Advanced Calculus fi..

666 Advanced Calculus, Fifth EditionIf kl = k2 = k3 and u3 = uo = 0, we obtain (10.33). It is physically clearthat when the end temperatures are maintained at 0, the temperatures of theinner pieces will also gradually approach 0.d) Equilibrium. We assume that uo and us have constant values u: and u; andthat constant forces F, and F2 are applied. Equations (10.25) then have anequilibrium solution, obtained by setting all derivatives with respect to tequal to 0:2- -k2(u2- 2 ~1 + u:) = FI , -k (u3*- 2 ~2 + UI) = F2. (10.38)Accordingly, the equilibrium values of u u2 are . .e) Approach to equilibrium. Let wl = ul - u;, w2 = u2 - u;, where u; and u;are defined by (10.39). Then substitution in (10.25), with uo = u;, u3 = uf,Fl = const, F2 = const, yields the equationsd2w1 dwlm1- +hl- -k (w2-2wl)=0,dt2 dt(10.40)d2w2 dw2m2 - dt2 + h2- - k (-2w2 + wI) = 0.dt, ,~If, for example, m 1= m2 = 0, then these are the same as (10.33), so thatAccordingly,Just as for one particle, the solution is an exponential appmach to equilibrium.f) Forced motion. External forces can be applied both by varying the wallpositions uo, u3 and through the forces FI , F2. We allow for both effects butneglect the masses by considering the equations