Introduction to Environment System Modeling

Department of Environment Systems,Graduate School of Frontier Sciences,

the University of TokyoMasaatsu AICHI

（Finite difference method～Diffusion equation & advection equation）

Example problem 1 of the diffusion equation

=

, 0 = 0.2

0, = 10.15, = 0.2

x [m]

C

= 1 − 0.8 ∗ 2

Initial concentration 0.2

x

Boundary concentration 1.0diffusivity D=2.5×10-9[m2/s]

time t=86400 [s]

Diffusion equation

Initial condition

Boundary condition

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0 0.05 0.1 0.15

Implicit method• Governing equation

− = ∆∆∆ − ∆

∆ − ∆ − ∆∆ − ∆

= ∆ − ∆ + ∆= ∆∆ , = 1 + ∆∆ , = ∆∆

• Boundary condition∆ = 1, ∆ − ∆ = 0• Initial condition = 0.2• Unconditionally stable

Implicit method• Governing equation

− = 2∆∆∆ − ∆

∆ + ∆ − ∆ −∆ + ∆ − ∆

= ∆ − ∆ + ∆= ∆

∆ ∆ ∆ , = 1 + ∆∆ ∆ ∆ + ∆ ∆ , = ∆

∆ ∆ ∆• Boundary condition

= 25 + 286400 , = 25• Initial condition = 25• Unconditionally stable

Example problem 2 of the diffusion equation

=

, 0 = 25

0, = 25 + 2864000.1, = 25

Initial condition

Boundary condition

= 0.6 / ∙1 2

1. concrete

= 1900 /= 0.88 / ∙

2. Insulator

Case A: Inner insulation

3cm 3cm

Case B: Outer insulation

2 1

3cm 3cm

= 0.04 / ∙= 50 /= 2.1 / ∙

RoomOutsideair

Calculate the time-dependent heat flux into the room

RoomOutsideair

Solving as simultaneous equations

•

∆ + ∆ + ∆ = 0∆ = 1∆ − ∆ = 0

↔ 1 0

⋱ ⋱ ⋱

−1 1

∆∆

⋮ ∆∆∆

=1−⋮−− 0

Tridiagonal matrix

= 2∆∆ 1∆ + ∆= 2∆∆ 1∆ + ∆= −1 − 2∆∆ 1∆ + ∆ + 1∆ + ∆

Solver for tridiagonal matrix

⋱ ⋱ ⋱

→

0 0⋱ ⋱ ⋱ 0 0

10 1⋱ ⋱ ⋱0 10 10 1

Is divided to lower and upper triangular matrices

= = → = /,

= + = → = −, , = → = / i=2～n

Sequential calculation finds all , , recursively

Solver for tridiagonal matrix0 0⋱ ⋱ ⋱ 0 0

10 1⋱ ⋱ ⋱0 10 10 1

⋮ = ⋮

⋮ =10 1⋱ ⋱ ⋱0 10 10 1

⋮Let

0 0⋱ ⋱ ⋱ 0 0⋮ = ⋮

= /= − / i=2～n

Sequential calculation finds all recursively

Solver for tridiagonal matrix10 1⋱ ⋱ ⋱0 10 10 1

⋮ = ⋮== − ′ i=N-1～1

Sequential calculation finds the solution recursively for all

Summarizing the procedure,

= = /= = − = / (i=2～n)

= /= − / (i=2～n)

== − ′ (i=N-1～1) (If , , , and are not necessarily stored in

the memory after calculation, they can be overwritten by , , , and ）

Solving the simultaneous equations with programming

• As a result the solution of simultaneous equations of tridiagonal matrix is necessary– This is applicable for various problems other than

the problem appeared in this lecture.– It is not effective if we write same code for each

occasion– Reusing the code is important.– Save the code as a module that solves the

tridiagonal matrix

Solver for tridiagonal matrixA core part of fortran code （overwriting the original matrix）

b(1)=b(1)/a(1)do i=2,n

a(i)=a(i)-c(i)*b(i-1)b(i)=b(i)/a(i)

enddor(1)=r(1)/a(1)do i=2,n

r(i)=(r(i)-c(i)*r(i-1))/a(i)enddodo i=n-1,1,-1

r(i)=r(i)-b(i)*r(i+1)enddo

LU decomposition

Forward substitution

Backward substitution

module m_Thomasimplicit nonecontains

subroutine Thomas(c,a,b,r)implicit nonereal(8), intent(inout) :: c(:),a(:),b(:),r(:)integer :: i,n

n=size(r)b(1)=b(1)/a(1)do i=2,n

a(i)=a(i)-c(i)*b(i-1)b(i)=b(i)/a(i)

enddor(1)=r(1)/a(1)do i=2,n

r(i)=(r(i)-c(i)*r(i-1))/a(i)enddodo i=n-1,1,-1

r(i)=r(i)-b(i)*r(i+1)enddo

returnend subroutine

end module m_Thomas

Module m_Thomas

Subroutine Thomas1D array c,a,b,r is input and r returns the solution

⋱ ⋱ ⋱ ⋮ = ⋮

Sub program

Main programprogram diffusionuse m_Thomasimplicit nonereal(8), parameter :: pi=4*atan(1d0)real(8), allocatable :: c(:),a(:),b(:),r(:),T(:,:),x(:),dx(:),k(:),rho(:),gm(:)real(8) :: dt,period,timeinteger :: i,n,nunit

open(newunit=nunit,file="input_diffusion.txt")read(nunit,*) n,dt,periodallocate(c(n),a(n),b(n),r(n),T(n,0:1),x(n),dx(n),k(n),rho(n),gm(n))do i=1,n

read(nunit,*) x(i),dx(i),k(i),rho(i),gm(i),T(i,0)enddoclose(nunit)

Main program diffusionm_Thomas is connected to this code

Calculation of π

(continued)

Main program（continued）open(newunit=nunit,file="output_diffusion.txt")time=0do while(time<period)

time=time+dtdo i=2,n-1

c(i)=2*dt/(rho(i)*gm(i)*dx(i))/(dx(i-1)/k(i-1)+dx(i)/k(i))a(i)=-1-2*dt/(rho(i)*gm(i)*dx(i))*(1d0/(dx(i+1)/k(i+1)+dx(i)/k(i))+1d0/(dx(i-1)/k(i-1)+dx(i)/k(i)))b(i)=2*dt/(rho(i)*gm(i)*dx(i))/(dx(i+1)/k(i+1)+dx(i)/k(i))r(i)=-T(i,0)

enddoa(1)=1; b(1)=0; r(1)=25+5*sin(2*pi*time/86400)c(n)=0; a(n)=1; r(n)=25call Thomas(c,a,b,r)T(1:n,1)=r(1:n)T(1:n,0)=T(1:n,1)write(nunit,*) time,-k(n)*(T(n,1)-T(n-1,1))/(x(n)-x(n-1))

enddoclose(nunit)

stopend program diffusion

Example problem 2 of the diffusion equation

=

, 0 = 25

0, = 25 + 2864000.1, = 25

Initial condition

Boundary condition

= 0.6 / ∙1 2

1. concrete

= 1900 /= 0.88 / ∙

2. Insulator

Case A: Inner insulation

3cm 3cm

Case B: Outer insulation

2 1

3cm 3cm

= 0.04 / ∙= 50 /= 2.1 / ∙

RoomOutsideair

Calculate the time-dependent heat flux into the room

RoomOutsideair

Training

• Calculate with MS-EXCEL– Space discretization：Refine the boundary

– Explicit method• What is the limitation of dt?• Calculate for the total period of 150,000 s

– Implicit method• Calculate for the total period of 150,000 s with dt=1500s

– Make a graph for heat flux vs time

Wall 1 Wall 2

22 3 3 4 4 4 3 3 2 22 3 3 4 4 4 3 223

Property boundary WallーAirAirーWall

ADVECTION EQUATION

Case of 1D advection equation

+ = 0Governing equation

− + −2Δ = 0Central difference equation

Analytical solution is a transport of initial distribution

One common ratio is negative and less than -1→The solution of this discretized equation is oscillating and diverges

Solving the characteristic equation for the recursive formula for three adjacent terms,

= −1 ± 1 + 4 = 2Δ

x+1-11 2 +1-1

0

∆

18

Stabilization for the discretized equation（upwind method）

Case of 1D advection equation

+ = 0Governing equation

− + −Δ = 0Upwind finite differential equation

Analytical solution is a transport of an initial distribution

Common ratios are positive and smaller than 1→The solution of this equation is smooth and finite.

= Δ

x+1-11 2 +1-1

0

∆

> 0= +

= 1 +19

Advantage and disadvantage of stabilization

Δx = 1 [m], = 1 [ / ], Δt = . [s]+ [ ]

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0 50 100 150

オリジナル

10m100m1km2km

Too smooth

20

original

Training

• Calculation with MS-EXCEL– Implicit central difference− + −2Δ = 0– Implicit first order upwind method− + −Δ = 0 ≥ 0

− + −Δ = 0 < 0

Why the upwind method finds a smooth solution?− + −Δ = 0 ≥ 0− + −Δ = 0 < 0

can be reformulated to+ − = 0

Central finite difference Diffusion term with finite difference method

• Artificial diffusion term is appeared in the equation• The diffusion coefficient is Δ /2, and then, the artificial diffusion effect is

important in rapid flow and large mesh• The countermeasure is refining the meshes→Computational effort

CIP method (Takewaki et al., 1985)• and are main variables ( = )

• Advection equation is separated to + = 0 and + = 0• Let = + + + , and the coefficients are determined by the upwind mesh

values and in the mesh i-1

= , = , = 2 +Δ − 3 −Δ , = +

Δ − 2 −Δ• Interpolating and • Because the solution of advection equation is a Δt transport of the distribution, Δt before,

the solution for the next step is obtained by upwind values based on the above interpolation functions

Δt

− Δt

Δt

Approximation by piecewise 三次 function

Δt

23

Advantage and disadvantage of stabilization

Δx = 1 [m], = 1 [ / ], Δt = . [s]+ [ ]

CIP method is much better but unphysical oscillation appears

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0 50 100 150

オリジナル

10m100m1km2km

24

original

Problem of advection term• There is no killer solution for the exact simulation of advection term.• Currently, it is important to choose an appropriate method for the

objective– Smoothness and stability→Upwind method family– Accuracy of the location of advection front→Lagrangian method family

• Advanced method– Higher order upwind method（cancelling numerical diffusion）– Higher order backward tracking（tracking upstream accurately）– Adaptive mesh refinement（Mesh is adaptively refined to follow the

advection front. ）– etc…

• For beginner, a first-order upwind method is used in this lecture because it is simple and stable. After this, you can study advanced methods.

Advection-diffusion-attenuationequation

, 0 = 0

0, = 1∞, = 0

Initial concentration 0

x

Boundary concentration 1.0 Diffusion coefficientD=2.5×10-5[m2/s]

time t=86400 [s]

Governing equation

Initial condition

Boundary condition

= − −

= 10 [m/s]

= ln 286400 [ ]

= 12 erfc − + 42

+ erfc + +2

Exact solution for comparison

= 4 +erfc = 1 − erf complementary error function

Practice

• Calculate with MS-Excel– Implicit first-order upwind finite difference

method− + −2Δ− Δ2 + − 2 +Δ + = 0