Disjoint sparsity for signal separation and applications to hybrid imaging inverse problems

Transcription

1 Disjoint sparsity for signal separation and applications to hybrid imaging inverse problems Giovanni S Alberti (joint with H Ammari) DMA, École Normale Supérieure, Paris June 16, 2015 Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

2 Photoacoustic tomography (From Wikipedia, 1. The image is H(x) = Γ(x)µ(x)u(x) 2. How to extract the unknowns from H? Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

3 Photoacoustic tomography (From Wikipedia, 1. The image is H(x) = Γ(x)µ(x)u(x) 2. How to extract the unknowns from H? Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

4 Photoacoustic tomography (From Wikipedia, 1. The image is H(x) = Γ(x)µ(x)u(x) 2. How to extract the unknowns from H? Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

5 PDE-based methods Works by Arridge, Bal, Beard, Cox, Gao, Jollivet, Jugnon, Kaipio, Köstli, Laufer, Naetar, Pulkkinen, Ren, Scherzer, Tarvainen, Uhlmann, Zhao A possible way to obtain Γ and µ from H = Γµu is based on the PDE satisfied by u. In the diffusive regime for light propagation there holds div(d u) + µu = 0 in Ω, where D is the diffusion parameter. This approach is sometimes very successful. However, there are some drawbacks PDE model non accurate (e.g. transport regime for light) Too many unknowns (e.g. if Γ 1 above) Several derivatives needed (problem with noise) Several measurements ui needed satisfying certain independence conditions. The focus of this talk is a new approach to this issue based on the separation of the unknowns from the fields via sparsity conditions: h = log H = log Γµ + log u = f + g. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

6 PDE-based methods Works by Arridge, Bal, Beard, Cox, Gao, Jollivet, Jugnon, Kaipio, Köstli, Laufer, Naetar, Pulkkinen, Ren, Scherzer, Tarvainen, Uhlmann, Zhao A possible way to obtain Γ and µ from H = Γµu is based on the PDE satisfied by u. In the diffusive regime for light propagation there holds div(d u) + µu = 0 in Ω, where D is the diffusion parameter. This approach is sometimes very successful. However, there are some drawbacks PDE model non accurate (e.g. transport regime for light) Too many unknowns (e.g. if Γ 1 above) Several derivatives needed (problem with noise) Several measurements ui needed satisfying certain independence conditions. The focus of this talk is a new approach to this issue based on the separation of the unknowns from the fields via sparsity conditions: h = log H = log Γµ + log u = f + g. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

7 PDE-based methods Works by Arridge, Bal, Beard, Cox, Gao, Jollivet, Jugnon, Kaipio, Köstli, Laufer, Naetar, Pulkkinen, Ren, Scherzer, Tarvainen, Uhlmann, Zhao A possible way to obtain Γ and µ from H = Γµu is based on the PDE satisfied by u. In the diffusive regime for light propagation there holds div(d u) + µu = 0 in Ω, where D is the diffusion parameter. This approach is sometimes very successful. However, there are some drawbacks PDE model non accurate (e.g. transport regime for light) Too many unknowns (e.g. if Γ 1 above) Several derivatives needed (problem with noise) Several measurements ui needed satisfying certain independence conditions. The focus of this talk is a new approach to this issue based on the separation of the unknowns from the fields via sparsity conditions: h = log H = log Γµ + log u = f + g. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

8 PDE-based methods Works by Arridge, Bal, Beard, Cox, Gao, Jollivet, Jugnon, Kaipio, Köstli, Laufer, Naetar, Pulkkinen, Ren, Scherzer, Tarvainen, Uhlmann, Zhao A possible way to obtain Γ and µ from H = Γµu is based on the PDE satisfied by u. In the diffusive regime for light propagation there holds div(d u) + µu = 0 in Ω, where D is the diffusion parameter. This approach is sometimes very successful. However, there are some drawbacks PDE model non accurate (e.g. transport regime for light) Too many unknowns (e.g. if Γ 1 above) Several derivatives needed (problem with noise) Several measurements ui needed satisfying certain independence conditions. The focus of this talk is a new approach to this issue based on the separation of the unknowns from the fields via sparsity conditions: h = log H = log Γµ + log u = f + g. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

9 PDE-based methods Works by Arridge, Bal, Beard, Cox, Gao, Jollivet, Jugnon, Kaipio, Köstli, Laufer, Naetar, Pulkkinen, Ren, Scherzer, Tarvainen, Uhlmann, Zhao A possible way to obtain Γ and µ from H = Γµu is based on the PDE satisfied by u. In the diffusive regime for light propagation there holds div(d u) + µu = 0 in Ω, where D is the diffusion parameter. This approach is sometimes very successful. However, there are some drawbacks PDE model non accurate (e.g. transport regime for light) Too many unknowns (e.g. if Γ 1 above) Several derivatives needed (problem with noise) Several measurements ui needed satisfying certain independence conditions. The focus of this talk is a new approach to this issue based on the separation of the unknowns from the fields via sparsity conditions: h = log H = log Γµ + log u = f + g. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

10 PDE-based methods Works by Arridge, Bal, Beard, Cox, Gao, Jollivet, Jugnon, Kaipio, Köstli, Laufer, Naetar, Pulkkinen, Ren, Scherzer, Tarvainen, Uhlmann, Zhao A possible way to obtain Γ and µ from H = Γµu is based on the PDE satisfied by u. In the diffusive regime for light propagation there holds div(d u) + µu = 0 in Ω, where D is the diffusion parameter. This approach is sometimes very successful. However, there are some drawbacks PDE model non accurate (e.g. transport regime for light) Too many unknowns (e.g. if Γ 1 above) Several derivatives needed (problem with noise) Several measurements ui needed satisfying certain independence conditions. The focus of this talk is a new approach to this issue based on the separation of the unknowns from the fields via sparsity conditions: h = log H = log Γµ + log u = f + g. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

11 Introduction to sparse representations Let h R n be a column vector (n = d d is the resolution of the image). Let A R n m be a dictionary of m atoms, which are used as building blocks: (1) h = Ay, for some coefficient vector y R m (weights). If m > n then (1) is in general underdetermined, and has many solutions y. Select the sparsest one, i.e. with fewest non-zero entries: min y R m y 0 subject to h = Ay, where y 0 := #supp y = #{α {1,..., m} : y(α) 0}. If the dictionary A is well chosen, it is possible to represent an n-dimensional vector with much fewer coefficients. In practice, we minimise min y R m y 0 + λ h Ay 2 for some λ > 0 (hard problem, but many available algorithms). Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

12 Introduction to sparse representations Let h R n be a column vector (n = d d is the resolution of the image). Let A R n m be a dictionary of m atoms, which are used as building blocks: (1) h = Ay, for some coefficient vector y R m (weights). If m > n then (1) is in general underdetermined, and has many solutions y. Select the sparsest one, i.e. with fewest non-zero entries: min y R m y 0 subject to h = Ay, where y 0 := #supp y = #{α {1,..., m} : y(α) 0}. If the dictionary A is well chosen, it is possible to represent an n-dimensional vector with much fewer coefficients. In practice, we minimise min y R m y 0 + λ h Ay 2 for some λ > 0 (hard problem, but many available algorithms). Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

13 Introduction to sparse representations Let h R n be a column vector (n = d d is the resolution of the image). Let A R n m be a dictionary of m atoms, which are used as building blocks: (1) h = Ay, for some coefficient vector y R m (weights). If m > n then (1) is in general underdetermined, and has many solutions y. Select the sparsest one, i.e. with fewest non-zero entries: min y R m y 0 subject to h = Ay, where y 0 := #supp y = #{α {1,..., m} : y(α) 0}. If the dictionary A is well chosen, it is possible to represent an n-dimensional vector with much fewer coefficients. In practice, we minimise min y R m y 0 + λ h Ay 2 for some λ > 0 (hard problem, but many available algorithms). Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

14 Introduction to sparse representations Let h R n be a column vector (n = d d is the resolution of the image). Let A R n m be a dictionary of m atoms, which are used as building blocks: (1) h = Ay, for some coefficient vector y R m (weights). If m > n then (1) is in general underdetermined, and has many solutions y. Select the sparsest one, i.e. with fewest non-zero entries: min y R m y 0 subject to h = Ay, where y 0 := #supp y = #{α {1,..., m} : y(α) 0}. If the dictionary A is well chosen, it is possible to represent an n-dimensional vector with much fewer coefficients. In practice, we minimise min y R m y 0 + λ h Ay 2 for some λ > 0 (hard problem, but many available algorithms). Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

15 Introduction to sparse representations Let h R n be a column vector (n = d d is the resolution of the image). Let A R n m be a dictionary of m atoms, which are used as building blocks: (1) h = Ay, for some coefficient vector y R m (weights). If m > n then (1) is in general underdetermined, and has many solutions y. Select the sparsest one, i.e. with fewest non-zero entries: min y R m y 0 subject to h = Ay, where y 0 := #supp y = #{α {1,..., m} : y(α) 0}. If the dictionary A is well chosen, it is possible to represent an n-dimensional vector with much fewer coefficients. In practice, we minimise min y R m y 0 + λ h Ay 2 for some λ > 0 (hard problem, but many available algorithms). Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

16 Introduction to sparse representations Let h R n be a column vector (n = d d is the resolution of the image). Let A R n m be a dictionary of m atoms, which are used as building blocks: (1) h = Ay, for some coefficient vector y R m (weights). If m > n then (1) is in general underdetermined, and has many solutions y. Select the sparsest one, i.e. with fewest non-zero entries: min y R m y 0 subject to h = Ay, where y 0 := #supp y = #{α {1,..., m} : y(α) 0}. If the dictionary A is well chosen, it is possible to represent an n-dimensional vector with much fewer coefficients. In practice, we minimise min y R m y 0 + λ h Ay 2 for some λ > 0 (hard problem, but many available algorithms). Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

17 Introduction to sparse representations Let h R n be a column vector (n = d d is the resolution of the image). Let A R n m be a dictionary of m atoms, which are used as building blocks: (1) h = Ay, for some coefficient vector y R m (weights). If m > n then (1) is in general underdetermined, and has many solutions y. Select the sparsest one, i.e. with fewest non-zero entries: min y R m y 0 subject to h = Ay, where y 0 := #supp y = #{α {1,..., m} : y(α) 0}. If the dictionary A is well chosen, it is possible to represent an n-dimensional vector with much fewer coefficients. In practice, we minimise min y R m y 0 + λ h Ay 2 for some λ > 0 (hard problem, but many available algorithms). Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

18 Morphological component analysis (Starck, Elad, Donoho, 2004,...) Back to the signal separation problem. Let h = f + g R n be the sum of two components. Let A f R n m f and A g R n mg two dictionaries such that: f can be sparsely represented w.r.t. Af but not w.r.t. A g; g can be sparsely represented w.r.t. Ag but not w.r.t. A f ; Decompose h w.r.t. the concatenated dictionary A = [A f, A g ]: min y R m f +mg y 0 subject to [A f, A g ] [ y f y g ] = h. Recover f A f y f, g A g y g. In other words, the original components can be recovered provided that they have different features. This is expressed in terms of the sparsity of their decompositions with different dictionaries. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

19 Morphological component analysis (Starck, Elad, Donoho, 2004,...) Back to the signal separation problem. Let h = f + g R n be the sum of two components. Let A f R n m f and A g R n mg two dictionaries such that: f can be sparsely represented w.r.t. Af but not w.r.t. A g; g can be sparsely represented w.r.t. Ag but not w.r.t. A f ; Decompose h w.r.t. the concatenated dictionary A = [A f, A g ]: min y R m f +mg y 0 subject to [A f, A g ] [ y f y g ] = h. Recover f A f y f, g A g y g. In other words, the original components can be recovered provided that they have different features. This is expressed in terms of the sparsity of their decompositions with different dictionaries. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

20 Morphological component analysis (Starck, Elad, Donoho, 2004,...) Back to the signal separation problem. Let h = f + g R n be the sum of two components. Let A f R n m f and A g R n mg two dictionaries such that: f can be sparsely represented w.r.t. Af but not w.r.t. A g; g can be sparsely represented w.r.t. Ag but not w.r.t. A f ; Decompose h w.r.t. the concatenated dictionary A = [A f, A g ]: min y R m f +mg y 0 subject to [A f, A g ] [ y f y g ] = h. Recover f A f y f, g A g y g. In other words, the original components can be recovered provided that they have different features. This is expressed in terms of the sparsity of their decompositions with different dictionaries. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

21 Morphological component analysis (Starck, Elad, Donoho, 2004,...) Back to the signal separation problem. Let h = f + g R n be the sum of two components. Let A f R n m f and A g R n mg two dictionaries such that: f can be sparsely represented w.r.t. Af but not w.r.t. A g; g can be sparsely represented w.r.t. Ag but not w.r.t. A f ; Decompose h w.r.t. the concatenated dictionary A = [A f, A g ]: min y R m f +mg y 0 subject to [A f, A g ] [ y f y g ] = h. Recover f A f y f, g A g y g. In other words, the original components can be recovered provided that they have different features. This is expressed in terms of the sparsity of their decompositions with different dictionaries. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

22 Morphological component analysis (Starck, Elad, Donoho, 2004,...) Back to the signal separation problem. Let h = f + g R n be the sum of two components. Let A f R n m f and A g R n mg two dictionaries such that: f can be sparsely represented w.r.t. Af but not w.r.t. A g; g can be sparsely represented w.r.t. Ag but not w.r.t. A f ; Decompose h w.r.t. the concatenated dictionary A = [A f, A g ]: min y R m f +mg y 0 subject to [A f, A g ] [ y f y g ] = h. Recover f A f y f, g A g y g. In other words, the original components can be recovered provided that they have different features. This is expressed in terms of the sparsity of their decompositions with different dictionaries. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

23 Morphological component analysis (Starck, Elad, Donoho, 2004,...) Back to the signal separation problem. Let h = f + g R n be the sum of two components. Let A f R n m f and A g R n mg two dictionaries such that: f can be sparsely represented w.r.t. Af but not w.r.t. A g; g can be sparsely represented w.r.t. Ag but not w.r.t. A f ; Decompose h w.r.t. the concatenated dictionary A = [A f, A g ]: min y R m f +mg y 0 subject to [A f, A g ] [ y f y g ] = h. Recover f A f y f, g A g y g. In other words, the original components can be recovered provided that they have different features. This is expressed in terms of the sparsity of their decompositions with different dictionaries. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

24 Toy example h = f + g Choose A f = A δ and A g = A F and let y = [ y f y g ] give the sparsest representation of h w.r.t. A = [A f, A g ]. This clearly provides the right reconstruction: only 4 atoms are used. f g Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

25 Toy example h = f + g Choose A f = A δ and A g = A F and let y = [ y f y g ] give the sparsest representation of h w.r.t. A = [A f, A g ]. This clearly provides the right reconstruction: only 4 atoms are used. f g Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

26 Theoretical justification (Elad & Bruckstein, 2002) Why does writing f = A f y f and g = A g y g give the correct reconstruction? (Uncertainty principle) If h 0 has the representations h = Ay A = By B w. r. t. two orthonormal bases A = [a 1,..., a n ] and B = [b 1,..., b n ], then y A 0 + y B 0 2/M, where M = max i,j (a i, b j ) 2 is the mutual coherence. If f and g have representations y f and y g satisfying y f 0 + y g 0 < 1/M, then the reconstruction is correct. In practice, the assumption y f 0 + y g 0 < 1/M is almost never satisfied, and so the above argument remains only a theoretical speculation. Focus of this new approach: extension to multiple measurements (many other generalisations by Georgiev, Theis, Cichocki, Bobin, Moudden, Starck,...). Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

27 Theoretical justification (Elad & Bruckstein, 2002) Why does writing f = A f y f and g = A g y g give the correct reconstruction? (Uncertainty principle) If h 0 has the representations h = Ay A = By B w. r. t. two orthonormal bases A = [a 1,..., a n ] and B = [b 1,..., b n ], then y A 0 + y B 0 2/M, where M = max i,j (a i, b j ) 2 is the mutual coherence. If f and g have representations y f and y g satisfying y f 0 + y g 0 < 1/M, then the reconstruction is correct. In practice, the assumption y f 0 + y g 0 < 1/M is almost never satisfied, and so the above argument remains only a theoretical speculation. Focus of this new approach: extension to multiple measurements (many other generalisations by Georgiev, Theis, Cichocki, Bobin, Moudden, Starck,...). Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

28 Theoretical justification (Elad & Bruckstein, 2002) Why does writing f = A f y f and g = A g y g give the correct reconstruction? (Uncertainty principle) If h 0 has the representations h = Ay A = By B w. r. t. two orthonormal bases A = [a 1,..., a n ] and B = [b 1,..., b n ], then y A 0 + y B 0 2/M, where M = max i,j (a i, b j ) 2 is the mutual coherence. If f and g have representations y f and y g satisfying y f 0 + y g 0 < 1/M, then the reconstruction is correct. In practice, the assumption y f 0 + y g 0 < 1/M is almost never satisfied, and so the above argument remains only a theoretical speculation. Focus of this new approach: extension to multiple measurements (many other generalisations by Georgiev, Theis, Cichocki, Bobin, Moudden, Starck,...). Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

29 Theoretical justification (Elad & Bruckstein, 2002) Why does writing f = A f y f and g = A g y g give the correct reconstruction? (Uncertainty principle) If h 0 has the representations h = Ay A = By B w. r. t. two orthonormal bases A = [a 1,..., a n ] and B = [b 1,..., b n ], then y A 0 + y B 0 2/M, where M = max i,j (a i, b j ) 2 is the mutual coherence. If f and g have representations y f and y g satisfying y f 0 + y g 0 < 1/M, then the reconstruction is correct. In practice, the assumption y f 0 + y g 0 < 1/M is almost never satisfied, and so the above argument remains only a theoretical speculation. Focus of this new approach: extension to multiple measurements (many other generalisations by Georgiev, Theis, Cichocki, Bobin, Moudden, Starck,...). Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

30 Theoretical justification (Elad & Bruckstein, 2002) Why does writing f = A f y f and g = A g y g give the correct reconstruction? (Uncertainty principle) If h 0 has the representations h = Ay A = By B w. r. t. two orthonormal bases A = [a 1,..., a n ] and B = [b 1,..., b n ], then y A 0 + y B 0 2/M, where M = max i,j (a i, b j ) 2 is the mutual coherence. If f and g have representations y f and y g satisfying y f 0 + y g 0 < 1/M, then the reconstruction is correct. In practice, the assumption y f 0 + y g 0 < 1/M is almost never satisfied, and so the above argument remains only a theoretical speculation. Focus of this new approach: extension to multiple measurements (many other generalisations by Georgiev, Theis, Cichocki, Bobin, Moudden, Starck,...). Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

31 Multi-measurement case Let h i = f + g i R n, i = 1,..., N be N measurements. The problem is to recover f and the g i s. Let A f R n m f and A g R n mg two dictionaries as before. Assume that A f and A g are left invertible. The reconstruction method applied here consists in the minimisation of min y 0 + λ y R m f +Nmg where y = t [ t y f, t y 1 g,..., t y N g ]. N [A f, A g ] i=1 [ yf y i g ] h i 2, The noisy case: h i = f + g i + n i, n i 2 ε for some ε > 0. Why does this provide a better reconstruction? Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

32 Multi-measurement case Let h i = f + g i R n, i = 1,..., N be N measurements. The problem is to recover f and the g i s. Let A f R n m f and A g R n mg two dictionaries as before. Assume that A f and A g are left invertible. The reconstruction method applied here consists in the minimisation of min y 0 + λ y R m f +Nmg where y = t [ t y f, t y 1 g,..., t y N g ]. N [A f, A g ] i=1 [ yf y i g ] h i 2, The noisy case: h i = f + g i + n i, n i 2 ε for some ε > 0. Why does this provide a better reconstruction? Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

33 Multi-measurement case Let h i = f + g i R n, i = 1,..., N be N measurements. The problem is to recover f and the g i s. Let A f R n m f and A g R n mg two dictionaries as before. Assume that A f and A g are left invertible. The reconstruction method applied here consists in the minimisation of min y 0 + λ y R m f +Nmg where y = t [ t y f, t y 1 g,..., t y N g ]. N [A f, A g ] i=1 [ yf y i g ] h i 2, The noisy case: h i = f + g i + n i, n i 2 ε for some ε > 0. Why does this provide a better reconstruction? Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

34 Multi-measurement case Let h i = f + g i R n, i = 1,..., N be N measurements. The problem is to recover f and the g i s. Let A f R n m f and A g R n mg two dictionaries as before. Assume that A f and A g are left invertible. The reconstruction method applied here consists in the minimisation of min y 0 + λ y R m f +Nmg where y = t [ t y f, t y 1 g,..., t y N g ]. N [A f, A g ] i=1 [ yf y i g ] h i 2, The noisy case: h i = f + g i + n i, n i 2 ε for some ε > 0. Why does this provide a better reconstruction? Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

35 Multi-measurement case Let h i = f + g i R n, i = 1,..., N be N measurements. The problem is to recover f and the g i s. Let A f R n m f and A g R n mg two dictionaries as before. Assume that A f and A g are left invertible. The reconstruction method applied here consists in the minimisation of min y 0 + λ y R m f +Nmg where y = t [ t y f, t y 1 g,..., t y N g ]. N [A f, A g ] i=1 [ yf y i g ] h i 2, The noisy case: h i = f + g i + n i, n i 2 ε for some ε > 0. Why does this provide a better reconstruction? Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

36 Assumptions In the applications we have in mind, we shall have (indirect) control on the g i s. We need enough incoherent data: this is measured by their disjoint sparsity. Write f A f ỹ f and g i A g ỹ i g. Definition Take β > 0, N N, ỹ f R m f and ỹ 1 g,..., ỹ N g R mg. We say that {ỹ 1 g,..., ỹ N g } is a complete set of measurements if these two conditions hold true: 1. if ỹ i g(α) ỹ j g(α) β for some α {1,..., m g } then i = j; 2. for every p R n such that {α : (A 1 g p)(α) ε} there holds N #supp (A 1 f p)\supp ỹ f + #{α : (A 1 g p)(α) ε}\supp ỹg i > 6+# N supp ỹg i + ỹ 0 f. i=1 i=1 1. is mainly a technical assumption. The uncertainty principle A 1 f p 0 + A 1 g p 0 2/M and a big enough N make 2. easy to satisfy, provided that the g i s are disjointly sparse. The bigger the noise level ε is, the bigger N needs to be. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

37 Assumptions In the applications we have in mind, we shall have (indirect) control on the g i s. We need enough incoherent data: this is measured by their disjoint sparsity. Write f A f ỹ f and g i A g ỹ i g. Definition Take β > 0, N N, ỹ f R m f and ỹ 1 g,..., ỹ N g R mg. We say that {ỹ 1 g,..., ỹ N g } is a complete set of measurements if these two conditions hold true: 1. if ỹ i g(α) ỹ j g(α) β for some α {1,..., m g } then i = j; 2. for every p R n such that {α : (A 1 g p)(α) ε} there holds N #supp (A 1 f p)\supp ỹ f + #{α : (A 1 g p)(α) ε}\supp ỹg i > 6+# N supp ỹg i + ỹ 0 f. i=1 i=1 1. is mainly a technical assumption. The uncertainty principle A 1 f p 0 + A 1 g p 0 2/M and a big enough N make 2. easy to satisfy, provided that the g i s are disjointly sparse. The bigger the noise level ε is, the bigger N needs to be. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

38 Assumptions In the applications we have in mind, we shall have (indirect) control on the g i s. We need enough incoherent data: this is measured by their disjoint sparsity. Write f A f ỹ f and g i A g ỹ i g. Definition Take β > 0, N N, ỹ f R m f and ỹ 1 g,..., ỹ N g R mg. We say that {ỹ 1 g,..., ỹ N g } is a complete set of measurements if these two conditions hold true: 1. if ỹ i g(α) ỹ j g(α) β for some α {1,..., m g } then i = j; 2. for every p R n such that {α : (A 1 g p)(α) ε} there holds N #supp (A 1 f p)\supp ỹ f + #{α : (A 1 g p)(α) ε}\supp ỹg i > 6+# N supp ỹg i + ỹ 0 f. i=1 i=1 1. is mainly a technical assumption. The uncertainty principle A 1 f p 0 + A 1 g p 0 2/M and a big enough N make 2. easy to satisfy, provided that the g i s are disjointly sparse. The bigger the noise level ε is, the bigger N needs to be. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

39 Assumptions In the applications we have in mind, we shall have (indirect) control on the g i s. We need enough incoherent data: this is measured by their disjoint sparsity. Write f A f ỹ f and g i A g ỹ i g. Definition Take β > 0, N N, ỹ f R m f and ỹ 1 g,..., ỹ N g R mg. We say that {ỹ 1 g,..., ỹ N g } is a complete set of measurements if these two conditions hold true: 1. if ỹ i g(α) ỹ j g(α) β for some α {1,..., m g } then i = j; 2. for every p R n such that {α : (A 1 g p)(α) ε} there holds N #supp (A 1 f p)\supp ỹ f + #{α : (A 1 g p)(α) ε}\supp ỹg i > 6+# N supp ỹg i + ỹ 0 f. i=1 i=1 1. is mainly a technical assumption. The uncertainty principle A 1 f p 0 + A 1 g p 0 2/M and a big enough N make 2. easy to satisfy, provided that the g i s are disjointly sparse. The bigger the noise level ε is, the bigger N needs to be. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

40 Assumptions In the applications we have in mind, we shall have (indirect) control on the g i s. We need enough incoherent data: this is measured by their disjoint sparsity. Write f A f ỹ f and g i A g ỹ i g. Definition Take β > 0, N N, ỹ f R m f and ỹ 1 g,..., ỹ N g R mg. We say that {ỹ 1 g,..., ỹ N g } is a complete set of measurements if these two conditions hold true: 1. if ỹ i g(α) ỹ j g(α) β for some α {1,..., m g } then i = j; 2. for every p R n such that {α : (A 1 g p)(α) ε} there holds N #supp (A 1 f p)\supp ỹ f + #{α : (A 1 g p)(α) ε}\supp ỹg i > 6+# N supp ỹg i + ỹ 0 f. i=1 i=1 1. is mainly a technical assumption. The uncertainty principle A 1 f p 0 + A 1 g p 0 2/M and a big enough N make 2. easy to satisfy, provided that the g i s are disjointly sparse. The bigger the noise level ε is, the bigger N needs to be. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

41 Assumptions In the applications we have in mind, we shall have (indirect) control on the g i s. We need enough incoherent data: this is measured by their disjoint sparsity. Write f A f ỹ f and g i A g ỹ i g. Definition Take β > 0, N N, ỹ f R m f and ỹ 1 g,..., ỹ N g R mg. We say that {ỹ 1 g,..., ỹ N g } is a complete set of measurements if these two conditions hold true: 1. if ỹ i g(α) ỹ j g(α) β for some α {1,..., m g } then i = j; 2. for every p R n such that supp (A 1 g ) N + # ( supp (A 1 f p) \ supp ỹ f i=1 p) there holds # ( supp (A 1 g p) \ supp ỹi g) > C. 1. is mainly a technical assumption. The uncertainty principle A 1 f p 0 + A 1 g p 0 2/M and a big enough N make 2. easy to satisfy, provided that the g i s are disjointly sparse. The bigger the noise level ε is, the bigger N needs to be. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

42 Assumptions In the applications we have in mind, we shall have (indirect) control on the g i s. We need enough incoherent data: this is measured by their disjoint sparsity. Write f A f ỹ f and g i A g ỹ i g. Definition Take β > 0, N N, ỹ f R m f and ỹ 1 g,..., ỹ N g R mg. We say that {ỹ 1 g,..., ỹ N g } is a complete set of measurements if these two conditions hold true: 1. if ỹ i g(α) ỹ j g(α) β for some α {1,..., m g } then i = j; 2. for every p R n such that supp (A 1 g ) N + # ( supp (A 1 f p) \ supp ỹ f i=1 p) there holds # ( supp (A 1 g p) \ supp ỹi g) > C. 1. is mainly a technical assumption. The uncertainty principle A 1 f p 0 + A 1 g p 0 2/M and a big enough N make 2. easy to satisfy, provided that the g i s are disjointly sparse. The bigger the noise level ε is, the bigger N needs to be. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

43 Assumptions In the applications we have in mind, we shall have (indirect) control on the g i s. We need enough incoherent data: this is measured by their disjoint sparsity. Write f A f ỹ f and g i A g ỹ i g. Definition Take β > 0, N N, ỹ f R m f and ỹ 1 g,..., ỹ N g R mg. We say that {ỹ 1 g,..., ỹ N g } is a complete set of measurements if these two conditions hold true: 1. if ỹ i g(α) ỹ j g(α) β for some α {1,..., m g } then i = j; 2. for every p R n such that supp (A 1 g ) N + # ( supp (A 1 f p) \ supp ỹ f i=1 p) there holds # ( supp (A 1 g p) \ supp ỹi g) > C. 1. is mainly a technical assumption. The uncertainty principle A 1 f p 0 + A 1 g p 0 2/M and a big enough N make 2. easy to satisfy, provided that the g i s are disjointly sparse. The bigger the noise level ε is, the bigger N needs to be. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

44 Assumptions In the applications we have in mind, we shall have (indirect) control on the g i s. We need enough incoherent data: this is measured by their disjoint sparsity. Write f A f ỹ f and g i A g ỹ i g. Definition Take β > 0, N N, ỹ f R m f and ỹ 1 g,..., ỹ N g R mg. We say that {ỹ 1 g,..., ỹ N g } is a complete set of measurements if these two conditions hold true: 1. if ỹ i g(α) ỹ j g(α) β for some α {1,..., m g } then i = j; 2. for every p R n such that {α : (A 1 g p)(α) ε} there holds N #supp (A 1 f p)\supp ỹ f + #{α : (A 1 g p)(α) ε}\supp ỹg i > 6+# N supp ỹg i + ỹ 0 f. i=1 i=1 1. is mainly a technical assumption. The uncertainty principle A 1 f p 0 + A 1 g p 0 2/M and a big enough N make 2. easy to satisfy, provided that the g i s are disjointly sparse. The bigger the noise level ε is, the bigger N needs to be. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

45 Assumptions In the applications we have in mind, we shall have (indirect) control on the g i s. We need enough incoherent data: this is measured by their disjoint sparsity. Write f A f ỹ f and g i A g ỹ i g. Definition Take β > 0, N N, ỹ f R m f and ỹ 1 g,..., ỹ N g R mg. We say that {ỹ 1 g,..., ỹ N g } is a complete set of measurements if these two conditions hold true: 1. if ỹ i g(α) ỹ j g(α) β for some α {1,..., m g } then i = j; 2. for every p R n such that {α : (A 1 g p)(α) ε} there holds N #supp (A 1 f p)\supp ỹ f + #{α : (A 1 g p)(α) ε}\supp ỹg i > 6+# N supp ỹg i + ỹ 0 f. i=1 i=1 1. is mainly a technical assumption. The uncertainty principle A 1 f p 0 + A 1 g p 0 2/M and a big enough N make 2. easy to satisfy, provided that the g i s are disjointly sparse. The bigger the noise level ε is, the bigger N needs to be. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

46 Main result Theorem Take β > 0, N N, x f R m f and ỹg, 1..., ỹg N R mg. Assume that {ỹg, 1..., ỹg N } is complete. There exists ε > 0 depending only on N, A g, ỹ f 0 and β such that for every ε ε the following is true. Assume that f, g i, n i R n satisfy n i 2 ε and Af ỹ f f 2 ε, Ag ỹ i g g i 2 ε, i = 1,..., N, and let y f R m f and yg i R mg realise the minimum of min y N Af y f + A g y i y R m f +Nmg g h 2 i, εn i=1 where h i = f + g i + n i.then A f y f f 2 C ε, A g y i g g i 2 C ε, i = 1,..., N for some C > 0 depending only on A g. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

47 Main result Theorem Take β > 0, N N, x f R m f and ỹg, 1..., ỹg N R mg. Assume that {ỹg, 1..., ỹg N } is complete. There exists ε > 0 depending only on N, A g, ỹ f 0 and β such that for every ε ε the following is true. Assume that f, g i, n i R n satisfy n i 2 ε and Af ỹ f f 2 ε, Ag ỹ i g g i 2 ε, i = 1,..., N, and let y f R m f and yg i R mg realise the minimum of min y N Af y f + A g y i y R m f +Nmg g h 2 i, εn i=1 where h i = f + g i + n i.then A f y f f 2 C ε, A g y i g g i 2 C ε, i = 1,..., N for some C > 0 depending only on A g. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

48 Main result Theorem Take β > 0, N N, x f R m f and ỹg, 1..., ỹg N R mg. Assume that {ỹg, 1..., ỹg N } is complete. There exists ε > 0 depending only on N, A g, ỹ f 0 and β such that for every ε ε the following is true. Assume that f, g i, n i R n satisfy n i 2 ε and Af ỹ f f 2 ε, Ag ỹ i g g i 2 ε, i = 1,..., N, and let y f R m f and yg i R mg realise the minimum of min y N Af y f + A g y i y R m f +Nmg g h 2 i, εn i=1 where h i = f + g i + n i.then A f y f f 2 C ε, A g y i g g i 2 C ε, i = 1,..., N for some C > 0 depending only on A g. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

49 Quantitative photoacoustic tomography, Γ = 1 It gives us the image H(x) = Γ(x)µ(x)u(x) = µ(x)u(x). Need to find µ. In the general case Γ 1, apply the same approach and find Γµ and u. Then by using the PDE approach all the unknowns can be reconstructed. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

50 Quantitative photoacoustic tomography, Γ = 1 It gives us the image H(x) = Γ(x)µ(x)u(x) = µ(x)u(x). Need to find µ. In the general case Γ 1, apply the same approach and find Γµ and u. Then by using the PDE approach all the unknowns can be reconstructed. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

51 Quantitative photoacoustic tomography, Γ = 1 It gives us the image H(x) = Γ(x)µ(x)u(x) = µ(x)u(x). Need to find µ. In the general case Γ 1, apply the same approach and find Γµ and u. Then by using the PDE approach all the unknowns can be reconstructed. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

52 Quantitative photoacoustic tomography, Γ = 1 Multi-measurement data: H(x) = µ(x)u i (x), where { ui + µu i = 0 in Ω, u i = ϕ i on Ω. Taking a log: h i = log µ + log u i. The above method may be used if log µ and log u i can be sparsely represented w.r.t. two different dictionaries. Following [Rosenthal, Razansky, Ntziachristos, 2009], observe that: the light absorption µ is a constitutive parameter of the tissue, and as such is discontinuous. Its discontinuities are typically the inclusions we are looking for; the light intensity ui is a solution of a PDE, and as such enjoys higher regularity properties. This motivates the following choices for the dictionaries: Af : wavelets, curvelets, ridgelets, shearlets,... (for simplicity, we shall use Haar wavelets) Ag: low frequency trigonometric polynomials The disjoint sparsity signal separation method may be used to recover µ. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

53 Quantitative photoacoustic tomography, Γ = 1 Multi-measurement data: H(x) = µ(x)u i (x), where { ui + µu i = 0 in Ω, u i = ϕ i on Ω. Taking a log: h i = log µ + log u i. The above method may be used if log µ and log u i can be sparsely represented w.r.t. two different dictionaries. Following [Rosenthal, Razansky, Ntziachristos, 2009], observe that: the light absorption µ is a constitutive parameter of the tissue, and as such is discontinuous. Its discontinuities are typically the inclusions we are looking for; the light intensity ui is a solution of a PDE, and as such enjoys higher regularity properties. This motivates the following choices for the dictionaries: Af : wavelets, curvelets, ridgelets, shearlets,... (for simplicity, we shall use Haar wavelets) Ag: low frequency trigonometric polynomials The disjoint sparsity signal separation method may be used to recover µ. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

54 Quantitative photoacoustic tomography, Γ = 1 Multi-measurement data: H(x) = µ(x)u i (x), where { ui + µu i = 0 in Ω, u i = ϕ i on Ω. Taking a log: h i = log µ + log u i. The above method may be used if log µ and log u i can be sparsely represented w.r.t. two different dictionaries. Following [Rosenthal, Razansky, Ntziachristos, 2009], observe that: the light absorption µ is a constitutive parameter of the tissue, and as such is discontinuous. Its discontinuities are typically the inclusions we are looking for; the light intensity ui is a solution of a PDE, and as such enjoys higher regularity properties. This motivates the following choices for the dictionaries: Af : wavelets, curvelets, ridgelets, shearlets,... (for simplicity, we shall use Haar wavelets) Ag: low frequency trigonometric polynomials The disjoint sparsity signal separation method may be used to recover µ. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

55 Quantitative photoacoustic tomography, Γ = 1 Multi-measurement data: H(x) = µ(x)u i (x), where { ui + µu i = 0 in Ω, u i = ϕ i on Ω. Taking a log: h i = log µ + log u i. The above method may be used if log µ and log u i can be sparsely represented w.r.t. two different dictionaries. Following [Rosenthal, Razansky, Ntziachristos, 2009], observe that: the light absorption µ is a constitutive parameter of the tissue, and as such is discontinuous. Its discontinuities are typically the inclusions we are looking for; the light intensity ui is a solution of a PDE, and as such enjoys higher regularity properties. This motivates the following choices for the dictionaries: Af : wavelets, curvelets, ridgelets, shearlets,... (for simplicity, we shall use Haar wavelets) Ag: low frequency trigonometric polynomials The disjoint sparsity signal separation method may be used to recover µ. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

56 Quantitative photoacoustic tomography, Γ = 1 Multi-measurement data: H(x) = µ(x)u i (x), where { ui + µu i = 0 in Ω, u i = ϕ i on Ω. Taking a log: h i = log µ + log u i. The above method may be used if log µ and log u i can be sparsely represented w.r.t. two different dictionaries. Following [Rosenthal, Razansky, Ntziachristos, 2009], observe that: the light absorption µ is a constitutive parameter of the tissue, and as such is discontinuous. Its discontinuities are typically the inclusions we are looking for; the light intensity ui is a solution of a PDE, and as such enjoys higher regularity properties. This motivates the following choices for the dictionaries: Af : wavelets, curvelets, ridgelets, shearlets,... (for simplicity, we shall use Haar wavelets) Ag: low frequency trigonometric polynomials The disjoint sparsity signal separation method may be used to recover µ. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

57 Quantitative photoacoustic tomography, Γ = 1 Multi-measurement data: H(x) = µ(x)u i (x), where { ui + µu i = 0 in Ω, u i = ϕ i on Ω. Taking a log: h i = log µ + log u i. The above method may be used if log µ and log u i can be sparsely represented w.r.t. two different dictionaries. Following [Rosenthal, Razansky, Ntziachristos, 2009], observe that: the light absorption µ is a constitutive parameter of the tissue, and as such is discontinuous. Its discontinuities are typically the inclusions we are looking for; the light intensity ui is a solution of a PDE, and as such enjoys higher regularity properties. This motivates the following choices for the dictionaries: Af : wavelets, curvelets, ridgelets, shearlets,... (for simplicity, we shall use Haar wavelets) Ag: low frequency trigonometric polynomials The disjoint sparsity signal separation method may be used to recover µ. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

58 Example 1: noise-free case, N = 1 (a) The true absorption µ (b) The true intensity ũ 1 (c) The datum H 1 (d) The reconstructed µ Vidéo: N = 1. Vidéo: N = 2. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

59 Example 1: noise-free case, N = 1 (a) The true absorption µ (b) The true intensity ũ 1 (c) The datum H 1 (d) The reconstructed µ Vidéo: N = 1. Vidéo: N = 2. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

60 Example 1: noisy case, N = 1, 3, 5 (a) The datum H 1. (b) The datum H 3. (c) The datum H 5. (d) Reconstructed µ, N = 1. (e) Reconstructed µ, N = 3. (f) Reconstructed µ, N = 5. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

61 Example 2: the Shepp-Logan Phantom (a) µ (b) H 2 (c) H 4 (d) µ, N = 1 (e) µ, N = 3 (f) µ, N = 5 Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

62 Example 2: the Shepp-Logan Phantom with noise (a) H 3 (b) H 5 (c) µ, N = 5 Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

63 Example 3: piecewise smooth (a) µ (b) µ, N = 1 (c) µ, N = 2 Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

64 Conclusions Past The reconstruction in many hybrid imaging inverse problems require the separation of several signals h i = f + g i. PDE techniques are often powerful, but sometimes they are not applicable. Present Multiple measurements and disjoint sparsity can be used to find f and g i. Uniqueness and stability proven. Orthogonal matching pursuit performs well in many numerical simulations related to quantitative photoacoustic tomography. Future How can we ensure that the solutions u i to the PDE yields the necessary incoherence, measured in terms of their disjoint sparsity? Random boundary conditions may help. Norm l 0 norm l 1. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

65 Conclusions Past The reconstruction in many hybrid imaging inverse problems require the separation of several signals h i = f + g i. PDE techniques are often powerful, but sometimes they are not applicable. Present Multiple measurements and disjoint sparsity can be used to find f and g i. Uniqueness and stability proven. Orthogonal matching pursuit performs well in many numerical simulations related to quantitative photoacoustic tomography. Future How can we ensure that the solutions u i to the PDE yields the necessary incoherence, measured in terms of their disjoint sparsity? Random boundary conditions may help. Norm l 0 norm l 1. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18

66 Conclusions Past The reconstruction in many hybrid imaging inverse problems require the separation of several signals h i = f + g i. PDE techniques are often powerful, but sometimes they are not applicable. Present Multiple measurements and disjoint sparsity can be used to find f and g i. Uniqueness and stability proven. Orthogonal matching pursuit performs well in many numerical simulations related to quantitative photoacoustic tomography. Future How can we ensure that the solutions u i to the PDE yields the necessary incoherence, measured in terms of their disjoint sparsity? Random boundary conditions may help. Norm l 0 norm l 1. Giovanni S Alberti (ENS, Paris) Disjoint sparsity and hybrid inverse problems June 16, / 18