This equation represents the propagation of heat in an infinite bar, all points of which were first at temperature 0, except those at the extremity which is maintained at the constant temperature 1. We suppose that heat cannot escape through the external surface of the bar; or, which is the same thing, that the thickness of the bar is infinitely great. This value of v indicates therefore the law according to which heat is propagated in a solid, terminated by an infinite plane, supposing that this infinitely thick wall has first at all parts a constant initial temperature 0, and that the surface is submitted to a constant temperature 1. It will not be quite useless to point out several results of this solution. r = R, we have, when R is a positive quantity, hence ↓ (R) = − $ (R) and (R) = } + $ (R),' ↓ (− R) — ↓ (R) = 24 (R) and v = 1 − 24 1 1 2 5 Paris, 1826. 4to. pp. 520-1. Table of the values of the integral fdx (log The first part for values of log of x from 0.80 to 0.00. from 0.00 to 0.50; the second part for values Encke. Astronomisches Jahrbuch für 1834. Berlin, 1832, 8vo. Table I. at the 2 end gives the values of e-"dt from t=0·00 to t=2.00. [A. F.] 1st, if we suppose x nothing, we find v=1; 2nd, if x not being nothing, we suppose t = 0, the sum of the terms which - taken from r=0 to r∞, (dre-r contain a represents the integral dreand consequently is equal to √; therefore v is nothing; 3rd, different points of the solid situated at different depths x,, x,,,, &c. arrive at the same temperature after different times t1, t,, t, &c. which are proportional to the squares of the lengths x1, x, x,, &c.; 4th, in order to compare the quantities of heat which during an infinitely small instant cross a section S situated in the interior of the solid at a distance x from the heated plane, we must take the integral sign. The preceding value at the surface of the at the surface varies with the quantities C, D, K, t; to find how much heat the source communicates to the solid during the lapse of the time t, we must take the integral thus the heat acquired increases proportionally to the square root of the time elapsed. 367. By a similar analysis we may treat the problem of the diffusion of heat, which also depends on the integration of the dv d'v equation dt da = k hv. Representing by f(x) the initial temperature of a point in the line situated at a distance a from the origin, we proceed to determine what ought to be the temperature of the same point after a time t. Making v=e-ht z, we have dz d'z k dt dt and consequently z = [__dqe-14 (x+2q √ht). When t = 0, we must have hence -8 To apply this general expression to the case in which a part of the line from x = a to xa is uniformly heated, all the rest of the solid being at the temperature 0, we must consider that the factor f(x+2q kt) which multiplies e-2 has, according to hypothesis, a constant value 1, when the quantity which is under the sign of the function is included between -a and a, and that all the other values of this factor are nothing. Hence the integral x+2q√kt a, faqe-a ought to be taken from x + 2q √kt = − a to x + 2g √kt = æ +. to q= 2kt the integral dre- taken from r = R to r = ∞, we have fdr to the case in which the infinite bar, heated by a source of constant intensity 1, has arrived at fixed temperatures and is then cooling freely in a medium maintained at the temperature 0. For this purpose it is sufficient to remark that the initial -X √ function denoted by f(x) is equivalent to e so long as the variable x which is under the sign of the function is positive, and that the same function is equivalent to e√ when the variable which is affected by the symbol ƒ is less than 0. Hence x+2q/kt∞ to x+2q/kt = 0. The first part of the value of v is making r=q+√ht. The integral should be taken from The second part of the value of v is e-ht ¿Sifaqeve1⁄2ï or e ̈vidre-r; making r=q-√t. The integral should be taken from x r=-∞ to r=-Jht -√ht- 2 √ht' 369. We have obtained (Art. 367) the equation to express the law of diffusion of heat in a bar of small thickness, heated uniformly at its middle point between the given limits x=-α, x = + α. We had previously solved the same problem by following a different method, and we had arrived, on supposing a=1, at the equation To compare these two results we shall suppose in each x=0; denoting again by ✈ (R) the integral fdr e- taken from r=0 |