Section 4.4Steady-State Temperature in a Disk61andnabn=1π2π0f(θ) sinnθdθ.Hencean=anπ2π0f(θ) cosnθdθandbn=anπ2π0f(θ) sinnθdθ.Note thata0is still arbitray. Indeed, the solution of a Neumann problem is notunique. It can be determined only up to an additive constant (which does not affectthe value of the normal derivative at the boundary).21.Using the fact that the solutions must be bounded asr→ ∞, we see thatc1= 0 in the first of the two equations in (3), andc2= 0 in the second of the twoequations in (3). ThusR(r) =Rn(r) =cnr-n=ranforn= 0,1,2,....The general solution becomesu(r,θ) =a0+∞n=1arn(ancosnθ+bnsinnθ), r>a.Settingr=aand using the boundary condition, we obtainf(θ) =a0+∞n=1(ancosnθ+bnsinnθ),which implies that theanandbnare the Fourier coefficients of
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Boundary value problem, Normal mode, Leonhard Euler, Boundary conditions, Dirichlet boundary condition, sin nθ