Chapter 1. Heat Equation

Section 1.2

1.2.9 (d) Circular cross section means that P = 2xr, A = r², and thus P/A2/r, where r is the radius. Also y = 0.

1.2.9 (e) u(x,t) = u(t) implies that

du

2h

dt

The solution of this first-order linear differential equation with constant coefficients, which satisfies the initial condition u(0) = uo, is

u(t) = o exp

2h

Section 1.3

1.3.2 du/dr is continuous if Ko(20) Ko(20+), that is, if the conductivity is continuous.

Section 1.4

1.4.1 (a) Equilibrium satisfies (1.4.14), du/dz² = 0, whose general solution is (1.4.17), uc + caz. The boundary condition u(0) = 0 implies c₁ = 0 and u(L) = T implies 2 T/L so that u = Tr/L.

1.4.1 (d) Equilibrium satisfies (1.4.14), du/dz² = 0, whose general solution (1.4.17), uc₁ + c2z. From the boundary conditions, u(0) T yields Tc and du/dz(L) a yields a cz. Thusu Taz.

1.4.1 (f) In equilibrium, (1.2.9) becomes du/dz²=-Q/Koz, whose general solution (by integrating twice) is u-z/12+c+caz. The boundary condition (0) T yields T, while du/dz(L) = 0 yields L³/3. Thus uz/12+Lz/3+T.

1.4.1 (h) Equilibrium satisfies d'u/dz20. One integration yields du/dz cz, the second integration yields the general solution ucC1+C22.

10:00-T) = 0 z=L:a and thus c₁=T+a.

Therefore, u (T+a)+azT+a(z+1).

1.4.7 (a) For equilibrium:

fu

dr = -1 implies 14- +c2+cand

du