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Transcript of PDEs - Slides (4)
-
8/3/2019 PDEs - Slides (4)
1/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
Part IV
20401
Tony Shardlow
1 / 5 7
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8/3/2019 PDEs - Slides (4)
2/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
Outline
1 Model parabolic problem
2 Uniqueness
3 Method of lines
4 Euler method
5 Finite differences for heat equation
2 / 5 7
-
8/3/2019 PDEs - Slides (4)
3/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
Outline
1 Model parabolic problem
2 Uniqueness
3 Method of lines
4 Euler method
5 Finite differences for heat equation
3 / 5 7
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8/3/2019 PDEs - Slides (4)
4/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
Model parabolic PDE
Consider the heat equation for t > 0 and x [0, 1],
ut =uxx
u(0, t) =u(1, t) = 0,
with initial condition u(x, 0) = u0(x).
We study
1 Uniqueness of soln
2 Finite difference method
3 Stability of time discretisation.
4 / 5 7
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8/3/2019 PDEs - Slides (4)
5/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
Outline
1 Model parabolic problem
2 Uniqueness
3 Method of lines
4 Euler method
5 Finite differences for heat equation
5 / 5 7
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8/3/2019 PDEs - Slides (4)
6/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
Uniqueness of soln
ut uxx = 0
u(0, t) = u(1, t) = 0u(x, 0) = u0(x)
(H)Multiply the PDE by u and integrate over x [0, 1],
Define the energy : E(t) =
1
0 u2(x, t) dx
and differentiate
E(t) =d
dt
10
u2 dx =
10
t(u2) dx = 2
10
uut dx
=21
0uuxx dx.
6 / 5 7
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8/3/2019 PDEs - Slides (4)
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Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
E(t) =
10
u2 dx; E(t) = 2
10
uuxx dx.
Integrate by parts
E(t) = 2
uux
10
2
10
(ux)2 dx
= 2
10
(ux)2 dx 0 ,
where we apply the boundary conditions u(0, t) = u(1, t) = 0.
We deduce that E(t) E(0) for all t > 0. That is,10
u2(x, t) dx = E(t) E(0) =
10
u20(x) dx.
7 / 5 7
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8/3/2019 PDEs - Slides (4)
8/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
E(t) =
10
u2 dx; E(t) = 2
10
uuxx dx.
Integrate by parts
E(t) = 2
uux
10
2
10
(ux)2 dx
= 2
10
(ux)2 dx 0 ,
where we apply the boundary conditions u(0, t) = u(1, t) = 0.
We deduce that E(t) E(0) for all t > 0. That is,10
u2(x, t) dx = E(t) E(0) =
10
u20(x) dx.
8 / 5 7
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8/3/2019 PDEs - Slides (4)
9/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
Uniqueness of soln
Theorem
The solution to (H) is unique.
Proof.
Let v, u be two solutions. By linearity, w = u v satisfies
wt =wxx, w(0, t) = w(1, t) = 0,
w(x, 0) =0.
E(t) = 1
0 w(x, t)2
dx
is a decreasing function.
But E(0) = 0, so E(t) = 0 for all t. Hence w = 0 for all t andu = v and the solution is unique.
9 / 5 7
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8/3/2019 PDEs - Slides (4)
10/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
A model parabolic PDE problem
HOMEWORK
You can now try Problem 1
Evolution of the temperature profiles for specific initialconditions ..
0 0.5 10
0.2
0.4
0.6
0.8
1
x
temperature
Problem 2.1
0 0.5 10
0.2
0.4
0.6
0.8
1
x
temperature
Problem 2.2
10/57
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8/3/2019 PDEs - Slides (4)
11/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
A model parabolic PDE problem
HOMEWORK
You can now try Problem 1
Evolution of the temperature profiles for specific initialconditions ..
0 0.5 10
0.2
0.4
0.6
0.8
1
x
temperature
Problem 2.1
0 0.5 10
0.2
0.4
0.6
0.8
1
x
temperature
Problem 2.2
11/57
-
8/3/2019 PDEs - Slides (4)
12/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
Outline
1 Model parabolic problem
2 Uniqueness
3 Method of lines
4 Euler method
5 Finite differences for heat equation
12/57
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8/3/2019 PDEs - Slides (4)
13/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
Heat equation
Consider the Fourier series
u(x, t) =
j=1
uj(t) sin(jx)
Notice that u(0, t) = u(1, t) and boundary conditions aresatisfied.Substitute into the heat equation ut = uxx. Then
ut(x, t) =
j=1uj(t) sin(jx)
uxx(x, t) =
j=1
uj(t) (j22)sin(jx).
13/57
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8/3/2019 PDEs - Slides (4)
14/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
Heat equation
Consider the Fourier series
u(x, t) =
j=1
uj(t) sin(jx)
Notice that u(0, t) = u(1, t) and boundary conditions aresatisfied.Substitute into the heat equation ut = uxx. Then
ut(x, t) =
j=1uj(t) sin(jx)
uxx(x, t) =
j=1
uj(t) (j22)sin(jx).
14/57
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8/3/2019 PDEs - Slides (4)
15/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
Decoupling the heat equation
After substitution into the heat equation,
j=1
uj(t)j22 sin(jx) =
j=1
uj(t) sin(jx).
and equating coefficients of sin(jx),
ODEs for Fourier coefficients uj(t)
uj = j22uj , j = 1, 2, . . . .
The system is decoupled, as the solution uj can be found inde-pendently of uk for j = k.
From u(0, x) =
j=1 uj(0) sin(jx) = u0(x), the initial data
uj(0) = 2
0u0(x)sinjx dx.
15/57
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8/3/2019 PDEs - Slides (4)
16/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
Method of lines
Replace infinite system of ODEs
u
j = j2
2
uj, j = 1, 2, . . . ,
with finite system of ODEs
uj = j22uj, j = 1, 2, . . . , J
where J is size of system and solve to determine uj(t),j = 1, . . . , J.
Method of lines approximation is
v(x, t) =J
j=1
uj
(t) sin(jx).
For large J, u(x, t) v(x, t).Finite difference methods are often used to solve the ODEs foruj(t). We study finite difference methods for ODEs.
16/57
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8/3/2019 PDEs - Slides (4)
17/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
Outline
1 Model parabolic problem
2 Uniqueness
3 Method of lines
4 Euler method
5 Finite differences for heat equation
17/57
Finite difference method for
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8/3/2019 PDEs - Slides (4)
18/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
Finite difference method forODE
Consider the following initial value problem for the ODE:
dXdt
= f(X), X(0) = X0
For k > 0 (the time step), we can approximate
dX
dt =
X(t+ k) X(t)
k + O(k)
Example: X(t) = et
dX
dt=
det
dt= et =
et+k et
k+ O(k)
As a numerical example, put t = 1 and k = 0.1, then X(t) =et = e = 2.7183 and
et+k et
k =e1.1 e1
0.1 = 2.8588 18/57
Finite difference method for
-
8/3/2019 PDEs - Slides (4)
19/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
Finite difference method forODE
Consider the following initial value problem for the ODE:
dXdt
= f(X), X(0) = X0
For k > 0 (the time step), we can approximate
dX
dt=
X(t+ k) X(t)
k+
O(k)
Example: X(t) = et
dX
dt=
det
dt= et =
et+k et
k+ O(k)
As a numerical example, put t = 1 and k = 0.1, then X(t) =et = e = 2.7183 and
et+k et
k =e1.1 e1
0.1 = 2.8588 19/57
Finite difference method for
-
8/3/2019 PDEs - Slides (4)
20/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
Finite difference method forODE
Consider the following initial value problem for the ODE:
dXdt
= f(X), X(0) = X0
For k > 0 (the time step), we can approximate
dX
dt=
X(t+ k) X(t)
k+ O(k)
Example: X(t) = et
dX
dt=
det
dt= et =
et+k et
k+ O(k)
As a numerical example, put t = 1 and k = 0.1, then X(t) =et = e = 2.7183 and
et+k et
k =e1.1 e1
0.1 = 2.8588 20/57
-
8/3/2019 PDEs - Slides (4)
21/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
Explicit Euler method
Put t = nk and substitute
X(t+ k) X(t)
k= f(X(t)) + O(k)
into ODE.
X((n + 1)k) X(nk)
k= f(X(nk)) + O(k)
Drop the error term and approximate X(nk) by Xn, gives the
Explicit Euler method
Given X0 and time step k, compute Xn X(nk) by
Xn+1 = Xn + kf(Xn)
It is possible to show that |X(nk) Xn| = O(k) for nk = T,under assumptions on f. The method is first order accurate.
We do not prove this. 21/57
-
8/3/2019 PDEs - Slides (4)
22/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
Example: explicit Euler method
Consider the ODE
dX
dt= X, X(0) = X0
with X0 = 2 and = 3. In this case, exact solution is
X(t) = e3t2 .
We find explicit Euler approximation for time step k = 0.1,
Xn+1 = Xn + kf(Xn), X0 = 2.
X1 = 2 + 0.1(3 2) = 1.4 ,
X2 = 1.4 + 0.1(3 1.4) = 0.98
The exact solution X1 X(0.1) = 2e0.3 = 1.4816,
X2 1.0976 = X(0.2).
Can get more accurate approximation by taking k smaller 22/57
E l li i E l h d
-
8/3/2019 PDEs - Slides (4)
23/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
Example: explicit Euler method
Consider the ODE
dX
dt= X, X(0) = X0
with X0 = 2 and = 3. In this case, exact solution is
X(t) = e3t2 .
We find explicit Euler approximation for time step k = 0.1,
Xn+1 = Xn + kf(Xn), X0 = 2.
X1 = 2 + 0.1(3 2) = 1.4 ,
X2 = 1.4 + 0.1(3 1.4) = 0.98
The exact solution X1 X(0.1) = 2e0.3 = 1.4816,
X2 1.0976 = X(0.2).
Can get more accurate approximation by taking k smaller 23/57
I li i E l h d
-
8/3/2019 PDEs - Slides (4)
24/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
Implicit Euler method
Approximating dX/dt by (X(t) X(t k))/k and repeating
gives X(nk) X((n 1)k)
k= f(X(nk)) + O(k)
Drop the error term and approximate X(nk) by Xn, gives the
Implicit Euler method
Given X0 and time step k, compute Xn X(nk) by
Xn = Xn1 + kf(Xn)
A nonlinear system of equations must be solved to determineXn. Hence the method is implicit.
Again possible to show that |X(nk) Xn| = O(k) for nk = T,under assumptions on f. The method is first order accurate.
Not proved. 24/57
Li ODE
-
8/3/2019 PDEs - Slides (4)
25/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
Linear test ODE
Linear test equation
dXdt
= X = X, X(0) = X0.
The solution is easily found to be
X(t) = e
t
X0.
Note that X(t) 0 as t if < 0 ( means real part).
What about numerical methods?For what values of , do the numerical approximations Xn 0
as n Hope that the set of such is also { R : < 0}.
25/57
Stability condition for Explicit
-
8/3/2019 PDEs - Slides (4)
26/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
Stab ty co d t o o p c tEuler
1 For the case f(X) = X, the explicit Euler method is
Xn+1 = Xn + kXn
2 Rearrange,Xn+1 = (1 + k)Xn.
and the solution Xn = (1 + k)nX0 .
3 Hence as n
Xn 0 |1 + k| < 1
if and only if k belongs tocircle of radius 1 in complex plane with centre at 1.
4 The set of k C such that Xn 0 as n is knownas the region of absolute stability.
26/57
Stability condition for Explicit
-
8/3/2019 PDEs - Slides (4)
27/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
y pEuler
1 For the case f(X) = X, the explicit Euler method is
Xn+1 = Xn + kXn
2 Rearrange,Xn+1 = (1 + k)Xn.
and the solution Xn = (1 + k)nX0 .
3 Hence as n
Xn 0 |1 + k| < 1
if and only if k belongs tocircle of radius 1 in complex plane with centre at 1.
4 The set of k C such that Xn 0 as n is knownas the region of absolute stability.
27/57
Time step restriction for explicit
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8/3/2019 PDEs - Slides (4)
28/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
p pEuler
1 Recall for the ODE, then
X(t) 0 as t
if < 0.2 The numerical method decays for all initial data if and
only if |1 + k| < 1 if and only if
timestep k < 2/|| .
That is, there is a restriction on the time step.3 For the method of lines method, = j22 where
j = 1, 2, . . . , J.4 Hence, the time step restriction is
timestep k < 2/J22 .
For large J, the time step condition is very restrictive!
Leads to inefficient method. 28/57
Time step restriction for explicit
-
8/3/2019 PDEs - Slides (4)
29/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
p pEuler
1 Recall for the ODE, then
X(t) 0 as t
if < 0.2 The numerical method decays for all initial data if and
only if |1 + k| < 1 if and only if
timestep k < 2/|| .
That is, there is a restriction on the time step.3 For the method of lines method, = j22 where
j = 1, 2, . . . , J.4 Hence, the time step restriction is
timestep k < 2/J22 .
For large J, the time step condition is very restrictive!
Leads to inefficient method. 29/57
Time step restriction for explicit
-
8/3/2019 PDEs - Slides (4)
30/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
p pEuler
1 Recall for the ODE, then
X(t) 0 as t
if < 0.2 The numerical method decays for all initial data if and
only if |1 + k| < 1 if and only if
timestep k < 2/|| .
That is, there is a restriction on the time step.3 For the method of lines method, = j22 where
j = 1, 2, . . . , J.4 Hence, the time step restriction is
timestep k < 2/J22 .
For large J, the time step condition is very restrictive!
Leads to inefficient method. 30/57
Implicit Euler
-
8/3/2019 PDEs - Slides (4)
31/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
Implicit Euler
Put f(X) = X into the implicit Euler method.
Implicit Euler method
Xn Xn1k
= Xn
It becomes
Xn =1
(1 k)Xn1 .
and the soln Xn = (1 k)nX0 .
HenceXn 0 |1 k| > 1
if and only if k isoutside the circle of radius 1 in complex plane with centre at 1
.
31/57
Implicit Euler
-
8/3/2019 PDEs - Slides (4)
32/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
Implicit Euler
Put f(X) = X into the implicit Euler method.
Implicit Euler method
Xn Xn1k
= Xn
It becomes
Xn =1
(1 k)Xn1 .
and the soln Xn = (1 k)nX0 .
HenceXn 0 |1 k| > 1
if and only if k isoutside the circle of radius 1 in complex plane with centre at 1
.
32/57
No stability restriction for
-
8/3/2019 PDEs - Slides (4)
33/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
implicit Euler
1 For implicit Euler, the region of absolute stability is
|1 k| > 1 and this holds for all with < 0.2 When the ODE is stable ( < 0), the implicit Euler is
stable too and |Xn| 0 as n .
3 There is no timestep restriction for the implicit Euler
method.This is important for the heat equation, where is large.
4 The implicit Euler is a better method for the timediscretisation of the heat equation.
33/57
No stability restriction for
-
8/3/2019 PDEs - Slides (4)
34/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
implicit Euler
1 For implicit Euler, the region of absolute stability is
|1 k| > 1 and this holds for all with < 0.2 When the ODE is stable ( < 0), the implicit Euler is
stable too and |Xn| 0 as n .
3 There is no timestep restriction for the implicit Euler
method.This is important for the heat equation, where is large.
4 The implicit Euler is a better method for the timediscretisation of the heat equation.
34/57
Explicit vs Implicit
-
8/3/2019 PDEs - Slides (4)
35/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
Explicit vs Implicit
To understand the names implicit and explicit, apply the
methods to dX
dt= f(X), X(0) = X0
for f(x) a nonlinear function. The explicit Euler method is
Xn+1 = Xn + kf(Xn)
so that Xn+1 is determined explicitly from Xn and evaluation off.The implicit Euler method is
Xn = Xn1 + kf(Xn)
so that Xn is determined implicitly by a solution of a nonlinearequation involving f.
HOMEWORK
You can now try Problem 24
35/57
Explicit vs Implicit
-
8/3/2019 PDEs - Slides (4)
36/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
Explicit vs Implicit
To understand the names implicit and explicit, apply the
methods to dX
dt= f(X), X(0) = X0
for f(x) a nonlinear function. The explicit Euler method is
Xn+1 = Xn + kf(Xn)
so that Xn+1 is determined explicitly from Xn and evaluation off.The implicit Euler method is
Xn = Xn1 + kf(Xn)
so that Xn is determined implicitly by a solution of a nonlinearequation involving f.
HOMEWORK
You can now try Problem 24
36/57
Examination
-
8/3/2019 PDEs - Slides (4)
37/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
Examination
The material from here on is not covered on the exam.
37/57
Outline
-
8/3/2019 PDEs - Slides (4)
38/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
Outline
1 Model parabolic problem
2 Uniqueness
3 Method of lines
4 Euler method
5 Finite differences for heat equation
38/57
Discretisation of (x, t) space :m ( t )
-
8/3/2019 PDEs - Slides (4)
39/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
um
j u(xj, tm)
Grid spacing in x direction is h.
Grid spacing in t direction is k.
6 6
s s s s s s
c
c
c
c
c
c
c
c
c
c
0 = x0 xj xn = 1j1 j j+1h
h =1
n -
k
mtm
m+1
6
?
umj
39/57
Explicit approximation
-
8/3/2019 PDEs - Slides (4)
40/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
p pp
ut (xj, tm) = uxx (xj, tm)
Approximate
uxx(xj, tm) 1
h22xu(xj, tm)
= 1h2
u(xj h, tm) 2u(xj, tm) + u(xj + h, tm)
ut(xj, tm) 1
k+t u(xj, tm)
=
1
ku(xj, tm + k) u(xj, tm).
40/57
Finite difference methodexplicit
-
8/3/2019 PDEs - Slides (4)
41/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method oflines
Euler method
Finite
differences forheat equation
ut (xj, tm) = uxx (xj, tm)j = 1, 2, . . . , n 1m = 0, 1, 2, . . .
.
Substitute difference approximations to derive the finitedifference method
1
k+t u
mj =
1
h22xu
mj
or1k
um+1j u
mj
= 1
h2
umj1 2um
j + um
j+1
s s s
j 1 j j + 1m
m + 1
Given umj1, um
j , um
j+1, um+1
j is given explicitly by
um+1j = um
j1 + (1 2)um
j + um
j+1, where = k/h2.
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Linear system of equations
-
8/3/2019 PDEs - Slides (4)
42/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method of
lines
Euler method
Finite
differences forheat equation
t t t
j 1 j j + 1
m
m + 1
For = k/h2
u
m+1
j = u
m
j1 + (1 2)u
m
j + u
m
j+1,
j = 1, 2, . . . , n 1
m = 0, 1, 2, . . .
This is a matrix-vector multiply at each time level:
um+11...
um+1j...
um+1n1
=
1 2 0.. .
.. .
1 2 . . .
. . .
0 1 2
um1...
umj...
umn1
.
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Consistency I
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8/3/2019 PDEs - Slides (4)
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Part IV
20401
Modelparabolic
problem
Uniqueness
Method of
lines
Euler method
Finite
differences forheat equation
Is the explicit approximation consistent?
Theorem (consistency)
The local error at the grid point (xj, tm),
Tmj =1
k
+t u(xj, tm) 1
h2
2xu(xj, tm) j = 1, 2, . . . , n,
is bounded from above:
Tmj
= O(k) + O(h2)
HOMEWORKYou can now try Problem 5
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Consistency I
-
8/3/2019 PDEs - Slides (4)
44/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method of
lines
Euler method
Finite
differences forheat equation
Is the explicit approximation consistent?
Theorem (consistency)
The local error at the grid point (xj, tm),
Tmj =1
k
+t u(xj, tm) 1
h2
2xu(xj, tm) j = 1, 2, . . . , n,
is bounded from above:
Tmj
= O(k) + O(h2)
HOMEWORKYou can now try Problem 5
44/57
Explicit Approximation II
-
8/3/2019 PDEs - Slides (4)
45/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method of
lines
Euler method
Finite
differences forheat equation
1 Is the explicit approximation stable?
It all depends on = r
0 0.5 10
0.2
0.4
0.6
0.8
1Explicit FD Scheme: r is 0.48
x
temperature
0 0.5 10
0.2
0.4
0.6
0.8
1Explicit FD Scheme: r is 0.52
x
temperature
45/57
-
8/3/2019 PDEs - Slides (4)
46/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method of
lines
Euler method
Finite
differences forheat equation
TheoremIf 1 2 0 1/2 then the explicit approximation
um+1j = um
j1 + (1 2)um
j + um
j+1,
is non-increasing:
maxj
um+1j max
j
umj . ()HOMEWORK
You can now try Problem 6
46/57
-
8/3/2019 PDEs - Slides (4)
47/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method of
lines
Euler method
Finite
differences forheat equation
TheoremIf 1 2 0 1/2 then the explicit approximation
um+1j = um
j1 + (1 2)um
j + um
j+1,
is non-increasing:
maxj
um+1j max
j
umj . ()HOMEWORK
You can now try Problem 6
47/57
-
8/3/2019 PDEs - Slides (4)
48/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method of
lines
Euler method
Finite
differences forheat equation
Proof.Let um = maxj |u
mj |.
|um+1j | = |um
j1 + (1 2)um
j + um
j+1|
|umj1| + |(1 2)um
j | + |um
j+1|
>0
um + (1 2) 0
um + ()>0
um
= um + (1 2)um + u
m = u
m .
This establishes () since |um+1
j | um for all j.
48/57
Implicit Approximation
-
8/3/2019 PDEs - Slides (4)
49/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method of
lines
Euler method
Finite
differences forheat equation
ut (xj, tm) = uxx (xj, tm)Approximate
uxx(xj, tm) 1
h22xu(xj, tm)
=1
h2u(xj h, tm) 2u(xj, tm) + u(xj + h, tm)
ut(xj, tm) 1
kt u(xj, tm)
=1
ku(xj, tm) u(xj, tm k).
where
2xu(xj, tm) = u(xj h, tm) 2u(xj, tm) + u(xj + h, tm)
t u(xj, tm) = u(xj, tm) u(xj, tm k).49/57
Implicit Approximation II
-
8/3/2019 PDEs - Slides (4)
50/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method of
lines
Euler method
Finite
differences forheat equation
ut (xj, tm) = uxx (xj, tm)j = 1, 2, . . . , n
1
m = 0, 1, 2, . . . .
1
kt u
mj =
1
h22xu
mj
1
kum
j um1
j =1
h2 um
j1
2umj + um
j+1
t
j 1 j j + 1
m 1
m
Thus umj1 + (1 + 2)um
j um
j+1 = um1
j ,
where = k/h2.50/57
Linear system of eqns
-
8/3/2019 PDEs - Slides (4)
51/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method of
lines
Euler method
Finite
differences forheat equation
t
j 1 j j + 1
m 1
m
For = k/h2
umj1 + (1 + 2)um
j um
j+1 = um1
j
j = 1, 2, . . . , n 1m = 0, 1, 2, . . .
This is a tridiagonal matrix solve at each time level:
1 + 2 0. . . . . .
1 + 2 . . .
. . .
0 1 + 2
um1...
umj...
umn1
=
um11...
um1j...
um1n1
51/57
Consistency
-
8/3/2019 PDEs - Slides (4)
52/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method of
lines
Euler method
Finite
differences forheat equation
Theorem
The local error at the grid point (xj, tm),
Tmj =1
kt u(xj, tm)
1
h22xu(xj, tm) j = 1, 2, . . . , n,
is bounded from above:Tmj = O(k) + O(h2).HOMEWORK
You can now try Problem 78
52/57
Consistency
-
8/3/2019 PDEs - Slides (4)
53/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method of
lines
Euler method
Finite
differences forheat equation
Theorem
The local error at the grid point (xj, tm),
Tmj =1
kt u(xj, tm)
1
h22xu(xj, tm) j = 1, 2, . . . , n,
is bounded from above:Tmj = O(k) + O(h2).HOMEWORK
You can now try Problem 78
53/57
Implicit Approximation II
-
8/3/2019 PDEs - Slides (4)
54/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method of
lines
Euler method
Finite
differences forheat equation
1 Is the implicit approximation stable?
Yes for any = r
0 0.5 10
0.2
0.4
0.6
0.8
1Implicit FD Scheme: r is 0.48
x
temperature
0 0.5 10
0.2
0.4
0.6
0.8
1Implicit FD Scheme: r is 0.52
x
temperature
54/57
-
8/3/2019 PDEs - Slides (4)
55/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method of
lines
Euler method
Finite
differences forheat equation
Theorem
The implicit approximation
umj1 + (1 + 2)um
j um
j+1 = um1
j
is non-increasing: maxj um+1j maxj umj . ()
55/57
Proof
-
8/3/2019 PDEs - Slides (4)
56/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method of
lines
Euler method
Finite
differences forheat equation
Proof.
Proof. Let um = maxj |um
j |.
| (1 + 2) >0
umj | = |umj1 + um
1j + umj+1|.
(1 + 2) |umj | = |um
j1 + um1
j + um
j+1|
|umj1| + |um1
j | + |um
j+1|
um + |um1
j | + um = 2u
m + |u
m1j |.
(1 + 2) um 2um + |u
m1k |
This establishes () since um |um1k | maxj |u
m1j |.
56/57
Summary
-
8/3/2019 PDEs - Slides (4)
57/57
Part IV
20401
Modelparabolic
problem
Uniqueness
Method of
lines
Euler method
Finite
differences forheat equation
1 We have looked at two finite difference approximations for
the heat equation. An explicit and an implicit method.2 The explicit method is easier to implement as it does NOT
involve solving a linear system of equations.
3 Both implicit and explicit methods are consistent with thesame order of accuracy (|Tm
j
| = O(k) + O(h2)). Secondorder accurate in space; first order accurate in time.
4 Stability depends on a step size condition for the explicitmethod ( = k/h2 1/2). The implicit method is alwaysstable.
5 Stable+consistency gives convergence for both methods.The implicit method is preferred in practice due to betterstability.
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