Explicit solutions in optimal design problems for ...

University of Osijek - Department of Mathematics

Explicit solutions in optimal designproblems for stationary diffusion equation

Kresimir Burazin

University J. J. Strossmayer of OsijekDepartment of MathematicsTrg Ljudevita Gaja 631000 Osijek, Croatia

http://www.mathos.unios.hr/˜[email protected]

Joint work with Marko Vrdoljak

Kresimir Burazin Benasque, September 2015 1/18


Outline

Compliance optimization, composite materials and relaxation

Multiple states - spherically symmetric case

Examples



Optimal design problem (single state)

� � Rd open and bounded, f 2 L2(�) given;




� � Rd open and bounded, f 2 L2(�) given; stationary diffusionequation with homogenous Dirichlet b. c.:

�� div (Ar u) = fu 2 H1

0(�) ; (1)





�� div (Ar u) = fu 2 H1

0(�) ; (1)

where A is a mixture of two isotropic materials with conductivities0 < � < �:





�� div (Ar u) = fu 2 H1

0(�) ; (1)

where A is a mixture of two isotropic materials with conductivities0 < � < �: A = ��I + (1 � �)�I, where � 2 L1 (�; f 0; 1g),R

� � dx = q� , for given 0 < q� < j�j.





�� div (Ar u) = fu 2 H1

0(�) ; (1)

where A is a mixture of two isotropic materials with conductivities0 < � < �: A = ��I + (1 � �)�I, where � 2 L1 (�; f 0; 1g),R

� � dx = q� , for given 0 < q� < j�j.For given � , �, �, q� and f we want to find such material A whichminimizes the compliance functional (total amount of heat/electrical energydissipated in � ):

J(�) =Z

�f (x)u(x)dx =

Z

�A(x)r u(x) �r u(x) dx �! min ;

where u is the solution of the state equation (1).Kresimir Burazin Benasque, September 2015 3/18


Relaxation by homogenisation

� 2 L1 (�; f 0; 1g) �� 2 L1 (�; [0; 1])A = ��I + (1 � �)�I A 2 K(�) a.e. on �

classical material composite material - relaxation



Relaxation by homogenisation

� 2 L1 (�; f 0; 1g) �� 2 L1 (�; [0; 1])A = ��I + (1 � �)�I A 2 K(�) a.e. on �

classical material composite material - relaxation

DefinitionA sequence of matrix functions A" is said to H-converge to A� if for every fthe sequence of solutions of

�� div (A" r u" ) = fu" 2 H1

0(�)

satisfies u" * u in H10(�), A" r u" * A�r u in L2(�;Rd ), where u is the

solution of the homogenised equation�

� div (A�r u) = fu 2 H1

0(�) :Kresimir Burazin Benasque, September 2015 4/18


Composite material

DefinitionIf a sequence of characteristic functions �" 2 L1 (�; f 0; 1g) andconductivities

A" (x) = �" (x)�I + (1 � �" (x))�I

satisfy �" * �weakly �and A" H-converges to A�, then it is said that A�

is homogenised tensor of two-phase composite material with proportions �of first material and microstructure defined by the sequence (�" ).



Composite material


A" (x) = �" (x)�I + (1 � �" (x))�I


is homogenised tensor of two-phase composite material with proportions �of first material and microstructure defined by the sequence (�" ).Example – simple laminates: if �" depend only on x1, then

A� = diag(�� ; �+

� ; �+� ; : : : ; �+

� ) ;where

�+� = �� + (1 � �)� ; 1

��

= �� + 1 � �

� :



Composite material


A" (x) = �" (x)�I + (1 � �" (x))�I


is homogenised tensor of two-phase composite material with proportions �of first material and microstructure defined by the sequence (�" ).Example – simple laminates: if �" depend only on x1, then

A� = diag(�� ; �+

� ; �+� ; : : : ; �+

� ) ;where

�+� = �� + (1 � �)� ; 1

��

= �� + 1 � �

� :

Set of all composites:

A := f (�;A) 2 L1 (�; [0; 1]� Md(R)) :Z

��dx = q� ; A 2 K(�) a.e.g



Effective conductivities – set K(�)

G–closure problem: for given �find allpossible homogenised (effective)tensors A�





K(�) is given in terms of eigenvalues(Murat & Tartar; Lurie & Cherkaev):

�� j � �+

� j = 1; : : : ; ddX

j= 1

1

�j � � � 1

��

+ d � 1

�+� � �

dX

j= 1

1

� � �j� 1

� � ��

+ d � 1

� � �+�

;

2D:

O �1

�2

�� = 1

�

�� = 0

�

�+�

�+�

��

��

K(�)






�� j � �+

� j = 1; : : : ; ddX

j= 1

1

�j � � � 1

��

+ d � 1

�+� � �

dX

j= 1

1

� � �j� 1

� � ��

+ d � 1

� � �+�

;

2D:

�1

�2

��

�+�

�� +

�






�� j � �+

� j = 1; : : : ; ddX

j= 1

1

�j � � � 1

��

+ d � 1

�+� � �

dX

j= 1

1

� � �j� 1

� � ��

+ d � 1

� � �+�

;

2D:

�1

�2

��

�+�

�� +

�

3D:

�1

�2

�3






�� j � �+

� j = 1; : : : ; ddX

j= 1

1

�j � � � 1

��

+ d � 1

�+� � �

dX

j= 1

1

� � �j� 1

� � ��

+ d � 1

� � �+�

;

minA J is a proper relaxation ofminL1 (� ;f 0;1g) I

2D:

�1

�2

��

�+�

�� +

�

3D:

�1

�2

�3



Multiple state optimal design problem

State equations�

� div (Ar ui ) = fiui 2 H1

0(�) i = 1; : : : ; m

State function u = (u1; : : : ; um)




State equations�


0(�) i = 1; : : : ; m

State function u = (u1; : : : ; um)8>>>><

>>>>:

I(�) =P m

i= 1 �iR

� fiui dx ! min

u = (u1; : : : ; um) state function for A = ��I + (1 � �)�I

� 2 L1 (�; f 0; 1g) ;Z

�� dx = q� ;

for some given weights �i > 0.




State equations�


0(�) i = 1; : : : ; m

State function u = (u1; : : : ; um)8>>>><

>>>>:

I(�) =P m

i= 1 �iR

� fiui dx ! min

u = (u1; : : : ; um) state function for A = ��I + (1 � �)�I

� 2 L1 (�; f 0; 1g) ;Z

�� dx = q� ;

for some given weights �i > 0. Proper relaxation:

J(�;A) =mX

i= 1

�i

Z

�fiui dx ! min on

A := f (�;A) 2 L1 (�; [0; 1] � Md(R)) :Z

��dx = q� ; A 2 K(�) a.e.g



How do we find a solution?

A. Single state equation: [Murat & Tartar,1985] This problem can be rewritten as asimpler convex minimization problem.





I(�) =Z

�fu dx �! min

T =�

� 2 L1 (�; [0; 1]) :R

� � = q��

� 2 T ; and u determined uniquely by8<

:� div (�+

� r u) = f

u 2 H10(�)





I(�) =Z

�fu dx �! min

T =�

� 2 L1 (�; [0; 1]) :R

� � = q��


:� div (�+

� r u) = f

u 2 H10(�)

B. Multiple state equations: Simplerrelaxation fails:





I(�) =Z

�fu dx �! min

T =�

� 2 L1 (�; [0; 1]) :R

� � = q��


:� div (�+

� r u) = f

u 2 H10(�)


I(�) =mX

i= 1

�i

Z

�fiui dx �! min

T =�

� 2 L1 (�; [0; 1]) :R

� � = q��

� 2 T ; and ui determined uniquely by8<

:� div (�+

� r ui ) = fi

ui 2 H10(�)

i = 1; : : : ; m ;





I(�) =Z

�fu dx �! min

T =�

� 2 L1 (�; [0; 1]) :R

� � = q��


:� div (�+

� r u) = f

u 2 H10(�)


I(�) =mX

i= 1

�i

Z

�fiui dx �! min

T =�

� 2 L1 (�; [0; 1]) :R

� � = q��

� 2 T ; and ui determined uniquely by8<

:� div (�+

� r ui ) = fi

ui 2 H10(�)

i = 1; : : : ; m ;

In spherically symmetric case the simplerrelaxation can be done!



Relaxed designs

A := f (�;A) 2 L1 (�; [0; 1]� Md(R)) :Z

��dx = q� ; A 2 K(�) a.e.g

O �1

�2

�� = 1

�

�� = 0

�

�+�

�+�

��

��

K(�)



Relaxed designs

A := f (�;A) 2 L1 (�; [0; 1]� Md(R)) :Z

��dx = q� ; A 2 K(�) a.e.g

Further relaxation:

B : : :R

� �dx = q�

�� min(A) ; �max(A) � �+

�

O �1

�2

�� = 1

�

�� = 0

�

�+�

�+�

��

��

K(�)



Relaxed designs

A := f (�;A) 2 L1 (�; [0; 1]� Md(R)) :Z

��dx = q� ; A 2 K(�) a.e.g

Further relaxation:

B : : :R

� �dx = q�

�� min(A) ; �max(A) � �+

�

B is convex and compact and J iscontinuous on B, so there is asolution of minB J.

O �1

�2

�� = 1

�

�� = 0

�

�+�

�+�

��

��

K(�)



Equivalence of minB J and minT I

Theorem

I There is unique u� 2 H10(�;Rm) which is the state for every solution

of minB J and minT I.




Theorem


of minB J and minT I.I If (��;A�) is an optimal design for the problem minB J, then �� is

optimal design for minT I.




Theorem



optimal design for minT I.I Conversely, if �� is a solution of optimal design problem minT I, then

any (��;A�) 2 B satisfying A�r u�i = �+ (��)r u�

i almosteverywhere on � (e.g. A� = �+ (��)I) is an optimal design for theproblem minB J.




Theorem



optimal design for minT I.I Conversely, if �� is a solution of optimal design problem minT I, then

any (��;A�) 2 B satisfying A�r u�i = �+ (��)r u�

i almosteverywhere on � (e.g. A� = �+ (��)I) is an optimal design for theproblem minB J.

I If m < d, then there exists minimizer (��;A�) for J on B, such that(��;A�) 2 A , and thus it is also minimizer for J on A .



Simpler relaxation in case of spherical symmetry

TheoremLet � � Rd be spherically symmetric, and let the right-hand sidesfi = fi (r ), r 2 ! , i = 1; : : : ; m be a radial function. Then there exists aminimizer (��;A�) of the optimal design problem minA J which is a radialfunction.



Simpler relaxation in case of spherical symmetry

TheoremLet � � Rd be spherically symmetric, and let the right-hand sidesfi = fi (r ), r 2 ! , i = 1; : : : ; m be a radial function. Then there exists aminimizer (��;A�) of the optimal design problem minA J which is a radialfunction. More preciselya) For any minimizer �of functional I over T , let us define a radial

function �� : � �! R as the average value over spheres of �: forr 2 ! we take

��(r) := �Z

@B(0;r )�dS ;

where S denotes the surface measure on a sphere. Then �� is alsominimizer for I over T .



Simpler relaxation in case of spherical symmetry. . . cont.

Theorem

b) For any radial minimizer �� of I over T , let us define A� as a simplelaminate with layers orthogonal to a radial direction er and localproportion of the first material ��. To be specific, we can defineA� : � �! Md(R) in the following way:




Theorem


I If x = re1 = (r; 0; 0; : : : ; 0), then

A�(x) := diag(�+ (��(r)); �� (��(r)); �+ (��(r)); : : : ; �+ (��(r))) :




Theorem


I If x = re1 = (r; 0; 0; : : : ; 0), then

A�(x) := diag(�+ (��(r)); �� (��(r)); �+ (��(r)); : : : ; �+ (��(r))) :

I For all other x 2 �, we take the unique rotation R(x) 2 SO(d) suchthat x = jxjR(x)e1, and define

A�(x) := R(x)A�(R�(x)x)R�(x) :




Theorem


I If x = re1 = (r; 0; 0; : : : ; 0), then

A�(x) := diag(�+ (��(r)); �� (��(r)); �+ (��(r)); : : : ; �+ (��(r))) :


A�(x) := R(x)A�(R�(x)x)R�(x) :

Then (��;A�) is a radial optimal design for minB J.




Theorem


I If x = re1 = (r; 0; 0; : : : ; 0), then

A�(x) := diag(�+ (��(r)); �� (��(r)); �+ (��(r)); : : : ; �+ (��(r))) :


A�(x) := R(x)A�(R�(x)x)R�(x) :

Then (��;A�) is a radial optimal design for minB J.Moreover, (��;A�) 2 A , and thus it is also a solution for minA J.



Optimality conditions for minT I

Lemma�� 2 T is a solution minT I if and only if there exists a Lagrange multiplierc � 0 such that

�� 2 h0; 1i )mX

i= 1

�i jr u�i j2 = c ;

�� = 0 )mX

i= 1

�i jr u�i j2 � c ;

�� = 1 )mX

i= 1

�i jr u�i j2 � c ;

or equivalentlymX

i= 1

�i jr u�i j2 > c ) �� = 0 ;

mX

i= 1

�i jr u�i j2 < c ) �� = 1 :



Ball with nonconstant right-hand side

In all examples � = 1, � = 2.

� = B(0; 2) � R2, one state equation, f (r ) = 1 � r






State equation in polar coordinates � 1

r

�r�+

�(r)u0�0

= 1 � r :

Integration gives ju0(r)j = (r)��(r)+ �(1� �(r)) ; where (r ) = j2r2� 3r j

6 :







r

�r�+

�(r)u0�0

= 1 � r :


6 :

Conditions of optimality: there exists a constant � :=p

c > 0 such thatfor optimal �� we have:

ju0(r)j > � ) ��(r) = 0

) g� := � > �

ju0(r)j < � ) ��(r) = 1

) g� := � < �

�� 2 h0; 1i ) ju0(r)j = �) ��(r) = �� (r)

�(�� )Kresimir Burazin Benasque, September 2015 14/18






r

�r�+

�(r)u0�0

= 1 � r :


6 :

Conditions of optimality: there exists a constant � :=p

c > 0 such thatfor optimal �� we have:

ju0(r)j > � ) ��(r) = 0

) g� := � > �

ju0(r)j < � ) ��(r) = 1

) g� := � < �

�� 2 h0; 1i ) ju0(r)j = �) ��(r) = �� (r)

�(�� )34

20

r

ju0j

�1

�2

g��3

�4g�




Lagrange multiplier � is uniquely determined by the constraint�R

� �� dx = � := q�j� j 2 [0; 1], which is algebraic equation for �.






�1 �2 �3 10 �

�

�1

�2

�3

�4






�1 �2 �3 10 �

�

�1

�2

�3

�4

20r

ju0j

�

g�g�

p�1 q�

1 p�2q�

2 p�3 q�

3

��

��



Multiple states

Two state equations on a ball � = B(0; 2)I f1 = �B(0;1) ; f2 � 1 ;



Multiple states


I

�� div (�+

� r ui ) = fiui 2 H1

0(�) i = 1; 2



Multiple states


I

�� div (�+


0(�) i = 1; 2

I �Z

�f1u1 dx +

Z

�f2u2 dx ! min



Multiple states


I

�� div (�+


0(�) i = 1; 2

I �Z

�f1u1 dx +

Z

�f2u2 dx ! min

Solving state equation

u0i (r ) = i (r )

�(r)� + (1 � �(r))� ; i = 1; 2 ;

with

1(r) =

8>><

>>:

� r2

; 0 � r < 1 ;

� 1

2r; 1 � r � 2 ;

and 2(r) = � r2

:

Similarly as in the first example: := � 21 + 2

2, g� := �2 , g� :=

�2 .Kresimir Burazin Benasque, September 2015 16/18


Geometric interpretation of optimality conditions

20 r

c

g�

g�

A: 0 < � � 120 r

c

4p �

g�

g�

B: 1 < � � 4

20 r

c

4p �

g�

g�

C: 4 < � � 16 20 r

c

g�

g�

D: 16 < �



Geometric interpretation of optimality conditions

20 r

c

g�

g�

A: 0 < � � 120 r

c

4p �

g�

g�

B: 1 < � � 4

20 r

c

4p �

g�

g�

C: 4 < � � 16 20 r

c

g�

g�

D: 16 < �

As before, Lagrange multiplier can be numerically calculated fromcorresponding algebraic equation �

R� �� dx = �.



Optimal �� for case B

20 r

c

cg�

g�

pc1 qc

1 qc2 qc

3

��

g�



Optimal �� for case B

20 r

c

cg�

g�

pc1 qc

1 qc2 qc

3

��

g�

In orange region:

��(r) = 1

� � �

� �

r (r)

c

!

20 r

��

1

pc1 qc

1 qc2 qc

3


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