Global Estimates for Kernels of Neumann Series and Greens Functions - PowerPoint PPT Presentation

Modifiable Kernels, continued � H 2 ( x , y ) = H ( x , z ) H ( z , y ) d ν ( z ) � K ( x , z ) K ( z , y ) m ( z ) m ( y ) m 2 ( z ) d ω ( z ) = m ( x ) m ( z ) � K ( x , z ) K ( z , y ) K 2 ( x , y ) = m ( y ) d ω ( z ) = m ( x ) m ( y ) . m ( x ) Similarly, get H j ( x , y ) = K j ( x , y ) / ( m ( x ) m ( y )) for all j .

Modifiable Kernels, continued � H 2 ( x , y ) = H ( x , z ) H ( z , y ) d ν ( z ) � K ( x , z ) K ( z , y ) m ( z ) m ( y ) m 2 ( z ) d ω ( z ) = m ( x ) m ( z ) � K ( x , z ) K ( z , y ) K 2 ( x , y ) = m ( y ) d ω ( z ) = m ( x ) m ( y ) . m ( x ) Similarly, get H j ( x , y ) = K j ( x , y ) / ( m ( x ) m ( y )) for all j . ∞ � H j ( x , y ) ≈ H ( x , y ) e CH 2 ( x , y ) / H ( x , y ) So j =1

Modifiable Kernels, continued � H 2 ( x , y ) = H ( x , z ) H ( z , y ) d ν ( z ) � K ( x , z ) K ( z , y ) m ( z ) m ( y ) m 2 ( z ) d ω ( z ) = m ( x ) m ( z ) � K ( x , z ) K ( z , y ) K 2 ( x , y ) = m ( y ) d ω ( z ) = m ( x ) m ( y ) . m ( x ) Similarly, get H j ( x , y ) = K j ( x , y ) / ( m ( x ) m ( y )) for all j . ∞ � H j ( x , y ) ≈ H ( x , y ) e CH 2 ( x , y ) / H ( x , y ) So j =1 ∞ K j ( x , y ) K ( x , y ) � m ( x ) m ( y ) e CK 2 ( x , y ) / K ( x , y ) . implies m ( x ) m ( y ) ≈ j =1

Background of Problem Kalton and Verbitsky (TAMS, 1999) studied the existence of solutions u ≥ 0 to −△ u − qu s = ϕ, for q ≥ 0 , ϕ ≥ 0 , s > 1.

Background of Problem Kalton and Verbitsky (TAMS, 1999) studied the existence of solutions u ≥ 0 to −△ u − qu s = ϕ, for q ≥ 0 , ϕ ≥ 0 , s > 1. Their methods fail for s = 1, the linear case.

Background of Problem Kalton and Verbitsky (TAMS, 1999) studied the existence of solutions u ≥ 0 to −△ u − qu s = ϕ, for q ≥ 0 , ϕ ≥ 0 , s > 1. Their methods fail for s = 1, the linear case. We considered −△ u − qu = ϕ , with q ≥ 0.

Background of Problem Kalton and Verbitsky (TAMS, 1999) studied the existence of solutions u ≥ 0 to −△ u − qu s = ϕ, for q ≥ 0 , ϕ ≥ 0 , s > 1. Their methods fail for s = 1, the linear case. We considered −△ u − qu = ϕ , with q ≥ 0. Eventually we generalized to ( −△ ) α/ 2 u − qu = ϕ , with 0 < α ≤ 2 ( α � = 2 in dimension 2), where ( −△ ) α/ 2 is defined probabilistically on a domain.

Green’s functions Let G ( x , y ) = G ( α ) ( x , y ) be the Green’s kernel for ( −△ ) α/ 2 on a domain Ω ⊆ R n , let G denote the Green’s operator.

Green’s functions Let G ( x , y ) = G ( α ) ( x , y ) be the Green’s kernel for ( −△ ) α/ 2 on a domain Ω ⊆ R n , let G denote the Green’s operator. For example, on R n , G ( x , y ) = c n | x − y | α − n , the kernel of the Riesz potential I α .

Green’s functions Let G ( x , y ) = G ( α ) ( x , y ) be the Green’s kernel for ( −△ ) α/ 2 on a domain Ω ⊆ R n , let G denote the Green’s operator. For example, on R n , G ( x , y ) = c n | x − y | α − n , the kernel of the Riesz potential I α . For a potential q ≥ 0 on Ω, let d ω ( x ) = q ( x ) dx .

Green’s functions Let G ( x , y ) = G ( α ) ( x , y ) be the Green’s kernel for ( −△ ) α/ 2 on a domain Ω ⊆ R n , let G denote the Green’s operator. For example, on R n , G ( x , y ) = c n | x − y | α − n , the kernel of the Riesz potential I α . For a potential q ≥ 0 on Ω, let d ω ( x ) = q ( x ) dx . Let G j be the j th iterate of G defined with respect to ω : G 1 = G and

Green’s functions Let G ( x , y ) = G ( α ) ( x , y ) be the Green’s kernel for ( −△ ) α/ 2 on a domain Ω ⊆ R n , let G denote the Green’s operator. For example, on R n , G ( x , y ) = c n | x − y | α − n , the kernel of the Riesz potential I α . For a potential q ≥ 0 on Ω, let d ω ( x ) = q ( x ) dx . Let G j be the j th iterate of G defined with respect to ω : G 1 = G and � G j ( x , y ) = G j − 1 ( x , z ) G ( z , y ) d ω ( z ) . Ω

Green’s functions Let G ( x , y ) = G ( α ) ( x , y ) be the Green’s kernel for ( −△ ) α/ 2 on a domain Ω ⊆ R n , let G denote the Green’s operator. For example, on R n , G ( x , y ) = c n | x − y | α − n , the kernel of the Riesz potential I α . For a potential q ≥ 0 on Ω, let d ω ( x ) = q ( x ) dx . Let G j be the j th iterate of G defined with respect to ω : G 1 = G and � G j ( x , y ) = G j − 1 ( x , z ) G ( z , y ) d ω ( z ) . Ω ∞ � Let G ( x , y ) = G j ( x , y ) . j =1

Schr¨ odinger equations ( ∗ ) : ( −△ ) α/ 2 u − qu = ϕ on Ω, u = 0 on ∂ Ω .

Schr¨ odinger equations ( ∗ ) : ( −△ ) α/ 2 u − qu = ϕ on Ω, u = 0 on ∂ Ω . G ( −△ ) α/ 2 u − G ( qu ) = G ( ϕ ). Apply G :

Schr¨ odinger equations ( ∗ ) : ( −△ ) α/ 2 u − qu = ϕ on Ω, u = 0 on ∂ Ω . G ( −△ ) α/ 2 u − G ( qu ) = G ( ϕ ). Apply G : � Let Tu ( x ) = G ( qu )( x ) = Ω G ( x , y ) u ( y ) d ω ( y ).

Schr¨ odinger equations ( ∗ ) : ( −△ ) α/ 2 u − qu = ϕ on Ω, u = 0 on ∂ Ω . G ( −△ ) α/ 2 u − G ( qu ) = G ( ϕ ). Apply G : � Let Tu ( x ) = G ( qu )( x ) = Ω G ( x , y ) u ( y ) d ω ( y ). Then we have u − Tu = G ( ϕ ), or ( I − T ) u = G ( ϕ ).

Schr¨ odinger equations ( ∗ ) : ( −△ ) α/ 2 u − qu = ϕ on Ω, u = 0 on ∂ Ω . G ( −△ ) α/ 2 u − G ( qu ) = G ( ϕ ). Apply G : � Let Tu ( x ) = G ( qu )( x ) = Ω G ( x , y ) u ( y ) d ω ( y ). Then we have u − Tu = G ( ϕ ), or ( I − T ) u = G ( ϕ ). u ( x ) = ( I − T ) − 1 G ( ϕ )( x ) Hence

Schr¨ odinger equations ( ∗ ) : ( −△ ) α/ 2 u − qu = ϕ on Ω, u = 0 on ∂ Ω . G ( −△ ) α/ 2 u − G ( qu ) = G ( ϕ ). Apply G : � Let Tu ( x ) = G ( qu )( x ) = Ω G ( x , y ) u ( y ) d ω ( y ). Then we have u − Tu = G ( ϕ ), or ( I − T ) u = G ( ϕ ). u ( x ) = ( I − T ) − 1 G ( ϕ )( x ) Hence ∞ ∞ � � � T j G ( ϕ )( x ) = = G j +1 ( x , y ) ϕ ( y ) dy Ω j =0 j =0

Schr¨ odinger equations ( ∗ ) : ( −△ ) α/ 2 u − qu = ϕ on Ω, u = 0 on ∂ Ω . G ( −△ ) α/ 2 u − G ( qu ) = G ( ϕ ). Apply G : � Let Tu ( x ) = G ( qu )( x ) = Ω G ( x , y ) u ( y ) d ω ( y ). Then we have u − Tu = G ( ϕ ), or ( I − T ) u = G ( ϕ ). u ( x ) = ( I − T ) − 1 G ( ϕ )( x ) Hence ∞ ∞ � � � T j G ( ϕ )( x ) = = G j +1 ( x , y ) ϕ ( y ) dy Ω j =0 j =0 ∞ � � � = G j ( x , y ) ϕ ( y ) dy = G ( x , y ) ϕ ( y ) dy . Ω Ω j =1

Schr¨ odinger equations ( ∗ ) : ( −△ ) α/ 2 u − qu = ϕ on Ω, u = 0 on ∂ Ω . G ( −△ ) α/ 2 u − G ( qu ) = G ( ϕ ). Apply G : � Let Tu ( x ) = G ( qu )( x ) = Ω G ( x , y ) u ( y ) d ω ( y ). Then we have u − Tu = G ( ϕ ), or ( I − T ) u = G ( ϕ ). u ( x ) = ( I − T ) − 1 G ( ϕ )( x ) Hence ∞ ∞ � � � T j G ( ϕ )( x ) = = G j +1 ( x , y ) ϕ ( y ) dy Ω j =0 j =0 ∞ � � � = G j ( x , y ) ϕ ( y ) dy = G ( x , y ) ϕ ( y ) dy . Ω Ω j =1 So G is the kernel of the solution operator for (*).

Green’s Function Estimates for Schr¨ odinger operators Hence we call G ( x , y ) = � ∞ j =1 G j ( x , y ) the Green’s function for the fractional Schr¨ odinger operator ( −△ ) α/ 2 − q .

Green’s Function Estimates for Schr¨ odinger operators Hence we call G ( x , y ) = � ∞ j =1 G j ( x , y ) the Green’s function for the fractional Schr¨ odinger operator ( −△ ) α/ 2 − q . Theorem: Let Ω = R n , or any bounded domain satisfying the boundary Harnack principle (e.g., any bounded Lipschitz domain).

Green’s Function Estimates for Schr¨ odinger operators Hence we call G ( x , y ) = � ∞ j =1 G j ( x , y ) the Green’s function for the fractional Schr¨ odinger operator ( −△ ) α/ 2 − q . Theorem: Let Ω = R n , or any bounded domain satisfying the boundary Harnack principle (e.g., any bounded Lipschitz domain). A.) (Lower bound) Then there exists c = c (Ω , α ) > 0 such that G ( x , y ) ≥ G ( x , y ) e cG 2 ( x , y ) / G ( x , y ) .

Upper bound B. (Upper bound) Let d ω ( y ) = q ( y ) dy , and � Tu ( x ) = G ( x , y ) u ( y ) d ω ( y ) . Ω

Upper bound B. (Upper bound) Let d ω ( y ) = q ( y ) dy , and � Tu ( x ) = G ( x , y ) u ( y ) d ω ( y ) . Ω If � T � L 2 ( ω ) → L 2 (Ω) < 1 , then there exists C = C (Ω , α, � T � ) such that G ( x , y ) ≤ G ( x , y ) e CG 2 ( x , y ) / G ( x , y ) .

About the proof If Ω = R n , then G ( x , y ) = c n | x − y | α − n is a quasi-metric kernal, and the result follows directly from the main theorem above.

About the proof If Ω = R n , then G ( x , y ) = c n | x − y | α − n is a quasi-metric kernal, and the result follows directly from the main theorem above. For a bounded domain, G may not be a quasi-metric kernel.

About the proof If Ω = R n , then G ( x , y ) = c n | x − y | α − n is a quasi-metric kernal, and the result follows directly from the main theorem above. For a bounded domain, G may not be a quasi-metric kernel. However, for a smooth enough domain, estimates for G are known: let δ ( x ) = dist ( x , ∂ Ω). Then

About the proof If Ω = R n , then G ( x , y ) = c n | x − y | α − n is a quasi-metric kernal, and the result follows directly from the main theorem above. For a bounded domain, G may not be a quasi-metric kernel. However, for a smooth enough domain, estimates for G are known: let δ ( x ) = dist ( x , ∂ Ω). Then δ ( x ) δ ( y ) G ( x , y ) ≈ | x − y | n − 2 ( | x − y | + δ ( x ) + δ ( y )) 2

About the proof If Ω = R n , then G ( x , y ) = c n | x − y | α − n is a quasi-metric kernal, and the result follows directly from the main theorem above. For a bounded domain, G may not be a quasi-metric kernel. However, for a smooth enough domain, estimates for G are known: let δ ( x ) = dist ( x , ∂ Ω). Then δ ( x ) δ ( y ) G ( x , y ) ≈ | x − y | n − 2 ( | x − y | + δ ( x ) + δ ( y )) 2 Then it is not difficult to see that G ( x , y ) / ( δ ( x ) δ ( y )) is a quasi-metric kernel.

About the proof, cont’d More generally, it is known (Hansen) that for bounded domains satisfying the boundary Harnack principle, G ( x , y ) / ( m ( x ) m ( y )) is a quasi-metric kernel for m ( x ) = min(1 , G ( x , x 0 )) , for x 0 ∈ Ω fixed.

About the proof, cont’d More generally, it is known (Hansen) that for bounded domains satisfying the boundary Harnack principle, G ( x , y ) / ( m ( x ) m ( y )) is a quasi-metric kernel for m ( x ) = min(1 , G ( x , x 0 )) , for x 0 ∈ Ω fixed. Hence the results follow from the remarks earlier about modifiable kernels.

Conditional Gauge For α = 2, there is a probabilistic formula � ζ 0 q ( X t ) dt , (1) G ( x , y ) / G ( x , y ) = E x , y e

Conditional Gauge For α = 2, there is a probabilistic formula � ζ 0 q ( X t ) dt , (1) G ( x , y ) / G ( x , y ) = E x , y e where X t is the Brownian path, with properly rescaled time, starting at x , E x , y is the conditional expectation conditioned on the event that X t hits y before exiting Ω, and ζ is the time when X t first hits y .

Conditional Gauge For α = 2, there is a probabilistic formula � ζ 0 q ( X t ) dt , (1) G ( x , y ) / G ( x , y ) = E x , y e where X t is the Brownian path, with properly rescaled time, starting at x , E x , y is the conditional expectation conditioned on the event that X t hits y before exiting Ω, and ζ is the time when X t first hits y . Hence our results give upper and lower bounds for the � ζ 0 q ( X t ) dt . conditional gauge E x , y e

Conditional Gauge For α = 2, there is a probabilistic formula � ζ 0 q ( X t ) dt , (1) G ( x , y ) / G ( x , y ) = E x , y e where X t is the Brownian path, with properly rescaled time, starting at x , E x , y is the conditional expectation conditioned on the event that X t hits y before exiting Ω, and ζ is the time when X t first hits y . Hence our results give upper and lower bounds for the � ζ 0 q ( X t ) dt . conditional gauge E x , y e For 0 < α < 2, similar estimates hold for the conditional gauge for α -stable processes.

Conditional Gauge For α = 2, there is a probabilistic formula � ζ 0 q ( X t ) dt , (1) G ( x , y ) / G ( x , y ) = E x , y e where X t is the Brownian path, with properly rescaled time, starting at x , E x , y is the conditional expectation conditioned on the event that X t hits y before exiting Ω, and ζ is the time when X t first hits y . Hence our results give upper and lower bounds for the � ζ 0 q ( X t ) dt . conditional gauge E x , y e For 0 < α < 2, similar estimates hold for the conditional gauge for α -stable processes. Using (1) and Jensen’s inequality, the lower bound G ( x , y ) ≥ G ( x , y ) e cG 2 ( x , y ) / G ( x , y ) with sharp constant ( c = 1?) follows.

Conditional Gauge Our upper bound G ( x , y ) ≤ G ( x , y ) e CG 2 ( x , y ) / G ( x , y ) seems to be new, even in the Schr¨ odinger case α = 2.

Conditional Gauge Our upper bound G ( x , y ) ≤ G ( x , y ) e CG 2 ( x , y ) / G ( x , y ) seems to be new, even in the Schr¨ odinger case α = 2. Question: Is there a probabilistic proof of the upper bound? With a sharper constant?

Applications Consider the following 3 problems on a smooth bounded domain Ω ⊆ R n :

Applications Consider the following 3 problems on a smooth bounded domain Ω ⊆ R n : � −△ u 0 − qu 0 = 1 on Ω , ( ∗ ) u 0 = 0 on ∂ Ω

Applications Consider the following 3 problems on a smooth bounded domain Ω ⊆ R n : � −△ u 0 − qu 0 = 1 on Ω , ( ∗ ) u 0 = 0 on ∂ Ω � −△ u 1 − qu 1 = 0 on Ω , ( ∗∗ ) u 1 = 1 on ∂ Ω

Applications Consider the following 3 problems on a smooth bounded domain Ω ⊆ R n : � −△ u 0 − qu 0 = 1 on Ω , ( ∗ ) u 0 = 0 on ∂ Ω � −△ u 1 − qu 1 = 0 on Ω , ( ∗∗ ) u 1 = 1 on ∂ Ω Remark: u 1 is called the Feynman-Kac gauge.

Applications Consider the following 3 problems on a smooth bounded domain Ω ⊆ R n : � −△ u 0 − qu 0 = 1 on Ω , ( ∗ ) u 0 = 0 on ∂ Ω � −△ u 1 − qu 1 = 0 on Ω , ( ∗∗ ) u 1 = 1 on ∂ Ω Remark: u 1 is called the Feynman-Kac gauge. −△ v − |∇ v | 2 = q on Ω , � ( ∗ ∗ ∗ ) v = 0 on ∂ Ω

Applications Consider the following 3 problems on a smooth bounded domain Ω ⊆ R n : � −△ u 0 − qu 0 = 1 on Ω , ( ∗ ) u 0 = 0 on ∂ Ω � −△ u 1 − qu 1 = 0 on Ω , ( ∗∗ ) u 1 = 1 on ∂ Ω Remark: u 1 is called the Feynman-Kac gauge. −△ v − |∇ v | 2 = q on Ω , � ( ∗ ∗ ∗ ) v = 0 on ∂ Ω Remark: this is an equation of Ricatti type

Application: solvability of (**) and (***) Let P be the Poisson kernel for a bounded C 2 domain Ω. Define the balyage operator P ∗ by � P ∗ f ( z ) = P ( x , z ) f ( x ) dx , Ω for z ∈ ∂ Ω.

Application: solvability of (**) and (***) Let P be the Poisson kernel for a bounded C 2 domain Ω. Define the balyage operator P ∗ by � P ∗ f ( z ) = P ( x , z ) f ( x ) dx , Ω for z ∈ ∂ Ω. Theorem: Suppose � T � < 1. Then there exists C > 0 such that if � e CP ∗ ( δ q ) d σ < ∞ , ∂ Ω where δ ( x ) = dist ( x , ∂ Ω), then ( ∗∗ ) and ( ∗ ∗ ∗ ) have solutions.

Application: solvability of (**) and (***) (cont’d) For a given C > 0, � e CP ∗ ( δ q ) d σ < ∞ ∂ Ω holds if

Application: solvability of (**) and (***) (cont’d) For a given C > 0, � e CP ∗ ( δ q ) d σ < ∞ ∂ Ω holds if � P ∗ ( δ q ) � BMO ( ∂ Ω) < ǫ, for ǫ small enough, which in turn holds if

Application: solvability of (**) and (***) (cont’d) For a given C > 0, � e CP ∗ ( δ q ) d σ < ∞ ∂ Ω holds if � P ∗ ( δ q ) � BMO ( ∂ Ω) < ǫ, for ǫ small enough, which in turn holds if � δ q � C < ǫ 1 , for ǫ 1 small enough, where � δ q � C denotes the Carleson norm of the measure δ ( x ) q ( x ) dx .

Brief outline of proof Have formal solutions u 0 , u 1 , need to show they are finite a.e.

Brief outline of proof Have formal solutions u 0 , u 1 , need to show they are finite a.e. Theorem about quasi-metric kernels implies: c 1 δ e c ( T δ ) /δ ≤ u 0 ≤ C 1 δ e C ( T δ ) /δ .

Brief outline of proof Have formal solutions u 0 , u 1 , need to show they are finite a.e. Theorem about quasi-metric kernels implies: c 1 δ e c ( T δ ) /δ ≤ u 0 ≤ C 1 δ e C ( T δ ) /δ . This in turn implies � � e CP ∗ ( δ q )( z ) d σ ( z ) ≥ c | ∂ Ω | + c u 0 ( y ) d ω ( y ) . ∂ Ω Ω

Brief outline of proof Have formal solutions u 0 , u 1 , need to show they are finite a.e. Theorem about quasi-metric kernels implies: c 1 δ e c ( T δ ) /δ ≤ u 0 ≤ C 1 δ e C ( T δ ) /δ . This in turn implies � � e CP ∗ ( δ q )( z ) d σ ( z ) ≥ c | ∂ Ω | + c u 0 ( y ) d ω ( y ) . ∂ Ω Ω Hence under assumption of theorem, get u 0 ∈ L 1 ( d ω ), which implies u 1 ∈ L 1 ( dx ), so u 1 solves ( ∗∗ ). Then v = log u satsifies ( ∗ ∗ ∗ ).

Remark Problems ( ∗ ) and ( ∗∗ ) have formal solutions. Recall that ∞ � � u ( x ) = G j ( x , y ) f ( y ) dy Ω j =1 formally satisfies ( −△ − q ) u = f on Ω, u = 0 on ∂ Ω.

Remark Problems ( ∗ ) and ( ∗∗ ) have formal solutions. Recall that ∞ � � u ( x ) = G j ( x , y ) f ( y ) dy Ω j =1 formally satisfies ( −△ − q ) u = f on Ω, u = 0 on ∂ Ω. Taking f = 1, the formal solution of ( ∗ ) is ∞ � � u 0 ( x ) = G j ( x , y ) dy . Ω j =1

Remark, cont’d We claim that the formal solution of ( ∗∗ ) is ∞ � � u 1 ( x ) = 1 + G j ( x , y ) d ω ( y ) Ω j =1

Remark, cont’d We claim that the formal solution of ( ∗∗ ) is ∞ � � u 1 ( x ) = 1 + G j ( x , y ) d ω ( y ) Ω j =1 Proof: Note that u 1 = 1 on ∂ Ω since G j ( x , y ) = 0 for all x ∈ ∂ Ω, for all j . Next, ( −△ − q )1 = − q , and

Remark, cont’d We claim that the formal solution of ( ∗∗ ) is ∞ � � u 1 ( x ) = 1 + G j ( x , y ) d ω ( y ) Ω j =1 Proof: Note that u 1 = 1 on ∂ Ω since G j ( x , y ) = 0 for all x ∈ ∂ Ω, for all j . Next, ( −△ − q )1 = − q , and ∞ ∞ � � � � w ( x ) = G j ( x , y ) d ω ( y ) = G j ( x , y ) q ( y ) dy Ω Ω j =1 j =1 solves ( −△ − q ) w = q on Ω,

Remark, cont’d We claim that the formal solution of ( ∗∗ ) is ∞ � � u 1 ( x ) = 1 + G j ( x , y ) d ω ( y ) Ω j =1 Proof: Note that u 1 = 1 on ∂ Ω since G j ( x , y ) = 0 for all x ∈ ∂ Ω, for all j . Next, ( −△ − q )1 = − q , and ∞ ∞ � � � � w ( x ) = G j ( x , y ) d ω ( y ) = G j ( x , y ) q ( y ) dy Ω Ω j =1 j =1 solves ( −△ − q ) w = q on Ω, so ( −△ − q ) u 1 = ( −△ − q )(1 + w ) = − q + q = 0 in Ω.

Remark, cont’d We claim that the formal solution of ( ∗∗ ) is ∞ � � u 1 ( x ) = 1 + G j ( x , y ) d ω ( y ) Ω j =1 Proof: Note that u 1 = 1 on ∂ Ω since G j ( x , y ) = 0 for all x ∈ ∂ Ω, for all j . Next, ( −△ − q )1 = − q , and ∞ ∞ � � � � w ( x ) = G j ( x , y ) d ω ( y ) = G j ( x , y ) q ( y ) dy Ω Ω j =1 j =1 solves ( −△ − q ) w = q on Ω, so ( −△ − q ) u 1 = ( −△ − q )(1 + w ) = − q + q = 0 in Ω. So the only question is whether the formal solutions are finite a.e.

Sketch of proof First, ( ∗∗ ) is related to ( ∗ ∗ ∗ ) as follows: if u 1 > 0 satisfies ( ∗∗ ), then v = log u satisfies ( ∗ ∗ ∗ ), and if v satisfies ( ∗ ∗ ∗ ) then u 1 = e v satisfies ( ∗∗ ).

Sketch of proof First, ( ∗∗ ) is related to ( ∗ ∗ ∗ ) as follows: if u 1 > 0 satisfies ( ∗∗ ), then v = log u satisfies ( ∗ ∗ ∗ ), and if v satisfies ( ∗ ∗ ∗ ) then u 1 = e v satisfies ( ∗∗ ). Second, ( ∗ ) is related to ( ∗∗ ) as follows:

Sketch of proof First, ( ∗∗ ) is related to ( ∗ ∗ ∗ ) as follows: if u 1 > 0 satisfies ( ∗∗ ), then v = log u satisfies ( ∗ ∗ ∗ ), and if v satisfies ( ∗ ∗ ∗ ) then u 1 = e v satisfies ( ∗∗ ). Second, ( ∗ ) is related to ( ∗∗ ) as follows: � ∞ � � � � � u 1 dx = 1 + G j ( x , y ) d ω ( y ) dx Ω Ω Ω j =1

Sketch of proof First, ( ∗∗ ) is related to ( ∗ ∗ ∗ ) as follows: if u 1 > 0 satisfies ( ∗∗ ), then v = log u satisfies ( ∗ ∗ ∗ ), and if v satisfies ( ∗ ∗ ∗ ) then u 1 = e v satisfies ( ∗∗ ). Second, ( ∗ ) is related to ( ∗∗ ) as follows: � ∞ � � � � � u 1 dx = 1 + G j ( x , y ) d ω ( y ) dx Ω Ω Ω j =1 ∞ � � � = | Ω | + G j ( x , y )( x , y ) dx d ω ( y ) Ω Ω j =1