Explanation Given: The given linear differential operator L is: L [ y ] ( x ) = 2 x 2 y ″ ( x ) + 3 x y ′ ( x ) − y ( x ) ; y ( 1 ) = 0 , y ( 4 ) = 0 Procedure used: Let L be the differential operator given by the L [ y ] = A 2 y ″ + A 1 y ′ + A 0 y . The formal adjoint of L is the differential operator L + define by L + [ y ] = ( A 2 y ) ′ ′ − ( A 1 y ) ′ + A 0 y provided A 2 ′ ′ , A 1 ′ and A 0 are continuous on [ a , b ] . Let L + be its formal adjoint. Then, L [ u ] v − u L + [ v ] = d d x [ P ( u , v ) ] Here P ( u , v ) , the bilinear concomitant associated with L , is define as P ( u , v ) = u A 1 v − u ( A 2 v ) ′ + u ′ A 2 v Linear differential operator L with domain D ( L ) and formal adjoint L + to a domain D ( L + ) . So that, ( L [ u ] , v ) = ( u , L + [ v ] ) Calculation: Consider the given linear differential operator L: L [ y ] ( x ) = 2 x 2 y ″ ( x ) + 3 x y ′ ( x ) − y ( x ) Here A 0 = − 1 , A 1 = 3 x and A 2 = 2 x 2 . Substitute in the formal adjoint formula: L + [ y ] = ( 2 x 2 ⋅ y ) ′ ′ − ( 3 x y ) ′ − y = [ 2 x 2 y ′ + 4 x y ] ′ − [ 3 x y ′ + 3 y ] − y = [ 2 x 2 y ″ + 4 x y ′ + 4 x y ′ + 4 y ] − 3 x y ′ − 3 y − y = 2 x 2 y ″ + 8 x y ′ + 4 y − 3 x y ′ − 4 y Further simplify as: 2 x 2 y ″ + 8 x y ′ + 4 y − 3 x y ′ − 4 y = 2 x 2 y ″ + 5 x y ′ In order to hold the equation ( L [ u ] , v ) = ( u , L + [ v ] )

Fundamentals of Differential Equations [With CDROM] - 7th Edition

7th Edition

ISBN: 9780321410481

Author: Saff, Edward B., Snider, Arthur David, Nagle, R. Kent

Publisher: Addison Wesley

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Chapter 11.4, Problem 9E

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The adjoint operator and its domain for the given linear differential operator L whose domain satisfy the given boundary condition L[y](x)=2x2y″(x)+3xy′(x)−y(x);y(1)=0, y(4)=0.

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Chapter 11.4, Problem 8E

Chapter 11.4, Problem 10E

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Fundamentals of Differential Equations [With CDROM] - 7th Edition

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The adjoint operator and its domain for the given linear differential operator L whose domain satisfy the given boundary condition L [ y ] ( x ) = 2 x 2 y ″ ( x ) + 3 x y ′ ( x ) − y ( x ) ; y ( 1 ) = 0 , y ( 4 ) = 0 .

Solution Summary: The author explains how to find the adjoint operator and its domain for the given linear differential operator L.

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To find:

The adjoint operator and its domain for the given linear differential operator L whose domain satisfy the given boundary condition L[y](x)=2x2y″(x)+3xy′(x)−y(x);y(1)=0, y(4)=0.

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Chapter 11 Solutions

The adjoint operator and its domain for the given linear differential operator L whose domain satisfy the given boundary condition L [ y ] ( x ) = 2 x 2 y ″ ( x ) + 3 x y ′ ( x ) − y ( x ) ; y ( 1 ) = 0 , y ( 4 ) = 0 .

Solution Summary: The author explains how to find the adjoint operator and its domain for the given linear differential operator L.

Concept explainers

Videos

To find: The adjoint operator and its domain for the given linear differential operator L whose domain satisfy the given boundary condition L[y](x)=2x2y″(x)+3xy′(x)−y(x);y(1)=0, y(4)=0.

Want to see the full answer?

Chapter 11 Solutions

To find:

The adjoint operator and its domain for the given linear differential operator L whose domain satisfy the given boundary condition L[y](x)=2x2y″(x)+3xy′(x)−y(x);y(1)=0, y(4)=0.