Download (direct link):
8a. Separating variables, as before, we obtain
X" + X2X = 0, X(0) = 0, X(a) = 0 and Y" - X2Y = 0, Y(y) bounded
as y ^ ^. Thus X(x) = sin(nnx/a), and X2 = (nn/a)2.
However, since neither sinhy nor coshy are bounded as y ^ ^, we must write the solution to Y" - (nn/a)2Y = 0 as Y(y) = c1exp[nny/a] + c2exp[-nny/a]. Thus we must choose
c1 = 0 so that u(x,y) = X(x)Y(y) ^ 0 as y ^ ro. The
fundamental solutions are then un(x,t) = e-nny/asin(nnx/a).
u(x,y) = cnun(x,y) then gives
V"1 2 fi nnx
u(x,0) = > cnsin(nnx/a) = f(x) so that cn = f(x)sin dx.
a J0 a
2 fa nnx 4 a
8b. cn = x(a-x)sin---------dx = - (1-cosnn)
aĞ*0 a n3n3
8c. Using just the first term and letting a = 5, we have 2 0 0 - nv/5 nx
u(x,y) = e y/5 sin , which, for a fixed y, has a maximum
at x = 5/2 and thus we need to find y such that
2 0 0 - ny/5
u(5/2,y) = e ny/5 = .1. Taking the logarithm of both n3
sides and solving for y yields y0 = 6.6315. With an equation solver, more terms can be used. However, to four decimal places, three terms yield the same result as above.
13a. Assuming that u(x,y) = X(x)Y(y) and substituting into
Eq.(1) leads to the two O.D.E. X" - OX = 0, Y" + OY = 0.
The B.C. u(x,0) = 0, uy(x,b) = 0 imply that Y(0) = 0 and Y'(b) = 0. For nontrivial solutions to exist for Y" + OY = 0 with these B.C. we find that O must take the values (2n-1)2n2/4b2, n = 1,2,...; the corresponding solutions for Y(y) are proportional to sin[(2n-1)ny/2b]. Solutions to X" - [(2n-1)2n2/4b2]X = 0 are of the form
X(x) = Asinh[(2n-1)nx/2b] + Bcosh[(2n-1)nx/2b]. However, the boundary condition u(0,y) = 0 implies that X(0) = B = 0. It follows that the fundamental solutions are un(x,y) = cnsinh[(2n-1)nx/2b]sin[(2n-1)ny/2b], n = 1,2,... . To satisfy the remaining B.C. at x = a we
assume that we can represent the solution u(x,y) in the
form u(x,y) = cnsinh[(2n-1)nx/2b]sin[(2n-1)ny/2b].
The coefficients cn are determined by the B.C.
u(a,y) = I cnsinh[(2n-1)na/2b]sin[(2n-1)ny/2b] = f(y).
By properly extending f as a periodic function of period 4b as in Problem 39, Section 10.4, we find that the coefficients cn are given by
cnsinh[(2n-1)na/2b] = (2/b)[bf(y)sin[(2n-1)ny/2b]dy,
n = 1,2,... .
Section 11.1, Page 626
2. Since the B.C. at x = 1 is nonhomogeneous, the B.V.P. is nonhomogeneous.
4. The D.E. may be written y" + (X-x2)y = 0 and is thus homogeneous, as are both B.C.
5. Since the D.E. contains the nonhomogeneous term 1, the B.V.P. is nonhomogeneous.
9. If X = 0, then y(x) = cix + c2 and thus y(0) = c2,
y(1) = c1 + c2, y'(0) = c1 and y'(1) = c1. Hence the B.C. yield the two equations c2 - c1 = 0 and c1 + c2 + c1 = 0
which give c1 = c2 = 0 and thus X = 0 is not an
If X > 0, the general solution of the D.E. is y = c1si^V"X x + c2cosV"X x. The B.C. require that
c2 - yfX c1 = 0, and
(sin \fX + yfX cos \[X ) c1 + (cosV"X - y/~X sin \[X )c2 = 0. In order to have nontrivial solutions X must satisfy (X-1)sinV"X - 2\[X cos^X = 0. In this case c2 = yfX c1
and thus ^n = sin^x + yjXn co^Xn x. If X ^ 1, the
eigenvalue equation is equivalent to tan^/X = 2y[X /(X-1) and thus by graphing f^yfX ) = tan^/X and
g(\J~X ) = 2\J~X / (X-1 ) we can estimate the eigenvalues. Since g(V"X) has a vertical asymptote at X = 1 and f(^/X ) has a vertical asymptote at = n/2, we see that 1 < yjX1 < n/2.
By interating numerically, we find yjX1 = 1.30655 and thus X1 = 1.7071. The second eigenvalue will lie to the right of n, the second zero of tan^/X . Again iterating numerically, we find yjX2 = 3.6732 and thus X2 = 13.4924. For large values
of n, we see from the graph that yjXn = (n-1)n, which are the zeros of tanV"X . Thus Xn = (n-1)2n2 for large n.
For X < 0, the discussion follows the pattern of Example 1 yielding y(x) = c1sin^X^ x + c2coshX|l x. The B.C. then yield c2 - yf^ c1 = 0 and
(sinhX^ + cos^V"^ )c1 + (coshX^ + sinhV"^ )c2 =0, which
have a non-zero solution if and only if
(p+1)sinh\/~p + 2\fp cosh^/"p = 0. By plotting y = tanh^"p and y = -2^/p /(p+1) we see that they intersect only at p. = 0, and
thus there are no negative eigenvalues.
10. If X = 0, the general solution of the D.E. is
y = c1 + c2x. The B.C. y(0) + y'(0) = 0 requires
c + c = 0 and the B.C. y(1) = 0 requires c + c = 0 1 2 1 2
and thus X = 0 is an eigenvalue with corresponding
eigenfunction 60(x) = 1-x.
If X < 0, set -X = p2 to obtain y = c^ospx + c2sinpx.
In this case the B.C. require c1 + pc2 = 0 and
c1cosp + c2sinp = 0 which yields nontrivial solutions for c1
and c2 (i.e., c1 = -pc2) if and only if tanp = p. By
plotting on the same graph f(p) = p and g(p) = tanp, we see that they intersect at p0 = 0 (p = 0 ^ X = 0, which has
already been discussed), p1 = 4.49341 (which is just to the
left of the vertical asymptote of tanp at p = 3n/2, p2 = 7.72525 (which is just to the left of the vertical asymptote of tanp at p = 5n/2) and for larger values p = (2n+1)n/2. Since X = -p2, we have X = -20.1907,