Chem Differential Eq HW Solutions Fall 2011 65

Chem Differential Eq HW Solutions Fall 2011 65 - Section...

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Section 4.6 The Helmholtz and Poisson Equations 65 Solutions to Exercises 4.6 1. Write (1) in polar coordinates: φ rr + 1 r φ r + 1 r 2 φ θθ = - φ ( a, θ )=0 . Consider a product solution φ ( r, θ )= R ( r )Θ( θ ). Since θ is a polar angle, it follows that Θ( θ +2 π )=Θ( θ ); in other words, Θ is 2 π -periodic. Plugging the product solution into the equation and simplifying, we Fnd R ±± Θ+ 1 r R ± 1 r 2 R Θ ±± = - kR Θ; ( R ±± + 1 r R ± + ) Θ= - 1 r 2 R Θ ±± ; r 2 R ±± R + r R ± R + kr 2 = - Θ ±± Θ ; hence r 2 R ±± R + r R ± R + 2 = λ, and - Θ ±± Θ = λ Θ ±± + λ Θ=0 , where λ is a separation constant. Our knowledge of solutions of second order linear ode’s tells us that the last equation has 2 π -periodic solutions if and only if λ = m 2 ,m =0 , ± 1 , ± 2 ,.... This leads to the equations Θ ±± + m 2 , and r 2 R ±± R + r R ± R + 2 = m 2 r 2 R ±± + rR ± +( 2 - m 2 ) R . These are equations (3) and (4). Note that the condition R ( a ) = 0 follows from φ ( a, θ R ( a )Θ( θ R ( a ) = 0 in order to avoid the constant 0 solution. 5. We proceed as in Example 1 and try u ( r, θ ± m =0 ± n =1 J m ( λ mn r )( A mn cos + B mn sin ± m =0 ± n =1 φ mn ( r, θ ) , where φ mn ( r, θ J m (
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