satisfies the equation in u and 0. We therefore assume the value

the integral being taken from r = 0 to r=π.

This value of v in

terms of x and m satisfies the differential equation, and retains a

finite value when x is nothing. Further, the equation hu+

=

du

dx

= 0

must be satisfied when x X the radius of the cylinder. This condition would not hold if we assigned to the quantity m any value whatever; we must necessarily have the equation

This definite equation, which is equivalent to the following,

gives to an infinity of real values denoted by 0, 0, 0, &c.; the

We can write, instead of 0,, one of the roots 0, 0, 0, &c., and compose by means of them a more general value expressed by the equation

a, a, a, &c. are arbitrary coefficients: the variable q disappears after the integrations, which should be taken from

=

315. To prove that this value of v satisfies all the conditions of the problem and contains the general solution, it remains only to determine the coefficients a1, a, a,, &c. from the initial state. Take the equation

v = a1e ̄m1t u1 + a ̧е ̄m ̧1μ‚ + α ̧Ã ̄m ̧Ð μ ̧ + &c.,

in which u1, u,, u,, &c. are the different values assumed by the function u, or

m x2 m2 x
+

k 22 k2 2242

- &c.

m

when, instead of

the values

k'

91, 92, 93, &c. are successively sub

stituted. Making in it t=0, we have the equation

√=α ̧μ ̧ +σ‚μ‚ +α ̧u3+ &c.,

in which V is a given function of x.

if we represent the function u, whose

have

Let p (x) be this function; index is i by ↓ (x √9.), we

$ (x) = a ̧¥ (x √g ̧) + a ̧¥ (x √ÿ2 +α ̧¥ (x √93) + &c.

To determine the first coefficient, multiply each member of the equation by σ, dx, σ, being a function of x, and integrate from x=0 to x = X. We then determine the function σ,, so that after the integrations the second member may reduce to the first term only, and the coefficient a, may be found, all the other integrals

having nul values. Similarly to determine the second coefficient a, we multiply both terms of the equation

2

02

2

by another factor σ, dr, and integrate from x = 0 to x = X. The factor must be such that all the integrals of the second member vanish, except one, namely that which is affected by the coefficient a. In general, we employ a series of functions of x denoted by σ, σ, σ,, &c. which correspond to the functions u1, u„, u„, &c.; each of the factors σ has the property of making all the terms which contain definite integrals disappear in integration except one; in this manner we obtain the value of each of the coefficients a1, a, a,, &c. We must now examine what functions enjoy the property in question.

316. Each of the terms of the second member of the equation

is a definite integral of the form a asœuda; u being a function of x

afoudæ;

d'u 1 du

u + +
da2 x dx

m x dx

Developing, by the method of integration by parts, the terms

The integrals must be taken between the limits = 0 and x= X, by this condition we determine the quantities which enter into the development, and are not under the integral signs. To indicate that we suppose x = 0 in any expression in x, we shall affect that expression with the suffix a; and we shall give it the suffix w to indicate the value which the function of x takes, when we give to the variable x its last value X.

Supposing x = 0 in the two preceding equations we have

the same equations, and supposing the integral to be taken from x=0 to x = X, we have

sign of integration in the second member were equal to the product of σ by a constant coefficient, the terms

would be collected into one, and we should obtain for the required

integral fouda

udx a value which would contain only determined quan

tities, with no sign of integration. It remains only to equate that value to zero.

Suppose then the factor o to satisfy the differential equation of

Between u and σ a very simple relation exists, which is dis

σ = xs; as the result of this substitution we have the equation

n

k

which shews that the function s depends on the function u given by the equation

d'u 1 du
+

m

u +

k dx2 x dx

To find s it is sufficient to change m into n in the value of u;

the two last terms destroy each other, it follows that on making x=0, which corresponds to the suffix a, the second member vanishes completely. We conclude from this the following equation