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 .. 500 501 502 503 504 505 < 506 > 507 508 509 510 511 512 .. 609 >> Next

second from the first, and integrate from 0 to 1 to
obtain
J1{J (/X"x)L[J (/X"x)] - J (/X"x)L[J (/X"x)]}dx =
J0 0 V m 0 V n ^Vn ^Vm
(X - X ) 11xJ h/X x)J h/X x)dx. Again the left side
n mj0^Vn 0 V m
is zero after each term is integrated by parts once, as
was done above. If X X , the result follows with
nm
(x) = J h[X~x).
T n 0 V n
2c. Since X = 0 is an eigenvalue we assume that y = b + > b J ( X x). Since
1 0 n 0 V n
n=1
-[xj" h/X x)]' = X xJ h/X x), n = 0,1,..., we find that
0 V n n 0 V n
250
Section 11.4
4a.
4b.
-(xy')' = x> X b J h/X x) [note that X = 0 and b are
n n 0 v n 0 0
n=1
missing on the right]. Now assume
f(x)/x = c0 + cnJ0 (^Xn x). Multiplying both sides by
n=1
xJ0(^1Xm x), integrating from 0 to 1 and using the
orthogonality relations of part b, we find
c = f1f(x)J h IX x)dx/f1xJ2 (/X x)dx, n = 0,1,2,... .
n J0 0 V n J0 H n
[Note that c = 2 f(x)dx since the denominator can be 00
integrated.] Substituting the series for y and f(x)/x into the D.E., using the above result for - (xy')', and simplifying we find that
( + c ) + X [c - b (X - ) h Fk~ x) = 0. Thus
r 0 0 n n n r 0 V n
n=1
b = -c / and b = c /(X -), n = 1,2,..., where i/X 00r n n n r Vn
are obtained from j' (/X ) = 0.
0 V n
Let L[y] = -[(1-x2)y']'. Then LM ] = X and
T n nT n
L[ ] = X . Multiply the first equation by , the
Tm mT m Tm
second by , subtract the second from the first, and
n
integrate from 0 to 1 to obtain
11( L ] - L^ ])dx = (X - X ) 11 dx. The integral
j0TmTn TnTm n m J0
on the left side can be shown to be 0 by integrating each term once by parts. Since X X if m n, the result
nm
follows. Note that the result may also be written as (x)P (x)dx = 0, m n.
j0 2m-1 2n-1
First let f(x) = c^ n(x), multiply both sides by m(x),
n=1
and integrate term by term from x = 0 to x = 1. The orthogonality condition yields
c = f(x^ (x)dx/ 2(x)dx, n = 1,2,... where it is n J0 Tn J0 Tn
understood that (x) = (x). Now assume
Tn 2n-1
y = > b (x). As in Problem 2 and in the text
nn
nn
n=1
25i
-[(i-x2)y'] ' = n since the n are eigenfunctions.
n=i
Thus, substitution of the series for y and f into the D.E. and simplification yields > [b (X -u) - c ] (x) = 0. Hence
n n n n
n=i
b = c /(X - u), n = i,2,... and the desired solution is
n n n *
obtained [after setting (x) = P (x)].
n 2n-i
Section ii.5, Page 666
ia. 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.
ib. 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? + uyy? = ux and
un = uxxn + uyyn = i/2ux + uy. Likewise
u?? = uxx - x? + uxyy? = uxx un = uxxxn + uxyyn = i/2uxx + uxy and unn = i/2uxxxn + i/2uxyyn + uyxxn + uyyyn = i/4uxx + uxy + uyy. Therefore,
5/4u?? - u?n + unn = uxx + uyy = 0.
ic. 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(?+i) (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(i,t) = 0, t > 0 and the finiteness condition are given by Eqs.(i5) and (i6). Thus assume u(r,t) is of the form given by Eq.(i7). The I.C. require
that u(r,0) = / c J (X r) = 0 and
n 0 n
n=i
252
Section 11.5
u (r,0) = > X ak J (X r) = g(r). From Eq.(26) of
t n n 0 n
n=1
Section 11.4 we obtain c = 0 and
n
X ka = J1rg(r)J (X r)dr/J1rJ2(X r)dr, n = 1,2,... . n n 0 0 n 0 0 n
Previous << 1 .. 500 501 502 503 504 505 < 506 > 507 508 509 510 511 512 .. 609 >> Next