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Sec. 17] Problems on Fourier's Method 363
3098*.
3099.
3100*.
3101*.
3102.
= 0; y = 0, y' = 1 for x = 0.
0\ 0=1, #' = for * = 0.
/' + */ = 0; y=l, */'=0 for Jt =
Sec. 17. Problems on Fourier's Method
+ */ = 0; 0=1, 0'=0 for x =
= 0; * = a; ~ = for * = 0.
To find the solutions of a linear homogeneous partial differential equation
by Fourier's method, first seek the particular solutions of this special-type
equation, each of which represents the product of functions that are dependent
on one argument only. In the simplest case, there is an infinite set of such
solutions M rt (tt=l, 2,...), which are linearly independent among themselves
in any finite number and which satisfy the given boundary conditions. The
desired solution u is represented in the form of a series arranged according
to these particular solutions:
u= C n u n . (1)
The coefficients C n which remain undetermined are found from the initial
conditions.
Problem. A transversal displacement u = u(x t t) of the points of a string
with abscissa x satisfies, at time *, the equation
_ ~~ a
dt* dx*
where a 2 = -? (TQ is the tensile force and Q is the linear density
(2)
of the
string). Find the form of the string at time t if its ends x = and * = / are
U
2
Fig. 107
fixed and at the initial instant, f = 0, the string had the form of a parabola
u =~* (/ x) (Fig. 107) and its points had zero velocity.
Solution. It is required to find the solution u = u(x, t) of equation (2)
that satisfies the boundary conditions
a(0, 0-0, !!(/, 0=0 (3)