SolSec 6.2

# SolSec 6.2 - 6-Chapter 6Problems and Solutions Section...

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Unformatted text preview: 6-Chapter 6Problems and Solutions Section 6.2 (6.1 through 6.7)6.1Prove the orthogonality condition of equation (6.28).Solution:Calculate the integrals directly. For n= n, let u= nπx/lso that du= (nπ/l)dxand the integral becomeslnπσιν2υδυ=λνπ12υ-14σιν2υνπ∫νπ=λνπ12νπ-14σιν4νπ- 0 =λ2where the first step used a table of integrals. For n≠mlet u= πx/lso that du = (π/l)dxand sinnπ ξλσινμπ ξλδξ=λπσινμ υσιννυδυλ∫λ∫which upon consulting a table of integrals islπσιν(μ-ν29π2(μ-ν29-σιν(ν+μ29π2(ν+μ29= 0.16-6.2Calculate the orthogonality of the modes in Example 6.2.3.Solution:One needs to show thatXn(x)Xm(x)dx= 0 φορμ≠ν, ϖηερεΞμ(τ29=ανσινσνξ.1∫But each mode Xn(x) must satisfy equation (6.14), i.e.′′Ξν= -σν2Ξν(1)Likewise′′Ξμ=σμ2Ξμ(2)Multiply (1) by Xmand integrate from 0 to l. Then multiply (2) by Xn(x) and integrate from 0 to l. This yields′′ΞνΞμδξ= -σν2ΞνΞμδξλ∫λ∫′′ΞμΞνδξ= -σμ2ΞμΞνδξλ∫λ∫Subtracting these two equations yields′′ΞνΞμ-′′ΞμΞν(29δξ=σν2-σμ2(29Ξν(ξ29Ξμ(ξ29δξλ∫λ∫Integrate by parts on the left side to get′Ξν′Ξμδξ-′Ξμ′Ξνδξ+′Ξνλ∫λ∫Ξμλ-Ξμ′Ξνλ=Ξμ(λ29κΞν(λ29-Ξν(λ29κΞμ(λ29= 0from the boundary condition given by eq. (6.50). Thusσν2-σμ2(29ΞνΞμδξ= 0....
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SolSec 6.2 - 6-Chapter 6Problems and Solutions Section...

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