scientific computing i - module 6: the 1d heat equation

ScientificComputing I

Michael Bader

OutlinesPart I: Analytic Solutions ofthe 1D Heat Equation

Part II: Numerical Solutionsof the 1D Heat Equation

Part III: EnergyConsiderations

Scientific Computing IModule 6: The 1D Heat Equation

Michael Bader

Lehrstuhl Informatik V

Winter 2006/2007


Michael Bader




Part I: Analytic Solutions of the 1DHeat Equation

1 The Heat Equation in 1D

2 Analytic SolutionsAnalytic SolutionsA Family of Solutions – Fourier’s Method


Michael Bader




Part II: Numerical Solutions of the 1DHeat Equation

3 Numerical Solution 1 – An Explicit SchemeDiscretisationAccuracyNeumann Stability

4 Numerical Solution 2 – An Implicit SchemeImplicit Time-SteppingStability of the Implicit Scheme


Michael Bader




Part III: Energy Considerations

5 Energy and StabilityEnergy of the analytic solutionUniqueness and StabilityEnergy of the numerical solutionEnergy for the Implicit Scheme


Michael Bader

The Heat Equationin 1D

Analytic SolutionsAnalytic Solutions

A Family of Solutions –Fourier’s Method

Part I

Analytic Solutions of the 1D HeatEquation


Michael Bader




The Heat Equation in 1D

remember the heat equation:

Tt = κ∆T

we examine the 1D case, and set κ = 1 to get:

ut = uxx for x ∈ (0,1), t > 0

using the following initial and boundaryconditions:

u(x,0) = f(x), x ∈ (0,1)

u(0, t) = u(1, t) = 0, t > 0


Michael Bader




Computing Analytic Solutions

First steps:

try to find some solution of the PDE

try to satisfy boundary conditions(but not the initial condition)

Ansatz: Separation of Variables

assumption:

u(x, t) = X(x) ·T(t)

insert this assumption into the heat equation


Michael Bader




Separation of Variablesinsert u(x, t) = X(x) ·T(t) into PDE:

∂

∂ t(X(x) ·T(t)) =

∂ 2

∂x2 (X(x) ·T(t))

or X(x) ·Tt(t) = T(t) ·Xxx(x)

divide by X(x)T(t), and get:

Tt(t)T(t)

=Xxx(x)

X(x)

true for all x and all t, only if:

Tt(t)T(t)

=Xxx(x)

X(x)=−λ


Michael Bader




Transforming the PDE into two ODEsseparation of variables leads to:

Tt(t)T(t)

=Xxx(x)

X(x)=−λ

thus, we obtain two ODEs:

Xxx(x)+λX(x) = 0 X(0) = X(1) = 0, (1)

Tt(t)+λT(t) = 0 (2)

solve X(x)-part:

Xk(x) = sin(kπx) λk = (kπ)2,k = 1,2, . . .

solve T(t)-part (compare Model of Maltus):

Tk(t) = e−λkt = e−(kπ)2t


Michael Bader




Fourier’s Methodthe functions

uk(x, t) := Tk(t)Xk(x) = e−(kπ)2t sin(kπx), k = 1,2, . . . ,

solve the 1D heat equation PDEfor the initial and boundary conditions:

uk(0, t) = uk(1, t) = 0, t > 0

uk(x,0) = sin(kπx), x ∈ (0,1).

use Fourier sine series for initial condition:

f(x) =∞

∑k=1

ck sin(kπx).

and obtain solution for uk(x,0) = f(x):

u(x, t) =∞

∑k=1

cke−(kπ)2t sin(kπx),


Michael Bader




Fourier’s Method – A Recipe1 Find coefficients ck such that the initial

condition f(x) can be represented as

f(x) =∞

∑k=1

ck sin(kπx).

2 Verify that the solution candidate

u(x, t) := ∑k

cke−(kπ)2t sin(kπx)

converges to a well-defined function u3 Verify that u solves the differential equation

ut = uxx4 Verify that u satisfies the boundary conditions

u(0, t) = u(1, t) = 05 Verify that u satisfies the initial condition

u(x,0) = f(x)


Michael Bader

NumericalSolution 1 – AnExplicit SchemeDiscretisation

Accuracy

Neumann Stability

NumericalSolution 2 – AnImplicit SchemeImplicit Time-Stepping

Stability of the ImplicitScheme

Part II

Numerical Solutions of the 1DHeat Equation


Michael Bader


Accuracy

Neumann Stability



Numerical Solution 1 – DiscretisationDiscretisation similar to ODEs:

compute approximations

v(m)j ≈ u(xj, tm)

at grid points xj and time points tm:

xj := j ·h tm := m · τ,

approximate equation ut = uxx by

v(m+1)j −v(m)

j

τ=

v(m)j−1 −2v(m)

j +v(m)j+1

h2 (3)

for j = 1, . . . ,n−1, and m≥ 0.


Michael Bader


Accuracy

Neumann Stability



An Explicit Scheme

add initial and boundary conditions:

v(m)0 = v(m)

n = 0, for all m≥ 0,

v(0)j = f(xj), for j = 1, . . . ,n−1.

and obtain an explicit scheme:

v(m+1)j = v(m)

j +τ

h2

(v(m)

j−1 −2v(m)j +v(m)

j+1

)we can, step by step, compute all values v(m)

jfor all time steps (m),starting with the initial conditions v(0)

j = f(xj).

“explicit time stepping scheme”


Michael Bader


Accuracy

Neumann Stability



Accuracy of the Explicit Scheme

Observations:

first order accurate in τ:

ut(x, t) =u(x, t+ τ)−u(x, t)

τ+O(τ)

second order accurate in h:

uxx(x, t) =u(x−h, t)−2u(x, t)+u(x +h, t)

h2 +O(h2)

stability condition of the step size!(similar to ODE)

⇒ examine conservation of “energy”or Neumann Stability Analysis


Michael Bader


Accuracy

Neumann Stability



Solutions of the Explicit Scheme

explicit scheme:

v(m+1)j −v(m)

j

τ=

v(m)j−1 −2v(m)

j +v(m)j+1

h2

assumption: there are solutions of the form

u(m)j = (ak)

m sin(πkxj), where xj := jh. (4)

compare with exact solution:

uk(x, t) = e−(kπ)2t sin(kπx)

(ak)m ∼ e−(kπ)2t, thus (ak)

m should decrease⇒ |ak|< 1 necessary


Michael Bader


Accuracy

Neumann Stability



Solutions of the Explicit Scheme (2)

insert u(m)j = (ak)

m sin(πkxj) into

v(m+1)j −v(m)

j

τ=

v(m)j−1 −2v(m)

j +v(m)j+1

h2

leads to condition (see separate worksheet)

ak = 1+τ

h2 (2cos(πkh)−2) = 1− 4τ

h2 sin2(

πkh2

).

for stability: |ak|< 1 only if

4τ

h2 < 2 or τ <h2

2.


Michael Bader


Accuracy

Neumann Stability



Solutions of the Explicit Scheme (3)

in practice: numerical solution is of the form

v(m)j :=

n−1

∑k=1

ck(ak)m sin(πkxj).

stability, because |ak|< 1 for all k

the coefficients ck result from a sine transformof the initial values fj:

u(0)j = fj =

n−1

∑k=1

ck sin(πkxj).


Michael Bader


Accuracy

Neumann Stability



Numerical Solution 2 – An ImplicitScheme

apply implicit Euler:

v(m+1)j −v(m)

j

τ=

v(m+1)j−1 −2v(m+1)

j +v(m+1)j+1

h2

for j = 1, . . . ,n−1, and m≥ 0.

boundary conditions:

v(m)0 = v(m)

n = 0, for all m≥ 0,

initial conditions:

v(0)j = f(xj), for j = 1, . . . ,n−1.


Michael Bader


Accuracy

Neumann Stability



Implicit Time Stepping

solve implicit scheme for v(m+1)j :

v(m+1)j − τ

h2

(v(m+1)

j−1 −2v(m+1)j +v(m+1)

j+1

)= v(m)

j .

with the ratio r := τ/h2, we can write it as

−rv(m+1)j−1 +(1+2r)v(m+1)

j − rv(m+1)j+1 = v(m)

j

for j = 1, . . . ,n−1, and m≥ 0

solve a system of linear equations to obtainv(m+1)

j in every step


Michael Bader


Accuracy

Neumann Stability



System of Linear Equations:

the matrix of the linear system of equations isgiven by I+ rA, where A = tridiag(−1,2,−1).

system is tridiagonal, solving requires O(n)

operations.

solution:

v(m) = (I+ rA)−1v(m−1);


Michael Bader


Accuracy

Neumann Stability



Solutions of the Implicit Scheme

implicit scheme:

v(m+1)j −v(m)

j

τ=

v(m+1)j−1 −2v(m+1)

j +v(m+1)j+1

h2

again: insert assumed solutions

u(m)j = (ak)

m sin(πkxj), where xj := jh.

into implicit scheme, and obtain:

ak−1 = −4τ

h2 sin2(

πkh2

)ak

ak =

(1+

4τ

h2 sin2(

πkh2

))−1

.


Michael Bader


Accuracy

Neumann Stability



Solutions of the Implicit Scheme (2)

solutions of the implicit scheme:

v(m)j :=

n−1

∑k=1

ck(ak)m sin(πkxj).

with

ak =

(1+

4τ

h2 sin2(

πkh2

))−1

.

0 < ak < 1 independent of k and h

⇒ unconditionally stable


Michael Bader

Energy andStabilityEnergy of the analyticsolution

Uniqueness and Stability

Energy of the numericalsolution

Energy for the ImplicitScheme

Part III

Energy Considerations(Additional Material – Not

Compulsory)


Michael Bader





Energy of the Analytic Solutionu(x,y) a solution of

(uk)t = (uk)xx, x ∈ (0,1), t > 0

uk(0, t) = uk(1, t) = 0, t > 0

uk(x,0) = f(x), x ∈ (0,1).

define the energy of the solution:

E(t) :=∫ 1

0u2(x, t)dx

for conservation of energy, analyse

E′(t) :=ddt

∫ 1

0u2(x, t)dx


Michael Bader





Energy of the Analytic Solution (2)

E′(t) =∫ 1

0

∂

∂ tu2(x, t)dx =

∫ 1

02u(x, t)ut(x, t)dx

= 2∫ 1

0u(x, t)uxx(x, t)dx

= 2 [u(x, t)ux(x, t)]10−2∫ 1

0(ux(x, t))2 dx

= −2∫ 1

0(ux(x, t))2 dx ≤ 0

Therefore:E(t)≤ E(0) (energy never increases)compare to initial condition u(x,0) = f(x):∫ 1

0u2(x, t)dx = E(t)≤ E(0) =

∫ 1

0f2(x)dx


Michael Bader





Corrollaries

assume: both u1(x, t) and u2(x, t) are solutionsfor initial conditions f1(x) and f2(x)

let w(x, t) := u1(x, t)−u2(x, t), then

wt(x, t) = (u1)t(x, t)− (u2)t(x, t)

= (u1)xx(x, t)− (u2)xx(x, t) = wxx(x, t)

w(0, t) = w(1, t) = 0

w(x,0) = u1(x,0)−u2(x,0) = f1(x)− f2(x)

therefore, w(x, t) is a solution of the heatequation for initial conditionfw(x) = f1(x)− f2(x)


Michael Bader





Corrollary 1 – Uniqueness

if f1 = f2, then fw(x) = 0

energy is decreasing:∫ 1

0(u1−u2)2(x, t)dx =

∫ 1

0w2(x, t)dx

≤∫ 1

0(f1− f2)2(x)dx = 0

therefore:∫ 1

0w2(x, t)dx ≤ 0 ⇔ w = 0 ⇔ u1 = u2.

proof of uniqueness of the solution!


Michael Bader





Corrollary 2 – Stability

now: f2 = f1 + ε (ε small), then∫ 1

0w2(x, t)dx ≤

∫ 1

0(f1− f2)2(x)dx

=∫ 1

0ε(x)dx ≤ ‖ε‖ ·1

therefore:if ε is small, the difference between u1 and u2

also has got to be small, i.e.small perturbations in the initial conditionslead to small perturbations in the solution.

stability estimate for the solution!


Michael Bader





Energy of the Numerical Solution

we introduce the discrete energy:

E(m) := hn−1

∑j=1

(v(m)

j

)2.

we would like to show that:

E(m+1) ≤ E(m) for m≥ 0.

thus, we will compute ∆E(m) := E(m+1)−E(m):

∆E(m) = hn−1

∑j=1

((v(m+1)

j

)2−

(v(m)

j

)2)


Michael Bader





Energy in the Explicit Scheme

lengthy computation (see separate worksheet)leads to stability condition:(

E(m+1)−E(m))≤ 0, or E(m+1) ≤ E(m)

is only correct, if

τ

h2 ≤12

or τ ≤ h2

2.

otherwise:increasing energy (physically incorrect), leadsto large oscillations in the solution


Michael Bader





Energy for the Implicit Schemeanalyse discrete energy

E(m) := hn−1

∑j=1

(v(m)

j

)2= h

(v(m)

j

)Tv(m)

j .

change of energy in each time step:

∆E(m) = h((

v(m+1))T

v(m+1)−(

v(m))T

v(m)

)= h

((Mv(m)

)TMv(m)−

(v(m)

)Tv(m)

)= h

((v(m)

)TMTMv(m)−

(v(m)

)Tv(m)

)= h

(v(m)

)T (MTM− I

)v(m)


Michael Bader





Energy for the Implicit Scheme (2)

energy for the implicit scheme:

∆E(m) = h(

v(m))T (

MTM− I)

v(m)

examine eigenvalues of matrix MTM− Iresult:

all eigenvalues < 0therefore ∆E(m) ≤ 0implicit scheme stable for any τ and h

scientific computing i - module 6: the 1d heat equation

Documents