Explanation Given: The given linear differential operator L is: L [ y ] = y ″ − 4 y ′ + 5 y ; y ( 0 ) = 0 , y ( 2 π ) = 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 ] = y ″ − 4 y ′ + 5 y Here A 0 = 5 , A 1 = − 4 and A 2 = 1 . Substitute in the formal adjoint formula: L + [ y ] = ( 1 ⋅ y ) ′ ′ − ( − 4 y ) ′ + 5 y = y ′ ′ + 4 y ′ + 5 y In order to hold the equation ( L [ u ] , v ) = ( u , L + [ v ] ) . We must have, P ( u , v ) ( x ) | 0 2 π = 0 for all u in D ( L ) and v in D ( L + )

Fundamentals Of Differential Equations And Boundary Value Problems, Books A La Carte Edition (7th Edition)

7th Edition

ISBN: 9780321977182

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

Publisher: PEARSON

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Chapter 11.4, Problem 8E

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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]=y″−4y′+5y;y(0)=0, y(2π)=0.

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Chapter 11.4, Problem 7E

Chapter 11.4, Problem 9E

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Fundamentals Of Differential Equations And Boundary Value Problems, Books A La Carte Edition (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 ] = y ″ − 4 y ′ + 5 y ; y ( 0 ) = 0 , y ( 2 π ) = 0 .

Solution Summary: The author explains how the adjoint operator and its domain satisfy the given boundary condition for the 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]=y″−4y′+5y;y(0)=0, y(2π)=0.

Want to see the full answer?

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 ] = y ″ − 4 y ′ + 5 y ; y ( 0 ) = 0 , y ( 2 π ) = 0 .

Solution Summary: The author explains how the adjoint operator and its domain satisfy the given boundary condition for the 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]=y″−4y′+5y;y(0)=0, y(2π)=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]=y″−4y′+5y;y(0)=0, y(2π)=0.