verification of gimp with manufactured solutions

Verification of GIMP with Manufactured Solutions

Philip Wallstedt – philip.wallstedt@utah.eduJim Guilkey – james.guilkey@utah.edu

Verification: Necessary Versus Sufficient

Eyeball Norms – no obvious error• Not predictive: you already know the answer

Symmetry – some coding mistakes exposed• Many mistakes are symmetric

Compare to existing code (Finite Element)• Existing code solves different problems• Existing code has unverified accuracy• When differences are found, are they errors or not?

Experimental results – scattered data shows same trends• Lack of data• Differences don’t show what’s wrong with code

Known Solutions to PDE’s• No solutions for large deformation

We need a better way: the Method of Manufactured Solutions (MMS)

Recently proposed as ASME standard

“V&V 10 - 2006 Guide for Verification and Validation in Computational Solid Mechanics”

Sufficient, not just necessary, if we test all modes:

• Boundary conditions

• Non-square cells and particles

• Time integration algorithms

• Shape functions

Each mode must be tested, but not all in the same test. Once a mode has “passed”, then further testing not needed.

Rate of convergence is very sensitive to errors and can be applied to individual pieces of a method

EXACTppu XuXx )(

N

L pp

max1 pL max

Displacement error compares current config to reference.

Average error Worst Error

Body Force on a 1D Bar

Zero Displacement

u(0,t) = 0

Zero Stress(L,t) = 0

Body Force b(x,t)

Body Force on a 1D Bar

Given

Momentum

Neo-Hookean Constitutive Model

Constitutive Model with assumptions: 1D, Poisson = 0

abσ

IFFIσ TlnJ

JJ

FFσ

1

2

E

Find displacement u(x) – in general this cannot be done.

Start with the answer and reformulate backwards

Given Displacement

1D Neo-Hookean with Poisson’s ratio = 0

Momentum

Solve for Gravity

)(Xu

abP 00

FFP

1

2

E

Now we just take derivatives . . .

Pab 0

1

What answer (displacement field) do we start with?

The chosen displacement field(s) must:

• exercise all features of the code (large deformation, translation, rotation, Dirichlet and Neumann boundaries

• be “smooth enough” – sufficiently differentiable in time and space

• Conform to assumptions made by the method. For GIMP this means zero normal stress at material boundaries.

For the 1D rod a parabolic form should work:

)(2210 tAXcXccu

Constants for the 1D bar

Zero stress at X = L

0 0.5 10

0.1

0.20.2

0

u X( )

L0 X

)(000 2210 tAccc

)(21

1)(21

2

1

20)(

2121 tALcc

tALccE

FF

ELP

Zero displacement at X = 0

Scale displacement at L

)()( 2210 tALcLcctA

)()2(

2tA

L

XLXu

Choose a convenient time function A(t)

t πcos2.0

)2(

02

E

L

XLXu

Trig function:

• Easy to differentiate

• Stays close to un-deformed shape

• Tests ability to preserve energy

• Can be made self-similar in time – same number of time steps per period, regardless of problem stiffness.

But other functions work just fine, provided:

• problem always has sufficient particles per cell

• displacement field is well-behaved (for us A(t) > -1/2)

A detour and a review: reference versus current configuration

Particles stationary in reference configuration

Grid stationary in current configuration

X = 4.75x = 5.2

X = 5x = 5

X = ?x = 5

X = 4.75x = 4.75

Reference

Current

Why manufacture solutions in the reference configuration?

)()2(

)( 20

0 tAL

xLxxu

– Because boundaries move in the current configuration.

How find the current length and apply boundary?

)())(2)((

)( 20

0000 tA

L

LLLLLLLuL

This is icky. We can avoid recursive / implicit definitions like the above by using the reference configuration.

Reference Configuration vs Current Configuration

Reference Configuration

“Total Lagrange”

Current Configuration

“Updated Lagrange”

Momentum

Deformation Gradient

Neo-Hookean

1D,

Poisson = 0

abσ

IFFIσ TlnJ

JJ

)(

1)(

2 xFxFσ

E

IFFFFP T11ln J

)(

1)(

2 XFXFP

E

abP 00

Stress Transformation: σFP 1J

)()( XuIXF 1)()( xuIxF

Return to the 1D Bar: Take Derivatives

Given

Displacement

Deformation Gradient

Divergence of Stress

Solve for b(X)

)()2(

2tA

L

XLXu

AL

XLF

2

)(21

AAL

XL

L

EP

2

22

)(211

AAL

XLEAXLX

Lb

2

20

2

)(211)2(

1

1D Bar: Restate the Problem

Zero Displacement

u(0,t) = 0

Zero Stress(L,t) = 0

Body Force

AAL

XLEAXLX

Lb

2

20

2

)(211)2(

1

)()2(

)(2

tAL

XLXXu

The answer is:

Solve with GIMP where

tπcos2.0)(

0E

tA

Now we can measure convergence under large deformation – the kind of problem MPM/GIMP is designed to solve

0.000001

0.00001

0.0001

0.001

0.01

0.1

10 100 1000mesh size

max

err

or

GIMP

UGIMP

Skeptical Questions

1st Piola-Kirchoff is neither objective nor fully Lagrangian – doesn’t that cause problems?

MPM is a first-order, fully non-linear method. It can’t be expected to agree with your manufactured solution due to its non-linearity.