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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.
Download (direct link): elementarydifferentialequations2001.pdf
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bn = (2/2L) 12Lf(x)sin(nnx/2L)dx, n = 1,2,... . The
0
Fourier sine series for f is f(x) =
X, bnsin(nnx/2L).
n=1
39. From Problem 38 we have bn = (1/L) 2Lf(x)sin(nnx/2L)dx
0
= (1/L)ILf(x)sin(nnx/2L)dx + (1/L)I2Lf(2L-x)sin(nnx/2L)dx
0L
= (1/L) ILf(x)sin(nnx/2L)dx - (1/L) I 0f(s)sin[nn(2L -s)/2L]ds
0L
= (1/L) ILf(x)sin(nnx/2L)dx-(1/L) ILf(s)cos(nn)sin(nns/2L)ds 00
and thus bn = 0 for n even and
bn = (2/L) Lf(x)sin(nnx/2L)dx for n odd. The Fourier 0
series for f is given in Problem 38, where the bn are given above.
Section 10.5, Page 579
3. We seek solutions of the form u(x,t) = X(x)T(t).
Substituting into the P.D.E. yields X"T + X' T' + XT' =
X"T + (X' + X)T' = 0. Formally dividing by the quantity
(X' + X)T gives the equation X"/(X' + X) = -T'/T in which
the variables are separated. In order for this equation to be valid on the domain of u it is necessary that both sides be equal to the same constant X. Hence
X"/(X' + X) = -T'/T = X or equivalently,
X" - X(X' + X) = 0 and T' + XT = 0.
5. We seek solutions of the form u(x,y) = X(x)Y(y).
Substituting into the P.D.E. yields X"Y + (x+y)XY" =
X"Y + xXY" + yXY" = 0. Formally dividing by XY yields X"/X + xY"/Y + yY"/Y = 0. From this equation we see that the presence of the independent variable x multiplying the term uyy in the original equation leads to the term xY"/Y when we attempt to separate the variables. It follows that the argument for a separation constant does not carry through and we cannot replace the P.D.E. by two
O.D.E.
218
Section 10.5
8. Following the procedures of Eqs.(5) through (8), we set u(x,y) = X(x)T(t) in the P.D.E. to obtain X"T = 4XT', or x"x = 4T'/T, which must be a constant. As stated in the text this separation constant must be -X2 (we choose -X2 so that when a square root is used later, the symbols are simpler) and thus X" + X 2X = 0 and T' + (X 2/4)T = 0. Now u(0,t) = X(0)T(t) = 0, for all t > 0, yields X(0) = 0, as discussed after Eq.(11) and similarly u(2,t) = X(2)T(t) = 0, for all t > 0, implies X(2) = 0. The D.E. for X has the solution X(x) = C1cosXx + C2sinXx and X(0) = 0 yields C1 = 0. Setting x = 2 in the remaining form of X yields X(2) = C2sin2X = 0, which has the solutions 2X = nn or X = nn/2, n = 1,2,... .
Note that we exclude n = 0 since then X = 0 would yield
X(x) = 0, which is unacceptable. Hence
X(x) = sin(nnx/2), n = 1,2,... . Finally, the solution of the
D.E. for T yields T(t) = exp(-X2t/4) = exp(-n2n2t/16). Thus we have found un(x,t) = exp(-n2n2t/16)sin(nnx/2). Setting t = 0 in this last expression indicates that un(x,0) has, for the correct choices of n, the same form as the terms in u(x,0), the initial condition. Using the principle of superposition we know that
u(x,t) = c1u1(x,t) + c2u2(x,t) + c4u4(x,t) satisfies the
P.D.E. and the B.C. and hence we let t = 0 to obtain u(x,0) = c1u1(x,0) + c2u2(x,0) + c4u4(x,0) =
c1sinnx/2 + c2sinnx + c4sin2nx. If we choose c1 = 2, c2 = -1 and c4 = 4 then u(x,0) here will match the given initial condition, and hence substituting these values in u(x,t) above then gives the desired solution.
10. Since the B.C. for this heat conduction problem are
u(0,t) = u(40,t) = 0, t > 0, the solution u(x,t) is given
by Eq.(19) with a2 = 1 cm2/sec, L = 40 cm, and the coefficients cn determined by Eq.(21) with the I.C. u(x,0) = f(x) = x, 0 < x < 20; = 40 - x, 20 < x < 40.
1 f 2 0 nnx f 4 0 nnx
Thus cn = ------ [ xsin-------dx + (40-x)sin--------dx]
20 0 40 20 40
n2n2 2
160
u(x,t) =
160 nn
sin. It follows that
= sin(nn/2) e-n2n2t/1-00sinnnx n2 ^ n2 40
n=1
Section 10.5
219
15a.
15b.
15c.
15d. As in Example 1, the maximum temperature will be at the midpoint, x = 20, and we use just the first term, since the others will be negligible for this temperature, since t is so large. Thus
u(20,t) = 1
160
- sin(n/2)e'
-n2t/16 0 0
sin(20n/40). Solving
for t, we obtain e 1600 160
t =
2
ln
2
re2t/1600 _ n2/160, or = 451.60 sec.
18a. Since the B.C. for this heat conduction problem are
u(0,t) = u(20,t) = 0, t > 0, the solution u(x,t) is given by
Eq.(19) with L = 20 cm, and the coefficients cn determined by Eq.(21) with the I.C. u(x,0) = f(x) = 100oC. Thus
cn = (1/10) 120 (10 0)sin(nnx/2 0)dx = -200 [(-1)n-1]/nn and hence
220
Section 10.5
c2n = 0 and c2n-1 = 400/(2n-1)n. Substituting these values into Eq.(19) yields
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