Section 4.5Steady-State Temperature in a Cylinder63Solutions to Exercises 4.51.Using (2) and (3), we have thatu(ρ, z)=∞±n=1AnJ0(λnρ) sinh(λnz),λn=αna,whereαn=α0,nis thenth positive zero ofJ0, andAn=2sinh(λnh)a2J1(αn)2²a0f(ρ)J0(λnρ)ρdρ=200sinh(2αn)J1(αn)2²10J0(αnρ)=200sinh(2αn)α2nJ1(αn)2²αn0J0(s)sds(lets=αnρ)=200sinh(2αn)α2nJ1(αn)2[J1(s)s]³³³³αn0=200sinh(2αn)αnJ1(αn).Sou(ρ, z)=200∞±n=1J0(αnρ) sinh(αnz)sinh(2αn)αnJ1(αn).5.(a) We proceed exactly as in the text and arrive at the conditionZ(h)=0which leads us to the solutionsZ(zZn(z) = sinh(λn(h-z)),whereλn=αna.So the solution of the problem isu(ρ, z∞±n=1CnJ0(λnρ) sinh(λn(h-z)),whereCn=2a2J1(αn)2sinh(λnh)²a0f(ρ)J0(λnρ)ρ dρ.(b) The problem can be decomposed into the sum of two subproblems, one
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.