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Scientific Computing
Partial Differential EquationsPoisson Equation
Partial Differential Equations
• Our study of PDE’s has focused on linear, second-order equations
• The following general form of such PDE’s can be classified by B2 - 4AC (Variables – x and y/t)
A2u
x 2 B2u
xyC
2u
y 2 D 0
Partial Differential Equations
B2-4AC Category Example
< 0 Elliptic Laplace equation (steady state with 2 spatial dimensions)
= 0 Parabolic Heat conduction equation (time variablewith one spatial dimension)
>0 Hyperbolic Wave equation (time-variable with onespatial dimension)
2
2
22
2 1
t
y
cx
y
2
2
2
20
T
x
T
y
t
T
x
T
2
2
Multi-dimensional Heat Equation
• Consider an object in which the temperature is a function of more than just the x-direction. Then the heat conduction equation can be written:– 2-D:
– 3-D:
tyyxx uuu )(2
tzzyyxx uuuu )(2
2-D Steady State
• The Steady State solution for the Heat Equation requires that the temperature reaches an equilibrium distribution, that is we have
• Then,
• This is called Laplace’s Equation or the Potential Equation.
uxx uyy 0
ut 0
The Laplace Equation
• Models a variety of problems involving the potential of an unknown variable u(x,y)
• The solution of this equation is used in thermodynamics (steady state of heat flow), fluid flow (finding potential function), and other problems where we are calculating the potential of some quantity.
uxx uyy 0
The Poisson Equation
• Models a variety of problems involving the potential of an unknown variable u(x,y) where the right side of the equality is non-zero.
• For example, in calculations involving an Electric Field.
uxx uyy f (x,y)
Electric Potential
• The electric field is related to the charge density bythe divergence relationship: (Gauss’s Law)• The electric field is related to the electric potential by a gradient relationship• Therefore the potential is related to the charge densityby Poisson's equation• In a charge-free region of space, this becomes Laplace's equationNote: is called the Laplacian operator = sum of 2nd
partials
E V
oE 0
E electric field
chargedensity
0 permittivity
oV 2V 0
2V 0
2
Poisson’s equation: Example
• Show that the 2D potential
distribution satisfies the Poisson’s equation
Let’s evaluate the Laplacian operator:
Thus,
V 4 0
(x 2 y 2)
2V
x 2 2
4 0
2 0
;2V
y 2 2
4 0
2 0
2V 0
2V (2V
x 2 2V
y 2 ) 0
Poisson vs Potential
• Laplace’s Equation or Potential Equation:
• Poisson Equation:
uxx uyy 0
uxx uyy f (x,y)
Solution Method
• Both the Laplace and Poisson Equations involve calculating the Laplacian Operator.
• We will approximate this sum of derivatives based on a grid in x and y and centered finite differences:
)()(2
2
22
2
21
22
nx
u
x
u
x
uu
Solution Method
• Here h = Δx and
k = Δy
(i,j)
(i+1,j)(i-1,j)
(i,j+1)
(i,j-1)
2u
x 2 ui1, j 2ui, j ui 1, j
h2
2u
y 2 ui, j1 2ui, j ui, j 1
k 2
Solution Method
• Thus, for the Poisson Equation
• For the Potential (Laplace) equation we have
ui1, j 2ui, j ui 1, j
h2 ui, j1 2ui, j ui, j 1
k 2 f (x i,y j )
ui1, j 2ui, j ui 1, j
h2 ui, j1 2ui, j ui, j 1
k 2 0
Solution Method
• The equation
is valid at all interior points of the grid. What about the boundary?
ui1, j 2ui, j ui 1, j
h2 ui, j1 2ui, j ui, j 1
k 2 f (x i,y j )
Boundary Conditions
• Simplest boundary conditions specify value of u(x,y) on each of the fours sides of the region (grid) 0 ≤ x ≤ 1, 0 ≤ y ≤ 1
)(),1(
(y)),0(
)()1,(
)()0,(
1
0
1
0
ygyu
gyu
xfxu
xfxu
Solution Method
• If we assume h = k, we get for the Poisson Equation (putting lower i, j terms first)
• For the Potential (Laplace) equation we have
04 ,1,1,,1,1 jijijijiji uuuuu
),(4 2,1,1,,1,1 jijijijijiji yxfhuuuuu
Solution Method
• If we negate the Poisson equation and re-write we get:
• Consider the interior node equations for j=1:
),(4 21,1,,1,,1 jijijijijiji yxfhuuuuu
),(4:)1,1(
),(4:)1,2(
),(4:)1,3(
),(4:)1,2(
),(4:)1,1(
122
1,0,12,11,11,2
122
0,22,21,11,21,3
132
0,32,31,41,31,2
122
0,22,21,31,21,1
112
1,00,12,11,21,1
yxfhuuuuun
yxfhuuuuun
yxfhuuuuu
yxfhuuuuu
yxfhuuuuu
nnnnnn
nnnnnn
Solution Method
• If we let
• Then, we can write the equations on the previous slide as where
tnnnnn
tnnnnn
ffffffF
uuuuuuu
],,,,,,,,[
],,,,,,,,[
1,11,12,12,11,11,1
1,11,12,12,11,11,1
€
B − I[ ] u = −h2F(:,1) +b1
1,0,1
0,2
0,2
1,00,1
1,
1
1
1
1
,
41
141
141
14
nn
n
uu
u
u
uu
bIB
Solution Method
• Now, consider the interior node equations for j=2: ),(4 2
23,1,2,12,2,1 yxfhuuuuu iiiiii
),(4:)2,1(
),(4:)2,2(
),(4:)2,3(
),(4:)2,2(
),(4:)2,1(
212
2,3,12,12,21,1
222
3,22,12,22,31,2
232
3,32,42,32,21,3
222
3,22,32,22,11,2
212
2,03,12,22,11,1
yxfhuuuuun
yxfhuuuuun
yxfhuuuuu
yxfhuuuuu
yxfhuuuuu
nnnnnn
nnnnnn
Solution Method
• If we let
• Then, we can write the equations on the previous slide as
tnnnnn
tnnnnn
ffffffF
uuuuuuu
],,,,,,,,[
],,,,,,,,[
1,11,12,12,11,11,1
1,11,12,12,11,11,1
€
−I B − I[ ] u = −h2F(:,2) +b2
2,
2,0
2
0
0
,
1
1
1
1
41
141
141
14
nu
u
bIandB
Solution Method
• Similarly, for j=3,4,…,n-2 we can write the equations as
where
€
−I B − I[ ] u = −h2F(:, j) +b j
jn
j
j
u
u
bIandB
,
,0
0
0
,
1
1
1
1
41
141
141
14
Solution Method
• Now, consider the interior node equations for j=n-1: ),(4 1
2,2,1,11,1,1 nininininini yxfhuuuuu
),(4:)1,1(
),(4:)1,2(
),(4:)1,3(
),(4:)1,2(
),(4:)1,1(
112
1,,11,11,22,1
122
,21,11,21,32,2
132
,31,41,31,22,3
122
,21,31,21,12,2
112
1,0,11,21,12,1
nnnnnnnnnnnn
nnnnnnnnnnnn
nnnnnn
nnnnnn
nnnnnn
yxfhuuuuunn
yxfhuuuuunn
yxfhuuuuun
yxfhuuuuun
yxfhuuuuun
Solution Method
• If we let
• Then, we can write the equations on the previous slide as
tnnnnn
tnnnnn
ffffffF
uuuuuuu
],,,,,,,,[
],,,,,,,,[
1,11,12,12,11,11,1
1,11,12,12,11,11,1
€
−I B[ ] u = −h2F(:,2) +bn−1
1,,1
,2
,2
1,0,1
1,
1
1
1
1
,
41
141
141
14
nnnn
nn
n
nn
n
uu
u
u
uu
bIB
Solution Method
• If we let
• Then, we can write the entire (grid interior) system of equations as:
tnnnnn
tnnnnn
ffffffF
uuuuuuu
],,,,,,,,[
],,,,,,,,[
1,11,12,12,11,11,1
1,11,12,12,11,11,1
bFhAu 2
Solution Method
BI
IBI
IBI
IBI
IB
A
41
141
141
14
B
bFhAu 2
1
2
2
1
n
n
b
b
b
b
b
1,0,1
0,2
0,2
1,00,1
1
nn
n
uu
u
u
uu
b
1,,1
,2
,2
1,0,1
1
nnnn
nn
n
nn
n
uu
u
u
uu
b )2,,2(
0
0
,
,0
nj
u
u
b
jn
j
j
Solution Method
• The matrix A is sparse, block-tridiagonal (for the way we set up the vector for u), symmetric, and positive definite (xt A x > 0 for all x).
• Also, A is a (n-1)2 x (n-1)2 matrix.
BI
IBI
IBI
IBI
IB
A
Solution Method
• Since A is positive definite, the eigenvalues of A will be positive, so A is invertible. The solution exists as
• Now, this solution is not ideal, as it is computationally expensive to find the inverse of A.
• We could instead use Gaussian Elimination or LU factorization to solve for u, but this is not optimal, as A has a lot of zeroes, so we are wasting operations.
)( 21 bFhAu
Iterative Solution Method
• It turns out (no proof) that our iterative methods from earlier in the course work well for this system.
• Review: The iterative methods we looked at to solve Ax=B were: – Jacobi Method– Gauss-Seidel Method– SOR (Successive Over-Relaxation)
Jacobi Iterative Formula
• Jacobi Iteration: D=diagonal of A. x(k+1) = D-1(B + (D-A) x(k)) Element-wise, this is
n
jij
kjiji
ii
ki xaB
ax
,1
)()1( )(1
Jacobi Iterative Formula
• In our case the x vector is
and xik+1 corresponds to solving for the diagonal terms in
Ax=B, thus in our case of the Jacobi equation becomes
• Now, after adding back the terms from the vector b, we get
tnnnnn uuuuuuu ],,,,,,,,[ 1,11,12,12,11,11,1
€
ui, j(k+1) =
1
4(−h2 f i, j +bi, j − sum(−1⋅ offdiagonal um,n )
)(4
11,1,,1,1,
2)1(,
kji
kji
kji
kjiji
kji uuuufhu
bFhAu 2
Jacobi Iterative Formula
• So, we initialize ui,j0 on the boundary of the grid, and
set the interior values to some start values, say uij= 1, then use the Jacobi iteration formula
repeatedly to converge to the solution, keeping tabs on the error in our approximation.
)(4
11,1,,1,1,
2)1(,
kji
kji
kji
kjiji
kji uuuufhu
Matlab Implementationfunction [x y u] = explicitPoisson(f, f0, f1, g0, g1, n)%Explicit solution of wave equation% Constants and vectors h = 1/n; x = 0:h:n*h; y = 0:h:n*h; % Initial guess at u is u=1 for all x,y u = ones(n+1,n+1); % Boundary conditions u(1:n+1, 1) = f0(x); u(1:n+1, n+1) = f1(x); u(1, 1:n+1) = g0(y); u(n+1, 1:n+1) = g1(y); % unew = next iterate, err will measure the error from % one iteration to next unew = u; err= unew-u;
Matlab Implementation% Iterative procedure epsilon = 1e-4; % tolerance for convergence imax = 1000; % maximum number of iterations allowed k = 1; % initial index value for iteration % Calculation loop while k<= imax for i = 2:n for j = 2:n unew(i,j)=(-h^2*f(i,j) + u(i-1,j)+u(i+1,j) +u(i,j-1)+u(i,j+1))/4.0; err(i,j) = abs(unew(i,j)-u(i,j)); end; end; u = unew; k = k + 1; errmax = max(max(err));
if errmax < epsilon u = u'; mesh(x,y,u); fprintf('Convergence for %i iterations.\n',k); return; end; end;end
