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Chem Differential Eq HW Solutions Fall 2011 61

# Chem Differential Eq HW Solutions Fall 2011 61 - Section...

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Section 4.4 Steady-State Temperature in a Disk 61 and n a b n = 1 π 2 π 0 f ( θ ) sin nθdθ. Hence a n = a 2 π 0 f ( θ ) cos nθdθ and b n = a 2 π 0 f ( θ ) sin nθdθ. Note that a 0 is still arbitray. Indeed, the solution of a Neumann problem is not unique. It can be determined only up to an additive constant (which does not affect the value of the normal derivative at the boundary). 21. Using the fact that the solutions must be bounded as r → ∞ , we see that c 1 = 0 in the first of the two equations in (3), and c 2 = 0 in the second of the two equations in (3). Thus R ( r ) = R n ( r ) = c n r - n = r a n for n = 0 , 1 , 2 , .... The general solution becomes u ( r,θ ) = a 0 + n =1 a r n ( a n cos + b n sin ) , r>a. Setting r = a and using the boundary condition, we obtain f ( θ ) = a 0 + n =1 ( a n cos + b n sin ) , which implies that the a n and b n are the Fourier coefficients of
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• Fall '11
• StuartChalk
• Boundary value problem, Normal mode, Leonhard Euler, Boundary conditions, Dirichlet boundary condition, sin nθ

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