ADJOINT METHOD FOR SOLVING INVERSE HEAT TRANSFER PROBLEMS (IHTP

3icipe - Port Ludlow - WA - June 13-18, 1999

__________________________________________________________________________________________Adjoint Method for solving IHTP - Examples

1

ADJOINT METHOD FOR SOLVING INVERSE HEAT TRANSFER PROBLEMS (IHTP)

EXAMPLES

Yvon JARNYLaboratoire de Thermocinétique, UMR CNRS 6607

ISITEM, Université de Nantes, 44306 Nantes Cedex 03E-mail [email protected]

_______________________________________________________________________

OUTLINE

I.- A LUMPED NON LINEAR HEAT TRANSFER PROBLEMDIRECT AND INVERSE FORMULATION

II.- ADJOINT METHOD AND CONJUGATE GRADIENTALGORITHM

III.- RESOLUTION OF THE CONTROL PROBLEM

IV.- RESOLUTION OF THE PARAMETER ESTIMATIONPROBLEM



2

I.- A LUMPED NON LINEAR HEATTRANSFER PROBLEMDIRECT AND INVERSE FORMULATION

1.-MODELING EQUATIONS OF THE MELTING PROCESS

2.- NUMERICAL RESOLUTION OF THE NON LINEARDIFFERENTIAL EQUATIONS

3.- INVERSE HEAT TRANSFER PROBLEMS



3

I.- A LUMPED NON LINEAR HEAT TRANSFERPROBLEM - MELTING PROCESS

1/7

1.- MODELING EQUATIONS

[ ]

m Cdy t

dtu t q t

m Cd t

dtq t

q t h y t t

1 1

2 2

( )( ) ( )

( )( )

( )

( ) ( ) ( )

= −

=

= −

θθ

θy y( ) = et 0 00 0θ θ( ) =

Figure 1 : Schematic of the system involving two heat capacities

0.00 0.20 0.40 0.60 0.80 1.00Température

0.00

0.20

0.40

0.60

0.80

1.00

Chaleurspécifique

Figure 2 : Heat capacity versus temperature

0.00 0.20 0.40 0.60 0.80 1.00temps

0.00

0.20

0.40

0.60

0.80

1.00

Tem

péra

ture

Figure 3 : Desired temperature versus time

( ): ( )C y t1

( ): ( )C t2 θ

u t( )



4


2/7

Reduced variables

tt

withm C

ha

m Cm C

uu

h Ty

y yT T

Tfixed

*

* , * , * ;

= = =

= =−

=−

ττ

θθ θ

1 1 2 2

1 1

0 0

;

∆ ∆ ∆

∆

Modeling Equations (* is omitted)

�( ) ( ) ( ) ( ),( ) �( ) ( ) ( ) ,

( ) ( )

y t y t t u t ta t t y t t

y

+ − = >+ − = >

��

= =

θθ θ θ

θ

00 0

0 0 0

Melting process with a constant heat flux

0.00 0.40 0.80 1.20 1.60 2.00t

0.00

0.40

0.80

1.20

u(t)

tet

a(t)

Figure 2a : Constant heat fluxu t t( ) . , .= < <088 0 14

0.00 0.40 0.80 1.20 1.60 2.00t

0.00

0.40

0.80

1.20

u(t)

tet

a(t)

Figure 2b : Constant heat fluxu t t( ) . , .= < <123 0 10



5


3/7

2.-NUMERICAL RESOLUTION OF THE NON LINEARDIFFERENTIAL EQUATIONS

∆t = time stept k t y y t t a ak k k k k k k= = = =. ; ( ); ( ); ( )∆ θ θ θ

Time discretization ( semi - implicit scheme )� Algebraic equations

11

1

1

1+ −− +�

��

�

�� =

++

+

+∆ ∆∆ ∆

∆t tt a t a

y y t u

k k

k

k

k k

k/ /.

θ θ

y

y t u tt a

Dt y t u

t aD

Dt t

t a t at t a

k

k k

k k

k

k

k k

k k

k

kk k

k

+

+

+

+

=

+ −+

�

��

�

��

=

+ +−�

��

�

��

=+ −

− +�

��

�

�� = + +

1

1

1

1

1

1

11

1

det.

/

det.

/

det/ /

/

∆ ∆∆

∆ ∆∆

∆ ∆∆ ∆

∆ ∆

θ

θθ



6

I.- A LUMPED NON LINEAR HEAT TRANSFERPROBLEM - THE MELTING PROCESS

4/7

Melting process with a constant heat flux

∆t = 0 02. , k = 1, ..., 100 (tf = 2.0)

x 0.0 0.25 0.30 0.35 1.0a(x) 0.1 0.15 1.0 0.25 0.2

table 1 : function a(x)

Determination of the constant heat flux u0 over ( , )0 tc

to reach the plateau θd = 1at the final time tf = 2.0

Eqs(1) ��( ) ( ) �( )

,,

y t a tu t t

t tc

c

+ =< << <

��

θ θ 0 00 2

� [ ]�( ) ( ) �( ) .y t a t dt u tc+ =� θ θ0

2

0

� [ ]ut

a x dxc

0 0

111≈ + � ( )

Numerical results

a x dx

t ut u

t

c

c

( ) .

. .

. .

0

1

0

0

0 23

10 12314 0 89

� =

= � == � =

, (Fig. 2.a), (Fig. 2.b)



7


5/7

Melting process with a time varying heat flux

0.00 0.40 0.80 1.20 1.60 2.00t

0.00

0.40

0.80

1.20

u(t)

tet

a(t)

Figure 3a : decreasing heat flux u t t t( ) . exp( / ), .= − < <1 43 2 0 1 0

0.00 0.40 0.80 1.20 1.60 2.00t

0.00

0.40

0.80

1.20

u(t)

tet

a(t)

Figure 3b : decreasing heat fluxu t t t( ) . exp( / ), .= − < <143 2 0 12

0.00 20.00 40.00 60.00 80.00 100.00time (x 5 )

-0.40

0.00

0.40

0.80

1.20

1.60

heat

flux

Figure 3C : time varying heat flux

← u(t) was computedwith the adjoint methodand a gradient algorithm



8


6/7

3.- INVERSE HEAT TRANSFER PROBLEMS (IHTP)

IHTP #1 : Parameter inverse problem

DATA : Heat flux history ~( ),u t t t f0 < <

Temperature history ~( ),y t t t f0 < <

UNKNOWN : Model Parameter a(θ), 0 < θ < 1

IHTP #2 : Control problem

DATA : Model Parameter a(θ), 0 < θ < 1

Desired Temperature θd(t), 0 < t < tf

UNKNOWN : Heat flux history u(t), 0 < t < tf

a

MODELu y

θ



9

1.- A LUMPED NON LINEAR HEAT TRANSFERPROBLEM - THE MELTING PROCESS

7/7

MATHEMATICAL FORMULATION OF THE INVERSE PROBLEMS

IHTP #1 : Parameter inverse problem

y t a( ; ) = solution of the modeling equations, computed with the parameter a(θ) ;

(the heat flux u(t) is fixed over the time interval)

Determine ( )a * ( ), ,θ θ ∈ 0 1 which minimizes the functional :

[ ]J a y t a y t dtt f( ) ( ; ) ~( )= −�

0

2


θ ( ; )t u = the solution of the modeling equations, computed with the heat flux u(t) ;

(the parameter a is fixed over the temperature range )

Determine ( )u t t t f*( ), ,∈ 0 which minimizes the functional :

[ ]J u t u t dt u t dtd

t tf f

µ θ θ µ

µ

( ) ( ; ) ( ) ( )

( )

= − +

>� �

0

22

0

0



10

II.- THE ADJOINT METHOD AND THECONJUGATE GRADIENT ALGORITHM

1.- MODELING EQUATIONS OF THE DIRECT PROBLEM

2.- MATHEMATICAL FORMULATION OF THE IHTP

3.- SENSITIVITY EQUATIONS

4.- ADJOINT EQUATIONS

5.- GRADIENT EQUATIONS

6.- LAGRANGIAN APPROACH

7.- CONJUGATE GRADIENT ALGORITHM



11

II - THE ADJOINT METHOD AND THECONJUGATE GRADIENT ALGORITHM

1/8

1.- MODELING EQUATIONS OF THE DIRECT PROBLEM

a) State equation� ( , ),

( )X E X u tX X

= >=

00 0

� X(u) = solution of the state equationcomputed with a given input uthe parameter model a being fixed

b) Observation equationy t CX t( ) ( )=

u

X(t0)

X yE � E C

a



12


2/8

2.- MATHEMATICAL FORMULATION OF THE IHTP


The desired output yd is given,the parameter model a is fixed,

Pb : FIND �u which minimizes Jα

J u J u CX u y udα α α : ( ) =→ − +( )2 2

under the constraint : X u( ) solution of the state equation� ( , ),

( )X E X u tX X

= >=

00 0

IHTP #1 : Parameter Inverse problem

The input heat flux u is fixed,the output history ~y is known,Pb : FIND a * which minimizes J

J a J a CX a y : =→ −( ) ( ) ~ 2

under the constraint : X a( ) solution of the state equation� ( , ),

( )X E X a tX X

= >=

00 0



13


3/8


3.- SENSITIVITY EQUATIONS

u u uε εδ= +�

X ε

δ δεε

εu XX X

� =−

→lim

0

Modeling Equations �� ( , ),

( )X E X u tX X

ε ε ε

ε

= >=

00 0

( E is supposed to be derivable )

� � ( , ) ( , ),

� � ,

( ) ( )

X X E X u E X u t

X X EX

X Eu

u t

X X

ε ε ε

ε

ε

ε ∂∂

δ ∂∂

δ

− = − >

− ≈ +�

��

��>

− =

0

0

0 0 0

δ∂

∂δ

∂∂

δ

δ

�( , )

.( , )

. ,

( )

XE X u

uu

E X uX

X t

X

= + >

=

0

0 0



14


4/8

4.- ADJOINT EQUATIONS

δ u � δ εδεε

JJ u u J u

=+ −

→lim

( ) ( )0

δεδ

εα

εδεε ε

JCX u u y CX u y u u ud d=

+ − − −+

+ −→ →

lim( ) ( )

lim0

2 2

0

2 2

Y Y Y Y

δ δ α δJ CX u y C X u ud= − +2 2( ) , ,

( )δ δ α δJ C CX u y X u ud= − +2 2* ( ) , ,C∗ = adjoint operator of C

Let P an adjoint variable

and ∂∂EX

∗

the adjoint operator of ∂∂

EX , i.e.

PEX

XEX

P X, . . ,∂∂

δ∂∂

δ=∗

Choose P solution of the following linear equations (= adjoint equation)

( )− = − − <

=

∗∗� . ( ) ,

( )

PEX

P C CX u y t t

P t

d f

f

∂∂

2

0



15


5/8


By definition, ∇J u( ) the gradient of J satisfies

δ δJ J u u= ∇ ( ),

Sensitivity and adjoint equations �

� , , �P X P Xδ δ+ = 0 ,

( )( )2 C CX u y Xd* ( ) ,− =δ � . ,P

EX

P X+∗∂

∂δ

( )

( ) ( )

= +�

��

�

��

= −�

��

�

�� +

= −�

��

�

��

� , . ,

� , . , , �

. ,*

P XEX

X P

P XEu

u P P X

Eu

P u

δ∂∂

δ

δ∂∂

δ δ

∂∂

δ

Finally

δ∂∂

δ α δJEu

P u u u= − +2 2*

, ,

and ∇J u( ) is obtained by

∇ = − +J uEu

P u( )*∂

∂α2



16


6/8

6.- LAGRANGIAN APPROACH

L u X P CX y u X E X u Pd( , , ) � , ),= − + −2

+ (α 2

P = Lagrange multiplier

a) Adjoint Equations

X and u are independent variables,P being fixed :

δ ∂∂

δ ∂∂

δLLu

uLX

X= +. .

Choose P in order to have :∂∂

δ δLX

X X. ,= ∀0

with� , , �P X P Xδ δ+ = 0

then

2 0CX y C X XEX

X P Xd− + − = ∀, � , ,δ δ∂∂

δ δ

2 0C CX y PEX

P X Xd∗

∗

− − − = ∀( ) � , ,∂∂

δ δ



17


7/8

b) Gradient Equations

δ∂

∂δL

L u X Pu

u=( , , )

.

δ∂∂

δ α δLEu

u P u u= − +. , ,2

δ∂∂

α δLEu

P u u= − +∗

. ,2

X solution of the modeling equations �

X X u L u X u P J u= � =( ) ( , ( ), ) ( )

then δ δL J=

and

∇ = − +∗

J uEu

P u( )∂∂

α2



18

THE ADJOINT METHOD AND THECONJUGATE GRADIENT ALGORITHM

8/8

7.- CONJUGATE GRADIENT ALGORITHM

n J T G V← ←0 0, . . . ; u0 =initial guess , β 0 0=

Repeat:

a) Compute the functional J un( )� Resolution of the modeling equations

b) Compute the gradient ∇J un( ) and define the search direction d J dn n n n= −∇ + −β 1

with β nn

n

JJ

=∇

∇ −

2

1 2

� Resolution of the adjoint equations

c) Compute the varied solution δX n with δu dn n=

and the scalar γδ

δn dC CX y X

C X=

−* ( ,2

� Resolution of the sensitivity equations

d) u u dn n n n+ ← +1 γ

e) n n← + 1

Until J u J un n( ) ( )+ − ≤1 ε



1

III - RESOLUTION OF THE CONTROLPROBLEM


2.- NUMERICAL IMPLEMENTATION OF THE CGA

3.- NUMERICAL RESULTS

IV - RESOLUTION OF THE INVERSEPARAMETER PROBLEM1.- GRADIENT EQUATIONS


Adjoint Method for solving IHTP - Examples

III - RESOLUTION OF THE CONTROLPROBLEM

1/5

Modeling Equations

�( ) ( ) ( ) ( ),( ) �( ) ( ) ( ) ,

( ) ( )


y

+ − = >+ − = >

��

= =

θθ θ θ

θ

00 0

0 0 0Sensitivity equations

u u uε εδ= + � yε εθ,

� ( )( ) �

y y u ua yε ε ε

ε ε ε ε

θ εδθ θ θ+ − = +

+ − =�� 0

yε εθ( ) = et 0 0 0 0( ) =then

( � � ) ( ) ( )( ) � ( ) � ( ) ( )

y y y y ua a y y

ε ε ε

ε ε ε ε

θ θ εδθ θ θ θ θ θ

− + − − − =− + − − − =

�� 0

when ε → 0, :θ θ εδθ θ θ ε θε ε θ≈ + ≈ + et a a a( ) ( ) '

then

[ ]

d ydt

y u t

d adt

y t

δ δ δθ δ

θ δθδθ δ

+ − = >

+ − = >

�

��

��

,

( ),

0

0 0

δ δθy( ) = et 0 0 0 0( ) =



3

III- RESOLUTION OF THE CONTROL PROBLEM 2/5

LAGRANGIAN

[ ]L( , , , ) ( ) ( ) ( )

( )

y p q u t t dt u t dt

pdydt

y u dt q addt

y dt

d

t t

t t

f f

f f

,θ θ θ µ

θ θθ

θ

= − +

+ + − −�

��

��+ + −�

��

��

� �

� �

2

0

2

0

0 0

If p and q are fixed , then

δ ∂∂

δ ∂∂θ

δθ ∂∂

δL L L L= + +y

yu

u

Choose p and q solution of the adjoint equations:

∂∂ δ

∂∂θ δθ δθ

Ly

y , dy

L,

= ∀

= ∀

�

��

��

0

0

∂∂

δ δ δ δLy

y p d ydt

y q y dtt f= + −�

��

�� ( )

0

[ ]∂∂θ

δθ θ θ δθ δθθ δθ

δθL = − − + +� � �20 0 0

( ) (( )

)d

t t tdt p dt q

d adt

dtf f f

then

− + − = <dpdt

p q t t f0,

− + − + − = <adqdt

q p t td f( ) ( ) ,θ θ θ2 0

p t q tf f( ) ( )= =0 0 et



III - RESOLUTION OF THE CONTROL PROBLEM 3/5

For p and q are solution of the adjoint equations �

δ∂∂ δLL

=u

u

[ ]δ µ δL = − +� p u udtt f 2

0

Gradient equations:

For y and θ are solution of the modeling equations �

L( ( ) ( ), , , ) ( )y u u p q u J u,θ µ=

then

δ δL J=and

[ ]∇ = − + <J t u p t u t t t fµ µ( ; ) ( ) ( ) ,2



5


2 - NUMERICAL IMPLEMENTATION OF THE CGA

To obtain u tn+1 ( ) , knowing u tn ( ) ,the numerical solutions of three problems are computed

• y and θθθθ solution of the modeling equations• p and q solution of the adjoint equations• δ δθy and solution of the sensitivity equations

At each iteration n, the time interval [ ]0,t f is scanned three times :

a) from t = 0 to t t f= , the forward resolution of the modeling equationsgives the values of

• the temperature y t u t un n( ; ) ( ; ) and θ .• the residual e t t u tn n

d( ) ( ; ) ( )= −θ θ• the functional J u n( )

b) from t t f= to t = 0 , the backward resolution of the adjoint equations givesthe values of

• the adjoint variables p t u q t un n( ; ) ( ; ) and ,• the gradient ∇J t u n( ; )• the direction of search d tn ( )

c) from t = 0 to t t f= , the forward numerical resolution of the sensitivityequations with δu dn n= , gives the values of

• the varied solution δθ n ,

• the scalar γ n

• the new approximation of the heat flux u tn+1 ( )

u t u t d t u k Nn k n kn n

k n t+ = + =1 1( ) ( ) ( ; ), , . . ,γ




3.- NUMERICAL RESULTS

Figures Captions

4a - 4b Computed heat flux u tn( ) with a fixed stepgradient algorithm( n = 50)

5a - 5b J un( ) computed with a Fixed step and aConjugate gradient algorithms

6a - 6b Computed heat flux u tn( ) with the conjugategradient algorithm ; n = 25 ; µ = 0 005.

7a- 7c Computed heat flux u tn( ) with the conjugategradient algorithm ; n = 25 ; µ = 0 001.

u u( ) ( ); .0 00 0 6= =

8a-8b J un( ) computed with the conjugate gradientalgorithm ; n = 25 ; µ = 0 001. ; u u( ) ( ); .0 00 0 6= =

9a-9b Computed heat flux u tn( ) with the conjugategradient algorithm ; n = 50; µ = 0 0. ; u( ) .0 0 6=

c Conjugate Gradient Algorithmc Inverse heat transfer problem - melting casec determination of the input heat flux (u) forc a prescribed temperature (xd)c 3icipe - June 13-18,1999, Port Ludlow, WAc Y Jarny - ISITEM - 44306 Nantesc

program deucapaparameter nn=101dimension x(nn),y(nn),u(nn),e(nn),xd(nn)dimension gun(nn),dd(nn),dd1(nn)

c initialisation



7

dt=0.02do n=1,nn

tn=(n-1)*dtu(n)=0.0xd(n)=1.0if (tn.lt.1.4) xd(n)=tn/1.4dd1(n)=0.

enddo

c optimisationitmax=25beta=0.sogn=0.do iter=0,itmax

c probleme directsom=0.x(1)=0.y(1)=0.xw=0.yw=0.do n=2,nn

tn=(n-1)*dtcall direct(yw,xw,u(n),dt)

c write(20,*) tn,u(n),yw,xw,xd(n)y(n)=ywx(n)=xwe(n)=x(n)-xd(n)som=som+e(n)*e(n)

enddowrite(25,*) iter,som

c probleme adjointpw=0.qw=0.sog1=sognsogn=0.do n=nn,2,-1

ew=-2*e(n)xw=x(n-1)call adjoint(pw,qw,ew,xw,dt)gun(n)=-pw+0.001*u(n)sogn=sogn+gun(n)*gun(n)

enddowrite(*,*)iter,'som=',som,'beta=',beta

c probleme de sensibilitebeta=0.if (iter.gt.0) beta =sogn/sog1dy=0.dx=0.sodx=0.som=0.do n=2,nn

dd(n)=-gun(n)+beta*dd1(n)du=dd(n)xw=x(n)call sensib(dy,dx,du,xw,dt)sodx=sodx+e(n)*dxsom=som+dx*dx

enddogama=-sodx/som

c nouvel itere



do n=2,nnu(n)=u(n)+gama*dd(n)dd1(n)=dd(n)write(30,*) n,u(n),x(n),xd(n)

enddo

enddoend

subroutine direct(y,x,u,dt)ax=a(x)d1=y+u*dtd2=xdet=1+dt+dt/axy=(d1*(1+dt/ax)+d2*dt)/detx=((1+dt)*d2+d1*dt/ax)/detreturnend

subroutine adjoint(p,q,e,x,dt)ax=a(x)d1=pd2=q+2*e*dt/axdet=1+dt+dt/axp=(d1*(1+dt/ax)+d2*dt)/detq=((1+dt)*d2+d1*dt/ax)/detreturnend

subroutine sensib(dy,dx,du,x,dt)ax=a(x)d1=dy+du*dtd2=dxdet=1+dt+dt/axdy=(d1*(1+dt/ax)+d2*dt)/detdx=((1+dt)*d2+d1*dt/ax)/detreturnend

function a(x)if (x.le.0) a=0.1if (x.le.0.25.and.x.ge.0.) a=0.1+0.05*x/0.25if (x.le.0.30.and.x.ge.0.25) a=0.15+0.85*(x-0.25)/0.05if (x.le.0.35.and.x.ge.0.30) a=1.0-0.75*(x-0.30)/0.05if (x.le.1.0.and.x.ge.0.35) a=0.25-0.05*(x-0.35)/0.65if (x.ge.1.0) a=0.2end



9

IV - RESOLUTION OF THE INVERSEPARAMETER PROBLEM

1/3

Modeling Equations

�( ) ( ) ( ) ( ),( ) �( ) ( ) ( ) ,

( ) ( )


y

+ − = >+ − = >

��

= =

θθ θ θ

θ

00 0

0 0 0

Approximation of the parameter function

a x a xj jj Np

( ) ( ),..,

==� ω1

Sensitivity equations

δ δ ωa x a xj jj Np

( ) ( ),..,

==�1

a a aε εδ= +� yε εθ,

�

( ) �y y u

a yε ε ε

ε ε ε ε

θθ θ θ

ε

+ − =+ − =

�� 0

yε εθ( ) = et 0 0 0 0( ) =then

( � � ) ( ) ( )( ) � ( ) � ( ) ( )

y y y ya a y y

ε ε ε

ε ε ε ε ε

θ θθ θ θ θ θ θ

− + − − − =− + − − − =

��

00

when ε → 0, :θ θ εδθ θ θ ε θε ε θ≈ + ≈ + et a a a( ) ( ) '

then



IV- RESOLUTION OF THE INVERSE PARAMETERPROBLEM

2/3

[ ]

d ydt

y t

d adt

y addt

t

δδ δθ

θ δθδθ δ δ θ

θ

+ − = >

+ − = − >

�

��

��

0 0

0

,

( )( ) ,

δ δθy( ) = et 0 0 0 0( ) =

LAGRANGIAN

[ ]L( , , , ) ( ) ~( )

( )

y p q a y t y t dt

pdydt

y u dt q addt

y dt

t

t t

f

f f

,θ

θ θθ

θ

= −

+ + − −�

��

��+ + −�

��

��

�

� �

2

0

0 0

p and q are fixed , then

δ∂∂ δ

∂∂θ δθ

∂∂ δL

L L L= + +

yy

aa

Choose p and q solution of the adjoint equations:

∂∂ δ δ

∂∂θ δθ δθ

Ly

y , y

L,

= ∀

= ∀

�

��

��

0

0

[ ]∂∂ δ δ

δδ δ

Ly

y y y ydt pd ydt

y q y dtt tf f

= − + + −�

��

�� 2

0 0~ ( )

[ ]∂∂θ

δθ δθθ δθ

δθL

= − + +�

��

�

��

� �p dt qd a

dtdt

t tf f

0 0

( )

then



11

III - RESOLUTION OF THE INVERSE PARAMETERPROBLEM

3/3

− + − + − = <dpdt

p q y y t t f2 0( ~ ) ,

− + − = <adqdt

q p t t f( ) ,θ 0

p t q tf f( ) ( )= =0 0 et

Gradient equations:

p and q are solution of the adjoint equations �

δ∂∂ δLL

=a

a

δθ

δL = � qddt

adtt f

0

y and θ are solution of the modeling equations �

L( ( ) ( ), , , ) ( )y a a p q a J a,θ =

then

δ δL J=

∇ = ��=

J a a qddt

dtj j

t

j Np

f( ) ( ),..,

δθ

ω θ01

and the Np components ∇J j of the gradient are computed by

∇ = =�J qddt

dt j Npj j

t f θ ω θ( ) , ..,0

1



CONCLUSIONS

• MATHEMATICAL FORMULATION OF NON LINEAR IHTP

→ Optimization problem of a least square functional

� Determination of a heat source u(t) for a prescribedTemperature evolution θd t( ) = Control problem

� Determination of a temperature varying parameter a( )θ froma measured Temperature history ~( )y t = Inverse parameterproblem

• ADJOINT METHOD

→ Iterative numerical resolution of three problems, similar to themodeling equations of the direct HTP

• Sensitivity and Adjoint equations are linear, even if themodeling equations are non linear

• Lagrangian formulation →Adjoint equations

� CONJUGATE GRADIENT ALGORITHM

→ The global structure of the IHTP solver is independent of

� the unknown function to be determined

� the structure of the modeling equations of the HTP( = set of algebraic equations, ordinary differential equations,

partial differential equations, 1-D, 2-D...)

Date post:	12-Sep-2021
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ADJOINT METHOD FOR SOLVING INVERSE HEAT TRANSFER PROBLEMS (IHTP

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