Explanation It is given that, u r ( a , θ ) = g ( θ ) , 0 ≤ θ ≤ 2 π . Note that, in polar coordinates ( r , θ ) Laplace equation has the form u r r + 1 r u r + 1 r 2 u θ θ = 0 . For u ( r , θ ) to be single-valued it is necessary that u ( r , θ ) be periodic in θ with period 2 π and u ( r , θ ) is bounded for r ≤ a . In order to apply the method of separation of variables, assume that u ( r , θ ) = R ( r ) Θ ( θ ) . Substitute the above equation in Laplace equation. R ″ Θ + 1 r R ′ Θ + 1 r 2 R Θ ″ = 0 r 2 R ″ R + r R ′ R = − Θ ″ Θ r 2 R ″ R + r R ′ R = λ [ λ is the separation constant ] Then, the two ordinary differential equations are, r 2 R ″ + r R ′ − λ R = 0 (1) Θ ″ − λ Θ = 0 (2) Consider three cases as follows. Case ( i ): If λ < 0 , let λ = − μ 2 , where μ > 0 . Then, Θ ″ − μ 2 Θ = 0 . Solve Θ ″ − μ 2 Θ = 0 as follows. Consider the characteristic equation as m 2 − μ 2 = 0 . m 2 − μ 2 = 0 m 2 = μ 2 m = ± μ That is, the solution of the differential equation is Θ ( θ ) = c 1 e μ θ + c 2 e − μ θ . Thus, Θ ( θ ) can be periodic only if c 1 = c 2 = 0 and λ cannot be negative. Case ( ii ): If λ = 0 , then Θ ″ = 0 and Θ ( θ ) = c 1 + c 2 θ . For Θ ( θ ) to be periodic, c 2 = 0 . So, Θ ( θ ) is a constant. r 2 R ″ + r R ′ = 0 This equation is of the Euler type and has the solution R ( r ) = k 1 + k 2 ln r . The logarithmic term cannot be accepted if u ( r , θ ) is to remain bounded as r → 0 . So, k 2 = 0 . Thus, corresponding to λ = 0 concluded that u ( r , θ ) must be constant. That is, proportional to the solution u 0 ( r , θ ) = 1 . Case ( iii ): If λ > 0 , then let λ = μ 2 , where μ > 0 . Then, the equations (1) and (2) become r 2 R ″ + r R ′ − μ 2 R = 0 and Θ ″ + μ 2 Θ = 0 . Solve Θ ″ + μ 2 Θ = 0 as follows. Consider the characteristic equation as m 2 + μ 2 = 0 . m 2 + μ 2 = 0 m 2 = − μ 2 m = ± μ i That is, the solution of the differential equation is Θ ( θ ) = c 1 sin μ θ + c 2 cos μ θ . Similarly, the solution of r 2 R ″ + r R ′ − μ 2 R = 0 is R ( r ) = k 1 r μ + k 2 r − μ . Note that, for Θ to be periodic with period 2 π , it is necessary for μ to be a positive integer n

Elementary Differential Equations and Boundary Value Problems, Enhanced

11th Edition

ISBN: 9781119381648

Author: Boyce

Publisher: WILEY

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Chapter 10.8, Problem 11P

To determine

The solution u(r,θ) of Laplace’s equation inside the circle r=a that satisfies the boundary condition on the circle ur(a,θ)=g(θ), 0<θ<2π.

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Chapter 10.8, Problem 10P

Chapter 10.8, Problem 12P

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Chapter 10 Solutions

Elementary Differential Equations and Boundary Value Problems, Enhanced

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The solution u ( r , θ ) of Laplace’s equation inside the circle r = a that satisfies the boundary condition on the circle u r ( a , θ ) = g ( θ ) , 0 &lt; θ &lt; 2 π .

The solution u(r,θ) of Laplace’s equation inside the circle r=a that satisfies the boundary condition on the circle ur(a,θ)=g(θ), 0<θ<2π.

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Chapter 10 Solutions

The solution u ( r , θ ) of Laplace’s equation inside the circle r = a that satisfies the boundary condition on the circle u r ( a , θ ) = g ( θ ) , 0 < θ < 2 π .