INSTRUCTOR'S MANUAL APPLIED PARTIAL DIFFERENTIAL EQUATIONS with Fourier Series and Boundary Value Problems Fourth Edition Richard Haberman (Complete And Verified Study material) (84pages) LEARNEXAMS

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