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Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
Download (direct link): elementarydifferentialequat2001.pdf
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n n n n
n=1
b = c /(X - u), n = 1,2,... and the desired solution is
n n n *
obtained [after setting 6 (x) = P (x)].
n 2n-1
Section 11.5, Page 666
1a. Since u(x,0) = 0 we have Y(0) = 0. However, since the other two boundaries are given by y = 2x and y = 2(x-2) we cannot separate x and y dependence and thus neither X nor Y satisfy homogeneous B.C. at both end points.
1b. The line y = 2x is transformed into ? = 0 and y = 2(x-2) is transformed into ? = 2. The lines y = 0 and y = 2 are transformed into n = 0 and n = 2 respectively, so the parallelogram is transformed into a square of side 2. From the given equations, we have x = ? + n/2 and y = n. Thus u? = uxx^ + u-yy^ = ux and
un = uxxn + u-yy-n = 1/2ux + uy. Likewise
u^ = uxx - xE, + uxyy^ = uxx un = uxxxn + uxyyn = 1/2uxx + uxy and unn = 1/2uxxxn + 1/2uxyyn + uyxxn + uyyyn = 1/4uxx + uxy + uyy. Therefore,
5/4u^^ - u^n + unn = uxx + uyy = 0.
1c. Substituting u(?,n) = U(?)V(n) into the equation of part b yields 5/4U"V - U'V' + UV" = 0 or upon dividing by UV 5 U" V" U'V'
+ = ------------, which is not separable. The
4 U V UV
B.C. become U(?,0) = 0, U(?,2) = f(^+1) (since x = ? + n/2), U(0,n) = 0, and U(2,n) = 0.
2. This problem is very similar to the example worked in the text. The fundamental solutions satisfying the P.D.E.(3), the B.C. u(1,t) = 0, t > 0 and the finiteness condition are given by Eqs.(15) and (16). Thus assume u(r,t) is of the form given by Eq.(17). The I.C. require
that u(r,0) = Vc J (X r) = 0 and
n 0 n
n=1
252
Section 11.5
u (r,0) = y1 ak J (1 r) = g(r). From Eq.(26) of
t n n 0 n
n=1
Section 11.4 we obtain c = 0 and
n
1 ka = [1rg(r)J (1 r)dr/ [ 1rJ2(1 r)dr, n = 1,2,... . n n 0 0 n 0 0 n
4. This problem is the same as Problem 22 of Section 10.7.
The periodicity condition requires that p. of that problem be an integer and thus substituting p2 = 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 = -12, and thus Z" - 12Z = 0 and, after some algebra, it follows that r2R"/R + rR'/R + 12r2 = -0"/0 = a. The periodicity condition 0(0) = 0(2n) requires that "\/~a be an integer n so a = n2. Thus r2R" + rR' + (12r2 - n2)R = 0,
0" + n20 = 0, and Z" - 12Z = 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 = -12, from which the desired result
follows.
6. Assuming that u(r,z) = R(r)Z(z) it follows from Problem 5 that R = cJ0(1r) + c2Y0(1r), from Eq.(13), and
Z = k1e-1z + k2e1z. 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(1) = 0 leading to an
infinite set of discrete positive eigenvalues 1 ,1 ,...1 ... . The fundamental solutions of the
1 2 n
-1 z
problem are then u (r,z) = J (1 r)e n, n = 1,2,... .
n 0 n
-1 z
Thus assume u(r,z) = 7 c J (1 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 (1 r) = f(r) so
n 0 n n=1
c = [ 1rf(r)J (1 r)dr/[1rJ2(1 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 = o,1,2... . For the
solution to remain bounded, c = o and thus
2
v(r,0) = (1/2)coJo(kr) + / Jm(kr)(b sinm0+c cosm0).
m m
m=1
Hence v(c,0) is then a Fourier Series of period 2n and
the coefficients are found as in Section 1o.2, Eqs.(13),
(14) and Problem 27.
9a. Substituting u(p,0,$) = P(p)0(0)O(^) into Laplace's equation leads to
p2P"/P + 2pP'/P = -(csc2^)0"/0 - O"/O - (cot^)O'/O = o.
In order to satisfy arbitrary B.C. it can be shown that o must be positive, so assume o = p2.
Thus p2P" + 2pP' - p2P = o. Then we have (sin2^)O"/O + (sin^cos^) O'/O + p2sin2^ = - 0"/0 = a.
The periodicity condition 0(o) = 0(2n) requires that "\/~a be an integer X so a = X2. Hence 0" + X 20 = o and (sin2^)O" + (sin^cos^)O' + (p2sin2^ - X2) O = o.
1o. Since u is independent of 0, only the first and third of the Eqs. in 9a hold. The general solution to the Euler equation is
r r I-----
P = c1p 1 + c2p 2 where r1 = (-1 + V 1+4p2 )/2 > o and
r2 = (-1-^/ 1+4p2 ) /2 < o. Since we want u to be bounded
as p ^ o, we set c2 = o. As found in Problem 22 of
Section 5.3, the solutions of Legendre's equation,
Problem 9c, are either singular at 1, at -1, or at both unless p2 = n(n+1), where n is an integer. In this case, one of the two linearly independent solutions is a polynomial denoted by Pn (Problems 23 and 24 of
Section 5.3). Since r1 = (-1 + \J 1+4n(n+1) )/2 = n, the fundamental solutions of this problem satisfying the
n
finiteness condition are un(p,$) = p P (s) = pnPn(cos^),
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