Books
in black and white
Main menu
Share a book About us Home
Books
Biology Business Chemistry Computers Culture Economics Fiction Games Guide History Management Mathematical Medicine Mental Fitnes Physics Psychology Scince Sport Technics
Ads

Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Download (direct link): elementarydifferentialequations2001.pdf
Previous << 1 .. 501 502 503 504 505 506 < 507 > 508 509 510 511 512 513 .. 609 >> Next

4. This problem is the same as Problem 22 of Section 10.7.
The periodicity condition requires that of that problem be an integer and thus substituting 2 = 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
follows.
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
-X z
problem are then u (r,z) = J(X r)e n, n = 1,2,... .
n 0 n
-X z
Thus assume u(r,z) = c J (X r)e n . The B.C.
n 0 n
n=1
u(r,0) = f(r), 0 < r < 1 requires that u(r,0) = c J (X r) = f(r) so
n 0 n
n=1
c = J1rf(r)J (X r)dr/|1rJ2(X r)dr, n = 1,2,... . n J0 0 n J0 0 n
253
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
2
v(r,0) = (1/2)c0J0(kr) + / Jm(kr)(b sinm0+c cosm0).
mm
m=1
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^) = ()0(0)() into Laplace's equation leads to
p2P"/P + 2pP'/P = -(csc^)0"/0 - "/ - (cot)'/ = .
In order to satisfy arbitrary B.C. it can be shown that must be positive, so assume = 2.
Thus p2P" + 2pP' - ^P = 0. Then we have 2)"/ + (sincos)'/ + 2sin2 = - 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 2)" + (sincos)' + (2sin2 - X 2) = 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+42 )/2 > 0 and
r2 = (-1-/1+42 )/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 2 = 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 + / 1+4n(n+1) )/2 = n, the fundamental solutions of this problem satisfying the
n
finiteness condition are ^(,) = p P (s) = pnPn(cos),
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
254
Section 11.6
(,) = L CnpnPn(cos). The B.C. (1,) = f^) requires
n=1
that (1,) = CnPn(cos) = :(), 0 < < n. From
n=1
Problem 28 of Section 5.3 we know that P (x) are
n
orthogonal. However here we have Pn(cos) and thus we
must rewrite the equation in Problem 9b to find _[^)']' = 2). Thus Pn(cos) and Pm(cos) are
orthogonal with weight function sin. Thus we must multiply the series expansion for f^) by sinPm(cos)
and integrate from 0 to n to obtain
cm = Jf()sinPm(cos)d/JsinPm(cos)d. To obtain the answer as given in the text let s = cos.
Section 11.6, Page 675
2a. bm = \p2 J1xsinmnxdx = \[2 (_1) m+1)/mn and thus
Previous << 1 .. 501 502 503 504 505 506 < 507 > 508 509 510 511 512 513 .. 609 >> Next