scientific computing
DESCRIPTION
Scientific Computing. Partial Differential Equations Explicit Solution of Heat Equation. The one-dimensional Heat Equation for the temperature u(x,t) in a rod at time t is given by: where c >0, T>0 are constants. One Dimensional Heat Equation. - PowerPoint PPT PresentationTRANSCRIPT
Scientific Computing
Partial Differential EquationsExplicit Solution of
Heat Equation
• The one-dimensional Heat Equation for the temperature u(x,t) in a rod at time t is given by:
where c >0, T>0 are constants
10 )()0,(
0 (t)),1(
0 )(),0(
010
1
0
xxfxu
Ttgtu
Tttgtu
Ttxcuu xxt
One Dimensional Heat Equation
• We will solve this equation for x and t values on a grid in x-t space:
One Dimensional Heat Equation
t
x00 x ix
nxh /1
jt
00 t
mTtk /
mtT
1nx
ApproximateSolution uij=u(xi, tj) at grid points
• To approximate the solution we use the finite difference approximation for the two derivatives in the heat equation. We use the forward-difference formula for the time derivative and the centered-difference formula for the space derivative:
Finite Difference Approximation
211
1
),(),(2),(),(
),(),(),(
h
txutxutxutxu
k
txutxutxu
jijijijixx
jijijit
• Then the heat equation (ut =cuxx ) can be approximated as
Or,
Let r = (ck/h2) Solving for ui,j+1 we get:
Finite Difference Approximation
€
u(x i, t j+1) − u(x i, t j )
k= cu(x i−1, t j ) − 2u(x i, t j ) + u(x i+1, t j )
h2
jiijjiijji uuuh
ckuu ,1,121, 2
jiijjiji ruurruu ,1,11, )21(
• Note how this formula uses info from the j-th time step to approximate the (j+1)-st time step:
One Dimensional Heat Equation
t
x
1ix ix 1ix
1jt
jt
mtT
00 t
00 x 1nx
jiijjiji ruurruu ,1,11, )21(
• On the left edge of this grid, u0,j = g0,j = g0(tj).
• On the right edge of this grid, un,j = g1,j = g1(tj).
• On the bottom row of the grid, ui,0 = fi = f(xi).
• Thus, the algorithm for finding the (explicit) solution to the heat equation is:
One Dimensional Heat Equation
nifu
jrgurruu
jruurrgu
nijruurruu
ii
jjnjnjn
jjjj
jiijjiji
0
0)21(
0)21(
2,20)21(
0,
,1,1,21,1
,2,1,01,1
,1,11,
function z = simpleHeat(f, g0, g1, T, n, m, c) %Simple Explicit solution of heat equation h = 1/n; k = T/m; r = c*k/h^2; % Constants x = 0:h:1; t = 0:k:T; % x and t vectors % Boundary conditions u(1:n+1, 1) = f(x)'; % Transpose, since it’s a row vector u(1, 1:m+1) = g0(t); u(n+1, 1:m+1) = g1(t); % compute solution forward in time for j = 1:m u(2:n,j+1) = r*u(1:n-1,j) + (1-2*r)*u(2:n,j) + r*u(3:n+1,j); end z=u'; mesh(x,t,z); % plot solution in 3-d end
Matlab Implementation
Usage: f = inline(‘x.^4’); g0 = inline(‘0*t’); g1 = inline(‘t.^0’); n=5; m=5; c=1; T=0.1; z = simpleHeat(f, g0, g1, T, n, m, c);
Matlab Implementation
Calculated solution appears correct:
Matlab Implementation
Try different T value: T=0.5
Values seem chaotic
Matlab Implementation
Why this chaotic behavior?
Matlab Implementation
Matrix Form of Solution
nifu
jrgurruu
jruurrgu
nijruurruu
ii
jjnjnjn
jjjj
jiijjiji
0
0)21(
0)21(
2,20)21(
0,
,1,1,21,1
,2,1,01,1
,1,11,
1,1
1,2
1,2
1,1
,1
,0
,1
,2
,2
,1
0
0
21
21
21
21
jn
jn
j
j
j
j
jn
jn
j
j
u
u
u
u
rg
rg
u
u
u
u
rr
rrr
rrr
rr
This is of the formStart: u(0)=u(:,0)=f(:) Iterate:
Matrix Form of Solution
1,1
1,2
1,2
1,1
,1
,0
,1
,2
,2
,1
0
0
21
21
21
21
jn
jn
j
j
j
j
jn
jn
j
j
u
u
u
u
rg
rg
u
u
u
u
rr
rrr
rrr
rr
)1(:,)(:, jubjAu
bIAAuAu
bIAuAbbAuAbAuu
bAuu
mmm )(
)()(
1)0()(
)0(2)0()1()2(
)0()1(
Now, suppose that we had an error in the initial value for u(0), say the initial u value was u(0)+e, where e is a small error.
Then, under the iteration, the error will grow like Ame.
For the error to stay bounded, we must have
Thus, the largest eigenvalue of A must be <= 1.
Matrix Form of Solution
eAemeeAm
Let A be a square nxn matrix. Around every elementaii on the diagonal of the matrix draw a circle with
radius
Such circles are known as Gerschgorin disks.
Theorem: Every eigenvalue of A lies in one of these Gerschgorin disks.
Gerschgorin Theorem
1,
j N
i ijj j i
r a
Example:
The circles that bound the Eigenvalues are:
C1: Center point (4,0) with radius r1 = |2|+|3|=5
C2: Center point (-5,0) with radius r2=|-2|+|8|=10
C3: Center Point (3,0) with radius r3=|1|+|0|=1
Gerschgorin Theorem
301
852
324
M
Example:
Actual eigenvalues in red
Gerschgorin Theorem
301
852
324
M
-15 -10 -5 5 10
-15
-10
-5
5
10
Example:
C1: Center point (1,0)
radius r1 = |0|+|7|=7
C2: Center point (-5,0)
radius r2=|2|+|0|=2
C3: Center Point (-3,0)
radius r3=|4|+|4|=8
Gerschgorin Theorem
344
052
701
M
-15 -10 -5 5 10
-15
-10
-5
5
10
Circles: center=(1-2r), radius = r or 2r. Take largest radius 2r. Then, if λ is an eigenvalue of A, we have
Matrix Form of Solution
rr
rrr
rrr
rr
A
21
21
21
21
141
221221
r
rrrr
0 1-2r
So, the eigenvalue satisfies: For the error to stay bounded, we needThus, we need
For our first example, we had T=0.1 andso, we expect the solution to be stable.For the second example, T=0.5, we have r =2.5, so the error in the iterates will grow in an unbounded
fashion, which we could see in the numerical solution.
Matrix Form of Solution
141 r
11
2
1411 rr
5.0
5
1
5
1.01
2
r