Explanation Given: The boundary value problem is ∂ 2 u ∂ r 2 + 2 r ∂ u ∂ r + 1 r 2 ∂ 2 u ∂ θ 2 + cot θ r 2 ∂ u ∂ θ = 0 , r > c , 0 < θ < π and the boundary conditions are u ( c , θ ) = 0 and lim r → ∞ u ( r , θ ) = − E z = − E r cos θ . Calculation: The given boundary value problem is, ∂ 2 u ∂ r 2 + 2 r ∂ u ∂ r + 1 r 2 ∂ 2 u ∂ θ 2 + cot θ r 2 ∂ u ∂ θ = 0 Consider the equation, u ( r , θ ) = R ( r ) Θ ( θ ) (1) Take separation of the variables of the above equation and separation constant is λ , r 2 R ″ + r R ′ R = λ and − Θ ″ + cot θ Θ ′ Θ = λ The equation of variable R is, r 2 R ″ + r R ′ − λ R = 0 (2) The equation of variable θ is, sin θ Θ ″ + cos θ Θ ′ + λ sin θ Θ = 0 (3) Substitute x = cos θ in the above equation, ( 1 − x 2 ) d 2 Θ d x 2 − 2 x d Θ d x + λ Θ = 0 , − 1 ≤ x ≤ 1 The above equation is the form of Legendre’s equation which has continuous derivatives on the closed interval [ − 1 , 1 ] is the Legendre polynomials P n ( x ) corresponding to the λ = n ( n + 1 ) where n is nonnegative integer. The general solution of the equation (3) is, Θ ( θ ) = P n ( cos θ ) (4) Equation (2) is the form of Cauchy Euler equation and substitute λ = n ( n + 1 ) then the general solution of the equation (2) is expressed as, R ( r ) = c 1 r n + c 2 r − ( n + 1 ) (5) Substitute the value of Θ ( θ ) from equation (4) and the value of R ( r ) from equation (5) in equation (1), u ( r , θ ) = ( c 1 r n + c 2 r − ( n + 1 ) ) P n ( cos θ ) Rewrite the term c 1 = A n and c 2 = B n in the above equation and apply superposition principle in the above equation, u ( r , θ ) = ∑ n = 0 ∞ ( A n r n + B n r − ( n + 1 ) ) P n ( cos θ ) (6) At given boundary condition u ( c , θ ) = 0 &

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Chapter 13.3, Problem 12E

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DIFF.EQUAT.W/BOUNDARY-VALUE...(LL)-TEXT

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The given solution of the boundary value problem.

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To show: The given solution of the boundary value problem.

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