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Analytical and numerical solutions of dissipative systems

Analytical and numerical solutions of dissipativesystems

R. Kovacsin collaboration with

Peter Van, Tamas Fulop, Gyula Grof, Matyas Szucs

Department of Energy Engineering, BMEDepartment of Theoretical Physics, Wigner RCP,

andMontavid Thermodynamic Research Group

Budapest, Hungary

October 11, 2018.

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Outline

Origin of dissipative systems:non-equilibrium thermodynamics

Structure of partial differential equations (PDE):parabolic vs. hyperbolic

Examples: heat conduction vs. kinetic theory

Benchmark: solve these PDEs with real boundary conditions

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Thermodynamics: physical aspects

Stability: direction of physical processes

Constitutive equations: material behavior

Universality: independent of the material type

Realization: II. law of thermodynamics

Entropy existsEntropy is a potential function of state variables(e.g. T , p, and so on)

Thermodynamic stability ⇐⇒ entropy is concave

Increasing entropy ⇒ entropy production is non-negative

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Thermodynamics: mathematical aspects

Resulting PDEquations:

Parabolic vs. hyperbolic ⇒ propagation speed

Structure: compatibility with other theories?

Boundary conditions: physically admissible solutions

Solution method: numerical or analytical?

Feedback: real or artificial phenomena?

Validation through EXPERIMENTS!

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The mathematics is the light.5 / 54


Motivation to extend the classical theory I.

Heat conduction

Paradox of the Fourier equation: infinite speed ofpropagation, parabolic type equation (”a”: thermal diffusivity)

∂tT = a∂xxT

First modification (1958): Maxwell-Cattaneo-Vernotte (MCV)equation → hyperbolic type, finite propagation speed →second sound, temperature: around 1-20 K

τq∂ttT + ∂tT = a∂xxT

TEST: heat pulse experiments

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Heat pulse experiments I.

Arrangement for solids

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Heat pulse experiments II.

Narayanamurti et al.: Experimental proof of second sound in Bi (’74)

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Heat pulse experiments III.

Further propagation mode at low temperatures!Presence of second sound and ballistic propagation!

Jackson, Walker andMcNelly (1968):NaF experiments

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Heat conduction: propagation modes

Diffusive propagation → Fourier: ∂tT = a∂xxT

Damped wave propagation (“Second sound”) →Maxwell-Cattaneo-Vernotte: τ∂ttT + ∂tT = a∂xxT

Ballistic propagation: goes with speed of sound.

Continuum theory: elastic wave propagation.Kinetic theory: phonon hydro, non-interacting particles,particle-wall interaction

An unified continuum theory is obtained!Speed of sound ⇒ mechanical coupling!

Leading theory: phonon hydrodynamics (kinetic theory)

Analogous with rarefied gases!

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Theory?

Goal:SIMPLE

INDEPENDENT OF MICRO / MESO / MACROSTRUCTURE

EASY TO USE AND IMPLEMENT

→ NON-EQUILIBRIUM THERMODYNAMICSWITH INTERNAL VARIABLES

See also: Berezovski - Van: Internal Variables in Thermoelasticity(2017, Springer)

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About the internal variables

Non-equilibrium description

Nonlocal extensions in space and time → generalizations

Rheology: Kluitenberg-Verhas bodyDual internal variables: beyond the Newton equationWave propagation: coupling between mechanical and thermaleffects

Non-equilibrium thermodynamics

Extension of entropy density and its fluxPre-assumption: only the tensorial orderII. law → evolution equation for internal variableCan be eliminated: does not require further physical knowledge⇒ Universality

Final (and simple) benchmark: EXPERIMENTS

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Classical approach

Fourier’s law (heat conduction only)

entropy density: s = se(e), e = cT

entropy current: J i = qi/T

local equilibrium

Balance: internal energy → ρc∂tT +∇ · q = 0,Entropy production:σs = ρs + ∂iJ

i = ρ (∂ese) + ∂i(qi/T

)=

��− 1T ∂iq

i+ ��1T ∂iq

i +qi∂i1T = qi∂i

1T ≥ 0

Constitutive equation in 1D: q = −λ∂xT .Together with the energy balance: ∂tT = a∂xxT .

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Generalizations of entropy density and entropy flux

Extended Thermodynamics:

entropy density: s(e, qi ) = se(e)− m12 qiq

i

entropy current: J i = qi/T

qi : heat flux ≡ vectorial internal variable

a hope of hyperbolic equations

Constitutive equation in 1D:τ∂tq + q = −λ∂xT .

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Kinetic theory: phonon hydrodynamics

Instead of solving the Boltzmann eq. (6+1 D)

Momentum series expansion + truncation closure

∂u〈n〉∂t

+n2

4n2 − 1c∂u〈n−1〉

∂x+ c

∂u〈n+1〉

∂x=

0 n = 0− 1

τRu〈1〉 n = 1

−(

1τR

+ 1τN

)u〈n〉 2 ≤ n ≤ N

It requires at least N=30 momentum equation to approximate thereal propagation speeds.

Coupling between different tensorial order quantities!

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Generalizations of entropy density and entropy flux

Non-equilibrium thermodynamics with current multiplier

entropy density: s(e, qi ,Q ij) = se(e)− m12 qiq

i − m22 QijQ

ij

entropy current: J i = bi j qj

qi : heat flux ≡ vectorial internal variable

Q jk ≡ tensorial internal variable: pressure!

bi j : current multiplier

→ coupling between qi and Q ij ⇒ CONSEQUENCE:

parabolic equations → simplified to hyperbolic ones

LET US SEE THE EXAMPLE!

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Generalization of heat conduction

Rigid body → ρ = const.

Entropy production: σs = s + ∂iJi ≥ 0 in 1 spatial dimension:(

b − 1T

)∂xq + (∂xb −m1∂tq) q − (∂xB −m2∂tQ)Q + B∂xQ ≥ 0

Linear relations between thermodynamic fluxes and forces, isotropy:

m1∂tq − ∂xb = −l1q,m2∂tQ − ∂xB = −k1Q + k12∂xq,

b − 1

T= −k21Q + k2∂xq.

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Generalization of heat conduction

Rigid body → ρ = const.

Entropy production: σs = s + ∂iJi ≥ 0 in 1 spatial dimension:(

b − 1T

)∂xq + (∂xb −m1∂tq) q − (∂xB −m2∂tQ)Q + B∂xQ ≥ 0

Linear relations between thermodynamic fluxes and forces:

m1∂tq − ∂xb = −l1q,m2∂tQ − ∂xB = −k1Q + k12∂xq,

b − 1

T= −k21Q +��k2∂xq.

Compatibility withkinetic theory,hyperbolic eq.

⇒ SAME structure as 3 momentum eqs. of phonon hydro.

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Ballistic-conductive system, tested on NaF experiments!

ρc∂tT + ∂xq = 0,

τq∂tq + q + λ∂xT + κ∂xQ = 0,

τQ∂tQ + Q + κ∂xq = 0.

Properties of the structure

Thermo-mechanical coupling:

Q ij → current density of the heat flux → pressure

Ballistic-conductive:τqτQ∂tttT + (τq + τQ)∂ttT + ∂tT = a∂xxT + (κ2 + τQ)∂txxT

Hyperbolicity is checked: eigenvalues of Aij , F ≡ 0

∂tui + Aij∂xu

i + B ijui = F ,

±√κ2+τQ√τqτQ

; 0→ strict hyperbolicity19 / 54


Solution: numerical method I.

Initial conditions

All fields are homogeneous at t=0.

Boundary conditions ⇐⇒ discretization method

Only for the field of heat flux ⇐⇒ given by the experiments

q(t, x = 0) =

{1− cos(2π · t

tpulse) if 0 < t ≤ tpulse

0 if t > tpulse

Shifted fields:One goes from x = 0to x = 1, the others

shifted by ∆x2 .

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Solution: numerical method II.

Explicit finite differences: quick, requires stability conditions !

Forward in time → first order only :(

(Weak) Consistency is checked: lim ∆t,∆x → 0⇒ ∂t , ∂x

Stability conditions: Neumann + Jury conditions

Neumann: φnj = ξne ikj∆x , ξ: growth factor; |ξ| ≤ 1Jury: what about the roots of the characteristic polynomial (f (ξ))?

Conditions:

1 f (ξ = 1) ≥ 0,

2 f (ξ = −1) ≥ 0 if n is even; f (ξ = −1) ≤ 0 if n is odd,

3 |a0| ≤ an and so on...

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Solution: numerical method III.

For higher order polynomials this method comes ”crazy”...1 f (ξ = 1) ≥ 0,2 f (ξ = −1) ≥ 0 if n is even; f (ξ = −1) ≤ 0 if n is odd,3 |a0| ≤ an,

4 |b0| > |b2|, where b0 =

∣∣∣∣ a0 a3

a3 a0

∣∣∣∣ and b2 =

∣∣∣∣ a0 a1

a3 a2

∣∣∣∣and so on...

Coefficients for ballistic-conductive equation:

a3 = 1

a2 =∆t

τq+

∆t

τQ− 3

a1 =∆t2

τqτQ+ 3−

2∆t

τq−

2∆t

τQ−

(κ2∆t2

τqτQ∆x2+

∆t2

τq∆x2

)(2 cos(k∆x)− 2)

a0 =∆t

τq+

∆t

τQ−

∆t2

τQτq− 1 +

(κ2∆t2

τqτQ∆x2−

∆t2

τq∆x2

(∆t

τQ− 1

))(2 cos(k∆x)− 2)

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Solution: numerical method IV.

Implicit method: robust stabilityMixed: Crank-Nicholson-type (Θ-method → linear combination)

E.g.: balance equation for internal energyρc∆t

(T k+1i − T k

i

)=

− 1∆x

((1−Θ)(qki+1 − qki ) + Θ(qk+1

i+1 − qk+1i )

)Θ = 0: completely explicitΘ = 1: completely implicitΘ = 1/2: Cranck-Nicholson-type discretization

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The ballistic-conductive model - Solutions

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Comparison: phonon hydro I.

Phonon hydrodynamics (RET) vs NET+IVRequire at least N=30 eqs., but only 3 is used vs 3 eqs.

No. of fitted parameters: 2 relaxation time vs. 2+1 parametersSolved on semi-infinite region vs real domain

Relative amplitudes: false vs trueSummary: RET results are more like model testing than fitting;

Wrong: heat pulse length, sample size, thermal conductivityIs the RET model appropriate? Can not be decided.

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Comparison: phonon hydro II.

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Comparison: phonon hydro III.

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Comparison: hybrid phonon gas theory

Hybrid phonon gas model of Y. Ma vs NET+IVLongitudinal signal: artificial extension vs simplified model

Fitted parameters: 2 relaxation time vs 2+1Boundary conditions: no information vs effective cooling

Wrong thermal conductivity, no information about the others.

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What about on room temperature?

Role of material heterogeneities?! Let’s see the experiments!

Arrangement of the measurement, DEE BME

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Samples I.

Capacitor: periodic layered structure, thickness = 3 mm

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Samples II.

5 different rock samples, e.g. limestone, thickness = 1.5 mm

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Samples III.

2 different metal foam samples, inclusions: mm scale; thickness =5 mm

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Over-diffusive phenomenon I.

Measurement at room temperature, metal foam sampleτq∂ttT + ∂tT = a∂xxT + κ2∂txxT , Fourier equation

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Over-diffusive phenomenon II.

Measurement at room temperature, metal foam sampleτq∂ttT + ∂tT = a∂xxT + κ2∂txxT , Guyer-Krumhansl equation,κ2/τq > a!! Enhanced diffusion?! Parabolic equation!

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Properties of Guyer-Krumhansl equation

τ∂tq + q = −λ∂xT + κ2∂xxq andρc∂tT + ∂xq = 0.

It is applicable for second sound (damped wave) andover-diffusive phenomena.

Parabolic equation.

Hierarchy: Fourier + its time derivativeτ∂ttT + ∂tT = a∂xxT + κ2∂txxT .

Fourier resonance condition: κ2

τ = a.

κ2

τ < a: under-damped region, “wave-like” propagation.

κ2

τ > a: over-damped region.

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Analytical solution of Guyer-Krumhansl equation I.

Temperature representation:τ∂ttT + ∂tT = a∂xxT + κ2∂txxT .

“Hybrid”: T and q together.Advantageous for numerical methods.

Heat pulse experiment: q(t, x = 0) and q(t, x = L) are given⇒ let us use only the q!

Heat flux representation: τ∂ttq + ∂tq = a∂xxq + κ2∂txxq .

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Analytical solution of Guyer-Krumhansl equation II.

Let it be dimensionless.

τ∂ttq + ∂tq = ∂xxq + κ2∂txxq and

BC:

q(x = 0, t) = q0(t) =

{ (1− cos

(2π · t

τ∆

))if 0 ≤ t ≤ τ∆,

0 if t > τ∆,

and q(t, x = 1) = 0 (adiabatic).

IC: homogeneous, q(t = 0, x) = 0, ∂tq(t = 0, x) = 0.Split the solution: section I. → 0 < t < τ∆, section II. → τ∆ ≤ t.

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Analytical solution of Guyer-Krumhansl equation III.

q(x , t) = w(x , t) + v(x , t), (1)

where

w(x , t) := q0(t) +x

L

(qL(t)− q0(t)

)=(

1− x

L

)q0(t), (2)

after substitution:

τq(w + v) + w + v = w ′′ + v ′′ + κ2(w ′′ + v ′′). (3)

τq v + v = v ′′ + κ2v ′′ − f (x , t), (4)

where f (x , t) = w + τqw .Preserves the IC for v , but now the BC is zero: v(x = 0, t) = 0.

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Analytical solution of Guyer-Krumhansl equation IV.

Separation of variables: “classical” but works,

v(x , t) = ϕ(t)X (x), (5)

and assume that X (x) (for f = 0) solves the inhomogeneous case(f 6= 0) too.

Xn(x) = sin(nπ

Lx)

(6)

with eigenvalues

βn =(nπ

L

)2. (7)

Calculating the Fourier series of f , and solving

τqϕn +(1 + κ2b

)ϕn + bϕn = −fn(t) (8)

ϕn(t) = Y1nex1t + Y2ne

x2t + A sin(gt) + B cos(gt). (9)

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Analytical solution of Guyer-Krumhansl equation V.

For section II, the IC are

qII (x , t = 0) = qI (x , t = τ∆), qII (x , t = 0) = qI (x , t = τ∆).(10)

with t = t − τ∆ with homogeneous BC.Separating the variables again...

qII (x , t) = γ(t)X (x), (11)

X (x) is the same.

γn(t) = C1ner1n t + C2ne

r2n t , (12)

with

r1,2 =1

2τq

(−1− βnκ2 ±

√(1 + βnκ2)2 − 4τqβn

). (13)

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Analytical solution of Guyer-Krumhansl equation VI.

C1n + C2n = ϕn(t = τ∆),

C1nr1n + C2nr2n = ϕn(t = τ∆). (14)

Temperature distribution is determined from the energy balance:

τ∆T + q′ = 0,⇒ T = − 1

τ∆q′ = − 1

τ∆

∞∑n=1

Γn(t)nπ

Lcos(nπ

Lx),

(15)

T = − 1

τ∆

t∫0

∞∑n=1

Γn(α)nπ

Lcos(nπ

Lx)dα, (16)

where Γn(t) could be ϕn or γn depending on which section isconsidered.

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Analytical solution of Guyer-Krumhansl equation VII.

IC for T: T (t = 0, x) = 0 for section I. (Dimensionless!)For section II: TII (x , t = 0) = TI (x , t = τ∆). Convergence:

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Analytical solution of Guyer-Krumhansl equation VIII.

The rear side temperature history considering τq = κ2 = 0.02,using 40 terms. Fourier solution!

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Analytical solution of Guyer-Krumhansl equation IX.

The rear side temperature history considering τq = 0.02, κ2 = 0,using 200 terms. MCV (under-damped) solution!

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Analytical solution of Guyer-Krumhansl equation X.

The rear side temperature history considering τq = 0.02, κ2 = 0.2,using 10 terms. Over-damped solution!

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Finite element solution of Guyer-Krumhansl equation

Implementing in COMSOL. Fourier solution:

Needs ca. 20 s. Our code: runs under 1 s.46 / 54



Implementing in COMSOL. MCV solution:

Needs ca. 45 s. Our code: runs under 1 s.47 / 54



Implementing in COMSOL. GK solution:

FALSE! Needs ca. 65 s. Our code: runs under 1 s.48 / 54


Finite element solution of ballistic equations

Implementing in COMSOL.

Needs ca. 3 hours! Our code: runs under 1 s.49 / 54


How it works for conservative systems?

Pure elastic body:conservative system with velocity v and deformation ε,(dimensionless) ε = −v ′ and v = −ε′. Shifted fields:

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How it works for conservative systems?

Shifted fields + semi-implicit Euler (symplectic in time):

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Summary

Thermodynamics = stability.Theory for dissipation: see the work of Peter Van.(Example: Dissipative Lotka-Volterra.)

Thermodynamics: hierarchy of parabolic and hyperbolic eqs.

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Summary

Under development:

Numerical solutions: shifted (staggered) in space +symplectic in time.

Analytical solutions: other methods, more applicable formulae.

Physics: understanding non-equilibrium thermodynamics.

Working on practical problems: experiments, industrialapplications.

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Thank you for your kind attention!

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