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4. This problem is the same as Problem 22 of Section 10.7.
The periodicity condition requires that ö of that problem be an integer and thus substituting ö2 = n2 into the previous results yields the given equations.
5a. Substituting u(r,0,z) = R(r)0(0)Z(z) into Laplace's equation yields R"0Z + R'0Z/r + R0"Z/r2 + R0Z" = 0 or equivalently R"/R + R'/rR + 0"/r20 = - Z"/Z = o. In order to satisfy arbitrary B.C. it can be shown that o must be negative, so assume o = -X2, and thus Z" - X2Z = 0 and, after some algebra, it follows that r2R"/R + rR'/R + X 2r2 = -0"/0 = a. The periodicity condition 0(0) = 0(2n) requires that yfa be an integer n so a = n2. Thus r2R" + rR' + (X 2r2 - n2)R = 0,
0" + n20 = 0, and Z" - X2Z = 0.
5b. If u(r,0,z) is independent of 0, then the 0"/r20 term
does not appear in the second equation of part a and thus
R"/R + R'/rR = - Z"/Z = -X2, from which the desired result
6. Assuming that u(r,z) = R(r)Z(z) it follows from Problem 5 that R = cj0(Xr) + c2Y0(Xr), from Eq.(13), and
Z = k1e-Xz + k2eXz. Since u(r,z) is bounded as r ^ 0 and
approaches zero as z ^ ^ we require that c2 = 0, k2 = 0.
The B.C. u(1,z) = 0 requires that J0(X) = 0 leading to an
infinite set of discrete positive eigenvalues X ,X ,...X ... . The fundamental solutions of the
1 2 n
problem are then u (r,z) = J(X r)e n, n = 1,2,... .
n 0 n
Thus assume u(r,z) = Ó c J (X r)e n . The B.C.
n 0 n
u(r,0) = f(r), 0 < r < 1 requires that u(r,0) = Ó c J (X r) = f(r) so
n 0 n
c = J1rf(r)J (X r)dr/|1rJ2(X r)dr, n = 1,2,... . n J0 0 n J0 0 n
7b. Again, 0 periodic of period 2n implies X2 = n2. Thus the solutions to the D.E. are R(r) = c J (kr) + c Y (kr) (note
1 n 2 n
that X and k here are the reverse of Problem 3 of Section
11.4) and 0(0) = d1cosn0 + d2sinn0, n = 0,1,2... . For the
solution to remain bounded, c = 0 and thus
v(r,0) = (1/2)c0J0(kr) + / Jm(kr)(b sinm0+c cosm0).
Hence v(c,0) is then a Fourier Series of period 2n and
the coefficients are found as in Section 10.2, Eqs.(13),
(14) and Problem 27.
9a. Substituting u(p,0^) = Ð(ð)0(0)Ô(ô) into Laplace's equation leads to
p2P"/P + 2pP'/P = -(csc^)0"/0 - Ô"/Ô - (cotô)Ô'/Ô = î.
In order to satisfy arbitrary B.C. it can be shown that î must be positive, so assume î = ö2.
Thus p2P" + 2pP' - ^P = 0. Then we have Ûò2ô)Ô"/Ô + (sinôcosô)Ô'/Ô + ö2sin2ô = - 0"/0 = a.
The periodicity condition 0(0) = 0(2n) requires that "\/~a be an integer X so a = X2. Hence 0" + X 20 = 0 and Ûò2ô)Ô" + (sinôcosô)Ô' + (ö2sin2ô - X 2)Ô = 0.
10. Since u is independent of 0, only the first and third of the Eqs. in 9a hold. The general solution to the Euler equation is
r r /------
P = c1p 1 + c2p 2 where r1 = (-1 + V 1+4Ö2 )/2 > 0 and
r2 = (-1-ä/1+4ö2 )/2 < 0. Since we want u to be bounded
as p ^ 0, we set c2 = 0. As found in Problem 22 of
Section 5.3, the solutions of Legendre's equation,
Problem 9c, are either singular at 1, at -1, or at both unless ö2 = n(n+1), where n is an integer. In this case, one of the two linearly independent solutions is a polynomial denoted by Pn (Problems 23 and 24 of
Section 5.3). Since r1 = (-1 + ä/ 1+4n(n+1) )/2 = n, the fundamental solutions of this problem satisfying the
finiteness condition are ^(ð,ô) = p P (s) = pnPn(cosô),
n = 1,2,... . It can be shown that an arbitrary
piecewise continuous function on [-1,1] can be expressed as a linear combination of Legendre polynomials. Hence we assume that
è(ð,ô) = L CnpnPn(cosô). The B.C. è(1,ô) = f^) requires
that è(1,ô) = CnPn(cosô) = :Å(ô), 0 < ô < n. From
Problem 28 of Section 5.3 we know that P (x) are
orthogonal. However here we have Pn(cosô) and thus we
must rewrite the equation in Problem 9b to find _[^ïô)Ô']' = ö2Ûòô)Ô. Thus Pn(cosô) and Pm(cosô) are
orthogonal with weight function sinô. Thus we must multiply the series expansion for f^) by sinôPm(cosô)
and integrate from 0 to n to obtain
cm = JÏf(ô)sinôPm(cosô)dô/JÏsinôPm(cosô)dô. To obtain the answer as given in the text let s = cosô.
Section 11.6, Page 675
2a. bm = \p2 J1xsinmnxdx = \[2 (_1) m+1)/mn and thus