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what is the state of the line after a given time. This problem may be made more general, by supposing, 1st, that the initial temperatures of the points included between a and b are unequal and represented by the ordinates of any line whatever, which we shall regard first as composed of two symmetrical parts (see fig. 16);

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2nd, that part of the heat is dispersed through the surface of the solid, which is a prism of very small thickness, and of infinite length.

The second problem consists in determining the successive states of a prismatic bar, infinite in length, one extremity of which is submitted to a constant temperature. The solution of these two problems depends on the integration of the equation

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(Article 105), which expresses the linear movement of heat. v is the temperature which the point at distance from the origin must have after the lapse of the time t; K, H, C, D, L, S, denote the internal and surface conducibilities, the specific capacity for heat, the density, the contour of the perpendicular section, and the area of this section.

345. Consider in the first instance the case in which heat is propagated freely in an infinite line, one part of which ab has received any initial temperatures; all other points having the initial temperature 0. If at each point of the bar we raise the ordinate of a plane curve so as to represent the actual temperature at that point, we see that after a certain value of the time t, the state of the solid is expressed by the form of the curve. Denote by v = F(x) the equation which corresponds to the given initial state, and first, for the sake of making the investigation

more simple, suppose the initial form of the curve to be composed of two symmetrical parts, so that we have the condition

F(x) = F (− x).

d'v

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k -hv, make ve-ht u, and we have

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du dt

Assume a particular value of u, namely, a cos qx e-kast; a and being arbitrary constants. Let 91, 92, 93, &c. be a series of any values whatever, and a,, a,, a,, &c. a series of corresponding values of the coefficient Q, we have

u = a, cos (q,x) e ̄k2 + a, cos (2,x) e-kt+a, cos (x) e-k22 + &c. Suppose first that the values q1, 12, 13, &c. increase by infinitely small degrees, as the abscissæ q of a certain curve; so that they become equal to dq, 2dq, 3dq, &c.; dq being the constant differential of the abscissa; next that the values a,, a,, a,, &c. are proportional to the ordinates Q of the same curve, and that they become equal to Q,dq, Q2dq, Q1q, &c., Q being a certain function of q. It follows from this that the value of u may be expressed thus:

u = dq Q cos qx € ̃ky2,

=fdq e-k3t,

Q is an arbitrary function f(q), and the integral may be taken from q=0 to q=. The difficulty is reduced to determining suitably the function Q.

346. To determine Q, we must suppose t0 in the expression for u, and equate u to F (x). We have therefore the equation of condition

F (x) = fdqQ cos qx.

If we substituted for Q any function of q, and conducted the integration from q=0 to q = c, we should find a function of æ: it is required to solve the inverse problem, that is to say, to ascertain what function of q, after being substituted for Q, gives as the result the function F(x), a remarkable problem whose lution demands attentive examination.

Developing the sign of the integral, we write as follows, the equation from which the value of Q must be derived:

F(x)=dqQ, cos q1x + dqQ, cos q2x + dqQ ̧ cos q ̧¤ + &c.

2

In order to make all the terms of the second member disappear, except one, multiply each side by dx cos rx, and then integrate with respect to x from x=0 to x = nπ, where n is an infinite number, and r represents a magnitude equal to any one. of 11, 12, 13, &c., or which is the same thing dq, 2dq, 3dq, &c. Let q, be any value whatever of the variable q, and q; another value, namely, that which we have taken for r; we shall have r=jdq, and q=idq. Consider then the infinite number n to express how many times unit of length contains the element dq, so that we Proceeding to the integration we find that the

have n =

1

dq

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value of the integral dx cos qx cos rx is nothing, whenever r and

Jax

q have different magnitudes; but its value is

1

when 2 пп, q=r. This follows from the fact that integration eliminates from the second member all the terms, except one; namely, that which The function which affects the same term

contains q; or r.

is Q, we have therefore

1

fdx F(x) cos qx = dq Q, nπ,

and substituting for ndq its value 1, we have

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determine the function Q which satisfies the proposed condition, we must multiply the given function F(x) by dx cos qx, and integrate from x nothing to x infinite, multiplying the result by ;

2

π

that is to say, from the equation F(x) = [dqƒ (4) cos q., we deduce

ƒ (9) = 2 [dx F(x) cos qr, the function F(x) representing the

f

π

F. H.

22

initial temperatures of an infinite prism, of which an intermediate part only is heated. Substituting the value of ƒ (q) in the expression for F(x), we obtain the general equation

= F(x) = ["dq cos qx ["dx F(x) cos q ...........................).

347. If we substitute in the expression for v the value which we have found for the function Q, we have the following integral, which contains the complete solution of the proposed problem,

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The integral, with respect to a, being taken from x nothing to a infinite, the result is a function of q; and taking then the integral with respect to q from q = 0 to q = ∞, we obtain for v a function of x and t, which represents the successive states of the solid. Since the integration with respect to x makes this variable disappear, it may be replaced in the expression of v by any variable a, the integral being taken between the same limits, namely from a = 0 to a = ∞. We have then

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or

πυ 2

=e-ht

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* ["dx F (a) [ "dq e-kit, cos qx cos qz.

The integration with respect to q will give a function of x, t and a, and taking the integral with respect to a we find a function of x and t only. In the last equation it would be easy to effect the integration with respect to q, and thus the expression of v would be changed. We can in general give different forms to the integral of the equation

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they all represent the same function of x and t.

348. Suppose in the first place that all the initial temperatures of points included between a and b, from x=-1, to x = 1, ave the common value 1, and that the temperatures of all the

other points are nothing, the function F'(x). will be given by this condition. It will then be necessary to integrate, with respect to x, from x=0 to 1, for the rest of the integral is nothing according to the hypothesis. We shall thus find

x =

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The second member may easily be converted into a convergent series, as will be seen presently; it represents exactly the state of the solid at a given instant, and if we make in it t=0, it expresses the initial state.

Thus the function

2

π 0

dq q

-

sin q cos qx is equivalent to unity, if we give to any value included between 1 and 1: but this function is nothing if to x any other value be given not included between 1 and 1. We see by this that discontinuous functions also may be expressed by definite integrals.

349. In order to give a second application of the preceding formula, let us suppose the bar to have been heated at one of its points by the constant action of the same source of heat, and that it has arrived at its permanent state which is known to be represented by a logarithmic curve.

It is required to ascertain according to what law the diffusion of heat is effected after the source of heat is withdrawn. Denoting by F(x) the initial value of the temperature, we shall have

-x

HL

F(x) = A e KS; A is the initial temperature of the point most heated. To simplify the investigation let us make A = 1,

and

HL

KS

= = 1. We have then F(x)=e, whence we deduce

Q=fdæ e*cos qo, and taking the integral from a nothing to x

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Thus the value of v in x and t is given by

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