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Equilibrium and Balayage A mini-tutorial A. Martnez-Finkelshtein (U. Almera) Optimal point configurations and orthogonal polynomials CIEM Castro Urdiales April 19, 2017 Equilibrium and Balayage A mini-tutorial A.


  1. Equilibrium and Balayage A mini-tutorial A. Martínez-Finkelshtein (U. Almería) “Optimal point configurations and orthogonal polynomials” CIEM Castro Urdiales April 19, 2017

  2. Equilibrium and Balayage A mini-tutorial A. Martínez-Finkelshtein (U. Almería) “Optimal point configurations and orthogonal polynomials” CIEM Castro Urdiales April 19, 2017

  3. LOGARITHMIC POTENTIAL AND LOGARITHMIC ENERGY

  4. POTENTIAL THEORY AND POLYNOMIALS

  5. LOGARITHMIC POTENTIALS

  6. LOGARITHMIC POTENTIALS 1 0.8 0.6 0.4 0.2 0 -0.2 -0.4 Example 1 -0.6 Example 2 -0.8 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2

  7. LOGARITHMIC POTENTIALS  − | z | 2 + 1 | z | ≤ 1 , 2 ,   2  V µ ( z ) = log 1 | z | > 1 . | z | ,   

  8. LOGARITHMIC ENERGY

  9. EQUILIBRIUM

  10. CLASSICAL EQUILIBRIUM For K ⊂ C compact, the Robin constant is κ = min { I ( µ ) : µ unit measure on K } 1 0.8 0.6 dx Example 1: dµ ( x ) = 1 − x 2 on [ − 1 , 1] 0.4 √ 0.2 π 0 -0.2 Example 2: dµ ( x ) = 1 2 dx on [ − 1 , 1] -0.4 Example 1 -0.6 Example 2 -0.8 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2

  11. CLASSICAL EQUILIBRIUM For K ⊂ C compact, the Robin constant is κ = min { I ( µ ) : µ unit measure on K } Classical application: asymptotic distribution of zeros of standard orthogonal polynomials 1 0.8 0.6 dx Example 1: dµ ( x ) = 1 − x 2 on [ − 1 , 1] 0.4 √ 0.2 π 0 -0.2 Example 2: dµ ( x ) = 1 2 dx on [ − 1 , 1] -0.4 Example 1 -0.6 Example 2 -0.8 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2

  12. WEIGHTED EQUILIBRIUM weighted logarithmic energy Z Z I Q ( µ ) := k µ k 2 + 2 Q dµ = I ( µ ) + 2 Q dµ For K ⊂ C compact and Q and admissible external field, κ Q = min { I Q ( µ ) : µ unit measure on K } The unique minimizer, λ Q , such that I Q ( λ Q ) = κ Q , is the weighted equilibrium measure on K. Characterization : Potential Density V λ Q + Q ≥ c Q q.e. on K, on supp( λ Q ) ≤ c Q (so, basically, V λ Q + Q = c Q q.e. on supp( λ Q )). New feature: we ignore a priori what is supp( λ Q )!

  13. WEIGHTED EQUILIBRIUM weighted logarithmic energy Z Z I Q ( µ ) := k µ k 2 + 2 Q dµ = I ( µ ) + 2 Q dµ For K ⊂ C compact and Q and admissible external field, κ Q = min { I Q ( µ ) : µ unit measure on K } The unique minimizer, λ Q , such that I Q ( λ Q ) = κ Q , is the Classical application: asymptotic distribution weighted equilibrium measure on K. of zeros of polynomials of varying orthogonality Characterization : Potential Density V λ Q + Q ≥ c Q q.e. on K, on supp( λ Q ) ≤ c Q (so, basically, V λ Q + Q = c Q q.e. on supp( λ Q )). New feature: we ignore a priori what is supp( λ Q )!

  14. CONSTRAINED WEIGHTED EQUILIBRIUM weighted logarithmic energy Z Z I Q ( µ ) := k µ k 2 + 2 Q dµ = I ( µ ) + 2 Q dµ For K ⊂ C compact, Q an admissible external field, and τ a measure with finite energy, supp( τ ) = K , and τ ( K ) > 1, Q = min { I Q ( µ ) : µ unit measure on K with µ ≤ τ } κ τ Again, the unique minimizer, λ τ Q , with I Q ( λ τ Q ) = κ τ Q , is the con- strained equilibrium measure on K. Characterization : Q + Q ≥ c τ V λ τ q.e. on supp( τ − λ τ Constraint Q ) , Q on supp( λ τ Q ) ≤ c Q Potential Density New feature: Void Band Saturated region

  15. CONSTRAINED WEIGHTED EQUILIBRIUM weighted logarithmic energy Z Z I Q ( µ ) := k µ k 2 + 2 Q dµ = I ( µ ) + 2 Q dµ For K ⊂ C compact, Q an admissible external field, and τ a measure with finite energy, supp( τ ) = K , and τ ( K ) > 1, Classical application: asymptotic distribution Q = min { I Q ( µ ) : µ unit measure on K with µ ≤ τ } κ τ of zeros of polynomials of discrete orthogonality Again, the unique minimizer, λ τ Q , with I Q ( λ τ Q ) = κ τ Q , is the con- strained equilibrium measure on K. Characterization : Q + Q ≥ c τ V λ τ q.e. on supp( τ − λ τ Constraint Q ) , Q on supp( λ τ Q ) ≤ c Q Potential Density New feature: Void Band Saturated region

  16. BALAYAGE

  17. MINIMAL ENERGY PARADIGM

  18. CLASSICAL BALAYAGE ν µ Since for reasonable ε and σ 2 M , k µ � ν k  k µ � (1 � ε ) ν � εσ k , we get Z ( V µ − V ν ) d ( ν − σ ) ≥ 0 , for all σ ∈ M In particular, supp( ν ) ⊂ ∂ D and ( “standard” definition q.e. in D c := C \ D of balayage V µ ( z ) = V ν ( z )

  19. CLASSICAL BALAYAGE Recall:  − | z | 2 + 1 | z | ≤ 1 , 2 ,   2  V µ ( z ) = log 1 | z | > 1 . | z | ,    Hence, the unit Lebesgue measure on | z | = 1 is the balayage of the unit plane Lebesgue measure on | z | ≤ 1

  20. CLASSICAL BALAYAGE Observations: • Since V ν − V µ is subharmonic in D , V ν ( z ) ≤ V µ ( z ) , z ∈ C (“balayage decreases the potential”). • ω a = Bal( δ a , ∂ D ) is the harmonic measure of ∂ D ν w.r.t. a . • If D is unbounded, we get instead µ V µ ( z ) = V ν ( z ) + c q.e. in D c • Extension: if supp( µ ) 6⇢ D , we we may take µ = µ out + µ in , � with µ out = µ D c , and define � ν = Bal( µ, D c ) := µ out + Bal( µ in , D c ) • Further extension: if µ is a signed measures, and µ = µ + − µ − is its Jordan decomposition, then Bal( µ, K ) := Bal( µ + , K ) − Bal( µ − , K )

  21. PARTIAL BALAYAGE Let µ be a positive measure, and τ a given measure on C such that µ ( C ) ≤ µ ( τ ). Take M := { ν : ν ≤ τ and ν ( C ) = µ ( C ) } is the partial balayage of µ under τ , denoted by ν = Bal( µ, τ ) A variational argument as before shows that • ν ≤ τ • V ν ≤ V µ on C • V ν = V µ on supp( τ − ν ) Moreover, min( µ, τ ) ≤ Bal( µ, τ ) ≤ τ

  22. PARTIAL BALAYAGE Example 1: a 2D case. Let µ = βδ a ( β > 0), and τ = α mes 2 on p a su ffi ciently large disk | z | ≤ R . If R ≥ | a | + β / ( πα ), then ⇣ ⌘ p � � Bal βδ a , α mes 2 = α mes 2 | z − a | ≤ r , with r = β / ( πα ) . � � | z | ≤ R Notice: here Bal( µ, τ ) either = τ or = 0 (saturation). τ µ Bal( µ, τ )

  23. PARTIAL BALAYAGE Example 1: a 2D case. Let µ = βδ a ( β > 0), and τ = α mes 2 on p a su ffi ciently large disk | z | ≤ R . If R ≥ | a | + β / ( πα ), then ⇣ ⌘ p � � Bal βδ a , α mes 2 = α mes 2 | z − a | ≤ r , with r = β / ( πα ) . � � | z | ≤ R Notice: here Bal( µ, τ ) either = τ or = 0 (saturation). 1.4 Robin measure on [-1,1] Its constrained balayage under =dx 1.2 1 0.8 0.6 0.4 0.2 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

  24. PARTIAL BALAYAGE Example 2: a generalization. Let µ = P n k =1 c k δ a k , c k > 0, and τ = mes 2 on C . Then � Bal ( µ, τ ) = τ � S with ∂ S being an algebraic curve. Moreover, n d mes 2 ( t ) Z c k X = t − z a k − z S k =1 ( S is a quadrature domain). Example 3: a Hele-Shaw flow. Given a domain S 0 3 a , let � µ t = t δ a + mes 2 S 0 , and τ = mes 2 on C . Then � � Bal ( µ t , τ ) = τ t > 0 S t , � In particular, Z Z z k d mes 2 ( z ) = z k d mes 2 ( z ) , k ∈ N , S t S 0 Z Z d mes 2 ( z ) = t + d mes 2 ( z ) S 0 S t S 0

  25. CONNECTIONS BETWEEN BALAYAGE AND EQUILIBRIUM

  26. Weighted Constrained Equilibrium Energy minimization Balayage Classical Partial

  27. CONSTRAINED & WEIGHTED EQUILIBRIUM Constraint Potential Density Potential Density Void Band Saturated region V λ Q + Q ≥ c Q Q + Q ≥ c τ V λ τ q.e. on supp( τ − λ τ Q ) , q.e. on K, Q on supp( λ τ on supp( λ Q ) ≤ c Q Q ) ≤ c Q

  28. BALAYAGE & EQUILIBRIUM ν µ

  29. BALAYAGE & EQUILIBRIUM 1.4 Robin measure on [-1,1] Its constrained balayage under =dx 1.2 1 0.8 0.6 0.4 0.2 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

  30. BALAYAGE & EQUILIBRIUM

  31. BALAYAGE & EQUILIBRIUM Example: consider the weighted equilibrium on C in the external field 1 Q ( z ) = α | z | 2 + β log α , β > 0 . | z − a | , Recall:  − | z | 2 + 1 | z | ≤ 1 , 2 ,   2  V µ ( z ) = log 1 | z | > 1 . | z | ,   

  32. BALAYAGE & EQUILIBRIUM Example: consider the weighted equilibrium on C in the external field 1 Q ( z ) = α | z | 2 + β log α , β > 0 . | z − a | , Hence, we can replace Q by V σ , with r σ = − 2 α β + 1 � | z | ≤ R + βδ a , R = π mes 2 � 2 α Thus, σ − = 2 α � σ + = βδ a , π mes 2 � | z | ≤ R Recall again: τ r Bal ( σ + , σ − ) = 2 α β � π mes 2 r = | z − a | ≤ r , 2 α , �

  33. BALAYAGE & EQUILIBRIUM Example: consider the weighted equilibrium on C in the external field 1 Q ( z ) = α | z | 2 + β log α , β > 0 . | z − a | , Hence, we can replace Q by V σ , with r σ = − 2 α β + 1 � | z | ≤ R + βδ a , R = π mes 2 � 2 α Thus, σ − = 2 α � σ + = βδ a , π mes 2 � | z | ≤ R q q β +1 β We conclude: if | a | ≤ 2 α , 2 α − λ Q = 2 α | z | ≤ R − 2 α a � � π mes 2 π mes 2 � � | z − a | ≤ r

  34. Thank you

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