On surjectivity of partial differential operators with a single - PowerPoint PPT Presentation

On surjectivity of partial differential operators with a single characteristic direction and on Runge pairs for such operators Thomas Kalmes TU Chemnitz Dortmund-Hagen-Wuppertal Analysis Meeting TU Dortmund, January 24, 2019 Thomas Kalmes PDO with a single characteristic direction 1 / 25

Surjectivity of differential operators 1 An approximation theorem of Runge type 2 The linear topological invariant (Ω) for kernels 3 Thomas Kalmes PDO with a single characteristic direction 2 / 25

Surjectivity of differential operators Thomas Kalmes PDO with a single characteristic direction 3 / 25

For P ∈ C [ X 1 , . . . , X d ] set P ( ∂ ) := P ( ∂ 1 , . . . , ∂ d ) := P ( ∂ ∂ , . . . , ) . ∂x 1 ∂x d (E.g. ∆ = P L ( ∂ ) for P L ( ξ ) = � d j =1 ξ 2 j (Laplace operator) ∂t − ∆ x = P H ( ∂ ) for P H ( ξ 1 , . . . , ξ d ) = ξ d − � d − 1 ∂ j =1 ξ 2 j (Heat operator) ∂t + ∆ x = P S ( ∂ ) for P S ( ξ 1 , . . . , ξ d ) = iξ d + � d − 1 i ∂ j =1 ξ 2 j (Schr¨ odinger operator) ∂ 2 d − � d − 1 ∂t 2 − ∆ x = P W ( ∂ ) for P W ( ξ 1 , . . . , ξ d ) = ξ 2 j =1 ξ 2 j (Wave operator) � ∂ 1 ∂x 1 + i ∂ z for P ( ξ 1 , ξ 2 ) = 1 � � � = ∂ ¯ ξ 1 + iξ 2 (Cauchy-Riemann operator).) 2 ∂x 2 2 Thomas Kalmes PDO with a single characteristic direction 4 / 25

For P ∈ C [ X 1 , . . . , X d ] set P ( ∂ ) := P ( ∂ 1 , . . . , ∂ d ) := P ( ∂ ∂ , . . . , ) . ∂x 1 ∂x d (E.g. ∆ = P L ( ∂ ) for P L ( ξ ) = � d j =1 ξ 2 j (Laplace operator) ∂t − ∆ x = P H ( ∂ ) for P H ( ξ 1 , . . . , ξ d ) = ξ d − � d − 1 ∂ j =1 ξ 2 j (Heat operator) ∂t + ∆ x = P S ( ∂ ) for P S ( ξ 1 , . . . , ξ d ) = iξ d + � d − 1 i ∂ j =1 ξ 2 j (Schr¨ odinger operator) ∂ 2 d − � d − 1 ∂t 2 − ∆ x = P W ( ∂ ) for P W ( ξ 1 , . . . , ξ d ) = ξ 2 j =1 ξ 2 j (Wave operator) � ∂ 1 ∂x 1 + i ∂ z for P ( ξ 1 , ξ 2 ) = 1 � � � = ∂ ¯ ξ 1 + iξ 2 (Cauchy-Riemann operator).) 2 ∂x 2 2 P ∈ C [ X 1 , . . . , X d ] \{ 0 } , X ⊆ R d open: for given f , solve P ( ∂ ) u = f in X ! Thomas Kalmes PDO with a single characteristic direction 4 / 25

For P ∈ C [ X 1 , . . . , X d ] set P ( ∂ ) := P ( ∂ 1 , . . . , ∂ d ) := P ( ∂ ∂ , . . . , ) . ∂x 1 ∂x d (E.g. ∆ = P L ( ∂ ) for P L ( ξ ) = � d j =1 ξ 2 j (Laplace operator) ∂t − ∆ x = P H ( ∂ ) for P H ( ξ 1 , . . . , ξ d ) = ξ d − � d − 1 ∂ j =1 ξ 2 j (Heat operator) ∂t + ∆ x = P S ( ∂ ) for P S ( ξ 1 , . . . , ξ d ) = iξ d + � d − 1 i ∂ j =1 ξ 2 j (Schr¨ odinger operator) ∂ 2 d − � d − 1 ∂t 2 − ∆ x = P W ( ∂ ) for P W ( ξ 1 , . . . , ξ d ) = ξ 2 j =1 ξ 2 j (Wave operator) � ∂ 1 ∂x 1 + i ∂ z for P ( ξ 1 , ξ 2 ) = 1 � � � = ∂ ¯ ξ 1 + iξ 2 (Cauchy-Riemann operator).) 2 ∂x 2 2 P ∈ C [ X 1 , . . . , X d ] \{ 0 } , X ⊆ R d open: for given f , solve P ( ∂ ) u = f in X ! Is this possible for every f from a fixed space of functions/distributions? ”Solution” in which sense; classical, distributional? Thomas Kalmes PDO with a single characteristic direction 4 / 25

� � � � � � (0 , 2) × ( − 4 , 4) ∪ ( − 1 , 1) × ( − 4 , − 2) ∪ ( − 1 , 1) × (2 , 4) Example: X = x 2 P 1 ( ξ 1 , ξ 2 ) = ξ 1 ⇒ P 1 ( ∂ ) = ∂ 1 ; 4 3 given f ∈ C ∞ ( X ) ⇒ ( x 1 , x 2) � x 1 (1 , x 2) f ( t, x 2 ) dt ∈ C ∞ ( X ) u ( x 1 , x 2 ) := 1 satisfies ∂ 1 u = f 1 x 1 0 − 1 2 ⇒ C ∞ ( X ) �⊆ P ( ∂ )( C 1 ( X )) − 1 − 3 − 4 Thomas Kalmes PDO with a single characteristic direction 5 / 25

� � � � � � (0 , 2) × ( − 4 , 4) ∪ ( − 1 , 1) × ( − 4 , − 2) ∪ ( − 1 , 1) × (2 , 4) Example: X = x 2 P 1 ( ξ 1 , ξ 2 ) = ξ 1 ⇒ P 1 ( ∂ ) = ∂ 1 ; 4 3 given f ∈ C ∞ ( X ) ⇒ ( x 1 , x 2) � x 1 (1 , x 2) f ( t, x 2 ) dt ∈ C ∞ ( X ) u ( x 1 , x 2 ) := 1 satisfies ∂ 1 u = f 1 x 1 0 − 1 2 ⇒ P 1 ( ∂ ) : C ∞ ( X ) → C ∞ ( X ) surjective − 1 ⇒ C ∞ ( X ) �⊆ P ( ∂ )( C 1 ( X )) − 3 − 4 Thomas Kalmes PDO with a single characteristic direction 5 / 25

� � � � � � (0 , 2) × ( − 4 , 4) ∪ ( − 1 , 1) × ( − 4 , − 2) ∪ ( − 1 , 1) × (2 , 4) Example: X = x 2 P 2 ( ξ 1 , ξ 2 ) = ξ 2 ⇒ P 2 ( ∂ ) = ∂ 2 ; 4 3 choose η ∈ C ∞ ( R ) with η ( t ) = 0 for � 1 t / ∈ [ − 1 , 1] and − 1 η ( t ) dt > 0 ; set � η ( x 2 ) 1 x 1 , if x 1 > 0 f ( x 1 , x 2 ) = 0 , if x 1 ≤ 0 x 1 0 − 1 2 − 1 ⇒ f ∈ C ∞ ( X ) ; assume ∃ u ∈ C 1 ( X ) : ∂ 2 u = f ; for x 1 ∈ (0 , 2) we then have − 3 � 3 u ( x 1 , 3) − u ( x 1 , − 3) = − 3 ∂ 2 u ( x 1 , t ) dt − 4 � 1 1 = − 1 η ( t ) dt → x 1 → 0 ∞ x 1 ⇒ C ∞ ( X ) �⊆ P ( ∂ )( C 1 ( X )) Thomas Kalmes PDO with a single characteristic direction 5 / 25

� � � � � � (0 , 2) × ( − 4 , 4) ∪ ( − 1 , 1) × ( − 4 , − 2) ∪ ( − 1 , 1) × (2 , 4) Example: X = x 2 P 2 ( ξ 1 , ξ 2 ) = ξ 2 ⇒ P 2 ( ∂ ) = ∂ 2 ; 4 3 ( x 1 , 3) choose η ∈ C ∞ ( R ) with η ( t ) = 0 for � 1 t / ∈ [ − 1 , 1] and − 1 η ( t ) dt > 0 ; set � η ( x 2 ) 1 x 1 , if x 1 > 0 f ( x 1 , x 2 ) = 0 , if x 1 ≤ 0 x 1 0 − 1 2 − 1 ⇒ f ∈ C ∞ ( X ) ; assume ∃ u ∈ C 1 ( X ) : ∂ 2 u = f ; for x 1 ∈ (0 , 2) we then have − 3 ( x 1 , − 3) � 3 u ( x 1 , 3) − u ( x 1 , − 3) = − 3 ∂ 2 u ( x 1 , t ) dt − 4 � 1 1 = − 1 η ( t ) dt → x 1 → 0 ∞ x 1 ⇒ C ∞ ( X ) �⊆ P ( ∂ )( C 1 ( X )) Thomas Kalmes PDO with a single characteristic direction 5 / 25

� � � � � � (0 , 2) × ( − 4 , 4) ∪ ( − 1 , 1) × ( − 4 , − 2) ∪ ( − 1 , 1) × (2 , 4) Example: X = x 2 P 2 ( ξ 1 , ξ 2 ) = ξ 2 ⇒ P 2 ( ∂ ) = ∂ 2 ; 4 3 ( x 1 , 3) choose η ∈ C ∞ ( R ) with η ( t ) = 0 for � 1 t / ∈ [ − 1 , 1] and − 1 η ( t ) dt > 0 ; set � η ( x 2 ) 1 x 1 , if x 1 > 0 f ( x 1 , x 2 ) = 0 , if x 1 ≤ 0 x 1 0 − 1 2 − 1 ⇒ f ∈ C ∞ ( X ) ; assume ∃ u ∈ C 1 ( X ) : ∂ 2 u = f ; for x 1 ∈ (0 , 2) we then have − 3 ( x 1 , − 3) � 3 u ( x 1 , 3) − u ( x 1 , − 3) = − 3 ∂ 2 u ( x 1 , t ) dt − 4 � 1 1 = − 1 η ( t ) dt → x 1 → 0 ∞ x 1 ⇒ C ∞ ( X ) �⊆ P ( ∂ )( C 1 ( X )) Thomas Kalmes PDO with a single characteristic direction 5 / 25

� � � � � � (0 , 2) × ( − 4 , 4) ∪ ( − 1 , 1) × ( − 4 , − 2) ∪ ( − 1 , 1) × (2 , 4) Example: X = x 2 For P 1 ( ξ 1 , ξ 2 ) = ξ 1 resp. P 2 ( ξ 1 , ξ 2 ) = ξ 2 is 4 3 P 1 ( ∂ ) : C ∞ ( X ) → C ∞ ( X ) surjective, P 2 ( ∂ ) : C ∞ ( X ) → C ∞ ( X ) not surjective. 1 x 1 0 − 1 2 Is it possible to ”see” this without − 1 calculation? What about P 2 ( ∂ ) if we allow for more general solutions of P 2 ( ∂ ) u = f, f ∈ C ∞ ( X ) , than − 3 u ∈ C 1 ( X ) ? − 4 ⇒ C ∞ ( X ) �⊆ P ( ∂ )( C 1 ( X )) Thomas Kalmes PDO with a single characteristic direction 5 / 25

Let P ∈ C [ X 1 , . . . , X d ] \{ 0 } and let X ⊆ R d be open. i) When is P ( ∂ ) : C ∞ ( X ) → C ∞ ( X ) surjective? ii) When is C ∞ ( X ) ⊆ P ( ∂ )( D ′ ( X )) ? Thomas Kalmes PDO with a single characteristic direction 6 / 25

Let P ∈ C [ X 1 , . . . , X d ] \{ 0 } and let X ⊆ R d be open. i) When is P ( ∂ ) : C ∞ ( X ) → C ∞ ( X ) surjective? ii) When is C ∞ ( X ) ⊆ P ( ∂ )( D ′ ( X )) ? � iii) When is P ( ∂ ) : D ′ ( X ) → D ′ ( X ) surjective? � Thomas Kalmes PDO with a single characteristic direction 6 / 25

Let P ∈ C [ X 1 , . . . , X d ] \{ 0 } and let X ⊆ R d be open. i) When is P ( ∂ ) : C ∞ ( X ) → C ∞ ( X ) surjective? ii) When is C ∞ ( X ) ⊆ P ( ∂ )( D ′ ( X )) ? � iii) When is P ( ∂ ) : D ′ ( X ) → D ′ ( X ) surjective? � Answers will depend on combined properties of P and X . Thomas Kalmes PDO with a single characteristic direction 6 / 25

Let P ∈ C [ X 1 , . . . , X d ] \{ 0 } and let X ⊆ R d be open. i) When is P ( ∂ ) : C ∞ ( X ) → C ∞ ( X ) surjective? ii) When is C ∞ ( X ) ⊆ P ( ∂ )( D ′ ( X )) ? � iii) When is P ( ∂ ) : D ′ ( X ) → D ′ ( X ) surjective? � Answers will depend on combined properties of P and X . Some thoughts on i): Equip C ∞ ( X ) with topology generated by the seminorms � · � l,K : C ∞ ( X ) → [0 , ∞ ) , f �→ x ∈ K | ∂ α f ( x ) | ( l ∈ N 0 , K ⋐ X ) max 0 , | α |≤ l max α ∈ N d ⇒ C ∞ ( X ) Fr´ echet space, P ( ∂ ) : C ∞ ( X ) → C ∞ ( X ) continuous, linear. Thomas Kalmes PDO with a single characteristic direction 6 / 25

Let P ∈ C [ X 1 , . . . , X d ] \{ 0 } and let X ⊆ R d be open. i) When is P ( ∂ ) : C ∞ ( X ) → C ∞ ( X ) surjective? ii) When is C ∞ ( X ) ⊆ P ( ∂ )( D ′ ( X )) ? � iii) When is P ( ∂ ) : D ′ ( X ) → D ′ ( X ) surjective? � Answers will depend on combined properties of P and X . Some thoughts on i): Equip C ∞ ( X ) with topology generated by the seminorms � · � l,K : C ∞ ( X ) → [0 , ∞ ) , f �→ x ∈ K | ∂ α f ( x ) | ( l ∈ N 0 , K ⋐ X ) max 0 , | α |≤ l max α ∈ N d ⇒ C ∞ ( X ) Fr´ echet space, P ( ∂ ) : C ∞ ( X ) → C ∞ ( X ) continuous, linear. echet, A : F → F continuous, linear Abstract theory: F Fr´ A surjective ⇔ A has dense range, A has closed range Thomas Kalmes PDO with a single characteristic direction 6 / 25