Inverse potential problems in divergence form: some uniqueness, - PowerPoint PPT Presentation

Characterization of silent distributions in divergence form Consider a 2-D potential in divergence form supported on Γ: P div V ( X ) = − 1 1 � ( div V )( X ′ ) log | X − X ′ | d | X ′ | , X / ∈ supp V , 2 π Γ with V = m ⊗ δ Γ and m = ( m 1 , m 2 ) t a vector field in L p (Γ). 8

Characterization of silent distributions in divergence form Consider a 2-D potential in divergence form supported on Γ: P div V ( X ) = − 1 1 � ( div V )( X ′ ) log | X − X ′ | d | X ′ | , X / ∈ supp V , 2 π Γ with V = m ⊗ δ Γ and m = ( m 1 , m 2 ) t a vector field in L p (Γ). Integrating by parts identifying vectors with complex numbers: 8

Characterization of silent distributions in divergence form Consider a 2-D potential in divergence form supported on Γ: P div V ( X ) = − 1 1 � ( div V )( X ′ ) log | X − X ′ | d | X ′ | , X / ∈ supp V , 2 π Γ with V = m ⊗ δ Γ and m = ( m 1 , m 2 ) t a vector field in L p (Γ). Integrating by parts identifying vectors with complex numbers: ( m 1 + im 2 )( X ′ ) / v ( X ′ ) P div V ( X ) = − 1 � dX ′ , X ′ − X 2 π Γ where v is the unit tangent to Γ. 8

Characterization of silent distributions in divergence form Consider a 2-D potential in divergence form supported on Γ: P div V ( X ) = − 1 1 � ( div V )( X ′ ) log | X − X ′ | d | X ′ | , X / ∈ supp V , 2 π Γ with V = m ⊗ δ Γ and m = ( m 1 , m 2 ) t a vector field in L p (Γ). Integrating by parts identifying vectors with complex numbers: ( m 1 + im 2 )( X ′ ) / v ( X ′ ) P div V ( X ) = − 1 � dX ′ , X ′ − X 2 π Γ where v is the unit tangent to Γ. By what precedes: Corollary div m has null potential in D − if, and only if ( m 1 + im 2 ) / v ∈ H p ( D + ) . 8

Higher dimensional generalization 9

Higher dimensional generalization The purpose of the talk is to carry over what precedes to higher dimension. 9

Higher dimensional generalization The purpose of the talk is to carry over what precedes to higher dimension. By higher dimension we mean higher real dimension, where analytic functions are replaced by gradient of harmonic functions (Stein-Weiss formalism). For simplicity we deal only with the case n = 3. 9

Higher dimensional generalization The purpose of the talk is to carry over what precedes to higher dimension. By higher dimension we mean higher real dimension, where analytic functions are replaced by gradient of harmonic functions (Stein-Weiss formalism). For simplicity we deal only with the case n = 3. The generalization of the previous Hardy decomposition stems from Hodge theory for currents supported on a surface in the ambient space, 9

Higher dimensional generalization The purpose of the talk is to carry over what precedes to higher dimension. By higher dimension we mean higher real dimension, where analytic functions are replaced by gradient of harmonic functions (Stein-Weiss formalism). For simplicity we deal only with the case n = 3. The generalization of the previous Hardy decomposition stems from Hodge theory for currents supported on a surface in the ambient space, but it is conveniently framed in terms of Clifford analysis that we use here as a tool. 9

Some motivation 10

Some motivation For n ≥ 3, we consider harmonic potentials in divergence form: x − y � P div V ( x ) = | x − y | n − 2 div V ( y ) for some vector distribution V = ( v 1 , v 2 , · · · , v n ) t on R n . 10

Some motivation For n ≥ 3, we consider harmonic potentials in divergence form: x − y � P div V ( x ) = | x − y | n − 2 div V ( y ) for some vector distribution V = ( v 1 , v 2 , · · · , v n ) t on R n . They solve ∆ u = div V on R n with “minimal growth” at infinity. 10

Some motivation For n ≥ 3, we consider harmonic potentials in divergence form: x − y � P div V ( x ) = | x − y | n − 2 div V ( y ) for some vector distribution V = ( v 1 , v 2 , · · · , v n ) t on R n . They solve ∆ u = div V on R n with “minimal growth” at infinity. They occur frequently when modeling electro-magnetic phenomena in the quasi-static approximation to Maxwell’s equations. 10

Examples 11

Examples EEG: 11

Examples EEG: Brain assumed non magnetic medium, 11

Examples EEG: Brain assumed non magnetic medium, with constant electric conductivity σ . 11

Examples EEG: Brain assumed non magnetic medium, with constant electric conductivity σ . Then the electric potential is u = P div J p /σ with J p the so-called primary current. 11

Examples EEG: Brain assumed non magnetic medium, with constant electric conductivity σ . Then the electric potential is u = P div J p /σ with J p the so-called primary current. Magnetization 11

Examples EEG: Brain assumed non magnetic medium, with constant electric conductivity σ . Then the electric potential is u = P div J p /σ with J p the so-called primary current. Magnetization If M is a magnetization, (density of magnetic moment), 11

Examples EEG: Brain assumed non magnetic medium, with constant electric conductivity σ . Then the electric potential is u = P div J p /σ with J p the so-called primary current. Magnetization If M is a magnetization, (density of magnetic moment), in the absence of sources, 11

Examples EEG: Brain assumed non magnetic medium, with constant electric conductivity σ . Then the electric potential is u = P div J p /σ with J p the so-called primary current. Magnetization If M is a magnetization, (density of magnetic moment), in the absence of sources, then the scalar magneic potential is u = P div M . 11

Inverse problems 12

Inverse problems The inverse potential problem in divergence form is to recover V from the knowledge of P div V away from the support of V . 12

Inverse problems The inverse potential problem in divergence form is to recover V from the knowledge of P div V away from the support of V . For instance the basic inverse problem in Electro-EncephaloGraphy is to recover the primary current J p (which shows the electrical activity in the brain) 12

Inverse problems The inverse potential problem in divergence form is to recover V from the knowledge of P div V away from the support of V . For instance the basic inverse problem in Electro-EncephaloGraphy is to recover the primary current J p (which shows the electrical activity in the brain) from measurements of the electric field E = −∇ u on the scalp. 12

Inverse problems The inverse potential problem in divergence form is to recover V from the knowledge of P div V away from the support of V . For instance the basic inverse problem in Electro-EncephaloGraphy is to recover the primary current J p (which shows the electrical activity in the brain) from measurements of the electric field E = −∇ u on the scalp. Likewise, the inverse magnetization problem is to recover the magnetization M on a given object, 12

Inverse problems The inverse potential problem in divergence form is to recover V from the knowledge of P div V away from the support of V . For instance the basic inverse problem in Electro-EncephaloGraphy is to recover the primary current J p (which shows the electrical activity in the brain) from measurements of the electric field E = −∇ u on the scalp. Likewise, the inverse magnetization problem is to recover the magnetization M on a given object, from measurements of the field H = −∇ φ near the object. 12

Inverse problems The inverse potential problem in divergence form is to recover V from the knowledge of P div V away from the support of V . For instance the basic inverse problem in Electro-EncephaloGraphy is to recover the primary current J p (which shows the electrical activity in the brain) from measurements of the electric field E = −∇ u on the scalp. Likewise, the inverse magnetization problem is to recover the magnetization M on a given object, from measurements of the field H = −∇ φ near the object. Today, inverse magnetization problems are a hot topic in Earth and Planetary Sciences. 12

Uniqueness issues 13

Uniqueness issues A basic question is: what are the densities V producing the zero field in a given component of R n \ Supp V ? 13

Uniqueness issues A basic question is: what are the densities V producing the zero field in a given component of R n \ Supp V ? Equivalently: when is it that Φ div ( V ) ( X ) = cst in a component (zero if the component is unbounded)? 13

Uniqueness issues A basic question is: what are the densities V producing the zero field in a given component of R n \ Supp V ? Equivalently: when is it that Φ div ( V ) ( X ) = cst in a component (zero if the component is unbounded)? In this case V is called silent from that component. 13

Uniqueness issues A basic question is: what are the densities V producing the zero field in a given component of R n \ Supp V ? Equivalently: when is it that Φ div ( V ) ( X ) = cst in a component (zero if the component is unbounded)? In this case V is called silent from that component. Let us look at the elementary case where V is supported on the horizontal plane with L p density there, 1 < p < ∞ . 13

Uniqueness issues A basic question is: what are the densities V producing the zero field in a given component of R n \ Supp V ? Equivalently: when is it that Φ div ( V ) ( X ) = cst in a component (zero if the component is unbounded)? In this case V is called silent from that component. Let us look at the elementary case where V is supported on the horizontal plane with L p density there, 1 < p < ∞ . This geometry is in fact realistic in scanning microscopy of rocks which are typically sanded down to thin slabs. 13

The thin plate case 14

The thin plate case Thin-plate : V = M ( x 1 , x 2 ) ⊗ δ 0 ( x 3 ) is supported on { x 3 = 0 } , 14

The thin plate case Thin-plate : V = M ( x 1 , x 2 ) ⊗ δ 0 ( x 3 ) is supported on { x 3 = 0 } , M ( x 1 , x 2 ) = ( m 1 ( x 1 , x 2 ) , m 2 ( x 1 , x 2 ) , m 3 ( x 1 , x 2 )) t . 14

The thin plate case Thin-plate : V = M ( x 1 , x 2 ) ⊗ δ 0 ( x 3 ) is supported on { x 3 = 0 } , M ( x 1 , x 2 ) = ( m 1 ( x 1 , x 2 ) , m 2 ( x 1 , x 2 ) , m 3 ( x 1 , x 2 )) t . � x 1 � At any X = , x 3 � = 0, the potential P div V is obtained by x 2 x 3 � x ′ � 1 letting M act on X ′ �→ ( X − X ′ ) / | X − X ′ | 3 , X ′ = : x ′ 2 0 14

The thin plate case Thin-plate : V = M ( x 1 , x 2 ) ⊗ δ 0 ( x 3 ) is supported on { x 3 = 0 } , M ( x 1 , x 2 ) = ( m 1 ( x 1 , x 2 ) , m 2 ( x 1 , x 2 ) , m 3 ( x 1 , x 2 )) t . � x 1 � At any X = , x 3 � = 0, the potential P div V is obtained by x 2 x 3 � x ′ � 1 letting M act on X ′ �→ ( X − X ′ ) / | X − X ′ | 3 , X ′ = : x ′ 2 0 � m 1 ( X ′ )( x 1 − x ′ 1 ) + m 2 ( X ′ )( x 2 − x ′ P div V = 1 2 ) � | X − X ′ | 3 4 π R n + m 3 ( X ′ ) x 3 � | X − X ′ | 3 d x ′ 1 d x ′ 2 14

The thin plate case Thin-plate : V = M ( x 1 , x 2 ) ⊗ δ 0 ( x 3 ) is supported on { x 3 = 0 } , M ( x 1 , x 2 ) = ( m 1 ( x 1 , x 2 ) , m 2 ( x 1 , x 2 ) , m 3 ( x 1 , x 2 )) t . � x 1 � At any X = , x 3 � = 0, the potential P div V is obtained by x 2 x 3 � x ′ � 1 letting M act on X ′ �→ ( X − X ′ ) / | X − X ′ | 3 , X ′ = : x ′ 2 0 � m 1 ( X ′ )( x 1 − x ′ 1 ) + m 2 ( X ′ )( x 2 − x ′ P div V = 1 2 ) � | X − X ′ | 3 4 π R n + m 3 ( X ′ ) x 3 � | X − X ′ | 3 d x ′ 1 d x ′ 2 14

The thin-plate case cont’d 15

The thin-plate case cont’d Thus P div M ( X ) = A 1 ( X ) + A 2 ( X ) + A 3 ( X ) where: 15

The thin-plate case cont’d Thus P div M ( X ) = A 1 ( X ) + A 2 ( X ) + A 3 ( X ) where: m 3 ( X ′ ) x 3 A 3 ( X ) = 1 � | X − X ′ | 3 d x ′ 1 d x ′ 2 . 4 π R n is sgn x 3 times half the harmonic (Poisson) extension of m 3 : A 3 ( X ) = sgn x 3 P X ( m 3 ) / 2, 15

The thin-plate case cont’d Thus P div M ( X ) = A 1 ( X ) + A 2 ( X ) + A 3 ( X ) where: m 3 ( X ′ ) x 3 A 3 ( X ) = 1 � | X − X ′ | 3 d x ′ 1 d x ′ 2 . 4 π R n is sgn x 3 times half the harmonic (Poisson) extension of m 3 : A 3 ( X ) = sgn x 3 P X ( m 3 ) / 2, and A j ( X ) = P X ( R j m j ) / 2 for j = 1 , 2, where 15

The thin-plate case cont’d Thus P div M ( X ) = A 1 ( X ) + A 2 ( X ) + A 3 ( X ) where: m 3 ( X ′ ) x 3 A 3 ( X ) = 1 � | X − X ′ | 3 d x ′ 1 d x ′ 2 . 4 π R n is sgn x 3 times half the harmonic (Poisson) extension of m 3 : A 3 ( X ) = sgn x 3 P X ( m 3 ) / 2, and A j ( X ) = P X ( R j m j ) / 2 for j = 1 , 2, where f ( X ′ ) ( y j − x ′ j ) 1 �� | Y − X ′ | 3 dX ′ , R j ( f )( Y ) := lim j = 1 , 2 , 2 π ǫ → 0 R 2 \ B ( Y ,ǫ ) are the Riesz transforms. 15

Silent planar distributions in divergence form 16

Silent planar distributions in divergence form Assume x 3 > 0. 16

Silent planar distributions in divergence form Assume x 3 > 0. We just saw that P div M ( X ) = A 1 ( X )+ A 2 ( X )+ A 3 ( X ) = 1 2 P X ( R 1 m 1 + R 2 m 2 + m 3 ) . 16

Silent planar distributions in divergence form Assume x 3 > 0. We just saw that P div M ( X ) = A 1 ( X )+ A 2 ( X )+ A 3 ( X ) = 1 2 P X ( R 1 m 1 + R 2 m 2 + m 3 ) . Since the Poisson extension of a function is zero iff the function is zero, M is silent from above iff R 1 m 1 + R 2 m 2 + m 3 = 0 . 16

Silent planar distributions in divergence form Assume x 3 > 0. We just saw that P div M ( X ) = A 1 ( X )+ A 2 ( X )+ A 3 ( X ) = 1 2 P X ( R 1 m 1 + R 2 m 2 + m 3 ) . Since the Poisson extension of a function is zero iff the function is zero, M is silent from above iff R 1 m 1 + R 2 m 2 + m 3 = 0 . Likewise M is silent from below iff R 1 m 1 + R 2 m 2 − m 3 = 0 . 16

Silent planar distributions in divergence form Assume x 3 > 0. We just saw that P div M ( X ) = A 1 ( X )+ A 2 ( X )+ A 3 ( X ) = 1 2 P X ( R 1 m 1 + R 2 m 2 + m 3 ) . Since the Poisson extension of a function is zero iff the function is zero, M is silent from above iff R 1 m 1 + R 2 m 2 + m 3 = 0 . Likewise M is silent from below iff R 1 m 1 + R 2 m 2 − m 3 = 0 . M is silent (from both sides) iff R 1 m 1 + R 2 m 2 = 0 and m 3 = 0. 16

Question What do these quantities mean? 17

Question What do these quantities mean? To approach it, we introduce some classical function spaces. 17

Hardy space of harmonic gradients 18

Hardy space of harmonic gradients Let H p + consist of ∇ u , u harmonic in { x 3 > 0 } , such that � R 2 |∇ u ( X ′ , x 3 ) | p dX ′ < ∞ . sup x 3 > 0 18

Hardy space of harmonic gradients Let H p + consist of ∇ u , u harmonic in { x 3 > 0 } , such that � R 2 |∇ u ( X ′ , x 3 ) | p dX ′ < ∞ . sup x 3 > 0 ∇ u has a nontangential limit on R 2 of the form ( R 1 f , R 2 f , f ) t , f ∈ L p ( R 2 ), 18

Hardy space of harmonic gradients Let H p + consist of ∇ u , u harmonic in { x 3 > 0 } , such that � R 2 |∇ u ( X ′ , x 3 ) | p dX ′ < ∞ . sup x 3 > 0 ∇ u has a nontangential limit on R 2 of the form ( R 1 f , R 2 f , f ) t , f ∈ L p ( R 2 ), and is the Poisson extension thereof. 18

Hardy space of harmonic gradients Let H p + consist of ∇ u , u harmonic in { x 3 > 0 } , such that � R 2 |∇ u ( X ′ , x 3 ) | p dX ′ < ∞ . sup x 3 > 0 ∇ u has a nontangential limit on R 2 of the form ( R 1 f , R 2 f , f ) t , f ∈ L p ( R 2 ), and is the Poisson extension thereof. In other words the R j are the maps sending the normal derivative to the tangential derivatives on the boundary of the solution to Neumann’s problem in the half space. 18

Hardy space of harmonic gradients Let H p + consist of ∇ u , u harmonic in { x 3 > 0 } , such that � R 2 |∇ u ( X ′ , x 3 ) | p dX ′ < ∞ . sup x 3 > 0 ∇ u has a nontangential limit on R 2 of the form ( R 1 f , R 2 f , f ) t , f ∈ L p ( R 2 ), and is the Poisson extension thereof. In other words the R j are the maps sending the normal derivative to the tangential derivatives on the boundary of the solution to Neumann’s problem in the half space. H p − is defined similarly on { x 3 < 0 } , 18

Hardy space of harmonic gradients Let H p + consist of ∇ u , u harmonic in { x 3 > 0 } , such that � R 2 |∇ u ( X ′ , x 3 ) | p dX ′ < ∞ . sup x 3 > 0 ∇ u has a nontangential limit on R 2 of the form ( R 1 f , R 2 f , f ) t , f ∈ L p ( R 2 ), and is the Poisson extension thereof. In other words the R j are the maps sending the normal derivative to the tangential derivatives on the boundary of the solution to Neumann’s problem in the half space. H p − is defined similarly on { x 3 < 0 } , with traces ( − R 1 f , − R 2 f , f ) t . 18

Hardy space of harmonic gradients Let H p + consist of ∇ u , u harmonic in { x 3 > 0 } , such that � R 2 |∇ u ( X ′ , x 3 ) | p dX ′ < ∞ . sup x 3 > 0 ∇ u has a nontangential limit on R 2 of the form ( R 1 f , R 2 f , f ) t , f ∈ L p ( R 2 ), and is the Poisson extension thereof. In other words the R j are the maps sending the normal derivative to the tangential derivatives on the boundary of the solution to Neumann’s problem in the half space. H p − is defined similarly on { x 3 < 0 } , with traces ± have L p nontangential ( − R 1 f , − R 2 f , f ) t . Functions in H p maximal function (Stein-Weiss). 18

Hardy space of harmonic gradients Let H p + consist of ∇ u , u harmonic in { x 3 > 0 } , such that � R 2 |∇ u ( X ′ , x 3 ) | p dX ′ < ∞ . sup x 3 > 0 ∇ u has a nontangential limit on R 2 of the form ( R 1 f , R 2 f , f ) t , f ∈ L p ( R 2 ), and is the Poisson extension thereof. In other words the R j are the maps sending the normal derivative to the tangential derivatives on the boundary of the solution to Neumann’s problem in the half space. H p − is defined similarly on { x 3 < 0 } , with traces ± have L p nontangential ( − R 1 f , − R 2 f , f ) t . Functions in H p maximal function (Stein-Weiss). We put D p for divergence-free vector fields in L p ( R 2 , R 2 ). 18

The Hardy-Hodge decomposition on R 2 Theorem (L.B., D. Hardin, E. Lima, E.B. Saff, B. Weiss) For 1 < p < ∞ one has the direct sum: ( L p ( R 2 )) 3 = H p − ⊕ ( D p × { 0 } ) . + ⊕ H p The decomposition is orthogonal if p = 2 . 19

The Hardy-Hodge decomposition on R 2 Theorem (L.B., D. Hardin, E. Lima, E.B. Saff, B. Weiss) For 1 < p < ∞ one has the direct sum: ( L p ( R 2 )) 3 = H p − ⊕ ( D p × { 0 } ) . + ⊕ H p The decomposition is orthogonal if p = 2 . Thus, every 3-D vector field of L p -class on R 2 is uniquely the sum of (the trace of) a harmonic gradient above, a harmonic gradient below, and a tangent divergence-free vector field. 19

The Hardy-Hodge decomposition on R 2 Theorem (L.B., D. Hardin, E. Lima, E.B. Saff, B. Weiss) For 1 < p < ∞ one has the direct sum: ( L p ( R 2 )) 3 = H p − ⊕ ( D p × { 0 } ) . + ⊕ H p The decomposition is orthogonal if p = 2 . Thus, every 3-D vector field of L p -class on R 2 is uniquely the sum of (the trace of) a harmonic gradient above, a harmonic gradient below, and a tangent divergence-free vector field. Analog to the decomposition of a complex function on R as the sum of two Hardy functions. 19

The Hardy-Hodge decomposition on R 2 Theorem (L.B., D. Hardin, E. Lima, E.B. Saff, B. Weiss) For 1 < p < ∞ one has the direct sum: ( L p ( R 2 )) 3 = H p − ⊕ ( D p × { 0 } ) . + ⊕ H p The decomposition is orthogonal if p = 2 . Thus, every 3-D vector field of L p -class on R 2 is uniquely the sum of (the trace of) a harmonic gradient above, a harmonic gradient below, and a tangent divergence-free vector field. Analog to the decomposition of a complex function on R as the sum of two Hardy functions. Divergence-free term is necessary for not every field is a gradient on R 2 . 19

The Hardy-Hodge decomposition on R 2 Theorem (L.B., D. Hardin, E. Lima, E.B. Saff, B. Weiss) For 1 < p < ∞ one has the direct sum: ( L p ( R 2 )) 3 = H p − ⊕ ( D p × { 0 } ) . + ⊕ H p The decomposition is orthogonal if p = 2 . Thus, every 3-D vector field of L p -class on R 2 is uniquely the sum of (the trace of) a harmonic gradient above, a harmonic gradient below, and a tangent divergence-free vector field. Analog to the decomposition of a complex function on R as the sum of two Hardy functions. Divergence-free term is necessary for not every field is a gradient on R 2 . Projecting on R 2 we get the standard Hodge decomposition � 2 = G p ⊕ D p , � L p ( R 2 ) 19