,�
��� ,�
⋮��� ,�� �
��� ,�
��� ,�� �
⋮��� ,�� �
��� ,�
=
Δ���� ,� + 2Δ���⋮⋮
△� ��� ,�� �
△� ��� ,�
△� ��� ,�� �
⋮⋮
Δ���� ,� − 2Δ���
The discretized equation can be represented as a system of linear equation:
Discretized equation for ��̂ �� = ��̂ ,� :
− 2 + ���Δ� ��̂ ,� + 2��̂ ,�� � = Δ���� ,� − 2Δ���
2��̂ ,� − 2 + ���Δ� ��̂ ,� = Δ���� ,� + 2Δ���
��̂ ,�� � − 2 + ���Δ� ��̂ ,� + ��̂ ,�� � = Δ���� ,�
− 2 + ���∆� 1
�
1 − 2 + ���∆� 1
1 − 2 + ���∆� 1
1 − 2 + ���∆� 1
�
1 − 2 + ���∆�
��� ,�
��� ,�
⋮��� ,�� �
��� ,�
��� ,�� �
⋮��� ,�� �
��� ,�� �
=
Δ���� ,� − ���⋮⋮
△� ��� ,�� �
△� ��� ,�
△� ��� ,�� �
⋮⋮
Δ���� ,�� � − ���
− 2 + ���∆� 2
�
1 − 2 + ���∆� 1
1 − 2 + ���∆� 1
1 − 2 + ���∆� 1
�
2 − 2 + ���∆�
��� ,�
��� ,�
⋮��� ,�� �
��� ,�
��� ,�� �
⋮��� ,�� �
��� ,�
=
Δ���� ,� + 2Δ���⋮⋮
△� ��� ,�� �
△� ��� ,�
△� ��� ,�� �
⋮⋮
Δ���� ,� − 2Δ���
• Dirichlet conditions on upper and lower boundaries:
• Neumann conditions on upper and lower boundaries:
− 2 + ���∆� 2
�
1 − 2 + ���∆� 1
1 − 2 + ���∆� 1
1 − 2 + ���∆� 1
�
2 − 2 + ���∆�
��� ,�
��� ,�
⋮��� ,�� �
��� ,�
��� ,�� �
⋮��� ,�� �
��� ,�
=
Δ���� ,� + 2Δ���⋮⋮
△� ��� ,�� �
△� ��� ,�
△� ��� ,�� �
⋮⋮
Δ���� ,� − 2Δ���
For � = 0, ��̂,� = ��̂ �� , i.e. the constant Fourier mode, the coefficient matrix � of the above system of equation � � = � is singular, i.e. det � = 0.
For example, � = 2, − 2 2 01 − 2 10 2 − 2
= − 8 + 4 + 4 = 0
� = 3,
− 2 21 − 2
0 01 0
0 10 0
− 2 12 − 2
= − 2 ×− 2 1 01 − 2 10 2 − 2
−2 0 01 − 2 10 2 − 2
= 4 − 4 = 0
���
���+���
���= � �,�
periodicperiodic
�
�
��
��= �(�)= given
��
��= �(�)= given
To fix the problem, i.e. to make the solution unique, a given constant is specified at the grid point.
For example, �(0,0)= ��,� ≡ 0.
�(0,0)= ��,� ≡ 0
��,� = � ��,�� = �� �� = � ��̂ �� �������
��� �
� � ���
= � ��̂ �� ���� ��
��� �
� � ���
�� =2�
��� �� = �
���
�� = ����
��,� = � 0,�� = � ��̂ ,����
����
��� �
� � ���
= � ��̂ ,�
��� �
� � ���
= 2 � ��̂ ,��
��� �
� � �
+ ��̂�,�
� + ��̂,��
��̂,�� = ��,� − 2 � ��̂ ,�
�
��� �
� � �
− ��̂�,�
�
��̂ ,� =1
�� ��,��
� � �����
� � �
�� �
⇒��̂,� =1
�� ��,�
� � �
�� �
∈ ℝ⇒��̂,�� = 0
To implement the solvability condition � 0,0 = ��,� ≡ 0:
For � = 0
��̂,�� = ��,� − 2 � ��̂ ,�
�
��� �
� � �
− ��̂�,�
�
solutions of � ≠ 0 modes
∴ ��̂,� are treated after solving for ��̂ ,�, � ≠ 0.
• After obtain ��̂,� = ��̂,�� , the other ��̂,� can be evaluated:
����̂ �����
− 0���̂ �� = ��� ��
��̂,�� � − 2��̂,� + ��̂,�� �△�
− 0���̂,� = ���,�
1
0
− 1
�
� + 1
� − 1
� + 1
�
� − 1
���̂��
= ���
���̂��
= ���
��̂,�� � − ��̂,�� �2Δ
= ��� ⇒ ��̂,�� � = ��̂,�� � + 2Δ���
��̂,� =��̂,�� � + Δ��� −1
2����,�
��̂,� − ��̂,� �2Δ
= ��� ⇒ ��̂,� � = ��̂,� − 2Δ���
��̂,�� � = 2��̂,� − ��̂,�� � + Δ����,�
��̂,�� = ��,� − 2 � ��̂ ,�
�
��� �
� � �
− ��̂�,�
�
��̂,� = 2��̂,�� � − ��̂,�� � + Δ����,�� �
��̂,� = 2��̂,� − ��̂,� � + Δ����,�
��̂,� =��̂,� + Δ��� +1
2����,�2
Numerical implementation
• Given the source function of the Poisson equation ��,�, and the lower and upper boundary
derivative values �� and ��, where � = 0~� − 1 and � = 0~�
• Call FFT to compute ��� , ���Loops of � for calling FFT to compute ��� ,�
• Loops of � = 1~�/2 for solving for ��̂ ,�
Loops of �=0~� to construct the tridiagonal matrix � for each �
end loops in �
Call suitable solver to solve � �� = �� and � �� = �� for ��̂ ,� of each �
end loops of �
• The constant modes ��̂,� need to be treated after solving ��̂ ,�,� ≠ 0
Compute ��̂,��
Loops of �=1~� to compute ��̂,� for � = 0
end loops in �
• Loops of � for calling inverse FFT to get ��̂,�
• Invent a known analytical function �� �,� and compute analytically the right-hand-side source function of the Poisson equation �(�,�):
The first approach:
Two approaches to test the Poisson solver
• Such an approach can be used to test the convergence of the solver by increase the grid points.
� �,� =���′
���+���′
���
• Given the analytical source function ��,�and the boundary conditions at ��and ��, the
Poisson equation is then solved for ��,�, and the numerical solution is compared with the
known function ��,�� .
���
�����,�
+���
�����,�
≈ ��,�
�� =����
�� =����
� = 0 1 2 �− 1 �� = 0
12
�− 1�
� �,0 = �′ �,0
� �,�� = �′ �,��
��
���,0 =
��′
���,0
��
���,�� =
��′
���,��
or
and the boundary conditions:
• Generate the solutions ��,�� using random numbers.
The second approach:
• Given the source function ��,� and the boundary conditions the Poisson equation is then solved
numerically:
• The source function ��,� and the boundary conditions are then computed numerically using
spectral scheme in the periodic direction and finite-difference scheme in the non-periodic direction.
��,� =���′
�����,�
+���′
�����,�
spectral finite-difference
• The numerical results ��,�should be equal to the original given random numbers ��,�� since the
forward (computing the source function) and inverse (solving the equation) operators are identical.
• Such an approach is for debugging the code.
���
�����,�
+���
�����,�
= ��,�
�� =����
�� =����
� = 0 1 2 �− 1 �� = 0
12
�− 1�
or
• Also generate the boundary conditions using random numbers:
��,� = ��,��
��,� = ��,��
��
���,�
=��′
���,�
��
���,�
=��′
���,�
finite-difference
■ Write a subroutine solving Poisson equation in a rectangular domain, as shown in the figure below, with
periodic boundary conditions in the � direction, Dirichlet condition on the upper boundary, and Neumann
condition on the lower boundary.
■ Use routine in Lapack to solve the tridiagonal system of linear equation (e,g, dgtsv).
■ Use FFFTW to do discrete Fourier transform.
■ In calling the subroutine, the following data are input :
• the lengths the rectangular domain in � and � directions:�� and ��
• the numbers of discrete grids in � and � directions: � and �
• the right-hand side of the Poisson equation: ��,�, � = 0~(� − 1), � = 0~�
• the values on the lower and upper boundary conditions: �� and �� , � = 0~(� − 1)
■ Write a test driver (use the second approach) to assess the maximum error calling the subroutine to confirm if
the Poisson solver has been implemented correctly. Use random numbers between − 1 and 1.
Homework
��
��
���
���+���
���= � �,�
periodic in �
� = �
��
��= �
periodic in �
� = 0 1 2 �(� − 1) � = 0
1
2
�
(� − 1)
�
�
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