Page 380 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 14] FOURIER INTEGRALS 371

ð
1 sin v

Also jI 3 j @ j f ðx þ 0Þj dv
v
L
ð ð
1 1 sin v
Since j f ðxÞj dx and dv both converge, we can choose L so large that jI 2 j @ =3, jI 3 j @ =3.
0 0 v
Also, we can choose so large that jI 1 j @ =3. Then from (4)we have jIj < for and L sufficiently large,
so that the required result follows.
This result follows by reasoning exactly analogous to that in part (a).
14.12. Prove Fourier’s integral formula where f ðxÞ satisfies the conditions stated on Page 364.
1 ð L ð 1
We must prove that lim f ðuÞ cos ðx uÞ du d ¼ f ðx þ 0Þþ f ðx 0Þ
2
L!1 ¼0 u¼ 1
ð ð
1 1

Since f ðuÞ cos ðx uÞ du @ j f ðuÞj du, which converges, it follows by the Weierstrass test

ð 1 1
1
that f ðuÞ cos ðx uÞ du converges absolutely and uniformly for all . Thus, we can reverse the
1
order of integration to obtain
1 ð L ð 1 1 ð 1 ð L
d f ðuÞ cos ðx uÞ du ¼ f ðuÞ du cos ðx uÞ d
¼0 ¼0
u¼ 1 u¼ 1
1 ð 1
du
sin Lðu xÞ
¼ f ðuÞ
u x
u¼ 1
1 ð 1 sin Lv
dv
v
¼ f ðx þ vÞ
u¼ 1
1 ð 0 sin Lv 1 ð 1 sin Lv
dv
¼ f ðx þ vÞ dv þ f ðx þ vÞ
v 0 v
1
where we have let u ¼ x þ v.
Letting L !1,we see by Problem 14.11 that the given integral converges to f ðx þ 0Þþ f ðx 0Þ as
required. 2
MISCELLANEOUS PROBLEMS
2
@U @ U 1 0 < x < 1
14.13. Solve ¼ subject to the conditions Uð0; tÞ¼ 0; Uðx; 0Þ¼ , Uðx; tÞ is
@t @x 2 0 x A 1
bounded where x > 0; t > 0.
We proceed as in Problem 13.24, Chapter 13. A solution satisfying the partial differential equation and
the first boundary condition is given by Be t 2 sin x. Unlike Problem 13.24, Chapter 13, the boundary
conditions do not prescribe the specific values for ,sowe must assume that all values of are possible. By
analogy with that problem we sum over all possible values of , which corresponds to an integration in this
case, and are led to the possible solution
ð
2
1 t
Bð Þ e sin xd
Uðx; tÞ¼ ð1Þ
0
where Bð Þ is undetermined. By the second condition, we have
ð
10 < x < 1
1
Bð Þ sin xd ¼ ¼ f ðxÞ ð2Þ
0 0 x A 1
from which we have by Fourier’s integral formula
2 ð 1 2 ð 1 2ð1 cos Þ
Bð Þ¼ f ðxÞ sin xdx ¼ sin xdx ¼ ð3Þ
0 0