Inverses problems and Regularized Solutions Lecture 9 Part A - Yvon - PowerPoint PPT Presentation

Inverses problems and Regularized Solutions Lecture 9 Part A - Yvon JARNY Laboratoire de Thermocinétique de Nantes (LTN), UMR CNRS 6607 ‐ Université de Nantes Polytech’Nantes – 44306 Nantes cedex 3 Metti5 – Roscoff 1 June 13-18, 2011

Introduction(1) Introduction(1) Resolution of Inverse (Heat Transfer) Problems = a specific methodology based on the development of interactive computational-experimental process Experiments Control problems C t l bl Inverse data Experimental processing design Mathematical modelling Metti5 – Roscoff June 13-18, 2011 2

Introduction(2) Introduction(2) • Inverse (heat transfer) problems are ill ‐ posed : • Solutions (theoretical and numerical) • might not exist for all data • might be not unique • might be unstable under data perturbations • Specific Regularization methods are required Specific Regularization methods are required Metti5 – Roscoff June 13-18, 2011 3

Introduction (3) Introduction (3) • This lecture will be devoted to present and to illustrate the need of regularization for g solving Inverse problems • And how to do it in practice • And how to do it in practice • It will be organized in three parts It ill be organi ed in three parts Metti5 – Roscoff June 13-18, 2011 4

Content • Introduction • Inverse Problems– 3 Basic examples of linear Ill ‐ posed problems – An initial state problem – A input heat source problem (semi ‐ infinite body) – A 2 ‐ D stationnary heat source problem A 2 D i h bl • Mathematical Analysis of the linear inverse problem – The Singular Value Decomposition (SVD) approach The Singular Value Decomposition (SVD) approach – Quasi ‐ solutions: existence and uniqueness conditions – Stability condition and regularized solutions ‐ • Regularization processes and Stability condition – Regularization by truncation – Regularization by parametrization R l i ti b t i ti – Regularization by penalization – Iterative Regularization Iterative Regularization • Conclusions Metti5 – Roscoff June 13-18, 2011 5

First part: Ill-posed problems First part: Ill posed problems • Example -1: initial state problem   S State equations i 2 U Y L ( 0 , 1 )   2 T T      ( x , t ) ( x , t ), 0 x 1 , 0 t t Direct problem: p     f 2 t t x x Linear mapping     T ( 0 , t ) T ( 1 , t ) 0 , 0 t t f     U U Y Y AU      ( ( , 0 0 ) ) ( ( ), ) 0 0 1 1 T T x U U x x Inverse problem: Output equation  Y    1 Y 1 Y Y ? ? Y Y U U A A    Y ( x ) T ( x , t ), 0 x 1 f Metti5 – Roscoff June 13-18, 2011 6

Example 1: initial state problem Example ‐ 1: initial state problem Numerical results of the Direct Problem   Y ( x ) T ( x , t 0 . 05 ) f   2 U ( x ) x ( 1 x ) Metti5 – Roscoff June 13-18, 2011 7

• Example1 : initial state problem   The solution of the direct problem is    2 2 n t T ( x , t ) c . e ( x ) n n  n 1    with ( x ) 2 sin( n x ) = Set of orthogonal functions n       2 2  f   n  n t so Y ( x ) c . e ( x ) and U(x) c ( x ) n n n n  n  1 1 1 1         2 2   f  n t because , Y ( x ) U , e ( x ) n m nm n n n  1 1 The solution of the inverse problem is   2  2  2 2    f  n t     n t - 1 U A ( Y ) Y , e ( x ) c e f Y , n n n n n   n 1 1 Metti5 – Roscoff June 13-18, 2011 8

• Example -1: initial state problem    Y Y Y The operator A is linear, then for any output error        2 2 2 2 The resulting variation of the solution will be       n t f , U e Y  n n               Suppose (for simplicity) Y Y Y Y ( ( x x ) ) 2 2 sin( sin( N N x x ) ) Y   2 2 2 2               N t N t Then Then f , and and f U U e e U U e e Y Y N U Y any arbitrarily small output error, may induce a great variation on the solution may induce a great variation on the solution The stability condition is violated, thi i this inverse initial state heat conduction problem is ill ‐ posed i iti l t t h t d ti bl i ill d Metti5 – Roscoff June 13-18, 2011 9

Example 1: initial state problem Example ‐ 1: initial state problem Modeling equations State equations   2 U Y L ( 0 , t ) f     2 T T     ( ( x , t ) ) ( ( x , t ) ), x 0 0 , 0 0 t t   f 2 t x Direct problem:  T     ( ( x 0 , , t ) ) U U ( ( t ), ), 0 t t Linear mapping Linear mapping  f f x (time convolution)       T ( x , 0 ) 0 x 0 U U Y Y AU Output equation    Y ( t ) T ( x , t ), 0 t t c f Metti5 – Roscoff June 13-18, 2011 10

Example 2 : heat source problem (semi infinite body)  t        Y ( t ) A ( U )( t ) f ( t ) U ( ) d , t ( 0 ,t ) s f à Semin-Infinite Body - Temperature response to a time varying heat flux - IE Model Semin-Infinite Body - Temperature response to a time varying heat flux - IE Model 20 The impulse response 15  0.5   2 1 x 10   s   exp( ) ( ) f t     s k t 4 t 5 5 T 0 -5 q(t) -10 0 0 2 2 4 4 6 6 8 8 10 10 12 12 14 14 16 16 18 18 20 20 time t*=t/tau Metti5 – Roscoff June 13-18, 2011 11

Example 3 : heat source problem (2-D steady state ) State equations    T 0.5,0 y 1 1     T 0 dans y        q y q y ( ) ( ) q q (sin( (sin( ) 1), ) 1) 0 0 y y 1 1   0 2 T T w 1  T  0 0  n   2 3  T      q q  n  4 O t Output equation t ti  Y T  3 3 Metti5 – Roscoff June 13-18, 2011 12

Example 3 : heat source problem (2 ‐ D steady state ) After standard discretization of the spatial variable, the resulting state and output equations take the form      n  R n q q q y q y ( ( ) )  i i       i i 1 1   A T B * T B q   1 w 4 n 1  T T      w i i   1 Y Y C T C T m     m Y Y x ( ) R   j  j 1                 1 1  Y C A ( B * T B q ) 1 w 4 the numerical solution of the inverse problem             X X will require the inversion of 1      Y C A B q X q 4 = the sensitivity matrix Metti5 – Roscoff June 13-18, 2011 13

Second Part: Linear Inverse Problem Analysis • The three conditions of Hadamard (existence, uniqueness, stability) are ( , q , y) investigated for the linear inverse problem in the finite dimensional case. • The mathematical analysis will show that the concept of quasi ‐ solution The mathematical analysis will show that the concept of quasi solution allows to satisfy the question of existence, but the uniqueness and stability conditions will remain unsatisfied. • Then some regularization is needed to build a unique and stable quasi ‐ solution. Metti5 – Roscoff June 13-18, 2011 14

Si Singular Value Decomposition (SVD) approach l V l D iti (SVD) h The linear inverse problem (finite dimensional case ) consists in finding p ( ) g U  R  n . Y A U solution of the matrix equation Y  R m and A matrix ( m x n) are given Wh When the matrix is rectangular , m ≠ n, h i i l the analysis of the Hadamard’s conditions is required. It is based on the singular value decomposition (SVD) of the matrix A : g p dim( W ) = m x m; dim( V ) = n x n; dim( S )= m x n  W V V and W are orthogonal V d W t t th l A A S S r = rank ( A A t ) = rank ( A t A ) ≤ inf (m,n) Metti5 – Roscoff June 13-18, 2011 15

Numerical example m = 3; n = 2         , if if i i j j 1,.., 1 r r   i S [W, S , V]= svd(A) ij  0, otherwise A A W W S S V V 1 0 0.1826 0.8944 -0.4082 2.4495 0 0.4472 0.8944 -1 2 -0.9129 0 -0.4082 0 1.0000 -0.8944 0.4472 0 1 -0.3651 0.4472 0.8165 0 0 Find U: Y = W S V t U or W t Y= W t W S V t U = S V t U or W Y= W W S V U = S V U Let us introduce the new variables     t t n n t t m m V V X X U U R R Z Z W W Y Y R R The linear inverse problem becomes: Find X : Z = S X Metti5 – Roscoff June 13-18, 2011 16