1. No category

Let L be a nonzero real number. (a) Show that the boundary

Let𝐿beanonzerorealnumber.
(a) Showthattheboundary-valueproblem𝑦 !! + πœ†π‘¦ = 0,𝑦 0 = 0,𝑦 𝐿 = 0has
onlythetrivialsolution𝑦 = 0forthecasesπœ† = 0andπœ† < 0.
For𝝀 = 𝟎:
𝑑 𝑑𝑦
𝑑
𝑦 !! + 0 𝑦 = 0 β‡’ 𝑦 !! = 0 β‡’
= 0 β‡’ 𝑑 𝑦! 𝑑 = 0 β‡’
𝑦 = 0 β‡’ 𝑑𝑦 = 0
𝑑𝑑 𝑑𝑑
𝑑𝑑
∴ 𝑦 = 0
For𝝀 < 𝟎:
𝑦 !! + βˆ’πœ† 𝑦 = 0 β‡’ π‘Ÿ ! βˆ’ πœ†π‘Ÿ = 0 β‡’ π‘Ÿ = 0 ∨ π‘Ÿ = πœ†
β‡’ 𝑦 = 𝑐! + 𝑐! 𝑒 !" Solvingforthegivenboundaryvalues:
𝑦 0 = 0 ∧ 𝑦 𝐿 = 0
𝑐! + 𝑐! = 0
1 1 𝑐!
0
β‡’
β‡’
=
!"
!" 𝑐
𝑐! + 𝑒 𝑐! = 0
1 𝑒
!
0
Onlythetrivial(obvious)solutionexistsi.e.𝑐! and𝑐! areequaltozero.
1 1
∡
~𝐼! ∴ 𝑐! = 0 ∧ 𝑐! = 0
1 𝑒 !"
∴ 𝑦 = 0
(b) Forthecaseπœ† > 0,findthevaluesofπœ†forwhichthisproblemhasa
nontrivialsolutionandgiventhecorrespondingsolution.
For𝝀 > 𝟎:
𝑦 !! + πœ†π‘¦ = 0 β‡’ π‘Ÿ ! + πœ† = 0 ⇔ π‘Ÿ = 0 ± πœ†π‘–
∴ 𝑦 = 𝑐! sin πœ†π‘‘ + 𝑐! cos πœ†π‘‘ Solvingforthegivenboundaryvalues:
𝑦 0 = 0 ∧ 𝑦 𝐿 = 0
β‡’
Solve:
𝑐! = 0
𝑐! sin
πœ†πΏ + 𝑐! cos
πœ†πΏ = 0
𝑐! sin
∴ 𝑐! sin
πœ†πΏ = 0 ∴ 𝑐! = 𝑐! = 0
πœ†πΏ = 0 β‡’ 𝑐! = 0
Thesolutionisagain𝑦 = 0fortheboundariesbutthequestionwantsustosolvefor
πœ†.
∴ sin πœ†πΏ = 0 β‡’ πœ†πΏ = arcsin 0 = π‘˜πœ‹ βˆ€! ∈ β„• Note:β„•isthesetofpositiveintegers.(Wecouldusenegativeintegersbutthat
wouldimply πœ†πΏ < 0andevenrootscanneverbenegative.)
π‘˜πœ‹ ! π‘˜ ! πœ‹ !
βˆ΄πœ†=
= ! 𝐿
𝐿
UsuallycalledtheEigenfunction:
π‘˜πœ‹
𝑐! sin πœ†π‘‘ = 0 β‡’ 𝑦 = 𝐢 sin
⋅𝑑 𝐿

 Download  Report

Paperzz.com

Your Paperzz

© Copyright 2026 Paperzz