Acoustic Scattering from a Sphere - Steve Turley

Substituting the value for s nm from Equation 56 into Equation 52, we have a solution for the normalderivative of u on the boundary.Equation 59 uses the identity∂u∂ˆn (aˆx′ ) = σ(ˆx ′ ) (57)= 4π ∞∑ i n−1 n∑Ykn=0 h (1)n m (ˆx ′ )Yn m (ˆk)n (ka) m=−n(58)= 1 ∞∑i n (2n + 1) P n(ˆx ′ · ˆk)ik(ka)(59)n=02n + 14π P n(ˆx · ŷ) =n∑m=−nh (1)nY m n (ˆx)Y m n (ŷ) . (60)Once σ is known, the far eld can be computed from Equation 14. Using the expansion in Equation 55,f(ˆx) = − 4π ∫ [∑ ∞i n−1 n∑Yk ∂D n=0 h (1)n m (ˆx ′ )Yn (ˆk)] ⎡ ⎤∞∑∑n ′m ⎣ (− n′ j n ′(ka) Y m′ m′n ′ (ˆx)Ynn (ka)) ⎦ dˆx (61)′m=−nn ′ =0m ′ =−n ′= 4πi ∑ j n (ka)km,n h (1)n (ka) Y n m (ˆx)Y n m (ˆk) (62)= i ∞∑(2n + 1) j n(ka)kh (1)n (ka) P n(ˆx · ˆk) . (63)n=0A more general solution good in the near and far eld can be computed from Equation 12 with the expansionin Equation 21 for the Green's function.u(x) =∞∑ i n−1n,n ′−4πin,n h (1)′ n (ka) j n ′(ka)h n ′(kx) ∑∫Yn m (ˆk)Y n m′Y ′ n m (ˆx ′ )Yn m′) dˆx ′m=−n,m ′ =−n ′ ∂D(64)=∞∑−4π i n h (1)n (kx) j n(ka)n∑Yn=0 h (1)n m (ˆk)Y n m (ˆx)n (ka) m=−n(65)=∞∑− (2n + 1)i n h (1)n (kx) j n(ka)h (1)n (ka) P n(ˆx · ˆk) . (66)n=0Series Approach A simpler, but less generalizable way to solve this problem is to apply the boundarycondition to series expansions of the incident plane wave and scattered wave at the surface os the sphere.An the surface,e ik·aˆx′ = 4π ∑ n,mi n j n (ka)Y m n (ˆx ′ )Y m n (ˆk) (67)u s (aˆx ′ ) = ∑ n,ma nm h (1)n (ka)Y m n (ˆx ′ ) . (68)For the soft sphere problem, the sum of these two series should be zero. Since the Ynmindependent, the terms in each series must be equal to minus each other.functions are linearlya nm = −4πi n j n(ka)h (1)n (ka) Y m n (ˆk) (69)u s (ˆx) = −4π ∑ i n j n(ka)h (1)n(ka) h(1)= − ∑ i n (2n + 1) j n(ka)h (1)nn (kx)Y m n (ˆk)Y m n (ˆx) (70)(ka) h(1)n (kx)P n (ˆk · ˆx) (71)8