Draft EE 8235: Lecture 12 1 Lecture 12: Waves, beams, . . . - PowerPoint PPT Presentation

Draft EE 8235: Lecture 12 1 Lecture 12: Waves, beams, . . . • Objective: study dynamics of waves and beams • Approach: identify commonalities between the two equations ⋆ Inner product that induces energy of wave/beam ⋆ Square-root of a positive self-adjoint operator

Draft EE 8235: Lecture 12 2 Wave equation φ tt ( x, t ) = φ xx ( x, t ) φ ( x, 0) = f ( x ) , φ t ( x, 0) = g ( x ) φ ( ± 1 , t ) = 0 � � � � ψ 1 ( t ) φ ( · , t ) Define ψ ( t ) = = and write an abstract evolution equation: ψ 2 ( t ) φ t ( · , t ) � ˙ � � � � � ψ 1 ( t ) 0 I ψ 1 ( t ) = ˙ d 2 / d x 2 0 ψ 2 ( t ) ψ 2 ( t ) � I 0 � � � ψ 1 ( t ) φ ( t ) = ψ 2 ( t ) • Dynamical generator A 0 = − d 2 � � 0 I A = , −A 0 0 d x 2 f ∈ L 2 [ − 1 , 1] , d 2 f � � D ( A 0 ) = d x 2 ∈ L 2 [ − 1 , 1] , f ( ± 1) = 0

Draft EE 8235: Lecture 12 3 Euler-Bernoulli beam φ tt ( x, t ) = − φ xxxx ( x, t ) φ ( x, 0) = f ( x ) , φ t ( x, 0) = g ( x ) φ ( ± 1 , t ) = 0 φ xx ( ± 1 , t ) = 0 � � � � ψ 1 ( t ) φ ( · , t ) Define ψ ( t ) = = and write an abstract evolution equation: ψ 2 ( t ) φ t ( · , t ) � ˙ � � � � � ψ 1 ( t ) 0 I ψ 1 ( t ) = ˙ − d 4 / d x 4 0 ψ 2 ( t ) ψ 2 ( t ) � I 0 � � � ψ 1 ( t ) φ ( t ) = ψ 2 ( t ) • Dynamical generator d 4 � � 0 I A = , A 0 = −A 0 0 d x 4 f ∈ L 2 [ − 1 , 1] , d 4 f � � d x 4 ∈ L 2 [ − 1 , 1] , f ( ± 1) = f ′′ ( ± 1) = 0 D ( A 0 ) =

Draft EE 8235: Lecture 12 4 Simply supported and cantilever beams • Simply supported beams φ (0 , t ) = φ ( L, t ) = 0 φ xx (0 , t ) = φ xx ( L, t ) = 0 • Cantilever beams φ (0 , t ) = 0 , φ x (0 , t ) = 0 φ xx ( L, t ) = 0 , φ xxx ( L, t ) = 0

Draft EE 8235: Lecture 12 5 Square-root of a positive operator • Self-adjoint operator A : H ⊃ D ( A ) − → H is ⋆ positive � ψ, A ψ � > 0 for all non-zero ψ ∈ D ( A ) ⋆ coercive: if there is ǫ > 0 such that � ψ, A ψ � > ǫ � ψ � 2 for all ψ ∈ D ( A ) 1 • Self-adjoint, non-negative A has a unique non-negative square-root A 2 1  D ( A 2 ) ⊃ D ( A )     1 1 2 ψ ∈ D ( A A 2 ) for all ψ ∈ D ( A )   1 1  2 A 2 ψ = A ψ A for all ψ ∈ D ( A )  1 positive A ⇒ positive A 2

Draft EE 8235: Lecture 12 6 • Examples of positive, self-adjoint operators: A 0 = − d 2 f ∈ L 2 [ − 1 , 1] , d 2 f � � d x 2 , D ( A 0 ) = d x 2 ∈ L 2 [ − 1 , 1] , f ( ± 1) = 0 d 4 f ∈ L 2 [ − 1 , 1] , d 4 f � � d x 4 ∈ L 2 [ − 1 , 1] , f ( ± 1) = f ′′ ( ± 1) = 0 A 0 = d x 4 , D ( A 0 ) = 1 D ( A 0 ) – determined from the following requirement: 2 1 1 � � A 0 f, A 2 0 g 2 = � f, A 0 g � , for all g ∈ D ( A 0 ) • For beam (wave left for homework): 0 = − d 2 f ∈ L 2 [ − 1 , 1] , d 2 f � � 1 1 A 2 d x 2 , D ( A 0 ) = 2 d x 2 ∈ L 2 [ − 1 , 1] , f ( ± 1) = 0

Draft EE 8235: Lecture 12 7 Abstract evolution equation � ˙ � � � � � ψ 1 ( t ) 0 I ψ 1 ( t ) = ˙ −A 0 − a 1 I ψ 2 ( t ) ψ 2 ( t ) Hilbert space: � � 1 D ( A 0 ) 2 H = L 2 [ − 1 , 1] Inner product: �� � � �� f 1 f 2 � φ 1 , φ 2 � e = , g 1 g 2 e 1 1 � � = A 0 f 1 , A 2 0 f 2 2 + � g 1 , g 2 � Energy:  2 � ψ 1 x , ψ 1 x � + 1 1 2 � ψ 2 , ψ 2 � wave     E ( t ) = 1 2 � ψ 1 xx , ψ 1 xx � + 1  2 � ψ 2 , ψ 2 � beam   

Draft EE 8235: Lecture 12 8 • Adjoint of A (w.r.t. �· , ·� e ): � � D ( A 0 ) � � � � 0 I 0 − I ⇒ A † = , D ( A † ) = D ( A ) = A = 1 −A 0 − a 1 I A 0 − a 1 I D ( A 0 ) 2 • In class: � � 1 D ( A 0 ) 2 ⋆ well-posedness on H = using Lumer-Phillips L 2 [ − 1 , 1] ⋆ spectral decomposition of A for the undamped wave equation ⋆ solution to the undamped wave equation ⋆ mention different forms of internal damping in beams

Draft EE 8235: Lecture 12 9 Spectral decomposition of the undamped wave equation  ψ 2 = λ ψ 1  � � � � � � 0 I ψ 1 ψ 1  ψ ′′ = λ ⇒ = λ ψ 2 1 ∂ xx 0 ψ 2 ψ 2  0 = ψ 1 ( ± 1)  • Showed:  � � λ n = + j nπ (1 /λ n ) φ n ( x ) 2 , v n ( x ) =   φ n ( x )     λ 2 ψ 1  � ψ ′′ =   � � λ − n = − j nπ n ∈ N (1 /λ n ) φ n ( x ) 1 − − − → 2 , v − n ( x ) = 0 = ψ 1 ( ± 1) − φ n ( x )     � nπ  �   φ n ( x ) = sin 2 ( x + 1)   ☞ { v n } n ∈ Z \ 0 – complete orthonormal basis (w.r.t. �· , ·� e )

Draft EE 8235: Lecture 12 10 Solution of the undamped wave equation • Represent the solution as ∞ ∞ � � ψ ( x, t ) = α n ( t ) v n ( x ) + α − n ( t ) v − n ( x ) n = 1 n = 1 � ( α n ( t ) + α − n ( t )) 1 ∞ � λ n φ n ( x ) � = ( α n ( t ) − α − n ( t )) φ n ( x ) n = 1 � a n ( t ) 1 ∞ � λ n φ n ( x ) � = ⇒ { a n ( t ) ∈ j R , b n ( t ) ∈ R } b n ( t ) φ n ( x ) n = 1 • Substitute into the evolution model � ˙ � � � � � a n ( t ) 0 j nπ/ 2 a n ( t ) α n ( t ) = +j nπ  = ˙ ˙ 2 α n ( t ) j nπ/ 2 0 b n ( t ) b n ( t )   ⇒ α − n ( t ) = − j nπ � nπ � nπ � � � � � � � � a n ( t ) cos 2 t j sin 2 t a n (0) ˙ 2 α − n ( t )  =  � nπ � nπ � � b n ( t ) j sin 2 t cos 2 t b n (0)