universit a degli studi di milano - università di paviarocca/roccawiasmay13.pdf · universit a...

$: Universit a degli Studi di Milano - Università di Paviarocca/RoccaWIASMay13.pdf · Universit a degli Studi di Milano IFIP TC 7.2 Workshop \Electromagnetics { Modelling, Simulation,$
Optimal control of multifrequency induction hardening

E Rocca

Universita degli Studi di Milano

IFIP TC 72 WorkshopldquoElectromagnetics ndash Modelling Simulation Control and Industrial Applicationsrdquo

WIAS Berlin May 13 ndash 17 2013

joint work with

Dietmar Homberg (WIAS Berlin)

Supported by the FP7-IDEAS-ERC-StG Grant ldquoEntroPhaserdquo 256872

E Rocca (Universita di Milano) Induction hardening May 14 2013 1 39

Outline

The process of Induction Hardening

- classical induction hardening

- multifrequency induction hardening

- aims simulation and optimization of the process

A thermodynamically consistent model

- the volume fraction relation

- the energy balance

- the induction heating (Maxwell equations)

Our main results on

existence of solution for the corresponding initial-boundary value problem

stability estimates

first-order optimality conditions

Comment on some numerics ([Homberg Petzold WIAS])


Outline









Our main results on


stability estimates




Outline









Our main results on


stability estimates




Outline









Our main results on


stability estimates




Outline









Our main results on


stability estimates




Outline









Our main results on


stability estimates




Hardening process

In most structural components in mechanicalengineering the surface is particularly stressedTherefore the aim of surface hardening is toincrease the hardness of the boundary layers of aworkpiece by rapid heating and subsequentquenching

This heat treatment leads to a change inmicrostructure which produces the desiredhardening effect Typical examples of applicationare gear-wheels

The mode of operation in induction hardeningfacilities relies on the transformer principle Agiven current density in the induction coil(inductor) Ω induces eddy currents inside theworkpiece Σ


Hardening process





Advantages and Disadvantages

Because of the Joule effect these eddy currentslead to an increase in temperature in theboundary layers of the workpiece

Then the current is switched off and theworkpiece is quenched by spray-water cooling



Since the magnitude of the eddy currentsdecreases with growing distance from theworkpiece surface because of the frequencydependent skin-effect induction heating issuitable for surface hardening if the currentfrequency is big enough

After heating the workpiece is quenched byspray-water cooling and another phase transitionleads to the desired hardening effect in theboundary layers of the workpiece



There is a growing demand in industry for amore precise process control

I for weight reduction especially in automotiveindustry leading to components made ofthinner and thinner steel sheets The surfacehardening of these sheets is a very delicate tasksince one must be careful not to harden thecomplete sheet which would lead toundesirable fatigue effects

I for tendency to use high quality steels with onlya small carbon content Since the hardenabilityof a steel is directly related to its carboncontent already from a metallurgical point ofview the treatment of these steels is extremelydifficult

Advantage Very fast and energy-efficient process

Drawback Difficult to generate desired close to contour hardening profile forcomplex work pieces such as gears

MF+HF Actually simultaneous supply of medium and high frequences (MF+HF)gives the best result


Multi-frequency induction hardening

Simultaneous supply of medium - and high frequency power on one induction coil

Close to contour hardening profile for gears and other complex-shaped parts

Only the bases heated Only the tips heated Both bases and tips heated


The model

The reason why one can change the hardness ofsteel by thermal treatment lies in the occurringphase transitions

In the case of surface hardening owing to highcooling rates most of the austenite istransformed to martensite by a diffusionlessphase transition leading to the desired increaseof hardness


The model

Hence a mathematical model for induction surface hardening has to account for

the electromagnetic effects that lead to the surface heating

the thermomechanical effects

the phase transitions that are caused by the enormous changes in temperatureduring the heat treatment


The phase transitions

The reason why one can change the hardness of steel by thermal treatment lies in theoccuring phase transitions

z0 ferritepearlitemartensite

z0(t0) = 1

heatingminusminusminusminusrarr

z0 initial com-position

z1 austenite

z0 + z1 = 1

coolingminusminusminusminusrarr


z1 austenitez2 ferritez3 pearlitez4 martensitesum

zi = 1

At room temperature in coil workpiece general steel is a mixture of ferrite pearliteand martensite

Upon heating these phases are transformed to austenite

Then during cooling austenite is transformed back to a mixture of ferrite pearliteand martensite ndash we do not consider this part of the process ndash less interesting

The actual phase distribution at the end of the heat treatment depends on the coolingstrategy In the case of surface hardening owing to high cooling rates most of theaustenite is transformed to martensite by a diffusionless phase transition leading to thedesired increase of hardness





z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1






The evolution of the phase fraction of austenite z = z1

According to [Leblond and Deveaux (1984)] theformation of austenite cannot be described bythe additivity rule since for fixed temperaturewithin the transformation range one can get anequilibrium volume fraction of austenite lessthan one

Therefore they propose to use the rate law

z(0) = 0 in Σ

zt(t) =1

τ(θ)max (zeq(θ)minus z(t)) 0

=1

τ(θ)(zeq(θ)minus z(t))+ in Q = Σtimes (0T )

zeq isin [0 1] is an equilibrium fraction of austenite


The magnetic field

In the eddy current problems we get the Maxwellrsquosequations in D times (0T )

curlH = J curlE = minusBt divB = 0

where

E is the electric field

B the magnetic induction

H the magnetic field

J the spatial current density

and we have neglected the electric displacementin the first relation (| partD

partt| ltlt |J|)


The magnetic field

Assume the Ohmrsquos law and a linear relation betweenthe magnetic induction and the magnetic field

J = σE B = microH

where the electrical conductivity σ and the magneticpermeability micro (sufficiently regular and bounded forbelow and above) may depend both on the spatial vari-ables and also on the phase parameter z

σ(x z) =

0 x isin D (Ω cup Σ)

σw (z) x isin Σ

σi x isin Ω

and

micro(x z) =

micro0 x isin D (Ω cup Σ)

microw (z) x isin Σ

microi x isin Ω


The magnetic vector potentialSince divB = 0 we can introduce the magnetic vector potential A such that

B = curlA in D

and since A is not uniquely defined we impose the Coulomb gauge

divA = 0 in D

Using curlE + Bt = 0 and B = curlA we define the scalar potential φ by

E + At = minusnablaφ in D times (0T )

and we get the total current density

J = σE = minusσAt︸︷︷︸Jeddy

minusσnablaφ︸︷︷︸Jsource

in D times (0T )

Since Jsource = 0 in D Ω we get

σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J

+ σχΩnablaφ = 0 in D times (0T ) and minus div(σnablaφ) = 0 in Ω



B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J



Boundary conditions

Introduction of boundary conditions on partD

(1) Perfect electric conductor E times n = 0

(2) Perfect magnetic conductor H times n = 0

(1) leads toAtimes n = 0 (tangential component vanish on partD)

(2) leads tomicrominus1 curlAtimes n = 0

Inside the inductor (a closed tube) we fix a section Γ and model the current densitywhich is generated by the hardening machine by an interface condition on Γ

σnablaφ middot n = 0 on partΩ

[σnablaφ middot n] = 0 and [φ] = U0 on Γ

where n n and n denote normal unit vectors to partD partΩ and Γ resp


Eliminating the scalar potential

For a given coil geometry (here a torus with rectangular cross-section) the sourcecurrent density Jsource = σnablaφ can be precomputed analytically

From div(σnablaφ) = 0 one obtains in cylindrical coordinates

φ = C1ϕ and consequently Jsource = σC1 (0 1r 0)T(rϕz)

where C1 = U0(2π) for a given voltage

For a given source current in the coil C1 is computed fromintΓ

Jsource middot n da = Icoil

In cartesian coordinates one obtains for a given source current

Jsource =Icoil

log(rArI )h

minusy(x2 + y 2)x(x2 + y 2)

0

It can be used as control for optimization Jsource = u(t)J0 where J0 is the spatialcurrent density prescribed in the induction coil Ω

J0(x) =

Ji (x) x isin Ω

0 x isin D Ω

and u = u(t) denotes a time-dependent control on [0T ]










Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω



The energy balance

Assuming to have as constant density ρ = 1 (for simplicity) the internal energy balanceresults as

et + div q = JE = σ|At +nablaφ|2 = σ|At |2 in Σtimes (0T )

where

e denotes the internal energy of the system

q the heat flux which accordingly to the standard Fourier law is assumed as follows

q = minusκnablaθ κ gt 0

Boundary conditions neglect the possible radiative heat transfer between the inductorand the workpiece assuming

κpartθ

partν+ ηθ = g in partΣtimes (0T )

where ν denotes the outward unit normal vector to partΣ η stands for an heat transfercoefficient and g is a given boundary source


The energy balance



where




Boundary conditions

neglect the possible radiative heat transfer between the inductorand the workpiece assuming

κpartθ




The energy balance



where





κpartθ




The thermodynamical consistencyFrom the Helmholtz relation e = ψ + θs where ψ = ψ(θ z) denotes the free energy ofthe system we have that the Clausius-Duhem inequality

θ(st + div

(qθ

))= et minus ψθθt minus ψzzt minus θts minus

q

θnablaθ + div q

= et + div q minus (ψθ + s)θt minus ψzzt +κ|nablaθ|2

θ

= σ|At |2 minus (ψθ + s)θt minus ψzzt +κ|nablaθ|2

θ

ge 0

is satisfied eg if we assume the standard relations ψθ + s = 0 and Lτ(θ)zt = minusψz andhence ψz = minusL(zeq(θ)minus z)+ for some positive constant L gt 0

Using the definition ofthe specific heat cv = θsθ we get

et = cvθt + (ψz + θsz)zt = cvθt + f (θ z)zt

where we have denoted for simplicity by

f (θ z) = ψz minus θ(ψz)θ = minusL(zeq(θ)minus z)+ + Lθz primeeq(θ)H(zeq(θ)minus z)

H=Heaviside function The internal energy balance results

cvθt + div q = σ|At |2 minus f (θ z)zt in Σtimes (0T )



θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0

is satisfied eg if we assume the standard relations ψθ + s = 0 and Lτ(θ)zt = minusψz andhence ψz = minusL(zeq(θ)minus z)+ for some positive constant L gt 0 Using the definition ofthe specific heat cv = θsθ we get




H=Heaviside function

The internal energy balance results




θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system

Model consists of vector-potential formulation ofMaxwellrsquos equation heat equation and rate law for phasefractions

Source term u can be used as control for optimization

σAt + curl1

microcurlA = J0u in D times (0T )

θt minus div κnablaθ = σ|At |2 + F(θ z)zt in Σtimes (0T )

zt =1

τ(θ)(zeq(θ)minus z)+ in Σtimes (0T )

where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0

)TF(θ z) = L(zeq(θ)minus θz primeeq(θ)minus z)


Analysis and control of Joule heating models ndash a few references

Resistance heating

heat source h = σ|nablaϕ|2 minusrarr thermistor problem

Cimatti Prodi (1988) Howison Rodrigues Shillor (1993) Antonsev Chipot(1994) Homberg Khludnev Sokolowski (2001) Homberg Meyer Rehberg (2010)Homberg R (2011)

Induction heating ndash time domain

heat source h = σ(θ)|nablaϕ+ At |2

Bossavit Rodrigues (1994) Homberg Sokolowski (2003) optimal shape designHomberg (2004) including mechanical effects minusrarr the L1-regularity of the rhs inthe internal energy balance minusrarr weak solutions via Boccardo-Gallouet estimates

Induction heating ndash frequency domain

Bachinger Langer Schoberl Numerical analysis of nonlinear multiharmonic eddy currentproblems Numer Math 100 (2005)

Druet Klein Sprekels Troltzsch Yousept Optimal control of 3D state-constrainedinduction heating problems with nonlocal radiation effects SICON 49 (2011)

Troltzsch Yousept PDEndashconstrained optimization of time-dependent 3D electromagneticinduction heating by alternating voltages ESAIM M2AN 46 (2012)



Resistance heating












Resistance heating












Resistance heating











Our results work in progess with D Hoemberg

Well-posedness of state system

Optimality conditions










Solution space for vector potential

X =v isin L2(D)

∣∣∣ curl v isin L2(D) div v = 0 n times v∣∣∣partD

= 0

Assume partD isin C 11 Then X equipped with the norm

vX = curl vL2(D)

is a closed subspace of H1(D)


Assumptions

(i) σ(x z) micro(x z) D times [0 1]rarr R are continuousand Lipschitz continuous (wrt z for almost allx isin D) function st

σ le σ(x z) le σ in D times [0 1]

micro le micro(x z) le micro in D times [0 1]

(ii) u isin H1(0T )

(iii) J0 D rarr R3 is an L2curl(D)-function

(iv) τ zeq isin C 2(R) and

τlowast le τ(θ) le τlowast 0 le zeq(θ) le 1

τC2(R) zeqC2(R) le M

| minus zeq(θ) + θz primeeq(θ)| |θz primeprimeeq(θ)| le M for all θ isin R

(v) g isin Linfin(0T Linfin(partΣ))

(vi) A0 isin X capH3(D) θ0 isinW 253(Σ)


Weak formulation

Find a triple (A θ z) with the regularity properties

A isin H2(0T L2(D)) capW 1infin(0T X) curlA isin Linfin(0T L6(D)) (1)

θ isinW 153(0T L53(Σ)) cap L53(0T W 253(Σ)) cap L2(0T H1(Σ)) cap Linfin(Q) (2)

z isinW 1infin(0T W 1infin(Σ)) 0 le z lt 1 ae in Q (3)

solving the following systemintΩcupΣ

σ(x z)At middot v dx +

intD

1

micro(x z)curlA middot curl v dx =

intΩ

J0(x)u(t) middot v dx (4)

for all v isin X ae in (0T )

θt minus∆θ = minusF(θ z)zt + σ(x z)|At |2 ae in Q (5)

zt =1

τ(θ)(zeq(θ)minus z)+ ae in Q (6)

partθ

partν+ θ = g ae on partΣtimes (0T ) (7)

A(0) = A0 ae in D θ(0) = θ0 z(0) = 0 ae in Σ (8)


Well-posedness result

Theorem 1 There exists a unique solution to (4)-(8) satisfying the regularities (1)-(3)and the following estimate

AH2(0T L2(D))capW 1infin(0T X) + curlALinfin(0T L6(D))

+ θW 153(0T L53(Σ))capL53(0T W 253(Σ))capL2(0T H1(Σ))capLinfin(Q) + zW 1infin(0T W 1infin(Σ)) le S

If we denote by (Ai θi zi ) (i = 1 2) two triples of solutions corresponding to data(A0i θ0i ui ) then there exists a positive constant C = C(S) such that the followingstability estimate holds true

(A1 minus A2)(t)2L2(D) + curl(A1 minus A2)2

L2(Dtimes(0T ))

+ partt(A1 minus A2)(t)2L2(D) + curl(partt(A1 minus A2))2

L2(Dtimes(0T ))

+ (θ1 minus θ2)(t)2L2(Σ) + θ1 minus θ22

L2(0T H1(Σ))

+ (z1 minus z2)(t)2H1(Σ) + partt(z1 minus z2)2

L2(0T H1(Σ))

le C(A01 minus A022

X + (partt(A1 minus A2))(0)2L2(D) + θ01 minus θ022

L2(Σ)

+u1J0 minus u2J02L2(0T ) + uprime1J0 minus uprime2J02

L2(0T )

)for all t isin [0T ]








L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )



Idea of the proof of Thm 1

Local existence via Shauder fixed point argument exploiting the Lipschitz continuityof the map θ 7rarr z from W 1p(0T0 Lp(Σ)) in itself

Global a-priori estimates on solutions1 Exploit the regularity and boundedness of the solution z to the ODE getting

zW 1infin(0T0Linfin(Σ)) le C 0 le z(x t) lt 1 ae in Q

2 Take first v = At then differentiate it (formally) in time and take v = At inintΩcupΣ


intD

1


intΩJ0(x)u(t) middot v dx

and use Est 1 and the assumptions on micro and σ getting

AH1(0T0X)capW 1infin(0T0L2(D)) + AtL103(Dtimes(0T0)) le C

3 Testθt minus∆θ = minusF(θ z)zt + σ(x z)|At |2

by θ and using maximal regularity results for parabolic equations we get

θL2(0T0H1(Σ))capLinfin(0T0L2(Σ))capL53(0T0W 253(Σ))capW 153(0T0L53(Σ)) le C

4 Differentiating (formally) the A-eq and taking v = Att arguing by comparison andusing the results of [Kawanago rsquo93] in order to prove the Linfin-estimate for θ we getthe desired regularity of solutions and we can prolpongate it over the whole timeinterval [0T ]








intD

1
















intD

1
















intD

1
















intD

1










Stability estimate

Differentiate in time equationintΩcupΣ


intD

1


intΩ

J0(x)u(t) middot v dx

and take the difference of the two differentiated relations in two solutions (Ai θi zi )and test the difference by (A1 minus A2)t

Use the Lipschitz continuity properties of the solution map to the ODE in z

Take the differences of the equations for θ and test by θ1 minus θ2 exploit the regularityof A and sum up the two relations


Optimal control problem ndash Existence

cost functional

J (A θ z u) =β1

2

Tint0

intΣ

(θ(x t)minus θd(x t))2dxdt +

β2

2

intΣ

(z(x T )minus zd)2dx +β3

2u2

H1(0T )

control problem (CP)minJ (A θ z u)such that A θ z satisfies (P) and u isin Uad sub H1(0T )

Theorem 2 (CP) has a solution u isin Uad



cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )




Optimal control problem ndash Differentiability

Lipschitz-continuity the control-to-state mapping

S u 7rarr (A θ z)

is Lipschitz continuous from H1(0T ) to Z by Theorem 1

Frechet differentiability the control-to-state mapping

S u 7rarr (A θ z)

is Frechet differentiable from H1(0T ) to Y Z sub Y


Optimal control problem ndash First-order necessary conditions of optimality

adjoint systemminusσαt minus curl

( 1

microcurlα

)minus σprime(x z)ztα = minus2(σAtϑ)t

minusϑt minus k∆ϑ+ fθ(θ z)ϑ = fθζ + β1(θ minus θd)

minusζt minus fz(θ z)ζ + σprimeAt middotαminus σprime|At |2ϑ =microprime

micro2curlA middot curlαminus fzϑ

αtimes n = 0 in partD times (0T )

kpartϑ

partν+ κϑ = 0 in partΣtimes (0T )

ϑ(T ) = 0 ζ(T ) = z(x T )minus zd(x) in Σ

α(T ) = 0 in D

variational inequalityTint

0

(β3u(t)minus

intD

α(x t) middot J0(x t)dx)

(u minus u)dt

+

Tint0

β3uprime(t)(uprime(t)minus uprime(t))dt ge 0 for all u isin Uad sub H1(0T )




( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0



Numerical results ndash Challenges [Homberg Petzold]

Multiple time scalesMagnetic vector potential and heat conductance live on different time scales(Averaging method)

Skin effectEddy currents are distributed in a small surface layer of the workpiece (Adaptivemesh generation)

Nonlinear material dataMagnetic permeability micro depends on temperature and magnetic field H(Linearization)

3DTime consuming simulation in 3D (Model reduction to tackle optimal controlproblem numerically)




















Numerical realization ndash multiple time scales

Time scale for Maxwellrsquos equation governed by frequency of source currentf asymp 10 kHzminus 100 kHz consequetly τ sim 10minus5s

Time scale for heat equation governed by heat diffusion

τ sim cpρL2

kasymp 1 s

Alternating computation

I Solve for A with fixed temperature on fast time-scale

I Compute Joule heat by averaging electric energy Q = 1T

int T0 σ

∣∣∣ partApartt ∣∣∣2 dt

I Solve heat equation with fixed magnetic potential on slow time-scale (one time step)I Update A since σ and micro change with temperature





τ sim cpρL2

kasymp 1 s




int T0 σ




Thermal and electrical conductivity heat capacity and density

Material data depend on temperature

Electrical conductivity σ(θ)Thermal conductivity κ(θ)

Density ρ(θ)Specific heat capacity cp(θ)

Nonlinear relation between magnetic induction B and magnetic field HMagnetization curve B = micro(θH)H


Example 1 HF temperature and growth of austenite

(Video austenitemp4)

Figure temperature and austenite growth at high frequency


cyl_hf_phasemp4
Media File (videomp4)

Example 1 HF vs MF Adaptive grid

Source current in induction coil I0 = 5000 A at f = 10 kHz

Heating time 10 s

Nonlinear data for σ cp ρ κ micror

Adaptive grid with approx 50000 DOF

Figure Comparison of the adaptive grid between HF (left) and MF (right)


Example 1 HF vs MF Temperature

Figure Comparison of the temperature profile after 1s between HF (left) and MF (right)


Example 2 HF temperature

(Video temperaturemp4)

Figure temperature evolution at high frequency


temp_hf_100kHzmp4

Example 2 HF growth of austenite


Figure austenite evolution at high frequency


phase_hf_100kHzmp4

Example 2 MF temperature


Figure temperature evolution at medium frequency


temp_mf_8kHz-1mp4

Example 2 MF growth of austenite


Figure austenite evolution at medium frequency


phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene

Figure Temperature and austenite growth f = 10 kHz 200 kHz


Open problems

The case of temperature-dependent electrical conductivity σ in that case (cf eg[Homberg (2004)]) we get only existence of weak solutions minusrarr parabolic equationfor θ with L1-rhs because we cannot differentiate A-equation

The case of magnetic permeability micro depending both on the temperature θ and onthe magnetic field H

Include Mechanical effects

Numerical Optimal Control


Thanks for your attention

cf httpwwwmatunimiitusersrocca

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International School on ldquoRecent advances in partial differential equations andapplicationsrdquo Milano (Italy) June 17 ndash 22 2013 Coordinators E R and EnricoValdinoci

Spring School on rdquoRate-independent evolutions and hysteresis modellingrdquo Milano(Italy) May 27 ndash 31 2013 Coordinators Stefano Bosia Michela Eleuteri E R andEnrico Valdinoci


Outline

Hardening

Modelling

Outline









Our main results on


stability estimates




Outline









Our main results on


stability estimates




Outline









Our main results on


stability estimates




Outline









Our main results on


stability estimates




Outline









Our main results on


stability estimates




Outline









Our main results on


stability estimates




Hardening process





Hardening process


























The model




The model









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z(0) = 0 in Σ

zt(t) =1


=1




The magnetic field



where






partt| ltlt |J|)


The magnetic field


J = σE B = microH


σ(x z) =


σw (z) x isin Σ

σi x isin Ω

and

micro(x z) =



microi x isin Ω



B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J



Boundary conditions



















Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω











Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω



The energy balance



where





κpartθ




The energy balance



where




Boundary conditions


κpartθ




The energy balance



where





κpartθ





θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling

Outline









Our main results on


stability estimates




Outline









Our main results on


stability estimates




Outline









Our main results on


stability estimates




Outline









Our main results on


stability estimates




Outline









Our main results on


stability estimates




Hardening process





Hardening process


























The model




The model









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z(0) = 0 in Σ

zt(t) =1


=1




The magnetic field



where






partt| ltlt |J|)


The magnetic field


J = σE B = microH


σ(x z) =


σw (z) x isin Σ

σi x isin Ω

and

micro(x z) =



microi x isin Ω



B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J



Boundary conditions



















Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω











Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω



The energy balance



where





κpartθ




The energy balance



where




Boundary conditions


κpartθ




The energy balance



where





κpartθ





θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling

Outline









Our main results on


stability estimates




Outline









Our main results on


stability estimates




Outline









Our main results on


stability estimates




Outline









Our main results on


stability estimates




Hardening process





Hardening process


























The model




The model









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z(0) = 0 in Σ

zt(t) =1


=1




The magnetic field



where






partt| ltlt |J|)


The magnetic field


J = σE B = microH


σ(x z) =


σw (z) x isin Σ

σi x isin Ω

and

micro(x z) =



microi x isin Ω



B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J



Boundary conditions



















Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω











Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω



The energy balance



where





κpartθ




The energy balance



where




Boundary conditions


κpartθ




The energy balance



where





κpartθ





θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling

Outline









Our main results on


stability estimates




Outline









Our main results on


stability estimates




Outline









Our main results on


stability estimates




Hardening process





Hardening process


























The model




The model









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z(0) = 0 in Σ

zt(t) =1


=1




The magnetic field



where






partt| ltlt |J|)


The magnetic field


J = σE B = microH


σ(x z) =


σw (z) x isin Σ

σi x isin Ω

and

micro(x z) =



microi x isin Ω



B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J



Boundary conditions



















Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω











Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω



The energy balance



where





κpartθ




The energy balance



where




Boundary conditions


κpartθ




The energy balance



where





κpartθ





θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling

Outline









Our main results on


stability estimates




Outline









Our main results on


stability estimates




Hardening process





Hardening process


























The model




The model









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z(0) = 0 in Σ

zt(t) =1


=1




The magnetic field



where






partt| ltlt |J|)


The magnetic field


J = σE B = microH


σ(x z) =


σw (z) x isin Σ

σi x isin Ω

and

micro(x z) =



microi x isin Ω



B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J



Boundary conditions



















Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω











Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω



The energy balance



where





κpartθ




The energy balance



where




Boundary conditions


κpartθ




The energy balance



where





κpartθ





θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling

Outline









Our main results on


stability estimates




Hardening process





Hardening process


























The model




The model









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z(0) = 0 in Σ

zt(t) =1


=1




The magnetic field



where






partt| ltlt |J|)


The magnetic field


J = σE B = microH


σ(x z) =


σw (z) x isin Σ

σi x isin Ω

and

micro(x z) =



microi x isin Ω



B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J



Boundary conditions



















Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω











Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω



The energy balance



where





κpartθ




The energy balance



where




Boundary conditions


κpartθ




The energy balance



where





κpartθ





θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling

Hardening process





Hardening process


























The model




The model









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z(0) = 0 in Σ

zt(t) =1


=1




The magnetic field



where






partt| ltlt |J|)


The magnetic field


J = σE B = microH


σ(x z) =


σw (z) x isin Σ

σi x isin Ω

and

micro(x z) =



microi x isin Ω



B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J



Boundary conditions



















Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω











Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω



The energy balance



where





κpartθ




The energy balance



where




Boundary conditions


κpartθ




The energy balance



where





κpartθ





θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling

Hardening process


























The model




The model









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z(0) = 0 in Σ

zt(t) =1


=1




The magnetic field



where






partt| ltlt |J|)


The magnetic field


J = σE B = microH


σ(x z) =


σw (z) x isin Σ

σi x isin Ω

and

micro(x z) =



microi x isin Ω



B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J



Boundary conditions



















Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω











Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω



The energy balance



where





κpartθ




The energy balance



where




Boundary conditions


κpartθ




The energy balance



where





κpartθ





θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling






















The model




The model









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z(0) = 0 in Σ

zt(t) =1


=1




The magnetic field



where






partt| ltlt |J|)


The magnetic field


J = σE B = microH


σ(x z) =


σw (z) x isin Σ

σi x isin Ω

and

micro(x z) =



microi x isin Ω



B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J



Boundary conditions



















Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω











Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω



The energy balance



where





κpartθ




The energy balance



where




Boundary conditions


κpartθ




The energy balance



where





κpartθ





θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling

The model









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z(0) = 0 in Σ

zt(t) =1


=1




The magnetic field



where






partt| ltlt |J|)


The magnetic field


J = σE B = microH


σ(x z) =


σw (z) x isin Σ

σi x isin Ω

and

micro(x z) =



microi x isin Ω



B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J



Boundary conditions



















Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω











Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω



The energy balance



where





κpartθ




The energy balance



where




Boundary conditions


κpartθ




The energy balance



where





κpartθ





θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling




z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z(0) = 0 in Σ

zt(t) =1


=1




The magnetic field



where






partt| ltlt |J|)


The magnetic field


J = σE B = microH


σ(x z) =


σw (z) x isin Σ

σi x isin Ω

and

micro(x z) =



microi x isin Ω



B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J



Boundary conditions



















Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω











Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω



The energy balance



where





κpartθ




The energy balance



where




Boundary conditions


κpartθ




The energy balance



where





κpartθ





θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling




z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z(0) = 0 in Σ

zt(t) =1


=1




The magnetic field



where






partt| ltlt |J|)


The magnetic field


J = σE B = microH


σ(x z) =


σw (z) x isin Σ

σi x isin Ω

and

micro(x z) =



microi x isin Ω



B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J



Boundary conditions



















Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω











Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω



The energy balance



where





κpartθ




The energy balance



where




Boundary conditions


κpartθ




The energy balance



where





κpartθ





θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling




z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z(0) = 0 in Σ

zt(t) =1


=1




The magnetic field



where






partt| ltlt |J|)


The magnetic field


J = σE B = microH


σ(x z) =


σw (z) x isin Σ

σi x isin Ω

and

micro(x z) =



microi x isin Ω



B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J



Boundary conditions



















Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω











Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω



The energy balance



where





κpartθ




The energy balance



where




Boundary conditions


κpartθ




The energy balance



where





κpartθ





θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling




z0(t0) = 1



z1 austenite

z0 + z1 = 1




zi = 1









z(0) = 0 in Σ

zt(t) =1


=1




The magnetic field



where






partt| ltlt |J|)


The magnetic field


J = σE B = microH


σ(x z) =


σw (z) x isin Σ

σi x isin Ω

and

micro(x z) =



microi x isin Ω



B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J



Boundary conditions



















Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω











Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω



The energy balance



where





κpartθ




The energy balance



where




Boundary conditions


κpartθ




The energy balance



where





κpartθ





θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling




z(0) = 0 in Σ

zt(t) =1


=1




The magnetic field



where






partt| ltlt |J|)


The magnetic field


J = σE B = microH


σ(x z) =


σw (z) x isin Σ

σi x isin Ω

and

micro(x z) =



microi x isin Ω



B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J



Boundary conditions



















Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω











Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω



The energy balance



where





κpartθ




The energy balance



where




Boundary conditions


κpartθ




The energy balance



where





κpartθ





θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling

The magnetic field



where






partt| ltlt |J|)


The magnetic field


J = σE B = microH


σ(x z) =


σw (z) x isin Σ

σi x isin Ω

and

micro(x z) =



microi x isin Ω



B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J



Boundary conditions



















Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω











Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω



The energy balance



where





κpartθ




The energy balance



where




Boundary conditions


κpartθ




The energy balance



where





κpartθ





θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling

The magnetic field


J = σE B = microH


σ(x z) =


σw (z) x isin Σ

σi x isin Ω

and

micro(x z) =



microi x isin Ω



B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J



Boundary conditions



















Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω











Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω



The energy balance



where





κpartθ




The energy balance



where




Boundary conditions


κpartθ




The energy balance



where





κpartθ





θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling


B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J



Boundary conditions



















Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω











Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω



The energy balance



where





κpartθ




The energy balance



where




Boundary conditions


κpartθ




The energy balance



where





κpartθ





θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling


B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J




B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J



Boundary conditions



















Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω











Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω



The energy balance



where





κpartθ




The energy balance



where




Boundary conditions


κpartθ




The energy balance



where





κpartθ





θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling


B = curlA in D


divA = 0 in D






in D times (0T )


σAt + curl

(1

microcurlA

)︸︷︷︸

H︸︷︷︸J



Boundary conditions



















Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω











Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω



The energy balance



where





κpartθ




The energy balance



where




Boundary conditions


κpartθ




The energy balance



where





κpartθ





θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling

Boundary conditions



















Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω











Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω



The energy balance



where





κpartθ




The energy balance



where




Boundary conditions


κpartθ




The energy balance



where





κpartθ





θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling









Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω











Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω



The energy balance



where





κpartθ




The energy balance



where




Boundary conditions


κpartθ




The energy balance



where





κpartθ





θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling









Jsource =Icoil

log(rArI )h


0


J0(x) =

Ji (x) x isin Ω

0 x isin D Ω



The energy balance



where





κpartθ




The energy balance



where




Boundary conditions


κpartθ




The energy balance



where





κpartθ





θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling

The energy balance



where





κpartθ




The energy balance



where




Boundary conditions


κpartθ




The energy balance



where





κpartθ





θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling

The energy balance



where




Boundary conditions


κpartθ




The energy balance



where





κpartθ





θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling

The energy balance



where





κpartθ





θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling


θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling


θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0










θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling


θ(st + div

(qθ


q

θnablaθ + div q


θ


θ

ge 0








The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling

The PDE system



σAt + curl1



zt =1


where

J0 =(minusy(x2 + y 2) x(x2 + y 2) 0




Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling


Resistance heating












Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling


Resistance heating












Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling


Resistance heating












Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling


Resistance heating
























X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling














X =v isin L2(D)


= 0


vX = curl vL2(D)



Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling

Assumptions




(ii) u isin H1(0T )









Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling

Weak formulation







intD

1


intΩ




zt =1


partθ










L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling







L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling







L2(Dtimes(0T ))


L2(Dtimes(0T ))


L2(0T H1(Σ))


L2(0T H1(Σ))

le C(A01 minus A022


L2(Σ)


L2(0T )









intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling







intD

1
















intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling







intD

1
















intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling







intD

1
















intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling







intD

1
















intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling







intD

1










Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling

Stability estimate



intD

1


intΩ







cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling


cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling


cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )





cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling


cost functional

J (A θ z u) =β1

2

Tint0

intΣ


β2

2

intΣ


2u2

H1(0T )






S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling



S u 7rarr (A θ z)



S u 7rarr (A θ z)





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling



( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0





( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling



( 1

microcurlα






kpartϑ



α(T ) = 0 in D


0

(β3u(t)minus

intD


(u minus u)dt

+

Tint0






























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling




























τ sim cpρL2

kasymp 1 s




int T0 σ







τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling




τ sim cpρL2

kasymp 1 s




int T0 σ














cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling











cyl_hf_phasemp4



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling



Heating time 10 s












temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling








temp_hf_100kHzmp4





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling





phase_hf_100kHzmp4





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling





temp_mf_8kHz-1mp4





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling





phase_mf_8kHz-1mp4

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling

Example 2 MF vs HF

uarrSymmetrieebene



Open problems








Avertisement





Outline

Hardening

Modelling

Open problems








Avertisement





Outline

Hardening

Modelling



Avertisement





Outline

Hardening

Modelling

universit a degli studi di milano - università di paviarocca/roccawiasmay13.pdf · universit a...

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