App H1

Do not edit,. this has been installed. Thus, equations (2.4.6) become [ A(τ)ch(u0τ) + B(τ)sh(u0τ) ] = V0 ch[τ(u0-π)] / ch(πτ) [ A(τ)ch(u0'τ) + B(τ)sh(u0'τ) ] = V0 ch[τ(u0-π)] / ch(πτ) (2.4.10) This is a standard 2x2 Cramer's Rule problem, but in order to get the solution into the form shown below, a certain amount of work is required, which we relegate to ******. The result from *** is then. A(τ) = V0 ch[(π–u0)τ] ch[(π+u0)τ] / ch2(πτ) B(τ) = – V0 ch[(π–u0)τ] sh[(π+u0)τ] / ch2(πτ) . (2.4.11) ************************************************ H.1 Solving for A(τ) and B(τ) in (2.4.10) The task is to solve the following equations for A and B, [ A(τ)ch(u0τ) + B(τ)sh(u0τ) ] = V0 ch[τ(u0-π)] / ch(πτ) u'0 ≡ u0 + 2π [ A(τ)ch(u0'τ) + B(τ)sh(u0'τ) ] = V0 ch[τ(u0-π)] / ch(πτ) (2.4.10) (H.1.1) Since both equations have the same right side, the left sides are also equal. [ A(τ)ch([u0 + 2π]τ) + B(τ)sh([u0 + 2π]τ) ] = [ A(τ)ch(u0τ) + B(τ)sh(u0τ) ] or A(τ) [ ch[(u0+2π)τ] – ch(u0τ) ] = - B(τ) [ sh[(u0+2π)τ] – sh(u0τ) ] . (H.1.2) Use the identities ch(a+x) - ch(x) = 2 sh(a/2+x) sh(a/2) sh(a+x) - sh(x) = 2 ch(a/2+x) sh(a/2) (H.1.3) with x = u0τ, a = 2πτ, a/2+x= (π+u0)τ to get ch[(u0+2π)τ] – ch(u0τ) = 2 sh[(π+u0)τ]sh(πτ) sh[(u0+2π)τ] – sh(u0τ) = 2 ch[(π+u0)τ]sh(πτ) . (H.1.4) Then (H.1.2) becomes, A(τ) [ sh[(π+u0)τ]sh(πτ) ] = - B(τ) [ ch[(π+u0)τ]sh(πτ) ] . so that B(τ) = – A(τ) th[(π+u0)τ] (H.1.5) providing a simple relationship between B and A. Insert this expression for B into the first equation of (H.1.1) to get [ A(τ)ch(u0τ) – A(τ) th[(π+u0)τ]sh(u0τ) ] = V0 ch[τ(u0-π)] / ch(πτ) or ch(πτ) A(τ) {ch(u0τ) – th[(π+u0)τ] sh(u0τ) } = V0 ch[τ(u0-π)] or ch(πτ) A(τ) {ch[(π+u0)τ]ch(u0τ) –sh[(π+u0)τ] sh(u0τ) } = V0 ch[τ(u0-π)] ch[(π+u0)τ] . (H.1.6) Now use identity chx chy - shx shy = ch(x-y) (H.1.7) with x = [(π+u0)τ] and y = (u0τ) to get ch[(π+u0)τ]ch(u0τ) –sh[(π+u0)τ] sh(u0τ) = ch(πτ) . (H.1.8) Then (H.1.6) may be written ch2(πτ) A(τ) = V0 ch[τ(u0-π)] ch[(π+u0)τ] (H.1.9) giving A(τ) = V0 ch[τ(u0-π)] ch[(π+u0)τ]/ ch2(πτ) . (H.1.10) Then from (H.1.5) that B(τ) = – A(τ) th[(π+u0)τ] , B(τ) = – V0 ch[τ(u0-π)] sh[(π+u0)τ]/ ch2(πτ) (H.1.11) so the solution is then A(τ) = V0 ch[τ(u0-π)] ch[(π+u0)τ] / ch2(πτ) B(τ) = – V0 ch[τ(u0-π)] sh[(π+u0)τ] / ch2(πτ) . (H.1.12)