# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

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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

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