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# 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.
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Section 10.7
229
2 -1 - at < x < 1 - at
f(x + at)
0 otherwise
18a. As in Problem 17a, we have u(x,0) = ôÛ + y(x) = 0 and ut(x,0) = ^ô'^) + ay'(x) = g(x).
18b. From part (a) we have y(x) = -ôÛ which yields
-2aô(x) = g(x) from the second equation in part a.
24. Substituting u(r,0,t) = R(r)0(0)T(t) into the P.D.E.
yields R"0T + R'0T/r + R0"T/r2 = R0T"/a2 or equivalently R"/R + R'/rR + 0"/0r2 = T"/a2T. In order for this
equation to be valid for 0 < r < r0, 0 < 0 < 2n, t > 0,
it is necessary that both sides of the equation be equal to the same constant -a. Otherwise, by keeping r and 0 fixed and varying t, one side would remain unchanged while the other side varied. Thus we arrive at the two equations T" + aa2T = 0 and r2R"/R + rR'/R + ar2 = -0"/0.
By an argument similar to the one above we conclude that both sides of the last equation must be equal to the same constant 5. This leads to the two equations r2R" + rR' + (ar2 - 5)R = 0 and 0" + 50 = 0. Since the
circular membrane is continuous, we must have 0(2n) = 0(0), which requires 5 = ö2, ö a non-negative integer. The condition 0(2n) = 0(0) is also known as the periodicity condition. Since we also desire solutions which vary periodically in time, it is clear that the separation constant a should be positive, a = X2. Thus we arrive at the three equations
Integration then yields ô^) - ô^0)
hence
y(x) = (1/2a) [xg(?)d? - ô^0). x0
x0
18c. u(x,t) = ô(x-at) + ty(x+at)
= (1/2a)[ [x+atg(?)d? - [x-atg(?)d?] x0x0
= (1/2a)[g(?)d? + [x0 g(?)d?]
0 x-at
= (1/2a[ x+atg(?)d?.
r2R" + rR' + (X2r2 - p.2)R = 0, 0" + ö20 = 0, and
T" + X2a2T = 0.
230
Section 10.8
Section 10.8, Page 611
1a. Assuming that u(x,y) = X(x)Y(y) leads to the two O.D.E. X" - GX = 0, Y" + GY = 0. The B.C. u(0,y) = 0,
u(a,y) = 0 imply that X(0) = 0 and X(a) = 0. Thus nontrivial solutions to X" - GX = 0 which satisfy these boundary conditions are possible only if G = -(nn/a) 2, n = 1,2...; the corresponding solutions for X(x) are proportional to sin(nnx/a). The B.C. u(x,0) = 0 implies that Y(0) = 0. Solving Y" - (nn/a) 2Y = 0 subject to this condition we find that Y(y) must be proportional to sinh(nny/a). The fundamental solutions are then un(x,y) = sin(nnx/a)sinh(nny/a), n = 1,2,..., which satisfy Laplace's equation and the homogeneous B.C. We
1b. Substituting for g(x) in the equation for cn we have
so cn = [4a sin(nn/2)]/[n2n 2sinh(nnb/a)]. Substituting
these values for cn in the above series yields the desired solution.
assume that u(x,y) = ^ cnsin(nnx/a)sinh(nny/a), where
n=1
the coefficients cn are determined from the B.C.
assume
u(x,b) = g(x) = ^ cnsin(nnx/a)sinh(nnb/a). It follows
n=1
1c.
Section 10.8
231
1d.
2. In solving the D.E. Y" - X2Y = 0, one normally writes Y(y) = c1sinhXy + c2coshXy. However, since we have Y(b) = 0, it is advantageous to rewrite Y as Y(y) = d1sinhX(b-y) + d2coshX(b-y), where d1, d2 are also arbitrary constants and can be related to c1, c2 using the appropriate hyperbolic trigonometric identities. The important thing, however, is that the second form also satisfies the D.E. and thus Y(y) = d1sinhX(b-y) satisfies the D.E. and the homogeneous B.C. Y(b) = 0. The rest of the problem follows the pattern of Problem 1.
3a. Let u(x,y) = v(x,y) + w(x,y), where u, v and w all
satisfy Laplace's Eq., v(x,y) satisfies the conditions in Eq. (4) and w(x,y) satisfies the conditions of Problem 2.
4. Following the pattern of Problem 3, one could consider
adding the solutions of four problems, each with only one non-homogeneous B.C. It is also possible to consider adding the solutions of only two problems, each with only two non-homogeneous B.C., as long as they involve the same variable. For instance, one such problem would be uxx + uyy = 0, u(x,0) = 0, u(x,b) = 0, u(0,y) = k(y),
u(a,y) = f(y), which has the fundamental solutions un(x,y) = [cnsinh(nnx/b) + dncosh(nnx/b)]sin(nny/b).
Assuming u(x,y) = un(x,y) and using the B.C.
n=1
u(0,y) = k(y) we obtain dn = (2/b) k(y)sin(nny/b)dy.
0
Using the B.C. u(a,y) = f(y) we obtain
cnsinh(nna/b) + dncosh(nna/b) = (2/b)Ibf(y)sin(nny/b)dy, which
0
can be solved for cn, since dn is already known. The second problem, in this approach, would be uxx + uyy = 0, u(x,0) =
h(x), u(x,b) = g(x), u(0,y) = 0 and u(a,y) = 0. This has the
fundamental solutions
232
Section 10.8
un(x,y) = [ansinh(nny/a) + bncosh(nny/a)]sin(nnx/a, so that u(x,y) = un(x,y). Thus u(x,0) = h(x) gives
n=1
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