Day2

Results

• Results = Postprocessing section

• Working with Data Sets

• Getting Derived Values

• Cross section plots

• Revolving solution

• Report = Exporting results

Menubar options

Point evaluationCut lines Cut planes • Slice

• Isosurface

• Volume

• Surface

• Line (edge)

• Arrow (in volume)

• Streamline

• Animate (transient

and parametric only)

Data Sets – Solution and Selection

Choose desired

geometric level

Select the desired

geometric entities

Results can be

visualized only on

the desired

geometric entities

1

2

3

4

Derived Values

• Results > Derived Values

• Allows you to evaluate and visualize numerical data

• Automatically creates table under Results > Table

Integrate any spatially

dependent variable on

appropriate geometric

entity level

Evaluates a variable at a point

or geometric vertices

Evaluates global variables,

lumped parameters and built-in

physical constants

Cross sections – Cut line

1Create a cut line

by selecting any

two points

2See the

coordinates under

Results > Data

Sets > Cut Line

3

Check the plot

along that cut line

in an automatically

created new 1D

Plot Group

Cross sections – Cut plane

1Create a cut plane

2

See the

coordinates

under

Results >

Data Sets >

Cut Plane

3

Check the plot along

that cut plane in an

automatically created

new 2D Plot Group

Revolving solution

2D solution

3D solution

See model file:

Example_results_revolve

Exporting results

Exported data can

be read into a

spreadsheet

Hands-on #6: Tuning Fork

• Structural model

• Eigenfrequency analysis

• Mode shapes

Equation-based Modeling in COMSOL

• PDEs, ODEs, and Algebraic Equations

• Use COMSOL’s GUI – no need for subroutines

• Why solve equations?

– Purely mathematical problems

– Combining predefined physics interface with additional equations

– Represent physics which is not available in the predefined options

• What does it mean to “solve” a PDE?

– Well-posed problems; existence, uniqueness, and smoothness

– Garbage in -> garbage out

• COMSOL uses the finite element method (FEM) to numerically

approximate solutions

Equations interface: Classical PDEs

• Built-in options for more common PDE forms

• Easy to use if you can fit your problem into any of these forms

• Available for 1D, 2D and 3D geometries

0 u

0)( uutt

uu )(

fu

Equations interface: More options

• Open-ended options for solving almost any PDE

• Options to enter in either differential or integral (weak) form

• Available for 1D, 2D and 3D geometries

• ODEs and DAEs can be solved on 0D

Coefficient Form

ru

guuc )( non boundary

mass damped mass

diffusion

convection

source

convection

absorption

source

fauuuuct

ud

t

ue aa

)(

2

2

In the domain

Example: Poisson’s equation

1 u

0u

inside subdomain

on subdomain boundary

Implies c = f = 1 and all other coefficients are 0.

mass damped mass

elastic stress initial/thermal stress

body force (gravitation)

a

ae

d

c

c

u

2

2( )a ae d c a f

t t

u uu u u u

Coefficient Form: Structural Analysis

density

damping coefficient

stress

Stiffness, “spring constant”

This is how

COMSOL solves

other physics

Information on equation interface

• PDEs can have as many variables (u1, u2, u3,…)

• Variables can be real or complex (u1 = a+j*b)

• Coefficients can be almost anything

- Scalar (numerical values)

- Vector or tensors (for multiple variables)

- Parameters and Variables (user-defined expressions)

- Functions of spatial coordinates and time (non-linear in space and/or

time)

- Functions of other physics variables (when PDE is coupled to another

predefined physics in a multiphysics model)

- Functions of variables being solved for (nonlinear in solution space)

- Derivatives of variables (highly nonlinear PDEs, also a way to

implement higher order equations)

The weak form is a variational statement of the problem in

which we integrate against a test function.

This has the effect of relaxing the problem; instead of finding

an exact solution everywhere, we are finding a solution that

satisfies the strong form on average over the domain.

Modeling using weak Form

Modeling using weak Form

Example1: Modeling using weak form Squeeze

film damping Velocity v

h

vdr

dphr

dr

d

r

12

1 3

Boundary Conditions:

0dr

dpat r = 0 and p = 0 at r = r0

pambient

p < pambient

lubricant

Example2: Damped harmonic oscillator

0)( tpkuucumForces

00 0,0 uuuu

k

p(t)

u

c m

00,10

0

uu

kuuu

dunderdampe ,2

overdamped ,2

damped critically ,2

k

k

k

Damped harmonic oscillator

overdamped, β=100

underdamped, β=1

critically damped, β=10

beta is a parameter

• 2nd order ODE

• 0D model

• Transient + Parametric

See model file:

Example_harmonic_oscillator

Example 2: CFD + Global equation

Set up and solve this Incompressible Steady State

Navier-Stokes fluid flow problem, (ρ=1e3, η=1e-3):

4 mm1 mm

Wall, no slip

Wall, no slip

Inlet, Pressure,

no viscous stress

P = p_in

Outlet, Pressure,

no viscous stress

P = 0 Pa

Solution, exhibiting

entrance effect

Integration coupling

operator on inlet:

spf.rho*u*1[m]

AvdfluxmassA

The mass flux integral is taken per unit

length out of the modeling plane 155

CFD + Global equation

mass_flux is a variable

• Combine CFD + equation

• Evaluate required pressure to get

desired mass flux at the inlet

• 2D model

• Stationary + Parametric

See model file:

Example_CFD_massflux

mass_flux_required is a parameter

Hands-on #7: Spherically Symmetric Transport

• Spherical symmetry

represented as 1D

• Equation-based model

• Time-dependent

Linear and Nonlinear FEA

• FEA formulation

• Linear problem

• Nonlinear problem

• Suggestions on convergence

A very simple finite element problem...

p= 2 Nk2 = 4 N/m k3 = 4 N/m

u1 u2u3

We can write a force balance on each node:

Node 1

c

k1(u3-u1)

Node 2

k3(u3-u2)-k2(u2-u1)

k1 = 6 N/m

k2(u2-u1)

Constraint Force

Node 3

p

-k1(u3-u1)

-k3(u3-u2)

Write the force balance as three equations:

However, we now have three equations in four unknowns, c, u1, u2, u3

k1(u3-u1) + k2(u2-u1) = c

k2(u2-u1) - k3(u3-u2) = 0

k1(u3-u1) + k3(u3-u2) = p

(k2)u2 + (k1)u3 = c

(k2+k3)u2 + (-k3)u3 = 0

(-k3)u2 + (k1+k3)u3 = p

Algebra

Constraint Equations

System Equations

k1(u3 - 0 ) + k2(u2 - 0 ) = c

k2(u2 - 0 ) - k3(u3 - u2) = 0

k1(u3 - 0 ) + k3(u3 - u2) = p

Three equations, three

unknowns: c,u2, u3

But we also have a constraint, u1=0

Can now solve the problem

(k2)u2 + (k1)u3 = c

(k2+k3)u2 + (-k3)u3 = 0

(-k3)u2 + (k1+k3)u3 = p

cu

ukk

3

2

12

pu

u

kkk

kkk 0

3

2

313

332

Solve the system

equations first

Ku=b

u=K-1b

k1= 6 N/m

k2=k3= 4 N/m

p=2 N

250.0

125.0

3

2

u

u

Can also solve the

constraint equations,

if desired

2250.0

125.064

Linear and nonlinear problems

A linear problem must have all three of these properties:

1) Applying zero loads results in a zero solution

2) Doubling the magnitude of the load doubles the magnitude of the solution

3) Changing the sign of the load changes only the sign of the solution

load

solution

load

solution

Linear Non-Linear

What makes a problem non-linear?

• Material properties that depend upon the solution, k(u)

• Loads that depend upon the solution, b(u)

u

k, b

Linear

u

k(u), b(u)

Weakly Non-linear

u

k(u), b(u)

Strongly Non-linear

u

k(u), b(u)

“Outrageously”

Non-linear

There are no clear distinctions

between these three cases

Not all non-linear problems have a (unique) solution

load

solution

No solution

load

solution

No unique solution

Two very simple finite element examples

k = 4 N/m

u

p = 2 N

Force balance on node:

f (u) = p - ku = 0

f (u) = 2 - 4 u = 0

f (u)

usolution= 0.5

u

k = exp(u) N/m

u

p = 2 N

Force balance on node:

f (u) = p - k u = 0

f (u) = 2 - exp(u) u = 0

u

f (u)

usolution≈ 0.853

First, solve the linear problem

f (u) = 2 - 4 u = 0

f (u)

usolution=0.5

Algorithm:

1) Start at a point, e.g. u0 = 0

2) Find the slope, f’(u0)

3) Solve the problem:

0

00

' uf

ufuusolution

5.04

20

solutionu

(1)(2)

(3)

i

iii

uf

ufuu

'1 This is a single Newton-Raphson iteration:

For a linear problem, the starting point does not matter

Solving a FE system matrix is equivalent to taking

a single Newton-Raphson iteration

f (u) = 2 - 4 u = 0

0

00

' uf

ufuusolution

5.04

20

solutionu

pu

u

kkk

kkk 0

3

2

313

332

0

00

3

2

313

332

u

u

kkk

kkk

puf

f (u) = b - Ku, u0 = 0

K

b

K

Kub0

uf

ufuu

0

0

00

'solution

bKu1solution

4 u = 2

The non-linear problem is solved the same way

Algorithm:

1) Start at a point, e.g. u0 = 0

2) Take repeated Newton-

Raphson iterations

3) Terminate when:

f (u) = 2 - exp(u) u

u

f (u)

toleranceuuff iiii /&/ 11

Similarly, if we have a non-linear system matrix

f (u) = b(u) - K(u)u

iiiii

i

iii uuKubuSu

uf

ufuu

1

1'

f′(u) = S(u)

Iterate until: toleranceff iiii uu /&/ 11

p= 2k2 = exp(u2-u1) k3 = 4

u1 u2u3

k1 = 6 + (u3-u1)2

S(u) is known as the JACOBIAN

The possibility of convergence of a non-linear

problem is dependent upon the starting point, u0

Non-linear problems have a radius of convergence

f (u) = 2 - exp(u) u

u

f (u)

Non-convergent

if starting on this

side of the line

u0= -1

Finding a good starting point, u0, can

improve convergence

f (u) = 2 - exp(u) u

u

f (u)

slow

convergence

fast

convergence

• It is not strictly possible to

define “slow” or “fast”

convergence

• Finding a good starting point is

a matter of experience and luck

A damping factor, α, is used to improve convergence

u

f (u)

i

iii

uf

ufuu

'1

choose (α < 1) & (α<αprevious)

while ( |fi+1| > |fi| )

recompute fi+1(ui+1)

end

|fi|

undamped |fi+1|

intermediate |fi+1|

final |fi+1|

f (u) = 2 - exp(u) u

A better solution is to ramp up the load

f (u)

f (u) = 2 - exp(u) u k = exp(u) N/m

u

p = 2 N

u

Recall that:Load

Load

f (u) = 1 - exp(u) u = 0

• Almost all problems have a

zero solution at zero load

• Ramping up the load is a

physically reasonable approach

Ramping up the non-linearity can also work

f (u)

f (u) = 2 - exp(u) u = 2 - ku

u

Non-linearity

f (u) = 2 - 1u

f (u) = 2 - {(1-0.5)+(0.5exp(u)}u

f (u) = 2 - exp(u) u

Identify non-linearity

- kNL = exp(u)

Linearize around chosen u0

- kLIN = exp(u0=0) = 1

Use an intermediate value

- k = (1-χ)kLIN+χkNL

- χ starts at 0

Use intermediate solution as new u0

Ramp χ from 0 to 1

Example - Heat flow through a square

T=0K

q = 100 W/m2

k, Thermal conductivity

1x1m square

Build this 2D model in Heat Transfer,

Save this model, we will build upon this

Mesh

Solution

Tinitial=0K

Insulated

Insulated

Study the following cases:

T

k

T

k = 0.1+exp(-(T/25[K])^2)[W/(m*K)]

= 1[W/(m*K)]

T

k = (1+T/200[K])[W/(m*K)]

T

= (0.1+10*(T>25[K]))[W/(m*K)]k

Questions:

• How many iterations did each case take? Was damping used?

• What does the temperature and thermal conductivity solution look like?

• Did they all converge? Why not?

• Try solving case #3 with an initial temperature of 50[K], how does this

affect the solution? What about the other cases?

– Set back to T=0 afterwards

Ramp up the heat load and monitor peak temperature

T

k =(1-T/200[K])[W/(m*K)]

Qin

Tmax

?

Why does this happen?

Use qflux as a parameter and ramp it up

as range(50,5,110)

The stationary adaptive mesher

Initial mesh

Final mesh

Meshes > Mesh 2

Go back to the square model, case 3

k = 0.1+exp(-(T/25[K])^2)[W/(m*K)]

Add Study 1 > Solver Sequences > Solver

Sequence 1 > Stationary 1 > Adaptive mesh

refinement

Triangular and tetrahedral

elements ONLY are subdivided

based on the L2 norm error

estimate.

The Study node

Assign the parameter, mesh

and physics to be used in this

study

This section contains more

information about the solver

settings

Stationary solver settings

…Solver Configurations > Stationary Solver 1

Stop condition for

Newton iterations

• COMSOL can inspect the

stiffness matrix and decide whether

the problem is linear or nonlinear

• We can also explicitly choose

linear or nonlinear

Solver settings – more information

…Solver Configurations > Stationary Solver 1 > Fully Coupled 1

Linear system of equations is

solved using the specified solver

Stop condition for

Newton iterations

Can choose different

damping methods

Convergence Criteria

Non-linear solver algorithm

Linearized about an initial guess point

Calculate relative error E:

If decrease

Stop iterations when weighted relative error < relative tolerance

0)( Uf

0U

00

' UfUUf UUU 01

is the damping factor 10

10

' UfEUf

1__ iterationcurrentiterationcurrent EE

Revisit: k = (0.1+10*(T>25[K]))

Instead of using a “jump” in k, use a smoothing function

Now try: k = 0.1+DK*flc2hs(T-25,STEP)[W/(m*K)]

Try: DK, STEP = 0 8 2.5 4 5 2 7.5 0.5 10 0.25

Achieving convergence for non-linear problems

Assuming that the problem is

well-posed, try solving it

1) Check the initial condition

2) Use parametric solver to ramp load

3) Use parametric solver to ramp non-linearity

4) Refine the mesh

Does the problem have a steady-state solution?

Are there additional effects? Check all assumptions.

Perform a mesh

refinement study.

Try a finer mesh

and check that the

solution is similar.

DONE

Not converging?

Not converging?

Not converging?

Not converging?

Not converging?

Converging?

Not converging?

Converging?

Hands on #8: H-Bend Waveguide

• Electromagnetic Waves

• Frequency domain study

Solving the Linear System Matrix

• Linear system matrix

• Direct methods

• Iterative methods

• Suggestions on solver selection

Let’s take another look at the system equations

0Kubuf

0

00

3

2

313

332

u

u

kkk

kkk

p

Define a quadratic function: r(u) = b·u-½Ku·u

u2

u3r(u)

SolutionThe solution, f(usolution) = 0, is

the point where r(u), is at a

minimum

Finding the minimum of a quadratic function

r (u) = 2u2 - 3u + 1

r (u)

u

Newton’s method: 0

00

u

uuu

r

rsol

For a quadratic function, this

converges in one iteration, for any u0

r′(u) = 4u - 3

r″(u) = 4

75.04

3040

0

00

ur

uruusol

u0 = 0

Via our choice of r(u), this reduces to the

same equation from the previous section:

r′(u) = b - Ku

r″(u) = - K

r(u) = b·u-½Ku·u

K

Kubu

u

uuu

0

0

0

00

r

rsol

0u 0

bKK

b0u

1sol

usol

Finding the minimum of the quadratic function,

r(u), by the direct method means solving u=K-1b

• This is known as Gaussian Elimination, or LU factorization

• The numerical algorithms are beyond the scope of this course,

but they have the following important properties:

– For 3D, requires O(n2)-O(n3) numeric operations, where n is the length of u

– Robustness of the algorithm is only very weakly dependent upon K

• The direct solvers in COMSOL are:

– MUMPS: fast, required for cluster computation

– PARDISO: fast, multi-core capable

– SPOOLES: slow, uses the least memory, multi-core capable

Introduction to iterative methods for finding the

minimum of a quadratic function, a naive approach

1) Start here

2) Search along coordinate axis

3) Find the minimum along that axis

4) Repeat until converged

This iterative method requires only

that we can repeatedly evaluate r(u)

as opposed to the direct methods,

which get to the minimum in one

step by evaluating [r″(u)]-1

The numerical values in K can affect the algorithm

u2

u3r(u)

Quadratic

surface

r(u) = b·u-½Ku·u

Condition

number = 1

Condition

number = 2

Condition

number = 10The higher the condition number the more

numerical error creeps into the solution

Ill-conditioned matrices require more iterations

Numerical error becomes significant

A better iterative method for finding the minimum:

The Conjugate Gradient (CG) method

3) Find the minimum along that vector

The CG method requires that we can

evaluate r(u), r′(u) and r″(u)

CG does NOT compute K-1

2) Initially, find the gradient vector

1) Start here

4) Find the conjugate gradient vector*

5) Repeat 3-4 until converged

*) See e.g. Wikipedia, or “Scientific Computing” by Michael Heath

Condition number matters for iterative methods

Numerical error becomes significant

for ill-conditioned matrices

A matrix with a condition number of 1

will converge in one iteration, (rare!)

All iterative methods in COMSOL are some

variation upon the CG method

Conjugate Gradient, GMRES, FGMRES and BiCGStab

The details of these algorithms are beyond the scope of this course

All these methods make use of PRECONDITIONERS

The system equation, Ku = b, is multiplied by a preconditioner matrix, M,

to improve the condition number

Exercise: Show that

the best possible

preconditioner is the

matrix M=K-1

Ku = b MKu = Mb

What you need to know about iterative solvers

• They converge in at most n iterations (good)

• Solution time is O(n1-n2) (good)– Solution time depends on condition number

• Memory requirements are O(n1-n2) (very good)

• They are less robust that the direct solvers (neutral)– Convergence depends upon condition number

– An ill-conditioned problem is often set up incorrectly

• Different physics require different iterative methods (bad)– This is an ongoing research topic

– Often cannot solve two physics with the same solver

– Improvements in methods are ongoing

– We have tried to find the best combination of iterative solvers and preconditioners for many physics, to find these settings, read the manuals or open a new model file, select a space dimension of 3D and the physics you want to solve

Solver selection for 1D and 2D problems

• Begin with the default solver, MUMPS, PARDISO or SPOOLES

• If running out of memory:

– Use MUMPS or PARDISO out-of-core

– Upgrade your computer, 64-bit & more memory

– Make sure that your problem is not overly complicated, typical 2D models

should solve easily on a 32-bit computer with 2GB of RAM

• If you are seeing differences between the solvers, check the

problem very carefully, this usually indicates a mistake in the model

Solver selection for 3D problems

• Set up a linear problem first

• Use the default solver settings for the physics you are solving

• If you run out of memory, upgrade to 64-bit and more RAM

• Monitor the memory requirements as you grow the problem size

• Setup a non-linear problem only after you have successfully solved

a linear problem

FEA formulation in Multiphysics problems

• Interpreting multiphysics FEA

• Fully-coupled and Segregated Solvers

• Suggestions on convergence of multiphysics problems

Consider two simple physics, on the same geometry

q=100 W/m2

T=0K

All other surfaces

insulated

0

0

100

0

2

T

KT

Tk

Tk

m

Wn

in the domain

right boundary

left boundary

other boundaries

fT = KTuT - bT

Heat Transfer Electric currents

All other surfaces

insulated

V=1VV=0V

0

0

1

0

V

VV

VV

V in the domain

right boundary

left boundary

other boundaries

fV = KVuV - bV

k = 1 W/mK σ = 1S/m

This part of the matrix is zero because: ∂fV / ∂uT = 0

This part of the matrix is zero because: ∂fT / ∂uV = 0

How to solve this linear problem

fT (uT) = KTuT - bT

fV (uV) = KVuV - bV

V

T

V

T

V

T

b

b

u

u

K0

0Kuf

Combine the two linear system equations:

bSub

b

u

u

uf0

0ufuf

V

T

V

T

VV

TT

Solve this linear problem with a

single Newton-Raphson iteration: bKuSuu

0

1

0sol

Now, consider a coupled multiphysics problem

σ = (1-0.001×T) S/m

The temperature solution now affects the

material properties of the electrical problem

fT = KTuT - bTfV = KV(uT)uV - bV

q=100 W/m2

T=0K

All other surfaces

insulated

All other surfaces

insulated

V=1VV=0Vk = 1 W/mK


How to solve this coupled multiphysics problem

fT (uT) = KTuT - bT

fV (uV) = KV(uT)uV - bV

Take the two system equations:

buuKb

b

u

u

ufuf

0ufuf

V

T

V

T

VVTV

TT

This coupled problem is solved via

Newton-Raphson iterations:

iiii

iiiii

ufuufuu

buuKuSuu

1

1

1

1

Combine these into a single system equation:

This coupled multiphysics problem is solved in the exact

same way that a non-linear single physics problem is solved

May no longer use the iterative solvers, since they are tuned

for single physics, and often can not handle the coupled terms

q=100 W/m2

T=0K

All other surfaces

insulated

V=1VV=0VQ(V)

k = 1 W/mK σ(T)

Consider more couplings between the physics

fT = KTuT - bT(uv) fV = KV(uT)uV - bV

Volumetric resistive

heating due to current

ubuuKb

ub

u

u

ufuf

ufufuf

V

VT

V

T

VVTV

VTTT

iiii

iiiiii

ufuufuu

ubuuKuSuu

1

1

1

1

Same equation

as before


σ(T,V)

Consider all possible couplings between the physics

q(T,V)

V=V(T,V)

fT = KT(uV,uT)uT - bT(uV,uT)

k(T,V)

),( VTQvol

ubuuKuub

uub

u

u

ufuf

ufufuf

TVV

TVT

V

T

VVTV

VTTT

,

,

iiii

iiiiii

ufuufuu

ubuuKuSuu

1

1

1

1

fV = KV(uV,uT)uV - bV(uV,uT)

V=V(T,V)T=T(T,V)

General Newton-

Raphson multi-

physics iteration:

Different two-way couplings have different

computational requirements

ubuuKuub

uub

u

u

ufuf

ufufuf

TVV

TVT

V

T

VVTV

VTTT

,

,

If one or both of these are zero, then the memory requirements are less

If the couplings are “weak” then the problem converges faster

σ(T,V)

q(T,V)

V=V(T,V)

k(T,V)

),( VTQvolV=V(T,V)

T=T(T,V)

Definitions of various types of couplings

• One-way coupled– Information passes from one physics to the next, in one direction

• Two-way coupled– Information gets passed back and forth between physics

• Load coupled– The results from one physics affect only the loading on the other physics

• Material coupled– The results from one physics affect the materials properties of other physics

• Non-linear coupled– The results of one physics affects both that, and other, physics

• Fully coupled– All of the above

• Weakly coupled– The physics do not strongly affect the loads/properties in other physics

• Strongly coupled– The opposite of weakly coupled

It is possible to solve multiphysics problems in a

segregated sense, solving each physics separately

TVV

TVT

V

T

VVTV

VTTT

uub

uub

u

u

ufuf

ufufuf

,

,

Assume that these are approximately zero and ignore them

initialize uT,i, uV,i

do {

uT,i+1= uT,i+αST(uT,i, uV,i)-1bT(uT,i, uV,i)

uV,i+1= uV,i+αSV(uT,i+1, uV,i)-1bV(uT,i+1, uV,i)

i=i+1

}

while ( not_converged )

α is damping

Solving in a segregated sense has some advantages

V

T

V

T

VVTV

VTTT

b

b

u

u

ufuf

ufufuf

V

T

V

T

VV

TT

b

b

u

u

uf0

0ufuf

Less memory to store matrices:

Less memory to solve:

iiiiii ubuuKuSuu

1

1

uT,i+1= uT,i+αST(uT,i, uV,i)-1bT(uT,i, uV,i)

uV,i+1= uV,i+αSV(uT,i+1, uV,i)-1bV(uT,i+1, uV,i)

Jacobian is exactly 2 times smaller in degrees of freedom

The optimal iterative solver can be used for each physics

If the problem is strongly coupled, then the segregated approach

will not work, and a direct solver is often necessary since the

iterative solvers are not tuned for a general Jacobian matrix.

Achieving convergence for multiphysics problems

• Set up the coupled problem and try solving it with a direct solver

• If it is not converging:

– Check initial conditions

– Ramp the loads up

– Ramp up the non-linear effects

– Make sure that the problem is well posed (this can be very difficult!)

• If you are running out of memory, or the solution time is very long:

– Use the segregated solver and select the optimal solver (direct or iterative) for

each physics, or group of physics, in the problem. FOR 3D, START HERE!

– Upgrade hardware

– Try the PARDISO out-of-core solver

• Perform a mesh refinement study

An uncoupled problem

J=10 A/m2

VL=0TL=0

All other surfaces insulated

Set up an Inward Current Flow

on the right side

k=1 W/mK σ=1 S/m


Identical to previous problem

q=100 W/m2


Consider the following multi-physics couplings:

Volumetric

Heat Load

Thermal

Conductivity

Electric

Conductivity

Normal

Current Iterations

Solution

Times (sec)

Linear 0 1 1 10 1 3.572

Non-linear

uncoupled0 1-(T/10000) 1-(V/1000) 10 2 9.422

One-way load

coupled

Joule

Heating1 1 10 3 27.846

Two-way load

coupled

Joule

Heating1 1 10+(T/100) 3 29.703

Fully coupledJoule

Heating1-(V/1000) 1-(T/10000) 10+(T/100) 3 29.469

Thermal problem Electric problem

Try yourself: heat and current flow through a

unit cube

J=10 A/m2

VL=0TL=0


Set up an Inward Current Flow

on the right side

k=1 W/mK σ=1 S/m


Identical to previous problem

Set up and solve this coupled problem

q=100 W/m2


Q(V) W/m3

Try yourself: Ramp up the applied current,

and monitor peak temperature

exp(-T/600) exp(-T/600)

Jin

Tmax

?

• Use the parametric solver

• Ramp up inward current density range(0,2,20)

• Plot the temperature distribution.

• What happens?

• Why?

Thermal

Conductivity

Electric

Conductivity

1 1Case 1

Case 2

Hands on exercises: Instead of current

density apply a voltage

exp(-T/600) exp(-T/600)

V

Tmax

?

• Ramp up applied voltage range(0,20,140)

• How high can it go?

• Why?

Thermal

Conductivity

Electric

Conductivity

1 1Case 1

Case 2

Same as before

Try boundary layer mesh with 50 layers where needed.

Keep mesh size to fine.

Using the segregated solver

Study > Solver Configurations > Solver

1> Stationary Solver1>Segregated

Can add multiple

Segregated steps as

required

Segregated step

Choose the

variable/s that you

want to solve in this

segregated step

Choose the linear

system solver and

the damping method

for each segregated

step

Study > Solver Configurations > Solver

1> Stationary Solver1>Segregated

When in doubt, set each physics to run to

convergence at each iteration

initialize uT,i, uV,i

do {

do {

non-linear solution for uT,i+1

use uT,i and uV,i as initial conditions

}

do {

non-linear solution for uV,i+1

use uT,i+1 and uV,i as initial conditions

}

i = i+1

}

while ( not_converged )

These settings are equivalent

to solving the complete non-

linear problem at each

iteration.

Recap: Working with solvers

• A problem could be Linear or Nonlinear

• The linear system of equations could be solved using a Direct

Solver or an Iterative Solver

• A multiphysics problem could be solved using a Fully-Coupled

Approach or Segregated Approach

You cannot model everything

• You can never model the

entire universe

• Boundary conditions

represent the outside

• Two types: physical walls and

artificial boundaries

• Use artificial boundaries to

model only the region of

interest

Model

Everything else

BC

BC

BC BC

Something else

Infinite Elements• Currently in ACDC and Heat Transfer

• Draw Rectangle, 2 Large Circles,

– Add Outer Inf Elem Geom

• Subdomain > Make Circles Not Active and

infinite elements (aluminum)

• Rectangle Boundaries Thermal Insulation

• Left Circle: T = 300[K] Right Circle T = 273[K]

• Compute

• subdmain – make active infinite elements

• Subdomain – Inf Elem Tab – Match Matls

• Inf Elems >Cartesian,

• compute Hide Inf Element Subdomains

See model file:

Example_Infinite Elements

Perfectly Matched Layers

Absorbing Subdomain

• Absorbs incident waves without reflection

• Artificial boundary for non-enclosed spaces

• Used instead of radiation boundary condition

• Make thickness at least one wavelength with

atleast 5 elements

PML

• Circle 1: R= 8 Center: (0,2)

• Circle 2: R=10 Center: (0,2)

• Circle 3: R= 4 Center: (0,0)

• Union C1+C2

• Difference all

• f =100 Hz c =343 m/sec (air)

• L=c/f = 3.4 Mesh Size = 3.4/5

• Inner Circle Acceleration BC, a0=1Sound Hard Wall Radiation - Cylindrical PML - Cylindrical

See model file: Example_PML

Hands-on #9: Piezoacoustic Transducer

• 3-physics: Piezoelectric

material (structural +

electrical) + Acoustics

• Uses symmetry

• Perfectly Matched Layer

• Custom mesh

• Frequency domain

• mm scale geometry

Time-dependent problems

• Time-dependent formulation

• Tolerance and other attributes of time-dependent solver

• Suggestions on convergence of time-dependent problems

Introduction to time-dependent solutions

bKuufu

)(

t

QTkt

TCp

ttttt

ttbuK

uuu

1

ttttt t ubuKu 1

1111

ttt

tt

ttbuK

uuu

111 tttt tt buuKI

Governing Equation

Time-dependent finite element formulation

Explicit time-stepping

Implicit time-steppingFaster

More stable

Slower

Simple

Backward Difference Formula (BDF) order

111 tttt tt buuKI A polynomial is fit through n

previous points to predict

properties at t+1

Here, the BDF order is n = 3

t+1

k(t)

tt-1t-2

• Keep Max order

to 2 for stable

Scheme

• Higher order

means more

accuracy but

smaller time steps

COMSOL picks its own optimum timesteps

These are the output

timesteps, the times

you want to see the

solution at.

They do not affect the

internal timesteps.

The internal timesteps

are controlled by the

TOLERANCES.

Absolute and Relative Tolerance

Solutions within

Relative tolerance

Solutions within

Absolute tolerance

t

u

t

u

True solution

t

uAt solutions near zero,

the relative tolerance

cannot be used, since it

would result in very

small timesteps

t

u

Tighten both the

absolute and relative

tolerance until you

are satisfied with the

results over time.

The absolute tolerance can be different for

each solution variable

Temperature:

Absolute tolerance: ±0.1K

Voltage:

Absolute tolerance: ±0.001V

Often need to set this when

solving multi-physics problems

Other timestepping options

Free - time stepping method

chooses time steps freely

Intermediate - force the

time stepping method to

take at least one step in

each subinterval of the

times specified

Strict - force the time

stepping method to take

steps that end at the times

specified

Selecting the physics at

startup will usually load

the appropriate settings.

For wave type problems,

Generalized-alpha is

often better.

BDF order, set max and

min to 2 and 1 for wave

problems.

More timestepping options

Specify for which steps

the solutions should be

stored

Make sure not to impose any “ringing” loads,

unless they are really there

Load

time

Response

time

A reasonable starting point is necessary

• The load applied at t = 0 must agree with the initial condition

• It is helpful to a have a load with time derivative zero at t = 0

• The Physics > Initial Values node is used to set initial conditions

• The Solver Configurations > Dependent Variables > Initial Valuesarea can be used to overwrite that information

Most common mistake:

Initial conditions for fluid flow: u = 0

Velocity/Pressure inlet condition

that does NOT ramp up from zero

Solution:

Use flc2hs or a smoothed step

function to ramp load from an initial

conditiont

u

t

b

u(tinitial) = [K]-1b(tinitial)

Convergence of a transient problem

t

u

Outside envelop is absolute

and relative tolerance

True solution

(unknown)

Observe the solution with respect to time and

monitor the magnitude of the changes as the

tolerances are increased.

The problem is converged when you say it is.

The mesh must be fine enough with respect to

space and time

Transient 1-D slab

Tfixed

x

Tinitq

T

increasing

time

Try yourself:

1) Heat transfer in a unit square model

2) Change the analysis type to Transient

3) Set material properties: ρ=1, Cp=1

4) Solve transiently for 1 second (default)

5) Store all timesteps taken by solver

6) Examine results at initial timesteps

There are two ways to minimize this problem

Modify the mesh

Use boundary layer

meshing to better resolve

the step in the solution

Modify the physics

Use a smoothed step

function to gradually ramp

up the heat load:

flc1hs(t-0.01,0.01)

t

qq

Solving transient problems

• Begin by assuming the problem is well posed, try solving it by

ramping all loads up from consistent initial conditions

– Tighten the tolerances and see if you get a converging answer

• If it is not converging:

– Refine the mesh

– Smooth out abrupt differences in loads and properties as reasonable

– Try linearized properties

• Some non-linear properties may require a very fine mesh and small time-step

– Check the physics carefully

• If the solution takes a long time, or runs out of memory

– Use the time-dependent segregated solver using all previous guidelines

– Upgrade hardware

Hands-on #10: Star Chip

• Laminar flow

• Multiple pulsating inlets

with phase difference

• Swept mesh

• Time-dependent boundary

conditions

• Setting absolute tolerance

Summary

• COMSOL graphical user interface

• Variables, functions, model couplings

• Geometry creation, sequencing, parametrization, CAD import

• Handling built-in and user-defined materials

• Meshing techniques

• Postprocessing techniques

• PDE interface

• Linear and nonlinear problems

• Direct and iterative solvers

• Fully-coupled and segregated approach

• Time-dependent study

End of Slides

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