Finite difference method for 2-D heat equation

Praveen. Cpraveen@math.tifrbng.res.in

Tata Institute of Fundamental ResearchCenter for Applicable Mathematics

Bangalore 560065http://math.tifrbng.res.in/~praveen

January 27, 2013

1 / 23

Heat equation

• Partial differential equation in Ω = (0, 1)× (0, 1), µ > 0

ut = µ(uxx + uyy), (x, y) ∈ Ω, t > 0

u(x, y, 0) = f(x, y), (x, y) ∈ Ω

u(x, y, t) = g(x, y, t), (x, y) ∈ Ω, t > 0

• Space mesh of (Mx + 1)× (My + 1) points in Ω

∆x =1

Mx, xj = j∆x, j = 0, 1, . . . ,Mx

∆y =1

My, yk = k∆y, k = 0, 1, . . . ,My

2 / 23

FTCS scheme

D+t u

nj,k = µ(D+

xD−x u

nj,k +D+

y D−y u

nj,k)

D+t u

nj,k =

un+1j,k − unj,k

∆t, D+

xD−x u

nj,k =

unj−1,k − 2unj,k + unj+1,k

∆x2

D+y D−y u

nj,k =

unj,k−1 − 2unj,k + unj,k+1

∆y2

Define

rx =µ∆t

∆x2, ry =

µ∆t

∆y2

Define undivided finite differences

δ2xu

nj,k = unj−1,k − 2unj,k + unj+1,k, δ2

yunj,k = unj,k−1 − 2unj,k + unj,k+1

FTCS schemeun+1j,k = unj,k + rxδ

2xu

nj,k + ryδ

2yu

nj,k

or

un+1j,k = (1 + rxδ

2x + ryδ

2y)unj,k

3 / 23

FTCS scheme: maximum stability

un+1j,k = (1− 2rx − 2ry)unj,k + rx(unj−1,k + unj+1,k) + ry(unj,k−1 + unj,k+1)

If

rx + ry ≤1

2

thenmin(unj±1,k, u

nj,k±1, u

nj,k) ≤ un+1

j,k ≤ max(unj±1,k, unj,k±1, u

nj,k)

so that the scheme is stable in maximum norm. If ∆x = ∆y = h, then

2µ∆t

h2≤ 1

2=⇒ ∆t ≤ h2

4µ

This time step restriction is half the value in one dimension.

4 / 23

FTCS scheme: Fourier stabilityTake Fourier mode with wave-number (ξ, η) ∈ (−π,+π)× (−π,+π)

unj,k = unei(jξ+kη)

Thenun+1 = ρun

where

ρ(ξ, η) = 1 + 2rx(cos ξ − 1) + 2ry(cos η − 1)

= 1− 4rx sin2(ξ/2)− 4ry sin2(η/2)

For stability we need |ρ(ξ, η)| ≤ 1. Maximum value of ρ

ρ = 1 at (ξ, η) = (0, 0)

and minimum value of ρ

ρ = 1− 4rx − 4ry at (ξ, η) = (π, π)

Hence we have stability provided

rx + ry ≤1

2

5 / 23

BTCS scheme

un+1j,k = unj,k + rxδ

2xu

n+1j,k + ryδ

2yu

n+1j,k

or

(1− rxδ2y − ryδ2

y)un+1j,k = unj,k

Amplification factor Amplification factor

ρ =1

1 + 4rx sin2(ξ/2) + 4ry sin2(η/2)≤ 1

BTCS scheme is unconditionally stable.

6 / 23

Crank-Nicholson scheme

un+1j,k = unj,k + rxδ

2x

unj,k + un+1j,k

2+ ryδ

2y

unj,k + un+1j,k

2or

(1− 1

2rxδ

2x −

1

2ryδ

2y)un+1

j,k = (1 +1

2rxδ

2x +

1

2ryδ

2y)unj,k

Amplification factor

ρ =1− 2rx sin2(ξ/2)− 2ry sin2(η/2)

1 + 2rx sin2(ξ/2) + 2ry sin2(η/2)≤ 1

Crank-Nicholson scheme is unconditionally stable.

7 / 23

Peaceman-Rachford schemeUpdate solution in two steps

tn → tn +1

2∆t→ tn+1, un → un+ 1

2 → un+1

Step 1: Implicit in x

un+ 1

2

j,k − unj,k12∆t

=µ

∆x2δ2xu

n+ 12

j,k +µ

∆y2δ2yu

nj,k

or

(1− 1

2rxδ

2x)u

n+ 12

j,k = (1 +1

2ryδ

2y)unj,k

Step 2: Implicit in y

un+1j,k − u

n+ 12

j,k12∆t

=µ

∆x2δ2xu

n+ 12

j,k +µ

∆y2δ2yu

n+1j,k

or

(1− 1

2ryδ

2y)un+1

j,k = (1 +1

2rxδ

2x)u

n+ 12

j,k

8 / 23

Peaceman-Rachford scheme: Fourier stability

Step 1:[1 + 2rx sin2(ξ/2)]un+ 1

2 = [1− 2ry sin2(η/2)]un

Step 2:[1 + 2ry sin2(η/2))un+1 = [1− 2rx sin2(ξ/2)]un+ 1

2

Combining the two

ρ =un+1

un=

[1− 2rx sin2(ξ/2)][1− 2ry sin2(η/2)]

[1 + 2rx sin2(ξ/2)][1 + 2ry sin2(η/2)]≤ 1

The scheme is unconditionally stable.

9 / 23

Peaceman-Rachford scheme: Consistency

From Step 1

(1− 1

2rxδ

2x)u

n+ 12

j,k = (1 +1

2ryδ

2y)unj,k

Multiply on both sides

(1 +1

2rxδ

2x)(1− 1

2rxδ

2x)u

n+ 12

j,k = (1 +1

2rxδ

2x)(1 +

1

2ryδ

2y)unj,k

Operators on left commute

(1− 1

2rxδ

2x) (1 +

1

2rxδ

2x)u

n+ 12

j,k︸ ︷︷ ︸use Step 2

= (1 +1

2rxδ

2x)(1 +

1

2ryδ

2y)unj,k

Using Step 2

(1− 1

2rxδ

2x)(1− 1

2ryδ

2y)un+1

j,k = (1 +1

2rxδ

2x)(1 +

1

2ryδ

2y)unj,k

10 / 23

Peaceman-Rachford scheme: Consistency

Expand difference operators on both sides

un+1j,k − uj,k

∆t=

µ

∆x2δ2x

unj,k + un+1j,k

2+

µ

∆y2δ2y

unj,k + un+1j,k

2

−µ2∆t

4

δ2x

∆x2

δ2y

∆y2(un+1j,k − u

nj,k)

This is Crank-Nicholson scheme with an extra term. This term is O(∆t2

)since

δ2x

∆x2

δ2y

∆y2(un+1j,k − u

nj,k) = ∆t

(∂5u

∂t∂2x∂2y

)nj,k

+O(∆t∆x2 + ∆t∆y2 + ∆t2

)Hence the Peaceman-Rachford scheme is second order accurate in space andtime.

11 / 23

D’Yakonov Scheme

Peaceman-Rachford scheme

(1− 1

2rxδ

2x) (1− 1

2ryδ

2y)un+1

j,k︸ ︷︷ ︸u∗j,k

= (1 +1

2rxδ

2x)(1 +

1

2ryδ

2y)unj,k

Step 1: Solve for u∗

(1− 1

2rxδ

2x)u∗j,k = (1 +

1

2rxδ

2x)(1 +

1

2ryδ

2y)unj,k

Step 2: Solve for un+1

(1− 1

2ryδ

2y)un+1

j,k = u∗j,k

12 / 23

Approximate factorizationPeaceman-Rachford scheme is close to Crank-Nicholson scheme

(1− 1

2rxδ

2x −

1

2ryδ

2y)un+1

j,k = (1 +1

2rxδ

2x +

1

2ryδ

2y)unj,k

Factorise operator on left hand side

(1− 1

2rxδ

2x −

1

2ryδ

2y) = (1− 1

2rxδ

2x)(1− 1

2ryδ

2y)− rxry

4δ2xδ

2y

We cannot neglect the last term since it is O(∆t2

)and the scheme would only

be first order accurate

rxry4

δ2xδ

2yu

n+1j,k =

∆t2

4

(∂4u

∂2x∂2y

)n+1

j,k

+ higher order terms

We factorise the right hand side also

(1−1

2rxδ

2x)(1−1

2ryδ

2y)un+1

j,k = (1+1

2rxδ

2x)(1+

1

2ryδ

2y)unj,k+

rxry4

δ2xδ

2y(un+1

j,k − unj,k)︸ ︷︷ ︸

O(∆t3)

13 / 23

Peaceman-Rachford Scheme: Dirichlet BCStep 1 is

(1− 1

2rxδ

2x)u

n+ 12

j,k = (1 +1

2ryδ

2y)unj,k

What boundary condition to use for un+ 1

2

j,k at j = 0, Mx ?

Option I: Since un+ 1

2

j,k ≈ u(xj , yk, tn + 1

2∆t)

un+ 1

2

j,k = g(xj , yk, tn +

1

2∆t), j = 0,Mx

Option II: Using Step 1 and 2

un+ 1

2

j,k =1

2(1− 1

2ryδ

2y)un+1

j,k +1

2(1 +

1

2ryδ

2y)unj,k

At j = 0,Mx

un+ 1

2

j,k =1

2(1− 1

2ryδ

2y)g(xj , yk, t

n +1

2∆t) +

1

2(1 +

1

2ryδ

2y)g(xj , yk, t

n)

If BC do not depend on t, then un+ 1

2

j,k = g(xj , yk)14 / 23

D’Yakonov Scheme: Dirichlet BC

What boundary condition to use for u∗j,k at j = 0,Mx ? Unlike un+ 12 , we do

not know what u∗ represents. Instead use the numerical scheme itself; fromStep 2

u∗j,k = (1− 1

2ryδ

2y)un+1

j,k

Boundary condition for u∗j,k

u∗j,k = (1− 1

2ryδ

2y)g(xj , yk, t

n+1), j = 0,Mx

E.g., at j = 0

u∗0,k = −ry2gn+1j,k−1 + (1 + ry)gn+1

j,k −ry2gn+1j,k+1

15 / 23

Peaceman-Rachford Scheme: ImplementationStep 1: For j = 1, . . . ,Mx − 1 and k = 1, . . . ,My − 1

(1− 1

2rxδ

2x)u

n+ 12

j,k = (1 +1

2ryδ

2y)unj,k

For k = 1, . . . ,My − 1, solve tridiagonal matrix

−rx2un+ 1

2

j−1,k + (1 + rx)un+ 1

2

j,k − rx2un+ 1

2

j+1,k = (1 +1

2ryδ

2y)unj,k, j = 1, . . . ,Mx − 1

Step 2: For j = 1, . . . ,Mx − 1 and k = 1, . . . ,My − 1

(1− 1

2ryδ

2y)un+1

j,k = (1 +1

2rxδ

2x)u

n+ 12

j,k

For j = 1, . . . ,Mx − 1, solve tridiagonal matrix

−ry2un+1j,k−1 + (1 + ry)un+1

j,k −ry2un+1j,k+1 = (1 +

1

2rxδ

2x)u

n+ 12

j,k , k = 1, . . . ,My − 1

Remark: Total work scales as O (MxMy) and hence is optimal.16 / 23

Douglas-Rachford SchemeThis is an approximate factorised version of the BTCS scheme.

(1− rxδ2y − ryδ2

y)un+1 = un

Factorise LHS operator

(1− rxδ2y − ryδ2

y) = (1− rxδ2x)(1− ryδ2

y)− rxryδ2xδ

2y

We can throw away the last term since we will get first order time accuracy.But we can make a better approximation as follows.

(1− rxδ2x)(1− ryδ2

y)un+1 = un + rxryδ2xδ

2yu

n+1

(1− rxδ2x)(1− ryδ2

y)un+1 = (1 + rxryδ2xδ

2y)un + rxryδ

2xδ

2y(un+1 − un)︸ ︷︷ ︸O(∆t3)

Douglas-Rachford scheme is given by

(1− rxδ2x)(1− ryδ2

y)un+1 = (1 + rxryδ2xδ

2y)un

17 / 23

Douglas-Rachford Scheme: Two step scheme

Introduce an intermediate state u∗

(1− rxδ2x) (1− ryδ2

y)un+1︸ ︷︷ ︸u∗

= (1 + rxryδ2xδ

2y)un

Two-step process

(1− rxδ2x)u∗ = (1 + ryδ

2y)un

(1− ryδ2y)un+1 = u∗ − ryδ2

yun

We avoid fourth differences δ2xδ

2y. The equivalence is seen as follows.

(1− rxδ2x)(1− ryδ2

y)un+1 = (1− rxδ2x)u∗ − (1− rxδ2

x)ryδ2yu

n

= (1 + ryδ2y)un − (1− rxδ2

x)ryδ2yu

n

= (1 + rxryδ2xδ

2y)un

18 / 23

Douglas-Rachford Scheme: Fourier stability

Amplification factor

ρ(ξ, η) =1 + 16rxry sin2(ξ/2) sin2(η/2)

[1 + 4rx sin2(ξ/2)][1 + 4ry sin2(η/2)]≤ 1

The scheme is unconditionally stable.

19 / 23

Douglas-Rachford Scheme: Dirichlet BC

We need BC for u∗ at j = 0,Mx. Using Step 2 of the scheme

u∗ = (1− ryδ2y)un+1 − ryδ2

yun

This gives the boundary conditions

u∗j,k = (1− ryδ2y)gn+1

j,k − ryδ2ygnj,k, j = 0,Mx

20 / 23

Source termsHeat equation with a forcing term

ut = µ(uxx + uyy) + F (x, y, t)

Crank-Nicholson scheme, second order in time and space

(1− 1

2rxδ

2x −

1

2ryδ

2y)un+1 = (1 +

1

2rxδ

2x +

1

2ryδ

2y)un +

∆t

2(Fn + Fn+1)

We can do a factorization as before

(1− 1

2rxδ

2x)(1− 1

2ryδ

2y)un+1 = (1 +

1

2rxδ

2x)(1 +

1

2ryδ

2y)un +

∆t

2(Fn + Fn+1)

− rxry4

δ2xδ

2y(un+1 − un)︸ ︷︷ ︸O(∆t3)

Ignoring last term, the resulting scheme is second order accurate

(1− 1

2rxδ

2x)(1− 1

2ryδ

2y)un+1 = (1 +

1

2rxδ

2x)(1 +

1

2ryδ

2y)un +

∆t

2(Fn + Fn+1)

21 / 23

Source terms: Peaceman-Rachford SchemeUse the viewpoint of two step process: tn → tn + 1

2∆t→ tn+1

(1− 1

2rxδ

2x)un+ 1

2 = (1 +1

2ryδ

2y)un +

∆t

2Fn

(1− 1

2ryδ

2y)un+1 = (1 +

1

2rxδ

2x)un+ 1

2 +∆t

2Fn+1

To check the consistency of the scheme, eliminate un+ 12

(1− 1

2rxδ

2x)(1− 1

2ryδ

2y)un+1 = (1− 1

2rxδ

2x)(1 +

1

2rxδ

2x)un+ 1

2

+∆t

2(1− 1

2rxδ

2x)Fn+1

= (1 +1

2rxδ

2x)(1 +

1

2ryδ

2y)un +

∆t

2(Fn + Fn+1)

− rx∆t

2δ2x(Fn+1 − Fn)︸ ︷︷ ︸O(∆t3)

Hence the two-step scheme is second order accurate in time and space.22 / 23

Source terms: D’Yakonov SchemeStart with approximately factored Crank-Nicholson scheme

(1− 1

2rxδ

2x) (1− 1

2ryδ

2y)un+1︸ ︷︷ ︸

u∗

= (1 +1

2rxδ

2x)(1 +

1

2ryδ

2y)un +

∆t

2(Fn + Fn+1)

Two-step D’Yakonov scheme

(1− 1

2rxδ

2x)u∗ = (1 +

1

2rxδ

2x)(1 +

1

2ryδ

2y)un +

∆t

2(Fn + Fn+1)

(1− 1

2ryδ

2y)un+1 = u∗

Remark: We cannot spread out the source term into the two steps like this

(1− 1

2rxδ

2x)u∗ = (1 +

1

2rxδ

2x)(1 +

1

2ryδ

2y)un +

∆t

2Fn

(1− 1

2ryδ

2y)un+1 = u∗ +

∆t

2Fn+1

This scheme is only first order accurate in time.23 / 23