Books in black and white
 Books Biology Business Chemistry Computers Culture Economics Fiction Games Guide History Management Mathematical Medicine Mental Fitnes Physics Psychology Scince Sport Technics

# 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.
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 (ä/X"x)L[J (ä/X"x)] - J (ä/X"x)L[J (ä/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 Ô X , the result follows with
nm
ô (x) = J h[X~x).
T n 0 V n
2c. Since X = 0 is an eigenvalue we assume that y = b + > b J (ó 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 (³/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
(öÜ + c ) + X [c - b (X - Ö)ÛÏ h Fk~ x) = 0. Thus
r 0 0 n n n r 0 V n
n=1
b = -c /ö and b = c /(X -ö), n = 1,2,..., where i/X 00r n n n r Vn
are obtained from j' (ë/X ) = 0.
0 V n
Let L[y] = -[(1-x2)y']'. Then LM ] = X ô and
T n nT n
L[ô ] = X ô . Multiply the first equation by ô , the
Tm mT m Tm
second by ô , subtract the second from the first, and
n
integrate from 0 to 1 to obtain
11(ô L¹ ] - ô L^ ])dx = (X - X ) 11ô ô 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 Ô X if m Ô n, the result
nm
follows. Note that the result may also be written as ÃÐ (x)P (x)dx = 0, m Ô n.
j0 2m-1 2n-1
First let f(x) = c^ n(x), multiply both sides by ôm(x),
n=1
and integrate term by term from x = 0 to x = 1. The orthogonality condition yields
c = f(x^ (x)dx/ Ðô2(x)dx, n = 1,2,... where it is n J0 Tn J0 Tn
understood that ô (x) = Ð (x). Now assume
Tn 2n-1
y = > b ô (x). As in Problem 2 and in the text
nn
nn
n=1
25i
-[(i-x2)y'] ' = Üïôn since the ôn are eigenfunctions.
n=i
Thus, substitution of the series for y and f into the D.E. and simplification yields > [b (X -u) - c ]ô (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 ô (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