Page 365 - Schaum's Outline of Theory and Problems of Advanced Calculus

356 FOURIER SERIES [CHAP. 13

Then from Problems 13.20 and 13.22, we have

f ðx þ 0Þþ f ðx 0Þ f ðx þ 0Þþ f ðx 0Þ
¼ 0or
2 2
lim S M ðxÞ lim S M ðxÞ¼
M!1 M!1

BOUNDARY-VALUE PROBLEMS

13.24. Find a solution Uðx; tÞ of the boundary-value problem
2
@U @ U
¼ 3 t > 0; 0 < x < 2
@t @x 2
Uð0; tÞ¼ 0; Uð2; tÞ¼ 0 t > 0
Uðx; 0Þ¼ x 0 < x < 2

Amethod commonly employed in practice is to assume the existence of a solution of the partial
differential equation having the particular form Uðx; tÞ¼ XðxÞ TðtÞ, where XðxÞ and TðtÞ are functions of
x and t, respectively, which we shall try to determine. For this reason the method is often called the method
of separation of variables.
Substitution in the differential equation yields
2
@ @ 2 dT d X
ðXTÞ¼ 3 or ð2Þ X ¼ 3T
@t @x dt dx
ð1Þ 2 ðXTÞ 2
where we have written X and T in place of XðxÞ and TðtÞ.
Equation (2)can be written as
2
1 dT 1 d X
3T dt ¼ X dx 2 ð3Þ
Since one side depends only on t and the other only on x, and since x and t are independent variables, it is
clear that each side must be a constant c.
In Problem 13.47 we see that if c A 0, a solution satisfying the given boundary conditions cannot exist.
2
Let us thus assume that c is a negative constant which we write as . Then from (3)weobtain two
ordinary differentiation equations
2
dT 2 d X 2
þ 3 T ¼ 0; þ X ¼ 0
dt dx 2 ð4Þ
whose solutions are respectively
2
T ¼ C 1 e 3 t ; X ¼ A 1 cos x þ B 1 sin x ð5Þ
Asolution is given by the product of X and T which can be written
2
Uðx; tÞ¼ e 3 t ðA cos x þ B sin xÞ ð6Þ
where A and B are constants.
We now seek to determine A and B so that (6)satisfies the given boundary conditions. To satisfy the
condition Uð0; tÞ¼ 0, we must have
2
s t
e ðAÞ¼ 0 or A ¼ 0 ð7Þ
so that (6)becomes
2
Uðx; tÞ¼ Be s t sin x ð8Þ
To satisfy the condition Uð2; tÞ¼ 0, we must then have
2
Be s t sin 2 ¼ 0 ð9Þ